IPMAT Indore 2024 · Quantitative Ability · Question 42

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 ✔

Read B’s row. {b}_{2}=5.

Why it cannot be anything else. Suppose {b}_{2}=6 (the value the published key gives). The constraint {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} would then force {b}_{3}\ge 7, {b}_{4}\ge 8, {b}_{5}\ge 9.

But C wins S3, so {c}_{3}>{b}_{3}\ge 7, i.e. {c}_{3}\ge 8; and C wins S2, so {c}_{2}>{a}_{2}=20-6-{c}_{2}, giving {c}_{2}\ge 8. Then {c}_{2}+{c}_{3}\ge 16, which exhausts C’s entire total and forces {c}_{1}={c}_{4}={c}_{5}=0 and {c}_{2}={c}_{3}=8. Hence {b}_{3}=7 exactly.

Now B must win S4 and S5 outright against A with no C votes present, so {b}_{4}\ge 11 and {b}_{5}\ge 12. Summing B’s row:

{b}_{1}+6+7+{b}_{4}+{b}_{5}\ge{b}_{1}+6+7+11+12={b}_{1}+36>35

which exceeds B’s total of 35 for any {b}_{1}\ge 0 — a contradiction. So {b}_{2}=6 is impossible.

The same elimination applied to every candidate value leaves {b}_{2}=5 as the only survivor.

Answer: Option 1 — 5

Note on the published key. The widely circulated answer key marks Option 4 (6) here, and 6 is arithmetically impossible once “distinct natural numbers in increasing order” is applied, as shown above.

The key contradicts its own worked solution. The solution table printed alongside it gives B’s row as 1, 5, 6, 11, 12 — identical to the grid above — and its accompanying text states that B “received 1, 5 and 6 votes in S1, S2 and S3”. By the source’s own working, {b}_{2}=5. The printed letter is a transcription error. The verified answer is 5.

Shortcut. Build B’s row from the top down. B must win S4 and S5 against 20-voter fields, so {b}_{4}\ge 11 and {b}_{5}\ge 12 — that is already 23 of B’s 35, leaving just 12 votes for S1, S2 and S3 combined.

With {b}_{1}<{b}_{2}<{b}_{3} summing to 12 and {b}_{3}<8 (C’s 8 must beat it), the only fit is 1+5+6=12. So {b}_{2}=5.

The 23-votes-at-the-top observation is the whole question — it collapses a five-variable problem to a three-variable one in a single line.