Question
There are 3 sets with multiples 7, 6 and 4. All multiples are less than 560, also consider only odd multiples of 7. What is total elements in
CAT 2021
Correct option is
Explanatory Answer :
Let A , B , C denotes the set of multiples of 6 and multiples of 4 and odd multiples of 7 respectively
Then , A = { 6 , 12 , 18 ,..., 498 }
498 = 6 × 83
∴ n(A) = 83
B = { 4 , 8, 12 ,.... 496 }
496 = 4 × 124
∴ n(B) = 124
Again ,C = { 7 , 21 , 35 ..., 497 }
497 = 7 × 71
∴ n(C) = 36
Now ,A ∩ B = { 12 , 24 , 36 ,.., 492 }
492 = 12 × 41
∴ n(A ∩ B) = 41
Clearly B ∩ C = Φ and A ∩ C = Φ
∴ n(B ∩ C) = 0 and n(A ∩ C) = 0
Also A ∩ B ∩ C = Φ
∴ n(A ∩ B ∩ C) = 0
Now the total elements in union of 3 sets
= n(A ∪ B ∪ C)
= n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)
= 83 + 124 + 36 - 41 - 0 - 0 + 0
= 202
Hence the correct option is B
