JEE Main 2024 — Mathematics Algebra

Given,

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

And $n(M\cap P)\leq 11,n(M\cap C)\leq 15,n(P\cap C)\leq 15$n(M∩P)≤11,n(M∩C)≤15,n(P∩C)≤15

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

Now, for maximum let $x=11$x=11.

So, the venn diagram will be,

Now, using the given conditions we get,

$$11+z\leq 15\Rightarrow z\leq 4$$11+z≤15⇒z≤4

$$11+y\leq 15\Rightarrow y\leq 4$$11+y≤15⇒y≤4

Now finding the sum of all we get,

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

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

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

We get, $10+z\leq 15\Rightarrow z\leq 5$10+z≤15⇒z≤5

$$10+y\leq 15\Rightarrow y\leq 5$$10+y≤15⇒y≤5

Plotting the venn diagram we get,

Now, finding the sum we get,

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

$\Rightarrow y+z=1$⇒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$10.