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Mirascope model extraction comparisons
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| import argparse | |
| import asyncio | |
| import inspect | |
| import json | |
| import logging | |
| import time | |
| from abc import ABC, abstractmethod | |
| from pathlib import Path | |
| from typing import Literal | |
| from anthropic.types import Usage as AnthropicUsage | |
| from mirascope import llm | |
| from mirascope.core import prompt_template | |
| from mirascope.core.base import BaseCallResponse | |
| from openai.types.completion_usage import CompletionUsage as OpenAIUsage | |
| from pydantic import BaseModel | |
| from tabulate import tabulate | |
| Model = str | |
| Provider = Literal["openai", "anthropic"] | |
| models: list[tuple[Provider, Model]] = [ | |
| ("openai", "gpt-4o-mini"), | |
| ("openai", "gpt-4o"), | |
| ("anthropic", "claude-3-5-haiku-latest"), | |
| ("anthropic", "claude-3-5-sonnet-latest"), | |
| ] | |
| puzzle_text = inspect.cleandoc( | |
| """ | |
| A factory in Standardtopia produces widgets on a very regular schedule. | |
| It outputs 4 widgets every weekday, 3 widgets per day on | |
| weekends, and two extra widgets on Mondays. | |
| Q1: How many widgets are produced in a week? | |
| Each month has four weeks, and one holiday. Holidays are uniformly distributed | |
| across days of the week, and the factory is closed on holidays. | |
| Q2: Taking monthly holidays into account, what's the average number of widgets | |
| produced per week? | |
| In this world, a year consists of exactly 10 months. Standardtopia celebrates | |
| fibonacci numbers, so each month whose 1-based index is a fibonacci number has | |
| two extra holidays. | |
| Q3: Taking these fibonacci holidays into account, what's the average number of | |
| widgets produced per week? | |
| """ | |
| ) | |
| class PuzzleResponse(BaseModel): | |
| q1: float | |
| q2: float | |
| q3: float | |
| def error(self) -> float: | |
| return abs(self.q1 - 28) + abs(self.q2 - 27) + abs(self.q3 - 26) | |
| class PuzzleResponseWithChainOfThought(BaseModel): | |
| chain_of_thought: str | |
| q1: float | |
| q2: float | |
| q3: float | |
| class PuzzleResponseWithThreeChainsOfThought(BaseModel): | |
| q1_chain_of_thought: str | |
| q1: float | |
| q2_chain_of_thought: str | |
| q2: float | |
| q3_chain_of_thought: str | |
| q3: float | |
| class PuzzleResult(BaseModel): | |
| q1: float | |
| q2: float | |
| q3: float | |
| reasoning: str | |
| error: float # sum of absolute errors across all questions | |
| correctness: float # proportion of answers that are correct, in [0, 1] | |
| total_tokens: float | |
| cost: float | |
| @classmethod | |
| def from_response( | |
| cls, | |
| response: PuzzleResponse, | |
| reasoning: str, | |
| total_tokens: float, | |
| cost: float, | |
| ) -> "PuzzleResult": | |
| q1_error = abs(response.q1 - 28) | |
| q2_error = abs(response.q2 - 27) | |
| q3_error = abs(response.q3 - 26) | |
| return cls( | |
| q1=response.q1, | |
| q2=response.q2, | |
| q3=response.q3, | |
| reasoning=reasoning, | |
| error=q1_error + q2_error + q3_error, | |
| correctness=((q1_error == 0) + (q2_error == 0) + (q3_error == 0)) / 3, | |
| total_tokens=total_tokens, | |
| cost=cost, | |
| ) | |
| def usage_to_tokens(usage: OpenAIUsage | AnthropicUsage) -> float: | |
| if isinstance(usage, OpenAIUsage): | |
| return usage.total_tokens or float("-Inf") | |
| elif isinstance(usage, AnthropicUsage): | |
| return usage.input_tokens + usage.output_tokens | |
| return float("-Inf") | |
| class Strategy(ABC): | |
| @property | |
| def name(self) -> str: | |
| base_name = self.__class__.__name__.removesuffix("Strategy") | |
| return base_name | |
| @abstractmethod | |
| async def solve(self, provider: Provider, model: Model) -> PuzzleResult: ... | |
| @prompt_template() | |
| async def puzzle_prompt(): | |
| return puzzle_text | |
| class ResponseModelStrategy(Strategy): | |
| async def solve(self, provider: Provider, model: Model) -> PuzzleResult: | |
| solution = await llm.call( | |
| provider=provider, model=model, response_model=PuzzleResponse | |
| )(puzzle_prompt)() | |
| bcr: BaseCallResponse = solution._response # pyright: ignore[reportAttributeAccessIssue] | |
| usage = bcr.usage | |
| return PuzzleResult.from_response( | |
| solution, "", usage_to_tokens(usage), bcr.cost or float("-Inf") | |
| ) | |
| class ModelWithCoTStrategy(Strategy): | |
| async def solve(self, provider: Provider, model: Model) -> PuzzleResult: | |
| solution: PuzzleResponseWithChainOfThought = await llm.call( | |
| provider=provider, | |
| model=model, | |
| response_model=PuzzleResponseWithChainOfThought, | |
| )(puzzle_prompt)() | |
| bcr: BaseCallResponse = solution._response # pyright: ignore[reportAttributeAccessIssue] | |
| return PuzzleResult.from_response( | |
| PuzzleResponse( | |
| q1=solution.q1, | |
| q2=solution.q2, | |
| q3=solution.q3, | |
| ), | |
| solution.chain_of_thought, | |
| usage_to_tokens(bcr.usage), | |
| bcr.cost or float("-Inf"), | |
| ) | |
| class ModelWithThreeCoTStrategy(Strategy): | |
| async def solve(self, provider: Provider, model: Model) -> PuzzleResult: | |
| solution: PuzzleResponseWithThreeChainsOfThought = await llm.call( | |
| provider=provider, | |
| model=model, | |
| response_model=PuzzleResponseWithThreeChainsOfThought, | |
| )(puzzle_prompt)() | |
| bcr: BaseCallResponse = solution._response # pyright: ignore[reportAttributeAccessIssue] | |
| return PuzzleResult.from_response( | |
| PuzzleResponse( | |
| q1=solution.q1, | |
| q2=solution.q2, | |
| q3=solution.q3, | |
| ), | |
| solution.q1_chain_of_thought | |
| + "\n" | |
| + solution.q2_chain_of_thought | |
| + "\n" | |
| + solution.q3_chain_of_thought, | |
| usage_to_tokens(bcr.usage), | |
| bcr.cost or float("-Inf"), | |
| ) | |
| class TwoStepTextExtractionStrategy(Strategy): | |
| async def solve(self, provider: Provider, model: Model) -> PuzzleResult: | |
| @prompt_template() | |
| async def extract_model(output: str): | |
| return f"Extract the solution from the following text:\n{output}" | |
| text = await llm.call(provider=provider, model=model)(puzzle_prompt)() | |
| bcr1: BaseCallResponse = text._response # pyright: ignore[reportAttributeAccessIssue] | |
| solution = await llm.call( | |
| provider=provider, model=model, response_model=PuzzleResponse | |
| )(extract_model)(text.content) | |
| bcr2: BaseCallResponse = solution._response # pyright: ignore[reportAttributeAccessIssue] | |
| return PuzzleResult.from_response( | |
| solution, | |
| text.content, | |
| usage_to_tokens(bcr1.usage) + usage_to_tokens(bcr2.usage), | |
| (bcr1.cost or float("-Inf")) + (bcr2.cost or float("-Inf")), | |
| ) | |
| class OutputParserStrategy(Strategy): | |
| async def solve(self, provider: Provider, model: Model) -> PuzzleResult: | |
| @prompt_template() | |
| async def puzzle_with_parser(): | |
| return f""" | |
| {puzzle_text} | |
| When solving this problem, first output all your reasoning inside | |
| <thinking></thinking> tags. Then, output the solution inside <answer> tags, | |
| as a JSON object with the following format: | |
| <answer> | |
| {{"q1": 42, "q2": 42,"q3": 42}} | |
| </answer> | |
| """ | |
| bcr: BaseCallResponse | None = None | |
| def parse_answer(response: BaseCallResponse) -> PuzzleResponse: | |
| nonlocal bcr # Hack to retrieve the BCR when using output_parser | |
| bcr = response | |
| answer = response.content.split("<answer>")[1].split("</answer>")[0] | |
| return PuzzleResponse.model_validate_json(answer) | |
| solution = await llm.call( | |
| provider=provider, model=model, output_parser=parse_answer | |
| )(puzzle_with_parser)() | |
| if bcr is None: | |
| raise ValueError("No BaseCallResponse found") | |
| return PuzzleResult.from_response( | |
| solution, | |
| bcr.content, | |
| usage_to_tokens(bcr.usage), | |
| (bcr.cost or float("-Inf")), | |
| ) | |
| strategies: list[Strategy] = [ | |
| ResponseModelStrategy(), | |
| ModelWithCoTStrategy(), | |
| ModelWithThreeCoTStrategy(), | |
| OutputParserStrategy(), | |
| TwoStepTextExtractionStrategy(), | |
| ] | |
| class TestResult(BaseModel): | |
| provider: Provider | |
| model: Model | |
| # Average distance from the correct answer across all questions | |
| avg_error: float | |
| # What proportion of the answers were correct | |
| avg_correctness: float | |
| # Average usage across all tests | |
| avg_tokens: float | |
| # Average cost across all tests | |
| avg_cost: float | |
| # How many trials were run | |
| n_tests: int | |
| # What proportion of attempts failed | |
| failure_rate: float | |
| results: list[PuzzleResult] | |
| class StrategyResult(BaseModel): | |
| strategy_name: str | |
| tests: list[TestResult] | |
| class ProviderLimits(BaseModel): | |
| openai_concurrency: int = 10 | |
| anthropic_concurrency: int = 10 | |
| class Progress: | |
| def __init__(self, total_trials: int): | |
| self.total = total_trials | |
| self.completed = 0 | |
| self.last_print = 0 | |
| self._lock = asyncio.Lock() | |
| async def increment(self): | |
| async with self._lock: | |
| self.completed += 1 | |
| # Only print if we've made 5% more progress since last print | |
| if (self.completed / self.total) - (self.last_print / self.total) >= 0.05: | |
| self._print_progress() | |
| self.last_print = self.completed | |
| # or if we finished | |
| elif self.completed == self.total: | |
| self._print_progress() | |
| print() | |
| def _print_progress(self): | |
| percent = (self.completed / self.total) * 100 | |
| print( | |
| f"\rCompleted {self.completed}/{self.total} trials ({percent:.1f}%)", end="" | |
| ) | |
| class TestRunner: | |
| def __init__(self, limits: ProviderLimits): | |
| self.openai_semaphore = asyncio.Semaphore(limits.openai_concurrency) | |
| self.anthropic_semaphore = asyncio.Semaphore(limits.anthropic_concurrency) | |
| self.n_retries = 3 | |
| self.progress: Progress | None = None | |
| async def run_with_provider_limit( | |
| self, provider: Provider, model: Model, strategy: Strategy | |
| ) -> PuzzleResult: | |
| semaphore = ( | |
| self.openai_semaphore if provider == "openai" else self.anthropic_semaphore | |
| ) | |
| async with semaphore: | |
| logging.info(f"Running {provider} {model} with {strategy.name}") | |
| return await strategy.solve(provider, model) | |
| async def attempt_with_retries( | |
| self, provider: Provider, model: Model, strategy: Strategy | |
| ) -> tuple[PuzzleResult | None, int]: | |
| n_failures = 0 | |
| for attempt in range(self.n_retries): | |
| try: | |
| result = await self.run_with_provider_limit(provider, model, strategy) | |
| if self.progress: | |
| await self.progress.increment() | |
| return result, n_failures | |
| except Exception as e: | |
| logging.info( | |
| f"Error for {provider} {model} (attempt {attempt + 1}):\n\t{str(e)}" | |
| ) | |
| n_failures += 1 | |
| if attempt == self.n_retries - 1: # Last attempt | |
| if self.progress: | |
| await self.progress.increment() | |
| return None, n_failures | |
| return None, n_failures | |
| async def test_model( | |
| self, | |
| provider: Provider, | |
| model: Model, | |
| strategy: Strategy, | |
| n_tests: int, | |
| ) -> TestResult: | |
| total_failures = 0 | |
| results = [] | |
| # Create all tasks | |
| tasks = [ | |
| self.attempt_with_retries(provider, model, strategy) for _ in range(n_tests) | |
| ] | |
| # Run tasks and handle failures | |
| for task in asyncio.as_completed(tasks): | |
| result, failures = await task | |
| total_failures += failures | |
| if result is not None: | |
| results.append(result) | |
| if not results: | |
| raise RuntimeError(f"All attempts failed for {provider} {model}") | |
| return TestResult( | |
| provider=provider, | |
| model=model, | |
| avg_error=sum(result.error for result in results) / len(results), | |
| avg_correctness=sum(result.correctness for result in results) | |
| / len(results), | |
| avg_tokens=sum(result.total_tokens for result in results) / len(results), | |
| avg_cost=sum(result.cost for result in results) / len(results), | |
| n_tests=len(results), | |
| failure_rate=total_failures / (len(results) + total_failures), | |
| results=results, | |
| ) | |
| async def test_strategy(self, strategy: Strategy, n_tests: int) -> StrategyResult: | |
| tests = await asyncio.gather( | |
| *( | |
| self.test_model(provider, model, strategy, n_tests) | |
| for provider, model in models | |
| ) | |
| ) | |
| return StrategyResult(strategy_name=strategy.name, tests=tests) | |
| async def run_all_strategies( | |
| self, strategies: list[Strategy], n_tests: int | |
| ) -> list[StrategyResult]: | |
| # Calculate total trials | |
| total_trials = len(strategies) * len(models) * n_tests | |
| self.progress = Progress(total_trials) | |
| # Run all strategies concurrently | |
| strategy_results = await asyncio.gather( | |
| *(self.test_strategy(strategy, n_tests) for strategy in strategies) | |
| ) | |
| return strategy_results | |
| def print_model_result( | |
| model_provider: Provider, model_name: Model, strategy_results: list[StrategyResult] | |
| ): | |
| model_data = [] | |
| for result in strategy_results: | |
| test = next( | |
| t | |
| for t in result.tests | |
| if t.provider == model_provider and t.model == model_name | |
| ) | |
| model_data.append( | |
| { | |
| "Strategy": result.strategy_name, | |
| "Error": f"{test.avg_error:.1f}", | |
| "Correct": f"{test.avg_correctness * 100:.0f}%", | |
| "Tokens": f"{test.avg_tokens:.0f}", | |
| "Cost": f"{test.avg_cost * 100:.2f}¢", | |
| "Failure Rate": f"{test.failure_rate * 100:.0f}%", | |
| } | |
| ) | |
| # Calculate averages | |
| n_strats = len(model_data) | |
| avg_error = sum(float(d["Error"]) for d in model_data) / n_strats | |
| avg_correct = sum(int(d["Correct"].rstrip("%")) for d in model_data) / n_strats | |
| avg_tokens = sum(float(d["Tokens"]) for d in model_data) / n_strats | |
| avg_cost = sum(float(d["Cost"].rstrip("¢")) for d in model_data) / n_strats | |
| avg_failure = sum(int(d["Failure Rate"].rstrip("%")) for d in model_data) / n_strats | |
| model_data.extend( | |
| [ | |
| { | |
| "Strategy": "---", | |
| "Error": "---", | |
| "Correct": "---", | |
| "Tokens": "---", | |
| "Cost": "---", | |
| "Failure Rate": "---", | |
| }, | |
| { | |
| "Strategy": "AVERAGE", | |
| "Error": f"{avg_error:.1f}", | |
| "Correct": f"{avg_correct:.0f}%", | |
| "Tokens": f"{avg_tokens:.0f}", | |
| "Cost": f"{avg_cost:.2f}¢", | |
| "Failure Rate": f"{avg_failure:.0f}%", | |
| }, | |
| ] | |
| ) | |
| print(f"\n{model_provider} {model_name}") | |
| print(tabulate(model_data, headers="keys", tablefmt="simple")) | |
| print("\n") | |
| async def main(): | |
| parser = argparse.ArgumentParser() | |
| parser.add_argument("--dump", type=Path, help="Path to dump detailed results") | |
| parser.add_argument( | |
| "-n", | |
| "--trials-per-model", | |
| type=int, | |
| default=5, | |
| help="Number of trials per model (default: 5)", | |
| ) | |
| parser.add_argument( | |
| "--openai-concurrency", | |
| type=int, | |
| default=10, | |
| help="Maximum concurrent OpenAI requests (default: 10)", | |
| ) | |
| parser.add_argument( | |
| "--anthropic-concurrency", | |
| type=int, | |
| default=10, | |
| help="Maximum concurrent Anthropic requests (default: 10)", | |
| ) | |
| args = parser.parse_args() | |
| limits = ProviderLimits( | |
| openai_concurrency=args.openai_concurrency, | |
| anthropic_concurrency=args.anthropic_concurrency, | |
| ) | |
| runner = TestRunner(limits) | |
| print("Running trials...") | |
| start = time.time() | |
| strategy_results = await runner.run_all_strategies( | |
| strategies, args.trials_per_model | |
| ) | |
| end = time.time() | |
| print(f"Time taken: {end - start:.2f} seconds") | |
| print("\n=== Results by Strategy ===") | |
| for result in strategy_results: | |
| print_result(result) | |
| print("\n=== Results by Model ===") | |
| for provider, model in models: | |
| print_model_result(provider, model, strategy_results) | |
| if args.dump: | |
| args.dump.write_text( | |
| json.dumps([result.model_dump() for result in strategy_results], indent=2) | |
| ) | |
| def print_result(result: StrategyResult): | |
| n_tests = len(result.tests) | |
| # Calculate averages | |
| avg_error = sum(t.avg_error for t in result.tests) / n_tests | |
| avg_correct = sum(t.avg_correctness for t in result.tests) / n_tests | |
| avg_tokens = sum(t.avg_tokens for t in result.tests) / n_tests | |
| avg_cost = sum(t.avg_cost for t in result.tests) / n_tests | |
| avg_failure_rate = sum(t.failure_rate for t in result.tests) / n_tests | |
| # Model results table with summary row | |
| model_data = [ | |
| { | |
| "Model": f"{test.provider} {test.model}", | |
| "Error": f"{test.avg_error:.1f}", | |
| "Correct": f"{test.avg_correctness * 100:.0f}%", | |
| "Tokens": f"{test.avg_tokens:.0f}", | |
| "Cost": f"{test.avg_cost * 100:.2f}¢", | |
| "Failure Rate": f"{test.failure_rate * 100:.0f}%", | |
| } | |
| for test in result.tests | |
| ] + [ | |
| { | |
| "Model": "---", | |
| "Error": "---", | |
| "Correct": "---", | |
| "Tokens": "---", | |
| "Cost": "---", | |
| "Failure Rate": "---", | |
| }, | |
| { | |
| "Model": "AVERAGE", | |
| "Error": f"{avg_error:.1f}", | |
| "Correct": f"{avg_correct * 100:.0f}%", | |
| "Tokens": f"{avg_tokens:.0f}", | |
| "Cost": f"{avg_cost * 100:.2f}¢", | |
| "Failure Rate": f"{avg_failure_rate * 100:.0f}%", | |
| }, | |
| ] | |
| print(f"{result.strategy_name}") | |
| print(tabulate(model_data, headers="keys", tablefmt="simple")) | |
| print("\n") | |
| if __name__ == "__main__": | |
| asyncio.run(main()) |
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| === Results by Strategy === | |
| ResponseModel | |
| Model Error Correct Tokens Cost Failure Rate | |
| ---------------------------------- ------- --------- -------- ------ -------------- | |
| openai gpt-4o-mini 16.5 10% 281 0.01¢ 0% | |
| openai gpt-4o 6.1 3% 277 0.10¢ 0% | |
| anthropic claude-3-5-haiku-latest 9.4 7% 730 0.56¢ 33% | |
| anthropic claude-3-5-sonnet-latest 7.1 3% 692 0.30¢ 0% | |
| --- --- --- --- --- --- | |
| AVERAGE 9.8 6% 495 0.24¢ 8% | |
| ModelWithCoT | |
| Model Error Correct Tokens Cost Failure Rate | |
| ---------------------------------- ------- --------- -------- ------ -------------- | |
| openai gpt-4o-mini 43.8 37% 1000 0.05¢ 17% | |
| openai gpt-4o 11.3 57% 787 0.60¢ 0% | |
| anthropic claude-3-5-haiku-latest 15.6 40% 1097 0.72¢ 23% | |
| anthropic claude-3-5-sonnet-latest 1.2 60% 1198 1.02¢ 0% | |
| --- --- --- --- --- --- | |
| AVERAGE 18.0 48% 1020 0.60¢ 10% | |
| ModelWithThreeCoT | |
| Model Error Correct Tokens Cost Failure Rate | |
| ---------------------------------- ------- --------- -------- ------ -------------- | |
| openai gpt-4o-mini 5.3 47% 816 0.04¢ 29% | |
| openai gpt-4o 2.7 43% 713 0.51¢ 0% | |
| anthropic claude-3-5-haiku-latest 8.7 47% 1297 0.82¢ 0% | |
| anthropic claude-3-5-sonnet-latest 0.1 77% 1367 1.19¢ 0% | |
| --- --- --- --- --- --- | |
| AVERAGE 4.2 53% 1048 0.64¢ 7% | |
| OutputParser | |
| Model Error Correct Tokens Cost Failure Rate | |
| ---------------------------------- ------- --------- -------- ------ -------------- | |
| openai gpt-4o-mini 9.1 33% 1035 0.05¢ 0% | |
| openai gpt-4o 1.2 47% 953 0.77¢ 0% | |
| anthropic claude-3-5-haiku-latest 7.5 43% 645 0.37¢ 0% | |
| anthropic claude-3-5-sonnet-latest 0.3 63% 766 0.82¢ 17% | |
| --- --- --- --- --- --- | |
| AVERAGE 4.5 47% 850 0.50¢ 4% | |
| TwoStepTextExtraction | |
| Model Error Correct Tokens Cost Failure Rate | |
| ---------------------------------- ------- --------- -------- ------ -------------- | |
| openai gpt-4o-mini 2.9 63% 2313 0.08¢ 0% | |
| openai gpt-4o 1.8 53% 1954 1.14¢ 0% | |
| anthropic claude-3-5-haiku-latest 9.0 30% 1396 0.96¢ 9% | |
| anthropic claude-3-5-sonnet-latest 0.6 70% 1637 1.14¢ 41% | |
| --- --- --- --- --- --- | |
| AVERAGE 3.6 54% 1825 0.83¢ 13% | |
| === Results by Model === | |
| openai gpt-4o-mini | |
| Strategy Error Correct Tokens Cost Failure Rate | |
| --------------------- ------- --------- -------- ------ -------------- | |
| ResponseModel 16.5 10% 281 0.01¢ 0% | |
| ModelWithCoT 43.8 37% 1000 0.05¢ 17% | |
| ModelWithThreeCoT 5.3 47% 816 0.04¢ 29% | |
| OutputParser 9.1 33% 1035 0.05¢ 0% | |
| TwoStepTextExtraction 2.9 63% 2313 0.08¢ 0% | |
| --- --- --- --- --- --- | |
| AVERAGE 15.5 38% 1089 0.05¢ 9% | |
| openai gpt-4o | |
| Strategy Error Correct Tokens Cost Failure Rate | |
| --------------------- ------- --------- -------- ------ -------------- | |
| ResponseModel 6.1 3% 277 0.10¢ 0% | |
| ModelWithCoT 11.3 57% 787 0.60¢ 0% | |
| ModelWithThreeCoT 2.7 43% 713 0.51¢ 0% | |
| OutputParser 1.2 47% 953 0.77¢ 0% | |
| TwoStepTextExtraction 1.8 53% 1954 1.14¢ 0% | |
| --- --- --- --- --- --- | |
| AVERAGE 4.6 41% 937 0.62¢ 0% | |
| anthropic claude-3-5-haiku-latest | |
| Strategy Error Correct Tokens Cost Failure Rate | |
| --------------------- ------- --------- -------- ------ -------------- | |
| ResponseModel 9.4 7% 730 0.56¢ 33% | |
| ModelWithCoT 15.6 40% 1097 0.72¢ 23% | |
| ModelWithThreeCoT 8.7 47% 1297 0.82¢ 0% | |
| OutputParser 7.5 43% 645 0.37¢ 0% | |
| TwoStepTextExtraction 9.0 30% 1396 0.96¢ 9% | |
| --- --- --- --- --- --- | |
| AVERAGE 10.0 33% 1033 0.69¢ 13% | |
| anthropic claude-3-5-sonnet-latest | |
| Strategy Error Correct Tokens Cost Failure Rate | |
| --------------------- ------- --------- -------- ------ -------------- | |
| ResponseModel 7.1 3% 692 0.30¢ 0% | |
| ModelWithCoT 1.2 60% 1198 1.02¢ 0% | |
| ModelWithThreeCoT 0.1 77% 1367 1.19¢ 0% | |
| OutputParser 0.3 63% 766 0.82¢ 17% | |
| TwoStepTextExtraction 0.6 70% 1637 1.14¢ 41% | |
| --- --- --- --- --- --- | |
| AVERAGE 1.9 55% 1132 0.89¢ 12% |
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| { | |
| "q1": 28.0, | |
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| "reasoning": "To find the number of widgets produced in a week, let's first calculate the production for each day. \n- Weekdays (Monday to Friday): 4 widgets each = 4 widgets * 5 weekdays = 20 widgets\n- Weekend (Saturday and Sunday): 3 widgets each = 3 widgets * 2 weekend days = 6 widgets\n- Extra widgets on Monday: 2 extra widgets\nTotal for the week = 20 + 6 + 2 = 28 widgets produced in a week.", | |
| "error": 53.0, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. The production schedule is: \n- Weekdays: 4 widgets/day, 5 weekdays = 5 * 4 = 20 widgets \n- Weekend days: 3 widgets/day, 2 weekend days = 2 * 3 = 6 widgets \n- Extra widgets on Monday: 2 widgets. \nSo the total widgets produced in a week = 20 (weekdays) + 6 (weekend) + 2 (extra on Monday) = 28 widgets.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 881.0, | |
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| }, | |
| { | |
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| "q2": 0.0, | |
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| "reasoning": "To calculate the total number of widgets produced in a week, we break down the production by weekdays and weekends. The factory produces: \n- 4 widgets every weekday (Monday to Friday) \n- 3 widgets each on Saturday and Sunday, plus 2 extra on Monday. \n\nCalculating the total for a week:\n- Monday: 4 (regular) + 2 (extra) = 6 \n- Tuesday: 4 \n- Wednesday: 4 \n- Thursday: 4 \n- Friday: 4 \n- Saturday: 3 \n- Sunday: 3 \n\nTotal = 6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 \n\nTherefore, Q1: The factory produces 28 widgets in a week.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
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| "reasoning": "To find the total widget production in a week, we calculate the daily production. \n\n- Weekdays (Monday to Friday): 4 widgets per day * 5 days = 20 widgets\n- Weekends (Saturday and Sunday): 3 widgets per day * 2 days = 6 widgets\n- Extra widgets on Monday: 2\n\nTotal widgets in a week = Weekday production + Weekend production + Extra widgets on Monday = 20 + 6 + 2 = 28.\n\nThus, the total number of widgets produced in a week is 28.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 964.0, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "In a week, the factory produces widgets as follows:\n- Weekdays (Monday to Friday): 4 widgets per day \u2192 4 widgets * 5 days = 20 widgets\n- Weekends (Saturday and Sunday): 3 widgets per day \u2192 3 widgets * 2 days = 6 widgets\n- On Monday, there are 2 extra widgets added \u2192 +2 widgets\nTotal for one week = 20 + 6 + 2 = 28 widgets.\nTherefore, Q1: The total number of widgets produced in a week is 28.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 955.0, | |
| "cost": 0.00046095 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "To calculate the total number of widgets produced in a week:\n\n1. **Weekdays Production**: \n - From Monday to Friday (5 days) the factory produces 4 widgets each day: \n - 5 days * 4 widgets/day = 20 widgets\n\n2. **Weekend Production**: \n - Saturday and Sunday (2 days) the factory produces 3 widgets each day:\n - 2 days * 3 widgets/day = 6 widgets\n\n3. **Extra Mondays**: \n - On Monday, there are 2 extra widgets:\n - 2 extra widgets on Monday\n\nTotal widgets produced in one week = 20 (weekdays) + 6 (weekend) + 2 (extra Monday) = 28 widgets.\n\nThus, Q1: 28 widgets are produced in a week.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1128.0, | |
| "cost": 0.00056475 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 21.0, | |
| "q3": 19.0, | |
| "reasoning": "In Standardtopia, the weekly production schedule for widgets is as follows:\n- Weekdays (Monday to Friday): 4 widgets/day * 5 days = 20 widgets\n- Weekends (Saturday and Sunday): 3 widgets/day * 2 days = 6 widgets\n- Extra on Monday: 2 widgets\n\nAdding these together will give the total weekly production:\nTotal weekly production = 20 (weekdays) + 6 (weekend) + 2 (extra) = 28 widgets.\nTherefore, the answer to Q1 is 28 widgets.", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1153.0, | |
| "cost": 0.00057975 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "To calculate the total widgets produced in a week, we first need to break down the production by day: \n- Weekdays (Monday to Friday): 4 widgets/day * 5 days = 20 widgets \n- Saturday and Sunday: 3 widgets/day * 2 days = 6 widgets \n- Extra widgets on Monday: 2 widgets \n\nTotal widgets produced in a week = 20 + 6 + 2 = 28 widgets.\n\nFor Q2, each month has 4 weeks, so the factory would typically produce 4 weeks * 28 widgets/week = 112 widgets per month. However, with one holiday each month, we need to remove the production loss due to the holiday. The holiday can be on any day, so we calculate the average loss: 28 widgets/week / 7 days/week \u2248 4 widgets per day lost due to the holiday. Hence, the production on a holiday is 4 widgets. So the adjusted monthly production is 112 - 4 = 108 widgets. This gives us an average weekly production of (108 widgets/month) / 4 weeks = 27 widgets/week.\n\nFor Q3, we identify the fibonacci numbers within 1 to 10 months: 1, 2, 3, 5, 8. Therefore, months 1, 2, 3, 5, and 8 will have 2 extra holidays adding a total of 10 extra holidays (2 holidays * 5 fibonacci months). \nThis totals to 10 holidays in a year. Each holiday represents an average of 4 widgets not produced. \nSo total widgets lost due to holidays in the year = 10 holidays * 4 widgets = 40 widgets. \nThe total proposed production without holidays in a year would be 28 widgets/week * 52 weeks = 1456 widgets.\nAdjusting for holidays, we'd have 1456 - 40 = 1416 widgets for the year. Thus, the average would be 1416 widgets/year \u00f7 52 weeks = ~27.23 widgets/week.\n\nOverall: \n- Q1: 28 \n- Q2: 27 \n- Q3: approximately 27.23", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
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| }, | |
| { | |
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| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "To find the total number of widgets produced in a week, we need to calculate the output for weekdays and weekends, then add the extra widgets produced on Mondays. \n\n- Weekdays (Monday to Friday): 5 days \u00d7 4 widgets = 20 widgets. \n- Weekend (Saturday and Sunday): 2 days \u00d7 3 widgets = 6 widgets. \n- Extra widgets on Monday: 2 widgets. \n\nTotal widgets produced in a week = 20 + 6 + 2 = 28 widgets.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 878.0, | |
| "cost": 0.00041474999999999997 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. On weekdays, 4 widgets are produced each day, and on Mondays, 2 extra widgets are produced. Thus, the total widgets produced in a week can be calculated as follows: \n\n- For 5 weekdays: 5 days * 4 widgets/day = 20 widgets \n- For 2 weekend days: 2 days * 3 widgets/day = 6 widgets \n- For extra widgets on Monday (counted in weekdays): 2 extra widgets \n\nTotal = 20 + 6 + 2 = 28 widgets. \n\nTherefore, the total number of widgets produced in a week is 28.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1000.0, | |
| "cost": 0.00048795 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "openai", | |
| "model": "gpt-4o", | |
| "avg_error": 11.2825, | |
| "avg_correctness": 0.5666666666666667, | |
| "avg_tokens": 786.8, | |
| "avg_cost": 0.006000500000000001, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.0, | |
| "reasoning": "To solve these questions, we must first understand the widget production schedule and consider different factors.\\n1. Widgets produced per week without holidays is: \\n- Weekdays: 5 days * 4 widgets = 20 widgets\\n- Weekends: 2 days * 3 widgets = 6 widgets\\n- Monday's extra widgets: 1 day * 2 widgets = 2 widgets\\nTherefore, total widgets in a week = 20 + 6 + 2 = 28 widgets\\n\\n2. Adjusting for one holiday per month within the 4-week period:\\n- Widgets lost due to one holiday (irrespective of day) =\\n\\n\\n\\n3. Adjusting for extra holidays on Fibonacci month indices: The Fibonacci months (choosing from the 1st month, 2nd month, 3rd month, 5th month, and 8th month out of 10 months) give 2 extra holidays in 5 months = 10 additional holidays \\n\\nCombining all these, we assess the impact on the widget production per week.", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 749.0, | |
| "cost": 0.0056225 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's break down each question to find the answers.\n\nFor Q1:\n- Weekdays: 5 days \u00d7 4 widgets/day = 20 widgets.\n- Weekend: 2 days \u00d7 3 widgets/day = 6 widgets.\n- Extra widgets on Monday: 2 widgets.\nTherefore, total widgets in a week = 20 + 6 + 2 = 28 widgets.\n\nFor Q2:\n- One holiday per month means 4 - 1 = 3 operational weeks per month after accounting for a holiday.\n- Total monthly widget production without holiday = 4 weeks \u00d7 28 widgets/week = 112 widgets.\n- Average widgets per week factoring in the holiday = 112 / 4 = 28 widgets (but without the holiday loss deduction). Let's consider one extra pause per month.\n- Adjusted monthly production: 112 - 4 (1 day's production) = 108 widgets.\n- Average weekly production = 108 / 4 = 27 widgets.\n\nFor Q3:\n- Months with 1-based Fibonacci indices are: 1, 2, 3, 5, 8 which totals to 5 months with 2 additional holidays each.\n- Thus, 5 months \u00d7 2 extra holidays = 10 extra holidays yearly.\n- Total holidays in a year = 10 + 10 (one per month) = 20 holidays.\n- Therefore, 20 holiday breaks within 40 weeks of operation over the 4 weeks/month pattern = 4 weeks \u00d7 10 months.\n- Yearly production without extra breaks: 40 weeks \u00d7 28 widgets = 1120 widgets.\n- Deduct for holiday effect: 20 holidays \u00d7 (4 daily widgets on average) = 80 widgets less per year.\n- Adjusted yearly output = 1120 - 80 = 1040.\n- Thus, average weekly production adjusting for all holidays = 1040 / 40 = 26 widgets.", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 702.0, | |
| "cost": 0.0051525 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.5, | |
| "q3": 26.5, | |
| "reasoning": "First, calculate the total number of widgets produced in a typical week. There are 5 weekdays, with 4 widgets produced each day, so that's 5 * 4 = 20 widgets. Additionally, 3 widgets are produced each day on the weekend, yielding 2 * 3 = 6 widgets. Therefore, without any additional considerations, the factory produces 20 + 6 = 26 widgets per week. However, since there are also 2 extra widgets produced on Mondays, we must add 2 more, resulting in 28 widgets per week total.\n\nQ1: A typical week results in 28 widgets without considering holidays.\n\nNext, consider the monthly holidays. Since there are 4 weeks in a month, with 1 holiday distributed randomly, on average we can assume that this is about 0.25 holidays/week.\n\nSo, the expected production per week, accounting for the average holiday, is 28 widgets/week - 4 widgets (if it's a weekday) + (possibly 3 on a weekend, so we'll just consider worst case: weekday) = 28 - 1 = 27.5 widgets per week.\n\nQ2: Considering holidays, the average becomes approximately 27.5 widgets per week.\n\nNext, account for the Fibonacci holidays. The months that are Fibonacci numbers (specifically months 1, 2, 3, 5, and 8 out of 10) have a total of 2 extra holidays per month, or 2/4 = 0.5 more holidays per week, effectively adding another 0.5 holidays per week for 5 months -- that's 2.5 additional holidays spread over a year (5 months) and thus this should average to about 0.25 more holidays ocurred in a month , so about 1/4 = 0.25 extra holidays per week over a year.\n\nThis reduces production by another approximately 28 widgets/4 = 1 widget making it 26.5.", | |
| "error": 1.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 706.0, | |
| "cost": 0.0051925 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.75, | |
| "q3": 27.5, | |
| "reasoning": "To solve these questions, let's calculate step by step.\n\nQ1: A week in Standardtopia consists of 5 weekdays and 2 weekend days. On weekdays, 4 widgets are produced, except Monday, where 2 additional widgets are also produced. So, the calculation is:\n- Monday produces 6 (4 + 2 extra) widgets.\n- Tuesday to Friday produce 4 each, so that's 4 days * 4 widgets = 16 widgets.\n- Saturday and Sunday produce 3 each, so that's 2 days * 3 widgets = 6 widgets.\nTotal weekly widgets = 6 (Monday) + 16 (Tue-Fri) + 6 (weekend) = 28 widgets.\n\nQ2: In a month, there are 4 weeks and 1 holiday, meaning there are 28 - 1/4 = 27.75 widgets produced per week, accounting for one holiday.\n\nQ3: Considering fibonacci indexed months (1, 2, 3, 5, 8) have extra 2 holidays, that's 5 months with 3 holidays total (1 + 2 extra) and 5 months with 1 holiday each. \nTotal holidays = 5 months * 3 holidays + 5 months * 1 holiday = 20 holidays in 10 months.\nTotal weeks in a year = 4 weeks/month * 10 months = 40 weeks.\nAverage holidays per week = 20 holidays / 40 weeks = 0.5 holidays.\nAverage widgets per week accounting for Fibonacci holidays = 28 - 0.5 = 27.5 widgets per week.", | |
| "error": 2.25, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 636.0, | |
| "cost": 0.0044925 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "Let's start by solving the first question. A week consists of 5 weekdays and 2 weekend days. Therefore, on weekdays (Monday to Friday), 4 widgets are produced each day. However, on Mondays, there are an additional 2 widgets produced, making it 6 on Monday. So, the total is 6 widgets on Monday, 4 widgets each on Tuesday through Friday, and 3 widgets each on Saturday and Sunday. Thus, in a typical week, these widgets are produced as follows: \n\nMonday: 6 widgets \nTuesday to Friday: 4 widgets each day, totaling 4 * 4 = 16 widgets\nSaturday & Sunday: 3 widgets each, totaling 3 * 2 = 6 widgets\n\nSo, total widgets produced per week = 6 + 16 + 6 = 28 widgets.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 878.0, | |
| "cost": 0.0069125 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 0.0, | |
| "q3": 0.0, | |
| "reasoning": "First, calculate the number of widgets produced in a standard week. There are 5 weekdays and 2 weekend days. Weekdays: 5 days * 4 widgets/day = 20 widgets. Weekends: 2 days * 3 widgets/day = 6 widgets. Additionally, 2 extra widgets are produced on Mondays. Total per week: 20 + 6 + 2 = 28 widgets. Thus, in a Standardtopia week, 28 widgets are produced.", | |
| "error": 53.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 852.0, | |
| "cost": 0.0066525 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.04, | |
| "q3": 27.035, | |
| "reasoning": "Let's break down the problem into parts.\n\nQ1: Weekly Production\n- Weekdays: 5 days \\* 4 widgets/day = 20 widgets\n- Weekend: 2 days \\* 3 widgets/day = 6 widgets\n- Extra widgets on Mondays: 2 widgets\n- Total weekly production = 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly Production with Holidays\n- 4 weeks = 4 \\* 28 = 112 widgets\n- One holiday reduces production by the average day's production:\n - Average day in a month (no holiday): 112 widgets / 29 days = ~3.86 widgets/day\n- New monthly production = 112 - 3.86 = 108.14 widgets\n- Weekly average: 108.14 widgets / 4 weeks = 27.04 widgets\n\nQ3: Accounting for Fibonacci Holidays\n- Fibonacci months: 1, 2, 3, 5, 8 = 5 out of 10 months\n- Each has 2 extra holidays = 5 months \\* 2 = 10 additional holidays in a year\n- Total holidays in a year: 10 (regular) + 10 (extra) = 20\n- Adjusted production loss for additional 10 holidays:\n - 10 holidays \\* ~3.86 widgets/day = 38.6 widgets total reduction\n- Annual production without any holidays: 10 months \\* 112 widgets = 1120 widgets\n- Adjusted annual production: 1120 - 38.6 = 1081.4 widgets\n- Average weekly production in a year: 1081.4 widgets / 40 weeks = 27.035 widgets", | |
| "error": 1.0749999999999993, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 654.0, | |
| "cost": 0.0046725000000000004 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "\n\nTo solve these problems, we first need to understand the production schedule and holidays.\n\n**Q1:** Weekly production calculation:\n- Weekdays (Monday to Friday): 4 widgets/day for 5 days = 20 widgets\n- Weekend (Saturday and Sunday): 3 widgets/day for 2 days = 6 widgets\n- Extra widgets on Monday: 2 widgets\n\nTotal weekly production without holidays = 20 (weekdays) + 6 (weekend) + 2 (extra Monday) = 28 widgets.\n\n**Q2:** Monthly production and holiday consideration:\n- Each month has 4 weeks = 4 * 7 = 28 days.\n- 1 holiday per month means one less production day, assuming holidays are uniformly distributed.\n- Average weekly production with 1 holiday per month = (Total monthly widgets - widgets missed on holiday)/4 weeks\n\n Calculating average weekly production:\n - Total days affected by a holiday (assuming it affects weekdays primarily):\n - If affect on weekend: 3 widgets/day\n - If affect on Monday: 6 widgets (4 regular + 2 extra)\n - If affect on any other weekday: 4 widgets/day\n\n Since holidays uniformly affect all days and majority are weekdays, average widget loss per holiday = 4.\n \n Total production per month without holidays = 28 widgets/week * 4 weeks = 112 widgets\n \n Total production per month with holidays = 112 - 4 (average loss per holiday) = 108 widgets\n \n Average weekly production with holidays = 108 widgets/month / 4 weeks = 27 widgets/week.\n\n**Q3:** Fibonacci holidays consideration:\n- Fibonacci month indices: 1, 2, 3, 5, 8 (so 5 out of 10 months)\n- Each such month will result in 2 additional holidays. \n- Calculation:\n Total additional holidays = 2 * 5 = 10 holidays/year additional due to Fibonacci number months\n \n With total regular holidays = 10 holidays/year + intermittent Fibonacci holidays (5 months * 2 holidays = 10)\n Total holidays per year = 20 holidays (original 10 + extra 10)\n \n Yearly production without any holidays = 28 widgets/week * 4 weeks/month * 10 months = 1120 widgets/year.\n\n Yearly widgets lost due to holidays (considering an average daily loss of 4 widgets based on uniform distribution):\n Loss = 20 holidays * 4 widgets = 80 widgets/year\n\n Yearly production with all holidays = 1120 - 80 = 1040 widgets.\n\n Average weekly production taking Fibonacci holidays into account = 1040 widgets/year / (10 months * 4 weeks/month) = 26 widgets/week.\n\nAnswers:\n1. 28 widgets\n2. 27 widgets\n3. 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 923.0, | |
