2013 RRB Allahabad

22 boys are standing in front of Kunal in a Queue 12 boys are standing to the back of Rohit
Total boys = 30.
So, rank of Kunal in the queue is 22+1 = 23
Rank of Rohit in the queue is 30 - 12 = 18
Hence, the number of boys between Rohit and Kunal in the queue is 23 - 18 - 1 = 4

There is only one a between the 7th 'a' from the left and 7 th 'a' from the right.

a a b a b a a b a b a a a b b a b a b b a a a a

A stands for ‘+’, B stands for ‘-’, C stands for ‘x’ and D for ‘$$\div$$’ then
$$\frac{1}{2} A \frac{1}{3} B \frac{1}{4} C \frac{1}{5} D \frac{1}{6}$$
= $$\frac{1}{2} + \frac{1}{3} - \frac{1}{4} \times \frac{1}{5} \div \frac{1}{6}=$$
= $$\frac{1}{2} + \frac{1}{3} - \frac{3}{10}$$
= $$\frac{15}{30} + \frac{10}{30} - \frac{9}{30}$$
= 16/30 = 8/15

There are 7 years between the two except two of them are leap years.

In a normal year, there are 365 days. 365 when divided by 7 leaves a reminder of 1. So, 1 July of a year and 1 July of the previous year are one weekday apart. If it is a leap year, 1 July of a year and 1 July of a previous year are separated by 2 weekdays. So, If July 1, 1977 was a Friday, 1st july 1976 is a Thursday and 1st july 1975-tuesday 1st july 1974-monday 1st july 1973-sunday 1st july 1972-saturday 1st july 1971-thrusday 1st july 1970- wednesday

The given equation means P is the father of Q and Q is the sister of R. So, P maybe the father of R.

In 278 and 853, 8 and fast are common.

In 853 and 376, 3 and come are common

So, back is represented by 5 in 853.

The conclusion can be drawn from "none but stout men are players" since it implies that all players have to be stout. 

Campaign is generally related to a military manoeuvre. Others are unrelated to military.

The opposite of Natural is artificial. The opposite of spontaneous is calculated. 

301 is divisible by 7 (43 times)

497 is also divisible by 7 (71 times)

So, the required sum is

7 ( 43 + 44 + .... + 71)

= 7 x (sum of numbers upto 71 - sum of numbers upto 42) = 7 * [(71x 72/2)-(42 x 43/2)]

= 11571