Yuvraj Khanna, Raghav Rastogi, Dhruv Kumar, Peter Relan, Hung Tran
As artificial intelligence progresses at an unprecedented pace, Large Language Models (LLMs) are redefining the boundaries of what machines can achieve. With the recent release of OpenAI’s o1-preview and o1-mini models, and building on insights from Apple’s research in “GSM-Symbolic: Understanding the Limitations of Mathematical Reasoning in Large Language Models,” prompted us to explore a key question: How much of an AI model’s mathematical ability stems from genuine reasoning versus mere recall of training data. Apple’s research, summarized here, indicates that even when focused on grade school math word problems, the most sophisticated of models are not completely driven by “reasoning.”
Our investigation builds on a key insight from Apple as mentioned in the conclusion of their paper:
“Hence, the current limitations of these models are likely to be more pronounced in more challenging mathematical benchmarks. We believe further research is essential to develop AI models capable of formal reasoning, moving beyond pattern recognition to achieve more robust and generalizable problem-solving skills. This remains a critical challenge for the field as we strive to create systems with human-like cognitive abilities or general intelligence.“
To further research in this area, we focused on three critical factors:
- Using more challenging mathematical benchmarks, we focused on high school level math (MATH test set vs GSM8K dataset)
- Exploring a 1-shot prompt with extreme closeness to the test problem
- Exploring an alternate approach to few-shot inference by using a best of “n” strategy
Drawing from Apple’s foundational research, our goal is to explore how these challenges affect the reasoning abilities of LLMs and to identify the areas that require the most improvements. Although our preliminary results are still undergoing validation, we are sharing our initial findings through a blog. As we continue to refine our methodology, we plan to adjust the sample size of our test set and enhance our testing strategy to ensure more robust and reliable outcomes.
Analysis of our work
Our study builds upon Apple’s methodology but differs in several key aspects:
Dataset Selection
- We utilize the MATH dataset, which presents more challenging problems compared to GSM8K. The MATH dataset involves questions from prealgebra, algebra, counting and probability, geometry, intermediate algebra, precalculus, and number theory.
- This choice allows us to better examine model performance across varying difficulty levels.
Problem Variations
- Apple used the GSM8K (Grade School Mathematics) dataset to generate a varied dataset called GSM-Symbolic.
- In GSM-Symbolic, the original question is converted into a template wherein all of the elements of the question (mathematical and non-mathematical) are detected and varied. These elements can be names, numbers, recurring words in the question, etc.
- We believe that changing the context in addition to names and variables used will lead to a more pronounced change in the “reasoning” performance.
- In contrast, our variations maintain the same numerical values and final answers, ensuring consistency in problem solutions. Our focus is on examining how changes in language, variables, and context affect model performance.
- For example, a “dog walking” scenario is transformed into a “dishwasher” problem. We employed this approach to mitigate the challenges the model faces due to the increased complexity of the MATH dataset.
We employ three distinct prompting approaches:
0-shot:
- Basic solve step-by-step instruction
1-shot (extremely similar problem vs 8-shot by Apple of a generally similar problem) with
- Basic solve step-by-step instruction
- Solve step-by-step instruction with Chain-of-Thought (CoT) reasoning and explicitly says that the problem is similar to the 1-shot given
Why we take a different Prompting approach
- The reason we have extremely similar content in our 1-shot prompts is that we give the original question and solution as 1-shot when solving the variant question. Our approach also diverges from Apple’s 8-shot CoT methodology, as we believe fewer examples better reflect real-world usage patterns. This design choice stems from the observation that most practical applications typically utilize at most one example – or often none at all.
Evaluation strategy
- We use a best-of-5 (n) evaluation strategy for statistical significance.
- Apple’s paper does not clarify their approach.
We do not use the NoOp strategy yet
- We agree with Apple’s finding of GSM-NoOp evaluation and that adding irrelevant information to the question will result in a drop in “reasoning.
- As mentioned in Apple paper: “The introduction of GSM-NoOp exposes a critical flaw in LLMs’ ability to genuinely understand mathematical concepts and discern relevant information for problem-solving. Adding seemingly relevant but ultimately inconsequential information to the logical reasoning of the problem led to substantial performance drops of up to 65% across all state-of-the-art models.”
Experiment:
We selected 82 problems from the MATH dataset, each containing both a mathematical variable and a context. From these, we crafted 4 types of variations, each designed to challenge the models in different ways. These ranged from simple variable changes to complete question reformulations.
Explanation of variants (The numbers used, solution logic and final answer always remain the same)
- Variant 1: Only the variables used in the question are changed. Nothing else changed in the question including the language and numbers used, and the final answer. Example ‘$x=3$’ becomes ‘$y=3$’
- Variant 2: Only the non-mathematical context used to build the question is changed. Nothing else changed in the question including the variable and numbers used, and the final answer. Example ‘4 dogs’ becomes ‘4 books’
- Variant 3: The full question is reformulated to change the language of the question completely. The variable and numbers used, and the final answer does not change. For example ‘Harry has x dogs’ becomes ‘Oscar sold x books’
- Variant 4: The full question language is reformulated and the variables used are changed. Only the numbers used and the final answer does not change. For example ‘Harry has x dogs’ becomes ‘Oscar sold y books’
Elements in a question to vary
- Variable: An explicit mathematical unknown (e.g., x, y, θ), excluding constants like $\pi$.
- Context: A non-mathematical, real-world element in the problem (e.g., “projectile”, “dog walking business”). We chose problems containing both elements to ensure consistent variation testing.
Example of all Variations
Original Question
Denali and Nate work for a dog walking business and are paid for each dog they walk. Denali is responsible for $16$ dogs and Nate is responsible for $12$ dogs. Under the company’s new policy, they will be assigned or unassigned new dogs in groups of $x$ dogs. The ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4x$ more dogs and Nate stayed at $12$ dogs or if $x$ of Nate’s dogs were reassigned to Denali. Find $x$ if $x\neq0$. Here $x$ is the variable used and “dog walking”, “dogs” and “company” are context.
Original Solution
Rewriting the sentence “the ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4x$ more dogs and Nate stayed at $12$ dogs or if $x$ of Nate’s dogs were reassigned to Denali” as an equation, we have
$$
\frac{16+4x}{12}=\frac{16+x}{12 – x}.
$$
Clearing denominators,
$$
\begin{align}
(16+4x)(12 – x) &= (16 + x)(12) \quad \Rightarrow \\
192 – 16x + 48x – 4x^2 &= 192 + 12x \quad \Rightarrow \\
32x – 4x^2 &= 12x \quad \Rightarrow \\
0 &= 4x^2 – 20x \quad \Rightarrow \\
0 &= 4x(x – 5).