| "cost": 0.0073625 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.5, | |
| "reasoning": "To solve these questions, we start by determining the production of widgets for each day of the week. Given the weekly production schedule, the factory produces 4 widgets each weekday and 3 on weekends, with an additional 2 widgets on Mondays. Observing these outputs reveals the best path to solving each question: \n\n- For Q1, the calculation is straightforward. Calculate the week's production by first determining the number of widgets produced on weekdays and weekends, then adding the extra 2 widgets made on Monday:\n - Weekdays (Mon-Fri): 5 days x 4 widgets = 20 widgets\n - Monday extra: 20 widgets + 2 = 22 widgets\n - Weekends (Sat & Sun): 2 days x 3 widgets = 6 widgets\n - Total production per week: 22 + 6 = 28 widgets\n \n- For Q2, consider the monthly setup where each month loses one day to a holiday, impacting the widget totals. Calculating production changes involves dividing the monthly widget count by four weeks and one holiday:\n - Monthly Potential: With 1 holiday and 28 weekly widgets, one week's worth of output is reduced by 1 day.\n - Weekly average production = (Monthly production - 1 day loss) / 4 weeks:\n - Monthly production = 4 weeks x 28 widgets/week = 112 widgets per month\n - Weekly average with 1 holiday = (112 - 4 widgets) / 4 weeks = 108 / 4 = 27 widgets/week\n \n- For Q3, modify the calendar to account for the 2 extra holidays in Fibonacci months (1, 1, 2, 3, 5, 8), adjusting the average production likewise.\n - Yearly holiday impact: 10 months x 1 holiday = 10 holidays + 5 additional (Fibonacci months) = 15 holidays\n - Effective monthly holidays = 15 / 10 months = 1.5 + 1 = 2.5 holidays monthly\n - Recalculate average weekly production with 2.5 holidays affecting monthly production:\n - Monthly production = 112 widgets - 2.5 days x 4 = 112 - 10 = 102 widgets\n - New weekly average = 102 widgets / 4 weeks = 25.5 widgets/week\n\nWe will now execute these calculations.", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 794.0, | |
| "cost": 0.006072500000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.0, | |
| "reasoning": "To solve these questions, let's break them down step by step.\n\n**Q1: Calculate weekly widget production without holidays.**\n- Weekdays (Monday to Friday): 5 days. Each day produces 4 widgets.\n- Weekends (Saturday and Sunday): 2 days. Each day produces 3 widgets.\n- Additional widgets on Mondays: 2 extra widgets.\n\nSo, the weekly production is:\n- Weekday production: 5 days * 4 widgets = 20 widgets\n- Weekend production: 2 days * 3 widgets = 6 widgets\n- Additional Monday production: 2 widgets\n\nTotal widgets per week = 20 + 6 + 2 = 28 widgets.\n\n**Q2: Calculate average weekly widget production with monthly holiday.**\n- Total widgets in 4 weeks (with Monday's additional widgets included): 4 weeks * 28 widgets = 112 widgets\n- 1 holiday per month means one less day of production. Since holidays are uniformly distributed, let's assume one holiday happens on a weekday.\n - If it's a weekday (loses 4 widgets), if it's on the weekend, it affects weekend production (loses 3 widgets), but since holidays are uniformly distributed, we take an average impact here.\n\nAssume the holiday equally impacts weekdays and weekends (weighted average), mostly it's likely a weekday off.\n- If holiday falls on a weekday (5/7 probability or 5 days), it costs an average of 4 widgets.\n\nSo, effectively a month with a holiday produces: 112 - 4 = 108 widgets.\nAverage per week with the holiday: 108 widgets / 4 weeks = 27 widgets/week.\n\n**Q3: Incorporate Fibonacci months with 2 more holidays each.**\n- Fibonacci numbers up to 10, the sequence: 1, 2, 3, 5, 8.\nThis indicates months 1, 2, 3, 5, and 8 have 2 additional holidays.\n- Total normal holidays in a year: 10 months * 1 holiday = 10 holidays\n- Additional for Fibonacci months: 5 months * 2 holidays = 10 holidays\n- Total holidays in year = 10 normal + 10 additional = 20 holidays\n- Months with Fibonacci holiday weeks impact: \n - Each added holiday in Fibonacci month loses 4 widgets (let's count avg as we did)\nThus effective monthly production without holidays: 112 widgets\n\nCalculating effective production per week over years: \n- Suppose Fibonacci holidays are each losing us the weekday average (weighted) of 4:2 (approximation per impact) with longer impacts on weekdays.\n- Probability strikes with weekdays: We'll assume a moderate loss-two days of weekend impact on Fibonacci holidays.\n- Annually, widgets produced when distributed = (10 months * 108 widgets - 10 * 4 + extra 10*4 from two days): 1008 - 22 (total lost during fib holidays)/4 weeks.\nEffectively 2 (+2 assuming extra 2x impacts) average annual production is 26 widgets.\n\nThus, it becomes ~26 widgets per week over a year inclusive of all holiday and Fibonacci impacts.", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 974.0, | |
| "cost": 0.007872500000000001 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-haiku-latest", | |
| "avg_error": 15.556999999999999, | |
| "avg_correctness": 0.4, | |
| "avg_tokens": 1096.7, | |
| "avg_cost": 0.0071548, | |
| "n_tests": 10, | |
| "failure_rate": 0.23076923076923078, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 25.5, | |
| "q3": 23.5, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Extra Monday widgets: 2 additional widgets\n- Weekend (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\nTotal: 20 + 2 + 6 = 28 widgets per week\n\nQ2: Average widgets per month with single holiday\n- Regular weeks: 28 widgets\n- One holiday distributed across the week reduces production by 1/7 on average\n- 28 * (6/7) = 24 widgets\n- Rounded to 25.5 to account for the fractional holiday impact\n\nQ3: Fibonacci holiday months\nFibonacci numbers \u2264 10 are: 1, 2, 3, 5, 8\nThese months get 2 extra holidays each\n- 5 months with 2 extra holidays\n- This reduces production further\n- 25.5 * (8/10) = 23.5 widgets per week on average", | |
| "error": 4.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1041.0, | |
| "cost": 0.006932 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 21.0, | |
| "q3": 19.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week calculation\n- Weekdays (Monday to Friday): 4 * 5 = 20 widgets\n- Weekends (Saturday and Sunday): 3 * 2 = 6 widgets\n- Extra widgets on Monday: 2\nTotal widgets per week: 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly widget production\n- 4 weeks per month\n- 1 holiday per month, distributed uniformly\n- On a holiday, 0 widgets are produced\n- So average widgets per week = (28 * 3 + 0 * 1) / 4 = 21 widgets\n\nQ3: Fibonacci holiday months\n- Fibonacci numbers: 1, 2, 3, 5, 8 (as 1-based month indices)\n- These months get 2 extra holidays\n- 10 months total\n- So 5 months have 2 extra holidays\n- This means 5 additional holiday days per year\n- These additional holidays will further reduce widget production\n- Average widgets per week will be slightly less than 21", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1048.0, | |
| "cost": 0.00696 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 21.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week calculation\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Extra Monday widgets: 2 additional widgets\n- Weekend (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\nTotal weekly widgets = 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly widget production with one holiday\n- Normal weeks: 28 widgets per week\n- One random holiday per month reduces production\n- Holiday occurs on an average of 1/7 of a week\n- Reduced weekly production: 28 \u00d7 (6/7) = 24 widgets per week\n\nQ3: Fibonacci holiday months\nFibonacci numbers (1-based month index): 1, 2, 3, 5, 8\nThese months get 2 extra holidays\n- Assume 5 months out of 10 are fibonacci months\n- Each fibonacci month has 2 extra holidays\n- Average reduction: (5/10) \u00d7 (2/7) = 1/7 of additional reduction\n- Final weekly production: 24 \u00d7 (6/7) = 20.57 widgets per week\n\nLet's round for the final submission.", | |
| "error": 8.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1072.0, | |
| "cost": 0.007056 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 16.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widget production per week\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Extra Monday widgets: 2 additional widgets\n- Weekend days (Saturday, Sunday): 3 widgets \u00d7 2 days = 6 widgets\n- Total weekly production: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly widget production with one holiday\n- In a month with four weeks and one holiday\n- First, calculate base weekly production: 28 widgets\n- Holiday reduces production by 1/7 (assuming uniform distribution)\n- Average weekly production: 28 \u00d7 (6/7) = 24 widgets\n\nQ3: Monthly widget production with fibonacci holidays\n- Fibonacci numbers from 1 to 10 are: 1, 2, 3, 5, 8\n- These months get 2 extra holidays\n- Holidays per month: 1 (standard) + 2 (if fibonacci month)\n- Reduction factor: (7 - 3)/7 = 4/7\n- Average weekly production: 28 \u00d7 (4/7) = 16 widgets", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1059.0, | |
| "cost": 0.007004 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week calculation\n- Weekdays (Monday to Friday): 4 widgets per day\n 5 days \u00d7 4 widgets = 20 widgets\n- Monday bonus: 2 extra widgets\n- Weekends: 3 widgets per day\n 2 days \u00d7 3 widgets = 6 widgets\nTotal weekly production: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly widget production with 1 holiday\n- 1 holiday randomly distributed means 3 weeks at full production, 1 week with reduced production\n- Full weeks: 3 \u00d7 28 = 84 widgets\n- Holiday week: Assume average reduction based on 1/7 of the week being closed\n- Reduction: 28 \u00d7 (1/7) = 4 widgets\n- Monthly total: 84 + (28 - 4) = 108 widgets\n- Average weekly production: 108 \u00f7 4 = 27 widgets\n\nQ3: Fibonacci holidays additional consideration\n- Fibonacci numbers in month indices: 1, 2, 3, 5, 8 (for 10 months)\n- These months get 2 extra holidays\n- Assume these holidays are evenly distributed\n- Each such month reduces production by 2 \u00d7 (28/7) = 8 widgets\n- Months with Fibonacci index holidays: 5 months\n- Total additional reduction: 5 \u00d7 8 = 40 widgets\n- Yearly total: (108 \u00d7 10) - 40 = 1040 widgets\n- Average weekly production: 1040 \u00f7 (10 \u00d7 4) = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1187.0, | |
| "cost": 0.007516 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 84.0, | |
| "q3": 56.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widget production in a week\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Extra widgets on Monday: 2 additional widgets\n- Weekend (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\n- Total weekly production: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly widget production with one holiday\n- Total weeks in a month: 4 weeks\n- Holiday distributed randomly across the week\n- On a holiday, no widgets are produced\n- Average weeks with full production: 3 weeks\n- Average production: 28 widgets \u00d7 3 = 84 widgets per month\n\nQ3: Fibonacci holiday months\n- Fibonacci numbers in 1-based month indexing: 1, 2, 3, 5, 8\n- These months get 2 extra holidays\n- In a 10-month year, these months are: 1st, 2nd, 3rd, 5th, and 8th months\n- Average weeks of full production reduced further\n- Estimated average weeks of production: 2 weeks\n- Average monthly production: 28 widgets \u00d7 2 = 56 widgets", | |
| "error": 87.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1073.0, | |
| "cost": 0.00706 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.43, | |
| "q3": 23.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Weekly widget production\n- Weekdays (Monday to Friday): 4 widgets per day\n * Monday has 2 extra widgets\n * So Monday: 4 + 2 = 6 widgets\n * Other weekdays: 4 widgets each\n- Weekends (Saturday and Sunday): 3 widgets per day\nCalculation:\n- Weekdays: (6 + 4 + 4 + 4 + 4) = 22 widgets\n- Weekends: (3 + 3) = 6 widgets\nTotal weekly production: 22 + 6 = 28 widgets\n\nQ2: Monthly production with one random holiday\n- Total weeks in a month: 4\n- One holiday reduces production by that day's widgets\n- Average holiday impact: \n * Weekday holiday: average 4 widgets lost\n * Weekend holiday: average 3 widgets lost\n- So average weekly production: (28 - (4/7 * 4 + 3/7 * 3)) = 28 - 3.57 = 24.43 widgets\n\nQ3: Fibonacci holidays\nFibonacci numbers up to 10 months: 1, 2, 3, 5, 8\nMonths with extra holidays: 1st, 2nd, 3rd, 5th, 8th months\nExtra impact: 2 additional holidays in these 5 months\n- Average holiday impact increases\n- Estimated average weekly production further reduced\n\nQ1: 28\nQ2: 24.43\nQ3: Around 22-23 widgets", | |
| "error": 5.57, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1167.0, | |
| "cost": 0.007436 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 20.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widget production per week calculation\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Monday bonus: 2 extra widgets\n- Weekend (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\n- Total weekly production: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly production with one holiday\n- 4 weeks per month\n- 1 holiday randomly distributed\n- Probability of holiday on any type of day is 1/7\n- Expected reduction in weekly production:\n - Weekday holiday: (5/7 \u00d7 6) + (2/7 \u00d7 0) = 20 - (1/7 \u00d7 20) = 17.14 widgets\n - Weekend holiday: (2/7 \u00d7 3) + (5/7 \u00d7 3) = 3 widgets\n- Average weekly production accounting for holiday: 28 - (28 \u00d7 1/7) = 24 widgets\n\nQ3: Fibonacci holiday months\n- Fibonacci numbers in 1-based month index: 1, 2, 3, 5, 8 (first 5 months)\n- These months get 2 extra holidays\n- 5 months with 2 extra holidays\n- Extra holiday distribution same as previous calculation\n- Additional reduction in production\n\nLet's calculate the final result.", | |
| "error": 9.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1118.0, | |
| "cost": 0.00724 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 20.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widget production in a week\n- Weekdays (Monday to Friday): 4 * 5 = 20 widgets\n- Extra Monday widgets: 2 \n- Weekend (Saturday and Sunday): 3 * 2 = 6 widgets\nTotal per week: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly widget production with one holiday\n- In a month with one holiday, we'll lose one day of production\n- Regular weekly production: 28 widgets\n- One holiday reduces production by the day's widgets\n- Holiday could be on any day, so on average, we'll lose the average day's production\n- Average daily production: 28 / 7 = 4 widgets\nSo average weekly production will be: 28 - 4 = 24 widgets\n\nQ3: Fibonacci holiday adjustments\nFibonacci numbers in 1-based month indexing: 1, 2, 3, 5, 8 \n(1st, 2nd, 3rd, 5th, and 8th months have two extra holidays)\n- Base monthly holiday: 1 holiday\n- These 5 months have an additional holiday\n- Total disruption increases\n- Average weekly production will be further reduced", | |
| "error": 9.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1077.0, | |
| "cost": 0.007076 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 22.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widget production per week\n- Weekdays (Monday to Friday): 4 widgets/day + 2 extra on Monday\n * Monday: 4 + 2 = 6 widgets\n * Tuesday-Friday: 4 widgets each\n- Calculation: (6 + 4 * 4) + (3 * 2) = 6 + 16 + 6 = 28 widgets per week\n\nQ2: Monthly widget production\n- Base weekly production: 28 widgets\n- One holiday per month (1/7 chance of being on any day)\n- On a holiday week, production reduces proportionally\n- Average weekly production: 28 * (6/7) = 24 widgets per week\n\nQ3: Fibonacci holiday months\n- Fibonacci numbers in 1-based month indexing: 1, 2, 3, 5, 8 \n- These months get 2 extra holidays\n- Months with 1 extra holiday: 5 months\n- Total extra holidays: (1 * 5) + (2 * 5) = 5 + 10 = 15 holidays per 10-month year\n- Effective holiday reduction: 15 / (10 * 7) \u2248 0.214 or 21.4%\n- Average weekly production: 28 * (1 - 0.214) \u2248 22 widgets per week", | |
| "error": 7.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1125.0, | |
| "cost": 0.007268 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-sonnet-latest", | |
| "avg_error": 1.1641571419999999, | |
| "avg_correctness": 0.6, | |
| "avg_tokens": 1198.5, | |
| "avg_cost": 0.0101655, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular week widget production\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Plus Monday bonus: 2 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\nTotal per week = 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly holiday impact\n- Each month has 4 weeks = 28 days\n- Plus 1 holiday where no production occurs\n- Average daily production without holidays:\n * Mondays: 6 widgets (4+2)\n * Tue-Fri: 4 widgets each\n * Sat-Sun: 3 widgets each\n- Holiday reduces production by its day's amount\n- Uniform distribution means 1/7 chance for each day type\n- Average production loss per holiday:\n (6 + 4\u00d74 + 3\u00d72)/7 = (6 + 16 + 6)/7 = 28/7 = 4 widgets\n- Monthly impact: -4 widgets\n- Per week impact: -4/4 = -1 widget\nTherefore weekly average = 28 - 1 = 27 widgets\n\nQ3: Fibonacci holiday impact\n- Fibonacci numbers up to 10: 1,1,2,3,5,8\n- Months with indices matching these get 2 extra holidays\n- In base-1 indexing, months 1,2,3,5,8 get extra holidays\n- That's 5 months with 2 extra holidays each = 10 extra holidays\n- These 10 holidays are spread across 10 months = 1 per month\n- Each holiday reduces weekly production by 1 widget\n- Therefore weekly average = 27 - 1 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1181.0, | |
| "cost": 0.009903 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Weekly widget production\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact\n- Each month has 4 weeks = 28 widgets \u00d7 4 = 112 widgets base\n- One holiday per month where no production occurs\n- Average daily production to subtract:\n * Weekday (5/7 chance): 4 widgets (+ 2/7 chance of being Monday's 2 extra)\n * Weekend (2/7 chance): 3 widgets\n * Average = (5/7 \u00d7 4 + 2/7 \u00d7 3 + 2/7 \u00d7 2) = 4 widgets/day\n- Subtract 4 widgets for holiday\n- 112 - 4 = 108 widgets per month\n- Average per week = 108 \u00f7 4 = 27 widgets\n\nQ3: Fibonacci impact\n- Fibonacci numbers up to 10: 1, 1, 2, 3, 5, 8\n- Months with extra holidays: 1st, 2nd, 3rd, 5th, 8th\n- 5 months have 2 extra holidays each = 10 extra holidays\n- These 10 extra holidays cost 4 widgets each = 40 widgets\n- 10 months \u00d7 108 widgets = 1080 base widgets\n- 1080 - 40 = 1040 widgets per year\n- 1040 \u00f7 (10 months \u00d7 4 weeks) = 26 widgets per week", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1185.0, | |
| "cost": 0.009963 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 20.0, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular weekly production\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Monday bonus: 2 extra widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\nTotal per week = 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly holidays impact\n- Each month has 4 weeks\n- One holiday per month means on average losing 1/7 of a day's production each week\n- If holiday falls on Monday: lose 6 widgets (4 regular + 2 bonus)\n- If holiday falls on Tue-Fri: lose 4 widgets\n- If holiday falls on weekend: lose 3 widgets\n- Average widgets lost = (6 + 4\u00d74 + 3\u00d72)/7 \u2248 4 widgets per week\nWeekly average = 28 - 4 = 24 widgets\n\nQ3: Fibonacci impact\n- Fibonacci numbers from 1 to 10 are: 1, 1, 2, 3, 5, 8\n- Months with indices matching these get 2 extra holidays\n- In a 10-month year, months 1, 2, 3, 5, 8 get extra holidays\n- That's 5 months with 3 holidays instead of 1\n- Additional holidays = 2 \u00d7 5 = 10 holidays per year\n- Extra holidays per month = 10/10 = 1 extra holiday per month on average\n- This means losing another 4 widgets per week on average\nWeekly average = 24 - 4 = 20 widgets", | |
| "error": 9.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1147.0, | |
| "cost": 0.009393 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step:\n\nQ1: Regular week production\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact\n- Each month has 4 weeks = 28 widgets \u00d7 4 = 112 widgets/month base\n- One holiday per month where no production occurs\n- Average daily production = (5 days \u00d7 4 widgets + 2 days \u00d7 3 widgets + 2 Monday widgets) \u00f7 7 days \u2248 4 widgets/day\n- So one holiday costs about 4 widgets\n- 112 - 4 = 108 widgets per month\n- Average per week = 108 \u00f7 4 = 27 widgets/week\n\nQ3: Fibonacci holiday impact\n- Fibonacci numbers up to 10 are: 1, 1, 2, 3, 5, 8\n- Months with indices matching these get 2 extra holidays\n- In a 10-month year:\n * Months 1, 2, 3, 5, 8 get 2 extra holidays (5 months)\n * Other months have 1 holiday (5 months)\n- Total extra holidays = 5 months \u00d7 1 extra holiday \u00d7 4 widgets = 20 widgets lost\n- Per month impact = 20 \u00f7 10 = 2 widgets lost per month\n- New monthly total = 108 - 2 = 106 widgets\n- New weekly average = 106 \u00f7 4 = 26.5 widgets/week", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1183.0, | |
| "cost": 0.009933 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07142857, | |
| "q3": 26.14285714, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular week production:\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: +2 widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact:\n- Regular weeks in a month: 4\n- One holiday per month\n- Since holidays are uniform across days, on average:\n * If holiday is on weekday: lose 4 widgets (5/7 chance)\n * If holiday is on weekend: lose 3 widgets (2/7 chance)\n * If holiday is on Monday: also lose 2 bonus widgets (1/7 chance)\nAverage widgets lost per holiday = (4\u00d75/7) + (3\u00d72/7) + (2\u00d71/7) = 3.71428...\nLost per week = 3.71428/4 (spreading holiday impact across 4 weeks)\nWeekly average = 28 - (3.71428/4) = 27.07142... widgets\n\nQ3: Fibonacci impact:\n- Fibonacci numbers from 1-10 are: 1,1,2,3,5,8\n- Months with extra holidays: 1,2,3,5,8 (5 months total)\n- Each of these months gets 2 additional holidays\n- Total extra holidays = 5 months \u00d7 2 holidays = 10 extra holidays per year\n- Extra holidays per month = 10/10 = 1 extra holiday on average\n- This doubles the holiday impact calculated in Q2\nWeekly average = 27.07142... - (3.71428/4) = 26.14285... widgets", | |
| "error": 0.21428570999999863, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1182.0, | |
| "cost": 0.009918 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07142857, | |
| "q3": 26.14285714, | |
| "reasoning": "Let me solve this step by step:\n\nQ1: Widgets per week calculation:\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact:\n- Each month has 4 weeks and 1 holiday\n- Since holidays are uniformly distributed, we can calculate the average impact\n- Each holiday reduces production by either:\n * 4 widgets if on weekday (5/7 chance)\n * 3 widgets if on weekend (2/7 chance)\nAverage widget loss per holiday = (4\u00d75/7 + 3\u00d72/7) = 3.71428...\nWeekly impact = 3.71428.../4 (since holiday affects one out of 4 weeks)\nNew weekly average = 28 - (3.71428.../4) = 27.07142... widgets\n\nQ3: Fibonacci holiday impact:\n- Fibonacci numbers from 1 to 10 are: 1, 1, 2, 3, 5, 8\n- Months with index matching these get 2 extra holidays\n- In positions 1, 2, 3, 5, 8 (5 months total)\n- 5 months have 2 extra holidays = 10 extra holidays per year\n- These 10 extra holidays are spread across 10 months (40 weeks)\n- Each holiday reduces production by 3.71428... widgets on average\nAdditional weekly impact = (10 \u00d7 3.71428...)/(40) = 0.92857...\nFinal weekly average = 27.07142... - 0.92857... = 26.14285...", | |
| "error": 0.21428570999999863, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1194.0, | |
| "cost": 0.010098 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular week widget production\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Month with one holiday\n- Regular production is 28 widgets/week \u00d7 4 weeks = 112 widgets/month\n- One holiday reduces production:\n * If on weekday (5/7 probability): lose 4 widgets\n * If on Monday (1/7 probability): lose 6 widgets (4 + 2 bonus)\n * If on weekend (2/7 probability): lose 3 widgets\n Average loss = (4\u00d74/7 + 6\u00d71/7 + 3\u00d72/7) = 4 widgets\n- Monthly production = 112 - 4 = 108 widgets\nAverage weekly = 108/4 = 27 widgets/week\n\nQ3: Fibonacci year effect\n- Fibonacci numbers up to 10 are: 1,1,2,3,5,8\n- Months with index matching these (excluding duplicates): 1,2,3,5,8\n- So 5 months have 2 extra holidays, 5 months have 1 holiday\n- Regular months (5): 108 widgets each\n- Double holiday months (5): lose another 4 widgets = 104 widgets each\n- Yearly production = (5\u00d7108 + 5\u00d7104) = 1060 widgets\n- Weeks per year = 10 months \u00d7 4 weeks = 40 weeks\nAverage weekly = 1060/40 = 26.5 widgets/week", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1181.0, | |
| "cost": 0.009903 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step.\n\nQ1 - Regular weekly production:\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2 - Accounting for monthly holiday:\n* Each month has 4 weeks = 28 days\n* Plus 1 holiday where no production occurs\n* Holiday could fall on any day with equal probability\n* If holiday falls on Monday: lose 6 widgets (4 regular + 2 bonus)\n* If holiday falls on Tue-Fri: lose 4 widgets\n* If holiday falls on Sat-Sun: lose 3 widgets\n* Average loss = (6 + 4\u00d74 + 3\u00d72) \u00f7 7 \u2248 4 widgets per month\n* 4 widgets \u00f7 4 weeks = 1 widget less per week\nTherefore: 28 - 1 = 27 widgets per week\n\nQ3 - Accounting for fibonacci holidays:\n* In a 10-month year, fibonacci month indices are: 1,2,3,5,8\n* So 5 months have 2 holidays instead of 1\n* This means 5 extra holidays per year\n* Extra holiday loss per year = 5 \u00d7 4 widgets (using average from Q2)\n* Per month: (5 \u00d7 4) \u00f7 10 = 2 widgets\n* Per week: 2 \u00f7 4 = 0.5 widgets\nTherefore: 27 - 0.5 = 26.5 widgets per week", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1173.0, | |
| "cost": 0.009783 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 26.929, | |
| "q3": 25.858, | |
| "reasoning": "For Q1, I'll calculate total widgets per week:\n\u2022 Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n\u2022 Plus Monday bonus: 2 widgets\n\u2022 Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\nTotal per week = 20 + 2 + 6 = 28 widgets\n\nFor Q2, I need to account for one holiday per month. With 4 weeks per month:\n\u2022 Normal production is 28 widgets \u00d7 4 weeks = 112 widgets per month\n\u2022 One holiday means losing that day's production\n\u2022 Since holidays are uniform across days, average loss is:\n - (5/7 chance of weekday \u00d7 4 widgets) + (2/7 chance of weekend \u00d7 3 widgets)\n - Plus 2/7 chance of losing Monday bonus (as Monday is 1/7 of days)\n - (5/7 \u00d7 4) + (2/7 \u00d7 3) + (2/7 \u00d7 2) = 2.857 + 0.857 + 0.571 = 4.285 widgets lost per holiday\n\u2022 Monthly production = 112 - 4.285 = 107.715 widgets\n\u2022 Weekly average = 107.715 \u00f7 4 = 26.929 widgets\n\nFor Q3, I need to consider Fibonacci holidays in certain months.\nFibonacci numbers for month indices (1-based): 1,1,2,3,5,8\nSo months 1,2,3,5,8 get extra holidays\nIn a 10-month year:\n\u2022 Regular months (4,6,7,9,10): 1 holiday each = 5 holidays\n\u2022 Fibonacci months (1,2,3,5,8): 3 holidays each = 15 holidays\nTotal 20 holidays per year vs 10 regular holidays\nThat's 2 average holidays per month instead of 1\nSo we lose 4.285 \u00d7 2 = 8.57 widgets per month\nMonthly production = 112 - 8.57 = 103.43 widgets\nWeekly average = 103.43 \u00f7 4 = 25.858 widgets", | |
| "error": 0.21300000000000097, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1279.0, | |
| "cost": 0.011373 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Widgets in a regular week\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Mondays get 2 extra widgets = +2\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\nTotal per week = 20 + 2 + 6 = 28 widgets\n\nQ2: Accounting for monthly holidays\n- Each month has 4 weeks = 28 \u00d7 4 = 112 widgets normally\n- One holiday per month where no production occurs\n- Average widgets lost per holiday:\n * If holiday on weekday (5/7 chance): 4 widgets (or 6 if Monday)\n * If holiday on weekend (2/7 chance): 3 widgets\n * Average loss = (5/7 \u00d7 4 + 2/7 \u00d7 3) = 3.71 widgets\n * (accounting for Monday's extra 2: (4/7 \u00d7 4 + 1/7 \u00d7 6 + 2/7 \u00d7 3) = 4 widgets)\n- 112 - 4 = 108 widgets per month\n- Per week average = 108 \u00f7 4 = 27 widgets\n\nQ3: Accounting for Fibonacci holidays\n- Fibonacci numbers up to 10: 1, 1, 2, 3, 5, 8\n- Months 1, 2, 3, 5, 8 get 2 extra holidays (5 months total)\n- Regular months (4, 6, 7, 9, 10) lose 4 widgets per month\n- Fibonacci months lose 8 widgets per month (2 holidays \u00d7 4 widgets)\n- Total monthly production: \n * Regular months (5): 5 \u00d7 108 = 540\n * Fibonacci months (5): 5 \u00d7 (112 - 8) = 520\n * Total per year = 1060 widgets\n- Weeks per year = 10 months \u00d7 4 weeks = 40 weeks\n- Average per week = 1060 \u00f7 40 = 26.5 widgets", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1280.0, | |
| "cost": 0.011388 | |
| } | |
| ] | |
| } | |
| ] | |
| }, | |
| { | |
| "strategy_name": "ModelWithThreeCoT", | |
| "tests": [ | |
| { | |
| "provider": "openai", | |
| "model": "gpt-4o-mini", | |
| "avg_error": 5.3104, | |
| "avg_correctness": 0.4666666666666666, | |
| "avg_tokens": 815.5, | |
| "avg_cost": 0.0003678, | |
| "n_tests": 10, | |
| "failure_rate": 0.2857142857142857, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.5, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. On weekdays, the factory produces 4 widgets each day, leading to a total of 5 * 4 = 20 widgets. On weekends, it produces 3 widgets each day, leading to a total of 2 * 3 = 6 widgets. Additionally, there are 2 extra widgets produced on Monday. Thus, the total production in a week is 20 (weekdays) + 6 (weekends) + 2 (extra widgets on Monday) = 28 widgets.\nIn a month, there are 4 weeks, producing 28 widgets each week, leading to a total of 4 * 28 = 112 widgets in a month. Each month also has 1 holiday, so on average, we lose 4 widgets from one holiday. Therefore, the monthly production becomes 112 - 4 = 108 widgets. Dividing this by 4 weeks gives an average weekly production of 108 / 4 = 27 widgets.\nIn the 10 months, the 1st, 2nd, 3rd, 5th, and 8th months are fibonacci months, leading to a total of 5 extra holidays (2 per fibonacci month). With 5 extra holidays, the total comes to 10 holidays in a year. Therefore, we have 112 (initial monthly production) - 10 (total holidays) = 102 widgets per year. With 10 months, the average monthly production becomes 102 / 10 = 10.2 widgets per month. Dividing this by 4 weeks gives an average production of 10.2 / 4 = 25.5 widgets per week, which rounds to 25.5 widgets.", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 702.0, | |
| "cost": 0.0002997 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.5, | |
| "reasoning": "The factory produces 4 widgets every weekday (Monday to Friday) and 3 widgets on weekends (Saturday and Sunday). Additionally, there are 2 extra widgets produced on Mondays. Calculating the total for each day of the week: Monday: 6 (4 + 2), Tuesday: 4, Wednesday: 4, Thursday: 4, Friday: 4, Saturday: 3, Sunday: 3. Summing these gives 6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 widgets per week.\nGiven that there are 4 weeks in a month and 1 holiday, we need to calculate the weekly production considering one closed day. The weekly production without holidays is 28 widgets. However, we need to account for 1 holiday affecting the production for the entire month. This means we have 27 days operation (28 days - 1 holiday), producing 28 widgets per week will mean losing 4 widgets for each week as the holiday will affect one week of out of total 4 weeks. Hence, total monthly production is 28 widgets/week * 4 weeks - (4 widgets/week for the closed holiday) = 112 - 4 = 108. Now, calculating the average widgets produced per week: 108 widgets/month / 4 weeks: 27 widgets/week on average.\nNow let's consider the fibonacci holidays. These months are 1, 2, 3, 5, 8 leading to 5 months (1-based) having 2 extra holidays. This gives us a total of 10 holidays (5 months * 2 holidays). So for the 10 months, each month has 4 weeks giving us 40 weeks, and so these extra holidays account for 40 widgets (10 holidays * 4 widgets/week). Average weekly production includes these reductions as well. Returning to the monthly production without weight of holidays, we still have 28 widgets/week on 40 weeks thus the production will be again 112 minus widgets lost due to holidays {108 from previous} - 10, then divide by total weeks now 40 weeks thus (112 - 10)/40 = 102/40 = 25.5 widgets/week.", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 797.0, | |
| "cost": 0.0003567 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.286, | |
| "q3": 20.57, | |
| "reasoning": "In a typical week, there are 5 weekdays and 2 weekend days. During weekdays, the factory produces 4 widgets each day, leading to a total of 5 * 4 = 20 widgets. On weekends, it produces 3 widgets each day, resulting in 2 * 3 = 6 widgets. Additionally, since there are 2 extra widgets produced on Monday, the total is adjusted to 20 + 6 + 2 = 28 widgets per week.\nWith one holiday uniformly distributed across the week, on average, there will be a decrease of widgets produced. The average widgets produced on a holiday is the average daily production. The weekdays produce 4 widgets and weekends produce 3 widgets. For a general average, considering all 7 days, the production per day is (5 weekdays * 4 + 2 weekend days * 3) / 7 = (20 + 6) / 7 = 3.714 widgets. Therefore, if one holiday reduces productivity, the average number of widgets produced per week would be 28 - 3.714 \u2248 24.286 widgets.\nFibonacci numbers up to 10 (the maximum month index) are 1, 2, 3, 5, and 8. This means that months 1, 2, 3, 5, and 8 have two extra holidays. Therefore, across the 10 months, there will be 5 extra holidays this way. Considering the previously calculated average production of 28 widgets per week, the average number of widgets produced decreased further by considering the daily average reduction during holidays. With these additional 5 holidays, the number of holidays in total becomes 1 (normal holiday) + 5 (fibonacci holidays) = 6. The average daily production calculation remains the same; thus, now we reduce by 6 days of the average daily production. Finally, the average will be adjusted to 28 - (6 * 3.714) \u2248 20.57 widgets per week after considering all holidays.", | |
| "error": 8.143999999999998, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 764.0, | |
| "cost": 0.0003369 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.125, | |
| "q3": 19.25, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. On weekdays, 4 widgets are produced each day, and on weekends, 3 widgets are produced each day. Additionally, there are 2 extra widgets produced on Mondays. Therefore, for weekdays: 5 days * 4 widgets/day = 20 widgets produced and for weekends: 2 days * 3 widgets/day = 6 widgets. Adding the extra widgets for Monday: 20 widgets + 6 widgets + 2 extra widgets = 28 widgets.\nEach month has 4 weeks, so total widgets produced in a month without holidays would be 28 widgets/week * 4 weeks = 112 widgets. However, there is 1 holiday each month, the production for that holiday would be: checking how a holiday falls affects widgets produced: 4 widgets for weekdays or 3 for weekends depending on day. So average loss is about 3.5 widgets. Monthly production accounting for holidays: 112 widgets - 3.5 widgets = 108.5 widgets. Average weekly production would be 108.5 widgets / 4 weeks = 27.125 widgets/week.\nIn Standardtopia, the months are indexed, and the Fibonacci sequence in the first 10 indices is (1, 1, 2, 3, 5, 8). The months 1, 2, 3, 5, and 8 will have 2 extra holidays each, totaling 10 additional holidays (2 holidays * 5 months). This adds to the average loss of widgets: assuming an average 3.5 widgets per holiday means a loss of 35 widgets over the year. Monthly production with Fibonacci holidays = 112 widgets - 35 widgets = 77 widgets. Average per week would then be 77 widgets / 40 weeks = 19.25 widgets/week.", | |
| "error": 6.875, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 724.0, | |
| "cost": 0.0003129 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. Each weekday, the factory produces 4 widgets, and on the weekends, it produces 3 widgets per day. Additionally, on Mondays, the factory produces 2 extra widgets. Calculating the total for the week: \n\\nWeekdays: 5 weekdays * 4 widgets/weekday = 20 widgets \nWeekend: 2 weekend days * 3 widgets/weekend day = 6 widgets \nExtra Monday widgets = 2 widgets \n\\nTotal widgets in a week = 20 + 6 + 2 = 28 widgets.\nIn a month, there are 4 weeks. The factory produces 28 widgets per week. Therefore, in 4 weeks, the total production is 4 weeks * 28 widgets/week = 112 widgets. Each month has one holiday, and the factory is closed on this holiday, meaning it would produce 112 - (28/7) = 112 - 4 = 108 widgets in a month. \n\\nNow, with the adjustments for the holiday, the average number of widgets produced per week is 108 widgets/month \u00f7 4 weeks/month = 27 widgets/week.\nIn Standardtopia, there are 10 months and each month is 4 weeks long, totaling 40 weeks. However, for the fibonacci holiday bonus, we need to verify which months are fibonacci indices: 1, 2, 3, 5, 8. Each of these months will have 2 extra holidays. \n\\nEach fibonacci holiday results in the factory being closed, which means subtracting more from total production. Each of the 5 fibonacci months has 28 widgets/week * 2 holidays/month = 10 widgets lost each from the months with fibonacci holidays, resulting in 50 extra widgets not produced due to holidays. Given 112 widgets produced normally in four weeks, we have 112 - 10 (or 2 weeks of holidays) = 102 widgets. Thus, the total production without holidays is: 40 weeks * 28 = 1120 widgets; \nSo, the effective production = 1120 - 50 = 1070. \n\\nTo find the average per week: 1070 total widgets \u00f7 40 weeks = 26.75., which rounds down to 26 widgets/week on average after adjustments for both the single holiday and the fibonacci holidays.", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 842.0, | |
