IPMAT Indore 2026 · Quantitative Ability · Question 9

Step 1 — list what is possible. With three activities, a person can like one, two or all three. The two-activity subsets are exactly three:

\{\text{CS},\text{SH}\}, \{\text{CS},\text{AS}\}, \{\text{SH},\text{AS}\}

Step 2 — place A, B and E. All three like exactly two activities, and no two friends share the same set. There are only three two-activity sets, so A, B and E take all three of them, one each (in some order we cannot determine — and never need to).

Step 3 — place D. Given directly: D likes only Countryside Sightseeing, a one-activity set.

Step 4 — place C. C likes more activities than D, so C likes two or three. But every two-activity set is already used by A, B and E, and no two friends may match. Therefore

\text{C likes all three activities.}

Step 5 — the complete picture.

Although we cannot say which pair each of A, B and E likes, every question below depends only on counts, which are fully determined.

Who likes Adventure Sports?

Of the three two-activity sets, two contain Adventure Sports: \{\text{CS},\text{AS}\} and \{\text{SH},\text{AS}\}. So two of A, B, E qualify.

C likes all three, so C qualifies.

D likes only Countryside Sightseeing, so D does not.

\text{Total}=2+1=3\text{friends}

Choosing a pair from them:

\left( \genfrac{}{}{0pt}{}{3}{2}\right)=3

Answer: 3

Shortcut. You never need to know which friend likes what — only how many like Adventure Sports. Two of the three pairs contain AS, plus C, giving 3 people and \left( \genfrac{}{}{0pt}{}{3}{2}\right)=3 pairs.