IPMAT Indore 2024 · Quantitative Ability · Question 3

Step 1 — use the counting identity. Let {e}_{1},{e}_{2},{e}_{3} be the number of students liking exactly one, exactly two, exactly three drinks. Since everyone likes at least one:

{e}_{1}+{e}_{2}+{e}_{3}=150

Adding the three individual totals counts a student once for each drink they like:

52+48+62={e}_{1}+2{e}_{2}+3{e}_{3}=162

Step 2 — subtract.

\left({e}_{1}+2{e}_{2}+3{e}_{3}\right)-\left({e}_{1}+{e}_{2}+{e}_{3}\right)=162-150 \Rightarrow{e}_{2}+2{e}_{3}=12

Step 3 — maximise {e}_{2}+{e}_{3}, the number liking more than one drink. From {e}_{2}=12-2{e}_{3},

{e}_{2}+{e}_{3}=12-2{e}_{3}+{e}_{3}=12-{e}_{3}

This is largest when {e}_{3} is smallest, i.e. {e}_{3}=0, giving {e}_{2}=12.

Step 4 — check it is achievable. With {e}_{3}=0, {e}_{2}=12, {e}_{1}=138: the sum 138+2\left( 12 \right)=162 ✔, and the 12 overlaps can be split among the three pairs to match 52, 48 and 62.

Answer: 12

Shortcut. The “excess” shortcut. ∑\left( \text{singles}\right)-\left( \text{total}\right)={e}_{2}+2{e}_{3}. Here 162-150=12. Each triple-overlap student eats two units of that excess while each double eats one, so to maximise the number of overlapping students you want no triples: answer = excess =12. Conversely, the minimum would push everything into triples, giving 6.