IPMAT Indore 2024 · Quantitative Ability · Question 32

This question was dropped by IIM Indore. The official answer key marks it with an asterisk: the institute accepted a candidate objection, acknowledged that none of the four options is correct, and awarded +4 to every test taker. The working below shows why.

Step 1 — formalise the condition. “No girl gets fewer balloons than a boy does” means every girl’s count is at least every boy’s count:

\min \left( \text{girls}\right) \ge \max \left( \text{boys}\right)

Step 2 — count by exhaustion. Let the girls receive {g}_{1},\ldots ,{g}_{5}\ge 1 and the boys {b}_{1},{b}_{2},{b}_{3}\ge 1, with

∑{g}_{i}+∑{b}_{j}=20, \min{g}_{i}\ge \max{b}_{j}

Enumerating every ordered pair of compositions subject to that constraint gives

n=2901

Step 3 — compare with the options. 2901 lies below every printed range — the smallest on offer starts at 6000. Hence no option is correct.

Why the options are so far out. A plausible intended reading drops the “\min \ge \max” restriction, giving the plain positive-composition count

\left( \genfrac{}{}{0pt}{}{19}{7}\right)=50388

which is larger than every option too. Whatever the setter intended, the four ranges do not match any natural reading, which is precisely what the objection established.

Answer: Officially cancelled — all candidates awarded full marks. (The marked option in the published key is Option 1, but it is not mathematically correct.)

Shortcut. Know this one as a “free 4 marks” question. In the real exam the right move was to attempt it (there was no penalty once it was cancelled) or skip it quickly — the balloon-with-ordering constraint is a 4–5 minute problem whose options do not fit.

The transferable lesson: when a counting question’s options cluster in a narrow band far from a quick estimate, sanity-check with a crude bound before investing time. Here even the unrestricted count \left( \genfrac{}{}{0pt}{}{19}{7}\right)=50,388 overshoots every option, which should trigger suspicion within 30 seconds.