Question 1

An automobile financier claims to be lending money at simple interest, but he includesthe interest every six months for calculating the principal. If he is charging an interest of10%, the effective rate of interest becomes:

Accepted Answer

Let the sum be Rs. 100. Then, S.I. for first 6 months = Rs.[ (100 x 10 x 1)/(100 x 2) ]= Rs.5 S.I. for last 6 months =Rs.[(102 x 10 x 1)/(100 x 2) ] = Rs.5.25 So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25 Effective rate = (110.25 - 100) = 10.25%

Question 2

What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?

Accepted Answer

Let the principal be P and rate of interest be R%. Required ratio = P * R * 6 100 P * R * 9 100 = 6 P R 9 P R = 6 9 = 2 3

Question 3

Rs.1500 is divided into two parts such that if one part is invested at 6% and the other at 5% the whole annual interest from both the sum is Rs.85. How much was lent at 5% ?

Accepted Answer

(kx5x1)/100 + [(1500 - k)x6x1]/100 = 85 5k/100 + 90 – 6k/100 = 85 k/100 = 5 => k = 500

Question 4

A sum of money becomes 7/6 of itself in 3 years at a certain rate of simple interest. The rate per annum is ?

Accepted Answer

Let sum = S. Then, amount = 7S/6 S.I. = 7S/6 - S = S/6; Time = 3 years. Rate = (100 x S) / (S x 6 x 3) = 5 5/9 = 50/9 %.

Question 5

Manu invested an amount of Rs. 14800 at the rate of 7% per annum. After how many years will she obtain a simple interest of Rs.15800?

Accepted Answer

We know that, I = PTR/100 ATQ, 15800 = 14800 x T x 6/100 T = 15800/148x7 T = 15800/1036 T = 15.25 yrs.

Question 6

A sum of money invested for a certain number of years at 8% p.a. simple interest grows to Rs.180. The same sum of money invested for the same number of years at 4% p.a. simple interest grows to Rs.120 only. For how many years was the sum invested?

Accepted Answer

From the information provided we know that, Principal + 8% p.a. interest on principal for n years = 180 …….. (1) Principal + 4% p.a. interest on principal for n years = 120 ……… (2) Subtracting equation (2) from equation (1), we get 4% p.a. interest on principal for n years = Rs.60. Now, we can substitute this value in equation (2), i.e Principal + 60 = 120 = Principal = Rs.60. We know that SI = , where p is the principal, n the number of years and r the rate percent of interest. In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60. Therefore, 60 =(60*n*4)/100 => n = 100/4 = 25 years.

Question 7

On Rs. 3500 invested at a simple interest rate of 7 per cent per annum, Rs. 500 is obtained as interest in certain years. In order to earn Rs. 800 as interest on Rs. 4900 in the same number of years, what should be the rate of simple interest?

Accepted Answer

From the given data, 3500x7xt/100 = 500 => t = 100/49 years Now, in the second case The interest per year = 49/100 x 800 = 392 => 4900 x 1 x r/100 = 392 => r = 8%

Question 8

A Rent 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:

Accepted Answer

Let the rate be R% p.a. Then, 5000 × r × 2 100 + 3000 × r × 4 100 = 2200 Rate = 10%

Question 9

What is the principal sum? I. The sum amounts to Rs. 690 in 3 years at S.I. II. The sum amounts to Rs. 750 in 5 years at S.I. III. The rate of interest is 5% p.a.

Accepted Answer

Clearly, any of the three will give us the answer

Question 10

How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

Accepted Answer

Time = (100 x 81)/(450 x 4.5) years = 4 years.

Question 11

If the rate increases by 2%, the simple interest received on a sum of money increases by Rs. 108. If the time period is increased by 2 years, the simple interest on the same sum increases by Rs.180. The sum is ?

Accepted Answer

Let the sum be Rs. p, rate be R% p.a. and time be T years. Then, P × T × ( R + 2 ) 100 - P × T × R 100 = 108 ⇒ 2 P T = 10800 . . . . . . ( 1 ) And, P × R × ( T + 2 ) 100 - P × R × T 100 = 180 ⇒ 2 P R = 18000 . . . . . . ( 2 ) Clearly, from (1) and (2), we cannot find the value of p So, the data is not sufficient.

Question 12

If J is the interest on K and K is the interest on L, the rate and time is the same on both the cases. What is the relation between J, K and L ?

Accepted Answer

Let the Time be 'N' and Rate be 'R' J = (K x NR)/100 K = (L x NR)/100 J/K = NR/100 K/L = NR/100 J/K = K/L K x K = JL

Question 13

A sum at simple interest at 1312 % per annum amounts to Rs. 2502.50 after 4 years. Find the sum

Accepted Answer

Let sum be Rs.x then, S.I. =Rs. [ x * (27/2) * 4 * (1/100) ] = Rs. 27x/50 Amount =Rs [ x + (27x/50)] = Rs.77x/50 => 77X/50 = 2502.50 =>X= (2502.50 * 50)/77 = 1625 Hence Sum = Rs.1625

Question 14

A car is purchased on hire-purchase. The cash price is $21 000 and the terms are a deposit of 10% of the price, then the balance to be paid off over 60 equal monthly instalments. Interest is charged at 12% p.a. What is the monthly instalment?

Accepted Answer

Cash price = $21 000 Deposit = 10% × $21 000 = $2100 Loan amount = $21000 − $2100 = $18900 I=p x r x t/100 I=11340 Total amount = 18900 + 11340 = $30240 Regular payment = total amount /number of payments

Question 15

On simple interest an amount becomes Rs. 812 in two years and Rs. 924 in four years. What is the rate of interest on that amount?

Accepted Answer

We know, S.I = PTR/100 where P = principal amount, T = time, R = rate of interest Here in the given data, Interest for two years S.I = 924 - 812 = Rs. 112 Now, Principal amount P = 812 - 112 = Rs. 700 Now, R = S.I x 100/PT R = 112 x 100/700 x 2 R = 11200/1400 R = 8% Hence, the rate of interest R = 8%.

Question 16

A sum of Rs. 2600 is invested at two different rates of interest. The difference between the simple interests got after 4 years is Rs 402.80 at these two rates, what is the difference between the rates of interest?

Accepted Answer

Let the two different rates of interests be r1 and r2 respectively. From the given data, 2600 x 4 x r 1 - r 2 100 = 402 . 80 r 1 - r 2 = 402 . 80 104 = 3.87 Hence, the difference in the interest rates = 3.87

Question 17

A simple interest earned on certain amount is triple the money when invested for 16 years.what is the interest rate offered ?

Accepted Answer

Given, S.I = 3 Principal Amount => 3A = A x 16 x R/100 By solving, we get => R = 18.75%

Question 18

Find the simple interest on Rs.600 for 11 months at 5 paisa per month ?

Accepted Answer

I = (600x11x5)/100 = 330 Paise = Rs. 3.30

Question 19

In how many years, Rs. 200 will produce the same interest @ 5 % as Rs. 900 produce in 2 years @ 3 ½ % ?

Accepted Answer

P = Rs. 900, R = 3 ½ % = 7/2 %, T = 2 years. Therefore, S.I. = PTR/100 S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63 Now, P = Rs. 200, S.I. = Rs. 63, R = 5 % Time = ((100 x 63) / (200 x 5) ) years = 6.3 years.

Simple Interest Questions