Quantitative Aptitude - Arithmetic Ability Questions

Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the numberof years for which it is invested, r is the rate of interest per annum

In this case, Rs. 1250 has become Rs.10,000.

Therefore, the interest earned = 10,000 – 1250 = 8750.

8750 = [(1250 x n x 12.5)/100]

=> n = 700 / 12.5 = 56 years.

Quantity of pure acid = 20% of 8 litres = ((20/100)*8) litres = 1.6 litres.

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )

Counting of odd days:

 1600 years have 0 odd day. 300 years have 1 odd day.

 47 years = (11 leap years + 36 ordinary years)= [(11 x 2) + (36 x 1) ]odd days = 58 odd days = 2 odd days.

Jan  Feb  Mar  Apr  May  Jun  Jul  Aug.

31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 days = (32 weeks + 3 days) = 3,

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was 'Saturday'.

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

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.