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

The correct answer is 42.

Let the number of coins collected per week by coin-collector A be $3x$ and by coin-collector B be $4x$, where $x$ is a positive integer.

According to the problem statement:
1. The total number of coins collected by A in 5 weeks is a multiple of 7.
The number of coins collected by A in 5 weeks is:
$5\times 3x=15x$
Since $15x$ is a multiple of 7, and 15 is not divisible by 7, $x$ must be a multiple of 7.
Thus, we can write $x=7k$ for some positive integer $k$.

2. The total number of coins collected by B in 3 weeks is a multiple of 24.
The number of coins collected by B in 3 weeks is:
$3\times 4x=12x$
Since $12x$ is a multiple of 24, we must have:
$12x=24m$ for some positive integer $m$.
Dividing both sides by 12, we get:
$x=2m$
This means $x$ must be an even number (a multiple of 2).

Combining both conditions:
- $x$ must be a multiple of 7.
- $x$ must be a multiple of 2.
Therefore, $x$ must be a multiple of the least common multiple of 2 and 7, which is:
$\text{LCM}(2,7)=14$
So, the minimum possible positive integer value for $x$ is 14.

We want to find the minimum possible number of coins collected by A in one week, which is $3x$:
${\left(3x\right)}_{\text{min}}=3\times 14=42$