Answer
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Hint: In this question first assume the first set to be A and the second set to be B, now we need to find what kind of relation does A has with B to do this it is always recommended to write the sets in roster form as it gives a clear picture of the elements and helps in deriving relation between them. Use this hint to find the proper relation between set A and set B.
Complete step-by-step answer:
Let set {x|x is a multiple of 12} be A and set {x|x is a multiple of 6} be B.
Let’s represent A and B in roster form.
A = {0,12,24,36,48,60,….}
B = {0,6,12,18,24,30,36,….}
Now, you can see that every element of A is a part of B.
This can also be established as
12= 6$ \times $2
Now, every element in A = 12$ \times $a where a$ \in $N.
so A can also be represented as A = 6$ \times $2$ \times $a, where a$ \in $N.
So, option B is correct.
A$ \subset $B i.e. A is a proper subset of B
Note: The sign $ \subset $ used in the above question stands for proper subset which means set A is a subset of B that is not equal to B. In other words, if A is a proper subset of B, then all elements of A are in B but B contains at least one element that is not present in A.
Complete step-by-step answer:
Let set {x|x is a multiple of 12} be A and set {x|x is a multiple of 6} be B.
Let’s represent A and B in roster form.
A = {0,12,24,36,48,60,….}
B = {0,6,12,18,24,30,36,….}
Now, you can see that every element of A is a part of B.
This can also be established as
12= 6$ \times $2
Now, every element in A = 12$ \times $a where a$ \in $N.
so A can also be represented as A = 6$ \times $2$ \times $a, where a$ \in $N.
So, option B is correct.
A$ \subset $B i.e. A is a proper subset of B
Note: The sign $ \subset $ used in the above question stands for proper subset which means set A is a subset of B that is not equal to B. In other words, if A is a proper subset of B, then all elements of A are in B but B contains at least one element that is not present in A.
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