IPMAT Indore 2026 · Quantitative Ability · Question 22

Take the class as 100 students.

Step 1 — the news readers. With I=25 and P=45,

\left| I\cup P \right|=25+45-\left| I\cap P \right|=70-\left| I\cap P \right|

Step 2 — the non-readers.

\text{non-readers}=100-\left| I\cup P \right|=30+\left| I\cap P \right|

Step 3 — who definitely does not play video games?

The 20 students who read both and do not play games.

70\% of the non-readers (since 30\% of them do play).

\text{don’t play}\ge 20+0.7\left( 30+\left| I\cap P \right| \right)

Step 4 — minimise. The expression grows with \left| I\cap P \right|, so take it as small as possible. We are told 20% read both and don’t play games, so

\left| I\cap P \right|\ge 20

Putting \left| I\cap P \right|=20:

20+0.7\left( 50 \right)=20+35=55

Answer: Option 3 — 55 percent

Shortcut. The phrase “minimum percentage” tells you to push the overlap to its smallest legal value, which the 20% clause pins at exactly 20. After that it is one line: the 20 guaranteed non-players plus 70\% of the 50 non-readers.