Quantitative Aptitude - Arithmetic Ability Questions

Let the weight of the empty tin = w

(2 - w) = weight of the cookies; and one quarter of the cookies weigh (2 - w)/4

One quarter of the cookies + tin = 0.8 = w + (2 - w)/4

Multiply through by 4; 3.2 = 4w + 2 - w

1.2 = 3w; w = 0.4

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

Let the age of sushma be x and
the age of her son is y
Then five years before x-5=5(y-5) ...(1)
Five years hence x+5 = 3(y+5)-8 .....(2)

By soving (1) & (2), we get
5y - 15 = 3y + 7
y = 11 => x = 35

Therefore, the age of Sushma = 35 and her son = 11.

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

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

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.

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.

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.