A and B have coins of Rs.1, Rs. 2, Rs. 5 and Rs. 10, in the ratio 4 ∶ 3 ∶ 6 ∶ 2 and 3 ∶ 5 ∶ 7 ∶ 3, respectively. A has Rs.6/‐ more than B. Which of the following can be the number of coins with A and B, respectively?
60, 54
Understanding the Coin Ratios and Values
The problem provides information about the coins held by two individuals, A and B, in different denominations (Rs.1, Rs.2, Rs.5, and Rs.10). The number of coins of each denomination is given in a ratio for both A and B.
Let's break down the ratios for A and B:
- For A: The ratio of coins of Rs.1, Rs.2, Rs.5, and Rs.10 is 4 ∶ 3 ∶ 6 ∶ 2.
- For B: The ratio of coins of Rs.1, Rs.2, Rs.5, and Rs.10 is 3 ∶ 5 ∶ 7 ∶ 3.
Total Parts in the Ratios
To find the total number of coins for a person based on the ratio, we first sum the parts in their respective ratio:
- Total parts for A's coins: \(4 + 3 + 6 + 2 = 15\) parts.
- Total parts for B's coins: \(3 + 5 + 7 + 3 = 18\) parts.
This means that the total number of coins A has must be a multiple of 15, and the total number of coins B has must be a multiple of 18.
Calculating the Total Value
Let's calculate the total value of coins for A and B based on their ratios. Suppose the number of coins for A is \(k_A\) times the ratio parts, and for B is \(k_B\) times the ratio parts, where \(k_A\) and \(k_B\) are positive integers.
Value for A:
- Number of Rs.1 coins: \(4k_A\)
- Number of Rs.2 coins: \(3k_A\)
- Number of Rs.5 coins: \(6k_A\)
- Number of Rs.10 coins: \(2k_A\)
Total value for A (\(V_A\)) = \((4k_A \times 1) + (3k_A \times 2) + (6k_A \times 5) + (2k_A \times 10)\)
\(V_A = 4k_A + 6k_A + 30k_A + 20k_A = 60k_A\)
The total number of coins for A is \(4k_A + 3k_A + 6k_A + 2k_A = 15k_A\).
Value for B:
- Number of Rs.1 coins: \(3k_B\)
- Number of Rs.2 coins: \(5k_B\)
- Number of Rs.5 coins: \(7k_B\)
- Number of Rs.10 coins: \(3k_B\)
Total value for B (\(V_B\)) = \((3k_B \times 1) + (5k_B \times 2) + (7k_B \times 5) + (3k_B \times 10)\)
\(V_B = 3k_B + 10k_B + 35k_B + 30k_B = 78k_B\)
The total number of coins for B is \(3k_B + 5k_B + 7k_B + 3k_B = 18k_B\).
Using the Value Difference Condition
We are given that A has Rs. 6 more than B. So, \(V_A - V_B = 6\).
Substituting the values of \(V_A\) and \(V_B\):
\(60k_A - 78k_B = 6\)
We can divide the entire equation by 6:
\(10k_A - 13k_B = 1\)
We need to find integer values for \(k_A\) and \(k_B\) that satisfy this equation. The possible number of coins for A is \(15k_A\) and for B is \(18k_B\). We will check the given options to see which pair satisfies both conditions: the total number of coins are multiples of 15 and 18 respectively, and the corresponding \(k_A\) and \(k_B\) values satisfy \(10k_A - 13k_B = 1\).
Evaluating the Options
| Option | A's Coins | B's Coins | Is A's coins multiple of 15? (\(15k_A\)) | Is B's coins multiple of 18? (\(18k_B\)) | Value of \(k_A\) (A's coins/15) | Value of \(k_B\) (B's coins/18) | Check \(10k_A - 13k_B = 1\) | Result |
|---|---|---|---|---|---|---|---|---|
| 1 | 42 | 36 | No (42 / 15 is not an integer) | Yes (36 / 18 = 2) | - | \(k_B = 2\) | Cannot evaluate \(k_A\) | Incorrect |
| 2 | 45 | 54 | Yes (45 / 15 = 3) | Yes (54 / 18 = 3) | \(k_A = 3\) | \(k_B = 3\) | \(10(3) - 13(3) = 30 - 39 = -9\) (\(-9 \neq 1\)) | Incorrect |
| 3 | 60 | 54 | Yes (60 / 15 = 4) | Yes (54 / 18 = 3) | \(k_A = 4\) | \(k_B = 3\) | \(10(4) - 13(3) = 40 - 39 = 1\) (\(1 = 1\)) | Correct |
| 4 | 60 | 72 | Yes (60 / 15 = 4) | Yes (72 / 18 = 4) | \(k_A = 4\) | \(k_B = 4\) | \(10(4) - 13(4) = 40 - 52 = -12\) (\(-12 \neq 1\)) | Incorrect |
From the evaluation, only Option 3 satisfies both conditions: the total number of coins for A (60) is a multiple of 15 (\(60 = 15 \times 4\)), the total number of coins for B (54) is a multiple of 18 (\(54 = 18 \times 3\)), and the corresponding values \(k_A = 4\) and \(k_B = 3\) satisfy the equation \(10k_A - 13k_B = 1\).
Therefore, the possible number of coins with A and B are 60 and 54, respectively.
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