JEE Main 2024 — Sets and Relations Question with Solution

Given,

$n\left(M\right)=20, n\left(P\right)=25, n\left(C\right)=16$

And $n\left(M\cap P\right)\le 11, n\left(M\cap C\right)\le 15, n\left(P\cap C\right)\le 15$

Now, plotting the diagram and taking all three subjects passed as $x$ we get,

Now, for maximum let $x =11$.

So, the venn diagram will be,

Now, using the given conditions we get,

$11+z\le 15\Rightarrow z\le 4$

$11+y\le 15\Rightarrow y\le 4$

Now finding the sum of all we get,

$9-z+0+14-y+z+11+y+5-y-z=40$

$\Rightarrow y+z=-1$ which is not possible hence the value of $x=11$ is not possible,

Now taking the maximum value as $x=10$,

We get, $10+z\le 15\Rightarrow z\le 5$

$10+y\le 15\Rightarrow y\le 5$

Plotting the venn diagram we get,

Now, finding the sum we get,

$10-z+0+15-y+z+10+y+6-y-z=40$

$\Rightarrow y+z=1$ which is possible,

Hence, possible venn diagram will be,

Hence, the maximum number of students passed in all the three subjects is $10$.