Bank PO Questions

In the given puzzle,

The logic is in the format of lets say, a/b

In that case b doesn't matter, because

2/2 = 1,

6/5 = 6/2 = 3,

10/11 = 10/2 = 5

This means that equation is a/2

So 16/19 is equal 16/2 = 8.

Let Mohan present age = p

ATQ,

5/6 x p = p - 5

p = 30

His wifes age = 30 - 3 = 27 yrs.

Px85100=120

=> p = 12000/85 = 141.17 %

As,

 A→+2C

 F→+2H

Similarly,

 $$\begin{gathered} K→+2M \\ T→+2V \end{gathered}$$

1113122110

In given series 2nd number gives number of  which digits present in the 1st number and so on continues for 3rd ,4th,...

Here 1st no. has only one zero, so  2nd no. is 10 and 2nd no. has one 1 and one 0,so 3rd no. is 1110, so like that 5th no. has one 1,one 3,one 2,two 1's and one 0.So the next no. is 1113122110. 

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

Total : C121×C320+C321×C120= 50540 ways.

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.