IPMAT Indore 2026 · Quantitative Ability · Question 33

Step 1 — sizes of the three sets.

\left| A \right|=\frac{180}{4}=45, \left| B \right|=\frac{180}{6}=30, \left| C \right|=\frac{180}{9}=20

Step 2 — the pairwise overlaps, using lcms:

\left| A\cap B \right|=\frac{180}{\text{lcm}\left( 4,6 \right)}=\frac{180}{12}=15

\left| B\cap C \right|=\frac{180}{\text{lcm}\left( 6,9 \right)}=\frac{180}{18}=10

\left| A\cap C \right|=\frac{180}{\text{lcm}\left( 4,9 \right)}=\frac{180}{36}=5

Step 3 — the triple overlap.

\left| A\cap B\cap C \right|=\frac{180}{\text{lcm}\left( 4,6,9 \right)}=\frac{180}{36}=5

Step 4 — apply the “exactly one” formula.

∑\left| A \right|-2∑\left| A\cap B \right|+3\left| A\cap B\cap C \right|

=\left( 45+30+20 \right)-2\left( 15+10+5 \right)+3\left( 5 \right)

=95-60+15=50

Answer: Option 1 — 50

Shortcut. Memorise the exactly one formula: singles -\,2\times pairs +\,3\times triple. Deriving it mid-exam costs minutes. Watch that \text{lcm}\left( 4,9 \right)=\text{lcm}\left( 4,6,9 \right)=36 here, so two of the counts coincide.