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 test series. Find the number of aspirants who have enrolled either only in NDA test series or only in Airforce group X test series.?

Option 4 : 29

Concept:

• Symmetric Difference of two Sets: Let A and B be two sets. The symmetric difference of sets A and B is the set (A - B) ∪ (B - A) and is denoted as A Δ B.

         i.e A Δ B = (A - B) (B – A)

• 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.

So, there are 6 defence aspirants who have enrolled in both the test series.

As we know that, the number of aspirants who have enrolled either only in NDA test series or only in Airforce group X test series is given by: n (M Δ A)

⇒ n (M Δ A) = n (M) + n (A) – 2 n (M ∩ A)

⇒ n (M Δ A) = 25 + 16 – 2 × 6 = 29

Hence, there are 29 aspirants who have enrolled either only in NDA test series or only in Airforce group X test series.