IPMAT Indore 2026 · Quantitative Ability · Set 1: Five friends – Countryside Sightseeing, Shopping and Adventure Sports

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.

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 both Shopping and Adventure Sports?

Among the two-activity sets, exactly one is \{\text{SH},\text{AS}\} — so one of A, B, E.

C likes all three, so C also likes both.

D likes only Countryside Sightseeing — no.

1+1=2

Answer: 2

Shortcut. “Both” includes the person who likes all three — forgetting C is the standard slip. One two-activity set plus C gives 2.

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 Countryside Sightseeing?

Two of the three two-activity sets contain CS: \{\text{CS},\text{SH}\} and \{\text{CS},\text{AS}\} — two of A, B, E.

C likes all three — yes.

D likes only Countryside Sightseeing — yes.

2+1+1=4

Answer: 4

Shortcut. Count by category: two of the pairs, plus C (all three), plus D (CS only). Only one of A, B, E is excluded — the one who likes \{\text{SH},\text{AS}\}.