In a group of 35 defence aspirants, 25 aspirants have enrolled in NDA test series and 16 have enrolled in Airforce Group X test series. Also each aspirant is enrolled in atleast one of the test series. Find the number of aspirants who have enrolled in both the test series.

In a group of 35 defence aspirants, 25 aspirants have enrolled in NDA test series and 16 have enrolled in Airforce Group X test series. Also each aspirant is enrolled in atleast one of the test series. Find the number of aspirants who have enrolled in both the test series. Correct Answer 6

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, M = No. of defence aspirants enrolled in NDA test series and A = No. of defence aspirants enrolled in Airforce Group X test series.

Given: n (M) = 25, n (A) = 16 and n (M ∪ A) = 35.

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 ∪ A) = n (M) + n (A) – n (M ∩ A)

⇒ 35 = 25 + 16 – n (M ∩ A)

⇒ n (M ∩ A) = 41 – 35 = 6.