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 weekly coin collection for A be $3x$ and for B be $4x$.

Given:

• A's coins in 5 weeks = $5 \times 3x = 15x$. This quantity must be divisible by 7.
• B's coins in 3 weeks = $3 \times 4x = 12x$. This quantity must be divisible by 24.

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

Substitute $x = 7k$ into the second condition:
$12x = 12 \times 7k = 84k$
For $84k$ to be a multiple of 24, $k$ must be selected to satisfy this.

Testing values for $k$:
If $k = 1$, $84k = 84$. 84 is divisible by 12 but not by 24.
If $k = 2$, $84k = 168$. 168 is divisible by 24.

The smallest valid value for $k$ is 2, thus $x = 7 \times 2 = 14$.

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