Quantitative Aptitude - Arithmetic Ability Questions

Let number of boys in the school = x

Let number of girls in the school = y

Number of students participated in the sports = 300

Out of which boys = 100

but given number of boys participated in sports = x/3

=> x/3 = 100

=> x = 300

Remaining girls = 300 - 100 = 200

=> y/2 = 200

=> y = 400

Therefore, total number of students in the school = x + y = 300 + 400 = 700

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

Given year 1856, when divided by 4 leaves a remainder of 0.

NOTE: When remainder is 0, 28 is added to the given year to get the result.

So, 1856 + 28 = 1884

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.

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.

P=2(l+b)

l=4b

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 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.