Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value of each Type...
GMAT Two Part Analysis : (TPA) Questions
Rufus has a total of 20 coins of 2 types, Type Q and Type R. The value of each Type Q coin is \(5\mathrm{q}\) currency units (cu), and the value of each Type R coin is \(5\mathrm{r}\) cu, where q and r are positive integers and \(\mathrm{q} < \mathrm{r}\). The total value of the Type Q coins is Q cu and the total value of the Type R coins is R cu. Additionally, \(\mathrm{Q} + \mathrm{R} = 130\), and there are more Type Q coins than Type R coins.
In the table, select a value for Q and a value for R that are jointly consistent with the given information. Make only two selections, one in each column.
Phase 1: Owning the Dataset
Visualization
Let's organize the coin information in a table:
| Coin Type | Count | Value per coin | Total Value |
| Type Q | \(\mathrm{x}\) | \(\mathrm{5q\ cu}\) | \(\mathrm{Q\ cu}\) |
| Type R | \(\mathrm{y}\) | \(\mathrm{5r\ cu}\) | \(\mathrm{R\ cu}\) |
| Total | 20 | - | 130 cu |
Constraints:
- \(\mathrm{x > y}\) (more Type Q coins)
- \(\mathrm{q < r}\) (Type Q coins have lower unit value)
- \(\mathrm{q, r}\) are positive integers
Phase 2: Understanding the Question
Key Relationships
From our table:
- \(\mathrm{x + y = 20}\) (total coins)
- \(\mathrm{Q = 5qx}\) (total value of Type Q coins)
- \(\mathrm{R = 5ry}\) (total value of Type R coins)
- \(\mathrm{Q + R = 130}\)
Substituting the value formulas:
\(\mathrm{5qx + 5ry = 130}\)
Dividing by 5:
\(\mathrm{qx + ry = 26}\)
What We're Looking For
We need to find specific values for Q and R from the choices [25, 60, 65, 70, 105] that:
- Sum to 130
- Satisfy all our constraints
Phase 3: Finding the Answer
Checking Valid Combinations
First, let's identify which pairs sum to 130:
- \(\mathrm{25 + 105 = 130}\) ✓
- \(\mathrm{60 + 70 = 130}\) ✓
- \(\mathrm{65 + 65 = 130}\) ✓
Testing Constraints
Since \(\mathrm{x > y}\) and \(\mathrm{x + y = 20}\), we know \(\mathrm{x > 10}\).
From \(\mathrm{qx + ry = 26}\) and \(\mathrm{x + y = 20}\), we can substitute \(\mathrm{y = 20 - x}\):
\(\mathrm{qx + r(20-x) = 26}\)
\(\mathrm{x(q-r) = 26 - 20r}\)
Since \(\mathrm{q < r}\), we have:
\(\mathrm{x = \frac{20r - 26}{r - q}}\)
Testing \(\mathrm{r = 2, q = 1}\):
\(\mathrm{x = \frac{40 - 26}{2 - 1} = 14}\)
\(\mathrm{y = 20 - 14 = 6}\)
Verification:
- \(\mathrm{qx + ry = 1 × 14 + 2 × 6 = 14 + 12 = 26}\) ✓
- \(\mathrm{x > y?}\) \(\mathrm{14 > 6}\) ✓
- \(\mathrm{q < r?}\) \(\mathrm{1 < 2}\) ✓
Calculating Q and R:
- \(\mathrm{Q = 5qx = 5 × 1 × 14 = 70}\)
- \(\mathrm{R = 5ry = 5 × 2 × 6 = 60}\)
- \(\mathrm{Q + R = 70 + 60 = 130}\) ✓
Phase 4: Solution
Final Answer:
- \(\mathrm{Q = 70}\)
- \(\mathrm{R = 60}\)
These values satisfy all constraints: we have 14 Type Q coins (each worth 5 cu) and 6 Type R coins (each worth 10 cu), giving us more Type Q coins than Type R coins, with Type Q coins having lower unit value, and a total value of 130 cu.