Aptitude and Reasoning Questions

From statement I. we can conclude that the Chairman visited Purchase department on Monday of the first week of September. So, I alone is sufficient, The time of visit of no department is mentioned in II. which is, therefore, insufficient 

First ten odd numbers form an arithmetic progression of the form
1,3,5,7,9......
here a = first term = 1
d = common difference = 2
Average of first n numbers = (2a + (n - 1)d)/2
n th term of the AP = a + (n - 1)d
Substituting a = 1, d = 2 and n = 10 in the above formulas
average of first 10 numbers = 10
10 th term of the AP = 19
Therefore average of first 10 terms is 19 - 10 = 9 greater than
the last term. Hence the average is greater than the sum by
9/19 X 100 = 900/19 %

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. 

Area of each tile = (41 x 41) sq.cm 

Required number of tiles = 1517 x (902/41) x 41 = 814

Since half of the four children are girls, two must be boys. It is not clear which children have blue or brown eyes. So only statement (b) follows.

Here 14×1521= 10 and 32×1428 = 16

In the same way  24×2624= 26.

With slight modifications, the basic formula can be made to deal with compounding at intervals other than annually.

Since the compounding is done at six-monthly intervals, 4 per cent (half of 8 per cent) will be added to the value on each occasion.

Hence we use r = 0.04. Further, there will be ten additions of interest during the five years, and so n = 10. The formula now gives:

V = P(1 + r)10 = 5,000 x (1.04)10 = 7,401.22

Thus the value in this instance will be £7,401.22.

In a case such as this, the 8 per cent is called a nominal annual

rate, and we are actually referring to 4 per cent per six months.