A certain adhesive is formed by mixing 2 components, a resin and a hardener, and allowing the mixture to dry....
GMAT Two Part Analysis : (TPA) Questions
A certain adhesive is formed by mixing 2 components, a resin and a hardener, and allowing the mixture to dry. For example, 2 grams of resin mixed with 2 grams of hardener will result in 4 grams of adhesive. A worker has 6 grams of resin and 8 grams of hardener and wishes to form adhesive using one of the following mixture formulas:
- Mixture has a ratio of \(3:2\) (3 parts resin to 2 parts hardener).
- Mixture has a ratio of \(2:4\) (2 parts resin to 4 parts hardener).
Select for Mixture A the maximum amount, in grams, that the worker can make of Mixture A, and select for Mixture B the maximum amount, in grams, the worker can make of Mixture B, according to the information provided. Make only two selections, one in each column.
Phase 1: Owning the Dataset
Visual Representation
Let's create a table to organize our information:
| Component | Available | Mixture A Ratio | Mixture B Ratio |
|---|---|---|---|
| Resin | 6 grams | 3 parts | 2 parts (or 1) |
| Hardener | 8 grams | 2 parts | 4 parts (or 2) |
| Total Parts | 5 parts | 6 parts (or 3) |
Key insight: Mixture B's ratio \(2:4\) simplifies to \(1:2\) for easier calculation.
Phase 2: Understanding the Question
We need to find the maximum amount of each mixture we can make given our limited resources. The limiting factor will be whichever component runs out first.
For Mixture A (\(3:2\) ratio):
- Every \(5\) grams of mixture requires \(3\mathrm{g}\) resin + \(2\mathrm{g}\) hardener
- We can think of this as: for every \(3\mathrm{g}\) resin, we need \(2\mathrm{g}\) hardener
For Mixture B (\(1:2\) ratio):
- Every \(3\) grams of mixture requires \(1\mathrm{g}\) resin + \(2\mathrm{g}\) hardener
- We can think of this as: for every \(1\mathrm{g}\) resin, we need \(2\mathrm{g}\) hardener
Phase 3: Finding the Answer
Mixture A Calculation
Let's check which component limits us:
- If we use all \(6\mathrm{g}\) resin: we need \(6 imes rac{2}{3} = 4\mathrm{g}\) hardener
- If we use all \(8\mathrm{g}\) hardener: we need \(8 imes rac{3}{2} = 12\mathrm{g}\) resin
Since we only have \(6\mathrm{g}\) resin (not \(12\mathrm{g}\)), resin is our limiting factor.
Maximum Mixture A:
- Resin used: \(6\) grams
- Hardener used: \(4\) grams
- Total mixture: \(6 + 4 = 10\) grams
Mixture B Calculation
Let's check which component limits us:
- If we use all \(6\mathrm{g}\) resin: we need \(6 imes 2 = 12\mathrm{g}\) hardener
- If we use all \(8\mathrm{g}\) hardener: we need \(8 imes rac{1}{2} = 4\mathrm{g}\) resin
Since we only have \(8\mathrm{g}\) hardener (not \(12\mathrm{g}\)), hardener is our limiting factor.
Maximum Mixture B:
- Resin used: \(4\) grams
- Hardener used: \(8\) grams
- Total mixture: \(4 + 8 = 12\) grams
Phase 4: Solution
Final Answer:
- Mixture A: \(10\) grams
- Mixture B: \(12\) grams
We've efficiently used our limited resources to maximize each mixture type, with resin limiting Mixture A production and hardener limiting Mixture B production.