CAT 2022 Slot 3 · Quant · Question 5

Let section A have a students, so section B has a+10.

The total marks are 32a from A and 60\left( a+10 \right) from B.

The combined average is \frac{32a+60\left( a+10 \right)}{a+\left( a+10 \right)}=\frac{92a+600}{2a+10}.

Dividing numerator and denominator by 2, this is \frac{46a+300}{a+5}.

Split the numerator around a+5: 46a+300=46\left( a+5 \right)+70.

So the average is 46+\frac{70}{a+5}.

For this to be an integer, a+5 must divide 70 exactly.

The divisors of 70 are 1, 2, 5, 7, 10, 14, 35 and 70.

Since section A has at least one student, a\ge 1 and so a+5\ge 6.

That leaves a+5\in \{7,10,14,35,70\}, giving a\in \{2,5,9,30,65\}.

The largest is 65 and the smallest is 2.

The difference is 65-2=63.

Shortcut. Rewriting the average as 46+\frac{70}{a+5} turns an averages question into a divisor hunt. Only the smallest and largest divisors of 70 above 5 are needed, so the middle cases never have to be listed.