| "cost": 0.0003837 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "In Standardtopia, the factory produces widgets every weekday and uses a different rate for weekends. Let's calculate the total widgets produced in a week: \n- Weekdays (Monday to Friday): 4 widgets/day x 5 days = 20 widgets. \n- Weekend (Saturday and Sunday): 3 widgets/day x 2 days = 6 widgets. \n- Extra widgets on Monday: 2 widgets. \nTotal for the week = 20 + 6 + 2 = 28.\nNow to consider the holidays. One holiday is taken into account each month, which means in every month of 4 weeks, there are 4 less widgets produced (1 for each week). \nOriginally, in a month, we would produce 28 widgets/week * 4 weeks = 112. \nSubtracting holidays: 112 - 4 for the holiday = 108 widgets produced in a month. \nTo find weekly production considering monthly holidays, average would be 108 / 4 weeks = 27 widgets/week.\nNext, we need to account for the extra holidays in months whose index is a Fibonacci number. The Fibonacci sequence starts with 1, 1, 2, 3, 5, 8,... Thus, weeks 1, 2, 3, 5, and 8 are Fibonacci months. \nIn those 5 months, there will be 2 extra holidays, meaning 5 x 2 = 10 additional holidays across the year. \nWith 10 extra holidays distributed across 4 weeks/month we find that is approximately 10/10 = 1 additional holiday per week. \nSo, originally 28 widgets per week becomes 28 - 1 = 27 widgets/week as average considering these extra holidays. \nThis leads to the final average production of widgets including Fibonacci holidays. Each month still maintains the holiday previously counted. Hence, the average still stands at 27 widgets/week.", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 743.0, | |
| "cost": 0.0003243 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 26.0, | |
| "q3": 24.0, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. On weekdays, the factory produces 4 widgets per day, so for 5 weekdays, the total is 5 * 4 = 20 widgets. On weekends, they produce 3 widgets each day for 2 days, totaling 2 * 3 = 6 widgets. Additionally, on Mondays, there are 2 extra widgets produced, adding 2 more to the total. Therefore, the total production in a week is 20 + 6 + 2 = 28 widgets.\nEach month has 4 weeks, which means the factory produces 28 widgets per week for 4 weeks, equaling 112 widgets before including holidays. Since there is 1 holiday per month, 1/4 of 28 widgets (7 widgets) would be lost during that holiday per month. Thus, after accounting for the holiday, the total production each month is 112 - 7 = 105 widgets. Averaging over the 4 weeks, we divide 105 by 4 to find the average production per week, resulting in 105 / 4 = 26.25 widgets per week.\nThere are 10 months in a year, and every month at indexes 1, 2, 3, 5, and 8 is a Fibonacci number, meaning these months will have 2 additional holidays. This means for 5 such months, there will be 5 * 2 = 10 extra holidays. Each of the extra holidays will again result in a loss of 7 widgets, equaling 10 * 7 = 70 widgets lost due to these holidays. The total production for the whole year would originally be 105 widgets * 10 months = 1050 widgets. Now, we subtract the losses: 1050 - 70 = 980 total widgets for the year. Dividing this by 10 months and then 4 weeks gives an average of 980 / 40 = 24.5 widgets per week.", | |
| "error": 3.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 752.0, | |
| "cost": 0.0003297 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 3.0, | |
| "q3": 23.0, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. Each weekday produces 4 widgets, so for 5 weekdays: 5 * 4 = 20 widgets. Each weekend day produces 3 widgets, so for 2 days: 2 * 3 = 6 widgets. Additionally, on Monday, there are 2 extra widgets produced. So the total for the week is 20 + 6 + 2 = 28 widgets.\nWith 4 weeks and one holiday in each month, that's 4 holidays per month on average. Over 10 months, this equals 40 holidays. However, since holidays are uniformly distributed, they would reduce the total widget production proportionally. The production without holidays is 28 widgets/week * 10 months = 280 widgets. With 40 holidays in total, the total days of production are 280 - (40 * (28/7)) = 280 - 160 = 120 widgets produced. Dividing this by 40 weeks gives an average of 3 widgets/week.\nThe fibonacci months (1, 2, 3, 5, 8) contribute 2 extra holidays each. This is a total of 5 fibonacci months * 2 holidays = 10 extra holidays. The total holidays now are 40 (previously calculated) + 10 = 50 holidays. Therefore, total production is now 28 widgets/week * 40 weeks - (50 * (28/7)). To calculate this, first find total production without holidays: 28 widgets/week * 40 = 1120. Then, find the reduction for holidays: 50 * 4 = 200 widgets. So the production with fibonacci holidays is 1120 - 200 = 920 widgets. Dividing 920 by 40 weeks gives an average of 23 widgets/week.", | |
| "error": 27.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 716.0, | |
| "cost": 0.0003081 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.125, | |
| "q3": 26.25, | |
| "reasoning": "The factory outputs 4 widgets every weekday, which makes 5 weekdays. So, the total for weekdays is 5 * 4 = 20 widgets. On weekends, there are 2 days, and the factory produces 3 widgets per day, resulting in 2 * 3 = 6 widgets. Additionally, there are 2 extra widgets produced on Monday. So, the total production per week is 20 + 6 + 2 = 28 widgets.\nIn a month, there are 4 weeks. So, weekly production without holidays is 28 widgets * 4 weeks = 112 widgets per month. However, there\u2019s one holiday that can be on any day of the week. If it falls on a weekday, the production for that day would be 4 widgets, and if it falls on a weekend, it would be 3 widgets. Therefore, on average we can assume the holiday affects the production by about 3.5 widgets. Thus, average production per month = 112 - 3.5 = 108.5, and the weekly production would be averaged to 108.5 / 4 = 27.125 widgets.\nIn Standardtopia, the months that are Fibonacci numbered have two extra holidays. The Fibonacci numbers within 1-10 are 1, 2, 3, 5, and 8, which means 5 out of 10 months have 2 extra holidays each. Therefore, in those 5 months, the extra holidays would reduce the production by 2 * 5 = 10 holidays impacting production on average. Since each holiday on average reduces production by 3.5 widgets, that\u2019s a reduction of 10 * 3.5 = 35 widgets over the total of 10 months. Without considering these holidays, the production would be 108.5 widgets per month * 10 months = 1085 widgets. Now accounting for 35 widgets lost to holidays leaves 1085 - 35 = 1050. The average weekly production over the year would then be 1050/40 = 26.25 widgets.", | |
| "error": 0.375, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 774.0, | |
| "cost": 0.0003429 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.29, | |
| "q3": 24.0, | |
| "reasoning": "In a week, Standardtopia has 5 weekdays and 2 weekend days. During weekdays, the factory produces 4 widgets per day. Thus, in 5 weekdays, it produces 4 * 5 = 20 widgets. On weekends, it produces 3 widgets per day for 2 days, which equals 3 * 2 = 6 widgets. Finally, on Mondays, there are 2 extra widgets produced. Therefore, the total weekly production is 20 (weekdays) + 6 (weekends) + 2 (extra widgets on Monday) = 28 widgets.\nWith one holiday uniformly distributed across the week, we assume that the holiday could fall on any day. The average weekly production without a holiday is 28 widgets. If a holiday lands on a weekday (which happens on 5 out of 7 days), then the production for that week will decrease by 4 widgets, giving 28 - 4 = 24 widgets. If the holiday lands on a weekend day (which happens on 2 out of 7 days), then the production drops by 3 widgets, giving 28 - 3 = 25 widgets. Thus, the expected production is the average of productions on holidays: (5 * 24 + 2 * 25) / 7 = (120 + 50) / 7 = 170 / 7 = 24.29 widgets, approximately.\nIn Standardtopia, there are 10 months, and the fibonacci numbers that are less than or equal to 10 are 1, 2, 3, 5, and 8, which account for 5 months having 2 extra holidays. Each month has 4 weeks, so with 5 months having 2 extra holidays, that results in 5 * 2 = 10 extra holidays in total for the year. When these holidays are accounted for, we have 40 weeks of production minus 10 holidays resulting in 30 weeks of production. Distribution for holidays is the same as above: out of 30 weeks, we should expect that 30*7=210 total holiday occurrences spread over 7 days. Conservation of expected production is taken into account as the average due to holidays remaining similar. Hence, average weekly production = Total weekly production (28 * 40 - total decrease due to holidays(210)) / effective production weeks (30) = (1120 - Total decrease from 210 holidays)/(30 weeks). Since 22 holiday occurrences fall on weekends means production is reduced by this amount and rest are weekdays. Average product would then be around 24 averaging out the above equations.", | |
| "error": 4.710000000000001, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1341.0, | |
| "cost": 0.0006831 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "openai", | |
| "model": "gpt-4o", | |
| "avg_error": 2.66537143, | |
| "avg_correctness": 0.4333333333333333, | |
| "avg_tokens": 712.8, | |
| "avg_cost": 0.005103000000000001, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 29.0, | |
| "q2": 28.75, | |
| "q3": 27.5, | |
| "reasoning": "Calculate weekly production by counting widgets produced on weekdays, weekends, and considering Monday's extra widgets. There are 5 weekdays and 2 weekend days.\nSubtract 1 day's production (due to the holiday) from the weekly total of widgets produced, and divide by 4 weeks to find the average widgets produced per week.\nCalculate total number of holidays in a month considering Fibonacci indices (months 1, 2, 3, 5, and 8 have 2 extra holidays). Subtract total monthly holidays from total widgets produced in 4 weeks and divide by 4 for the average weekly production.", | |
| "error": 4.25, | |
| "correctness": 0.0, | |
| "total_tokens": 465.0, | |
| "cost": 0.002625 | |
| }, | |
| { | |
| "q1": 31.0, | |
| "q2": 29.893, | |
| "q3": 28.7857143, | |
| "reasoning": "The factory operates on a system where it produces 4 widgets on each weekday and 3 widgets on each weekend day. Additionally, on Mondays, it produces 2 extra widgets. Calculating the number of widgets produced in one week considering: \n\nWeekdays (5 days): 4 widgets per day\nWeekends (2 days): 3 widgets per day\nMonday (1 day): 2 extra widgets\n\nThe total widgets produced per week = (4 widgets * 5 weekdays) + (3 widgets * 2 weekend days) + 2 extra widgets (on Monday).\nConsidering the factory operates for 4 weeks in a month and there is 1 holiday each month. The holidays are uniformly distributed, so on average, each week could lose 0.25 days of production due to a holiday (as 1 day per month / 4 weeks per month = 0.25). \n\nThe loss of production is therefore (31 / 7 days per week) * 0.25 days lost = 1.107 widgets per week approximately.\n\nSo, the average widgets produced per week accounting for holidays = 31 widgets per week - 1.107 widgets lost per week.\nIn a year, consisting of 10 months, the months that have a Fibonacci number in their index are month 1, 2, 3, 5, and 8 (Fibonacci numbers up to 10). Each of these months has two extra holidays.\n\nThis means there are 2 * 5 = 10 extra holidays per year, in addition to the 10 regular holidays (one per month). Total holidays in a year = 20.\n\nSpread over 40 weeks in 10 months, each week on average loses 20/40 = 0.5 days to holidays. \n\nTherefore, the average widgets produced per week = Total weekly production 31 - ((31/7) * 0.5) widgets lost per week.\n", | |
| "error": 8.6787143, | |
| "correctness": 0.0, | |
| "total_tokens": 756.0, | |
| "cost": 0.005535000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "A week has 5 weekdays and 2 weekend days. The factory produces 4 widgets per weekday and 3 widgets per weekend day. Additionally, on Mondays, it produces 2 extra widgets. Therefore, the total weekly production is calculated as follows: (5 weekdays * 4 widgets/weekday) + (2 weekend days * 3 widgets/weekend day) + 2 extra widgets on Monday = 20 widgets (weekdays) + 6 widgets (weekend) + 2 extra = 28 widgets.\nThere are 4 weeks in a month and 1 holiday, meaning there are 4 * 7 = 28 days in a month and 1 day as a holiday. The production without holidays is 4 weeks * 28 widgets per week = 112 widgets per month. With 1 holiday, one day of production is lost; we assume the factory loses the average day production. Average daily production = 112 widgets / 28 days = 4 widgets/day. Thus, losing 4 widgets due to 1 holiday means 112 - 4 = 108 widgets/month for production adjusted for the holiday. Therefore, the weekly average is 108 widgets/month / 4 weeks = 27 widgets.\nIn a 10-month year, the months with Fibonacci indices (1, 2, 3, 5, 8) have 2 additional holidays each. There are 5 such months. Normal production is 112 widgets/month, with regular 1 holiday/month adjusted production being 27 widgets/week. The additional holidays for Fibonacci months means (5 months * 2 extra holidays = 10 additional holidays). Total holidays in a year is therefore 10 months * 1 holiday/month + 10 additional holidays = 20 holidays. Total days available for production in a year is 280 days (10 months * 28 days) minus 20 holidays, which is 260 days. Total production in those 260 days is 260 days * 4 widgets/day = 1040 widgets/year. Therefore, the weekly average with additional holidays is 1040 widgets/year / 40 weeks/year (10 months * 4 weeks/month) = 26 widgets.", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 785.0, | |
| "cost": 0.005825 | |
| }, | |
| { | |
| "q1": 29.0, | |
| "q2": 28.75, | |
| "q3": 27.7, | |
| "reasoning": "Calculate the total number of widgets produced in a week by accounting for both weekdays and weekends, including the extra widgets on Mondays.\nConsider that one holiday occurs per month, reducing a weekday or weekend day of production. Hence, adjust the weekly production accordingly while considering one less production day per month in the calculation.\nRecognize that months indexed by Fibonacci numbers will have 3 holidays instead of 1. Distribute these additional holidays over the weekly production, adjusting for the extra holidays in certain months.", | |
| "error": 4.449999999999999, | |
| "correctness": 0.0, | |
| "total_tokens": 441.0, | |
| "cost": 0.002385 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "The factory produces widgets as follows: 4 widgets on each of the 5 weekdays (Monday to Friday), 3 widgets on each of the 2 weekend days (Saturday and Sunday), and 2 extra widgets on Monday. So, in a week, the total = (4 widgets/weekday * 5 weekdays) + (3 widgets/weekend day * 2 weekend days) + 2 extra widgets on Monday = 20 + 6 + 2 = 28 widgets.\nEach month has exactly 4 weeks and one holiday. A week without the holiday would have 28 widgets. Since the holiday can be any day, the average loss for a week's worth of production is 1 widget. Thus, on average, each week of 4-week long month contributes 27 widgets.\nMonths that are Fibonacci months have 2 extra holidays. The Fibonacci months are: 1, 2, 3, 5, 8. There are 5 such months out of 10. For these months, they will have 3 holidays, thus reducing weekly widgets by (3 days / 4 weeks = 0.75 days per week). Normal average production is reduced by 0.25 days due to the regular holiday from Q2, and by an additional 0.75 days here, yielding 1 day of lost production on average, so the average number of widgets is 28 - 1 = 27 per week in Fibonacci months. The rest 5 normal months, the average per week remains 27 from Q2. Thus, the yearly average = (5 * 27(avg in Fibonacci months) + 5 * 27(avg in normal months)) / 10 = 27 widgets.", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 689.0, | |
| "cost": 0.004865 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.6, | |
| "reasoning": "The factory operates 5 weekdays plus 2 weekend days, totaling 7 days a week. Each weekday, 4 widgets are produced, so 5x4 = 20 widgets. Each weekend day, 3 widgets are produced, so 2x3 = 6 widgets. Additionally, 2 extra widgets are produced on Monday, so total widgets in a week are 20 + 6 + 2 = 28 widgets.\nIn a month, there are 4 weeks, and each month has one holiday where no widgets are produced. Without the holiday, the factory produces 4 weeks x 28 widgets/week = 112 widgets per month. Taking away one day of widget production (choose an average day rate), we subtract 4 widgets (average weekday). So, 112 - 4 = 108 widgets per month, or the weekly average is 108 widgets / 4 weeks = 27 widgets per week.\nFibonacci months indices are 1, 2, 3, 5, 8 (within a 10-month year). During these months, an extra 2 holidays occur, for a total of 3 holidays in those months. This means there are 2 months with 3 holidays and 8 months with 1 holiday each. Total holidays = (2x3) + (8x1) = 6 + 8 = 14 holidays per year. Normally, there are 10 holidays (1 per month), so 4 additional holidays are spread over 10 months. On average, the factory closes 14 holidays per 40 production weeks (10 months). So, total production without holidays: 40 weeks x 28 widgets/week = 1120 widgets per year. Losing 14 days of production (mostly weekdays), we lose up to 14x4 = 56 widgets. So, 1120 - 56 = 1064 widgets per year. Average weekly production: 1064 widgets / 40 weeks = 26.6 widgets/week.", | |
| "error": 0.6000000000000014, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 751.0, | |
| "cost": 0.005485 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.125, | |
| "q3": 27.5, | |
| "reasoning": "The factory produces 4 widgets per weekday, so for 5 weekdays it produces 4 * 5 = 20 widgets. On weekends, it produces 3 widgets per day, so 3 * 2 = 6 widgets for Saturday and Sunday. On Mondays, it produces 2 extra widgets, so 20 + 6 + 2 = 28 widgets per week.\nEach month has four weeks and one distributed holiday, meaning there are 4 * 7 = 28 days a month - 1 holiday = 27 operational days. Considering all 4 weeks, we lose 1/4 of a week to the holiday. So, (28 widgets per week * 4 weeks - 1 holiday * average widgets per day)/4 weeks: a holiday on weekday 4 widgets, on the weekend 3 widgets. Let's average holiday at 3.5 widgets day = (28 * 4 - 3.5) / 4 = (112 - 3.5) / 4 = 108.5 / 4 = 27.125 avg. per week.\nFibonacci month index (1, 2, 3, 5, 8) each having two extra holidays per month. In 10 months, 5 months have extra 2 holidays. Monthly, a typical month has one holiday, but these 5 have an additional two: 3 holidays * 5 months, 1 holiday * 5 months = 20 total holidays/year => 12 more than original; spread over a year = 1.2 extra holidays per month: 5 holidays/month averaged = 5 holidays/4 weeks = 1.25 holidays/week, Week's output (28 widgets for 5 operational days), losing both widgets for day (2-4) or weekends (2-3 widgets): avg loss: 3 widgets per holiday; (28*4*10 months/year - 30 holidays *3 widgets)/40 weeks = 1100 total widgets/year/40 = 27.5 avg/week", | |
| "error": 1.625, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 763.0, | |
| "cost": 0.005605000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 22.5, | |
| "reasoning": "To find the total number of widgets produced in a week, calculate the widgets produced on weekdays and weekends separately, including extra widgets on Monday. There are 5 weekdays and 2 weekend days. On weekdays, the factory produces 4 widgets each day, resulting in a total of 5 * 4 = 20 widgets. Additionally, on weekends, 3 widgets are produced each day, so 2 * 3 = 6 widgets are produced. Moreover, 2 extra widgets are produced on Mondays. Therefore, the total number of widgets produced in a week is 20 + 6 + 2 = 28.\nTo calculate the average number of widgets produced per week considering the holiday, we should first find the total number of widgets produced in a month and then reduce the widgets by one week's worth of production (since one day is a holiday in each four-week period). The total widgets per month without holidays is 4 weeks * 28 widgets/week = 112 widgets. With one holiday, production drops by the average daily production (total weekly widgets / 7). The average daily production is 28 widgets / 7 days = 4 widgets/day. Hence, the monthly production after one holiday is 112 - 4 = 108 widgets, giving a weekly average of 108 / 4 weeks = 27 widgets/week.\nFor each 1-based month index that's a Fibonacci number (1, 2, 3, 5, 8, 13, etc.), add two extra holidays, which leads to more days off with no production. In 10 months, Fibonacci indices are 1, 2, 3, 5, 8, totaling 5 months with extra holidays. Thus, in a year, there are 5 extra holidays. This means 15 holidays per year in 10 months (5 Fibonacci months with 2 extra holidays each plus 10 regular holidays) which adds to a total of 25 holidays annually. This gives 250 weekdays total in 10 months minus 25 holidays = 225 production days annually. Since the average production day has 4 widgets, the annual production equals 225 * 4 = 900 widgets. The weekly average over 40 weeks (10 months of 4 weeks each) is 900 / 40 = 22.5 widgets/week.", | |
| "error": 3.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 816.0, | |
| "cost": 0.006135000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 26.25, | |
| "q3": 24.5, | |
| "reasoning": "The factory produces 4 widgets every weekday (Monday through Friday) and 3 widgets each day on the weekend (Saturday and Sunday). Additionally, the factory produces 2 extra widgets on Monday. Hence, the weekly production is calculated as follows: On weekdays (Monday to Friday), the factory produces 4 widgets/day * 5 days = 20 widgets. On Saturday and Sunday, it produces 3 widgets/day * 2 days = 6 widgets. Including the extra 2 widgets on Monday, the total becomes: 20 + 6 + 2 = 28 widgets per week.\nThere are 4 weeks in a month and 1 holiday per month, so we distribute this holiday across the 4 weeks. On average, we have 4 weeks in a month, but with 1 holiday affecting production. This means 1 out of 4 weeks will have one less production day. Therefore, the number of widgets reduced is the average of (1 holiday / 4 weeks) * 7 (as each day potentially loses production). In a week with an average holiday factor, the production is thus 28 - (7/4) on average per week. Calculating that, we have approximately 28 - 1.75 = 26.25 widgets per week on average with distributed holidays.\nA month has 4 weeks and there are 10 months in a year. The total holidays in a year are 10 (one per month) plus additional holidays for fibonacci-indexed months. The fibonacci sequence within 10 months is 1, 2, 3, 5, 8. So, months 1, 2, 3, 5, and 8 have 2 additional holidays each, amounting to 10 extra holidays. Total holidays = 10 + 10 = 20 holidays a year. Therefore, instead of 40 weeks, 20 of these days in the year will not have production. Thus, 20 days have 0 production, so the annual production days reduce per these 20 holidays/40 weeks. That's essentially removing half a week\u2019s worth of production each month, meaning: (28 widgets/week - (20 holidays/40 weeks *7 days of production re-distributed)). The effective average is 28 - (20/40 * 28/4 weeks) = 28 - 3.5 = 24.5 widgets per week on average accounting for fibonacci holidays.", | |
| "error": 2.25, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 841.0, | |
| "cost": 0.006385000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.3, | |
| "reasoning": "In a week, there are 5 weekdays and 2 weekend days. Each weekday, 4 widgets are produced, and on weekends, 3 widgets are produced per day. Additionally, on Monday, 2 extra widgets are produced. The calculation is as follows: For weekdays: 5 days * 4 widgets/day = 20 widgets; For weekend: 2 days * 3 widgets/day = 6 widgets; For extra Monday widgets: 2 widgets. Therefore, the total production in a week is 20 + 6 + 2 = 28 widgets.\nEach month has four weeks and one holiday. With one random holiday, on average, a week loses 1/7 of a working day. Therefore, we need to calculate the loss in production due to this one holiday day per month. The unadjusted weekly production is 28 widgets, so approximately, we lose 1/7 of that, which averages to 4 widgets over a month. This loss gets distributed over four weeks, which makes a decrement of 1 widget per week due to the holiday. Thus, the average production per week, adjusted for holidays, is 28 - 1 = 27 widgets.\nThere are 10 months in a year in Standardtopia, which introduces extra holidays for Fibonacci-indexed months: 1, 2, 3, 5, 8 (5 months). These months have 3 holidays each, as opposed to 1 in others. Thus, we calculate the effective average holiday impact over the year. With regular months: 7 holidays (5 regulars * 1) + With extra holidays in Fibonacci months: 5 months * 2 additional holidays = 10 extra holidays. In total: 12 holidays per year (7 regular + 10 extra). Dividing these 17 holidays evenly across 10 months leads to 1.7 holidays per month on average. Subtracting this from the unadjusted weekly production of 28 widgets: approximately, 1.7/7 of a workweek lost to holidays, meaning roughly 0.24 of a week's production (28 * (1.7/7) = 6.8 widgets lost per month, approximately 1.7 per week). Therefore, average production per week is 28 - 1.7 = 26.3 widgets.", | |
| "error": 0.3000000000000007, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 821.0, | |
| "cost": 0.006185 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-haiku-latest", | |
| "avg_error": 8.7, | |
| "avg_correctness": 0.4666666666666666, | |
| "avg_tokens": 1296.9, | |
| "avg_cost": 0.0082196, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 17.0, | |
| "reasoning": "Calculation for widgets per week:\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Weekends (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\n- Extra widgets on Monday: 2 additional widgets\nTotal: 20 + 6 + 2 = 28 widgets per week\nCalculation for widgets per week with one monthly holiday:\n- Normally 28 widgets per week\n- One random holiday reduces production\n- Holiday reduces weekly production by 1/7 on average\n- 28 \u00d7 (6/7) = 24 widgets per week on average\nFibonacci numbers in 1-based indexing: 1, 2, 3, 5, 8\nIn a 10-month year, these would be months 1, 2, 3, 5, and 8\nEach of these months gets 2 extra holidays\nExtra holidays per month: 2\nMonths with extra holidays: 5\nProbability of a day being a holiday is now higher\nRough calculation:\n- Base average from Q2: 24 widgets\n- Increased holiday probability will further reduce production\n- Estimate: 24 \u00d7 (5/7) = 17.14 widgets per week on average", | |
| "error": 12.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1192.0, | |
| "cost": 0.0078 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 16.0, | |
| "reasoning": "Let's calculate the widgets per week:\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Mondays have 2 extra widgets: 20 + 2 = 22 widgets\n- Weekends (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\n- Total widgets per week: 22 + 6 = 28 widgets\nMonthly calculation:\n- Normal week: 28 widgets\n- One holiday per month (distributed uniformly)\n- Average probability of holiday on each day of the week is 1/7\n- Holiday reduces widgets by (7/7 - 1/7) = 6/7 of production\n- Average weekly production: 28 \u00d7 (6/7) = 24 widgets\nFibonacci numbers from 1-based month index are: 1, 2, 3, 5, 8\n(1st, 2nd, 3rd, 5th, and 8th months get 2 extra holidays)\n- Basic monthly holiday: 1 day\n- Fibonacci month extra holidays: 2 days\n- Total monthly holidays: 3 days\n- Probability of holiday: 3/7 per week\n- Adjusted weekly production: 28 \u00d7 (4/7) = 16 widgets", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1203.0, | |
| "cost": 0.007844 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 21.0, | |
| "reasoning": "Let's calculate the widgets produced in a week:\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Extra widgets on Monday: 2 additional widgets\n- Weekend (Saturday and Sunday): 3 widgets per day \u00d7 2 days = 6 widgets\nTotal weekly production: 20 + 2 + 6 = 28 widgets\nIn a month with 4 weeks and 1 holiday:\n- Holidays are uniformly distributed, so on average, 1/7 of a week will be lost\n- Normal week production: 28 widgets\n- Holiday impact: 28 \u00d7 (1/7) = 4 widgets lost per month\n- Average weekly production: 28 - 4 = 24 widgets\nFibonacci numbers are 1, 2, 3, 5, 8 \nFibonacci months (1-based index): 1st, 2nd, 3rd, 5th, 8th months\nThese months get 2 extra holidays each\n- Previous average (from Q2): 24 widgets\n- Extra holidays in fibonacci months: 2 \u00d7 5 = 10 additional lost widget days\n- Total lost production: 4 (from Q2) + 10 = 14 widgets\n- New average weekly production: 28 - (14/4) = 21 widgets", | |
| "error": 8.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1215.0, | |
| "cost": 0.007892 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 17.0, | |
| "reasoning": "Weekdays (Monday to Friday): \n- 4 widgets per day\n- 2 extra widgets on Monday\n- So Monday: 4 + 2 = 6 widgets\n- Tuesday to Friday: 4 widgets each\nCalculation:\n6 + (4 * 4) = 6 + 16 = 22 widgets on weekdays\n\nWeekend (Saturday and Sunday):\n- 3 widgets per day\nCalculation:\n3 * 2 = 6 widgets on weekend\n\nTotal weekly production:\n22 + 6 = 28 widgets\nMonthly calculation:\n- 4 weeks per month\n- 1 holiday distributed uniformly\n- Holiday means 0 widgets produced that day\n\nAdjust weekly production:\n- One holiday reduces production by the daily average\n- Daily average across week = 28 / 7 = 4 widgets per day\n- So one holiday reduces monthly production by 4 widgets\n- 4 weeks * 28 widgets - 4 = 108 total monthly widgets\n- Average weekly production: 108 / 4 = 27 widgets\nFibonacci numbers in 1-based month index: 1, 2, 3, 5, 8\nSo months 1, 2, 3, 5, and 8 get 2 extra holidays\n\nAdjust calculation:\n- Base monthly calculation from Q2: 27 widgets per week\n- 5 months out of 10 have 2 extra holidays\n- Extra holiday impact: 5 * 2 * 4 = 40 additional lost production days\n- 108 total monthly widgets - 40 = 68 total monthly widgets\n- Average weekly production: 68 / 4 = 17 widgets", | |
| "error": 9.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1308.0, | |
| "cost": 0.008264 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 16.0, | |
| "reasoning": "Calculating weekday and weekend widget production:\n- Weekdays (Monday to Friday): 4 widgets per day + 2 extra on Monday\n * Monday: 4 + 2 = 6 widgets\n * Tuesday to Friday: 4 widgets each\n- Weekend days (Saturday and Sunday): 3 widgets per day\n\nDetailed calculation:\n- Monday: 6 widgets\n- Tuesday: 4 widgets\n- Wednesday: 4 widgets\n- Thursday: 4 widgets\n- Friday: 4 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nTotal weekly production: 6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 widgets\nMonthly calculation with one holiday:\n- Normal weeks still produce 28 widgets\n- One holiday per month reduces production\n- Holidays are uniformly distributed, so on average, one day of production is lost\n\nAverage weekly calculation:\n- One day lost means losing the average daily production\n- Average daily production = 28 widgets / 7 days = 4 widgets per day\n- So with one holiday, weekly production becomes: 28 - 4 = 24 widgets\nFibonacci numbers are: 1, 2, 3, 5, 8\nMonths with extra holidays:\n- Month 1 (first month): +2 holidays\n- Month 2 (second month): +2 holidays\n- Month 3 (third month): +2 holidays\n- Month 5 (fifth month): +2 holidays\n- Month 8 (eighth month): +2 holidays\n\nTotal additional holidays: 5 \u00d7 2 = 10 holidays across 10 months\nAverage holidays per month: 1 (from previous calculation) + 2 = 3 holidays\n\nCalculation:\n- Normal week production: 28 widgets\n- Average daily production: 28 / 7 = 4 widgets per day\n- Total days lost per week: 3 days\n- Reduced weekly production: 28 - (3 \u00d7 4) = 28 - 12 = 16 widgets", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1393.0, | |
| "cost": 0.008604 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Weekdays (Monday to Friday): \n- 4 widgets per day\n- 5 days: 4 * 5 = 20 widgets\nWeekends (Saturday and Sunday):\n- 3 widgets per day\n- 2 days: 3 * 2 = 6 widgets\nExtra Monday widgets:\n- 2 additional widgets\nTotal weekly widgets: 20 + 6 + 2 = 28 widgets\nRegular weekly production: 28 widgets\nMonthly holidays: \n- 1 holiday per 4-week month\n- Holiday is closed (0 widgets)\n- Holidays are uniformly distributed\nAverage impact:\n- 1 holiday per 4 weeks = 1/4 of a holiday per week\n- This means on average, we reduce the weekly production by 1/4 of a day's production\n- A typical day produces around 4 widgets\n- Average reduction: 4 * (1/4) = 1 widget per week\nAverage weekly production: 28 - 1 = 27 widgets\nFibonacci numbers (1-based index): 1, 2, 3, 5, 8\nMonths with these indexes get 2 extra holidays\nIn a 10-month year, this means months 1, 2, 3, 5, and 8 get 2 extra holidays\nTotal extra holidays: 5 * 2 = 10 holidays\n\nRegular calculation from Q2:\n- Base weekly production: 28 widgets\n- Average monthly holiday reduction: 1 widget per week\n\nAdditional fibonacci holiday reduction:\n- 10 extra holidays distributed across the year\n- In a 10-month year, that's 1 extra holiday per month on average\n- A typical day produces 4 widgets\n- Additional average reduction: 4 * (1/4) = 1 more widget per week\n\nTotal average weekly production: 28 - 1 - 1 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1358.0, | |
| "cost": 0.008464 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 21.0, | |
| "q3": 21.0, | |
| "reasoning": "Weekdays (Monday to Friday): \n- 4 widgets per day\n- 2 extra widgets on Monday\nSo per weekday: 4 + (2 on Monday) = 6 or 4 on other weekdays\nWeekend days (Saturday and Sunday): 3 widgets per day\n\nCalculation:\n- Weekdays (Mon-Fri): (6 + 4 + 4 + 4 + 4) = 22 widgets\n- Weekend days (Sat-Sun): 3 * 2 = 6 widgets\nTotal widgets in a week: 22 + 6 = 28 widgets\nIn a 4-week month with 1 holiday:\n- Holidays are uniformly distributed\n- Factory is closed on holidays\n- In 4 weeks, 1 holiday means on average 1/4 of a week has a holiday\n\nCalculation:\n- Normal week: 28 widgets\n- Holiday week: Close to 0 widgets\nAverage: 28 * (3/4) + 0 * (1/4) = 21 widgets per week\nFibonacci numbers are: 1, 2, 3, 5, 8 \nIn a 10-month year, fibonacci months are: \n1st, 2nd, 3rd, 5th, and 8th months\n\nCalculation:\nFibonacci months have 2 extra holidays\n- 5 fibonacci months out of 10 get extra holidays\n- 2 extra holidays per such month\n- Total extra holidays: 5 * 2 = 10 holidays across the year\n\nPer month impact:\n- 1 base holiday \n- 2 extra fibonacci holidays\nTotal: 3 holidays per month on average\nIn a 4-week month: 3/4 of a week lost\n\nCalculation similar to Q2:\n28 * (3/4) = 21 widgets per week", | |
| "error": 11.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1347.0, | |
| "cost": 0.00842 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 20.0, | |
| "reasoning": "Weekday widgets:\n- 4 widgets per weekday\n- 5 weekdays in a week\n- 5 * 4 = 20 widgets from weekdays\n\nWeekend widgets:\n- 3 widgets per weekend day\n- 2 weekend days\n- 2 * 3 = 6 widgets from weekend days\n\nExtra Monday widgets:\n- 2 extra widgets on Monday\n- Total calculation: 20 + 6 + 2 = 28 widgets per week\nRegular weekly production: 28 widgets\n\nMonthly holidays:\n- 1 holiday per month\n- Holiday is uniformly distributed (can fall on any day)\n- With one holiday, we lose 1/7 of weekly production\n\nCalculation:\n- Reduction factor: 6/7 of regular production\n- 28 * (6/7) = 24 widgets per week on average\nFibonacci numbers in 1-based month indexing: 1, 2, 3, 5, 8 (months)\nThese months get 2 extra holidays each\n\nFibonacci holiday months:\n1st month: 2 holidays\n2nd month: 2 holidays\n3rd month: 2 holidays\n5th month: 2 holidays\n8th month: 2 holidays\n\nTotal extra holidays: 5 * 2 = 10 holidays per 10-month year\nAverage extra holidays per month: 1\n\nCombined with previous monthly holiday:\n- Total monthly holidays: 2\n- Production reduction: 2/7 of regular weekly production\n\nCalculation:\n- Regular weekly production: 28 widgets\n- Production reduction: 28 * (5/7) = 20 widgets per week on average", | |
| "error": 9.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1295.0, | |
| "cost": 0.008212 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 21.0, | |
| "q3": 21.0, | |
| "reasoning": "Let's calculate the weekly widget production:\n- Weekdays (Monday to Friday): 4 widgets per day\n 5 days * 4 widgets = 20 widgets\n- Weekends (Saturday and Sunday): 3 widgets per day\n 2 days * 3 widgets = 6 widgets\n- Extra widgets on Monday: 2 additional widgets\nTotal weekly production = 20 + 6 + 2 = 28 widgets\nMonthly calculation:\n- Regular weeks: 4 weeks\n- One holiday per month, uniformly distributed\n- On a holiday, the factory is closed, so 0 widgets produced\nCalculation:\n- 3 regular weeks: 3 * 28 = 84 widgets\n- 1 holiday week: 0 widgets\nAverage weekly production = (84 + 0) / 4 = 21 widgets\nFibonacci numbers are 1, 2, 3, 5, 8\nMonths with extra holidays:\n1st, 2nd, 3rd, 5th, 8th months get 2 extra holidays\nCalculation:\n- These 5 months have 2 extra holidays each\n- Total additional holiday weeks in 10 months: 5 * 2 = 10 holiday weeks\n- Total weeks: 10 months * 4 weeks = 40 weeks\n- Regular weeks: 40 - 10 = 30 weeks\nAverage weekly production = (30 * 28) / 40 = 21 widgets", | |
| "error": 11.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1247.0, | |
| "cost": 0.00802 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "Let's break down the widget production:\n- Weekdays (Monday to Friday): 4 widgets per day, plus 2 extra on Monday\n * Monday: 4 + 2 = 6 widgets\n * Tuesday to Friday: 4 widgets each\n- Weekends (Saturday and Sunday): 3 widgets per day\n\nCalculation:\n- Monday: 6 widgets\n- Tuesday to Friday: 4 * 4 = 16 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nTotal widgets per week: 6 + 16 + 3 + 3 = 28 widgets\nIn a month with four weeks and one holiday:\n- The holiday is randomly distributed across days\n- Assume this means the factory loses production for one full day\n\nCalculation:\n- Normal week production: 28 widgets\n- Adjusted for monthly holiday: \n * Total monthly production: 28 * 4 = 112\n * Subtract one day's production (let's use the average daily production)\n * Average daily production: 28 / 7 = 4 widgets per day\n * Monthly production with holiday: 112 - 4 = 108\n * Average weekly production: 108 / 4 = 27 widgets\nFibonacci numbers are: 1, 2, 3, 5, 8 (1-based month indices)\nMonths with extra holidays: 1st, 2nd, 3rd, 5th, and 8th months\n\nExtra holidays calculation:\n- 5 months have two extra holidays each\n- Total extra holidays: 5 * 2 = 10 holidays per 10-month year\n- Average holidays per month: 1 additional holiday\n\nSimilar to Q2 calculation:\n- Normal week production: 28 widgets\n- Subtract average daily production per extra holiday\n- Average daily production: 28 / 7 = 4 widgets per day\n- Monthly production: 28 * 4 = 112\n- Adjusted for extra holidays: 112 - (4 * 1) = 108\n- Average weekly production: 108 / 4 = 27 widgets", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1411.0, | |
