SSC CGL 2022 PYQ — Question 63

The correct answer is option 4 i.e. 0

99 = 9 × 11

So the number should be divisible by both 9 and 11

Now,

4 + 8 + k + 2 + 0 + 4 + 8 + p + 6 = 32 + k + p

Now,

Possible values of k + p = 4, 13

Again,

(4 + k + 0 + 8 + 6) - (8 + 2 + 4 + p) = 0 or multiple of 11

⇒ 4 + k - p = 0 or multiple of 11

Now,

k - p can be - 4 or 7

Now,

For k + p = 4 and k - p = - 4

We can say by adding both 2k = 0

⇒ k = 0

So, p = 4 (Both possible)

Again,

For k + p = 4 and k - p = 7

We can say by adding both 2k = 11

⇒ k = 5.5 (Not possible)

Again,

For k + p = 13 and k + p = - 4 

We can say by adding both 2k= 9

⇒ k = 4.5  (Not possible)

Again,

k + p = 13 and k - p = 7

We can say by adding both 2k = 20

k = 10 (not possible)

So the only possible value of k and p is 0 and 4

Now k × p = 0 × 4 = 0