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A= { x: x is a factor of 12 } ; B = { x: x is a factor of 6 } , then A - B =​

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CBSE

Mathematics

Grade 10

Sets

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If A = {x: x ∈ N, x is a factor of 12} and B = {x: x ∈ N, x is a factor of 18}. Find A∩B?

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Hint: The set A and set B given in the question is in set – builder form. So, first we need to convert set – builder form into a set of elements form then find A∩B which is the number of elements common in both the sets A and B.

Complete step-by-step answer:

Converting the set – builder form into the set of elements form:

A = {x: x ∈ N, x is a factor of 12}

The above set A is written in the set – builder form. The property of x here is that it is a natural number and is a factor of 12. So, the natural factors of 12 are the elements that contain in set A.

The factors of 12 are: 1, 2, 3, 4, 6 and 12

So, set A can be written as: n (A) = {1, 2, 3, 4, 6, 12}.

Now, writing the elements of set B.

B = {x: x ∈ N, x is a factor of 18}

The set – builder form of the above set B is similar to A, the only difference is that here x is a factor of 18. So, we are going to write the factors of 18.

The factors of 18 are: 1, 2, 3, 6, 9 and 18.

So, set B can be written as: n (B) = {1, 2, 3, 6, 9, 18}.

As you can see form the set A and set B, the common elements among both the sets are 1, 2, 3 and 6.

Hence, A∩B = {1, 2, 3, 6}.