Question

A nd B are two sets such that n(A-B)= 14+x,n(B-A)=3x and n(AnB)=x.if n(A)=n(B) find x and n(AuB)

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the value of x and n(A ∪ B)

$\large\bf\underline{Given:-}$

• n(A - B) = 14 + x
• n(B - A) = 3x
• n(A ∩ B) = x
• n(A) = n(B)

$\huge\bf\underline{Solution:-}$

We can write n(A) as :-

➛ n(A) = n(A - B) + n(A ∩ B)

➛ n(A) = 14 + x + x

➛ n(A) = 14 + 2x ......1)

we can write n(B) as :-

➛n(B) = n(B - A) + n(A ∩ B)

➛ n(B) = 3x + x

➛ n(B) = 4x ......2)

It is given in the Question that n(A) = n(B)

then , From 1) and 2)

➛ 14 + 2x = 4x

➛ 14 = 4x - 2x

➛ 14 = 2x

➛ x = 14/2

• ➛ x = 7

hence value of x = 7

Now,

➛ n(A - B) = 14 + x

➛ n(A - B) = 14 + 7

• ➛ n(A - B) = 21

➛ n(B - A) = 3x

➛ n(B - A) = 3 × 7

• ➛ n(B - A) = 21

➛ n(A ∩ B) = x

• ➛ n(A ∩ B) = 7

So,

• n(A ∩ B) = 7
• n(B - A) = 21
• n(A - B) = 21
• n(A ∪ B) = ?

By using Formula

$\underline{\boxed{ \bf \:n(A \cup \: B) + n(A \cap \: B) = n(A) + n(B) }}$

➛ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

➛ n(A ∪ B) = [n(A - B) + n(A ∩ B)] + [n(B - A) + n(A ∩ B)] - n(A ∩ B)

➛n(A ∪ B) =[ 21 + 7 ] + [ 21 + 7] - 7

➛n(A ∪ B) = 42 + 14 - 7

➛ n(A ∪ B) = 56 - 7

➛ n(A ∪ B) = 49

Hence,

• Value of x = 7
• n(A ∪ B) = 49

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Answer 2

Answer:

Value of x = 7

n(A ∪ B) = 49