Question

If X is the universal set and A, B are subsets of X such that n(X) = 100, n(A') = 60, n(B') = 45 and n[(AUB)'] = 10, find (i) n(A n B) (ii) n(A'U B').​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm \: n(X) = 100$

$\rm \: n(A') = 60$

$\rm\implies \:n(A) = n(X) - n(A') = 100 - 60 = 40$

Also,

$\rm \: n(B') = 45$

$\rm\implies \:n(B) = n(X) - n(B') = 100 - 45 = 55$

Also, given that

$\rm \: n[(A\cup B)'] = 10$

$\rm\implies \:\rm \: n[(A\cup B)] = n(X) - n[(A\cup B)'] = 100 - 10 = 90$

So, we have now

$\rm \: n(A) = 40$

$\rm \: n(B) = 55$

$\rm \: n(A\cup B) = 90$

Now, we know

$\rm \: n(A\cup B) = n(A) + n(B) - n(A\cap B)$

So,

$\rm \: n(A\cap B) = n(A) + n(B) - n(A\cup B)$

On substituting the values, we get

$\rm \: n(A\cap B) = 40 + 55 - 90$

$\rm \: n(A\cap B) = 95 - 90$

$\rm\implies \:\boxed{\sf{  \:\rm \: n(A\cap B) = 5 \: \: }}$

Also,

$\rm \: n(A'\cup B')$

$\rm \: =  \:n(X) - n(A\cap B)$

$\rm \: =  \:100 - 5$

$\rm \: =  \:95$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: n(A'\cup B') = 95 \: \: }}$

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Additional Information :-

$\rm \: n(A - B) = n(A) - n(A\cap B)$

$\rm \: n(B - A) = n(B) - n(A\cap B)$

$\rm \: n(A\cup B) = n(A - B) + n(A\cap B) + n(B - A)$

$\rm \: n(A'\cap B') = n(U) - n(A\cup B)$

Answer 2

$\sf \fbox \red{Answer}$

$\sf \: = n(x) = 100 \: \: \: \: \: \: \: \: \: \: \: \\ \sf \:= n ('A') = 60\: \: \: \: \: \: \: \: \: \: \\ \sf \: = n(B') = 45 \: \: \: \: \: \: \: \: \: \: \\ \sf \: = n(A \: \cup \ \: B)' = 10$

$1) \: \sf \: = n(A \: \cap \: B)' = \: n(A') + n(B') - n(A \: \cup \: B)' \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 60 + 45 - 10 \\ \\ \sf \: = n(A \: \cap \: B) = 100 - ( \: A \: \cap \: B)' \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: = 100 - 95 \\ \sf \: = 5$

$2) \sf \: = (A \: \cup \: B) = (A \: \cap \: B)' \\ \sf \: = 95$

therefore

1. n(A∩B) = 5
2. n('A∪B') = 95