\end{align}
$$
Because $x$ cannot be $0$, $x = \boxed{5}$.
Variation 1 Question
Denali and Nate work for a dog walking business and are paid for each dog they walk. Denali is responsible for $16$ dogs and Nate is responsible for $12$ dogs. Under the company’s new policy, they will be assigned or unassigned new dogs in groups of $y$ dogs. The ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4y$ more dogs and Nate stayed at $12$ dogs or if $y$ of Nate’s dogs were reassigned to Denali. Find $y$ if $y\neq0$.
Variation 1 Solution
Rewriting the sentence “the ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4y$ more dogs and Nate stayed at $12$ dogs or if $y$ of Nate’s dogs were reassigned to Denali” as an equation, we have
$$
\frac{16+4y}{12}=\frac{16+y}{12 – y}.
$$
Clearing denominators,
$$
\begin{align}
(16+4y)(12 – y) &= (16 + y)(12) \quad \Rightarrow \\
192 – 16y + 48y – 4y^2 &= 192 + 12y \quad \Rightarrow \\
32y – 4y^2 &= 12y \quad \Rightarrow \\
0 &= 4y^2 – 20y \quad \Rightarrow \\
0 &= 4y(y – 5).
\end{align}
$$
Because $y$ cannot be $0$, $y = \boxed{5}$.
Variation 2 Question
Denali and Nate work in a library and are responsible for cataloging a certain number of books. Denali is responsible for $16$ books and Nate is responsible for $12$ books. Under the library’s new policy, they will be assigned or unassigned new books in groups of $x$ books. The ratio of Denali’s work to Nate’s work would be the same if Denali started cataloging $4x$ more books and Nate stayed at $12$ books or if $x$ of Nate’s books were reassigned to Denali. Find $x$ if $x\neq0$.
Variation 2 Solution
Rewriting the sentence “the ratio of Denali’s work to Nate’s work would be the same if Denali started cataloging $4x$ more books and Nate stayed at $12$ books or if $x$ of Nate’s books were reassigned to Denali” as an equation, we have
$$
\frac{16+4x}{12}=\frac{16+x}{12 – x}.
$$
Clearing denominators,
$$
\begin{align}
(16+4x)(12 – x) &= (16 + x)(12) \quad \Rightarrow \\
192 – 16x + 48x – 4x^2 &= 192 + 12x \quad \Rightarrow \\
32x – 4x^2 &= 12x \quad \Rightarrow \\
0 &= 4x^2 – 20x \quad \Rightarrow \\
0 &= 4x(x – 5).
\end{align}
$$
Because $x$ cannot be $0$, $x = \boxed{5}$.
Variation 3 Question
Farmer Alice and Farmer Bob each grow a different type of crop. Alice currently has $16$ units of her crop, and Bob has $12$ units of his. Under certain agricultural changes, the ratio of Alice’s crop to Bob’s crop would remain the same if Alice were to harvest $4$ times $x$ more units of her crop and Bob’s crop yield stayed the same, or if Bob were to give away $x$ units of his crop to Alice. Find the value of $x$ if $x$ is not equal to zero.
Variation 3 Solution
Rewriting the sentence “the ratio of Alice’s crop to Bob’s crop would remain the same if Alice were to harvest $4$ times $x$ more units of her crop and Bob’s crop yield stayed the same, or if Bob were to give away $x$ units of his crop to Alice” as an equation, we have
$$
\frac{16+4x}{12}=\frac{16+x}{12 – x}.
$$
Clearing denominators,
$$
\begin{align}
(16+4x)(12 – x) &= (16 + x)(12) \quad \Rightarrow \\
192 – 16x + 48x – 4x^2 &= 192 + 12x \quad \Rightarrow \\
32x – 4x^2 &= 12x \quad \Rightarrow \\
0 &= 4x^2 – 20x \quad \Rightarrow \\
0 &= 4x(x – 5).
\end{align}
$$
Because $x$ cannot be $0$, $x = \boxed{5}$.
Here we show the different language elements of a question explored as Names, Context, and Variables used in the question. Here is an example for variation 4. In this case we determine the context to be the names “Denali” and “Nate”, and any text used around “dogs”. We change the names “Denali” with “Alice”, “Nate” with “Bob”, and any idea related to “dogs walking” with “book collection”
| Original question: | Variation 4 Question: |
|---|---|
| Alice and Bob are avid book collectors. Alice currently has 16 rare first editions, while Bob has 12 signed copies. They have agreed to trade books in groups of $y$ books , where $y$ is a positive integer. The ratio of Alice’s first editions to Bob’s signed copies would remain the same if Alice acquired $4y$ more first editions and Bob kept his collection unchanged, or if Bob gave away $y$ of his signed copies to Alice. Find the value of $y$ if $y$$≠0$. |
Variation 4 Solution
Rewriting the sentence “The ratio of Alice’s first editions to Bob’s signed copies would remain the same if Alice acquired $4y$ more first editions and Bob kept his collection unchanged, or if Bob gave away $y$ of his signed copies to Alice” as an equation, we have:
$$
\frac{16 + 4y}{12} = \frac{16 + y}{12 – y}
$$
Clearing denominators, we get:
$$
\begin{align}
(16 + 4y)(12 – y) &= (16 + y)(12) \\
192 – 16y + 48y – 4y^2 &= 192 + 12y \\
32y – 4y^2 &= 12y \\
0 &= 4y^2 – 20y \\
0 &= 4y(y – 5)
\end{align}
$$
Since $y$ cannot be $0$, we have $y = \boxed{5}$.
Results
For example, here is the best-of-5 evaluation on a problem from the MATH test set (index = 9287) which is level 5 (highest difficulty) algebra problem. Not only do we include all models that Apple had in its evaluation, but we also include some open-source models for completeness.
| Model | Original | Variation 1 | Variation 2 | Variation 3 | Variation 4 |
|---|---|---|---|---|---|
| o1-mini | 100% | 100% | 100% | 0% | 0% |
| o1-preview | 100% | 100% | 100% | 100% | 60% |
| GPT-4o-mini | 100% | 100% | 100% | 80% | 20% |
| GPT-4o | 100% | 100% | 100% | 0% | 20% |
| GPT-4 | 0% | 0% | 0% | 0% | 0% |
| Llama3-8B | 0% | 0% | 0% | 0% | 0% |
| Llama3-70B | 40% | 0% | 0% | 0% | 0% |
| Llama 3.1-8B | 100% | 0% | 0% | 0% | 0% |
| Llama 3.1-70B | 40% | 60% | 100% | 100% | 0% |
| Llama 3.1-405B | 100% | 80% | 0% | 0% | 0% |
| Deepseek-coder-v2 | 60% | 80% | 40% | 20% | 0% |
| Deepseek-chat-v2 | 40% | 40% | 0% | 60% | 0% |
| Deepseek-v2.5 | 60% | 80% | 40% | 0% | 0% |
| Phi-3.5-mini | 40% | 0% | 40% | 60% | 0% |
| Phi3-medium-128k | 0% | 0% | 20% | 0% | 0% |
| Mistral-Nemo-12B | 20% | 0% | 0% | 0% | 0% |
Look at Appendix for examples of correct and incorrect solutions
Analysis
We tested these variations across a range of models, including GPT-4, GPT-4o, GPT-4o-mini, o1-mini, and o1-preview. We also experimented with different prompt types to gauge the impact of in-context learning on model performance.
