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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

A
B
C
D
##### 237
Solution

Correct option is

(B)

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