| "cost": 0.008676 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-sonnet-latest", | |
| "avg_error": 0.1493000000000002, | |
| "avg_correctness": 0.7666666666666667, | |
| "avg_tokens": 1367.2, | |
| "avg_cost": 0.011904000000000001, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.035, | |
| "q3": 26.07, | |
| "reasoning": "Weekdays (Mon-Fri): \n- Monday: 4 + 2 = 6 widgets\n- Tuesday-Friday: 4 widgets each = 4 \u00d7 4 = 16 widgets\nWeekends (Sat-Sun): 3 widgets each = 3 \u00d7 2 = 6 widgets\nTotal per week = 6 + 16 + 6 = 28 widgets\nBase widgets per week = 28\nEach month has 4 weeks and 1 holiday\nHoliday reduces production:\n- If on weekday (5/7 chance): loses 4 widgets (or 6 if Monday)\n- If on weekend (2/7 chance): loses 3 widgets\nAverage widgets lost per holiday:\n(5/7 \u00d7 4.2 + 2/7 \u00d7 3) = 3.86 widgets\n[Note: 4.2 accounts for probability of Monday vs other weekdays]\nLost per week = 3.86 \u00f7 4 = 0.965\nWeekly average = 28 - 0.965 = 27.035\nFibonacci numbers in range 1-10 are: 1,2,3,5,8\nMonths with indices matching these get 2 extra holidays\nRegular months (indices 4,6,7,9,10): 1 holiday each = 5 holidays\nFibonacci months (indices 1,2,3,5,8): 3 holidays each = 15 holidays\nTotal holidays per year = 20 holidays\nPer month average = 20/10 = 2 holidays\nPer week average = 2/4 = 0.5 holidays\nLost production per week = 0.5 \u00d7 3.86 = 1.93\nNew weekly average = 28 - 1.93 = 26.07", | |
| "error": 0.10500000000000043, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1299.0, | |
| "cost": 0.010881 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "For Q1, let's count:\nWeekdays (Mon-Fri):\n- Monday: 4 + 2 = 6 widgets\n- Tuesday-Friday: 4 widgets \u00d7 4 days = 16 widgets\nWeekends (Sat-Sun):\n- 3 widgets \u00d7 2 days = 6 widgets\nTotal per week = 6 + 16 + 6 = 28 widgets\nFor Q2:\n- Each month has 4 weeks = 28 days\n- Plus 1 holiday where no production occurs\n- The holiday could fall on any day with equal probability\nAverage widget loss per holiday:\n- If on Monday: lose 6 widgets (1/7 probability)\n- If on Tue-Fri: lose 4 widgets (4/7 probability)\n- If on Sat-Sun: lose 3 widgets (2/7 probability)\nAverage widgets lost = (6\u00d71 + 4\u00d74 + 3\u00d72) \u00f7 7 = 4 widgets per holiday\nMonthly production = (4 weeks \u00d7 28 widgets) - 4 = 108 widgets\nAverage weekly production = 108 \u00f7 4 = 27 widgets per week\nFor Q3:\nFibonacci numbers from 1 to 10 are: 1, 2, 3, 5, 8\nSo months 1, 2, 3, 5, and 8 get 2 extra holidays (3 total)\nOther months (4, 6, 7, 9, 10) have 1 holiday\n\nTotal holidays per year:\n- 5 months with 3 holidays = 15 holidays\n- 5 months with 1 holiday = 5 holidays\n- Total = 20 holidays per year\n\nIn a 10-month year:\n- Total weeks = 10 months \u00d7 4 weeks = 40 weeks\n- Average holidays per week = 20 holidays \u00f7 40 weeks = 0.5 holidays/week\n- Average widgets lost per holiday = 4 (from previous calculation)\n- Weekly widget loss = 0.5 holidays \u00d7 4 widgets = 2 widgets\nTherefore, average weekly production = 28 - 2 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1393.0, | |
| "cost": 0.012291 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's calculate widgets per week:\nWeekdays (Mon-Fri):\n- Monday: 4 + 2 = 6 widgets\n- Tuesday-Friday: 4 widgets \u00d7 4 days = 16 widgets\nWeekends (Sat-Sun):\n- 3 widgets \u00d7 2 days = 6 widgets\nTotal per week: 6 + 16 + 6 = 28 widgets\nWith holidays:\n- Each month has 4 weeks = 28 days\n- Plus 1 holiday where no production occurs\n- The holiday could fall on any day with equal probability\n- If holiday falls on Monday: lose 6 widgets\n- If holiday falls on Tue-Fri: lose 4 widgets\n- If holiday falls on weekend: lose 3 widgets\nAverage widgets lost per holiday:\n((6\u00d71) + (4\u00d74) + (3\u00d72)) \u00f7 7 = (6 + 16 + 6) \u00f7 7 = 4 widgets\nSo we lose 4 widgets per month on average\nPer week that's: 4 \u00f7 4 = 1 widget\nTherefore average weekly production: 28 - 1 = 27 widgets\nFibonacci numbers in range 1-10 are: 1,2,3,5,8\nMonths with indices matching these numbers get 2 extra holidays\nIn a 10-month year:\n- 5 months have 1 holiday (normal)\n- 5 months have 3 holidays (fibonacci months)\nAverage holidays per month: (5\u00d71 + 5\u00d73) \u00f7 10 = 2 holidays\nThis means we lose 2 \u00d7 4 = 8 widgets per month on average\nPer week that's: 8 \u00f7 4 = 2 widgets\nTherefore average weekly production: 27 - 1 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1309.0, | |
| "cost": 0.011031000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.071, | |
| "q3": 26.607, | |
| "reasoning": "Let's break this down:\n1. Weekdays (Mon-Fri): 5 days\n - Monday: 4 + 2 = 6 widgets\n - Tue-Fri: 4 widgets each = 4 \u00d7 4 = 16 widgets\n2. Weekends (Sat-Sun): 2 days\n - 3 widgets per day = 3 \u00d7 2 = 6 widgets\n3. Total for one week = 6 + 16 + 6 = 28 widgets\nLet's calculate:\n1. Base weekly production = 28 widgets\n2. One holiday per month means 1 holiday per 4 weeks\n3. Average loss per week due to holiday = 1/4 of a day's production\n4. If holidays are uniformly distributed:\n - 5/7 chance it's a weekday (losing 4 widgets, or 6 if Monday)\n - 2/7 chance it's a weekend (losing 3 widgets)\n5. Average daily loss on holiday = (5/7 \u00d7 4 + 2/7 \u00d7 3) = 3.714 widgets\n (Note: The extra Monday widgets average out to (1/7 \u00d7 2) = 0.286 additional widgets lost if it's a Monday)\n6. Weekly loss = 3.714/4 = 0.929 widgets\n7. Average weekly production = 28 - 0.929 = 27.071 widgets\nLet's solve this step by step:\n1. First, identify Fibonacci months (1-based index):\n - Fibonacci numbers up to 10 are: 1, 1, 2, 3, 5, 8\n - So months 1, 2, 3, 5, 8 get extra holidays\n2. In a 10-month year:\n - 5 months have 2 holidays (months 1,2,3,5,8)\n - 5 months have 1 holiday (months 4,6,7,9,10)\n3. Total holidays per year = (5 \u00d7 2) + (5 \u00d7 1) = 15 holidays\n4. Average holidays per month = 15/10 = 1.5 holidays\n5. Average holidays per 4 weeks = 1.5 holidays\n6. Using previous calculations:\n - Each holiday costs 3.714 widgets on average\n - Weekly loss = (1.5 \u00d7 3.714)/4 = 1.393 widgets\n7. Average weekly production = 28 - 1.393 = 26.607 widgets", | |
| "error": 0.6780000000000008, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1485.0, | |
| "cost": 0.013671 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's solve this step by step:\n1. Weekdays (Mon-Fri): 5 days with 4 widgets each = 5 \u00d7 4 = 20 widgets\n2. Weekends (Sat-Sun): 2 days with 3 widgets each = 2 \u00d7 3 = 6 widgets\n3. Monday bonus: 2 extra widgets\n4. Total per week = 20 + 6 + 2 = 28 widgets\nLet's calculate:\n1. Normal weekly production is 28 widgets\n2. Each month has 4 weeks and 1 holiday\n3. Since holidays are uniformly distributed, on average 1/7 of holidays fall on each day\n4. When holiday falls on Mon-Fri: lose 4 widgets (or 6 if Monday)\n5. When holiday falls on Sat-Sun: lose 3 widgets\n6. Monday holiday (1/7 chance): lose 6 widgets\n7. Other weekday holiday (4/7 chance): lose 4 widgets\n8. Weekend holiday (2/7 chance): lose 3 widgets\n9. Average widgets lost per holiday = (6\u00d71/7 + 4\u00d74/7 + 3\u00d72/7) = 4\n10. Per month: 4 weeks \u00d7 28 widgets - 4 widgets = 108 widgets\n11. Average per week = 108/4 = 27 widgets\nLet's solve this final part:\n1. First identify fibonacci months (1-based index): 1,2,3,5,8\n2. In a 10-month year:\n - Months 1,2,3,5,8 have 3 holidays (1 regular + 2 extra)\n - Months 4,6,7,9,10 have 1 holiday\n3. Total holidays per year = (5 months \u00d7 3 holidays) + (5 months \u00d7 1 holiday) = 20 holidays\n4. Average holidays per month = 20/10 = 2 holidays\n5. Each holiday costs 4 widgets on average (from previous calculation)\n6. Monthly production = 4 weeks \u00d7 28 widgets - (2 holidays \u00d7 4 widgets) = 112 - 8 = 104 widgets\n7. Average weekly production = 104/4 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1411.0, | |
| "cost": 0.012561000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's calculate widgets per week:\n* Weekdays (Mon-Fri): 5 days\n* Weekends (Sat-Sun): 2 days\n* Monday bonus: +2 widgets\nRegular weekday production: 5 days \u00d7 4 widgets = 20 widgets\nWeekend production: 2 days \u00d7 3 widgets = 6 widgets\nMonday bonus: 2 widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\nStarting with 28 widgets/week from Q1\nEach month has:\n* 4 weeks = 28 \u00d7 4 = 112 widgets normally\n* 1 holiday where no production occurs\nOn average, the holiday reduces production by:\n* If on weekday (5/7 chance): lose 4 widgets\n* If on Monday (1/7 chance): lose 6 widgets (4 + 2 bonus)\n* If on weekend (2/7 chance): lose 3 widgets\nAverage loss = (5/7 \u00d7 4 + 1/7 \u00d7 6 + 2/7 \u00d7 3) = 4 widgets/month\nMonthly production = 112 - 4 = 108 widgets\nWeekly average = 108 \u00f7 4 = 27 widgets/week\nFibonacci numbers from 1-10 are: 1,2,3,5,8\nSo months 1,2,3,5,8 get extra holidays\nIn a 10-month year:\n* 5 fibonacci months with 2 extra holidays each = 10 extra holidays\n* These 10 extra holidays are spread across the year\n* Each month still has its regular holiday\nTotal holidays per year = 10 regular + 10 extra = 20 holidays\nAverage holidays per month = 20/10 = 2 holidays\nUsing same logic as Q2, but with 2 holidays per month:\n* Regular production = 112 widgets/month\n* Average loss per holiday = 4 widgets\n* Monthly loss = 2 holidays \u00d7 4 widgets = 8 widgets\nMonthly production = 112 - 8 = 104 widgets\nWeekly average = 104 \u00f7 4 = 26 widgets/week", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1383.0, | |
| "cost": 0.012141 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's calculate widgets per week:\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\nLet's adjust for monthly holidays:\n1. Normal week production = 28 widgets\n2. Each month has 4 weeks = 112 widgets potential\n3. One holiday per month means one day of production lost\n4. Since holidays are uniform across days:\n - If weekday (5/7 chance): lose 4 widgets (or 6 if Monday)\n - If weekend (2/7 chance): lose 3 widgets\n5. Average production loss per holiday:\n (5/7 \u00d7 4 + 2/7 \u00d7 3) = 3.714 widgets\n Plus (1/7 \u00d7 2) = 0.286 for Monday possibility\n Total average loss = 4 widgets\n6. Monthly production = 112 - 4 = 108 widgets\n7. Weekly average = 108 \u00f7 4 = 27 widgets\nLet's calculate with Fibonacci holidays:\n1. Fibonacci numbers for month indices (1-10): 1,2,3,5,8\n2. So months 1,2,3,5,8 get 2 extra holidays\n3. In a year:\n - 5 Fibonacci months: 3 holidays each\n - 5 regular months: 1 holiday each\n4. Total holidays per year = (5\u00d73) + (5\u00d71) = 20 holidays\n5. Average holidays per month = 20/10 = 2 holidays\n6. Using previous calculation method:\n - Base weekly production = 28 widgets\n - Monthly potential = 112 widgets\n - Average loss per holiday = 4 widgets\n - Monthly loss = 2 holidays \u00d7 4 widgets = 8 widgets\n - Monthly production = 112 - 8 = 104 widgets\n7. Weekly average = 104 \u00f7 4 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1383.0, | |
| "cost": 0.012141 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's break this down:\n1. Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n2. Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n3. Monday bonus: +2 widgets\n\nTotal for one week = 20 + 6 + 2 = 28 widgets\nLet's calculate:\n1. Regular weekly production is 28 widgets\n2. Each month has 4 weeks = 28 \u00d7 4 = 112 widgets normally\n3. One holiday per month means one day of production is lost\n4. Since holidays are uniform across days:\n - If on weekday (5/7 chance): lose 4 widgets\n - If on Monday (1/7 chance): lose 6 widgets (4+2 bonus)\n - If on weekend (2/7 chance): lose 3 widgets\n Average production loss = (5/7\u00d74 + 1/7\u00d76 + 2/7\u00d73) = 4 widgets\n5. Monthly production = 112 - 4 = 108 widgets\n6. Weekly average = 108 \u00f7 4 = 27 widgets\nLet's solve step by step:\n1. First, identify Fibonacci months (1-based index):\n Fibonacci numbers up to 10 are: 1,1,2,3,5,8\n So months 1,2,3,5,8 get extra holidays\n2. In a year (10 months):\n - 5 months have 2 extra holidays (months 1,2,3,5,8)\n - 5 months have 1 holiday\n3. Total holidays per year = (5\u00d73 + 5\u00d71) = 20 holidays\n4. Per month average = 20/10 = 2 holidays\n5. Using same loss calculation as before:\n Average production loss per holiday = 4 widgets\n6. Monthly production = 112 - (2 \u00d7 4) = 104 widgets\n7. Weekly average = 104 \u00f7 4 = 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1363.0, | |
| "cost": 0.011841 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "For a normal week:\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Mondays get 2 extra widgets = +2\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\nTotal per week = 20 + 2 + 6 = 28 widgets\nFirst, find baseline widgets per month:\n- 4 weeks \u00d7 28 widgets = 112 widgets per month\nNow account for one holiday per month:\n- Holidays are uniformly distributed, so on average:\n * 5/7 chance it's a weekday (4 widgets lost)\n * 1/7 chance it's a Monday (6 widgets lost - 4 regular + 2 extra)\n * 2/7 chance it's a weekend (3 widgets lost)\nAverage widgets lost per holiday = (5/7\u00d74 + 1/7\u00d76 + 2/7\u00d73) = 4 widgets\nMonthly total = 112 - 4 = 108 widgets\nWeekly average = 108 \u00f7 4 = 27 widgets per week\nFirst 10 Fibonacci numbers: 1,1,2,3,5,8\nMonths with indices matching these (1-based): 1,2,3,5,8\nSo 5 months have extra holidays\nRegular months (5 months): \n- 4 weeks \u00d7 27 widgets = 108 widgets/month\nFibonacci months (5 months):\n- Lose an additional 4 widgets (average) for second holiday\n- 108 - 4 = 104 widgets/month\nTotal yearly production:\n- (5 regular \u00d7 108) + (5 fibonacci \u00d7 104) = 540 + 520 = 1060 widgets\nAverage per month = 1060 \u00f7 10 = 106 widgets\nAverage per week = 106 \u00f7 4 = 26.5 widgets", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1336.0, | |
| "cost": 0.011436 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.14, | |
| "reasoning": "For a regular week:\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n1. First calculate impact of one holiday per month:\n- One holiday per month = one holiday per 4 weeks\n- Average impact per week = 1/4 of a holiday\n- If holiday falls on weekday (5/7 chance): lose 4 widgets\n- If holiday falls on weekend (2/7 chance): lose 3 widgets\n- Average widgets lost per holiday = (4\u00d75/7 + 3\u00d72/7) = 3.71428...\n- Average weekly loss = 3.71428.../4 = 0.928571...\nRegular weekly production (28) - Average weekly loss (0.928571)\n= 27.071429... \u2248 27.07 widgets per week\n1. Identify Fibonacci months in positions 1,2,3,5,8\n2. In a 10-month year, months 1,2,3,5,8 get 2 extra holidays each\n3. Total extra holidays = 5 months \u00d7 2 holidays = 10 extra holidays\n4. These 10 holidays are spread across 10 months (40 weeks)\n5. Additional average weekly loss:\n - 10 holidays cause loss of 10 \u00d7 3.71428... = 37.1428... widgets\n - Spread over 40 weeks = 37.1428.../40 = 0.928571... widgets/week\n6. Previous average (27.07) - Additional loss (0.928571...)\n= 26.142857... \u2248 26.14 widgets per week", | |
| "error": 0.21000000000000085, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1310.0, | |
| "cost": 0.011046 | |
| } | |
| ] | |
| } | |
| ] | |
| }, | |
| { | |
| "strategy_name": "OutputParser", | |
| "tests": [ | |
| { | |
| "provider": "openai", | |
| "model": "gpt-4o-mini", | |
| "avg_error": 9.05, | |
| "avg_correctness": 0.3333333333333333, | |
| "avg_tokens": 1035.0, | |
| "avg_cost": 0.00051075, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 21.0, | |
| "reasoning": "<thinking>\nTo solve the problems, I'll calculate the number of widgets produced under the specified conditions.\n\n**Q1: How many widgets are produced in a week?**\n\n- **Weekdays (Monday to Friday)**: The factory produces 4 widgets each day.\n - There are 5 weekdays: \n - Monday: 4 + 2 (extra) = 6 widgets \n - Tuesday: 4 widgets \n - Wednesday: 4 widgets \n - Thursday: 4 widgets \n - Friday: 4 widgets \n - Total widgets from weekdays = 6 + 4 + 4 + 4 + 4 = 22 widgets.\n\n- **Weekends (Saturday and Sunday)**: The factory produces 3 widgets each day.\n - Total widgets from the weekend = 3 + 3 = 6 widgets.\n\nCombining the two gives:\nTotal widgets in a week = 22 (weekdays) + 6 (weekend) = 28.\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\nGiven that one holiday is uniformly distributed across the week, on average, this will result in 0.14 holidays each week (1 holiday per month distributed over 4 weeks). \n\n- Each week, the factory misses 0.14 * 28 average production = approximately 3.92 widgets due to a holiday.\n\nThus, the average production is:\n28 - 3.92 \u2248 24.08. \n\nTo find an integer average per week, we might consider rounding down to 24 widgets produced on average per week.\n\n**Q3: Taking these fibonacci holidays into account, what's the average number of widgets produced per week?**\n\nThe indices of the months that are Fibonacci numbers up to 10 are:\n1, 2, 3, 5, and 8. There are 5 Fibonacci numbers within the range of 1 to 10.\n- Each of these 5 months has an additional 2 holidays, leading to a total of 10 extra holidays per year due to fibonacci holidays.\n\nAssuming these are distributed across the 10 months, the average is:\n- 10 holidays spread evenly over 40 weeks = 10/40 = 0.25 holidays lost per week.\n\nCalculating the adjustment for this:\n0.25 * 28 = 7 widgets lost per week due to these extra holidays.\n\nSo on average, the number of widgets produced per week is:\n28 - 7 = 21.\n\nNow I will format and provide the answers.\n\n</thinking>\n<answer>\n{\"q1\": 28, \"q2\": 24, \"q3\": 21}\n</answer>", | |
| "error": 8.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 803.0, | |
| "cost": 0.00037155 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 21.0, | |
| "q3": 19.0, | |
| "reasoning": "<thinking>\nTo solve the problem, we will break down each question step by step.\n\n**Q1: How many widgets are produced in a week?**\n\n- Each weekday (Monday to Friday) the factory produces 4 widgets.\n- On weekends (Saturday and Sunday), the factory produces 3 widgets per day.\n- There are two extra widgets produced on Mondays.\n\nThus, the widget production per week can be calculated as follows:\n- For Monday: 4 (regular) + 2 (extra) = 6 widgets\n- For Tuesday, Wednesday, Thursday, Friday: 4 widgets each\n- For Saturday and Sunday: 3 widgets each\n\nCalculating the total production for the week:\n- Monday: 6\n- Tuesday: 4\n- Wednesday: 4\n- Thursday: 4\n- Friday: 4\n- Saturday: 3\n- Sunday: 3\n\nTotal widgets produced in a week:\n6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 widgets\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\nEach month has four weeks and one holiday. \nThis means that in a month, 4 weeks x 7 days = 28 days, minus 1 holiday = 27 operational days.\n\nThe number of widgets produced in a month without holidays is:\n- 4 weekdays (22 weekdays) = 4 widgets x 22 = 88 widgets\n- 2 weekend days (8 weekend days) = 3 widgets x 8 = 24 widgets\n\nTotal before holidays:\n88 + 24 = 112 widgets\n\nAfter accounting for 1 holiday:\n112 - (7/28) * 28 = 112 - 28 = 84 widgets.\n\nSo, the weekly average becomes:\n84 / 4 weeks = 21 widgets per week.\n\n**Q3: Taking the fibonacci numbers into account, what's the average number of widgets produced per week?**\n\nIn Standardtopia, months whose 1-based index is a Fibonacci number are 1, 2, 3, 5, and 8. There are 2 extra holidays for these months.\n\nTotal months with holidays: 5 months with 2 holidays each = 10 holidays.\n\nNow, we need to account for these into our average:\n- 10 holidays throughout the year.\n- Total working days in a year would be 10 months x 27 working days per month - 10 holidays = 270 - 10 = 260 working days.\n\nNow, with the same weekly widget calculation of 28 widgets before holiday count for a year:\n28 widgets/week x 10 months = 280 widgets.\n\nBut we have now taken out the holidays, so we need to keep in mind how these holidays distribute through the weeks as well.\n \nThus, for 10 months, we'll have:\n280 - (10 days from 10 holidays)*(average widgets/day)\nand since there would be 5 weekdays and 2 weekend (1/7 = 0.4286) average, take into account into weekly count each.\n \nEach complete week: 28 widgets = 4 weekdays and 2 weekend days will reduce from total outlook.\n\nFinally, on an average the total number of valid working weeks in 10 months will be less based on holiday breakdown.\n\nWe can conclude that:\n\nThe average widgets produced per week after fibbonaci consideration is 19.\n\n</thinking>\n<answer>\n{\"q1\": 28, \"q2\": 21, \"q3\": 19}\n</answer>", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 983.0, | |
| "cost": 0.00047955 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 26.0, | |
| "q3": 25.0, | |
| "reasoning": "<thinking>\nTo find the number of widgets produced at the factory in Standardtopia, we will first calculate the weekly output based on the given production schedule.\n\n1. **Weekly Production Calculation (Q1)**:\n - **Weekdays (Monday - Friday)**:\n - Monday: 4 (regular) + 2 (extra) = 6 widgets\n - Tuesday: 4 widgets\n - Wednesday: 4 widgets\n - Thursday: 4 widgets\n - Friday: 4 widgets\n - Total for weekdays = 6 + 4 + 4 + 4 + 4 = 22 widgets\n\n - **Weekends (Saturday - Sunday)**: \n - Saturday: 3 widgets\n - Sunday: 3 widgets\n - Total for weekends = 3 + 3 = 6 widgets\n\n - **Total weekly production**:\n - Total widgets = Weekdays + Weekends = 22 + 6 = 28 widgets\n\n2. **Taking Holidays into Account (Q2)**:\n - Each month has 4 weeks and one holiday, equating to 28 days.\n - Therefore, there are 4 production weeks per month (totaling 28 widgets without holidays) and 1 holiday, resulting in 27 operational days (28 - 1).\n - Over the month, we will recalculate:\n - Total production per month (4 weeks) = 28 widgets\n - Since there are 4 weekdays producing 4 widgets each and 2 weekend days producing 3 widgets each:\n - With one holiday, we are effectively losing a day of production:\n - 22 widgets (from Weekdays) + 6 widgets (from Weekends) - 4 (average loss from weekday production since on holiday, we lose one weekday) = 26 widgets.\n - Thus, the average widgets produced per week taking into account one holiday = Total Production for a month / 4 weeks = 26 / 4 = 6.5 widgets per week (which will need to be rounded since widgets can't be fractional; but let's keep it as a decimal for accuracy).\n\n3. **Fibonacci Months and Additional Holidays (Q3)**:\n - Fibonacci numbers start with 1, 1, 2, 3, 5 etc. For the months indexed by fib numbers 1, 2, 3, 5:\n - - Months 1 and 2 have 2 extra holidays each \u2192 total of 4 extra holidays in months 1 and 2\n - Month 3 has 2 holidays \u2192 total of 2 extra holidays \n - Month 5 has 2 holidays \u2192 total of 2 extra holidays \n \n - This adds up to an extra 10 holidays throughout the year.\n - Each of these holidays would further reduce production by 1 day, thus reducing the total production over the whole year. \n - 10 total holidays means 28 * 10 = 280 widgets NOT produced. \n - Given a total original capacity (widgets in 10 months) = 10 * (28 - 1) = 270.\n - Total production loss from holidays = 270 widgets \u2013 10 holidays = 260 widgets.\n - Therefore, average widgets produced after taking holidays into account = 260 widgets / 40 weeks (since there are 10 months at 4 weeks each) = 6.5 widgets per week (maintaining consistency with earlier results).\n\nThus to sum up:\n- Q1: 28 widgets per week from regular production.\n- Q2: Average during a month taking holiday into account is slightly lower at 27 widgets / 4 weeks = 6.75 ~ 6.5 taking into account rounding.\n- Q3: With fib holidays accounted, it averages to 6.5 now too long-term.\n\nFinal answers summary:\n</thinking>\n<answer>\n{\"q1\": 28, \"q2\": 26, \"q3\": 25}\n</answer>", | |
| "error": 2.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1076.0, | |
| "cost": 0.00053535 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 20.0, | |
| "reasoning": "<thinking>\nTo solve the problems, we first need to clarify the production rates:\n\n1. The factory produces:\n - 4 widgets every weekday.\n - 3 widgets on weekends.\n - Additionally, it produces 2 extra widgets on Mondays.\n\nTo determine the total number of widgets produced in a week, we first identify the days in a week:\n- Monday: 4 (weekday) + 2 (extra) = 6 widgets\n- Tuesday: 4 widgets\n- Wednesday: 4 widgets\n- Thursday: 4 widgets\n- Friday: 4 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nNow, let's add those up:\n- Total for Monday: 6\n- Total for Tuesday: 4\n- Total for Wednesday: 4\n- Total for Thursday: 4\n- Total for Friday: 4\n- Total for Saturday: 3\n- Total for Sunday: 3\n\nSo the weekly production is:\n6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 widgets\n\nNow, for Q1, the total number of widgets produced in a week is 28.\n\nNow for Q2, we consider the monthly holiday. Each month has four weeks, and there's one holiday uniformly distributed across the days of the week. Therefore, if a month has 28 widgets per week, in 4 weeks it would be 28 * 4 = 112 widgets. However, with one holiday, we lose the widget production for that day.\n\nSince the factory produces an average of 4 widgets on weekdays and 3 on weekends, we need to estimate how many holidays fall on weekdays versus weekends. There are 5 weekdays and 2 weekend days in a week.\n\n1. The average for a holiday falling on a weekday would reduce the production by 4 widgets.\n2. The average for a holiday falling on a weekend would reduce the production by 3 widgets.\n\nThe number of weekdays in a month is 20 (4 weeks * 5 days) and weekends are 8 (4 weeks * 2 days). Hence, the average holiday impacts:\n- Assuming a uniformly random day, the fraction of weekdays is 20 out of 28, approximately 0.714 and weekends 8 out of 28, approximately 0.286.\n\nThe effective reduction on holidays can be calculated as:\n(0.714 * 4 + 0.286 * 3) = 2.857\n\nThus, with one holiday, the average widget production for a month would be:\n112 - 2.857 = 109.143 widgets\n\nNow, dividing by the 4 weeks gives us an average:\n109.143 / 4 = 27.2853 widgets a week.\n\nSo for Q2, the average number of widgets produced per week after accounting for one holiday is approximately 27.\n\nFinally, for Q3, we need to account for the extra holidays in the months that correspond to the Fibonacci sequence. The Fibonacci sequence of months (1-based) is: 1, 2, 3, 5, 8, and 13, but since there are only 10 months in Standardtopia, the months 1, 2, 3, 5, and 8 will receive two additional holidays.\n\nWith 5 months having an extra 2 holidays, we incur 10 additional holidays. The production loss for these extra holidays, once again calculated similarly, needs to be accounted for the same way. Over a week, 10 holidays spread evenly results in an added average of -2.857 over the 4-week assessment proportionally across weeks:\n\nThat gives us a total of another 10 extra holidays impacting:\n10 * 2.857 = 28.57 widgets lost.\n\nThus, in 40 weeks we reduce -28.57 from previously calculated 109.143 widgets, thus, the average production drops to:\n(109.143 - 28.57) = 80.573 widgets over the 40 weeks.\n\nAverage per week:\n80.573 / 40 = 20.143 widgets.\n\nTherefore:\nQ1: 28 widgets per week\nQ2: 27 widgets per week (rounded)\nQ3: 20 widgets per week (rounded)\n\nSummarizing:\n- Q1: 28\n- Q2: 27\n- Q3: 20\n</thinking>\n\n<answer>\n{\"q1\":28, \"q2\":27,\"q3\":20}\n</answer>", | |
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| "q1": 32.0, | |
| "q2": 28.29, | |
| "q3": 24.29, | |
| "reasoning": "<thinking>\nTo solve the problems, we need to calculate the number of widgets produced in a week, taking into account the factory's production schedule, holidays, and extra holidays related to Fibonacci numbers.\n\n**Q1: Calculate total widgets produced in a week.**\n\nThe factory has the following production schedule:\n- Weekdays (Monday to Friday): 4 widgets each day\n- Weekends (Saturday and Sunday): 3 widgets each day\n- Additional 2 widgets on Monday\n\nNow, let's calculate the total number of widgets produced in one week:\n- Monday: 4 + 2 = 6 widgets\n- Tuesday: 4 widgets\n- Wednesday: 4 widgets\n- Thursday: 4 widgets\n- Friday: 4 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nAdding these together:\n6 + 4 + 4 + 4 + 4 + 3 + 3 = 32 widgets\n\nSo, the total number of widgets produced in one week is 32.\n\n**Q2: Average number of widgets produced per week considering one holiday per month.**\n\nEach month has 4 weeks, with one holiday. If the factory is closed on a holiday, we need to find the average number of holidays per week over a month.\n\nMonths in Standardtopia: 4 weeks in a month (4 * 4 = 16 days), 1 holiday.\nTherefore, the number of operational days = 16 - 1 = 15 days.\n\nThe number of widgets produced in 15 days can be calculated. Since there is one holiday, this holiday could fall on any weekday or weekend:\n- If the holiday falls on a weekday, we would lose 4 widgets.\n- If the holiday falls on a weekend, we would lose 3 widgets.\n\nOn average, since there are 5 weekdays and 2 weekend days, we can assume the holiday could be uniformly distributed:\n- Average loss = (5/7) * 4 + (2/7) * 3\n= (20/7) + (6/7) = 26/7 \u2248 3.71 widgets lost (rounding has implications, but we assume this is acceptable for average).\n\nThus,\nWidgets produced in 15 operational days on average = 32 - (26/7) = 32 - 3.71 = 28.29.\n\nMonthly average production = 28.29 widgets/month / 4 weeks = 7.07 widgets/week (approx).\n\n**Q3: Calculate average number of widgets produced per week considering Fibonacci numbers.**\n\nThe Fibonacci numbers in the first 10 are: 1, 1, 2, 3, 5, 8, 13, ...\n\nThe Fibonacci indexed months in Standardtopia are March (3) and May (5).\nEach of these months gets 2 extra holidays. Thus in March and May, we have:\n- March: 4 weeks - 3 holidays \u2192 1 holiday per week\n- May: similar calculation: 4 weeks - 3 holidays\n\nTaking the extra holidays into account:\n- Loss due to holidays in the Fibonacci months = 2 (March) + 2 (May),\n- So, Production in 4 weeks across 10 months would have 2 months with (32 - 3.71 - 3.71).\n\nCalculating the average considering the 2 extra holidays on those two months:\nIf we assume they both fell on weekdays, we lose an additional 4 widgets (2 days * 4). Thus average number of widgets lost per month = 3.71 + 4 = 7.71.\n\nReducing the monthly average: 32 - 7.71 = 24.29.\n\nFinal average per week after holiday adjustment due to the Fibonacci holiday consideration:\n(20 / 7) / 4 = 24.29 widgets / 4 = 6.07 widgets/week approximately.\n\nTaking into account the rounding, the average production over 10 months due to two extra holidays in each Fibonacci month would be approximately equal to the previous step's values.\n\nFinal answer:\n{\"q1\": 32, \"q2\": 28.29, \"q3\": 24.29}\n\n</thinking>\n\n<answer>\n{\"q1\": 32, \"q2\": 28.29, \"q3\": 24.29}\n</answer>", | |
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| "q1": 28.0, | |
| "q2": 26.0, | |
| "q3": 38.0, | |
| "reasoning": "<thinking>\nTo solve this problem, I will first calculate the total number of widgets produced in a week without any holidays. \n\n1. **Calculating weekly production:**\n - Monday: 4 (weekdays) + 2 (extra) = 6 widgets\n - Tuesday: 4 widgets\n - Wednesday: 4 widgets\n - Thursday: 4 widgets\n - Friday: 4 widgets\n - Saturday: 3 widgets\n - Sunday: 3 widgets\n\n Total widgets produced in a week:\n 6 (Monday) + 4 (Tuesday) + 4 (Wednesday) + 4 (Thursday) + 4 (Friday) + 3 (Saturday) + 3 (Sunday) = 28 widgets.\n\n2. **Q1: Total weekly production without holidays: 28 widgets.**\n\nNext, I will determine the average number of widgets produced per week while accounting for one holiday per month.\n\n3. **Calculating holidays:**\n Each month has 4 weeks and 1 holiday, leading to:\n - 4 weeks = 28 days, and on 1 holiday, there are 28 - 1 = 27 days of production.\n\n Therefore, the monthly widget production before holidays:\n - Per weekday (Monday to Friday), 4 widgets for 20 days (4 weeks).\n - Weekend (Saturday and Sunday) yields 3 widgets each for 8 days.\n\n Therefore, the total per month (before accounting for the holiday):\n - Weekdays: 20 * 4 = 80 widgets\n - Weekends: 8 * 3 = 24 widgets\n - Monthly production = 80 + 24 = 104 widgets.\n\n Now, considering the holiday, we effectively have 27 days of production. The production on the holiday should not be counted:\n - Total widgets produced in 27 working days would be slightly less than 104 due to the holiday.\n\n The average weekly production now becomes:\n 104 widgets produced in 28 days = 104/4 weeks = 26 widgets/week average.\n\n4. **Q2: Average weekly production accounting for 1 holiday per month: 26 widgets.**\n\nNext step is to address the effect of Fibonacci-based holidays. The months indexed by Fibonacci numbers (1, 2, 3, 5, 8) within 10 months are 1, 2, 3, 5, and 8. So that's 5 months out of 10 that each receive 2 extra holidays, which totals to 10 extra holidays across the year (5 months * 2 extra holidays).\n\n5. **Adjusting for holidays:**\n With 2 extra holidays in 5 months remaining months would have just 1 holiday each. Therefore, there are 5 months with 2 holidays and 5 months with 1 holiday:\n - Total holidays = (5 months * 2 holidays) + (5 months * 1 holiday) = 10 + 5 = 15 holidays.\n\n The production days available within the year (40 weeks considering 10 months) would be 40 * 7 = 280 days total - 15 holidays = 265 production days.\n So, the calculation for total widgets now is 265 days produced = (6 + 4 + 4 + 4 + 4 + 3 + 3) = 28 widgets per week. \n\n Therefore, after redistributing the reduction evenly across the weeks, average weekly production would become:\n Total production in 265 days = 265/7 = 37.857 \u2248 38 per week average over the course of a year now with these holidays. \n\n6. **Q3: Average weekly production accounting for Fibonacci holidays and regular holidays: 38 widgets.**\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 26, \"q3\": 38}\n</answer>", | |
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| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "<thinking>\nTo determine the number of widgets produced in a week, we calculate the output for each day of the week according to the production rules given.\n\n1. **Daily Production**:\n - Weekdays: The factory produces 4 widgets every weekday.\n - There are 5 weekdays (Monday to Friday): \n \\[ 4 \\text{ widgets/day} \\times 5 \\text{ days} = 20 \\text{ widgets} \\]\n - Weekends: The factory produces 3 widgets per day on weekends.\n - There are 2 weekend days (Saturday and Sunday): \n \\[ 3 \\text{ widgets/day} \\times 2 \\text{ days} = 6 \\text{ widgets} \\]\n - Extra on Monday: The factory produces 2 extra widgets on Mondays, which we have already counted but will explicitly add: \n \\[ +2 \\text{ extra widgets} \\]\n\n2. **Total Production in a Week**:\n - Thus, for a week:\n \\[ 20 \\text{ (weekdays)} + 6 \\text{ (weekend)} + 2 \\text{ (extra on Monday)} = 28 \\text{ widgets} \\]\n\nSo, the total production in a week (Q1) is **28 widgets**.\n\nNext, we consider the effect of one holiday per month:\n\n1. Each month has 4 weeks, and the factory is closed on holidays. A month has a total output of:\n \\[ 28 \\times 4 = 112 \\text{ widgets} \\]\n\n2. With the holiday included, each month yields:\n \\[ 112 - \\text{widgets lost on a holiday} \\]\n\nSince the holiday can occur on any day of the week, the number of widgets lost will be a function of the day on which the holiday falls:\n - If a holiday falls on a weekday, the loss is 4 widgets.\n - If it falls on a weekend, the loss is 3 widgets.\n\nConsidering uniform distribution, we will average our losses. There are 7 days:\n- 5 weekdays losing 4 widgets = 20 widgets total loss if holiday on weekdays\n- 2 weekend days losing 3 widgets = 6 widgets total loss if holiday on weekends.\n\nAverage loss across these days:\n \\[ \\text{Average loss} = \\frac{5 \\times 4 + 2 \\times 3}{7} = \\frac{20 + 6}{7} = \\frac{26}{7} \\approx 3.71 \\text{ widgets} \\]\n\nMonthly production then becomes:\n \\[ 112 - 3.71 \\approx 108.29 \\text{ widgets per month.} \\]\n \nNow calculate weekly production taking into account the loss.\n \\[ \\frac{108.29}{4} \\approx 27.07 \\text{ widgets/week} \\]\n\nSo, the average production taking holidays into account (Q2) is approximately **27.07 widgets**.\n\nNow, we consider the effect of Fibonacci holidays. The months corresponding to Fibonacci numbers (1, 2, 3, 5, 8) have two extra holidays each.\n\n1. These months are: 1, 2, 3, 5, and 8. This results in 5 months with 2 extra holidays, totaling 10 additional holidays.\n2. The 10 extra holidays further reduce the number of widgets produced.\n3. For each of the 10 extra holidays, we can again average the loss: it will be around 3.71 widgets lost per holiday.\n \\[ 10 \\cdot 3.71 = 37.1 \\text{ widgets lost annually due to extra holidays} \\]\n\nConsidering annually:\n \\[ 112 \\times 10 = 1120 \\text{ widgets for 10 months} \\]\n Reduced by holiday losses:\n \\[ 1120 - 37.1 = 1082.9 \\text{ widgets (total)} \\]\n\nAverage over weeks across 10 months:\n \\[ \\frac{1082.9}{40} = 27.07 \\text{ widgets/week} \\]\n\nHence, the average number of widgets produced per week (Q3) remains approximately **27.07 widgets**.\n\nFinal answers:\n- Q1: 28\n- Q2: 27.07\n- Q3: 27.07\n\nConsidering integer values for practical production counts, round where necessary.\n</thinking>\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 27}\n</answer>", | |
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| "q1": 32.0, | |
| "q2": 24.0, | |
| "q3": 16.0, | |
| "reasoning": "<thinking>\nTo solve the problems, let's first break down the widget production schedule.\n\n1. **Widgets produced in a week:**\n - Weekdays (Monday to Friday): 4 widgets per day\n - Saturday and Sunday: 3 widgets per day\n - On Mondays, the factory also produces 2 extra widgets.\n\nCalculate the total production for each day in a week:\n- Monday: 4 widgets + 2 extra widgets = 6 widgets\n- Tuesday: 4 widgets\n- Wednesday: 4 widgets\n- Thursday: 4 widgets\n- Friday: 4 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nTotal production in a week:\n6 (Mon) + 4 (Tue) + 4 (Wed) + 4 (Thu) + 4 (Fri) + 3 (Sat) + 3 (Sun) = 6 + 20 + 6 = 32 widgets.\n\n2. **Average number of widgets produced per week taking into account one monthly holiday:**\n - Each month has 4 weeks and 1 holiday (the factory is closed on this holiday). \n - With 4 weeks, if we remove one holiday, this means there will be a total of 4 x 32 - 32 = 128 - 32 = 96 widgets produced in a month.\n - The average number of widgets per week after accounting for one holiday would be:\n Total monthly production (96) / 4 weeks = 24 widgets/week.\n\n3. **Average number of widgets produced per week with fibonacci holidays:**\n - Months that have 1-based index as fibonacci numbers are the 1st, 2nd, 3rd, 5th, and 8th months which total up to 5 months.\n - Each of these 5 months has 2 extra holidays.\n - Therefore, for these months, the total holidays would be 5 months x 2 holidays = 10 extra holidays in a year.\n - In total, there are 10 months with 1 holiday each = 10 holidays + 10 extra holidays = 20 holidays.\n - Total weeks in a year = 10 months x 4 weeks = 40 weeks. Each holiday is missed in one week on average.\n - Average widgets lost due to holidays: 20 holidays x 32 widgets = 640 widgets.\n - Total production without holidays: 40 weeks x 32 widgets = 1280 widgets.\n - Adjusted production: 1280 - 640 = 640.\n - Average production per week after considering fibonacci holidays: 640 / 40 = 16 widgets/week.\n\nNow, summarizing the answers for each question:\nQ1: 32 widgets\nQ2: 24 widgets\nQ3: 16 widgets\n</thinking>\n<answer>\n{\"q1\": 32, \"q2\": 24, \"q3\": 16}\n</answer>", | |
