A = {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84}

B = {9, 18, 27, 36, 45, 54, 63, 72, 81}

C = {21, 42, 63, 84}

Only one number is in common:63

I do not think there is other way than to enlist the numbers.

Sometimes is so to solve a question. Sometimes is trial and error...............it depends

By doing so you can solve the question in 30 seconds or something. After all. it is not so time consuming

Regards

Still we know that the most restrictive set is C so we can only take those four numbers in C and check whether they are multiples of 7 and 9.

The fastest and easiest way is to consider thinking in terms of LCM,what is the lowest number that would be common to 7,9,21?Ofcourse,it would be LCM.When you calculate LCM for these 3,it would be 63,and since its the lowest common multiple,there is no chance of any common number occuring before 63 and the next common is 84,which takes the count to 2.

Therefore, A is 2 and B is 4,hence B is the answer

84 is not a multiple of 9

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A quick way to do this one is to find out the number of multiples in each set:

Set A:

We know 70 is a multiple of 7. So 77 and 84 are as well, making a total of 12 multiples of 7 in the range 1-89. That means there are 12 numbers in Set A.

Set B:

By the same logic above, we know that 9*9 = 81, so are there are 9 numbers in Set B.

Set C:

21*4 = 84, so there are 4 numbers in Set C.


Now notice Quantity B is 4.

Quantity A can't be more than 4, because Set C only has 4 numbers.

Quantity A also can't be equal to 4 because that would mean that the 4 numbers in Set C are also in Set A and B. Now the 4 numbers are in Set A because 21 is a multiple of 7, but not all of the four multiples of 21 in Set C are in Set B (only 63 is). So only 1 number is in common with all three sets.

To think about this quickly without writing out the multiples of the sets, you can just notice that 21 is not a multiple of 9, which means the numbers in common can't be 4 and, subsequently, must be less than 4.

So Quantity B is greater.

Just take LCM of 3 numbers together, you will get 3*3*7. Not sure if we have to consider 3 twice or once but in any case 3 numbers are less than 4 numbers so option B is the answer.

by considering the mutliples of 7,9,21 in set of integers between 1 and 89 only 63 will be common

so quant A:1
quant B:4

quant A < quant B

answer will be B