IPMAT Indore 2024 · Quantitative Ability · Question 41

Step 1 — check the totals. Five constituencies with 20 voters each is 100 votes, and 49+35+16=100 ✔ Every voter is accounted for, so in each seat

{a}_{i}+{b}_{i}+{c}_{i}=20

Step 2 — who won S4 and S5? A won only S1, and C won S2 and S3. So S4 and S5 went to B or C.

To win a 20-voter seat outright a party needs more than both others combined would leave it, i.e. strictly more than half of what remains. Concretely, to win seat i a party needs at least 7 votes (if it had 6, the other two would share 14 and one of them would have at least 7).

Now, C won S2 and S3, so {c}_{2}\ge 7 and {c}_{3}\ge 7, using up at least 14 of C’s 16 votes. That leaves C at most 2 votes for S1, S4 and S5 combined — far too few to win anything. Therefore

\text{B won S4 and S5.}

Step 3 — pin down C. Since {c}_{2}+{c}_{3}\le 16 and both are \ge 7, and C must beat B in those seats, the only arrangement that also leaves A and B enough votes elsewhere is

{c}_{2}={c}_{3}=8,{c}_{1}={c}_{4}={c}_{5}=0

Step 4 — B’s votes. {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} are distinct natural numbers summing to 35. B wins S4 and S5 with {c}_{4}={c}_{5}=0, so {b}_{4}>{a}_{4}=20-{b}_{4}, forcing {b}_{4}\ge 11 and likewise {b}_{5}\ge 12. Also {b}_{2}<{c}_{2}=8 and {b}_{3}<{c}_{3}=8.

An exhaustive check of every possibility leaves exactly one distribution:

Verify: each column sums to 20 ✔, each row to its stated total ✔, B’s row is strictly increasing ✔, and every seat has a unique top scorer ✔

Now read the question. We need the seat where B’s tally is below both A’s and C’s.

Only S2 qualifies: 5<7 and 5<8.

Answer: Option 4 — S2

Shortcut. B is last only where C is strong and B is weak. C has votes in just two seats (S2 and S3), so B can possibly be third only there. In S3 B ties A at 6, so it must be S2.

You can reach this without the full grid: B finishes third only where a third party is actually competitive, which narrows the field to two seats before any arithmetic.