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Question: Solve , If we have \(n\left( A \right)=15,n\left( A\cup B \right)=29,n\left( A\cap B \right)=7\) fin...

Solve , If we have n(A)=15,n(AB)=29,n(AB)=7n\left( A \right)=15,n\left( A\cup B \right)=29,n\left( A\cap B \right)=7 find n(B)n\left( B \right).

Explanation

Solution

Hint: Use the data given in the question , n(A)=15,n(AB)=29n\left( A \right)=15,n\left( A\cup B \right)=29 and n(AB)=7n\left( A\cap B \right)=7 and substitute in the formula , n(AB)=n(A)+n(B)n(AB)n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right) and thus find n(B)n\left( B \right) .

Complete step-by-step solution -
In the question we are provided with values of n(A),n(AB)n\left( A \right),n\left( A\cup B \right) AND n(AB)n\left( A\cap B \right) which is 15, 29, 7 respectively and we have to find value of n(B)n\left( B \right) .
Here in the question n(A)n\left( A \right) represent number of elements of A, n(AB)n\left( A\cup B \right) represent number of elements of A and B collectively and lastly n(AB)n\left( A\cap B \right) represent number of elements common to both A and B .
Here to find n(B)n\left( B \right) which means number of elements of B we can use the formula that is, n(AB)=n(A)+n(B)n(AB)n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right) .
So, on rearranging we can rewrite it as,
n(B)=n(AB)+n(AB)n(A)n\left( B \right)=n\left( A\cup B \right)+n\left( A\cap B \right)-n\left( A \right) .
On substituting values n(AB)n\left( A\cup B \right) as 29, n(AB)n\left( A\cap B \right) as 7 and n(A)=15n\left( A \right)=15 we get, n(B)=29+715n\left( B \right)=29+7-15 .
=21.
So the value of n(B)n\left( B \right) is 21.

Note: Students generally have confusion between the sign \cup and \cap . The sign \cup means union which means the ABA\cup B then it contains all the elements of A and B collectively and \cap means intersection like the ABA\cap B contains common elements of respective two sets A & B. At the time of using these signs, students need to be careful with their meaning.