CUET PG 2026 — Mathematics PYQ

Solution

Let the three sets be $M$ (Mathematics), $S$ (Statistics), and $P$ (Physics).

1. Given Data:

• Total students $n(U)=100$

• Mathematics only $=15$

• Statistics only $=12$

• Physics only $=8$

• Mathematics and Statistics $n(M\cap S)=10$

• Physics and Mathematics $n(P\cap M)=40$

• Physics and Statistics $n(P\cap S)=20$

• Total Physics $n(P)=65$

2. Finding the Intersection of all three ($M\cap S\cap P$):

Let the number of students who offered all three subjects be $x$.

From the data of the Physics set, we can break down the total (65) into its components:

$n(P)=(\text{Physics only})+(\text{Physics and Math only})+(\text{Physics and Stats only})+(\text{All three})$

• Physics and Math only $=n(P\cap M)-x=40-x$

• Physics and Stats only $=n(P\cap S)-x=20-x$

Substitute these into the Physics total:

$65=8+(40-x)+(20-x)+x$

$65=68-x$

$x=68-65=3$

So, 3 students offered all three subjects.

3. Finding the Total number of students who offered at least one subject:

Now we calculate the other intersection parts:

• Math and Stats only $=n(M\cap S)-x=10-3=7$

• Physics and Math only $=40-3=37$

• Physics and Stats only $=20-3=17$

Total students who offered at least one subject $n(M\cup S\cup P)$:

$=(\text{Math only})+(\text{Stats only})+(\text{Physics only})+(\text{Math \& Stats only})+(\text{Phys \& Math only})+(\text{Phys \& Stats only})+(\text{All three})$

$=15+12+8+7+37+17+3$

$=99$

4. Students not offered any of the three:

Correct Option: (a)