Question

The number of coins collected per week by two coin-collectors A and B are in the ratio 3 : 4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is

• A. 20
• B. 42
• C. 66
• D. None of Above

Answer 1

The Correct Option is B

Let, the number of coins collected by \(A\) and \(B\) in one week as \(3x\) and \(4x\) respectively.
The total number of coins collected by \(A\) in \(5\) weeks \(=15x\)
For \(15x\) to be a multiple of \(7\), \(x\) must be a multiple of \(7\).
Similarly, the total number of coins collected by \(B\) in \(3\) weeks \(=12x\)
For \(12x\) to be a multiple of \(24\), \(x\) must be a multiple of \(2\).
Thus, \(x\) must be a multiple of \(7×2 = 14\)
The minimum value for \(x\) \(=14\).
Therefore, the minimum number of coins collected by \(A\) in one week \(=3x = 3 × 14 = 42\).

So, the correct option is (B): \(42\)