In a group of 100 higher secondary students, 50 students have cleared NDA exam, 40 students have cleared JEE exam and 30 students have cleared both the exams. How many such students are there who have neither qualified NDA nor JEE exam.?

Option 4 : 40

Concept:

Let A, B and C be three finite sets and U is the finite universal set, then

1. n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
2. n (A ∪ B) = n (A) + n (B) ⇔ A ∩ B = ϕ
3. n (A - B) = n (A) – n (A ∩ B) = n (A ∩ B’)
4. n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n (B ∩ C) – n (A ∩ C) + n (A ∩ B ∩ C)
5. n (A’ ∪ B’) = n = n (U) – n (A ∩ B)
6. n (A’ ∩ B’) = n = n (U) – n (A ∪ B)
7. n (A Δ B) = n (A) + n (B) – 2 n (A ∩ B)
8. n (A’) = n (U) – n (A)

Calculation:

Let U = No. of students in higher secondary group, M = No. of higher secondary students who qualified NDA exam and J = No. of higher secondary students who qualified JEE exam.

Given: n (U) = 100, n (M) = 50, n (J) = 40 and n (M ∩ J) = 30

As we know that, if A and B are two finite sets, then n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

⇒ n (M ∪ J) = n (M) + n (J) – n (M ∩ J)

⇒ n (M ∪ J) = 50 + 40 – 30

⇒ n (M ∪ J) = 60

∴ There are 60 higher secondary students who have qualified either NDA or JEE exam.

As, we know that the number of higher secondary students who have neither qualified NDA nor JEE exam is given by: n

⇒ n = n (U) – n (M ∪ J) = 100 – 60 = 40