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Question

Answers

A. $16$

B. $12$

C. $8$

D. $4$

Answer

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Hint: Write the multiples of 4 and 6 for different sets A and B in roster form to make simplification easier.

As mentioned in the question, $A = \left\{ {x{\text{ : x is a multiple of 4}}} \right\}$

So the elements belonging to set A would be

$\left\{ {4,8,12,16,20,..............} \right\}$

Now it is also given in the problem statement that $B = \left\{ {x:x{\text{ is a multiple of 6}}} \right\}$

So elements of set B will be

$\left\{ {6,12,18,24............} \right\}$

Now we have to find a set that has common elements from both the above sets that is we have to find $A \cap B$

So from above two sets we can say that $A \cap B$$ = ${All the multiples which are common for both set A and set B]

$A \cap B = \left\{ {12,24,36...............} \right\}$

Now, clearly $A \cap B = \left\{ {x{\text{ : x is a multiple of 12}}} \right\}$

Note: Whenever we come across such problems with sets we simply need to think of all the possible elements of the set that satisfy the given relationship. This set wise approach ensures that any element belonging to a particular set is not missed.

As mentioned in the question, $A = \left\{ {x{\text{ : x is a multiple of 4}}} \right\}$

So the elements belonging to set A would be

$\left\{ {4,8,12,16,20,..............} \right\}$

Now it is also given in the problem statement that $B = \left\{ {x:x{\text{ is a multiple of 6}}} \right\}$

So elements of set B will be

$\left\{ {6,12,18,24............} \right\}$

Now we have to find a set that has common elements from both the above sets that is we have to find $A \cap B$

So from above two sets we can say that $A \cap B$$ = ${All the multiples which are common for both set A and set B]

$A \cap B = \left\{ {12,24,36...............} \right\}$

Now, clearly $A \cap B = \left\{ {x{\text{ : x is a multiple of 12}}} \right\}$

Note: Whenever we come across such problems with sets we simply need to think of all the possible elements of the set that satisfy the given relationship. This set wise approach ensures that any element belonging to a particular set is not missed.

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