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?

Question

Prepp Team · Accepted Answer

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.