CAT 2022 Slot 3 · Quant · Question 20

Let the counts be a cheques of ₹100, b of ₹250 and c of ₹500.

The cheque count gives a+b+c=100.

The money total gives 100a+250b+500c=15250.

Dividing the second equation by 50 keeps the numbers small: 2a+5b+10c=305.

Substituting a=100-b-c gives 2\left( 100-b-c \right)+5b+10c=305.

Expanding, 200-2b-2c+5b+10c=305.

Collecting terms, 3b+8c=105.

To maximise c, note that 3b=105-8c must be non-negative, so c\le 13.1.

It must also be a multiple of 3, and since 105 already is, 8c must be a multiple of 3 — which forces c to be a multiple of 3.

The largest multiple of 3 not exceeding 13 is c=12.

Then 3b=105-96=9, so b=3, and a=100-3-12=85.

Checking the money: 8500+750+6000=15250, exactly right.

So the maximum is 12 cheques of ₹500.

Shortcut. Reducing to 3b+8c=105 makes the divisibility visible at once: 105 and 3b are both multiples of 3, so 8c must be too, hence c is. Counting down from 13 in threes lands on 12 immediately.