IPMAT Indore 2024 · Quantitative Ability · Question 43

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 ✔

Step 1 — reduce to a two-party contest. If A and C merge, each seat becomes alliance versus B. Since every seat has exactly 20 voters,

\text{Alliance votes in seat}i={a}_{i}+{c}_{i}=20-{b}_{i}

Step 2 — state the winning condition. The alliance wins seat i when

20-{b}_{i}>{b}_{i}\Leftrightarrow{b}_{i}<10

So the question reduces entirely to: how many of B’s five tallies are below 10?

Step 3 — count.

\text{3 seats}

A stronger statement. Even without pinning down the full grid, B’s row must contain exactly three values below 10: B needs {b}_{4}\ge 11 and {b}_{5}\ge 12 to win its two seats, which leaves at most 35-23=12 votes for the other three seats — so all three of {b}_{1},{b}_{2},{b}_{3} are necessarily below 10. The answer is 3 under every reading of the constraints.

Answer: Option 1 — 3

Note on the published key. The circulated key marks Option 4 (2). For the alliance to win only 2 seats, three of B’s tallies would have to be at least 10, which would require {b}_{3}+{b}_{4}+{b}_{5}\ge 10+11+12=33, leaving under 2 votes for {b}_{1}+{b}_{2} — incompatible with distinct increasing values and with C’s ability to win S2 and S3.

As with Q42, the key contradicts its own worked solution: the table printed in the same document gives B’s row as 1, 5, 6, 11, 12, from which the alliance totals are 19, 15, 14, 9, 8 and it wins S1, S2 and S3 — three seats. The verified answer is 3.

Shortcut. One inequality replaces the whole table. With only two contestants and 20 voters, the alliance wins seat i exactly when {b}_{i}<10. So count how many of B’s numbers are single-digit.

B spends at least 11+12=23 votes winning S4 and S5, leaving at most 12 for the remaining three seats — so all three must be under 10. Answer 3, reachable in two lines without ever constructing A’s or C’s rows.

The general trick: in an alliance question, convert “who wins” into a threshold on the one party left out. It is almost always a shorter route than recomputing every seat.