Our findings offer a nuanced understanding of how different AI models tackle mathematical problem-solving. Consistent with Apple’s research, we observed a steady decline in accuracy as problems shifted from original questions to more challenging variations. This indicates that models struggle to maintain their performance when confronted with tasks that diverge from their training data. Additionally, we found that performance drops were more pronounced not only with harder mathematical benchmarks but also with problem variations. For instance, the o1-mini model’s accuracy fell from 93.66% on original questions to 88.54% on Variation 4, while the o1-preview model experienced a similar decline, dropping from 91.22% to 82.93%.
| Model | Apple Research (8-shot CoT on GSM8k) | MathGPT Research (0-shot best of 5 on MATH) | ||
|---|---|---|---|---|
| Original Accuracy | Drop | Original Accuracy | Drop | |
| GPT 4o | 95.20% | -0.3% | 78.77% | -5.44%% |
| GPT 4o-mini | 94.20% | -2.4% | 78.42% | -6.58% |
| o1-mini | 95.10% | -0.6% | 93.66% | -5.12% |
| o1-preview | 94.90% | -2.2% | 91.22% | -8.29% |
| Llama 3-8B | 76.00% | -1.3% | 25.50% | 2.25%* |
| Phi 3.5 mini | 84.90% | -2.8% | 42.93% | -5.13% |
Obviously, results are worse when the test is on MATH (MATH dataset is a more advanced mathematical testing benchmark than GSM8k)
*Why did llama3-8b increase? As observed in the Apple research, Llama3-8B has an equal distribution on both sides of the accuracy spectrum. So testing on multiple samples of variation 4 will help us understand the variability of the accuracy of models like llama3-8b better. We plan to explore this further in our next blog.
| Model | Apple Research (8-shot CoT on GSM8k) | MathGPT (1-shot extreme similarity best of 5 on MATH) | ||
|---|---|---|---|---|
| Original Accuracy | Drop | Original Accuracy | Increase | |
| GPT 4o | 95.20% | -0.3% | 78.77% | 9.38% |
| GPT 4o-mini | 94.20% | -2.4% | 78.42% | 3.16% |
| o1-mini | 95.10% | -0.6% | 93.66% | 1.95% |
| o1-preview | 94.90% | -2.2% | 91.22% | 3.66% |
| Llama 3-8B | 76.00% | -1.3% | 25.50% | 46.10% |
| Phi 3.5 mini | 84.90% | -2.8% | 42.93% | 24.14% |
How do we achieve an increase in accuracy?
We noticed that by including a very similar problem and its solution in the prompt, we can significantly improve the model’s accuracy. The model leverages this in-context example to enhance its reasoning process, effectively using the provided question and solution as a few-shot learning opportunity. This approach requires human intelligence to select an extremely similar problem that guides the model towards a correct solution, demonstrating the importance of carefully crafted prompts in optimizing performance.
Our performance analysis across various language models reveals intriguing patterns in their mathematical problem-solving capabilities. While Apple’s GSM-Symbolic dataset showed minimal performance degradation in OpenAI models (ranging from -0.3% to -2.4%), we observed significant variations in outcomes based on the solution methods employed. Notably, introducing Chain-of-Thought (CoT) with extremely similar 1-shot prompting improved performance across all models. However, when models were provided with the original question as a one-shot example along with CoT instructions, their performance improved substantially. This improvement likely stems from the models leveraging the provided example to mimic the solution process, rather than engaging in genuine reasoning. These findings align with observations from prior studies, suggesting that large language models struggle with deeper problem-solving challenges that cannot be resolved through few-shot prompting or fine-tuning alone.
Prompt types and accuracy
Only the o1 series of models (o1-mini and o1-preview) are Reasoning models. All other models evaluated are Non-Reasoning models.
Basic solve Prompt
Non-Reasoning models: “Solve the question step by step” + question
Reasoning models: only question
Original shot solve Prompt
Non-Reasoning models: Original question as 1-shot + “Solve the question step by step” + question
Reasoning models: Original question as 1-shot + question
Similar instruction CoT with Original shot solve Prompt
Non Reasoning models: Original shot + “The question given above is similar. Math and logic used above are similar. Solve the question step by step” + question
Reasoning models: Original shot + “The question given above is similar. Math and logic used above are similar.” + question
| Prompt type | Basic solve | Original shot solve | Similar instruction with Original shot solve | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Model used | GPT-4o | GPT-4o-mini | o1-mini | o1-preview | GPT-4o | GPT-4o-mini | o1-mini | o1-preview | GPT-4o | GPT-4o-mini | o1-mini | o1-preview |
| Original Question | 78.77% | 78.42% | 93.66% | 91.22% | ||||||||
| Variation 1 Only change variables | 79.01% | 77.63% | 93.20% | 87.32% | 86.91% | 77.89% | 97.07% | 97.56% | 91.36% | 81.84% | 97.56% | 95.61% |
| Variation 2 Only change context | 72.84% | 73.68% | 85.61% | 80.49% | 82.72% | 80.00% | 91.46% | 93.66% | 88.64% | 80.26% | 93.66% | 92.20% |
| Variation 3 Reformulate question, variable same | 72.84% | 74.47% | 89.51% | 88.05% | 84.69% | 79.21% | 94.63% | 96.34% | 85.19% | 80.00% | 96.34% | 95.61% |
| Variation 4 Reformulate question, change variable | 73.33% | 71.84% | 88.54% | 82.93% | 84.20% | 76.05% | 95.85% | 95.61% | 88.15% | 81.58% | 95.61% | 94.88% |
Summary
Our findings highlight a critical limitation of current large language models (LLMs): their performance is heavily reliant on the similarity between test problems and training data. This underscores the urgent need to develop models capable of genuine reasoning, rather than mere pattern recognition and recall.
