Aptitude and Reasoning Questions

2 and 3 are the satements involving the extreme terms of the last 2 premises and form a complementary pair. Thus either 2 or 3 follows

The given number sequence is 101, 98, 93, 86, 75, ?

101

101 - 3 = 98

98 - 5 = 93

93 - 7 = 86

86 - 11 = 75

75 - 13 = 62

Here the series follows a pattern that consecutive subtraction of prime numbers.

Hence, the next number in the sequence is 62.

Clearly, A’s new position is 15th from the left. But this is the same as B’s earlier position which is 9th from the right.

Here the letter series is alternative reverse order series. So we get P, N.

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 

Jan   Feb   Mar   Apr   May   Jun   Jul  

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 

Total number of odd days = (0 + 2) = 2.  

Required day was 'Tuesday'.

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

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

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

Given equation can be written as

$$\begin{gathered} 215 x 103 + ? = 24 \\ => 14 + ? = 24 \\ => ? = 24 - 14 \\ => ? = 10 \end{gathered}$$

A : B = 100 : 90.

A : C = 100 : 87.

B/C = (B/A*A/C) = (90/100*100/87) = 30/29

When B runs 30 m, C runs 29 m.

When B runs 180 m, C runs (29/30*180)m =174m

 B beats C by (180 - 174) m = 6 m.