Aptitude and Reasoning Questions

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March. April. May. June.
(31 + 28 + 31 + 30 + 31 + 17) = 168 days
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

P, Q, R are children of same parents. So. S who is R's mother and T, who is R's father will be mother and father of all three. 

However, it is not mentioned whether Q is male or female So, D cannot be definitely true. 

Here the given series 28, 12, 4, 2, -2 follows a pattern that

$$\begin{gathered} 28 \\ 28 - 24 = 28 - 16 = 12 \\ 12 - 23 = 12 - 8 = 4 \\ 4 - 22 = 4 - 4 = 0 (NOT 2) \\ 0 - 21 = 0 - 2 = -2 \end{gathered}$$

Hence, the odd man in the given series is 2.

D is father of A and grandfather of F. So, A is father of F. 

Thus. D and A are the two fathers. C is the sister of F So. C is the daughter of A. 

Since there is only one mother, it is evident that E is the wife of A and hence the mother of C and F. 

So, B is brother of A There are three brothers. So. F is the brother of C. 

 Clearly, A is E's Husband.

Here the given number series is 3, 14, 36, 67, 113, 168

It follows a pattern that

3

3 + 1 x 11 = 3 + 11 = 14

14 + 2 x 11 = 14 + 22 = 36

36 + 3 x 11 = 36 + 33 = 69

69 + 4 x 11 = 69 + 44 = 113

113 + 5 x 11 = 113 + 55 = 168

Hence, the wrong number is 67.

Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.