While our study provides valuable insights into the mathematical reasoning capabilities of LLMs, it is important to acknowledge certain limitations. Our analysis, based on a sample of 82 carefully selected problems, has a relatively narrow scope. Moreover, our focus on open-source models leaves room for broader exploration across a wider range of LLMs. It’s also worth noting the methodological differences between our approach and Apple’s GSM-Symbolic research. Apple employed an 8-shot Chain of Thought (CoT) prompting strategy, guiding models through step-by-step problem-solving. In contrast, we adopted a “best-of-five” approach to generate multiple solutions for improving accuracy and mitigating variability. We believe our method is more reflective of real-world scenarios, where fewer-shot examples are often more practical than extensive in-context demonstrations.
Future Work
Looking ahead, we aim to collaborate with Apple research and other researchers to leverage the strengths of diverse methodologies. As part of this effort, we are closely examining the sample generation process used to assess accuracy. While Apple generated 50 samples per question, our approach involved testing 4 variations of each question across 5 trials. In future research, we plan to expand the variety of sample variations per question to capture the distribution of model accuracy to better identify subtle nuances in performance across different contexts.
Additionally, we are independently working on developing contamination resistant datasets to ensure robustness of evaluation results. We have collected a very large proprietary set of real-world problems and human-verified solutions spanning all math subjects, from grade school to college, giving us a unique benchmark for testing models on novel scenarios without relying on publicly available data. Our goal is to tackle the key challenge of distinguishing true reasoning from mere memorization of training data. Since many models are trained on open-source datasets, there is an inherent risk of unintentional overlap with publicly available problems. Recent research, such as the GSM1K paper, has highlighted the impact of data contamination on model performance. The MATH dataset, for instance, includes content from well-known sources like AMC 10, AMC 12, and AIME, underscoring this issue. We look forward to working with prominent AI researchers to validate our approach to contamination resistant datasets.
As we push the boundaries of AI, enhancing large language models’ (LLMs) reasoning capabilities will remain a primary focus. To address these challenges, we are actively exploring both pre-training strategies and continuous fine-tuning methods as well as leveraging agentic flows with tool uses to enhance performance on complex reasoning tasks. Our work with challenging datasets, such as MATH, has revealed that models still struggle with fewer-shot examples, exposing limitations in their reasoning depth. We are actively investigating and deploying techniques like CoT—which we have found to improve model performance—to enhance reasoning abilities.
Ultimately, our goal is to develop AI systems that transcend mere recall, advancing towards reasoning mechanisms that exhibit creativity, insight, and adaptive problem-solving skills. By refining our evaluation methods and pioneering new training techniques, we aim to bridge the gap between current models and the next generation of AI systems capable of robust, human-like reasoning.
Appendix
(Examples of Solutions)
Original (No change)
Problem
Denali and Nate work for a dog walking business and are paid for each dog they walk. Denali is responsible for $16$ dogs and Nate is responsible for $12$ dogs. Under the company’s new policy, they will be assigned or unassigned new dogs in groups of $x$ dogs. The ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4x$ more dogs and Nate stayed at $12$ dogs or if $x$ of Nate’s dogs were reassigned to Denali. Find $x$ if $x\neq0$. Here $x$ is the variable used and “dog walking”, “dogs” and “company” are context.
Ground Truth Solution
Rewriting the sentence “the ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4x$ more dogs and Nate stayed at $12$ dogs or if $x$ of Nate’s dogs were reassigned to Denali” as an equation, we have
$$\frac{16+4x}{12}=\frac{16+x}{12-x}.$$
Clearing denominators,
$$
\begin{align}
(16+4x)(12-x)&=(16+x)(12)\quad \Rightarrow\\
192-16x+48x-4x^2&=192+12x\quad \Rightarrow\\
32x-4x^2&=12x\quad \Rightarrow\\
0&=4x^2-20x\quad \Rightarrow\\
0&=4x(x-5).
\end{align}
$$
Because $x$ cannot be $0$, $x=\boxed{5}$.
Correct Solutions
o1-mini
Let’s analyze the problem step-by-step. Given:
- Denali is initially responsible for 16 dogs.
- Nate is initially responsible for 12 dogs.
- The ratio of their pay is initially $\frac{16}{12} = \frac{4}{3}$.
Scenarios Under New Policy:
- Scenario A: Denali starts walking 4x more dogs, making her total $16 + 4x$ dogs. Nate remains at 12 dogs. The new ratio of their pay would be: $$\frac{16 + 4x}{12}$$
- Scenario B: x of Nate’s dogs are reassigned to Denali. This means Denali now walks $16 + x$ dogs and Nate walks $12 – x$ dogs. The new ratio of their pay would be: $$\frac{16 + x}{12 – x}$$
Setting the Ratios Equal:
For the ratio to remain consistent across both scenarios, set the two expressions equal to each other:
$$\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{16 + x}{12 – x}$}$$
Solving for x:
- Cross-multiply to eliminate the fractions: $$(16 + 4x)(12 – x) = 12(16 + x)$$
- Expand both sides: $$192 + 32x – 4x^2 = 192 + 12x$$
- Simplify the equation:
$$
32x – 4x^2 = 12x
20x – 4x^2 = 0
4x(5 – x) = 0
$$ - Solve for $x$:$$
x = 0 \quad \text{or} \quad x = 5.$$
Since $x \neq 0$, the solution is:
$$x = 5
$$
Conclusion: The value of $x$ is 5.
$\colorbox{green}{Answer: 5}$
o1-preview
To solve this problem, we’ll set up the equations based on the given scenarios. Let’s let $k$ represent the pay per dog (since they are paid per dog, the exact amount doesn’t affect the ratio).
First scenario: Denali walks $16 + 4x$ dogs. Nate walks $12$ dogs. The ratio of their pays is: $$\frac{16 + 4x}{12}$$
Second scenario: Denali walks $16 + x$ dogs (since $x$ of Nate’s dogs are reassigned to Denali). Nate walks $12 – x$ dogs. The ratio of their pays is: $$\frac{16 + x}{12 – x}$$
According to the problem, these two ratios are equal:
$$
\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{16 + x}{12 – x}$}
$$
Cross-multiplying:
$$(16 + 4x)(12 – x) = 12(16 + x)$$
Expanding both sides:
$$
\begin{align}
(16)(12) – (16)(x) + (4x)(12) – (4x)(x) &= 12 \times 16 + 12x \\
192 – 16x + 48x – 4x^2 &= 192 + 12x \\
192 + 32x – 4x^2 &= 192 + 12x
\end{align}
$$
Subtract $192$ from both sides and simplify:
$$
\begin{align}
32x – 4x^2 &= 12x \\
32x – 4x^2 – 12x &= 0 \\
20x – 4x^2 &= 0
\end{align}
$$
Divide by $4$:
$$
\begin{align}
5x – x^2 &= 0 \\
x^2 – 5x &= 0 \\
x(x – 5) &= 0
\end{align}
$$
This gives $x = 0$ or $x = 5$. Since $x \neq 0 $, we conclude $x = 5 $.
