A group of 40 students appeared in an examination of 3 subjects -...

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,

$11 + z \le 15 ⇒ z \le 4$

$11 + y \le 15 ⇒ y \le 4$

Now finding the sum of all we get,

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

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

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

We get, $10 + z \le 15 ⇒ z \le 5$

$10 + y \le 15 ⇒ y \le 5$

Plotting the venn diagram we get,

Now, finding the sum we get,

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

$⇒ y + z = 1$ which is possible,

Hence, possible venn diagram will be,

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