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 in one week be $3x$ and by B be $4x$. 

We are given:

• Coins collected by A in 5 weeks = $5 \times 3x = 15x$ (must be a multiple of 7)
• Coins collected by B in 3 weeks = $3 \times 4x = 12x$ (must be a multiple of 24)

For $15x$ to be divisible by 7, $x$ must be a multiple of 7.
So, let $x = 7k$, where $k$ is a positive integer.

Substituting into the second condition:
$12x = 12 \times 7k = 84k$
For $84k$ to be a multiple of 24, $k$ must be chosen appropriately.

Check with $k = 1$: $84 \times 1 = 84$, which is divisible by 12 but not by 24.
Try $k = 2$: $84 \times 2 = 168$, which is divisible by 24.

So, the smallest suitable value is $k = 2$, and hence $x = 14$.

Therefore, coins collected by A in one week = $3x = 3 \times 14 = \mathbf{42}$.