$\colorbox{green}{Answer: 5}$
GPT-4o
To solve the problem, we need to set up equations based on the given conditions and solve for $x$. First, let’s denote Denali’s initial number of dogs as $D$ and Nate’s initial number of dogs as $N$:
$$
\begin{align}
D &= 16 \\
N &= 12
\end{align}
$$
Condition 1: Denali walks $4x$ more dogs
If Denali walks $4x$ more dogs, the number of dogs Denali walks becomes: $$D + 4x = 16 + 4x$$
Nate’s number of dogs remains the same: $$N = 12$$
Condition 2: $x$ of Nate’s dogs are reassigned to Denali
If $x$ of Nate’s dogs are reassigned to Denali, the number of dogs Denali walks becomes: $$D + x = 16 + x$$ The number of dogs Nate walks becomes: $$N – x = 12 – x$$
Setting up the ratio conditions
The problem states that the ratio of Denali’s pay to Nate’s pay remains the same under both conditions. Since the pay is proportional to the number of dogs walked, we can set up the following ratio equality:
$$
\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{16 + x}{12 – x}$}
$$
Solving the equation
Cross-multiplying to eliminate the fractions, we get: $$(16 + 4x)(12 – x) = 12(16 + x)$$
Expanding both sides:
$$192 – 16x + 48x – 4x^2 = 192 + 12x$$
Combining like terms:
$$192 + 32x – 4x^2 = 192 + 12x$$
Subtracting 192 from both sides:
$$32x – 4x^2 = 12x$$
Rearranging the equation:
$$
\begin{align}
32x – 12x – 4x^2 &= 0 \\
20x – 4x^2 &= 0
\end{align}
$$
Factoring out $4x$: $$4x(5 – x) = 0$$
Setting each factor to zero: $$4x = 0 \quad \text{or} \quad 5 – x = 0$$
Since $x \neq 0$:
$$
\begin{align}
5 – x &= 0 \\
x &= 5
\end{align}
$$
Thus, the value of $x$ is: $$\boxed{5}$$
$\colorbox{green}{Answer: 5}$
GPT-4o-mini
Let the pay per dog be $p$.
Denali’s initial pay for walking 16 dogs is:
$$
\text{Denali’s pay} = 16p
$$
Nate’s pay for walking 12 dogs is:
$$
\text{Nate’s pay} = 12p
$$
According to the problem, we have two scenarios to consider:
- Denali walks $4x$ more dogs, while Nate continues to walk 12 dogs. Denali’s new pay becomes:
$$
\text{Denali’s new pay} = (16 + 4x)p = (16 + 4x)p
$$
Nate’s pay remains:
$$
\text{Nate’s pay} = 12p
$$
The ratio of Denali’s pay to Nate’s pay in this scenario is:
$$
\frac{(16 + 4x)p}{12p} = \frac{16 + 4x}{12}
$$
- If $x$ of Nate’s dogs are reassigned to Denali, then Denali will have:
$$
16 + x \text{ dogs}
$$
and Nate will have:
$$
12 – x \text{ dogs}
$$
Denali’s pay in this case is:
$$
\text{Denali’s pay} = (16 + x)p
$$
Nate’s pay becomes:
$$
\text{Nate’s pay} = (12 – x)p
$$
The ratio of Denali’s pay to Nate’s pay in this scenario is:
$$
\frac{(16 + x)p}{(12 – x)p} = \frac{16 + x}{12 – x}
$$
Since the ratios must be equal, we set them equal to each other:
$$
\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{16 + x}{12 – x}$}
$$
Cross-multiplying gives:
$$
(16 + 4x)(12 – x) = (16 + x)(12)
$$
Expanding both sides:
$$
192 – 16x + 48x – 4x^2 = 192 + 12x
$$
Combining like terms:
$$
192 + 32x – 4x^2 = 192 + 12x
$$
Subtracting $192 + 12x$ from both sides:
$$
32x – 12x – 4x^2 = 0
$$
$$
20x – 4x^2 = 0
$$
Factoring out $4x$:
$$
4x(5 – x) = 0
$$
Setting each factor to zero gives:
$$
4x = 0 \quad \text{or} \quad 5 – x = 0
$$
Since $x \neq 0$, we have:
$$
5 – x = 0 \implies x = 5
$$
Thus, the value of $x$ is:
$$\colorbox{green}{Answer: 5}$$
Variation 1 (Only change variable used but keep context used the same)
Problem
Denali and Nate work for a dog walking business and are paid for each dog they walk. Denali is responsible for $16$ dogs and Nate is responsible for $12$ dogs. Under the company’s new policy, they will be assigned or unassigned new dogs in groups of $y$ dogs. The ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4y$ more dogs and Nate stayed at $12$ dogs or if $y$ of Nate’s dogs were reassigned to Denali. Find $y$ if $y\neq0$.
Ground Truth Solution
Rewriting the sentence “the ratio of Denali’s pay to Nate’s pay would be the same if Denali started walking $4y$ more dogs and Nate stayed at $12$ dogs or if $y$ of Nate’s dogs were reassigned to Denali” as an equation, we have $$\frac{16+4y}{12}=\frac{16+y}{12-y}.$$
Clearing denominators,
$$
\begin{align}
(16+4y)(12-y)&=(16+y)(12)\quad \Rightarrow\\
192-16y+48y-4y^2&=192+12y\quad \Rightarrow\\
32y-4y^2&=12y\quad \Rightarrow\\
0&=4y^2-20y\quad \Rightarrow\\
0&=4y(y-5).
\end{align}
$$
Because $y$ cannot be $0$, $y=\boxed{5}$.
Variation 2 (Only change in context but keep variable used same)
Problem
Denali and Nate work in a library and are responsible for cataloging a certain number of books. Denali is responsible for 16 books and Nate is responsible for 12 books. Under the library’s new policy, they will be assigned or unassigned new books in groups of $x$ books. The ratio of Denali’s work to Nate’s work would be the same if Denali started cataloging $4x$ more books and Nate stayed at 12 books or if $x$ of Nate’s books were reassigned to Denali. Find $x$ if $x≠0$.
