CAT 2023 Slot 3 · Quant · Question 15

Let A collect 3x coins a week and B collect 4x, so the ratio 3:4 is automatic.

A's total over 5 weeks is 5\times 3x=15x, and this must be a multiple of 7.

Since 15 and 7 share no factor, the 7 must come from x, so x is a multiple of 7.

B's total over 3 weeks is 3\times 4x=12x, and this must be a multiple of 24.

Dividing, \frac{12x}{24}=\frac{x}{2} must be a whole number, so x is even.

So x is a multiple of both 7 and 2, hence a multiple of 14.

The smallest such value is x=14.

A then collects 3\times 14=42 coins in one week.

Checking: A's five-week total is 15\times 14=210=7\times 30, and B's three-week total is 12\times 14=168=24\times 7 — both conditions hold.

Shortcut. Each condition reduces to a divisibility on x alone once the constant is cancelled: 15 contributes nothing against 7, and 12 leaves only a factor of 2 against 24. So x is a multiple of \text{LCM}\left( 7,2 \right)=14, and the answer is 3\times 14.