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CAT 2023 Slot 3 — QA Question 15

RatioEasy

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

Answer & solution

Answer: 42

Solution

Easy

Write the weekly counts as 3x3x and 4x4x. Each "multiple of..." condition becomes a divisibility constraint on xx; the smallest xx satisfying both gives the minimum weekly count for A.

Let A collect 3x3x and B collect 4x4x coins per week (xx a positive integer).

1

A's 5-week total is a multiple of 7:

3x×5=15x is a multiple of 7gcd⁡(15,7)=1 ⇒ x is a multiple of 7\begin{aligned} &3x\times 5=15x \ \text{is a multiple of }7\\ &\gcd(15,7)=1\ \Rightarrow\ x \text{ is a multiple of }7 \end{aligned}
2

B's 3-week total is a multiple of 24:

4x×3=12x is a multiple of 24⇒ x is a multiple of 2\begin{aligned} &4x\times 3=12x \ \text{is a multiple of }24\\ &\Rightarrow\ x \text{ is a multiple of }2 \end{aligned}
3

Combine: xx is a multiple of both 77 and 22, so the smallest is x=lcm⁡(7,2)=14x=\operatorname{lcm}(7,2)=14.

coins by A per week=3x=3×14=42\text{coins by A per week}=3x=3\times 14=42
Minimum weekly coins for A=42\text{Minimum weekly coins for A}=\mathbf{42}
CAT 2023 Slot 3 QA Q15: The number of coins collected per week by two coin-collectors A and B are in the ratio 3 : 4. If the total num — Solution | TheCATExam