Q. 15

# A and B are two s

Given that

n(A–B) = 32 + x

n(B–A) = 5x

n(AB) = x

and n(A) = n(B)

using these results

n(A) = n(A–B) + n(AB)

n(B) = n(B–A) + n(AB)

n(AB) = n(A–B) + n(AB) + n(B–A)

But given that n(A) = n(B)

So, n(A–B) + n(AB) = n(B–A) + n(AB)

n(A–B) = n(B–A)

32 + x = 5x

5x–x = 32

4x = 32

(i) x = 32/4 = 8

(ii) n(AB) = n(A–B) + n(AB) + n(B–A)

= 32 + x + x + 5x

= 32 + 7x

= 32 + 7(8)

= 88

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