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Set $U = \left\{ {1,2,3,4,5,6,7,8,9} \right\},$ $A = \left\{ {x:x \in N,30 \leqslant {x^2} \leqslant 70} \right\},$$B = \left\{ {x:x{\text{ is a prime number < 10}}} \right\}$.
Which of the following does not belong to the set $\left( {A - B} \right)'$ ?
$
  {\text{A}}{\text{. 4}} \\
  {\text{B}}{\text{. 5}} \\
  {\text{C}}{\text{. 6}} \\
  {\text{D}}{\text{. 9}} \\
$

seo-qna
Last updated date: 22nd Mar 2024
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Views today: 11.12k
MVSAT 2024
Answer
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Hint: To solve this question first we have to find how many elements of set ‘U’ will satisfy the condition by set ‘A’ by squaring all elements of ‘U’ and checking condition of ‘A’ and also check how many elements will satisfy the condition of set ‘B’ . then proceed using properties of set theory.

Complete step-by-step answer:
We have
$U = \left\{ {1,2,3,4,5,6,7,8,9} \right\}$
We have set $A = \left\{ {x:x \in N,30 \leqslant {x^2} \leqslant 70} \right\}$
So $A = $ $\left\{ {6,7,8} \right\}$will be in the set ‘A’ according to condition.
We have set $B\left\{ {x:x{\text{ is a prime number < 10}}} \right\}$
So $\left\{ {2,3,5,7} \right\}$ will be in set B according to condition.
$\because B = \left\{ {2,3,5,7} \right\},\therefore B' = \left\{ {1,4,6,8,9} \right\}$
$\left( {A - B} \right) = A \cap B'$ ( property of sets and relation)
$
  \therefore \left( {A - B} \right) = \left\{ {6,7,8} \right\} \cap \left\{ {1,4,6,8,9} \right\} \\
   \Rightarrow \left( {A - B} \right) = \left\{ {6,8} \right\} \\
   \Rightarrow \left( {A - B} \right)' = \left\{ {1,2,3,4,5,7,9} \right\} \\
$
Hence option C is the correct option.

Note: Whenever we get this type of question the key concept of solving is we should have remembered properties of sets and relation like $\left( {\left( {A - B} \right) = \left( {A \cap B} \right)'} \right)$ then we can solve easily. And should have knowledge of taking intersections and union.