Ground Truth Solution
Rewriting the sentence “the ratio of Denali’s work to Nate’s work would be the same if Denali started cataloging $4x$ more books and Nate stayed at 12 books or if $x$ of Nate’s books were reassigned to Denali” as an equation, we have $$\frac{16+4x}{12}=\frac{16+x}{12-x}.$$
Clearing denominators,
$$
\begin{align}
(16+4x)(12-x)&=(16+x)(12)\quad \Rightarrow\\
192-16x+48x-4x^2&=192+12x\quad \Rightarrow\\
32x-4x^2&=12x\quad \Rightarrow\\
0&=4x^2-20x\quad \Rightarrow\\
0&=4x(x-5).
\end{align}
$$
Because $x$ cannot be $0$, $x=\boxed{5}$.
Variation 3 (Reformulate question language and all context but keep variable used same)
Problem
Farmer Alice and Farmer Bob each grow a different type of crop. Alice currently has 16 units of her crop, and Bob has 12 units of his. Under certain agricultural changes, the ratio of Alice’s crop to Bob’s crop would remain the same if Alice were to harvest 4 times $x$ more units of her crop and Bob’s crop yield stayed the same, or if Bob were to give away $x$ units of his crop to Alice. Find the value of $x$ if $x$ is not equal to zero.
Ground Truth Solution
Rewriting the sentence “the ratio of Alice’s crop to Bob’s crop would remain the same if Alice were to harvest 4 times $x$ more units of her crop and Bob’s crop yield stayed the same, or if Bob were to give away $x$ units of his crop to Alice” as an equation, we have $$\frac{16+4x}{12}=\frac{16+x}{12-x}.$$
Clearing denominators,
$$
\begin{align}
(16+4x)(12-x)&=(16+x)(12)\quad \Rightarrow\\
192-16x+48x-4x^2&=192+12x\quad \Rightarrow\\
32x-4x^2&=12x\quad \Rightarrow\\
0&=4x^2-20x\quad \Rightarrow\\
0&=4x(x-5).
\end{align}
$$
Because $x$ cannot be $0$, $x=\boxed{5}$.
Incorrect Solutions
o1-mini
Let’s analyze the problem step by step.
Initial Conditions:
- Alice: 16 units of her crop.
- Bob: 12 units of his crop.
- Initial Ratio (Alice : Bob): $\frac{16}{12} = \frac{4}{3}$.
Scenario 1: Alice Harvests 4x More Units
- New Amount for Alice: $16 + 4x$.
- Bob’s Amount Remains: 12 units.
- Ratio After Harvest: $\frac{16 + 4x}{12}$.
To maintain the original ratio:$$
\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{4}{3}$}
$$
$$
\begin{align}
3(16 + 4x) &= 4 \times 12 \\
48 + 12x &= 48 \\
12x &= 0 \\
x &= 0
\end{align}
$$
Scenario 2: Bob Gives Away x Units to Alice
- New Amount for Alice: $16 + x$.
- New Amount for Bob: $12 – x$.
- Ratio After Transfer: $\frac{16 + x}{12 – x}$.
To maintain the original ratio:$$
\colorbox{yellow}{$\frac{16 + x}{12 – x} = \frac{4}{3}$} \\
$$
$$\begin{align}
3(16 + x) &= 4(12 – x)\\
48 + 3x &= 48 – 4x\\
7x &= 0\\
x &= 0 \\
\end{align}$$
Conclusion:
Both scenarios require $x = 0$ to maintain the original ratio of $\frac{4}{3}$. However, the problem specifies that $x \neq 0 $. Therefore, there is no non-zero value of $x$ that satisfies the given conditions.
Answer:There is no valid nonzero solution for $x$; no such $x$ satisfies the given conditions.
GPT-4o-mini
Let’s denote the current amounts of crops that Alice and Bob have as follows:
- Alice’s crop: $A = 16$
- Bob’s crop: $B = 12$
We are given two scenarios:
- If Alice harvests $4x$ more units of her crop, her new amount of crop will be $A + 4x$ and Bob’s amount remains $B$.
- If Bob gives away $x$ units of his crop to Alice, Alice’s new amount will be $A + x$ and Bob’s new amount will be $B – x$.
In both scenarios, the ratio of Alice’s crop to Bob’s crop remains the same. The current ratio is:
$$
\frac{A}{B} = \frac{16}{12} = \frac{4}{3}
$$
Scenario 1: Alice harvests $4x$
The new ratio becomes:
$$
\frac{A + 4x}{B} = \frac{16 + 4x}{12}
$$
Setting this equal to the original ratio:
$$
\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{4}{3}$}
$$
Cross-multiplying gives:
$$
3(16 + 4x) = 4 \cdot 12
$$
Calculating the right side:
$$
3(16 + 4x) = 48
$$
Expanding the left side:
$$
48 + 12x = 48
$$
Subtracting $48$ from both sides:
$$
12x = 0
$$
This implies $x = 0$, which we know is not allowed. Therefore, we will check the second scenario.
Scenario 2: Bob gives away $x$
The new ratio becomes:
$$
\frac{A + x}{B – x} = \frac{16 + x}{12 – x}
$$
Setting this equal to the original ratio:
$$
\colorbox{yellow}{$\frac{16 + x}{12 – x} = \frac{4}{3}$}
$$
Cross-multiplying gives:
$$
3(16 + x) = 4(12 – x)
$$
Expanding both sides:
$$
48 + 3x = 48 – 4x
$$
Adding $4x$ to both sides:
$$
48 + 3x + 4x = 48
$$
This simplifies to:
$$
48 + 7x = 48
$$
Subtracting 48 from both sides:
$$
7x = 0
$$
Again, this implies $x = 0$, which is not allowed.
Re-evaluating the equations
Let’s go back to the first scenario and check for any mistakes.
From the first scenario, we had:
$$
\colorbox{yellow}{$3(16 + 4x) = 48$}
$$
This simplifies to:
$$
48 + 12x = 48
$$
This means $12x = 0$, which is consistent.
Now, let’s check the second scenario again:
From the second scenario, we had:
$$
3(16 + x) = 4(12 – x)
$$
This expands to:
$$
48 + 3x = 48 – 4x
$$
Adding $4x$ to both sides gives:
$$
48 + 7x = 48
$$
This leads to $7x = 0$, which is consistent.
Conclusion
Both scenarios lead to $x = 0$, which contradicts the problem statement that $x$ is not equal to zero. Let’s try to find a non-zero solution by re-evaluating the ratios.
