Question

Q. 9 Suppose the number of elements in set A is p, the number of elements in set B is q and the number of elements in AXB is 7. then p2 +q2 equal to
1 O
49
2) O
50
3) O
42
4)
51​

Answer 1

Answer:

50

Step-by-step explanation:

prime factorisation of 7 is 7*1 which is the only way in which the no. of elements will be in the sets a and b . so let p be 7 and q be 1 . therefore 7^2+1^2= 49+1 = 50

.

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Answer 2

The value of p² + q² = 50

Given :

Suppose 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 × B is 7

To find :

The value of p² + q² is

1) 49

2) 50

3) 42

4) 51

Solution :

Step 1 of 3 :

Define Cartesian product

Let A and B are two sets. Then the Cartesian product of A and B is denoted by A × B and defined as

$\sf{A \times B = \{(x, y) : x \in A \: \: and \: \: y \in B \}}$

Step 2 of 3 :

Find the possible value of p and q

The number of elements in set A is p

The number of elements in set B is q

Then the number of elements in A × B

= n( A × B )

= n(A) × n(B)

= p × q

= pq

Now it is given that the number of elements in A × B is 7

So we have pq = 7

Since 7 is a prime number

Then two cases arise

p = 7 , q = 1

p = 1 , q = 7

Step 3 of 3 :

Find the value of p² + q²

Case : I

p = 7 , q = 1

p² + q² = 7² + 1² = 49 + 1 = 50

Case : II

p = 1 , q = 7

p² + q² = 1² + 7² = 1 + 49 = 50

Therefore in both cases p² + q² = 50

Hence the correct option is 2) 50

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