SSC CHSL 2019 PYQ — Question 66

The correct answer is option 3 i.e. 20

For divisibility by 88, number should be divisible by 11 and 8 both.

6220x558y2 is divisible by 88

For divisible by 8 last 3 digit of the number (8y2) must be divisible by 8.

So, the value of y can be 3 or 7.

For divisibility by 11, difference of sum of digits at odd place and sum of digits at even place must be zero or divisible by 11.

So, (6 + 2 + x + 5 + y) - (2 + 0 + 5 + 8 + 2) = 0 or 11

Now, x + y– 4 = 0

x + 3 – 4 =0 (for y=3)

x = 1 ……digit cannot be negative

And,

x + 7 – 4 = 0 (for y = 7)

x = -3 ……digit cannot be negative

Hence,

We have x = 1 and y = 3

5x + 5y = 5 + 15 = 20