We can set up the equations again:
- From the first scenario:$$
3(16 + 4x) = 4 \cdot 12
$$ - From the second scenario:$$
3(16 + x) = 4(12 – x)
$$
Both lead to $x = 0$.
After checking the calculations, it seems that the problem may not have a valid solution under the constraints given.
However, if we assume $x$ can be any non-zero value, we can set $x = 1$ for simplicity, but that does not satisfy the original conditions.
Thus, we conclude that the problem as stated does not yield a valid non-zero solution for $x$.
$\colorbox{red}{Answer: 0}$
GPT-4o
To solve the problem, we need to set up two scenarios and use the given information to find the value of $x$.
Scenario 1: Alice harvests 4 times $x$ more units of her crop
- Alice’s new crop amount: $16 + 4x$
- Bob’s crop amount remains the same: $12$
The ratio of Alice’s crop to Bob’s crop remains the same:
$$
\frac{16 + 4x}{12} = \frac{16}{12}
$$
Simplify the ratio on the right-hand side:
$$
\frac{16}{12} = \frac{4}{3}
$$
So, we have:
$$
\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{4}{3}$}
$$
Cross-multiply to solve for $x$:
$$
3(16 + 4x) = 4 \cdot 12
$$
$$
48 + 12x = 48
$$
$$
12x = 0
$$
$$
x = 0
$$
Since $x \neq 0$, this scenario does not provide a valid solution.
Scenario 2: Bob gives away $x$ units of his crop to Alice
- Alice’s new crop amount: $16 + x$
- Bob’s new crop amount: $12 – x$
The ratio of Alice’s crop to Bob’s crop remains the same:
$$
\frac{16 + x}{12 – x} = \frac{16}{12}
$$
Simplify the ratio on the right-hand side:
$$
\frac{16}{12} = \frac{4}{3}
$$
So, we have:
$$
\colorbox{yellow}{$\frac{16 + x}{12 – x} = \frac{4}{3}$}
$$
Cross-multiply to solve for $x$:
$$
\begin{align}
3(16 + x) &= 4(12 – x) \\
48 + 3x &= 48 – 4x \\
3x + 4x &= 48 – 48 \\
7x &= 0 \\
x &= 0
\end{align}
$$
Since $x \neq 0$, this scenario also does not provide a valid solution.
Re-evaluation
Since both scenarios lead to $x = 0$ which contradicts the condition $x \neq 0$, we need to re-evaluate the problem setup. However, based on the given conditions and the algebraic steps, the problem seems to have no valid solution for $x \neq 0$.
$\colorbox{red}{Answer: No valid solution}$
Variation 4 (Reformulate question language and all context and variable used)
Problem
Alice and Bob are avid book collectors. Alice currently has $16$ rare first editions, while Bob has $12$ signed copies. They have agreed to trade books in groups of $y$ books, where $y$ is a positive integer. The ratio of Alice’s first editions to Bob’s signed copies would remain the same if Alice acquired $4y$ more first editions and Bob kept his collection unchanged, or if Bob gave away $y$ of his signed copies to Alice. Find the value of $y$ if $y$ is not equal to zero.
Ground Truth Solution
Rewriting the sentence “The ratio of Alice’s first editions to Bob’s signed copies would remain the same if Alice acquired $4y$ more first editions and Bob kept his collection unchanged, or if Bob gave away $y$ of his signed copies to Alice” as an equation, we have:
$$
\frac{16 + 4y}{12} = \frac{16 + y}{12 – y}
$$
Clearing denominators, we get:
$$
\begin{align}
(16 + 4y)(12 – y) &= (16 + y)(12) \\
192 – 16y + 48y – 4y^2 &= 192 + 12y \\
32y – 4y^2 &= 12y \\
0 &= 4y^2 – 20y \\
0 &= 4y(y – 5)
\end{align}
$$
Since $y$ cannot be $0$, we have $y = \boxed{5}$.
Incorrect Solutions
o1-mini
Let’s break down the problem step by step to find the value of $y$.
Initial Conditions:
- Alice has 16 rare first editions.
- Bob has 12 signed copies.
- The original ratio of Alice’s first editions to Bob’s signed copies is $\frac{16}{12} = \frac{4}{3}$.
Scenario 1:
- Alice acquires $4y$ more first editions.
- Bob‘s collection remains unchanged.
After this change:
- Alice has $16 + 4y$ first editions.
- Bob still has 12 signed copies.
The new ratio should remain the same:
$$
\colorbox{yellow}{$\frac{16 + 4y}{12} = \frac{4}{3}$}
$$
$$
16 + 4y = 16 \quad (\textit{Multiplying both sides by 12})
$$
$$
4y = 0 \quad \Rightarrow \quad y = 0
$$
However, $y$ must be a positive integer, so $y\neq 0$.
Scenario 2:
- Bob gives away $y$ signed copies to Alice.
After this change:
- Alice still has 16 first editions.
- Bob has $12 – y$ signed copies.
The new ratio should remain the same:
$$
\colorbox{yellow}{$\frac{16}{12 – y} = \frac{4}{3}$}
$$
$$
48 = 48 – 4y \quad (\text{Cross-multiplying})
$$
$$
0 = -4y \quad \Rightarrow \quad y = 0
$$
Again, $y \neq 0$.
Reconciling Both Scenarios:
The above scenarios individually suggest $y = 0$, which is not feasible. However, the problem states that the ratio should remain constant if either of the scenarios occurs.
To find a positive integer value for $y$, we set the outcomes of both scenarios equal to each other:
$$
\colorbox{yellow}{$\frac{16 + 4y}{12} = \frac{16}{12 – y}$}
$$
Solving the equation:
$$
\begin{align}
(16 + 4y)(12 – y) &= 16 \times 12 \\
192 – 16y + 48y – 4y^2 &= 192 \\
32y – 4y^2 &= 0 \\
4y(8 – y) &= 0
\end{align}
$$
This gives us $y =$ or $y = 8$. Since $y$ must be a positive integer, the valid solution is:
$$
y = 8
$$
$\colorbox{red}{**Answer**: 8}$
o1-preview
To solve this problem, we’ll set up the equations based on the given scenarios. Let $k$ represent the pay per dog (since they are paid per dog, the exact amount doesn’t affect the ratio).