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| "q1": 28.0, | |
| "q2": 26.0, | |
| "q3": 26.0, | |
| "reasoning": "<thinking>\nTo find the number of widgets produced in a week, we first need to summarize the production schedule for a single week.\n\nDuring a week:\n- The factory operates 5 weekdays (Monday to Friday) and produces 4 widgets each day.\n- Additionally, Monday produces 2 extra widgets.\n- The two weekend days (Saturday and Sunday) produce 3 widgets each.\n\nCalculating the widgets produced in a week:\n- From Monday to Friday: \n - 5 days * 4 widgets/day = 20 widgets\n - Plus 2 extra widgets from Monday = 20 + 2 = 22 widgets\n\n- For the weekend (Saturday and Sunday): \n - 2 days * 3 widgets/day = 6 widgets\n\nNow, add the widgets from the weekdays and the weekend together for total production in a week:\n- Total widgets = 22 (weekdays) + 6 (weekend) = 28 widgets\n\nSo, the total number of widgets produced in a week is 28.\n\nQ1 results in 28 widgets.\n\nNow, let's account for the holiday in each month:\nEach month has 4 weeks (28 days) and one holiday. So, there are 28 - 1 = 27 production days in a month.\nIn a week of 7 days, one of the days is a holiday, which means only 6 days produce widgets.\n\nThe production days in a week with a holiday:\n- In a week: 6 days of production.\n- Weekdays total: 6 * 4 = 24 on weekdays and 0 on the holiday.\n- Weekend: 3 * 2 = 6 widgets (Saturday + Sunday).\n\nThus, in a holiday week:\n- Total widgets = 24 (weekdays) + 6 (weekend) = 30 widgets.\n\nSince every month has a holiday distributed over the weekdays, we calculate the average number of widgets produced per week as follows:\n- Total widgets in 4 weeks and 1 holiday: 28 widgets/week * 4 weeks = 112\n- After accounting for a holiday: 112 - 7 (the holiday won't have widgets) = 105\n\nNow divide by the 4 weeks to find the average:\n- Average = 105 / 4 = 26.25 widgets per week.\n\nSince there are no half widgets, we can round it down to 26 widgets for count.\n\nNow for Q2, the average number of widgets produced per week with one holiday is 26 (rounded).\n\nNext, we consider the Fibonacci numbered months (1, 2, 3, 5, 8) which have two additional holidays each. The months indexed with Fibonacci numbers within 10 (months) are 1, 2, 3, 5, and 8. This adds 5 additional holidays.\n\nSo, instead of 1 holiday per month, we have a total of:\n- 5 regular holidays + 5 additional = 10 holidays in 10 months\n\nUpon distribution across the weeks:\nThe total production days in a month with 10 holidays = \n30 - 10 = 20 production days total in 10 months\n\nAverage monthly production is 27 * 10 - 10 holidays = 260 for the year.\nThe average weekly production is 260 / 40 (total weeks in 10 months) = 6.5 production per week.\n\nThus Q3 will be 26 (from Q2) after the holidays are looked at as a count, within the weeks.\n\nFinal values:\n- Q1: 28\n- Q2: 26\n- Q3: 26\n\nSo, the final answer is as follows:\n</thinking>\n<answer>\n{\"q1\": 28, \"q2\": 26, \"q3\": 26}\n</answer>", | |
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| "q1": 26.0, | |
| "q2": 19.5, | |
| "q3": 13.0, | |
| "reasoning": "<thinking>\nTo solve these questions, we need to break down the information given into manageable parts.\n\n**Q1: Total widget production in one week.**\n- The factory produces:\n - 4 widgets for each weekday (Monday to Friday) which is 5 days: \n \\( 4 \\text{ widgets} \\times 5 \\text{ days} = 20 \\text{ widgets} \\)\n - 3 widgets for each weekend day (Saturday and Sunday), which is 2 days: \n \\( 3 \\text{ widgets} \\times 2 \\text{ days} = 6 \\text{ widgets} \\)\n - On Mondays, there are 2 extra widgets produced.\n\nCalculating Monday's production:\n- Monday has 4 regular widgets + 2 extra widgets = 6 widgets.\n\nNow, adding all widgets produced in a week:\n- Total widgets = 20 (weekdays) + 6 (weekend) + 2 (extra on Monday) = 26 widgets.\n\n**So, the answer for Q1 is 26 widgets per week.**\n\n**Q2: Average production with one holiday per month.**\nA month has 4 weeks on average, so that\u2019s equivalent to 4 holidays in a year (one for each month within 10 months).\n- Each week has 26 widgets produced typically.\n- If the factory is closed on the holiday, that\u2019s 26 widgets lost for that week.\n\nIf a month has 4 weeks, that would mean there are \\(10 \\text{ months} \\times 1 \\text{ holiday per month} = 10 \\text{ holidays total}\\). \nHowever, these holidays fall across weeks, not perfectly impacting one full week.\n- Assuming even distribution, averaging those 10 holidays across the weeks gives roughly 10/40 total weeks = 0.25 weeks where a holiday lands. So that means:\n \nEffect of holidays on total production:\n- Each week is reduced by 0.25 weeks with 26 widgets affected, meaning:\n \nAverage production = \\( 26 - 26 \\times 0.25 = 26 - 6.5 = 19.5 \\) widgets per week.\n\n**So, the answer for Q2 is approximately 19.5 widgets per week.**\n\n**Q3: Now considering extra holidays in Fibonacci months.**\nFibonacci months based on the index (1, 1, 2, 3, 5, 8) gives us: \n- 1st Month\n- 2nd Month\n- 3rd Month\n- 5th Month\n- 8th Month\n\nThat totals 5 Fibonacci months. Each of these has 2 additional holidays, so we have:\n- 5 months x 2 extra holidays = 10 holidays.\n\nStill averaging across the week, with 20 total holidays (10 from Q2 and 10 from Fibonacci), we now simplify to look across 40 weeks (10 months):\n- Again distribute those 20 holidays evenly across 40 weeks, resulting in 0.5 holiday weeks lost on average.\n\nAverage production = \\( 26 - (26 \\times 0.5) = 26 - 13 = 13 \\) widgets per week.\n\n**So, the answer for Q3 is 13 widgets per week.**\n\nBased on this reasoning:\n- Q1: 26\n- Q2: 19.5\n- Q3: 13\n</thinking>\n<answer>\n{\"q1\":26, \"q2\":19.5, \"q3\":13}\n</answer>", | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 25.86, | |
| "reasoning": "<thinking>\nLet's break down the problem step by step:\n\nQ1: First, we need to calculate how many widgets are produced in a typical week:\n\n- Weekdays (Monday to Friday): 5 days x 4 widgets = 20 widgets\n- Weekends (Saturday and Sunday): 2 days x 3 widgets = 6 widgets\n- Additional widgets on Mondays: 2 additional widgets\n\nTotal widgets in a week = 20 (weekdays) + 6 (weekends) + 2 (extra on Monday) = 28 widgets\n\nQ2: In a month, there are four weeks and one holiday. We assume that holidays can happen on any day of the week with equal probability. Thus, the number of widgets produced on the holiday is deducted from the weekly average.\n\nWeeks in a month = 4\nWidgets produced in 4 weeks (without holiday) = 4 weeks x 28 widgets/week = 112 widgets\n\nNow, we calculate the expected widget production lost due to one holiday per month:\n- Probability of holiday on a weekday: 5/7\n Loss if a holiday is on a weekday = 4 widgets\n- Probability of holiday on a weekend: 2/7\n Loss if a holiday is on a weekend = 3 widgets\n\nExpected loss due to holiday per month = (5/7 * 4) + (2/7 * 3) = 20/7 + 6/7 = 26/7 \u2248 3.71 widgets\n\nAverage weekly production considering monthly holiday:\nTotal production in a month, accounting for holiday = 112 - 26/7\nAverage weekly production with holiday = (112 - 26/7) / 4 \u2248 27.07 widgets\n\nQ3: We account for Fibonacci month holidays. The first few Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ... so in a year with 10 months, Fibonacci-indexed months are 1, 2, 3, 5, and 8. These months have 3 holidays instead of 1.\n\nCalculate the annual holiday impact:\n- Regular months: 5 (non-Fibonacci months) x 1 = 5 holidays\n- Fibonacci months: 5 (Fibonacci months) x 3 = 15 holidays\n\nTotal holidays in a 10-month year = 5 + 15 = 20 holidays\n\nExpected widget production lost due to Fibonacci holidays:\nExpected loss per holiday = 26/7 widgets (from previous calculation)\n\nTotal annual loss = 20 holidays x 26/7 per holiday = 520/7 widgets\n\nAverage weekly production considering Fibonacci holidays:\nTotal annual production without holidays = 10 months x 112 widgets/month = 1120 widgets\n\nTotal production considering holidays = 1120 - 520/7\nAverage weekly production in a year with holidays = (1120 - 520/7) / (10 months x 4 weeks/month) \u2248 25.86 widgets\n\nThus, we have the answers:\n- Q1: 28 widgets per week\n- Q2: Approximately 27.07 widgets per week\n- Q3: Approximately 25.86 widgets per week\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.07, \"q3\": 25.86}\n</answer>", | |
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| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "<thinking>\nLet's tackle this step by step.\n\nQ1: How many widgets are produced in a week?\n\n1. Count the widgets produced each weekday (Monday to Friday).\n - Monday produces 4 widgets plus 2 extra widgets, totaling 6 widgets.\n - Tuesday, Wednesday, Thursday, and Friday each produce 4 widgets.\n - Total widgets produced during weekdays = 6 (Monday) + 4 (Tuesday) + 4 (Wednesday) + 4 (Thursday) + 4 (Friday) = 22 widgets.\n2. Count widgets produced on weekends (Saturday and Sunday):\n - Each weekend day produces 3 widgets.\n - Total widgets produced on weekends = 3 (Saturday) + 3 (Sunday) = 6 widgets.\n3. Total widgets in a week = 22 (weekdays) + 6 (weekends) = 28 widgets.\n\nQ2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\n1. Each month has one holiday which is uniformly distributed across the days of the week. This means that over many weeks, the average impact can be seen across all days. \n2. Each week consists of 7 days, and each month has 4 weeks, which means:\n - 28 days in total in a month and 1 additional holiday.\n3. Total widgets potentially produced in a month = 4 weeks * 28 widgets/week = 112 widgets.\n4. One holiday means one day where the production (3 or 4 or 6 widget day) is completely removed, averaged over weeks.\n5. Average reduction per week due to 1 holiday distributed over 4 weeks is \\(1/4\\) holiday worth of widgets.\n6. Assume average daily production of 4 widgets per day (across weekdays and weekends):\n - Average weekly produce per week considering one holiday per month = 28 widgets - (average daily production) * \\(1/4\\).\n - Average weekly produce per week = 28 - 4/4 = 28 - 1 = 27.\n\nQ3: Taking these Fibonacci holidays into account, what's the average number of widgets produced per week?\n\n1. Fibonacci numbers up to 10 (number of months) are: 1, 2, 3, 5, 8\n2. Months: January, February, March, May, and August are Fibonacci months, each having 2 extra holidays.\n3. Each year has 5 Fibonacci months with 2 extra holidays each = 5 * 2 = 10 extra holidays per year.\n4. Monthly holidays = 1 per month * 10 months = 10 regular holidays.\n5. Total holidays = 10 (regular from all months) + 10 (extra from Fibonacci months) = 20 holidays per year.\n6. Average weekly reduction considering 20 holidays per 40 weeks (10 months * 4 weeks/month):\n - Average holidays per week = \\(20/40\\) = 0.5 holidays per week.\n7. Average weekly widget production = 28 - 0.5 * 4 = 28 - 2 = 26.\n\nThus, our answers are:\n- Q1: The number of widgets produced in a week is 28.\n- Q2: The average number of widgets produced per week with monthly holidays is 27.\n- Q3: The average number of widgets produced per week with Fibonacci holidays is 26.\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26}\n</answer>", | |
| "error": 0.0, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.0, | |
| "reasoning": "<thinking>\nLet's calculate the solutions step by step.\n\nQ1: Weekly Widget Production\n- Weekdays: There are 5 weekdays (Monday to Friday).\n- Weekend: Saturday and Sunday.\n- Monday: An extra 2 widgets are produced.\n\nCalculating the total widgets for each:\n- Weekday production: 5 days \u00d7 4 widgets = 20 widgets.\n- Weekend production: 2 days \u00d7 3 widgets = 6 widgets.\n- Monday extra widgets: 2 widgets.\n\nTotal weekly production: 20 + 6 + 2 = 28 widgets.\n\nQ2: Considering Monthly Holidays\n- Each month has one holiday. Thus, in a 4-week month, one day would be a non-production day.\n- Average weekly impact: 1 holiday per 4 weeks = 1/4 holiday per week.\n\nCalculating average weekly widgets with respect to holidays:\n- Expected widget loss per holiday across the week (average daily production):\n Total weekly production without holidays = 28.\n Average daily production = 28 widgets / 7 days \u2248 4 widgets/day.\n \nThus, the expected widget loss per week due to holidays is 1/4 \u00d7 4 = 1 widget.\n\nAverage weekly production considering holidays:\n28 - 1 = 27 widgets.\n\nQ3: Calculating Fibonacci Month Holidays\n- Fibonacci numbers up to 10: 1, 2, 3, 5, 8 (since the year has 10 months).\n- There are 5 Fibonacci-indexed months in a 10-month year.\n- Each of these months has 2 extra holidays, totaling an additional 2 \u00d7 5 = 10 holidays in a year.\n\nCalculating the total number of holidays per year:\n- Regular monthly holidays: 10 months \u00d7 1 holiday/month = 10 holidays.\n- Extra Fibonacci holidays: 10 holidays.\n- Total holidays in a year = 10 + 10 = 20 holidays.\n\nCalculating average weekly impact over a year:\n- Total weeks in a year = 10 months \u00d7 4 weeks/month = 40 weeks.\n- Average holidays per week considering Fibonacci holidays:\n 20 holidays / 40 weeks = 0.5 holidays/week.\n\nAverage weekly widgets accounting for Fibonacci holidays:\n- Expected widget loss per holiday: 4 widgets (average daily production).\n- Expected weekly widget loss: 0.5 \u00d7 4 = 2 widgets.\n\nAverage weekly production considering all holidays:\n27 - 2 = 25 widgets.\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 25}\n</answer>", | |
| "error": 1.0, | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.15, | |
| "reasoning": "<thinking>\nTo solve each part of the problem, we need to carefully consider the production schedule, holidays, and how they affect widget output.\n\nQ1: In a week, we have 5 weekdays and 2 weekend days.\n- Weekdays: 4 widgets/day for 5 days = 4 * 5 = 20 widgets\n- Weekends: 3 widgets/day for 2 days = 3 * 2 = 6 widgets\n- Monday bonus: 2 extra widgets on Monday\n\nTotal weekly production = 20 (weekdays) + 6 (weekends) + 2 (Monday bonus) = 28 widgets.\n\nQ2: Each month has 4 weeks and 1 holiday. Holidays are uniformly distributed, which suggests that, over a long period, a holiday could appear on any day of the week with equal probability. Importantly, every month has 1 holiday, which means we'll need to calculate the average reduction of widgets due to the holiday's uncertainty and regular 4-week production schedule.\n\nNormally, without holidays:\n- Weekly production would be 28 widgets.\n- Monthly production without holidays (4 weeks) = 28 * 4 = 112 widgets.\n\nThe average loss due to a regular holiday is 3.5 widgets because:\n1. If the holiday is on a weekday: loss = 4 widgets (5/7 chance)\n2. If the holiday is on a weekend: loss = 3 widgets (2/7 chance)\n\nExpected average widget production loss = (4 widgets * 5/7) + (3 widgets * 2/7) = (20/7 + 6/7) = 26/7 \u2248 3.71 widgets.\n\nSo, for 1 regular holiday per month, monthly production loss \u2248 3.71 widgets, translating to:\n- Adjusted monthly production: 112 - 3.71 \u2248 108.29 widgets\n- Average weekly production = 108.29 / 4 \u2248 27.07 widgets.\n\nQ3: A year in Standardtopia has 10 months. Fibonacci months among these 10 are: months 1, 2, 3, 5, and 8. These months have 2 extra holidays on top of the regular one.\n- There are 5 months with an additional 2 holidays.\n- Total annual holidays due to Fibonacci months = 5 months * 2 extra holidays/month = 10 extra holidays.\n- Total annual regular holidays = 10 months * 1 holiday/month = 10 regular holidays.\n- Total annual holidays = 10 regular + 10 Fibonacci = 20 holidays.\n\nMonthly: 10 months * 112 widgets (if no holidays) = 1120 widgets/year.\nSince each month with extra holidays (Fibonacci) loses additional widgets:\n- Average annual widget loss per holiday is as previously calculated, 3.71 widgets per holiday.\n- Total annual widget loss = 3.71 widgets/holiday * 20 holidays = 74.2 widgets.\n\nAdjusted total annual production = 1120 widgets - 74.2 widgets \u2248 1045.8 widgets.\nAverage weekly widget production over 10 months (40 weeks) = 1045.8 / 40 = 26.145 widgets/week \u2248 26.15 widgets/week.\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.07,\"q3\": 26.15}\n</answer>", | |
| "error": 0.21999999999999886, | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.0725, | |
| "q3": 26.145, | |
| "reasoning": "<thinking>\nLet's begin solving each question one by one.\n\nQ1: Calculate the number of widgets produced in a standard week.\n- From Monday to Friday, the factory produces 4 widgets per day, but on Monday, it produces 2 extra widgets.\n Therefore, the weekly weekday production is:\n - Monday: 4 + 2 = 6 widgets\n - Tuesday to Friday: 4 widgets each = 4*4 = 16\n - Thus, total weekday production = 6 + 16 = 22 widgets\n\n- On weekends (Saturday and Sunday), the factory produces 3 widgets each day:\n - Weekend production: 3+3 = 6 widgets\n\nSo, the total weekly production is: 22 + 6 = 28 widgets.\n\nQ2: Calculate the average number of widgets produced per week accounting for one monthly holiday.\n- The month consists of 4 weeks + 1 holiday, meaning each week can be affected by the holiday.\n- If each holiday falls once a month and can randomly fall on any day of the week, on average, it impacts one production day in a 4-week span.\n\nAnnual monthly cycle without extra holidays:\n- Total number of weeks: 4 per month * 10 months = 40 weeks\n- Total monthly holidays: 1 per month * 10 months = 10 holidays\n\nNow distribute these holidays over the 40 weekly cycles.\n- Average loss due to holiday per week: 10 holidays / 40 weeks = 0.25 holiday per week.\n\nFor each weekday or weekend, losing a day matters as follows:\n- Losing a weekday: Average loss is 4 widgets.\n- Losing a weekend day: Average loss is 3 widgets.\n \nSince the holidays are assumed to be uniformly distributed and there's a 5/7 chance of a holiday falling on a weekday and a 2/7 chance for it to fall on a weekend:\n- Expected impact per holiday:\n (5/7 * 4) + (2/7 * 3) = 20/7 + 6/7 = 26/7 = ~3.71 widgets loss per holiday\n\n- Weekly average widget production loss: 0.25 * 3.71 \u2248 0.9275\n\nSo, the average number of widgets per week taking holidays into account: \n28 - 0.9275 \u2248 27.0725 widgets\n\nQ3: Calculate the average weekly production considering Fibonacci month holidays.\n\nFibonacci indexed months are 1, 2, 3, 5, 8 (months that are Fibonacci numbers within the range of 10). There are 5 months meeting this criterion.\n- Each such month has 2 additional holidays: \n Additional holidays = 5 months * 2 = 10 additional holidays\n\n- Total holidays (the original plus extra): 10 (original) + 10 (Fibonacci) = 20 holidays\n\nAverage holiday impact recalculated:\n- Distribute these holidays over 40 weekly cycles: 20 holidays / 40 weeks = 0.5 holiday per week\n\nWidget loss given holiday distribution:\n- Weekly average number of widgets produced considering holidays:\n\n- Weekly loss per classified week: 0.5 * 3.71 = 1.855 widgets lost weekly.\n\nAverage weekly production with both holiday types: \n28 - 1.855 \u2248 26.145 widgets.\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.0725, \"q3\": 26.145}\n</answer>", | |
| "error": 0.21750000000000114, | |
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| "total_tokens": 983.0, | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.8575, | |
| "q3": 20.735, | |
| "reasoning": "<thinking>\nTo solve this problem, let's address each question one by one.\n\nQ1: Number of widgets produced in a week:\n- Weekdays (Monday to Friday): The factory produces 4 widgets each weekday.\n - Total widgets from weekdays: 4 widgets/day * 5 days = 20 widgets.\n- Saturdays and Sundays: The factory produces 3 widgets per day.\n - Total widgets from weekends: 3 widgets/day * 2 days = 6 widgets.\n- On Mondays, the factory produces 2 extra widgets.\n - Extra widgets on Monday: 2 widgets.\n- Total widgets in a week = Widgets from weekdays + widgets from weekends + extra widgets on Monday = 20 + 6 + 2 = 28 widgets.\n\nQ2: Average number of widgets per week considering holidays:\n- Each month has 4 weeks and 1 holiday.\n- Typically, there are 4 weeks/week * 7 days/week = 28 days + 1 holiday, i.e., 29 days in a month.\n- Of the 29 days, 20 are weekdays, and 9 are weekend days. \n- Widgets for a complete month excluding holidays:\n - Weekdays: 20 days * 4 widgets = 80 widgets.\n - Weekend: 9 days * 3 widgets = 27 widgets.\n - Mondays (accounting for 4 Mondays per month): 4 days * 2 extra widgets = 8 extra widgets.\n - Total without holidays = 80 + 27 + 8 = 115 widgets that would be produced monthly without holidays.\n- Subtract widgets lost due to the holiday:\n - If a holiday occurs on a weekday, 4 widgets are lost. If on a weekend, 3 widgets are lost.\n - On average, each month has 4 weekdays and 3 weekends per holiday distribution.\n - Therefore, (4 * 4 widgets + 3 * 3 widgets)/7 days = average loss of approximately 3.57 widgets per holiday.\n - Average widgets produced per month = 115 - 3.57 = 111.43 widgets.\n- Average widgets produced per week = 111.43 widgets/month / 4 weeks/month = 27.8575 widgets.\n\nQ3: Average production considering Fibonacci holidays:\n- In Standardtopia, a year has 10 months, and Fibonacci number months are 1, 2, 3, 5, and 8.\n- These months have 2 extra holidays, making a total of 3 holidays for these months.\n- Therefore, 5 Fibonacci months have 3 holidays each, and the remaining 5 months have 1 holiday each.\n- Total holidays in a year = (5 months * 3 holidays) + (5 months * 1 holiday) = 15 + 5 = 20 holidays per year.\n- Total widgets lost due to Fibonacci holidays considering monthly average loss:\n - Yearly widget loss due to Fibonacci holidays = 20 holidays * 3.57 widgets/holiday = 71.4 widgets\n- Monthly production minus holidays = (115 * 10 months/year - 71.4) widgets/year = 1078.6 widgets/year.\n- Average widgets per week over a year = 1078.6 widgets/year / (52 weeks/year) \u2248 20.735 widgets/week.\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.8575, \"q3\": 20.735}\n</answer>", | |
| "error": 6.122500000000002, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 971.0, | |
| "cost": 0.007872500000000001 | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 25.0, | |
| "reasoning": "<thinking>\nTo solve these questions, we need to break down the widget production schedule and incorporate the effect of holidays, including special Fibonacci holidays.\n\n**Q1: Widgets produced in a week:**\n\n- Monday production: 4 widgets for a weekday + 2 extra widgets = 6 widgets\n- Tuesday to Friday production: 4 widgets each day = 4 days * 4 widgets = 16 widgets\n- Weekend production (Saturday and Sunday): 3 widgets each day = 2 days * 3 widgets = 6 widgets\n\nTotal widgets in a week: 6 (Monday) + 16 (Tuesday to Friday) + 6 (weekend) = 28 widgets.\n\n**Q2: Taking a monthly holiday into account:**\n\nA month typically has 28 days (4 weeks), and one holiday. This holiday is on average distributed across weekdays first (Monday to Friday), because that's when most working days occur. Without loss of generality, we assume the holiday can fall on any weekday or weekend, each having an equal chance.\n\nWhen accounting for holidays:\n- A holiday during a weekday reduces production by 4 widgets because weekdays output 4 widgets.\n- A holiday during the weekend reduces production by 3 widgets.\n\nTo compute the average widget drop per week due to one holiday per month:\n- Number of days: 28 days divided into 5 weekdays, 2 weekend days, assume uniform distribution over 7 days = 1/7 probability for each day.\n- Average widget loss per week due to one holiday: (5/7) * 4 (loss on a weekday) + (2/7) * 3 (loss on a weekend) = 20/7 + 6/7 = 26/7.\n\nGiven that every month has one holiday, for 4 weeks, there is 26/7 widgets lost due to one holiday per month. Average production adjustments over 4 weeks is (26/7) / 4 per week \u2248 0.93 widgets.\n\nThe original widgets per week are 28, subtracting out this reduction:\nAverage widgets per week considering one holiday per month = 28 - 0.93 = 27.07 \u2248 27 widgets.\n\n**Q3: Taking Fibonacci holidays into account:**\n\nAs months are indexed from 1 to 10, the Fibonacci months indices are 1, 2, 3, 5, 8, which results in 5 months having 2 extra holidays each.\n\nTotal extra holidays in a year: 5 months * 2 extra holidays = 10 holidays.\n\nWith already 10 regular holidays, total holidays in a year = 10 + 10 = 20 holidays.\n\nNew total holidays = 20 in 10 months => 20 holidays / (10 months * 4 weeks) = 0.5 extra holidays per week on average.\n\nAverage widget production drop with the new holidays:\n- Average widget loss per week due to 0.5 extra holidays per week = 0.5 \u00d7 average cost of a holiday (3.71 from Step 2) = 1.85.\n\nAdjust widgets production gain: 27 (from Step 2) - 1.85 = 25.15 \u2248 25 widgets.\n\nTherefore, with Fibonacci holidays, the average number of widgets produced per week across the year is approximately 25.\n\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 25}\n</answer>", | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.215, | |
| "q3": 26.429, | |
| "reasoning": "<thinking>\nFirst, let's address each question separately.\n\nQ1: Number of Widgets Produced in a Week\n- Standard weekdays: 4 widgets per day \u00d7 5 days = 20 widgets.\n- Weekends: 3 widgets per day \u00d7 2 days = 6 widgets.\n- Additional widgets on Mondays: 2 extra widgets on Monday (already considered in the weekday calculation).\n\nTotal for the week = 20 (weekday) + 6 (weekend) + 2 (Monday) = 28 widgets.\n\nQ2: Average Number of Widgets Produced Per Week with Monthly Holidays\n- Each month has a single holiday, which means one less workday per month.\n- A month typically has 4 full weeks, i.e., 28 days.\n- The holiday can fall on any of the 28 days of an ideal month:\n - Weekday (Monday to Friday): 20 widgets - 4 if it's a single weekday, -2 additional if Monday.\n - Saturday or Sunday: 6 widgets - 3 if it's a weekend day.\n\nThe holidays are uniformly distributed, so:\n- If a holiday occurs on a weekday, the average loss is (4*5 - 2) / 7 = 2.857 widgets.\n- If on weekend: average loss of (3*2) / 7 = 0.857 widgets.\n\nAverage loss per month (considering holidays uniformly distributed):\n(5 * 2.857 + 2 * 0.857) / 7 \u2248 3.143 widgets.\n\nSo, in a month, expected production without holidays is 4 * 28 = 112 widgets, and the expected drop is 3.143 widgets.\n\nFor 4 weeks, total expected drop = 3.143/4 \u2248 0.786 widgets (rounded).\n\nWidgets produced per week (on holiday average):\n112/4 - 0.786 \u2248 27.215 widgets.\n\nQ3: Average Number of Widgets Produced Per Week with Fibonacci Holidays\n- Fibonacci months indices that have 2 extra holidays: 1, 2, 3, 5, and 8.\n- These months experience 2 more holidays, i.e., total holidays in those months becomes 3.\n\nIn these months, the average loss of production due to holidays becomes:\n[(5 * 2.857 + 2 * 0.857) / 7] * 3 \u2248 9.429 widgets per such month.\n\nNon-Fibonacci months still see a loss of 3.143 widgets.\n\nAverage holiday loss for the year is (5 * 9.429 + 5 * 3.143) / 10 \u2248 6.286 widgets per month.\n\nExpected decrease per week = 6.286 / 4 \u2248 1.5715 widgets.\n\nAverage number of widgets per week, taking Fibonacci holidays into account:\n112/4 - 1.5715 \u2248 26.429 widgets.\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.215,\"q3\": 26.429}\n</answer>", | |
| "error": 0.6439999999999984, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 889.0, | |
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| "q1": 28.0, | |
| "q2": 27.5, | |
| "q3": 28.0, | |
| "reasoning": "<thinking>\n\nTo solve these questions, let's break them down step by step:\n\n**Q1: How many widgets are produced in a week?**\n\n1. **Weekdays Production**: \n - There are 5 weekdays in a week: Monday, Tuesday, Wednesday, Thursday, and Friday.\n - The factory produces 4 widgets every weekday.\n - Therefore, the total production from weekdays is \\(5 \\times 4 = 20\\) widgets.\n\n2. **Weekend Production**:\n - There are 2 weekend days in a week: Saturday and Sunday.\n - The factory produces 3 widgets each day of the weekend.\n - Therefore, the total production from weekends is \\(2 \\times 3 = 6\\) widgets.\n\n3. **Extra Production on Monday**:\n - On Monday, the factory produces 2 extra widgets.\n - Therefore, the total Monday production is increased to \\(4 + 2 = 6\\) widgets.\n\nTotal widgets produced in a week = Weekday production + Weekend production + Extra Monday production adjustment = \\(20 + 6 + 2 = 28\\) widgets.\n\n**Q2: Widgets produced per week accounting for holidays**\n\n1. **Monthly Holidays Impact**:\n - Each month has 4 weeks and 1 holiday.\n - Since the holiday can fall on any day, on average, the holiday will disrupt the average production of one day.\n - There are 28 days in a (4-week) month. An average of 1 holiday leads to an average weekly holiday impact of \\(\\frac{1}{4}\\) day per week.\n - Without knowing if the holiday falls on a weekday or the weekend, let's calculate the average:\n - Weekly production is 28 widgets across 7 days, so daily production is \\( \\frac{28}{7} = 4 \\) widgets per day.\n - Lost production from a holiday (average under uniform distribution) = \\(\\frac{4}{8} = 0.5\\) widgets per week.\n\nAverage weekly production = 28 - 0.5 = 27.5 widgets.\n\n**Q3: Widgets produced per week accounting for extra holidays in Fibonacci months**\n\n1. **Fibonacci Months in a 10-Month Year**:\n - The 1st-5th Fibonacci numbers are: 1, 1, 2, 3, 5. Thus, months with indices 1, 2, 3, 5, and 8 are Fibonacci months.\n - In a 10-month year, 5 months are Fibonacci months.\n\n2. **Extra Holidays in Fibonacci Months**:\n - In each Fibonacci month, there are 2 extra holidays, making a total of 3 holidays for each Fibonacci month (1 standard + 2 extra).\n - This leads to \\(3 \\times 5 = 15\\) holidays from Fibonacci months alone plus \\(1 \\times (10 - 5) = 5\\) holidays from non-Fibonacci months.\n - Total holidays over a 10-month year = \\(15 + 5 = 20\\) holidays. \n - Over a 40-week year, the average holidays per week = \\(\\frac{20}{40} = 0.5\\) holidays per week.\n - Lost production from holidays is \\(\\frac{0.5}{7} \\times 28 \\approx 2\\) (rounded down for practical widget impact due to discrete daily production).\n\nAdjusted weekly production = 28 - 2 = 26 widgets.\n\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.5, \"q3\": 28}\n</answer>", | |
| "error": 2.5, | |
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| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.14, | |
| "reasoning": "<thinking>\nQ1: Calculate how many widgets are produced in a regular week.\n\n- The factory produces 4 widgets on each weekday (Monday to Friday), which totals: \n 4 widgets/weekday * 5 weekdays = 20 widgets.\n- On weekends (Saturday and Sunday), the factory produces 3 widgets per day:\n 3 widgets/day * 2 days = 6 widgets.\n- Additionally, on Mondays, the factory produces 2 extra widgets, so:\n 20 widgets (weekday) + 6 widgets (weekend) + 2 extra widgets (Monday) = 28 widgets.\n\nTherefore, 28 widgets are produced in a regular week.\n\nQ2: Calculate the average number of widgets produced per week, considering monthly holidays.\n\n- Each month has 4 weeks and 1 holiday.\n- The weekly production without holidays is 28 widgets.\n- There are 4 weeks * 28 widgets/week = 112 widgets produced in 4 weeks.\n- The holiday means 1 less day of production. Calculate how this affects the average weekly production:\n The holiday can fall on any day of the week. Let's consider its impact on production on average.\n\nAverage weekday production without extra widgets: 4 widgets\nAverage weekends production: 3 widgets\nProbability of holiday on a weekday: 5/7\nProbability of holiday on a weekend: 2/7\n\nImpact on production due to the holiday:\nAverage impact = (Probability of holiday on weekday * Average weekday production) + (Probability of holiday on weekend * Average weekend production) \n = ((5/7) * 4) + ((2/7) * 3)\n = 20/7 + 6/7 = 26/7 \u2248 3.71 widgets\n\nTherefore, during a 4-week month, the impact of the holiday is approximately 3.71 widgets less production.\nHence, adjusted 4-week production is 112 widgets - 3.71 widgets \u2248 108.29 widgets.\nAverage weekly production considering the holiday = 108.29 widgets / 4 weeks \u2248 27.07 widgets.\n\nQ3: Calculate the average number of widgets produced per week, considering Fibonacci holidays.\n\n- The Fibonacci sequence starts as 1, 1, 2, 3, 5, 8, 13, ...\n- The months with index: 1, 2, 3, 5, 8 are Fibonacci months for the 10-month year.\n- In each Fibonacci month, there are 2 extra holidays.\n- Total extra holidays in 10 months: 5 months * 2 extra holidays = 10 extra holidays.\n- Calculate the impact of these extra holidays on the average weekly production:\n Total holidays in the year due to monthly holidays: 10 months * 1 holiday = 10 holidays.\n Total holidays in the year including extra Fibonacci holidays: 10 + 10 = 20 holidays.\n- Adjusted production for the year due to all holidays:\n Impact of all holidays on production per week:\n Average yearly impact = (20 holidays/year * (Average impact of a holiday)) = 20 * 26/7 = 74.29 widgets per year\n \n Original yearly production = 10 months * 112 widgets/month = 1120 widgets/year\n Adjusted yearly production = 1120 widgets/year - 74.29 widgets = 1045.71 widgets\n Average weekly production with extra Fibonacci holidays:\n 1045.71 widgets / (10 months * 4 weeks/month) \u2248 26.14 widgets/per week\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.07, \"q3\": 26.14}\n</answer>", | |
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| { | |
| "q1": 28.0, | |
| "q2": 25.0, | |
| "q3": 24.0, | |
| "reasoning": "Here's my solution with detailed reasoning:\n\n<thinking>\nQ1: Widget Production per Week\n- Weekdays (Mon-Fri): 4 widgets/day\n- Extra 2 widgets on Monday\n- Weekends (Sat-Sun): 3 widgets/day\n\nCalculation:\n- Weekdays: (4 * 5) + 2 = 22 widgets\n- Weekends: 3 * 2 = 6 widgets\n- Total per week: 22 + 6 = 28 widgets\n\nQ2: Accounting for Monthly Holiday\n- 4 weeks per month\n- 1 holiday randomly distributed\n- Holiday reduces production by that day's typical output\n- Average impact depends on which day is randomly selected\n\nAverage weekly reduction:\n- If holiday falls on weekday: Average reduction is (4 + 2) = 6 widgets\n- If holiday falls on weekend: Average reduction is 3 widgets\n- Probability of weekday/weekend holiday is equal\n\nAverage weekly production: (28 - ((6 + 3)/2)) = 25.5 widgets\n\nQ3: Fibonacci Holidays\nFibonacci numbers: 1, 2, 3, 5, 8\nMonths with 2 extra holidays: 1st, 2nd, 3rd, 5th, 8th months\n- Additional holiday reduction in these months\n- Reduces production further\n\nEstimated average: Around 23-24 widgets per week\n</thinking>\n\n<answer>\n{\n \"q1\": 28,\n \"q2\": 25,\n \"q3\": 24\n}\n</answer>", | |
| "error": 4.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 656.0, | |
| "cost": 0.0037159999999999997 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 22.0, | |
| "reasoning": "Here's my solution:\n\n<thinking>\nQ1: Widgets per week calculation\n- Weekdays (Monday-Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Weekends (Saturday-Sunday): 3 widgets \u00d7 2 days = 6 widgets\n- Extra Monday widgets: 2 additional widgets\nTotal weekly widgets: 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact\n- 4 weeks per month\n- 1 holiday per month, randomly distributed\n- Holiday reduces widget production\n- Assuming holiday equally likely on any day:\n * Holiday probability on weekday: 5/7\n * Holiday probability on weekend: 2/7\nAverage weekly reduction:\n * Weekday holiday: (4 \u00d7 5/7) = 2.86 fewer widgets\n * Weekend holiday: (3 \u00d7 2/7) = 0.86 fewer widgets\nTotal average weekly widgets: 28 - (2.86 + 0.86) = 24.28 widgets\n\nQ3: Fibonacci holiday impact\n- Fibonacci month indices: 1, 2, 3, 5, 8\n- These months get 2 extra holidays\n- Two extra holidays further reduce widget production\nAdditional reduction beyond Q2 calculation\n</thinking>\n\n<answer>\n{\n \"q1\": 28,\n \"q2\": 24,\n \"q3\": 22\n}\n</answer>", | |
| "error": 7.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 621.0, | |
| "cost": 0.0035759999999999998 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 16.0, | |
| "reasoning": "Here's my solution with detailed reasoning:\n\n<thinking>\nQ1: Widgets per Week Calculation\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Mondays get 2 extra widgets, so: 20 + 2 = 22 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Total weekly widgets: 22 + 6 = 28 widgets\n\nQ2: Monthly Widget Calculation\n- One holiday per month, randomly distributed\n- In 4 weeks, 1 day is lost\n- Lost day means reduced production\n- Each lost day reduces: 4 widgets (weekday) or 3 widgets (weekend)\n- Average reduction: (4 \u00d7 5/7) + (3 \u00d7 2/7) = ~3.57 widgets\n- Average weekly production: 28 - 3.57 = 24.43 widgets\n\nQ3: Fibonacci Holiday Calculation\n- Fibonacci month indices: 1, 2, 3, 5, 8 (1-based indexing)\n- 5 such months get 2 extra holidays\n- Additional holiday reduction: 5 \u00d7 2 = 10 days\n- Additional production loss: ~8.57 widgets\n- Average weekly production: 24.43 - 8.57 = 15.86 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 24, \"q3\": 16}\n</answer>", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 645.0, | |
| "cost": 0.003672 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let's solve this step by step:\n\n<thinking>\nQ1: Widgets per week calculation\n* Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n* Mondays have 2 extra widgets: 20 + 2 = 22 widgets\n* Weekend days (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\n* Total weekly widgets: 22 + 6 = 28 widgets\n\nQ2: Monthly holiday consideration\n* 4 weeks per month\n* 1 holiday per month, randomly distributed\n* If holiday falls on a weekday, reduce weekday production\n* If holiday falls on a weekend, reduce weekend production\n* Average reduction: 4 widgets per month across 4 weeks\n* Average weekly reduction: 1 widget\n* Average weekly production: 28 - 1 = 27 widgets\n\nQ3: Fibonacci holidays additional consideration\n* Fibonacci numbers in 1-based month index are: 1, 2, 3, 5, 8\n* These months get 2 extra holidays\n* Additional holiday reduction per month: 2 extra holidays\n* Additional average weekly reduction: 0.5 widgets\n* Average weekly production: 27 - 0.5 = 26.5 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26.5}\n</answer>", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 613.0, | |
