Quantitative Aptitude - Arithmetic Ability Questions

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

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

P.W = Rs. (1760 - 160) = Rs.1600  

S.I on Rs.1600 at 12% is Rs. 160 

Time = (100 x 160)/(1600 x 12)  = 5/6 years = (5/6) x 12 months = 10 months

The time taken by the leak to empty the tank = 18 - 110 = 5 - 440 = 140

Therefore, the leak empties the tank in 40 hours.

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 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

Let F and C denotes the temparature in Fahrenheit anid Celcsius respectively.

Then, (F - 32) /(212 - 32) = (C - 0)/(100 - 0),  if c = 35, then F = 95.

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.

As now, the time 4 : 48 pm after 45 min it will be

4:48 + 45 min = (4 hrs : 48 + 12 min) pm + 33 min = 5 hrs : 33 min pm.

Hence, it is 5:33 pm.