Aptitude and Reasoning Questions

Similar small figures are increasing by one in each of the bigger figure.

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.

For an income of Rs. 10, investment = Rs. 96. 

For an income of Rs. 12, investment = Rs. (96/10)*12 = Rs. 115.20.

Number of square units in 13 by 9 is given by the area it forms with length and breadth as 13 & 9

Area = 13 x 9 = 117

Hence, number of square units in 13 by 9 is 117 sq.units.

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

Total number of odd days = (0 + 0 + 3 ) = 3 odd day.

Given day is Wednesday.

area of the plot = 120m x 60m = 7200 sq.m
area of plot excluding gravel = 117m x 57m = 6669 sq.m
area of gravel = 7200 sq.m - 6669 sq.m = 531 sq.m
cost of building it = 531 sq.m x 65 ps = 34515 ps
In Rs. = 34515/100 = Rs. 345.15.

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

⇒14+112=13

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours

Thus, work will be completed at 3pm.