| "cost": 0.0035439999999999994 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 24.0, | |
| "reasoning": "Let's solve this step by step:\n\n<thinking>\nQ1 Widget Calculation:\n- Weekdays (Monday-Friday): 4 widgets per day, but +2 extra on Monday\n- Monday: 4 + 2 = 6 widgets\n- Tuesday-Friday: 4 widgets per day\n- Weekend days (Saturday-Sunday): 3 widgets per day\n- Total weekly calculation:\n * Weekdays: (6 + 4 + 4 + 4 + 4) = 22 widgets\n * Weekend days: (3 + 3) = 6 widgets\n * Total per week = 28 widgets\n\nQ2 Monthly Holiday Calculation:\n- 4 weeks in a month, but 1 holiday distributed randomly\n- This means on average, the holiday replaces one of the existing days\n- Assuming the holiday most likely replaces a weekday (for simplicity)\n- Reduces widget production by the average daily widget count\n- Average daily widget count is: 28 widgets / 7 days \u2248 4 widgets\n- So monthly widget production will be: (28 * 4 - 4) = 108 widgets\n- Average weekly production: 108 / 4 = 27 widgets\n\nQ3 Fibonacci Holiday Calculation:\n- Fibonacci month indices: 1, 2, 3, 5, 8 (in a year of 10 months)\n- These months get 2 extra holidays\n- So overall holiday reduction increases\n- Estimated weekly reduction might be around 4-5 widgets per such month\n- Estimated average weekly production: around 23-24 widgets\n</thinking>\n\n<answer>\n{\n \"q1\": 28,\n \"q2\": 27,\n \"q3\": 24\n}\n</answer>", | |
| "error": 2.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 703.0, | |
| "cost": 0.0039039999999999995 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 21.0, | |
| "q3": 14.0, | |
| "reasoning": "Let's solve this step by step:\n\n<thinking>\nQ1: Widget production calculation\n- Weekdays (Monday-Friday): 4 widgets per day\n- Weekend days (Saturday-Sunday): 3 widgets per day\n- Extra 2 widgets on Mondays\nCalculation:\n- Weekdays: 5 days * 4 widgets = 20 widgets\n- Weekend days: 2 days * 3 widgets = 6 widgets\n- Monday extra: 2 widgets\nTotal weekly production: 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact\n- 4 weeks per month\n- 1 holiday randomly distributed\n- Holiday reduces production for that day\n- Probability of holiday on each day type varies\nAverage: ((28 * 3) + (4 * 0)) / 4 = 21 widgets per week\n\nQ3: Fibonacci holiday impact\n- Fibonacci numbers: 1, 2, 3, 5, 8\n- These months get 2 extra holidays\n- This will further reduce monthly widget production\n- Estimated average reduction: ((28 * 2) + (4 * 0)) / 4 = 14 widgets per week\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 21, \"q3\": 14}\n</answer>", | |
| "error": 18.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 595.0, | |
| "cost": 0.0034719999999999994 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 24.0, | |
| "reasoning": "Here's my solution with detailed reasoning:\n\n<thinking>\nQ1: Widget Production per Week\n- Weekdays (Monday to Friday): 4 widgets/day\n * Monday gets 2 extra widgets\n * So Monday: 6 widgets\n * Other weekdays: 4 widgets each\n- Weekends (Saturday, Sunday): 3 widgets/day\n- Calculation:\n * Monday: 6 widgets\n * Tuesday-Friday: 4 * 4 = 16 widgets\n * Saturday-Sunday: 3 * 2 = 6 widgets\n * Total per week: 6 + 16 + 6 = 28 widgets\n\nQ2: Monthly Production with Random Holiday\n- 4 weeks per month\n- 1 holiday randomly distributed\n- Probability of holiday on each day: 1/7\n- Widgets lost per holiday: \n * If holiday on weekday: 4 widgets\n * If holiday on weekend: 3 widgets\n- Average widgets lost per week: (4 * 5/7) + (3 * 2/7) = 3.57\n- Average weekly production: 28 - 3.57 = 24.43 widgets\n\nQ3: Fibonacci Holidays\n- Fibonacci month indices: 1, 2, 3, 5, 8\n- These months get 2 extra holidays\n- Total extra holidays: 5 * 2 = 10 holidays per year\n- 10 months in a year, so extra holiday probability increases\n- Additional widgets lost per week will be slightly higher\n- Rough estimate: ~24 widgets per week\n</thinking>\n\n<answer>\n{\n \"q1\": 28,\n \"q2\": 24,\n \"q3\": 24\n}\n</answer>", | |
| "error": 5.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 701.0, | |
| "cost": 0.0038959999999999997 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 17.0, | |
| "reasoning": "Let's solve this step by step:\n\n<thinking>\nQ1: Widget production per week\n- Weekdays (Monday to Friday): 4 widgets per day\n * Monday has 2 extra widgets\n * So, Monday: 6 widgets, other weekdays: 4 widgets\n- Weekends (Saturday, Sunday): 3 widgets per day\n- Calculation:\n * Weekdays: (6 + 4 * 4) = 22 widgets\n * Weekends: (3 * 2) = 6 widgets\n * Total per week: 22 + 6 = 28 widgets\n\nQ2: Monthly holiday impact\n- 4 weeks per month\n- 1 holiday per month, randomly distributed\n- Some weeks might lose 1 day of production\n- Average impact: 1/7 of weekly production reduced\n- Average weekly production: 28 * (6/7) = 24 widgets\n\nQ3: Fibonacci holidays additional impact\n- Fibonacci numbers are: 1, 2, 3, 5, 8\n- Months with 2 extra holidays: 1st, 2nd, 3rd, 5th, 8th months\n- Each of these months loses 2 more production days\n- Additional reduction in production\n- Estimated average reduction from base Q2 calculation\n * ~2/7 additional reduction\n- Average weekly production: 24 * (5/7) \u2248 17.14 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 24, \"q3\": 17}\n</answer>", | |
| "error": 12.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 657.0, | |
| "cost": 0.0037199999999999998 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 24.0, | |
| "reasoning": "Let's solve this step by step:\n\n<thinking>\nQ1: Widget production per week\n* Weekdays (Monday-Friday): 4 widgets \u00d7 5 days = 20 widgets\n* Monday gets 2 extra widgets: 20 + 2 = 22\n* Weekends (Saturday-Sunday): 3 widgets \u00d7 2 days = 6 widgets\n* Total per week: 22 + 6 = 28 widgets\n\nQ2: Monthly widget production\n* A month has 4 weeks and 1 holiday\n* Holiday is uniformly distributed, reducing weekly production\n* Average weekly production = (28 \u00d7 4 - (1 \u00d7 4)) \u00f7 4\n* So holiday reduces weekly production by 1 widget\n* New average = 27 widgets per week\n\nQ3: Fibonacci holidays\n* Fibonacci numbers (1-based index): 1, 2, 3, 5, 8\n* These months get 2 additional holidays\n* 5 months with 2 extra holidays\n* Total holiday reduction: 4 + (5 \u00d7 2) = 14 widgets per month\n* Average weekly production = (28 \u00d7 4 - 14) \u00f7 4 = 24 widgets per week\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 24}\n</answer>", | |
| "error": 2.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 607.0, | |
| "cost": 0.0035199999999999997 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 17.0, | |
| "reasoning": "Let's solve this step by step:\n\n<thinking>\nQ1: Widget Production in a Week\n- Weekdays (Monday to Friday): 4 widgets/day\n * Extra 2 widgets on Monday\n * So Monday = 6 widgets\n * Tuesday-Friday = 4 widgets/day\n- Weekends (Saturday-Sunday): 3 widgets/day\n- Calculation:\n * Monday: 6 widgets\n * Tuesday-Friday: 4 * 4 = 16 widgets\n * Saturday-Sunday: 3 * 2 = 6 widgets\n * Total per week: 6 + 16 + 6 = 28 widgets\n\nQ2: Monthly Production with Holidays\n- 4 weeks per month\n- 1 holiday randomly distributed\n- If holiday occurs on a work day, that day's production is lost\n- Average impact = 1/7 reduction in weekly production\n- So average weekly production = 28 * (6/7) = 24 widgets\n\nQ3: Fibonacci Holidays\n- Fibonacci numbers: 1, 2, 3, 5, 8\n- Months with 2 extra holidays: months 1, 2, 3, 5, 8\n- Extra holidays reduce production further\n- Each month with fibonacci index loses 2/7 of production\n- So reduce Q2 result by 2/7\n- 24 * (5/7) = 17.14 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 24, \"q3\": 17}\n</answer>", | |
| "error": 12.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 654.0, | |
| "cost": 0.0037079999999999995 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-sonnet-latest", | |
| "avg_error": 0.32828571399999973, | |
| "avg_correctness": 0.6333333333333333, | |
| "avg_tokens": 765.9, | |
| "avg_cost": 0.008212500000000001, | |
| "n_tests": 10, | |
| "failure_rate": 0.16666666666666666, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\nQ1: Widgets in a regular week\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Accounting for monthly holidays\n- Each month has 4 weeks and 1 holiday\n- Holidays are uniformly distributed, so on average:\n * 5/7 chance it's on a weekday (losing 4 widgets)\n * 1/7 chance it's on Monday (losing 6 widgets)\n * 2/7 chance it's on weekend (losing 3 widgets)\nAverage widgets lost per holiday = (5/7 \u00d7 4) + (1/7 \u00d7 6) + (2/7 \u00d7 3) = 4 widgets\nLost per week = 4/4 = 1 widget\nNew weekly average = 28 - 1 = 27 widgets\n\nQ3: Accounting for Fibonacci holidays\n- Fibonacci numbers up to 10: 1,1,2,3,5,8\n- Months with indices 1,2,3,5,8 get 2 extra holidays\n- In a 10-month year:\n * Regular months (5 months): 1 holiday each\n * Fibonacci months (5 months): 3 holidays each\n- Total holidays per year = (5 \u00d7 1) + (5 \u00d7 3) = 20 holidays\n- Holidays per month = 20/10 = 2 holidays\n- Lost widgets per week = 2 \u00d7 4/4 = 2 widgets\nNew weekly average = 28 - 2 = 26 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26}\n</answer>", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 762.0, | |
| "cost": 0.008154 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.14, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\nQ1: Widgets per week\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\n* Total per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holidays impact\n* Normal month has 4 weeks = 28 \u00d7 4 = 112 widgets\n* One holiday per month where no widgets are produced\n* Each day has equal probability of being a holiday\n* Average widgets lost per holiday:\n * If weekday (5/7 probability): 4 widgets (plus 2 if Monday, 2/7 probability)\n * If weekend (2/7 probability): 3 widgets\n * Average widgets lost = (5/7 \u00d7 4) + (2/7 \u00d7 3) + (1/7 \u00d7 2) = 3.71 widgets\n* Monthly production = 112 - 3.71 widgets\n* Weekly average = (112 - 3.71) / 4 = 27.07 widgets\n\nQ3: Fibonacci impact\n* Fibonacci numbers in range 1-10 are: 1, 2, 3, 5, 8\n* So months 1, 2, 3, 5, and 8 have two extra holidays\n* Regular months (4, 6, 7, 9, 10) have one holiday = 5 months\n* Fibonacci months have three holidays = 5 months\n* Total extra holiday impact per year = 5 \u00d7 2 \u00d7 3.71 = 37.1 widgets\n* Yearly production = 10 months \u00d7 (112 - 3.71) - 37.1\n* Weekly average = (10 \u00d7 (112 - 3.71) - 37.1) / (10 \u00d7 4) = 26.14 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.07, \"q3\": 26.14}\n</answer>", | |
| "error": 0.21000000000000085, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 798.0, | |
| "cost": 0.008694 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07142857, | |
| "q3": 26.14285714, | |
| "reasoning": "<thinking>\nLet me solve this step by step.\n\nQ1: Widgets per week\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Accounting for monthly holidays\n* Regular production is 28 widgets/week\n* Each month has 4 weeks + 1 holiday where no production occurs\n* Holiday could fall on any day, so let's calculate average loss:\n * If holiday on weekday (5/7 chance): lose 4 widgets\n * If holiday on weekend (2/7 chance): lose 3 widgets\n * If holiday on Monday (1/7 chance): lose additional 2 widgets\nAverage widgets lost per holiday = (5/7 \u00d7 4) + (2/7 \u00d7 3) + (1/7 \u00d7 2) = 3.714285714\n* Per week loss = 3.714285714 \u00f7 4 = 0.928571429\nWeekly average = 28 - 0.928571429 = 27.07142857\n\nQ3: Accounting for fibonacci holidays\n* Fibonacci numbers in range 1-10 are: 1, 2, 3, 5, 8\n* These months get 2 extra holidays each\n* So out of 10 months, 5 months have 1 holiday, 5 months have 3 holidays\n* Average holidays per month = (5\u00d71 + 5\u00d73)/10 = 2\n* Average widgets lost per week due to holidays = 2 \u00d7 3.714285714 \u00f7 4 = 1.857142857\nWeekly average = 28 - 1.857142857 = 26.14285714\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.07142857, \"q3\": 26.14285714}\n</answer>", | |
| "error": 0.21428570999999863, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 782.0, | |
| "cost": 0.008454 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\nQ1: Widgets per week\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\n* Total per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Accounting for monthly holidays\n* Normal week production = 28 widgets\n* Each month has 4 weeks and 1 holiday\n* A holiday reduces production based on which day it falls on:\n * If on weekday (5/7 chance): loses 4 widgets\n * If on weekend (2/7 chance): loses 3 widgets\n * If on Monday (1/7 chance): loses 6 widgets (4+2)\n* Average widgets lost per holiday:\n * (5/7 \u00d7 4 + 2/7 \u00d7 3 + 1/7 \u00d7 6) = 4 widgets\n* Monthly loss = 4 widgets\n* Weekly loss = 4/4 = 1 widget\n* Average weekly production = 28 - 1 = 27 widgets\n\nQ3: Accounting for fibonacci holidays\n* Fibonacci numbers from 1 to 10: 1,1,2,3,5,8\n* Months with extra holidays: 1,2,3,5,8\n* That's 5 months with extra holidays\n* Extra holidays = 5 \u00d7 2 = 10 additional holidays per year\n* Additional weekly loss = (10 holidays \u00d7 4 widgets)/(10 months \u00d7 4 weeks)\n* = 40/(40) = 1 widget per week\n* Final average = 27 - 1 = 26 widgets per week\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26}\n</answer>", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 749.0, | |
| "cost": 0.007959 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "<thinking>\nLet me break this down step by step:\n\nQ1: Widgets per week:\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\n* Total per week: 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holiday impact:\n* Normal week produces 28 widgets\n* Each month has 4 weeks and 1 holiday\n* On a holiday, production is 0\n* A holiday could fall on any day:\n - If on Monday: Lose 4 + 2 = 6 widgets\n - If on Tue-Fri: Lose 4 widgets\n - If on Sat-Sun: Lose 3 widgets\n* Average loss per holiday = (6 + 4*4 + 3*2)/7 \u2248 4 widgets\n* Monthly loss = 4 widgets\n* Per week loss = 4/4 = 1 widget\n* New weekly average = 28 - 1 = 27 widgets\n\nQ3: Fibonacci impact:\n* Fibonacci numbers in range 1-10 are: 1,1,2,3,5,8\n* Months 1,2,3,5,8 get extra holidays\n* That's 5 months with extra holidays\n* Each extra holiday causes same average loss as before (4 widgets)\n* Total extra loss = 5 \u00d7 4 = 20 widgets per year\n* Per month = 20/10 = 2 widgets\n* Per week = 2/4 = 0.5 widgets\n* Final weekly average = 27 - 0.5 = 26.5 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26.5}\n</answer>", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 749.0, | |
| "cost": 0.007959 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07142857, | |
| "q3": 26.60714286, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\nQ1) For a regular week:\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2) For a month with one holiday:\n- There are 4 weeks = 28 \u00d7 4 = 112 widgets base\n- One holiday removes:\n * If on weekday: 4 widgets (plus 2 if Monday)\n * If on weekend: 3 widgets\n * Since holidays are uniform, average loss is:\n (5\u00d74 + 2\u00d73 + 2)/7 = 3.714285714 widgets\n- So average weekly production = (112 - 3.714285714)/4 = 27.07142857 widgets\n\nQ3) In a 10-month year:\n- Fibonacci numbers up to 10 are: 1,1,2,3,5,8\n- So months 1,2,3,5,8 get 2 extra holidays\n- Regular months (4,6,7,9,10) have 1 holiday = 5 months\n- Fibonacci months have 3 holidays = 5 months\n- Total extra holiday loss per year = 5 months \u00d7 2 holidays \u00d7 3.714285714\n- This means an additional loss of 37.14285714 widgets per fibonacci month\n- Per week loss = (37.14285714 \u00d7 5)/(10 months \u00d7 4 weeks) = 0.464285714\n- Final weekly average = 27.07142857 - 0.464285714 = 26.60714286\n\n</thinking>\n<answer>\n{\"q1\": 28, \"q2\": 27.07142857, \"q3\": 26.60714286}\n</answer>", | |
| "error": 0.6785714299999981, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 761.0, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\n1. For Q1 (widgets per week):\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n- Monday bonus: 2 widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\n2. For Q2 (accounting for monthly holiday):\n- Each month has 4 weeks plus 1 holiday\n- Holidays are distributed uniformly across days\n- If holiday falls on Monday: lose 6 widgets (4 regular + 2 bonus)\n- If holiday falls on Tue-Fri: lose 4 widgets\n- If holiday falls on Sat-Sun: lose 3 widgets\nAverage widgets lost per holiday = (6 + 4\u00d74 + 3\u00d72)/7 = 4 widgets\nLoss per week = 4/4 = 1 widget\nNew weekly average = 28 - 1 = 27 widgets\n\n3. For Q3 (accounting for fibonacci holidays):\n- Fibonacci numbers from 1 to 10 are: 1,1,2,3,5,8\n- Months 1,2,3,5,8 get extra holidays\n- That's 5 months with extra holidays\n- Total extra holidays = 5 per year\n- Extra holidays per month = 5/10 = 0.5\n- Extra widgets lost per week due to fibonacci holidays = (0.5 \u00d7 4)/4 = 0.5\nNew weekly average = 27 - 0.5 = 26.5 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26.5}\n</answer>", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 716.0, | |
| "cost": 0.007464 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\nQ1: Widgets in a regular week\n- Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n- Mondays get 2 extra widgets = +2\n- Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\nTotal per week = 20 + 2 + 6 = 28 widgets\n\nQ2: Taking monthly holidays into account\n- Each month has 4 weeks but also 1 holiday\n- That means per 4 weeks, we lose 1 day of production\n- The holiday could fall on any day:\n * If on Monday: lose 6 widgets (4 regular + 2 extra)\n * If on Tue-Fri: lose 4 widgets\n * If on weekend: lose 3 widgets\n- Average widgets lost = (6 + 4\u00d74 + 3\u00d72)/7 \u2248 4 widgets\n- So per 4 weeks we lose 4 widgets\n- Average per week = (28\u00d74 - 4)/4 = 27 widgets\n\nQ3: Fibonacci holidays\n- In a 10-month year, fibonacci numbers are: 1,2,3,5,8\n- So months 1,2,3,5,8 have 2 holidays instead of 1\n- That's 5 months with extra holidays\n- In a year:\n * Regular months (4,6,7,9,10): 5 months \u00d7 1 holiday\n * Fibonacci months (1,2,3,5,8): 5 months \u00d7 2 holidays\n * Total holidays = 5 + 10 = 15 holidays per year\n- Per month average = 15/10 = 1.5 holidays\n- So per 4 weeks we lose 1.5 \u00d7 4 = 6 widgets\n- Average per week = (28\u00d74 - 6)/4 = 26.5 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26.5}\n</answer>", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 797.0, | |
| "cost": 0.008679000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.61, | |
| "reasoning": "<thinking>\nLet me solve this step by step:\n\nQ1: Widgets per week\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\n* Total per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Adding monthly holidays\n* Normal 4-week month has 28 \u00d7 4 = 112 widgets\n* One holiday per month where no widgets are produced\n* Holiday could fall on:\n * Weekday (prob 5/7): Lose 4 widgets\n * Weekend (prob 2/7): Lose 3 widgets\n * If Monday (prob 1/7): Also lose 2 bonus widgets\n* Average widgets lost per holiday:\n * (5/7 \u00d7 4) + (2/7 \u00d7 3) + (1/7 \u00d7 2) = 3.71 widgets\n* Lost widgets per week = 3.71 \u00f7 4 = 0.93\n* Weekly average = 28 - 0.93 = 27.07 widgets\n\nQ3: Adding fibonacci holidays\n* Fibonacci numbers up to 10: 1,1,2,3,5,8\n* Months with index 1,2,3,5,8 get 2 extra holidays\n* In a 10-month year:\n * 5 months have 2 holidays (fibonacci months)\n * 5 months have 1 holiday (regular months)\n* Total holidays per year = (5 \u00d7 2) + (5 \u00d7 1) = 15\n* Average holidays per month = 15/10 = 1.5\n* Weekly widget loss = (3.71 \u00d7 1.5) \u00f7 4 = 1.39\n* Weekly average = 28 - 1.39 = 26.61 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27.07, \"q3\": 26.61}\n</answer>", | |
| "error": 0.6799999999999997, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 795.0, | |
| "cost": 0.008649 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "<thinking>\nLet me break this down step by step:\n\nQ1:\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\n* Total per week = 20 + 6 + 2 = 28 widgets\n\nQ2:\n* Each month has 4 weeks and 1 holiday\n* A holiday removes production for that day\n* Since holidays are uniformly distributed:\n * 5/7 chance it falls on a weekday (losing 4 widgets)\n * 2/7 chance it falls on weekend (losing 3 widgets)\n * 1/7 chance it falls on Monday (losing additional 2 widgets)\n* Average widgets lost per holiday:\n * (5/7 \u00d7 4) + (2/7 \u00d7 3) + (1/7 \u00d7 2) = 20/7 + 6/7 + 2/7 = 28/7 = 4\n* One holiday per month = one holiday per 4 weeks\n* Average widgets lost per week = 1 widget\n* Weekly production = 28 - 1 = 27 widgets\n\nQ3:\n* Fibonacci numbers from 1 to 10 are: 1, 2, 3, 5, 8\n* 5 months have extra holidays (months 1,2,3,5,8)\n* Total extra holidays = 5 \u00d7 2 = 10 extra holidays per year\n* Extra holidays per month = 1 additional holiday\n* Now each month averages 2 holidays instead of 1\n* Average widgets lost per week = 2 widgets\n* Weekly production = 28 - 2 = 26 widgets\n</thinking>\n\n<answer>\n{\"q1\": 28, \"q2\": 27, \"q3\": 26}\n</answer>", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 750.0, | |
| "cost": 0.007974 | |
| } | |
| ] | |
| } | |
| ] | |
| }, | |
| { | |
| "strategy_name": "TwoStepTextExtraction", | |
| "tests": [ | |
| { | |
| "provider": "openai", | |
| "model": "gpt-4o-mini", | |
| "avg_error": 2.885, | |
| "avg_correctness": 0.6333333333333333, | |
| "avg_tokens": 2313.1, | |
| "avg_cost": 0.000817035, | |
| "n_tests": 10, | |
| "failure_rate": 0.0, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.75, | |
| "q3": 26.65, | |
| "reasoning": "Let's break down the calculations step by step for each question.\n\n### Q1: How many widgets are produced in a week?\n\nWe first outline the widget production schedule:\n\n- **Weekdays (Monday to Friday)**:\n - **Monday**: 4 widgets + 2 extra widgets = 6 widgets\n - **Tuesday**: 4 widgets\n - **Wednesday**: 4 widgets\n - **Thursday**: 4 widgets\n - **Friday**: 4 widgets\n- **Weekends (Saturday and Sunday)**:\n - **Saturday**: 3 widgets\n - **Sunday**: 3 widgets\n\nNow, calculating the total production for each day of the week:\n- Monday: 6 widgets\n- Tuesday: 4 widgets\n- Wednesday: 4 widgets\n- Thursday: 4 widgets\n- Friday: 4 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nNow sum these amounts:\n\\[\n6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 \\text{ widgets}\n\\]\nSo, the total widget production per week is **28 widgets**.\n\n---\n\n### Q2: Average number of widgets produced per week accounting for one monthly holiday\n\nIf each month has one holiday uniformly distributed across the week, that means on average, one day will have no production. Since there are 4 weeks in a month, this suggests that throughout the month, we lose one day of production.\n\nSince there are 10 months in total, and each month has one holiday:\n- Total holidays in a year = 10 holidays.\n\nNow, the total widget production for the year without holidays is:\n\\[\n28 \\text{ widgets/week} \\times 52 \\text{ weeks} = 1456 \\text{ widgets}\n\\]\nHowever, since there are 10 holidays, we need to calculate the total widget production lost due to holidays. On average, we\u2019ll lose 4 widgets for each holiday (since a weekday produces 4 widgets and a weekend day produces 3, averaging to around 3.5 widgets per holiday for uniformity). But for simplicity and history of question context, we can calculate directly by one widget per week for straightforward consideration of the holistic scenario. \n\nThus overtime for 10 holidays, based on 28 widgets per week over 52 weeks:\n- Widgets lost due to holidays = 10\n- Total production accounting for holidays: \n\\[\n1456 - 10 = 1446 \\text{ widgets}\n\\]\n\nNow to find the average widgets produced in a week, divide this amount by 52 weeks:\n\\[\n\\frac{1446}{52} \\approx 27.75 \\text{ widgets}\n\\]\nThus, the average number of widgets produced per week, accounting for the one holiday each month, is approximately **27.75 widgets**.\n\n---\n\n### Q3: Average number of widgets produced per week accounting for Fibonacci-related holidays\n\nThe Fibonacci months (1st, 2nd, 3rd, 5th, 8th) are the following with an index of month:\n- Month 1, Month 2, Month 3, Month 5, Month 8 = 5 months.\n\nEach of these months has 2 extra holidays. Thus, the total additional holidays due to Fibonacci is:\n- Total extra holidays = \\(5 \\text{ months} \\times 2 \\text{ holidays/month} = 10 \\text{ holidays}\\).\n\nSo now, combined with previous holidays from all months:\n- Total holidays = 10 (normal) + 10 (fibonacci) = 20 holidays in a week.\n\nCalculating:\n- The average widget loss per holiday still considers a moderate daily outlay of 3.5, thus: \n```\n20 holidays leads to: 20 * 3.5 = 70 widgets lost overall.\n```\nThus, finding production:\n\\[\n1456 - 70 = 1386 \\text{ widgets}\n\\]\nFinding the average production per week:\n\\[\n\\frac{1386}{52} \\approx 26.65 \\text{ widgets}\n\\]\nThus, the average number of widgets produced per week considering both sets of holidays is approximately **26.65 widgets**.", | |
| "error": 1.3999999999999986, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 2051.0, | |
| "cost": 0.0007194 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 27.0, | |
| "reasoning": "To answer your questions, let's calculate the number of widgets produced based on the provided production schedule.\n\n### Q1: How many widgets are produced in a week?\n\n1. **Weekday Production**: There are 5 weekdays and the factory produces 4 widgets each weekday. \n \\[\n 5 \\text{ weekdays} \\times 4 \\text{ widgets/weekday} = 20 \\text{ widgets}\n \\]\n\n2. **Weekend Production**: There are 2 weekend days and the factory produces 3 widgets each weekend day. \n \\[\n 2 \\text{ weekend days} \\times 3 \\text{ widgets/weekend day} = 6 \\text{ widgets}\n \\]\n\n3. **Extra Production on Monday**: On Monday, there are 2 extra widgets. Since Monday is a weekday, this adds to the weekday production output:\n \\[\n 2 \\text{ extra widgets (Monday)} = 2 \\text{ widgets}\n \\]\n\n4. **Total Production in a Week**: Now sum all the widgets produced.\n \\[\n \\text{Total Production} = 20 \\text{ (weekdays)} + 6 \\text{ (weekend)} + 2 \\text{ (extra on Monday)} = 28 \\text{ widgets}\n \\]\n\n**Answer to Q1**: The factory produces **28 widgets** in a week.\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has 4 weeks and one holiday, and the factory is closed on holidays:\n\n1. **Monthly Holidays**: There is 1 holiday in each month, which means in 4 weeks, 4 widgets will not be produced (1 widget production estimate for each weekday).\n - The production per week will decrease due to the holidays:\n \\[\n \\text{Weekly Production without Holidays} = 28 \\text{ widgets}\n \\]\n \\[\n \\text{Holidays in a month} = 1 \\text{ holiday causes the loss of 1 weekday; thus, average loss in 4 weeks = 1 widget.}\n \\]\n \\[\n \\text{Adjusted Monthly Production} = 28 \\text{ widgets/week} \\times 4 \\text{ weeks} - 4 \\text{ widgets (1 for each month)} = 112 - 4 = 108 \\text{ widgets}\n \\]\n\n2. **Average Weekly Production with Holidays**: There are 4 weeks in a month, therefore:\n \\[\n \\text{Average Weekly Production} = \\frac{108 \\text{ widgets}}{4 \\text{ weeks}} = 27 \\text{ widgets/week}\n \\]\n\n**Answer to Q2**: The average number of widgets produced per week is **27 widgets**.\n\n### Q3: Taking these Fibonacci holidays into account, what's the average number of widgets produced per week?\n\nThe months whose indices are Fibonacci numbers (1, 2, 3, 5, 8) will have 2 extra holidays. In 10 months, the Fibonacci months are:\n\n- Month 1: 2 extra holidays (total 3 holidays)\n- Month 2: 2 extra holidays (total 3 holidays)\n- Month 3: 2 extra holidays (total 3 holidays)\n- Month 4: 1 holiday (total 1 holiday)\n- Month 5: 2 extra holidays (total 3 holidays)\n- Month 6: 1 holiday (total 1 holiday)\n- Month 7: 1 holiday (total 1 holiday)\n- Month 8: 2 extra holidays (total 3 holidays)\n- Month 9: 1 holiday (total 1 holiday)\n- Month 10: 1 holiday (total 1 holiday)\n\nCounting the holidays:\n\n- Months with 3 holidays: 4 months (1, 2, 3, and 5) \u2192 4 months \u00d7 3 holidays = 12 holidays\n- Months with 1 holiday: 6 months (4, 6, 7, 9, 10) \u2192 6 months \u00d7 1 holiday = 6 holidays\n\n**Total Holidays**: \\(12 + 6 = 18\\)\n\nNow, take these holidays into account for production:\n\n1. Total widgets produced without any holidays in a year:\n \\[\n 28 \\text{ widgets/week} \\times 40 \\text{ weeks (10 months)} = 1120 \\text{ widgets}\n \\]\n\n2. Adjusted total production after holidays:\n \\[\n \\text{Total Widgets Produced} = 1120 - 18 = 1102 \\text{ widgets}\n \\]\n\n3. The average weekly production:\n \\[\n \\text{Average Weekly Production} = \\frac{1102}{40} = 27.55 \n \\]\n\nRounding down gives us 27. \n\n**Answer to Q3**: The average number of widgets produced per week, taking into account Fibonacci holidays, is approximately **27 widgets**.", | |
| "error": 1.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 2405.0, | |
| "cost": 0.00085125 | |
| }, | |
| { | |
| "q1": 24.0, | |
| "q2": 23.07, | |
| "q3": 13.8, | |
| "reasoning": "Let's break down the production of widgets step by step for each question.\n\n### Q1: Number of widgets produced in a week\n\n1. **Weekdays (Monday to Friday):** \n - Each weekday, the factory produces 4 widgets.\n - Total for the weekdays:\n \\[\n 4 \\text{ widgets/day} \\times 5 \\text{ days} = 20 \\text{ widgets}\n \\]\n\n2. **Weekends (Saturday and Sunday):**\n - Each weekend day, the factory produces 3 widgets.\n - Total for the weekend:\n \\[\n 3 \\text{ widgets/day} \\times 2 \\text{ days} = 6 \\text{ widgets}\n \\]\n\n3. **Extra widgets on Mondays:**\n - The factory produces 2 extra widgets on Monday.\n - Total for Monday:\n \\[\n 4 \\text{ widgets} + 2 \\text{ extra} = 6 \\text{ widgets}\n \\]\n\n4. **Total widgets produced in a week:**\n - Total from weekdays, weekends, and extra Monday widgets:\n - From weekdays (adjusted for Monday):\n \\[\n 20 \\text{ (weekdays)} - 4 + 6 \\text{ (Monday)} + 6 \\text{ (weekend)} = 24 \\text{ widgets/week}\n \\]\n\nSo, the total number of widgets produced in a week is:\n\\[\n\\text{Total} = 24 \\text{ widgets}\n\\]\n\n### Q2: Average number of widgets produced per week (with one holiday)\n\n1. **Total weeks in a month:** 4 weeks\n - First, calculate total production in a month without holidays:\n \\[\n 24 \\text{ widgets/week} \\times 4 \\text{ weeks} = 96 \\text{ widgets}\n \\]\n\n2. **Account for the holiday:** \n - When the factory is closed on one holiday, that holiday could fall on any day of the week. To find the average impact of one holiday:\n - The average widget production on a holiday is:\n - If the holiday is on a weekday (4 days), the factory would lose 4 widgets.\n - If the holiday is on a weekend (2 days), the factory would lose 3 widgets.\n - Considering uniform distribution:\n \\[\n \\text{Average loss} = \\left(\\frac{4 \\text{ widgets} \\times 5 + 3 \\text{ widgets} \\times 2}{7} \\right) = \\left(\\frac{20 + 6}{7} \\right) = 3.71 \\text{ widgets (approx.)}\n \\]\n\n3. **Widgets produced in a month after accounting for the holiday:**\n \\[\n 96 - 3.71 \\approx 92.29 \\text{ widgets}\n \\]\n\n4. **Average widgets produced per week:**\n\\[\n\\frac{92.29 \\text{ widgets}}{4 \\text{ weeks}} \\approx 23.07 \\text{ widgets/week}\n\\]\n\nSo, the average number of widgets produced per week, taking the holiday into account, is approximately:\n\\[\n\\text{Average} \\approx 23.07 \\text{ widgets}\n\\]\n\n### Q3: Average number of widgets produced per week (with fibonacci holidays)\n\n1. **Identify the months that have extra holidays** (Fibonacci sequence indices): \n - Month indices: 1, 2, 3, 5, 8 (5 months in a year)\n - Total months in a year = 10 months, so 5 months with 2 extra holidays will have:\n - Total holidays due to Fibonacci numbers:\n \\[\n 5 \\text{ months} \\times 2 = 10 \\text{ extra holidays}\n \\]\n\n2. **Calculate total holidays including the monthly holiday:**\n - Total holidays now = Regular holiday + Fibonacci holidays = 1 + 10 = 11 holidays.\n\n3. **Adjusting monthly production for these holidays:**\n - Total calculated widgets in a month without holidays previously was 96. \n - Profit loss:\n \\[\n 11 \\text{ Holidays} \\text{ each averaging to around } 3.71 \\approx 40.81 \\text{ total loss}\n \\]\n\n4. **Total production after accounting for holidays:**\n \\[\n 96 - 40.81 \\approx 55.19 \\text{ widgets/month}\n \\]\n\n5. **Average number of widgets produced per week:**\n\\[\n\\frac{55.19 \\text{ widgets}}{4 \\text{ weeks}} \\approx 13.80 \\text{ widgets/week}\n\\]\n\nSo, taking into account the fibonacci holidays as well, the average number of widgets produced per week is approximately:\n\\[\n\\text{Average} \\approx 13.80 \\text{ widgets}\n\\]", | |
| "error": 20.13, | |
| "correctness": 0.0, | |
| "total_tokens": 2371.0, | |
| "cost": 0.0008393999999999999 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 23.0, | |
| "reasoning": "To solve this problem step-by-step, we'll begin by calculating the total number of widgets produced in a week without any holidays, then factor in the holidays, and finally account for the extra holidays in months that are indexed by Fibonacci numbers.\n\n### Q1: How many widgets are produced in a week?\n\nThe production schedule is as follows:\n- Every weekday (Monday to Friday), the factory produces 4 widgets per day.\n- Every weekend day (Saturday and Sunday), the factory produces 3 widgets per day.\n- Additionally, on Mondays, the factory produces 2 extra widgets.\n\nCalculating the total production for each day of the week:\n- **Monday**: 4 + 2 = 6 widgets\n- **Tuesday**: 4 widgets\n- **Wednesday**: 4 widgets\n- **Thursday**: 4 widgets\n- **Friday**: 4 widgets\n- **Saturday**: 3 widgets\n- **Sunday**: 3 widgets\n\nNow, we will sum these values to find the weekly production:\n\\[\n\\text{Total widgets} = 6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 \\text{ widgets per week}\n\\]\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has one holiday, and since there are 4 weeks in a month, that means:\n- The factory is closed on that holiday, which means it will not produce widgets on that day.\n\nCalculating the number of widgets produced in a month without holidays:\n\\[\n\\text{Widgets in a month} = 28 \\text{ widgets/week} \\times 4 \\text{ weeks} = 112 \\text{ widgets/month}\n\\]\n\nIf there is one holiday in a month, we need to determine on which day the holiday falls:\n- If the holiday lands on a weekday (Monday to Friday), the production for that day is 4 widgets (or 6 widgets if it's Monday).\n- If it falls on a weekend (Saturday or Sunday), the production for that day is 3 widgets.\n\nLet's consider the average holiday production value. Since holidays are uniformly distributed, we can assume:\n\n- If it\u2019s a weekday, average widgets lost = 4. \n- The average for a holiday on a weekend = 3.\n \nThus, the average holiday loss is:\n\\[\n\\text{Average loss per holiday} = \\frac{5 \\times 4 + 2 \\times 3}{7} = \\frac{20 + 6}{7} = \\frac{26}{7} \\approx 3.71\n\\]\n\nNow, adjusting the monthly production for the holiday loss:\n\\[\n\\text{Adjusted widgets in a month} = 112 - \\left(\\frac{26}{7}\\right) \\approx 112 - 3.71 \\approx 108.29 \n\\]\n\nTo find the average weekly production after accounting for the holiday:\n\\[\n\\text{Average weekly production} = \\frac{108.29}{4} \\approx 27.07 \\text{ widgets/week}\n\\]\n\n### Q3: Taking the Fibonacci holidays into account, what's the average number of widgets produced per week?\n\nThe months with 1-based Fibonacci indices up to 10 are Month 1 (1), Month 2 (1), Month 3 (2), Month 5 (5), and Month 8 (21). Thus, months 1, 2, 3, 5, and 8 will have 2 extra holidays each.\n\nSince a year consists of 10 months:\n- Number of months with holidays = 5\n- Each of those months has 2 extra holidays, resulting in 10 extra holidays during the year.\n\nNow, we'll calculate the average production loss for these extra holidays:\nAverage holiday loss due to the two extra holidays for the Fibonacci months:\n- Assuming, similar analysis as before, if they fall on a weekday, average = 4, and weekend = 3. We can find that the expected loss from two holidays can be approximated based on the same reasoning (losing at least 3 or 4 widgets) roughly translates to about 7.42 (7.41 being from 2 holidays at half of their average widget loss).\n \nThus, each of the 5 Fibonacci months would lose an additional 7.42,\nleaving us with a total loss across the year that we must account for. \n10 months with a quantified loss of 3.71 retains their average lost productivity from holidays.\n\nIn total, we have:\n\\[\n10 \\text{ original holidays} + 10 \\text{(extra) site workplace losses} = 20 \\text{ new losses } \\times \\frac{4 + 3}{2\\times2} \\approx 20\n\\]\n\nAdjusting the previously calculated monthly production:\n\\[\n\\text{Average loss total} = 20 - 20 \\text{ weekly widgets produced} \\approx 104.36 \\left(20 - 3 \\text{ average per holiday}\\right) = 93.43 \n\\]\n\nThus dividing for average per week:\n\\[\n\\text{Average weekly production} \\approx \\frac{93.43}{4} \\approx 23.36 \\text{ widgets/week}\n\\]\n\nPutting it all together, the answers to the questions are:\n- **Q1:** 28 widgets\n- **Q2:** Approximately 27.07 widgets/week.\n- **Q3:** Approximately 23.36 widgets/week.", | |
| "error": 3.0, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 2580.0, | |
| "cost": 0.00091665 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 27.25, | |
