KCET · Maths · Sets and Relations
Suppose that the number of elements in set \(A\) is \(p\), the number of elements in set \(B\) is \(q\) and the number of elements in \(A \times B\) is 7 , then \(p^2+q^2=\)
- A 50
- B 51
- C 42
- D 49
Answer & Solution
Correct Answer
(A) 50
Step-by-step Solution
Detailed explanation
Given, \(n(A)=p, n(B)=q\) and \(n(A \times B)=7\)
Since, \(n(A \times B)=n(A) \times n(B)\)
\[
\Rightarrow \quad 7=p \times q \Rightarrow p q=7
\]
So, possible values of \(p\) and \(q\) are 7,1 respectively.
\[
\Rightarrow p^2+q^2=7^2+1^2=49+1=50
\]
Since, \(n(A \times B)=n(A) \times n(B)\)
\[
\Rightarrow \quad 7=p \times q \Rightarrow p q=7
\]
So, possible values of \(p\) and \(q\) are 7,1 respectively.
\[
\Rightarrow p^2+q^2=7^2+1^2=49+1=50
\]
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