# Simple Interest - Aptitude Questions and Answers - RejinpaulPlacement

57.
A person borrowed Rs. 500 @ 3% per annum S.I. and Rs. 600 @ 4    1/2% per annum on the agreement that the whole sum will be returned only when the total interest becomes Rs. 126. The number of years, after which the borrowed sum is to be returned, is:

Solution:   Let the time be x years.

Then, [(500 * 3 * x)/100] + [(600 * 9 * x)/100] = 126.
=> 15x + 27x = 126
=> 42x = 126
=> x = 3.

Required time = 3 years.
58.
A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is: [C.B.I. 2003]

Solution:   Let the rat be R% p.a.

Then, [(5000 * R * 2)/100] + [(3000 * R * 4)/100] = 2200
100R + 120R = 2200
R = 2200/220 = 10

Rate = 10%
59.
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest? [Bank P.O 2003]

Solution:   Let the original rate be R%. Then, new rate = (2R)%

[(725 * R * 1)/100] + [(365.50 * 2R * 1)/(100 * 3)] = 33.50
=> (2175 + 725)R = 33.50 * 100 * 3 = 10050
=> R = 10050/2900 = 3.46.

Original rate = 3.46%.

The correct answer is none of these.
60.
The difference between the simple interest received from two different sources on Rs. 1500 for 3 years is Rs. 13.50. The difference between their rate of interest is: [S.S.C. 1999]

Solution:   [(1500 * R1 * 3)/100] - [(1500 * R2 * 3)/100] = 13.50
=> 4500(R1 - R2) = 1350
=> R1 - R2 = 1350/4500 = 0.3%
61.
Peter invested an amount of Rs. 12,000 at the rate of 10 p.c.p.a. simple interest and another amount at the rate of 20 p.c.p.a. simple interest. The total interest earned at the end of one year on the total amount invested became 14 p.c.p.a. Find the total amount invested. [S.B.I.P.O. 1999]

Solution:   Let the second amount be Rs. x. Then,

[(12000 * 10 * 1)/100] + [(x * 20 * 1)/100] = [(12000 + x) * 14 * 1]/100
=> 12000 + 20x = 168000 + 14x
=> 6x = 48000
=> x = 8000.

Total investment = Rs. (12000 + 8000) = Rs. 20000.
62.
What should be the least number of years in which the simple interest on Rs. 2600 at 6   2/3 % will be an exact number of rupees?