| "reasoning": "To calculate the widgets produced by the factory in Standardtopia and account for the holidays, let's break it down step by step.\n\n### Q1: Total Widgets Produced in a Week\n\nThe production schedule is as follows:\n\n- **Weekdays (Monday to Friday)**: 4 widgets each\n- **Saturdays and Sundays**: 3 widgets each\n- **Extra widgets on Mondays**: 2 additional widgets\n\nLet's calculate the widgets produced:\n\n1. **Weekdays (Monday to Friday)**:\n - 5 weekdays \u00d7 4 widgets = 20 widgets\n\n2. **Weekend (Saturday and Sunday)**:\n - 2 weekend days \u00d7 3 widgets = 6 widgets\n\n3. **Extra widgets on Monday**:\n - 2 extra widgets on Monday\n\nNow, let's combine all these:\n\n\\[\n\\text{Total widgets per week} = 20 \\text{ (weekdays)} + 6 \\text{ (weekend)} + 2 \\text{ (extra Monday)} = 28\n\\]\n\nSo, the factory produces **28 widgets in a week**.\n\n### Q2: Average Widgets Produced Per Week Taking Monthly Holidays into Account\n\nEach month has 4 weeks and 1 holiday. Given that holidays fall on different days of the week and the factory is closed on holidays, we need to account for one holiday each month.\n\nCalculating the production loss due to one holiday per month:\n\nSince there are 4 weeks in a month:\n\n\\[\n4 \\text{ weeks} \\times 28 \\text{ widgets/week} = 112 \\text{ total widgets in a month}\n\\]\n\nIf the holiday takes away a day\u2019s worth of production, we should calculate how many widgets are lost on a day:\n\nTo calculate the number of widgets lost due to the holiday:\n- **Daily widget production**:\n - For a weekday (e.g., holiday on a weekday): 4 widgets\n - For a weekend day: 3 widgets\n\nAssuming the holiday could be any day in equal likelihood, since it\u2019s uniformly distributed across 7 days:\n\n\\[\n\\text{Average widget loss} = \\frac{5 \\text{ weekdays} \\times 4 \\text{ widgets} + 2 \\text{ weekend days} \\times 3 \\text{ widgets}}{7} = \\frac{20 + 6}{7} = \\frac{26}{7} \\approx 3.71\n\\]\n\nNow, subtract the average daily loss from the monthly total:\n\n\\[\n\\text{Remaining widgets after one holiday} = 112 - 3.71 \\approx 108.29\n\\]\n\nTo find the average weekly production accounting for the holiday:\n\n\\[\n\\text{Average weekly production} = \\frac{108.29}{4} \\approx 27.07\n\\]\n\nThus, after factoring in typical holidays, the average number of widgets produced per week is approximately **27.07**.\n\n### Q3: Average Widgets Produced Per Week Considering Fibonacci Holidays\n\nMonths indexed by Fibonacci numbers (1, 2, 3, 5, 8):\n\n1. **Months 1, 2, 3, 5, 8 have extra 2 holidays each.**\n2. **Total holidays = 5 (from Fibonacci) + 5 (from normal months) = 10 holidays total over 10 months.**\n\nThe total number of holidays = 10 (standard holidays) + 10 (fibonacci holidays) = 10 total holidays in a year.\n\nCalculating the loss for 10 total holidays: \n\nUsing the average loss calculation as before (~3.71 per holiday) for the average daily loss over the year:\n\nFor the additional 5 holidays in the Fibonacci months, consider the total holiday impact:\n\n1. 5 holidays account for an average of \\(\\approx 3.71\\) each while the extra 5 account for some repeating between 1-3 widgets each depending on weekday/weekend.\n\nSo effectively, with 10 holidays we use the earlier \\(4 \\times 5 + 3 \\times 2 = 26\\) heuristic: \n\n\\[\n\\text{Total average loss over 10 holidays} = 10 \\times 3.71 = 37.1\n\\]\nTotal widgets in a year originally calculated would be:\n\\[\n28 \\text{ widgets per week} \\times 52 weeks = 1456 - 37.1 = 1418.9\n\\]\n\nDividing this by \\(52 weeks\\) yields approximately:\n\\[\n\\text{Average release per week} = \\frac{1418.9}{52} \\approx 27.25 \\text{ widgets}\n\\]\n\nThus, the average number of widgets produced per week considering Fibonacci holiday effects is approximately **27.25**.", | |
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| "reasoning": "To answer these questions about widget production in Standardtopia, let's break it down step by step.\n\n### Q1: How many widgets are produced in a week?\n\nThe production schedule is as follows:\n- **Weekdays (Monday to Friday)**: 4 widgets per day\n- **Saturday and Sunday**: 3 widgets per day\n- **Extra widgets on Monday**: 2 extra widgets\n\nCalculating the total production for one week (7 days):\n\n1. **Monday**: \n - 4 widgets (regular) + 2 widgets (extra) = 6 widgets\n2. **Tuesday**: 4 widgets\n3. **Wednesday**: 4 widgets\n4. **Thursday**: 4 widgets\n5. **Friday**: 4 widgets\n6. **Saturday**: 3 widgets\n7. **Sunday**: 3 widgets\n\nNow summing these up:\n- Total for weekdays: \\(6 + 4 + 4 + 4 + 4 = 22\\)\n- Total for the weekend: \\(3 + 3 = 6\\)\n\nTotal production in one week = \\(22 + 6 = 28\\) widgets.\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has 4 weeks and 1 holiday. If the holiday falls on a weekday, we lose some production as follows:\n- If the holiday is on a weekday, we lose 4 widgets.\n- If the holiday is on Saturday, we lose 3 widgets.\n- If the holiday is on Sunday, we lose 3 widgets.\n\nAssuming the holidays are uniformly distributed, we can expect:\n- 5 weekdays + 2 weekend days = 7 total days in a week.\n\nThe average widget production loss per holiday scenario:\n- Holidays are uniformly distributed, so the loss per holiday can be calculated as follows:\n\nAverage loss on weekdays (5 days) = \\(\\frac{5 \\times 4 + 2 \\times 3}{7} = \\frac{20 + 6}{7} \\approx 3.71\\) widgets.\n\nTherefore, the total number of widgets produced in a month (4 weeks) without holidays:\n- Total monthly production = \\(28 \\times 4 = 112\\) widgets.\n- Total holiday loss = \\(3.71 \\times 1 \\text{ (holiday/month)} = 3.71\\) widgets.\n\nFinal monthly production = \\(112 - 3.71 = 108.29\\) (approximately rounded down to 108 for practical purposes).\nThus, the average weekly production would be:\n- \\(108.29 \\div 4 \\approx 27.07\\) widgets (about 27 widgets).\n\n### Q3: Taking these Fibonacci holidays into account, what's the average number of widgets produced per week?\n\nThe months with 1-based indices that are Fibonacci numbers up to 10 are: 1 (1), 2 (1), 3 (2), 5 (5), 8 (21). Thus, months 1, 2, 3, 5, and 8 each have an additional 2 holidays for a total of 5 additional holidays.\n\nTotal holidays in these 5 months:\n- 5 months with 1 holiday + 5 months with 2 holidays = 5 + 10 = 15 holidays.\n\nNow, we must account for the average loss per holiday as calculated before (3.71 widgets).\n\nTotal widget production loss = \\(3.71 \\times 15 = 55.65\\) widgets.\n\nTotal widget production for the 10 months without considering any holidays:\n- \\(28 \\text{ (weekly production)} \\times 4 \\text{ (weeks/month)} \\times 10 \\text{ (months)} = 1120\\) widgets.\n\nFinal total after subtracting losses:\n- Total production = \\(1120 - 55.65 = 1064.35\\).\n\nNow we can calculate the average number of widgets produced per week:\n- \\(1064.35 \\div (10 \\times 4) = 1064.35 \\div 40 = 26.61\\) widgets (about 27 widgets rounded).\n\n### Final summary of answers:\n- **Q1**: 28 widgets produced in a week.\n- **Q2**: Average of about 27 widgets produced per week after 1 holiday.\n- **Q3**: Average of about 27 widgets produced per week after considering Fibonacci holidays.", | |
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| "reasoning": "To solve these problems, let's break down the widget production schedule.\n\n### Q1: How many widgets are produced in a week?\n\n1. **Weekdays (Monday to Friday)**:\n - Each weekday (5 days) produces 4 widgets.\n - Total from weekdays = 4 widgets/day \u00d7 5 days = 20 widgets.\n\n2. **Weekends (Saturday and Sunday)**:\n - Each weekend day (2 days) produces 3 widgets.\n - Total from weekends = 3 widgets/day \u00d7 2 days = 6 widgets.\n\n3. **Extra widgets on Monday**:\n - On Mondays, the factory produces 2 extra widgets.\n - So, total production on Monday = 4 widgets (regular) + 2 extra = 6 widgets.\n\nNow let's summarize the weekly production:\n\n- **Monday**: 6 widgets\n- **Tuesday**: 4 widgets\n- **Wednesday**: 4 widgets\n- **Thursday**: 4 widgets\n- **Friday**: 4 widgets\n- **Saturday**: 3 widgets\n- **Sunday**: 3 widgets\n\nTotal production in a week:\n\n\\( 6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 \\)\n\nTherefore, **Q1: A total of 28 widgets are produced in a week.**\n\n---\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has 4 weeks and 1 holiday. This means that per month:\n\n- Total weeks in a month = 4\n- Total holidays in a month = 1 holiday\n\nGiven that the factory is closed on holidays, we can say that during this holiday (which is 1 day), production is lost.\n\nTotal production in one month without holidays:\n\n- Monthly production = 28 widgets/week \u00d7 4 weeks = 112 widgets.\n\nNow subtract the lost production on the holiday (which is a weekday and therefore produces 4 widgets):\n\n- Therefore, average monthly production considering 1 holiday = 112 - 4 = 108 widgets.\n\nNow, distribute this monthly production over the total number of weeks in the month:\n\n- Average production per week = Total monthly production / 4 weeks = 108 / 4 = 27.\n\nThus, **Q2: The average number of widgets produced per week is 27.**\n\n---\n\n### Q3: Taking these fibonacci holidays into account, what's the average number of widgets produced per week?\n\nThe months are numbered from 1 to 10. The Fibonacci sequence indexes relevant to our months are: 1, 2, 3, 5, and 8. Thus, these months (1st, 2nd, 3rd, 5th, and 8th) will each have 2 extra holidays.\n\nThis means out of the 10 months, 5 months will have:\n\n- Regular holidays: 1\n- Fibonacci holidays: 2\n- Total holidays in these 5 months = 1 + 2 = 3 holidays.\n\nFor the other 5 months, there is only 1 holiday.\n\nCalculating total holidays:\n\n- Total holidays from Fibonacci months: 5 months \u00d7 3 holidays/month = 15 holidays.\n- Total holidays from non-Fibonacci months: 5 months \u00d7 1 holiday/month = 5 holidays.\n\nTotal holidays in the year = 15 + 5 = 20 holidays.\n\n**Production loss due to holidays**:\n\n- Since each holiday results in losing 4 widgets (maximum on a regular weekday), the production lost from holidays = 20 holidays \u00d7 4 widgets/holiday = 80 widgets.\n\n**Total production in a year without any holidays**:\n\n- Yearly production = 28 widgets/week \u00d7 52 weeks (10 months, 4 weeks/month) = 28 \u00d7 40 = 1120 widgets.\n\n**Total production after accounting for holidays**:\n\n- Total annual production = 1120 - 80 = 1040 widgets.\n\nTo find the average weekly production:\n\n- Total weeks in a year = 40 (since 10 months with 4 weeks each).\n \nAverage production per week:\n\n- \\( 1040 \\, \\text{widgets} / 40 \\, \\text{weeks} = 26 \\, \\text{widgets/week} \\).\n\nTherefore, **Q3: The average number of widgets produced per week is 26.**", | |
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| "reasoning": "Let's break down each question step by step.\n\n### Q1: How many widgets are produced in a week?\n\n**Production schedule:**\n- Weekdays (Monday to Friday): 4 widgets/day\n- Weekend (Saturday and Sunday): 3 widgets/day\n- Additional widgets on Monday: 2 extra widgets\n\n**Calculating weekly production:**\n1. Calculate the production for weekdays:\n - Monday: 4 widgets + 2 extra = 6 widgets\n - Tuesday: 4 widgets\n - Wednesday: 4 widgets\n - Thursday: 4 widgets\n - Friday: 4 widgets\n\n Total for weekdays: \n \\( 6 + 4 + 4 + 4 + 4 = 22 \\) widgets\n\n2. Calculate the production for the weekend:\n - Saturday: 3 widgets\n - Sunday: 3 widgets\n\n Total for the weekend: \n \\( 3 + 3 = 6 \\) widgets\n\n3. Combine weekday and weekend totals:\n Total for the week:\n \\( 22 + 6 = 28 \\) widgets\n\n**Answer for Q1:**\nThe factory produces **28 widgets in a week**.\n\n---\n\n### Q2: Average number of widgets produced per week taking holidays into account\n\nEach month has 4 weeks and 1 holiday, thus:\n- Total widgets produced in a month without holidays: \\( 28 \\times 4 = 112 \\) widgets\n- 1 holiday means no widgets are produced that day.\n\nCalculating the average:\n- Each month has 1 holiday, so total production for the month with 1 holiday is:\n \\( 112 - 4 = 108 \\) widgets (since the holiday is on a weekday, we remove the daily average for 1 day which is 4 widgets).\n\nTotal production across the 4 weeks:\n- Total widgets produced with holidays: \\( 4 \\times 28 - 4 = 108 \\)\n\nAverage production per week accounting for holidays:\n\\[\n\\text{Average production per week} = \\frac{108}{4} = 27 \\, \\text{widgets}\n\\]\n\n**Answer for Q2:**\nTaking into account the holidays, the average number of widgets produced per week is **27 widgets**.\n\n---\n\n### Q3: Average number of widgets produced per week taking Fibonacci holidays into account\n\nThe Fibonacci numbers (starting from index 1) are 1, 2, 3, 5, 8, 13... For 10 months, the Fibonacci indices are:\n- Month 1 (Fibonacci 1): 2 holidays\n- Month 2 (Fibonacci 1): 2 holidays\n- Month 3 (Fibonacci 2): 2 holidays\n- Month 4 (Fibonacci 3): 2 holidays\n- Month 5 (Fibonacci 5): 2 holidays\n- Month 6 (not Fibonacci): 1 holiday\n- Month 7 (not Fibonacci): 1 holiday\n- Month 8 (not Fibonacci): 1 holiday\n- Month 9 (not Fibonacci): 1 holiday\n- Month 10 (not Fibonacci): 1 holiday\n\n**Count total holidays:**\n- Months with 2 holidays: 5 (months 1, 2, 3, 4, 5) => \\( 5 \\times 2 = 10 \\) holidays \n- Months with 1 holiday: 5 => \\( 5 \\times 1 = 5 \\) holidays \n\nTotal holidays in a year:\n\\[\n10 + 5 = 15 \\, \\text{holidays}\n\\]\n\nThe production considered for 10 months without holidays:\n\\[\n10 \\times 112 = 1120 \\, \\text{widgets}\n\\]\n\nNow, we need to remove the widgets corresponding to the holidays:\nEach holiday (assumed uniformly distributed) will remove 4 widgets for weekdays and 3 for weekends. \n\nSince holidays are uniformly distributed, we divide the holidays by 7 to estimate how many weekdays and weekend days will be on average in 15 holidays. We can calculate:\n\nLet\u2019s say there are \\(h_w\\) weekdays and \\(h_w\\) weekends over 15 holidays. Given there are:\n\\[\n\\frac{15}{7} \\approx 2.14 \\, \\text{weekdays (rounded down to 2)} , \\text{ and } \\frac{15}{7} \\approx 2.14 \\, \\text{weekends (rounded down to 2)}\n\\]\n\nAssuming 12 weekdays and 3 weekends:\n- Widgets lost on weekdays: \\( 4 \\times 12 = 48 \\)\n- Widgets lost on weekends: \\( 3 \\times 3 = 9 \\)\n\nTotal widgets lost to holidays: \n\\[\n48 + 9 = 57\n\\]\n\nFinal production after holidays:\n\\[\n1120 - 57 = 1063\n\\]\n\n**Average production per week**:\n\\[\n\\text{Average production per week} = \\frac{1063}{40} \\approx 26.575\n\\]\n\n**Answer for Q3:**\nTaking the Fibonacci holidays into account, the average number of widgets produced per week is approximately **26.6 widgets**.", | |
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| "reasoning": "To solve the questions step-by-step, let's define the widget production schedule for each type of day:\n\n1. **Production Schedule Recap:**\n - **Weekdays (Monday to Friday):** \n - 4 widgets each day\n - On Monday, an additional 2 widgets are produced, making it 6 widgets total for Monday.\n - **Weekends (Saturday and Sunday):** \n - 3 widgets each day.\n\n2. **Calculating Weekly Production (Q1):**\n\n **Monday:** 6 widgets \n **Tuesday:** 4 widgets \n **Wednesday:** 4 widgets \n **Thursday:** 4 widgets \n **Friday:** 4 widgets \n **Saturday:** 3 widgets \n **Sunday:** 3 widgets \n\n Now, let's sum these up:\n\n \\[\n \\text{Total widgets per week} = 6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 \\text{ widgets}\n \\]\n\n **Answer to Q1:** 28 widgets are produced in a week.\n\n3. **Calculating Average Weekly Production with One Holiday (Q2)**\n\n Each month has 4 weeks and one holiday, meaning out of 4 weeks, one day is not produced per month.\n\n \\[\n \\text{Monthly production without holiday} = 28 \\text{ widgets/week} \\times 4 \\text{ weeks} = 112 \\text{ widgets}\n \\]\n\n Since one holiday means production for that day is lost, we need to evaluate the average widgets lost in a month due to a holiday:\n\n - **Assuming the holiday could fall on any day of the week:**\n - If the holiday falls on a weekday (Monday to Friday), the loss would be 4 widgets.\n - If it falls on a weekend (Saturday/Sunday), the loss would be 3 widgets.\n\n The average production lost due to a holiday can be calculated as follows:\n - 5 weekdays \u00d7 4 widgets + 2 weekend days \u00d7 3 widgets = 20 + 6 = 26 widgets total weekly production.\n - Average lost production = \\( \\frac{26\\text{ widgets}}{7\\text{ days}} = \\frac{26}{7} \\approx 3.71 \\text{ widgets} \\) per holiday.\n\n To account for the holiday in the monthly production:\n\n \\[\n \\text{Adjusted monthly production} = 112 \\text{ widgets} - 3.71 \\text{ widgets} \\approx 108.29 \\text{ widgets}\n \\]\n\n Now dividing this by 4 weeks gives:\n\n \\[\n \\text{Average monthly widgets per week} = \\frac{108.29}{4} \\approx 27.07\n \\]\n\n **Rounded down:** For simplicity, it rounds to about **27 widgets per week**.\n\n **Answer to Q2:** Approximately 27 widgets produced per week after accounting for the holiday.\n\n4. **Calculating Average Weekly Production with Fibonacci Holidays (Q3)**\n\n In this scenario, the following months are considered as \"fibonacci months\": 1, 2, 3, 5, 8. That's 5 months out of 10.\n\n Each of these fibonacci months has **2 extra holidays**. So we need to account for additional production loss from these holidays.\n\n For the 5 fibonacci months:\n - Each has 2 extra holidays, resulting in 10 extra holidays.\n - For 5 non-fibonacci months, each has 1 holiday resulting in 5 extra holidays.\n\n Thus, total holidays = 10 (fibonacci) + 5 (non-fibonacci) = **15 holidays** throughout the year.\n\n The average weekly production loss due to holidays remains the same as earlier.\n\n Total weeks in a year is \\( 10 \\times 4 = 40 \\, \\text{ weeks} \\).\n\n The total lost production due to holidays is:\n\n \\[\n \\text{Total losses} = 15 \\text{ holidays} \\times 3.71 \\text{ widgets/holiday} = 55.65 \\text{ widgets}.\n \\]\n\n Hence adjusted yearly production is:\n\n \\[\n \\text{Yearly production without holidays} = 28 \\text{ widgets/week} \\times 40 \\text{ weeks} = 1120 \\text{ widgets}.\n \\]\n\n Adjusted production = \\( 1120 - 55.65 = 1064.35 \\text{ widgets} \\).\n\n Therefore, the average widgets produced per week is:\n\n \\[\n \\text{Average per week} = \\frac{1064.35}{40} \\approx 26.61.\n \\]\n\n **Rounded down:** About **26 widgets per week** after accounting for all holidays.\n\n **Answer to Q3:** Approximately 26 widgets produced per week after accounting for fibonacci holidays.", | |
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| "reasoning": "To answer your questions, let's analyze the widget production step by step.\n\n### Q1: How many widgets are produced in a week?\n\n- **Weekdays:** There are 5 weekdays (Monday to Friday).\n - Each weekday produces 4 widgets: \\( 5 \\times 4 = 20 \\) widgets.\n \n- **Weekend:** There are 2 weekend days (Saturday and Sunday).\n - Each weekend day produces 3 widgets: \\( 2 \\times 3 = 6 \\) widgets. \n\n- **Extra widgets on Monday:** There are 2 extra widgets produced on Monday.\n\nNow let's calculate the total widgets produced in a week:\n\n\\[\n\\text{Total widgets} = \\text{Weekday widgets} + \\text{Weekend widgets} + \\text{Extra Monday widgets}\n\\]\n\\[\n\\text{Total widgets} = 20 + 6 + 2 = 28 \\text{ widgets}\n\\]\n\nSo, **the factory produces 28 widgets in a week.**\n\n---\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has 4 weeks, and there is 1 holiday each month:\n\n- Holiday impact on a week: Since there is one holiday, in one of the weeks of the month, we'll be losing the production of that day. The average loss incurred by having one holiday can be considered as follows:\n - If the holiday is on a weekday: The total production loss would be 4 (the number of widgets produced on that day).\n - If the holiday is on a weekend: The total production loss would be 3 (the number of widgets produced on that day).\n\nGiven that holidays are uniformly distributed, we average these out over the course of a month. \n\n1. **Weekdays lost on holidays:** If the holiday falls on a weekday, the average widget loss is \\( \\frac{4}{7} \\) days over one week multiplied by the number of weekdays, which in 4 weeks translates to a loss of approximately:\n \\[ \n \\text{Loss on weekdays} = 4 \\text{ (holiday on weekday)}\n \\]\n\n2. **Weekends lost on holidays:** If the holiday falls on a weekend, the average widget loss is \\( \\frac{3}{7} \\) days over the weekends.\n Apply the same for weekends:\n \\[ \n \\text{Loss on weekends} = 3 \\text{ (holiday on weekend)}\n \\]\n\nHowever, given there are 5 weekdays and 2 weekend days, we can reason that over time, typically we lose approximately half a widget on weekdays/go across the spectrum of the multiplication. So, let's take \\( 4 \\) on average on weekdays since working days summon 5 weekdays, we can average \\( \\frac{4+3}{2} \\) because holidays are scattered evenly.\n\nSo, on average, the total production will be decreased by about 4 widgets per month. Thus:\n\\[\n\\text{Tokens for holidays} = \\frac{28 - \\frac{4+3}{2}}{4} = 27 \\text{ (average per week)}\n\\]\nThe production becomes:\n\\[\n\\text{Avg per week} = 27 \\text{ widgets.}\n\\]\n\n**The average number of widgets produced per week, taking into account the monthly holidays, is approximately 27.**\n\n---\n\n### Q3: Taking these Fibonacci holidays into account, what's the average number of widgets produced per week?\n\nThe months which are Fibonacci numbers in the range of 1 to 10 are 1, 2, 3, 5, and 8. Therefore, for these 5 months, there are **2 extra holidays per month**.\n\nCalculating the total holidays:\n- Each of the 5 Fibonacci months has 1 regular holiday + 2 Fibonacci holidays = 3 holidays.\n- Therefore, there are \\( 5 \\) Fibonacci months.\n\n**Total holidays in non-Fibonacci months**: \n- 5 non-Fibonacci months = 1 holiday each = 5 holidays.\n\nSo, the total holidays in a year:\n\\[\n\\text{Total holidays} = 5 \\text{ (from fibs)} + 5 \\text{ (non-fibs)} = 10 \\text{ holidays in total.}\n\\]\n\nNow for total widgets lost, if the holiday is assumed to be random uniformly, for simplicity we will take the most conservative loss widget day (weekend) of 3 where it cancels on average across the days:\nSo: \\( \\text{Average Widget loss} \\) = \\( 10 \\times 3 = 30 \\text{ throughout the year,}\\)\ndistributing \\( 30/10 = 3 \\text{ widgets lost weekly.}\\)\n\nAdjust the total average weekly production:\n\\[\n\\text{Average weekly production} = 28 - 3 = 25.\n\\]\nThus, the factory produces around:\n**The average number of widgets produced per week, considering both monthly and Fibonacci holidays, is approximately 25 widgets.**", | |
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| "q2": 27.07, | |
| "q3": 26.145, | |
| "reasoning": "To solve the problems, we need to break down the production and holiday schedules.\n\n### Q1: Total Widgets Produced per Week\n\nUnder normal circumstances (without holidays):\n- Weekdays (Monday to Friday): 4 widgets/day x 5 days = 20 widgets\n- Weekends (Saturday and Sunday): 3 widgets/day x 2 days = 6 widgets\n- Extra widgets on Mondays: 2 widgets\n\nTotal per week = 20 (weekdays) + 6 (weekends) + 2 (extra on Monday) = 28 widgets\n\n### Q2: Adjusting for Uniform Holidays\n\nEach month has:\n- 4 weeks x 7 days/week = 28 days\n\nThere is one holiday per month, which means one less day of work per month.\n\nSince a month typically has 28 - 1 = 27 active days, the distribution of a holiday can affect any type of day. A uniform distribution implies the holiday equally removes the expected production of that day:\n\n- **Weekdays (Monday to Friday) chances:** 5/7 chance\n - Reduction: (5/7 chance) x (4 widgets/day) = 20/7 widgets\n- **Weekend (Saturday or Sunday) chances:** 2/7 chance\n - Reduction: (2/7 chance) x (3 widgets/day) = 6/7 widgets\n\nExpected reduction per holiday = 20/7 + 6/7 = 26/7 \u2248 3.71 widgets lost on average due to one holiday per month.\n\nThe holiday reduces the total weekly output, on average, by a quarter of its monthly holiday impact (since there are 4 weeks in a month):\n\nAverage production lost per week due to holidays = (3.71 widgets/month) / 4 weeks/month \u2248 0.9275 widgets/week\n\nHence, average weekly production considering monthly holidays:\n= 28 - 0.9275 \u2248 27.07 widgets per week\n\n### Q3: Adjusting for Fibonacci Holidays\n\nFirst, identify months with Fibonacci indices in a 10-month year: \n- Indices are: 1, 2, 3, 5, 8 (as per Fibonacci sequence)\nThus, 5 months have 2 additional holidays due to Fibonacci celebration.\n\nTotal additional holidays in a year = 5 months x 2 holidays/month = 10 holidays.\n\nNow, we calculate the total number of holidays each year:\n- Regular holidays: 1/month x 10 months = 10 holidays\n- Extra Fibonacci holidays: 10 holidays\n\nTotal holidays per year = 10 (regular) + 10 (Fibonacci) = 20 holidays/year.\n\nAverage holidays per month = 20 holidays / 10 months = 2 holidays/month\n\nRecompute the average weekly production with now 2 holidays per month:\n- Reduction per holiday remains ~3.71 widgets, as calculated\n- With 2 holidays per month, the monthly loss = 2 holiday/month x 3.71 widgets/holiday = 7.42 widgets/month\n\nAverage production lost per week with this new average of 2 holidays:\n= 7.42 widgets/month / 4 weeks/month = 1.855 widgets/week\n\nFinal average widgets produced per week:\n= 28 - 1.855 \u2248 26.145 widgets per week\n\nThus, after accounting for the uniform monthly holidays and extra holidays in Fibonacci months, the average number of widgets produced per week rounds to approximately 26.15 widgets.", | |
| "error": 0.21499999999999986, | |
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| { | |
| "q1": 30.0, | |
| "q2": 29.07, | |
| "q3": 28.15, | |
| "reasoning": "To solve these questions, we first need to determine the base production of widgets under the provided schedule and then adjust for holidays.\n\n**Q1: How many widgets are produced in a week?**\n\n1. **Weekdays (Monday to Friday)**: \n - 4 widgets per day\n - 5 weekdays in a week\n - Total = 4 widgets/day * 5 days = 20 widgets\n\n2. **Weekends (Saturday and Sunday)**:\n - 3 widgets per day\n - 2 weekend days\n - Total = 3 widgets/day * 2 days = 6 widgets\n\n3. **Extra widgets on Monday**:\n - 2 extra widgets\n - Therefore, on Monday = 4 + 2 = 6 widgets\n - Adjusting the previous weekday total: 18 (Tuesday-Friday) + 6 (Monday) = 24 widgets\n\nAdding everything:\n - Total weekly widgets produced = 24 (weekdays) + 6 (weekends) = 30 widgets\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\n- Each month has four weeks and one holiday (on average).\n- A holiday reduces production based on which day it falls on:\n - If it falls on a weekday where 4 widgets were planned, production reduces by 4.\n - If it falls on a weekend, production reduces by 3.\n - Using the uniform distribution of the holiday over 7 days, the expected production loss per holiday is: \\\n \\((5/7) \\times 4 + (2/7) \\times 3 = \\frac{20}{7} + \\frac{6}{7} = \\frac{26}{7}\\) \u2248 3.71 widgets\n\nSo, in a month with one holiday, production is reduced by approximately 3.71 widgets.\n\n- Monthly Production without Holidays = \\(4 \\times 30 = 120\\) widgets\n- Monthly Production Loss due to 1 Holiday \u2248 3.71 widgets\n- Adjusted Monthly Production \u2248 120 - 3.71 = 116.29 widgets\n- Average weekly production: \\\n \\( \\frac{116.29}{4} \\) \u2248 29.07 widgets\n\n**Q3: Taking the Fibonacci holidays into account, what's the average number of widgets produced per week?**\n\n- The Fibonacci sequence begins: 1, 1, 2, 3, 5, 8, 13, ...\n- Relevant months under a 1-based index: Months 1, 2, 3, 5, and 8. (5 such months)\n\n- After considering two extra holidays for these Fibonacci months:\n - There are 10 (total months) - 5 (Fibonacci months) = 5 normal months\n - Each normal month still loses ~3.71 widgets to its one holiday\n\n- Each Fibonacci month now has three total holidays:\n - Production loss for the Fibonacci months = 3 \\times 3.71 \u2248 11.13 widgets per such month\n - Total production loss over 10 months = 5 (normal months) \\times 3.71 + 5 (Fibonacci months) \\times 11.13 \u2248 18.55 + 55.65 = 74.20 widgets\n\n- Adjusted production over the entire year:\n - Total potential monthly production = 120 * 10 = 1200\n - Adjusted yearly production = 1200 - 74.20 = 1125.8 widgets\n\n So, average weekly production for the year = \\( \\frac{1125.8}{4 \\times 10} \\) = 28.15 widgets \n\nThe average number of widgets produced per week, accounting for all holidays, is approximately **28.15** widgets.", | |
| "error": 6.219999999999999, | |
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| "total_tokens": 1923.0, | |
| "cost": 0.011250000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.145, | |
| "reasoning": "To solve the problem, we'll address each question one by one:\n\n### Q1: How many widgets are produced in a week?\n\n1. **Weekdays (Monday to Friday):**\n - Each weekday, the factory produces 4 widgets.\n - In total for the weekdays: \\(5 \\times 4 = 20\\) widgets.\n\n2. **Weekends (Saturday and Sunday):**\n - Each weekend day, the factory produces 3 widgets.\n - In total for the weekend: \\(2 \\times 3 = 6\\) widgets.\n\n3. **Extra Monday widgets:**\n - 2 extra widgets are produced on Monday.\n\nAdding all these together, the total number of widgets produced in a regular week is:\n\\[ 20 + 6 + 2 = 28 \\text{ widgets.} \\]\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has four weeks and one holiday:\n\n1. **Total widgets produced in 4 weeks without holidays:**\n - \\(4 \\times 28 = 112\\) widgets.\n\n2. **Holiday adjustment:**\n - One holiday reduces the widget production by 4 if it's a weekday (in most cases) or by 3 if it's a weekend. \n - However, since holidays are uniformly distributed, we average this out:\n - The probability of the holiday being on a weekday: \\(5/7\\), and on a weekend: \\(2/7\\).\n - Average widgets not produced due to a holiday:\n \\[\n \\text{Expected widgets lost} = \\left(\\frac{5}{7} \\times 4\\right) + \\left(\\frac{2}{7} \\times 3\\right) = \\frac{20}{7} + \\frac{6}{7} = \\frac{26}{7} \\approx 3.71\n \\]\n\n3. **Average weekly production adjusted for holidays:**\n \\[\n 112 - \\frac{26}{7} = 112 - 3.71 \\approx 108.29 \\text{ widgets per month.}\n \\]\n - Divide by 4 weeks to find the average weekly production:\n \\[\n \\frac{108.29}{4} \\approx 27.07 \\text{ widgets per week.}\n \\]\n\n### Q3: Taking Fibonacci holidays into account, what's the average number of widgets produced per week?\n\nFibonacci sequence numbers relevant to this problem are: 1, 2, 3, 5, and 8 (since months in Standardtopia are indexed 1\u201310).\n\n1. **Extra holidays in a year:**\n - The Fibonacci-indexed months are 1, 2, 3, 5, and 8.\n - There are 5 months with 2 extra holidays each: \\(5 \\times 2 = 10\\) extra holidays.\n\n2. **Total holidays in a year:**\n - Originally 10 holidays (1 per month) + 10 extra holidays = 20 holidays per year.\n\n3. **Total months in a year:**\n - 10 months, meaning \\(10 \\times 4 = 40\\) weeks.\n\n4. **Average reduction in widgets per extra holiday:**\n - Using the distribution of holidays as before, we lose approximately 3.71 widgets per holiday.\n - Total widgets lost due to 20 holidays: \\(20 \\times 3.71 = 74.2\\).\n\n5. **Yearly production without holidays:**\n - \\(40 \\times 28 = 1120\\) widgets.\n\n6. **Yearly production including all holidays:**\n - \\(1120 - 74.2 = 1045.8 \\text{ widgets.}\\)\n\n7. **Average weekly production including Fibonacci holidays:**\n \\[\n \\frac{1045.8}{40} \\approx 26.145 \\text{ widgets per week.}\n \\]\n\nIn conclusion, the factory produces approximately 26.145 widgets per week when accounting for all holidays, including extra Fibonacci month holidays.", | |
| "error": 0.21499999999999986, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 2029.0, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let's solve these questions step by step.\n\n**Q1: How many widgets are produced in a week?**\n\nThe factory produces widgets according to the following schedule:\n- Monday: 4 widgets + 2 extra widgets = 6 widgets\n- Tuesday: 4 widgets\n- Wednesday: 4 widgets\n- Thursday: 4 widgets\n- Friday: 4 widgets\n- Saturday: 3 widgets\n- Sunday: 3 widgets\n\nNow, let's add up the widgets for each day of the week:\n\n\\[\n6 + 4 + 4 + 4 + 4 + 3 + 3 = 28 \\text{ widgets per week}\n\\]\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\nEach month has 4 weeks and one holiday. With one holiday per month, one day per month does not produce widgets. The effect of the holiday can vary depending on which day it falls. Since holidays are uniformly distributed across the week, we'll calculate the expected reduction in widgets per week due to the holiday.\n\nThe average loss in widgets per day due to the holiday can be calculated as follows:\n\n\\[\n\\text{Average widgets on a holiday} = \\frac{6 + 4 + 4 + 4 + 4 + 3 + 3}{7} = 4 \\text{ widgets}\n\\]\n\nSince there is 1 holiday per month (every 4 weeks), we distribute this holiday's impact over 4 weeks:\n\n\\[\n\\text{Average widgets lost per week} = \\frac{4}{4} = 1 \\text{ widget per week}\n\\]\n\nSo, the average number of widgets produced per week, accounting for the holiday, is:\n\n\\[\n28 - 1 = 27 \\text{ widgets per week}\n\\]\n\n**Q3: Taking these Fibonacci holidays into account, what's the average number of widgets produced per week?**\n\nFibonacci numbers within 10 are 1, 2, 3, 5, and 8. Corresponding months are January, February, March, May, and August, which means these months have 2 extra holidays each. This results in 5 months having 3 holidays and 5 months having 1 holiday.\n\nThe total number of extra holidays due to Fibonacci months is:\n\n\\[ \n5 \\text{ months} \\times 2 \\text{ extra holidays each} = 10 \\text{ extra holidays}\n\\]\n\nAdding the monthly holidays, the total number of holidays in a year is:\n\n\\[\n10 \\text{ months} \\times 1 \\text{ regular holiday per month} + 10 \\text{ extra holidays} = 20 \\text{ holidays}\n\\]\n\nThus, the total number of holidays in a year (10 months) is 20 holidays. \n\nDistributing these holidays over the year, each month has:\n\n\\[\n\\frac{20}{10} = 2 \\text{ holidays per month}\n\\]\n\nGiven each month has 4 weeks, the average number of holidays per week is:\n\n\\[\n\\frac{2 \\text{ holidays per month}}{4 \\text{ weeks per month}} = 0.5 \\text{ holidays per week}\n\\]\n\nSince each holiday averages 4 widgets that are not produced:\n\n\\[\n0.5 \\times 4 = 2 \\text{ widgets not produced per week}\n\\]\n\nThus, considering Fibonacci holidays, the average number of widgets produced per week is:\n\n\\[\n28 - 2 = 26 \\text{ widgets per week}\n\\]\n\nTherefore, the average number of widgets produced per week, considering all holidays, is **26 widgets per week**.", | |
| "error": 0.0, | |
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| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "To solve these questions, we first need to understand the weekly production schedule of the factory and then adjust it based on holidays.\n\n### Q1: How many widgets are produced in a week?\n\n- **Monday:** 4 widgets + 2 extra = 6 widgets\n- **Tuesday to Friday (each day):** 4 widgets \u00d7 4 days = 16 widgets\n- **Saturday and Sunday (each day):** 3 widgets \u00d7 2 days = 6 widgets\n\nTotal widgets per week: \n\\[ 6 + 16 + 6 = 28 \\text{ widgets} \\]\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\nEach month has 4 weeks and 1 holiday. Thus, the factory is effectively open for 4 weeks of production minus 1 day.\n\nTo calculate the expected weekly widget production, we account for the monthly holiday and how it averaging out to weekly:\n\n- Total production in a month without holidays: \\( 28 \\text{ widgets/week} \\times 4 \\text{ weeks} = 112 \\text{ widgets/month} \\)\n\n- Since the holiday distribution is uniform across different days of the week, we'll assume that a holiday occurs on a weekday or weekend with equal probability. An average loss will be formulated based on the average widget production for any arbitrary day in the week:\n\n - Average production per day in a typical week without holidays = Total weekly production / Total days = \\( 28 \\text{ widgets/week} \\,/\\, 7 \\text{ days/week} = 4 \\text{ widgets per day} \\)\n\n- If we lose 1 production day (holiday) in a month due to a holiday, we lose 4 widgets.\n\n- Adjusted total production for the month: \\( 112 \\text{ widgets} - 4 \\text{ widgets} = 108 \\text{ widgets/month} \\)\n\n- Average weekly production with 1 monthly holiday: \\( 108 \\text{ widgets/month} \\,/\\, 4 \\text{ weeks} = 27 \\text{ widgets/week} \\)\n\n### Q3: Taking Fibonacci holidays into account, what's the average number of widgets produced per week?\n\nThe Fibonacci sequence starts as 0, 1, 1, 2, 3, 5, 8, etc. The 1-based index Fibonacci months in a 10-month year are 1, 2, 3, 5, and 8. So there are 5 months with 2 extra holidays each.\n\nFor these months:\n- **Holiday loss per such month:** 2 additional holidays results in \\( 2\\text{ days} \\times 4 \\text{ widgets/day} = 8 \\text{ widgets} \\) lost\n\n- Total holiday loss in a 10-month year from Fibonacci holidays: \\( 5 \\text{ months} \\times 8 \\text{ widgets/month} = 40 \\text{ widgets/year} \\)\n\nNow, over the year (10 months): \n- Yearly production loss from regular holidays = \\( 10 \\text{ months} \\times 4 \\text{ widgets/month} = 40 \\text{ widgets/year} \\)\n- Plus the Fibonacci holiday loss = 40 widgets\n- Total loss = \\( 40+40 = 80 \\text{ widgets/year} \\)\n\n- So annual production without holidays: \\( 112 \\text{ widgets/month} \\times 10 \\text{ months} = 1120 \\text{ widgets/year} \\)\n\n- Actual annual production: \\( 1120 - 80 = 1040 \\text{ widgets/year} \\)\n\nFinally, average yearly production per week adjusting for all holidays over the 10-month year:\n\nTotal weeks in a year = \\( 4 \\text{ weeks/month} \\times 10 \\text{ months/year} = 40 \\text{ weeks/year} \\)\n\n\\[ \\text{Average weekly production} = 1040 \\text{ widgets/year} \\,/\\, 40 \\text{ weeks/year} = 26 \\text{ widgets/week} \\]\n\nTherefore, accounting for regular and Fibonacci holidays, the average number of widgets produced per week is 26.", | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.14, | |
| "reasoning": "To solve these problems, let's break down each question step by step.\n\n**Q1: How many widgets are produced in a week?**\n\nThe factory's widget production is as follows:\n- **Weekdays (Monday to Friday):** 4 widgets per day.\n- **Weekends (Saturday and Sunday):** 3 widgets per day.\n- **Extra widgets on Monday:** 2 additional widgets.\n\nCalculating the weekly production:\n- **Monday:** 4 widgets + 2 extra = 6 widgets\n- **Tuesday to Friday (4 days):** 4 widgets/day = 16 widgets\n- **Saturday and Sunday (2 days):** 3 widgets/day = 6 widgets\n\nTotal weekly widget production = 6 (Monday) + 16 (Tuesday to Friday) + 6 (Saturday and Sunday) = 28 widgets.\n\n**Answer to Q1:** The factory produces 28 widgets per week.\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\nEach month has four weeks (28 days) and one holiday. The holiday can fall on any day of the week, affecting production for that day. Let's account for this:\n\n- A month has an average of \\( \\frac{28}{7} = 4 \\) weeks.\n- 1 holiday reduces production by 4 widgets if on a weekday, or by 3 widgets if on a weekend.\n- The holiday distribution is uniform across the week. There is a \\(\\frac{5}{7}\\) chance it's a weekday and a \\(\\frac{2}{7}\\) chance it's a weekend.\n - Expected weekday holiday widget loss = \\( \\frac{5}{7} \\times 4 \\) = \\(\\frac{20}{7}\\)\n - Expected weekend holiday widget loss = \\( \\frac{2}{7} \\times 3 \\) = \\(\\frac{6}{7}\\)\n\nExpected total widget loss due to one holiday = \\(\\frac{20}{7} + \\frac{6}{7} = \\frac{26}{7} \\approx 3.71\\) widgets.\n\nSo, the monthly production adjusted for one holiday is \\( (4 \\times 28) - \\frac{26}{7} \\).\n\nDivide this monthly production by 4 to find the average weekly production:\n\\[\n\\text{Average weekly production} = 28 - \\frac{26}{28} = 28 - 0.93 \\approx 27.07\n\\]\n\n**Answer to Q2:** The average number of widgets produced per week, accounting for one holiday per month, is approximately 27.07 widgets.\n\n**Q3: Taking Fibonacci holidays into account, what's the average number of widgets produced per week?**\n\nThe months with Fibonacci indices (1, 2, 3, 5, and 8) each have 3 holidays (1 regular + 2 additional). Therefore:\n- These months have 3 holidays.\n- The remaining months have 1 holiday.\n\nCalculate the average duration of the entire year (10 months), considering 3 and 1 holiday months:\n\nLet \\(x = \\text{Average monthly widget loss due to holidays}\\).\n\n1. **Months with Fibonacci indices (5 months):**\n - 3 holidays x (average loss of 3.71 each holiday) = 3 x 3.71 = 11.13 widgets lost\n\n2. **Months without Fibonacci indices (5 months):**\n - 1 holiday x 3.71 = 3.71 widgets lost\n\nTotal widget loss per year = (5 x 11.13) + (5 x 3.71) = 55.65 + 18.55 = 74.2 widgets.\n\nAverage monthly widget production after all holidays = \\( 4 \\times 28 \\times 10 - 74.2 \\).\n\nConvert annual loss to weekly production impact:\n\\[\n\\text{Weekly average widget loss} = \\frac{74.2}{4 \\times 10} = \\frac{74.2}{40} = 1.855\n\\]\n\nSo, average weekly production = \\( 28 - 1.855 = 26.145 \\).\n\n**Answer to Q3:** The average number of widgets produced per week, with Fibonacci holidays considered, is approximately 26.14 widgets.", | |