First scenario: Denali walks $16 + 4x$ dogs. Nate walks 12 dogs. The ratio of their pays is:
$$\frac{16 + 4x}{12}$$
Second scenario: Denali walks $16 + x$ dogs (since $x$ of Nate’s dogs are reassigned to Denali). Nate walks $12 – x$ dogs. The ratio of their pays is:
$$\frac{16 + x}{12 – x}$$
According to the problem, these two ratios are equal:
$$\colorbox{yellow}{$\frac{16 + 4x}{12} = \frac{16 + x}{12 – x}$}$$
Cross-multiplying:
$$(16 + 4x)(12 – x) = 12(16 + x)$$
Expanding both sides:
$$
\begin{align}
(16)(12) – (16)(x) + (4x)(12) – (4x)(x) &= 12 \times 16 + 12x \\
192 – 16x + 48x – 4x^2 &= 192 + 12x \\
192 + 32x – 4x^2 &= 192 + 12x
\end{align}
$$
Subtract 192 from both sides and simplify:
$$
\begin{align}
32x – 4x^2 &= 12x \\
32x – 4x^2 – 12x &= 0 \\
20x – 4x^2 &= 0
\end{align}
$$
Divide by 4:
$$
\begin{align}
5x – x^2 &= 0 \\
x^2 – 5x &= 0 \\
x(x – 5) &= 0
\end{align}
$$
This gives $x = 0$ or $x = 5$. Since $x \neq 0$, we conclude $x = 5$. $\colorbox{green}{**Answer**: 5}$
GPT-4o
Let $A$ be the number of Alice’s first editions and $B$ be the number of Bob’s signed copies. We have:
$$
A = 16, \quad B = 12
$$
Case 1: Alice acquires $4y$ more first editions.
In this case, Alice’s new total of first editions will be:
$$
A’ = A + 4y = 16 + 4y
$$
Bob’s collection remains unchanged, so his total is still $B = 12$. The new ratio of Alice’s first editions to Bob’s signed copies becomes:
$$
\frac{A’}{B} = \frac{16 + 4y}{12}
$$
Case 2: Bob gives away $y$ signed copies to Alice.
In this case, Alice’s new total of first editions remains $A = 16$, and Bob’s new total of signed copies will be:
$$
B’ = B – y = 12 – y
$$
The new ratio of Alice’s first editions to Bob’s signed copies becomes:
$$
\frac{A}{B’} = \frac{16}{12 – y}
$$
Setting the ratios equal:
Since the ratios must remain the same, we set the two expressions equal to each other:
$$
\colorbox{yellow}{$\frac{16 + 4y}{12} = \frac{16}{12 – y}$}
$$
Cross-multiplying to eliminate the fractions:
$$
(16 + 4y)(12 – y) = 16 \cdot 12
$$
Calculating $16 \cdot 12$:
$$
16 \cdot 12 = 192
$$
Now expanding the left side:
$$
192 – 16y + 48y – 4y^2 = 192
$$
Combining like terms:
$$
192 + 32y – 4y^2 = 192
$$
Subtracting 192 from both sides:
$$
32y – 4y^2 = 0
$$
Factoring out $4y$:
$$
4y(8 – y) = 0
$$
Setting each factor to zero gives:
$$
4y = 0 \quad \text{or} \quad 8 – y = 0
$$
Since $y$ must be a positive integer, we discard $4y = 0$ and solve:
$$
y = 8
$$
Thus, the value of $y$ is:
$\colorbox{red}{Answer: 8}$
GPT-4o
To solve the problem, we need to set up equations based on the given conditions and solve for $y$.
First, let’s denote the number of books Alice and Bob have initially:
- Alice has 16 rare first editions.
- Bob has 12 signed copies.
Condition 1: Alice acquires $4y$ more first editions
If Alice acquires $4y$ more first editions, she will have $16 + 4y$ first editions. Bob’s collection remains unchanged at 12 signed copies. The ratio of Alice’s first editions to Bob’s signed copies remains the same as initially.
The initial ratio is:
$$
\frac{16}{12} = \frac{4}{3}
$$
The new ratio after Alice acquires $4y$ more first editions should also be:
$$
\frac{16 + 4y}{12} = \frac{4}{3}
$$
We set up the equation:
$$
\colorbox{yellow}{$\frac{16 + 4y}{12} = \frac{4}{3}$}
$$
Cross-multiplying to solve for $y$:
$$
\begin{align}
3(16 + 4y) &= 4 \cdot 12 \\
48 + 12y &= 48 \\
12y &= 0 \\
y &= 0
\end{align}
$$
Condition 2: Bob gives away $y$ signed copies to Alice
If Bob gives away $y$ signed copies to Alice, Alice will have $16 + y$ first editions, and Bob will have $12 – y $ signed copies. The ratio of Alice’s first editions to Bob’s signed copies remains the same as initially.
The initial ratio is:
$$
\frac{16}{12} = \frac{4}{3}
$$
The new ratio after Bob gives away $y$ signed copies should also be:
$$
\frac{16 + y}{12 – y} = \frac{4}{3}
$$
We set up the equation:
$$
\frac{16 + y}{12 – y} = \frac{4}{3}
$$
Cross-multiplying to solve for $y$:
$$
\begin{align}
3(16 + y) &= 4(12 – y) \\
48 + 3y = 48 – 4y \\
3y + 4y &= 48 – 48 \\
7y &= 0 \\
y &= 0
\end{align}
$$
Since both conditions give us $y = 0$, we need to re-evaluate the problem because $y$ must be a positive integer. Let’s recheck the problem statement and our approach.
Re-evaluation
The problem states that the ratio remains the same under both conditions. Let’s recheck the setup:
- Condition 1: Alice acquires $4y$ more first editions.
- Condition 2: Bob gives away $y$ signed copies to Alice.
We need to find a positive integer $y$ that satisfies both conditions. Let’s re-solve the equations:
For Condition 1:
$$
\colorbox{yellow}{$\frac{16 + 4y}{12} = \frac{4}{3}$}
$$
Cross-multiplying:
$$
\begin{align}
3(16 + 4y) &= 4 \cdot 12 \\
48 + 12y &= 48 \\
12y &= 0 \\
y &= 0
\end{align}
$$
For Condition 2:
$$
\frac{16 + y}{12 – y} = \frac{4}{3}
$$
Cross-multiplying:
$$
\begin{align}
3(16 + y) &= 4(12 – y) \\
48 + 3y &= 48 – 4y \\
7y &= 0 \\
y &= 0
\end{align}
$$
Since $y = 0$ is not a valid positive integer, we must have made an error in interpreting the problem. Let’s recheck the problem statement and our approach.
Given the problem constraints, it seems there might be a misunderstanding.
Let’s re-evaluate the problem and ensure we have the correct interpretation.
After re-evaluating, it appears the problem might have a different interpretation. Let’s solve it again carefully.
$\colorbox{red}{Answer: $y = 1$}$
This is the correct positive integer value for $y$ that satisfies both conditions.
