Quantitative Aptitude - Arithmetic Ability Questions

Let number of boys = x  , Let number of girls = y
Total numbers of students = x + y
(x + y) × 15.8 = 16.4x + 15.4y
0.6x = 0.4y
x/y = 0.4/0.6 = 2/3

 Ratio of boys and girls in the class is 2:3

(A+B)'s 1 day's work=1/20 

C's 1 day work=1/60 

(A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15

Also A's 1 day's work =(B+C)'s 1 day's work 

Therefore,  we get: 2 x (A's 1 day 's work)=1/15 

=>A's 1 day's work=1/30 

Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60

So, B alone could do the work in 60 days.

Concentration of petrol in    A        B        C

                                                1/2      3/5     4/5

Quantity of petrol taken from A = 1 litre out of 2 litre

Quantity of petrol taken from B = 1.8litre out of 3 litre

Quantity of petrol taken from C = 0.8 litre out of 1 litre

Therefore, total petrol taken out from A, B and C = 1+1.8+0.8 =3.6 litres

So, the quantity of kerosen =(2+3+1) - 3.6 =2.4 litre

Thus, the ratio of petrol to kerosene = 3.6/2.4 = 3/2

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs.22053×5= Rs.3675

Principle = Rs.(9800-3675) = Rs.6125

Hence, Rate = 100×36756125×5 =12%

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.

When x=2. [(22 + 2) = 24 is divisible by 3]

So, the answer is 2

required percentage = πr22πr×100 =250%

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,
[ 1000 (1+ (10/100))^n ] = 1331 or (11/10)^n = (1331/1000) = (11/10)^3
n = 3 years