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Question
If the sets A and B are defined as
A={x:x=2n, n∈N, n<100}B={x :x=3n, n∈N, n<100}
then, the number of elements in A U B is
If the sets A and B are defined as
A={x:x=2n, n∈N, n<100}B={x :x=3n, n∈N, n<100}
then, the number of elements in A U B is
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Solution
The correct option is A 165
A={x:x=2n, n∈N, n<100}={2,4,6,8,.....,198}B={x :x=3n, n∈N, n<100}={3,6,9,12,....,297}n(A)=n(B)=99The elements in A∪ B will be all the elements of A and Btaken together, but the common elements will written only once. As common elements of A and B will be written twice while writing elements of A and B. Therefore, we subtract the number of common elements from the sum of number of elements in A and elements in B to get the number of elements in A∪B.n(A∪B)=n(A)+n(B)−n(A∩B)The elements in A∩B will be the common multiples of 2 and 3 which are less than 199.Common multiples of 2 and 3 are multiple of 6. So, A∩B will contain multiples of 6 less than 199.∴A∩B=(x: x=6n, n∈N, x<199}A∩B=(6, 12, 18, 24, ...., 198}n(A∩B)=33n(A∪B)=n(A)+n(B)−n(A∩B)⇒n(A∪B)=99+99−33=165
A={x:x=2n, n∈N, n<100}={2,4,6,8,.....,198}B={x :x=3n, n∈N, n<100}={3,6,9,12,....,297}n(A)=n(B)=99The elements in A∪ B will be all the elements of A and Btaken together, but the common elements will written only once. As common elements of A and B will be written twice while writing elements of A and B. Therefore, we subtract the number of common elements from the sum of number of elements in A and elements in B to get the number of elements in A∪B.n(A∪B)=n(A)+n(B)−n(A∩B)The elements in A∩B will be the common multiples of 2 and 3 which are less than 199.Common multiples of 2 and 3 are multiple of 6. So, A∩B will contain multiples of 6 less than 199.∴A∩B=(x: x=6n, n∈N, x<199}A∩B=(6, 12, 18, 24, ...., 198}n(A∩B)=33n(A∪B)=n(A)+n(B)−n(A∩B)⇒n(A∪B)=99+99−33=165