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| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "To solve these questions, let's first break down the widget production schedule:\n\n1. **Weekday Production**: 4 widgets per day\n2. **Weekend Production**: 3 widgets per day\n3. **Additional Monday Widgets**: 2 extra widgets\n\n### Q1: How many widgets are produced in a week?\n\n1. **Weekday Production** (Monday to Friday):\n - 5 weekdays \u00d7 4 widgets = 20 widgets\n\n2. **Weekend Production** (Saturday and Sunday):\n - 2 weekend days \u00d7 3 widgets = 6 widgets\n\n3. **Extra Monday Widgets**:\n - 2 extra widgets on Monday\n\nTotal weekly production:\n\\[ 20 \\text{ (weekday)} + 6 \\text{ (weekend)} + 2 \\text{ (extra Monday)} = 28 \\text{ widgets} \\]\n\n### Q2: Average widgets produced per week considering monthly holidays\n\nEach month has exactly 4 weeks and 1 holiday, making it 29 days (weekday or weekend). \n\n- A holiday uniformly distributed means each weekday or weekend day has an equal chance of being a holiday. \n\nTo find the average production per week accounting for one monthly holiday impact:\n- Average monthly widgets without holiday: \\(28 \\text{ widgets/week} \u00d7 4 \\text{ weeks} = 112 \\text{ widgets/month}\\)\n\nThe holiday can fall on any of the seven days, averaging:\n\n- If on a weekday: Lose 4 widgets\n- If on a weekend: Lose 3 widgets\n- Extra impact if on a Monday: loss of 4 widgets + 2 extra = 6 widgets\n\nTwo weekdays without Monday and two weekend days, each having 1/7 chance and Monday having 1/7:\n\n\\[ \n\\text{Average loss} = \\left(\\frac{1}{7} \\times 6\\right) + \\left(\\frac{4}{7} \\times 4\\right) + \\left(\\frac{2}{7} \\times 3\\right) = \\frac{6}{7} + \\frac{16}{7} + \\frac{6}{7} = \\frac{28}{7} = 4 \n\\]\n\nSo, monthly widgets with holiday:\n\\[ 112 \\text{ widgets (without holiday)} - 4 \\text{ widgets (average holiday loss)} = 108 \\text{ widgets}\\]\n\nNow, average weekly production considering 1 holiday impact:\n\\[ \\frac{108 \\text{ widgets/month}}{4} \\approx 27 \\text{ widgets/week} \\]\n\n### Q3: Average widgets produced per week considering monthly and fibonacci holidays\n\nTen months have specific Fibonacci-index months (1, 2, 3, 5, 8). These months have two extra holidays, totaling 3 holidays for these months. The other five months maintain one holiday. \n\n**For Fibonacci months:**\n- Total widgets loss:\n - 3 holidays each losing an average of 4 widgets: \\(3 \\times 4 = 12\\) widget loss.\n\n- Adjusted production for Fibonacci months:\n - \\(112 - 12 = 100\\) widgets/month\n - Average weekly (Fibonacci months): \\( \\frac{100}{4} = 25 \\) widgets/week\n\n**Non-Fibonacci months are the same** as Q2:\n- Average weekly (non-Fibonacci months): 27 widgets/week\n\n**Average weekly production throughout the year** (10 months: 5 Fibonacci + 5 non-Fibonacci):\n- \\(\\frac{(5 \\times 25) + (5 \\times 27)}{10} = \\frac{125 + 135}{10} = 26 \\) widgets/week\n\nThus, considering monthly and Fibonacci holidays, the average number of widgets produced per week is approximately **26** widgets.", | |
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| "q2": 27.0, | |
| "q3": 25.5, | |
| "reasoning": "To answer these questions, we need to calculate the widget production based on the given schedule and adjust for holidays as described.\n\n**Q1: How many widgets are produced in a week?**\n\n- During weekdays (Monday to Friday), the factory produces 4 widgets per day.\n- On weekends (Saturday and Sunday), it produces 3 widgets per day.\n- Furthermore, the factory produces 2 extra widgets on Mondays.\n\nSo, let's calculate the total weekly production:\n\n1. **Weekdays (Monday to Friday):**\n - Total widgets from Tuesday to Friday = 4 widgets \u00d7 4 days = 16 widgets\n - On Monday, the factory produces extra 2 widgets, so it makes 4 + 2 = 6 widgets.\n\n2. **Weekend (Saturday and Sunday):**\n - Total widgets from Saturday and Sunday = 3 widgets \u00d7 2 days = 6 widgets\n\n**Total widgets in a week:**\n - \\(6 \\text{ (Monday)} + 16 \\text{ (Tuesday to Friday)} + 6 \\text{ (Weekend)} = 28\\) widgets.\n\nSo, the factory produces **28 widgets in a week**.\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\nEach month has exactly 4 weeks plus 1 holiday. Since holidays are uniformly distributed across all days, each week in a month of 29 days with one holiday will have the expectation of:\n\n- **4 months have 28 production days + 1 holiday: 28 total + 1 non-working day.**\n\nThis means that on average during a month, one week will have 6 working days and 1 holiday. On a weekly basis, this means calculating the production for an average week over the month:\n\nAverage production in a month over 4 weeks (28 production days + something lost to holiday):\n\n- **Holidays can occur on any day affecting production:** \n- Assume evenly distributed, reduction would be \\(\\frac{28 \\text{ widgets}}{7 \\text{ days}} = 4 \\text{ widgets} \\text{ per holiday day}\\)\n\nWe lose 2 widgets on average because holidays can affect a high work day (like a Monday)\n\nTherefore, you need to subtract 1 additional widget (since holidays aren't exactly distributed over weeks, but rather across the whole schema) in **long-term average accounting**:\n\n- So effectively, each 4 week month loses 1 widget for 1 holiday, leaving:\n \\(\\frac{28 \\text{ widgets/week} \\times 4 \\text{ weeks} - 4 \\text{ widgets}}{4 \\text{ weeks}} = 27 \\text{ widgets/week average}\\).\n \nThe average number of widgets produced per week when accounting for regular holidays is approximately **27 widgets**.\n\n**Q3: Taking these Fibonacci holidays into account, what's the average number of widgets produced per week?**\n\nThe Fibonacci months in a 10 month year are those whose indices are Fibonacci numbers. Thus, the Fibonacci months are months 1, 2, 3, 5, and 8.\n\n- There are 5 Fibonacci months in a year, each with 2 extra holidays, so that's 10 more holidays in total each year.\n\nNow, accounting for these additional holidays:\n\n1. **Annual Holidays = 10 (regular holidays, one per month) + 10 (fibonacci holidays) = 20 holidays**\n\n2. **Work weeks counted:** 10 months \u00d7 4 weeks/month = 40 weeks.\n\nThe average total holidays per month = 2 (one regular + on average a fraction due to fibonacci spread).\n\nThen, in the scope of production on a weekly basis over the year:\n\n- On average each week loses 0.5 days to a holiday:\n - **Reduction** on 5:\\[ 28 \\text{ widgets/week} \u00d7 40 weeks - 20 days \u00d7 4 \\text{ average widget loss} / 40 weeks \\]\n\nThus, the average production is reduced each time the production affects a heavier widget-day:\n\n- Approximation of 1 full-day equivalent lost typically due to these added holidays.\n\nOver the weekly balance, The effective average number when accounting these extra Fibonacci holidays is:\n\\[ 5060 \\div 40 \\approx 25.5 \\]\n\nTherefore, the average number of widgets produced per week when accounting for both regular and Fibonacci holidays is approximately **25.5 widgets**.", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 2131.0, | |
| "cost": 0.012542500000000002 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.07, | |
| "q3": 26.14, | |
| "reasoning": "To solve these problems, let's break down each question.\n\n**Q1: How many widgets are produced in a week?**\n\n1. For Monday:\n - Regular widgets: 4\n - Extra widgets: 2\n - Total for Monday: \\(4 + 2 = 6\\)\n\n2. For each weekday (Tuesday to Friday), the factory produces 4 widgets:\n - Total for weekdays excluding Monday: \\(4 \\times 4 = 16\\)\n\n3. For Saturday and Sunday, the factory produces 3 widgets per day:\n - Total for weekends: \\(3 \\times 2 = 6\\)\n\n4. Total widgets in a week:\n \\[\n 6 \\, (\\text{Monday}) + 16 \\, (\\text{Tuesday to Friday}) + 6 \\, (\\text{weekends}) = 28\n \\]\n\nSo, **28 widgets are produced in a regular week.**\n\n**Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?**\n\n1. Each month has 4 weeks and 1 holiday. Therefore, there are 4 weeks + 1 holiday per month.\n2. The holiday reduces production on its respective day:\n - If a holiday falls on a weekday (Monday to Friday), 4 widgets are not produced.\n - If a holiday falls on a weekend, 3 widgets are not produced.\n3. Each day of the week is equally likely for the holiday, thus:\n - Probability holiday is on a weekday: \\( \\frac{5}{7} \\)\n - Probability holiday is on a weekend: \\( \\frac{2}{7} \\)\n4. Expected loss of widgets due to holiday:\n \\[\n \\frac{5}{7} \\times 4 + \\frac{2}{7} \\times 3 = \\frac{20}{7} + \\frac{6}{7} = \\frac{26}{7} \\approx 3.71\n \\]\n5. Monthly production, accounting for holiday:\n \\[\n (28 \\times 4) - \\frac{26}{7} = 112 - \\frac{26}{7} = \\frac{784}{7} - \\frac{26}{7} = \\frac{758}{7}\n \\]\n Monthly average production per week:\n \\[\n \\frac{758}{7} \\div 4 = \\frac{758}{28} \\approx 27.07\n \\]\n\nSo, on average, approximately **27.07 widgets are produced per week** after accounting for a monthly holiday.\n\n**Q3: Taking Fibonacci holidays into account, what's the average number of widgets produced per week?**\n\n1. Identify Fibonacci-indexed months within a 10-month year: Months 1, 2, 3, 5, 8 are Fibonacci-indexed.\n\n2. These months have 2 extra holidays. This results in 3 holidays for each Fibonacci month.\n - Total number of holidays in a Fibonacci month: 3\n - Other months still have 1 holiday.\n\n3. In one year (10 months), there are:\n - \\(5 \\times 3 = 15\\) holidays in Fibonacci months\n - \\(5 \\times 1 = 5\\) holidays in non-Fibonacci months\n - Total holidays in a year: \\(15 + 5 = 20\\)\n\n4. Average monthly holidays in a year:\n \\[\n \\frac{20 \\text{ holidays}}{10 \\months} = 2 \\text{ holidays per month}\n \\]\n\n5. Average weekly production:\n - Expectation reduces production as described before, so:\n \\[\n 28 - 2 \\times \\frac{26}{7 \\times 4} = 28 - \\frac{52}{28} \\approx 28 - 1.86 = 26.14\n \\]\n\nSo, the average number of widgets produced per week, considering Fibonacci holidays, is approximately **26.14 widgets**.", | |
| "error": 0.21000000000000085, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1979.0, | |
| "cost": 0.011540000000000002 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 23.64, | |
| "q3": 19.29, | |
| "reasoning": "Let's tackle each question step by step.\n\n### Q1: How many widgets are produced in a week?\n\n- **Weekdays (Monday to Friday):**\n The factory produces 4 widgets each weekday.\n \\[\n \\text{Total widgets on weekdays} = 4 \\times 5 = 20 \n \\]\n Additionally, on Mondays, 2 extra widgets are produced.\n \\[\n \\text{Total widgets on Monday} = 4 + 2 = 6\n \\]\n Adjusting for the extra widgets on Monday:\n \\[\n \\text{Adjusted total for weekdays} = 20 - 4 + 6 = 22\n \\]\n\n- **Weekends (Saturday and Sunday):**\n The factory produces 3 widgets each day of the weekend.\n \\[\n \\text{Total widgets on weekends} = 3 \\times 2 = 6 \n \\]\n\n- **Total widgets produced in a week:**\n \\[\n \\text{Total widgets per week} = 22 + 6 = 28\n \\]\n\n### Answer to Q1:\nThe factory produces 28 widgets in a week.\n\n### Q2: Taking monthly holidays into account, what's the average number of widgets produced per week?\n\n- **Monthly structure:** Each month has 4 weeks and 1 holiday.\n- **Monthly widgets without holidays:** \n \\[\n \\text{4 weeks} \\times \\text{28 widgets/week} = 112\n \\]\n- **Holidays:** On a random day, no widgets are produced.\n Given that a holiday occurs in all 5 weekdays or 2 weekend days, the average daily production lost on the holiday:\n \\[\n \\text{Average daily production} = \\frac{22 \\times 5 + 6 \\times 2}{7} = \\frac{110 + 12}{7} = \\frac{122}{7} \\approx 17.43\n \\]\n- **Monthly widgets with holidays:** \n \\[\n 112 - 17.43 \\approx 94.57\n \\]\n\n- **Average widgets per week considering one holiday:** \n \\[\n \\frac{94.57}{4} \\approx 23.64\n \\]\n\n### Answer to Q2:\nThe average number of widgets produced per week, accounting for one holiday, is approximately 23.64 widgets.\n\n### Q3: Taking fibonacci holidays into account, what's the average number of widgets produced per week?\n\n- **Fibonacci numbers:** 1, 2, 3, 5, and 8 (within a 10-month year). Thus, 5 months have 2 extra holidays each.\n- **Total holidays per year:** \n - 10 regular holidays (1 per month)\n - 5 months with 2 extra holidays: \\( 5 \\times 2 = 10 \\) extra holidays\n- **Total holidays:** \\( 10 + 10 = 20 \\)\n- **Total weeks in a year:** \\( 10 \\times 4 = 40 \\)\n\n- **Yearly widget production without holidays:**\n \\[\n 28 \\times 40 = 1120\n \\]\n\n- **Average loss due to holidays per year:**\n \\[\n 17.43 \\times 20 = 348.6\n \\]\n\n- **Yearly widgets considering all holidays:**\n \\[\n 1120 - 348.6 = 771.4\n \\]\n\n- **Average weekly widget production over the year:**\n \\[\n \\frac{771.4}{40} \\approx 19.29\n \\]\n\n### Answer to Q3:\nThe average number of widgets produced per week, accounting for all holidays, is approximately 19.29 widgets.", | |
| "error": 10.07, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1883.0, | |
| "cost": 0.010940000000000002 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-haiku-latest", | |
| "avg_error": 8.964, | |
| "avg_correctness": 0.3, | |
| "avg_tokens": 1395.8, | |
| "avg_cost": 0.0095772, | |
| "n_tests": 10, | |
| "failure_rate": 0.09090909090909091, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 20.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets in a week\n\u2022 Weekdays (5 days): 4 widgets \u00d7 5 days = 20 widgets\n\u2022 Weekend days (2 days): 3 widgets \u00d7 2 days = 6 widgets\n\u2022 Extra Monday widgets: 2\n\u2022 Total per week: 20 + 6 + 2 = 28 widgets\n\nQ2: Average widgets per week with one random holiday\n\u2022 A holiday removes one day of production\n\u2022 If average day is 4, then average week becomes:\n 28 - 4 = 24 widgets per week\n\nQ3: Fibonacci holidays\n\u2022 Fibonacci numbers: 1, 2, 3, 5, 8\n\u2022 Months with extra holidays: 1st, 2nd, 3rd, 5th, 8th\n\u2022 Extra holidays per year: 2 \u00d7 5 = 10 total holidays\n\u2022 Average extra holiday per month: 1\n\u2022 So now weekly total becomes: 24 - 4 = 20 widgets per week\n\nThe answers are:\nQ1: 28\nQ2: 24\nQ3: 20", | |
| "error": 9.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1294.0, | |
| "cost": 0.008968 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 20.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets in a week\n- Weekdays (Monday to Friday): 4 widgets \u00d7 5 days = 20 widgets\n- Extra Monday widgets: 2 additional widgets\n- Weekend (Saturday and Sunday): 3 widgets \u00d7 2 days = 6 widgets\nTotal weekly widgets = 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly holiday calculation\n- 4 weeks in a month\n- 1 holiday per month, randomly distributed\n- When a holiday occurs, no widgets are produced that day\n\nProbability of holiday:\n- Weekday holiday (5/7 chance): Changes daily output to 0\n- Weekend holiday (2/7 chance): Changes daily output to 0\n\nAverage weekly widgets:\n- Normal weeks: 28 widgets\n- Holiday week: Expected reduction = 28 \u00d7 (1/7) = 4 widgets\nAverage weekly widgets = 28 - 4 = 24 widgets\n\nQ3: Fibonacci holiday calculation\nFibonacci month indices are 1, 2, 3, 5, 8\nThese months have 2 extra holidays\n\nFibonacci monthly holidays = 2 per month\nAdditional holiday impact = 28 \u00d7 (2/7) = 8 widgets reduction\n\nAverage weekly widgets = 28 - 8 = 20 widgets\n\nAnswer:\nQ1: 28\nQ2: 24\nQ3: 20", | |
| "error": 9.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1420.0, | |
| "cost": 0.009724 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 16.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: How many widgets are produced in a week?\n\u2022 Weekdays (Mon-Fri): 5 days\n - Normally 4 widgets per day\n - Monday has 2 extra widgets\n\u2022 Weekend (Sat-Sun): 2 days\n - 3 widgets per day\nCalculation:\n(4 \u00d7 5) + 2 + (3 \u00d7 2) = 20 + 2 + 6 = 28 widgets per week\n\nQ2: Accounting for one random holiday per month\n\u2022 A holiday reduces widget production for that day\n\u2022 So widgets per week = previous calculation - average daily production\n\u2022 Average daily production = 28/7 = 4 widgets per day\n\u2022 Average weekly widgets = 28 - 4 = 24 widgets\n\nQ3: Fibonacci holidays\n\u2022 Fibonacci sequence: 1, 2, 3, 5, 8\n\u2022 Fibonacci months get 2 extra holidays\n\u2022 1st, 2nd, 3rd, 5th, and 8th months get 2 holidays each\n\u2022 This means more production reduction\n\u2022 New average weekly widgets = 24 - (2 \u00d7 4) = 16 widgets\n\nThe answers are:\nQ1: 28\nQ2: 24\nQ3: 16", | |
| "error": 13.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1374.0, | |
| "cost": 0.009448 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 24.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n\u2022 Weekdays (Mon-Fri): 4 widgets \u00d7 5 days = 20 widgets\n\u2022 Mondays have 2 extra widgets: +2\n\u2022 Weekends (Sat-Sun): 3 widgets \u00d7 2 days = 6 widgets\nTotal: 20 + 2 + 6 = 28 widgets per week\n\nQ2: Monthly holidays\n\u2022 One holiday per month means one day closed\n\u2022 Average probability of holiday is 1/7 for any given day\n\u2022 This reduces weekly production proportionally\n\u2022 Average widgets per week: 28 \u00d7 (6/7) = 24 widgets\n\nQ3: Fibonacci holiday months\n\u2022 Fibonacci numbers less than 11 are: 1, 2, 3, 5, 8\n\u2022 5 months out of 10 have two extra holidays\n\u2022 This means 5 months have 2 additional closed days each\n\u2022 Additional closed days: 5 \u00d7 2 = 10 days per year\n\u2022 10 days out of 70 total working days\n\u2022 Average weekly production: 28 \u00d7 (60/70) \u2248 24 widgets\n\nThe answer is 24 widgets per week.", | |
| "error": 5.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1338.0, | |
| "cost": 0.009231999999999999 | |
| }, | |
| { | |
| "q1": 36.0, | |
| "q2": 32.42, | |
| "q3": 29.42, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n\u2022 Weekdays (Monday-Friday): 5 days \u00d7 (4 widgets + 2 extra Monday widgets) = 5 \u00d7 6 = 30 widgets\n\u2022 Weekend (Saturday-Sunday): 2 days \u00d7 3 widgets = 6 widgets\n\u2022 Total per week: 30 + 6 = 36 widgets\n\nQ2: Monthly holidays\n\u2022 Each month has 4 weeks and 1 holiday\n\u2022 Holiday is randomly distributed across 7 days\n\u2022 Average holiday scenario:\n - Weekday holiday: Reduces daily production by 4 widgets\n - Weekend holiday: Reduces daily production by 3 widgets\n\u2022 Average impact per month: (4/7 \u00d7 4) + (3/7 \u00d7 3) = 2.29 + 1.29 = 3.58 widgets less per week\n\u2022 Average weekly production: 36 - 3.58 = 32.42 widgets\n\nQ3: Fibonacci month holidays\n\u2022 Fibonacci numbers: 1, 2, 3, 5, 8\n\u2022 In first 10 months, months 1, 2, 3, 5, 8 get 2 extra holidays\n\u2022 These extra holidays further reduce production\n\u2022 Calculation would be similar to Q2, with additional holiday reductions\n\nThe answer is 32.42 for Q2.\n\nWould you like me to elaborate on any part of the solution?", | |
| "error": 16.840000000000003, | |
| "correctness": 0.0, | |
| "total_tokens": 1442.0, | |
| "cost": 0.009847999999999999 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 21.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n\u2022 Weekdays (Mon-Fri): 4 widgets \u00d7 5 days + 2 extra Monday widgets = 22 widgets\n\u2022 Weekend (Sat-Sun): 3 widgets \u00d7 2 days = 6 widgets\n\u2022 Total per week: 22 + 6 = 28 widgets\n\nQ2: Monthly holidays\n\u2022 1 holiday per month (randomly distributed)\n\u2022 Probabilities of holiday on each day type:\n Weekday: 5/7 chance of losing 4 widgets\n Weekend: 2/7 chance of losing 3 widgets\n\u2022 Average weekly widget reduction:\n (5/7 \u00d7 4) + (2/7 \u00d7 3) = (20/7) + (6/7) = 26/7 \u2248 3.71 widgets\n\u2022 Average weekly production: 28 - (26/7) = 170/7 \u2248 24.29 widgets\n\nQ3: Fibonacci Holidays\n\u2022 Fibonacci numbers less than or equal to 10: 1, 2, 3, 5, 8\n\u2022 These months get two extra holidays\n\u2022 These months reduce production by: \n 2 \u00d7 ((5/7 \u00d7 4) + (2/7 \u00d7 3)) = 2 \u00d7 (26/7) = 52/7 \u2248 7.43 widgets\n\u2022 Average weekly production: 28 - (52/7) = 144/7 \u2248 20.57 widgets", | |
| "error": 8.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1494.0, | |
| "cost": 0.010168 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 24.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n\u2022 Weekdays (Monday-Friday): 4 widgets/day \u00d7 5 days = 20 widgets\n\u2022 Mondays have 2 extra widgets: 20 + 2 = 22 widgets\n\u2022 Weekend days (Saturday-Sunday): 3 widgets/day \u00d7 2 days = 6 widgets\n\u2022 Total per week: 22 + 6 = 28 widgets\n\nQ2: Monthly holidays\n\u2022 4 weeks per month\n\u2022 1 holiday per month\n\u2022 Holiday is randomly distributed, so on average:\n - 1/7 of a week will be missed\n\u2022 Average widgets per week:\n - 28 \u00d7 (6/7) = 24 widgets\n\nQ3: Fibonacci holidays\n\u2022 Fibonacci month indices are: 1, 2, 3, 5, 8\n\u2022 These months get 2 additional holidays\n\u2022 Per month with fibonacci index: 2 extra holidays\n\u2022 Total extra holidays: 2 \u00d7 5 = 10 holidays out of 70 total days\n\u2022 Adjusted weekly production:\n - 28 \u00d7 (60/70) = 24 widgets\n\nThe answer for Q3 is 24 widgets.", | |
| "error": 5.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1336.0, | |
| "cost": 0.009219999999999999 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.43, | |
| "q3": 27.37, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n\u2022 Weekdays (Monday-Friday): 4 * 5 = 20 widgets\n\u2022 Mondays have 2 extra widgets: +2\n\u2022 Weekend days (Saturday-Sunday): 3 * 2 = 6 widgets\n\u2022 Total per week: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly holidays\n\u2022 4 weeks + 1 holiday\n\u2022 Holiday is randomly distributed, so on average will replace 1 day's production\n\u2022 Average widgets lost per week = (1/7) * average daily production\n\u2022 Daily production during week = (20 + 2)/5 = 4.4\n\u2022 Daily weekend production = 3\n\u2022 Average daily production = ((20 + 2) + (3 * 2)) / 7 = 4\n\u2022 Widgets lost per week = 4 * (1/7) = ~0.57\n\u2022 Average weekly widgets = 28 - 0.57 = 27.43\n\nQ3: Fibonacci holidays\n\u2022 Fibonacci month indices are 1, 2, 3, 5, 8 \n\u2022 These months get 2 extra holidays\n\u2022 Total extra holidays = 2 * 5 = 10\n\u2022 Total holidays = 1 (standard) + 10 (fibonacci) = 11\n\u2022 Average widgets lost = 4 * (11/70) = ~0.63\n\u2022 Average weekly widgets = 28 - 0.63 = 27.37", | |
| "error": 1.8000000000000007, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1476.0, | |
| "cost": 0.010052 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 20.0, | |
| "q3": 18.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets in a week\n\u2022 Weekdays (Mon-Fri): 4 \u00d7 5 = 20 widgets\n\u2022 Extra Monday widgets: 2\n\u2022 Weekend (Sat-Sun): 3 \u00d7 2 = 6 widgets\n\u2022 Total per week: 20 + 2 + 6 = 28 widgets\n\nQ2: Monthly holiday considerations\n\u2022 4 weeks per month\n\u2022 1 holiday uniformly distributed\n\u2022 Probability of holiday on workday: 5/7\n\u2022 Probability of holiday on weekend: 2/7\n\nAverage widgets = (28 \u00d7 (5/7)) + (0 \u00d7 (2/7))\n = 20 widgets\n\nQ3: Fibonacci holidays\n\u2022 Fibonacci month indices: 1, 2, 3, 5, 8\n\u2022 These months get 2 extra holidays\n\u2022 5 months with 2 extra holidays: (5/10) months\n\u2022 Extra reduction: 28 \u00d7 (5/10) \u00d7 (1/7)\n\u2022 Final average: 20 - (28 \u00d7 (5/10) \u00d7 (1/7))\n = 20 - 2 = 18 widgets\n\nAnswer: 18 widgets per week on average.", | |
| "error": 15.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1348.0, | |
| "cost": 0.009292 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 24.0, | |
| "q3": 22.0, | |
| "reasoning": "Let's solve this step by step:\n\nQ1: Widgets per week\n- Weekdays (5 days): 4 \u00d7 5 = 20 widgets\n- Weekend (2 days): 3 \u00d7 2 = 6 widgets\n- Extra Monday widgets: 2\nTotal per week: 20 + 6 + 2 = 28 widgets\n\nQ2: Monthly holidays\n- 4 weeks per month\n- 1 holiday per month\n- Holiday is on a random day\n- On holiday days, no widgets are produced\n\nAverage impact:\n- 1 day out of 28 days is a holiday\n- Proportional reduction: (28 - 4) / 28 = 24/28 = 6/7\n- Average weekly widgets: 28 \u00d7 (6/7) = 24 widgets\n\nQ3: Fibonacci holidays\n- Fibonacci month indexes: 1, 2, 3, 5, 8\n- These months get 2 extra holidays\n- Total months: 10\n- Extra holiday months: 5\n\nAverage impact:\n- From Q2 baseline: 24 widgets\n- Extra reduction due to multiple holidays\n- Rough estimate: 24 \u00d7 (6/7)^(5/10) = 24 \u00d7 (6/7)^0.5 \u2248 22 widgets\n\nAnswer: \nQ1: 28 widgets\nQ2: 24 widgets\nQ3: 22 widgets", | |
| "error": 7.0, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1436.0, | |
| "cost": 0.009819999999999999 | |
| } | |
| ] | |
| }, | |
| { | |
| "provider": "anthropic", | |
| "model": "claude-3-5-sonnet-latest", | |
| "avg_error": 0.6123999999999998, | |
| "avg_correctness": 0.7, | |
| "avg_tokens": 1637.4, | |
| "avg_cost": 0.0113814, | |
| "n_tests": 10, | |
| "failure_rate": 0.4117647058823529, | |
| "results": [ | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let me help solve this step by step.\n\nQ1: Regular weekly production\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Accounting for monthly holidays\n* One holiday per month means 1 holiday per 4 weeks\n* Average loss per week due to holiday = 1/4 of a day's production\n* If holidays are uniform across days:\n * 5/7 chance it's a weekday (losing 4 widgets)\n * 1/7 chance it's a Saturday (losing 3 widgets)\n * 1/7 chance it's a Sunday (losing 3 widgets)\n * 1/7 chance it's a Monday (losing 6 widgets - regular 4 plus bonus 2)\n* Average loss per holiday = (5/7 \u00d7 4 + 1/7 \u00d7 3 + 1/7 \u00d7 3 + 1/7 \u00d7 6) = 4 widgets\n* Average weekly loss = 4/4 = 1 widget\n* New weekly average = 28 - 1 = 27 widgets\n\nQ3: Accounting for fibonacci holidays\n* Fibonacci numbers from 1 to 10 are: 1, 1, 2, 3, 5, 8\n* So months 1, 2, 3, 5, 8 get two extra holidays\n* In a 10-month year:\n * 5 months have 3 holidays\n * 5 months have 1 holiday\n* Average holidays per month = (5\u00d73 + 5\u00d71)/10 = 2\n* Average weekly loss = 2 holidays/month \u00d7 4 widgets/holiday \u00f7 4 weeks/month = 2 widgets\n* Final weekly average = 28 - 2 = 26 widgets\n\nTherefore:\nQ1: 28 widgets\nQ2: 27 widgets\nQ3: 26 widgets", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1757.0, | |
| "cost": 0.012447 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me help solve this step by step.\n\nQ1: Widgets per week\n* Weekdays (Mon-Fri):\n * Monday: 4 + 2 = 6 widgets\n * Tuesday-Friday: 4 \u00d7 4 = 16 widgets\n* Weekends (Sat-Sun): 3 \u00d7 2 = 6 widgets\n* Total per week = 6 + 16 + 6 = 28 widgets\n\nQ2: With monthly holidays\n* Regular days per month = (4 weeks \u00d7 7 days) - 1 holiday = 27 days\n* Need to find average impact of one holiday per 28 days\n* If holiday lands on Monday: lose 6 widgets\n* If holiday lands on Tue-Fri: lose 4 widgets\n* If holiday lands on Sat-Sun: lose 3 widgets\n* Average widgets lost per holiday: \n * (6\u00d71 + 4\u00d74 + 3\u00d72) \u00f7 7 = (6 + 16 + 6) \u00f7 7 = 28 \u00f7 7 = 4 widgets\n* Lost widgets per week = 4 \u00f7 4 = 1\n* Average widgets per week = 28 - 1 = 27 widgets\n\nQ3: With fibonacci holidays\n* Fibonacci numbers up to 10: 1,1,2,3,5,8\n* Therefore extra holidays in months 1,2,3,5,8\n* In a year (10 months):\n * 5 months have 2 holidays\n * 5 months have 1 holiday\n * Total yearly holidays = (5\u00d72) + (5\u00d71) = 15 holidays\n * Average holidays per month = 15 \u00f7 10 = 1.5\n* Additional 0.5 holidays per month vs Q2\n* Additional widgets lost per week = (0.5 \u00d7 4) \u00f7 4 = 0.5\n* Average widgets per week = 27 - 0.5 = 26.5 widgets\n\nAnswer: 26.5 widgets per week on average", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1747.0, | |
| "cost": 0.012369000000000002 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.875, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular weekly production\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\nWeekly total = 20 + 6 + 2 = 28 widgets\n\nQ2: Accounting for monthly holidays\n* In a 4-week month, there's 1 holiday where no production occurs\n* Holidays are uniformly distributed, so on average:\n * 5/7 chance it's on a weekday (losing 4 widgets)\n * 2/7 chance it's on a weekend (losing 3 widgets)\n * 1/7 chance it's on Monday (also losing 2 extra widgets)\n* Average production lost per holiday:\n * (5/7 \u00d7 4) + (2/7 \u00d7 3) + (1/7 \u00d7 2) = 20/7 + 6/7 + 2/7 = 28/7\n* This is lost once per month, or 0.25 times per week\n* Weekly loss = (28/7) \u00d7 0.25 = 1 widget\nNew weekly average = 28 - 1 = 27 widgets\n\nQ3: Accounting for fibonacci month holidays\n* Fibonacci numbers up to 10: 1,1,2,3,5,8\n* Months 1,2,3,5,8 get an extra holiday (5 months total)\n* 5 extra holidays over 10 months = 0.5 extra holidays per month\n* Extra weekly loss = 1 \u00d7 0.5 \u00d7 0.25 = 0.125 widgets\nFinal weekly average = 27 - 0.125 = 26.875 widgets\n\nThe answer is 26.875 widgets per week on average.", | |
| "error": 0.875, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1637.0, | |
| "cost": 0.011379 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Weekly production without holidays\n* Monday: 4 + 2 = 6 widgets\n* Tuesday to Friday: 4 widgets/day \u00d7 4 days = 16 widgets\n* Saturday and Sunday: 3 widgets/day \u00d7 2 days = 6 widgets\nTotal per week = 6 + 16 + 6 = 28 widgets\n\nQ2: Monthly holiday impact\n* Each month has 4 weeks = 28 days\n* One holiday per month where no widgets are produced\n* Holiday can fall on any day with equal probability\n* Average loss:\n * If on Monday: lose 6 widgets\n * If Tuesday-Friday: lose 4 widgets\n * If weekend: lose 3 widgets\n * Average loss = (6 + 4\u00d74 + 3\u00d72)/7 \u2248 4 widgets per month\n* Monthly loss spread over 4 weeks = 1 widget per week\nNew weekly average = 28 - 1 = 27 widgets\n\nQ3: Fibonacci impact\n* Fibonacci numbers up to 10: 1, 1, 2, 3, 5, 8\n* Months with index in this series get 2 extra holidays\n* In a 10-month year, months 1, 2, 3, 5, 8 get 2 extra holidays\n* That's 5 months with double holidays\n* Additional monthly loss = 4 widgets \u00d7 5 months / 10 months = 2 widgets\n* Additional weekly loss = 2/4 = 0.5 widgets\nFinal weekly average = 27 - 0.5 = 26.5 widgets\n\nAnswer: 26.5 widgets per week", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1539.0, | |
| "cost": 0.010497 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular weekly production\n* Weekdays (Mon-Fri): 4 widgets/day\n * Monday has 2 extra\n* Weekends (Sat-Sun): 3 widgets/day\n\nCalculation:\n* Monday: 4 + 2 = 6 widgets\n* Tues-Fri: 4 \u00d7 4 = 16 widgets\n* Sat-Sun: 3 \u00d7 2 = 6 widgets\nTotal per week = 6 + 16 + 6 = 28 widgets\n\nQ2: Monthly holiday impact\n* Each month has 4 weeks but loses one day to holiday\n* Over 4 weeks, each day has equal chance of being holiday\n* When holiday falls on:\n * Monday: Lose 6 widgets\n * Tue-Fri: Lose 4 widgets\n * Sat-Sun: Lose 3 widgets\n* Average loss = (6 + 4\u00d74 + 3\u00d72)\u00f77 = (6 + 16 + 6)\u00f77 = 28\u00f77 = 4 widgets per month\n* 4 widgets \u00f7 4 weeks = 1 widget per week lost\nTherefore, average weekly production = 28 - 1 = 27 widgets\n\nQ3: Fibonacci holiday impact\n* Fibonacci numbers up to 10: 1,1,2,3,5,8\n* Months with extra holidays: 1,2,3,5,8\n* 5 months have one extra holiday\n* Average additional holiday loss per year = 5 \u00d7 4 = 20 widgets\n* 20 widgets \u00f7 10 months \u00f7 4 weeks = 0.5 widgets per week lost\nTherefore, final average = 27 - 0.5 = 26.5 widgets per week\n\nThe answer is 26.5 widgets per week", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1641.0, | |
| "cost": 0.011415000000000002 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let me solve this step by step.\n\nQ1:\n* Weekdays (Mon-Fri): 4 widgets per day\n* Weekend (Sat-Sun): 3 widgets per day\n* Extra Monday widgets: 2\n\nWeekly calculation:\n* Monday: 4 + 2 = 6 widgets\n* Tuesday to Friday: 4 \u00d7 4 = 16 widgets\n* Saturday and Sunday: 3 \u00d7 2 = 6 widgets\nTotal: 6 + 16 + 6 = 28 widgets per week\n\nQ2:\n* One holiday per month = one holiday per 4 weeks\n* Average widgets lost per week due to holiday = (probability of day being a holiday) \u00d7 (widgets lost on that day)\n* Weekday holiday (5/7 probability): loses 4 widgets\n* Weekend holiday (2/7 probability): loses 3 widgets\n* Monday holiday (1/7 probability): loses 6 widgets\nAverage widgets lost per week = (5/7 \u00d7 4 + 2/7 \u00d7 3 + 1/7 \u00d7 6) \u00d7 (1/4) = 1\nTherefore: 28 - 1 = 27 widgets per week average\n\nQ3:\nFirst 8 fibonacci numbers: 1,1,2,3,5,8,13,21\nMonths with extra holidays (indices 1-10): 1,2,3,5,8\n* 5 months have 2 extra holidays each = 10 extra holidays\n* These 10 extra holidays are spread across the year (10 months)\n* 10 extra holidays / (10 months \u00d7 4 weeks) = 0.25 extra holidays per week\n* Using same calculation as Q2 for widget loss per holiday\nAverage extra widgets lost per week = 0.25 \u00d7 (5/7 \u00d7 4 + 2/7 \u00d7 3 + 1/7 \u00d7 6) = 1\nTherefore: 27 - 1 = 26 widgets per week average\n\nAnswer: 26 widgets per week", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1687.0, | |
| "cost": 0.011817000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.5, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Regular weekly production\n* Weekdays (Mon-Fri):\n * Regular weekday output = 5 days \u00d7 4 widgets = 20 widgets\n * Monday bonus = 2 widgets\n* Weekends (Sat-Sun):\n * Weekend output = 2 days \u00d7 3 widgets = 6 widgets\n* Total per week = 20 + 2 + 6 = 28 widgets\n\nQ2: Accounting for monthly holidays\n* Each month has 4 weeks and 1 holiday\n* Since holidays are evenly distributed:\n * 5/7 chance holiday is on weekday (losing 4 + 2 if Monday, 4 if other weekday)\n * 2/7 chance holiday is on weekend (losing 3 widgets)\n* Average widgets lost per holiday:\n * (5/7 \u00d7 4.4 + 2/7 \u00d7 3) = 4 widgets per holiday\n* Monthly loss = 4 widgets\n* Weekly loss = 4 widgets \u00f7 4 weeks = 1 widget per week\n* New weekly average = 28 - 1 = 27 widgets\n\nQ3: Accounting for fibonacci holidays\n* Fibonacci numbers up to 10 are: 1, 1, 2, 3, 5, 8\n* Months with extra holidays: 1st, 2nd, 3rd, 5th, 8th\n* Five months have an extra holiday\n* Extra holidays per year = 5\n* Extra weekly loss = (5 holidays \u00d7 4 widgets) \u00f7 (10 months \u00d7 4 weeks)\n = 20 \u00f7 40 = 0.5 widgets per week\n* Final weekly average = 27 - 0.5 = 26.5 widgets\n\nThe answer to Q3 is 26.5 widgets per week.", | |
| "error": 0.5, | |
| "correctness": 0.6666666666666666, | |
| "total_tokens": 1629.0, | |
| "cost": 0.011307000000000001 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.071, | |
| "q3": 26.607, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: First, let's calculate widgets per week (no holidays)\n* Monday: 4 + 2 = 6 widgets\n* Tuesday-Friday: 4 widgets each = 16 widgets\n* Saturday-Sunday: 3 widgets each = 6 widgets\nTotal per week = 6 + 16 + 6 = 28 widgets\n\nQ2: For monthly holidays:\n* Each 4-week month has one holiday where no widgets are produced\n* Daily average production without holiday = 28/7 = 4 widgets\n* One holiday removes 4 widgets per month (if on weekday) or 3 widgets (if on weekend)\n* Assuming uniform distribution: (5/7 \u00d7 4) + (2/7 \u00d7 3) = 20/7 + 6/7 = 26/7 \u2248 3.714 widgets lost per holiday\n* Per week impact = 3.714/4 = 0.929 widgets lost per week\n* Weekly production with monthly holidays = 28 - 0.929 = 27.071 widgets\n\nQ3: Fibonacci months in base 1 up to 10 are: 1,2,3,5,8\n* These months get one extra holiday each\n* Additional holidays per year = 5\n* Additional impact per year = 5 \u00d7 3.714 = 18.57 widgets\n* Weekly impact = 18.57/(10\u00d74) = 0.464 widgets per week\n* Final weekly production = 27.071 - 0.464 = 26.607 widgets\n\nThe answer is 26.607 widgets per week.", | |
| "error": 0.6780000000000008, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1503.0, | |
| "cost": 0.010185 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.857, | |
| "q3": 27.714, | |
| "reasoning": "Let me solve this step by step.\n\nQ1: Widgets per week\n* Weekdays (Mon-Fri): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (Sat-Sun): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\n* Total per week = 20 + 6 + 2 = 28 widgets\n\nQ2: Accounting for monthly holidays\n* Each month has 4 weeks = 28 days\n* Plus 1 holiday where no widgets are produced\n* In a week, probability of a holiday is 1/28\n* If a holiday falls on a:\n * Monday: Lose 4 + 2 = 6 widgets\n * Regular weekday (Tue-Fri): Lose 4 widgets\n * Weekend: Lose 3 widgets\n* Average loss per holiday = (6 + 4\u00d73 + 3\u00d72)/(7) = 4 widgets\n* Weekly impact of holidays = 4 \u00d7 (1/28) = 0.143 widgets\n* New weekly average = 28 - 0.143 = 27.857 widgets\n\nQ3: Accounting for fibonacci holidays\n* Fibonacci numbers in range 1-10: 1,1,2,3,5,8\n* Months with extra holidays: 1,2,3,5,8\n* These months get 3 holidays instead of 1\n* Out of 10 months, 5 have 2 extra holidays\n* Additional holidays = 10 holidays per year\n* Total holidays = 20 per year instead of 10\n* Weekly impact doubles: 0.286 widgets\n* Final weekly average = 28 - 0.286 = 27.714 widgets\n\nThe answers are:\nQ1: 28 widgets\nQ2: 27.857 widgets\nQ3: 27.714 widgets", | |
| "error": 2.570999999999998, | |
| "correctness": 0.3333333333333333, | |
| "total_tokens": 1621.0, | |
| "cost": 0.011247 | |
| }, | |
| { | |
| "q1": 28.0, | |
| "q2": 27.0, | |
| "q3": 26.0, | |
| "reasoning": "Let me solve this step by step.\n\nQ1:\n* Weekdays (5 days): 5 days \u00d7 4 widgets = 20 widgets\n* Weekends (2 days): 2 days \u00d7 3 widgets = 6 widgets\n* Monday bonus: 2 extra widgets\nTotal per week = 20 + 6 + 2 = 28 widgets\n\nQ2:\n* Each month has 4 weeks and 1 holiday\n* Since holidays are uniformly distributed, on average:\n * 5/7 chance holiday is on weekday (losing 4 widgets)\n * 2/7 chance holiday is on weekend (losing 3 widgets)\n * 1/7 chance holiday is on Monday (also losing 2 bonus widgets)\n* Average widgets lost per holiday:\n * (5/7 \u00d7 4) + (2/7 \u00d7 3) + (1/7 \u00d7 2) = 20/7 + 6/7 + 2/7 = 28/7 \u2248 4\n* Lost widgets per month = 28/7\n* Lost widgets per week = (28/7)/4 = 1\nTherefore, weekly average = 28 - 1 = 27 widgets\n\nQ3:\n* Fibonacci numbers from 1 to 10 are: 1, 2, 3, 5, 8\n* So months 1, 2, 3, 5, and 8 have 2 additional holidays\n* In 10 months: 10 regular holidays + (5 \u00d7 2) = 20 total holidays\n* Lost widgets per year = 20 \u00d7 (28/7) = 80\n* Lost widgets per week = 80/(10 \u00d7 4) = 2\nTherefore, weekly average = 28 - 2 = 26 widgets\n\nThe answer is 26 widgets per week.", | |
| "error": 0.0, | |
| "correctness": 1.0, | |
| "total_tokens": 1613.0, | |
| "cost": 0.011151000000000001 | |
| } | |
| ] | |
| } | |
| ] | |
| } | |
| ] |
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