Question 1

Rs. 4200 amounts to Rs. 5712 in 4 years. If rate of interest is increased by 3%. What will be the amount?

Accepted Answer

S.I = Principal x Rate x Time / 100 ∴ Amount - Principal = Principal x Rate x Time / 100 ⇒ 5712 - 4200 = (4200 x Rate x 4) /100 ⇒ 1512 = 42 x 4 x Rate Rate = 1512 / (42 x 4) = 9% On increasing 3%, new rate of interest = 9 + 3 = 12% Simple interest on new rate = (4200 x 12 x 4) / 100 =Rs. 2016 Amount = Principal + Interest = 4200 + 2016 = Rs. 6216

Question 2

In how many years, a sum will become four times itself at the rate of 12% per annum?

Accepted Answer

Let the sum be ₹ P . Then, SI = 4P - P = ₹ 3P (P x 12 x T)/100 = 3P ⇒ T = (3 x 100)/12 = 25 yr

Question 3

At simple interest of 5%, 6% and 8% for three consecutive years, the interest earned is ₹ 760. find the principle.?

Accepted Answer

Let the principle be ₹ P . Then, (P x 1 x 5)/100 + (P x 1 x 6)/100 + (P x 1 x 8)/100 = 760 ⇒ 5P/100 + 6P/100 + 8P/100 = 760 ⇒ 19P = 760 x 100 ∴ P = (760 x 100)/19 = ₹ 4000 Hence, the principle is ₹ 4000.

Question 4

What would be the simple interest accrued in 4 yr on a principle of ₹ 18440 at the rate 15% per annum?

Accepted Answer

Given, P = 18440, R = 15%, T = 4 ∴ Simple interest (SI) = (P x R x T) / 100 = (4 x 18440 x 15) / 100 = ₹ 11064

Question 5

What will be the simple interest on ₹ 4000 at 12 1/ 2% per annum for the period from 4th February, 2005 to 18th April, 2005?

Accepted Answer

Here, P = ₹ 4000, R = 121/2% = 25 / 2% T = (24 + 31 + 18) days = 73 days = 73/365 yr = 1/5 yr ∴ SI = (P x R x T)/100 = (4000 x 25/2 x 1/5 )/100 = ₹ 100

Question 6

Rakesh lent out ₹ 8750 at 7% annual interest, Find the simple interest in 3 yr.

Accepted Answer

Given, P = ₹ 8750., R = 7% T = 3 yr According to the formula. SI = (P x R x T)/100 = (8750 x 7 x 3)/100 = ₹ 1837.50

Question 7

What will be simple interest for 1 yr and 4 months on a sum of ₹ 25800 at the rate of 14% per annum?

Accepted Answer

Here, P = ₹ 25800, R = 14% T = 1 yr 4 months = 1 + (4/12) years = 1 + (1/3) = 4/3 yr. According to the formula, SI = (P x R x T) / 100 = {25800 x 14 x (4/3)}/100 = (258 x 14 x 4)/3 = ₹ 4816

Question 8

A sum at simple interest of 13 1/ 2% per annum amounts to ₹ 3080 in 4 yr Find the sum.

Accepted Answer

Let sum = P Then, SI = (P x R x T)/100 = {P x (27/2) x 4} / 100 = {P x 54} / 100 = 27P / 50 ∴ Amount = P + 27P / 50 = 77P/50 According to the question 77P/50 = 3080 ∴ P = (3080 x 50)/77 = ₹ 2000

Question 9

The sum which amount to ₹ 364.80 in 8 yr at 3.5% simple interest per annum is

Accepted Answer

Given, t = 8 yr, r = 3.5%, A = ₹ 364.80 Let amount = ₹ P Since, A =P [1 + (RT/100)] ∴ 364.80 = P[1 + (3.5 x 8)/100] ⇒ 364.80 = P[ 1+ (35 x 8)/100] ⇒ 3648/10 = P x (128/100) ⇒ P = 3648/128 = ₹ 285

Question 10

A sum of ₹ 2668 amount to ₹ 4669 in 5 yr at the rate of simple interest . Find the rate per cent.

Accepted Answer

Here, P = ₹ 2668, T = 5 yr, A = ₹ 4669 We know that, Amount (A) = Principal (P) + (Simple Interest) SI 4669 = 2668 + SI ⇒ SI = 4669 - 2668 = ₹ 2001 Again, SI = (P x R x T)/100 ∴ 2001 = (2668 x R x 5)/100 ⇒ R= (2001 x 100) / (2668 x 5) = (2001 x 5)/667 = 15%

Question 11

Find the difference in amount and principal for ₹ 4000 at the rate of 5% annual interest in 4 yr.

Accepted Answer

The required difference in amount and principal is SI = A - P Here, P = ₹ 4000, R = 5% T = 4 yr According to the formula. SI = (P x R x T) /100 = (4000 x 5 x 4)/100 = ₹ 800

Question 12

A sum becomes its double in 10 yr. Find the annual rate of simple interest.

Accepted Answer

Here, n = 2, T = 10 yr ∴ R = 100 (n - 1)/T = 100(2 - 1)/10 = 100/10 = 10%

Question 13

How long will a sum of money invested at 5% per annum SI take to increase its value by 50%

Accepted Answer

Let sum be P. ∴ 50% of P = P/2 = SI Now, P/2 = (P x 5 x T) / 100 [ as time = 10 yr] ⇒ P/2 = 5PT / 100 ⇒ 1/2 = T/20 ∴ = T = 10 yr

Question 14

A certain sum becomes 8 fold in 15 yr at simple interest. What will be the rate interest?

Accepted Answer

Let sum = p Then, after 15 yr Sum = 8p ∴ SI = 8p - P = 7P Now, 7P = (P x R x 15)/100 ⇒ 7 = 15R/100 = 3R/20 ∴ R = (20 x 7)/3 = 140/3 = 462/3%

Question 15

At a simple interest, a sum amounts to ₹ 1012 in 2 1/ 2 and becomes ₹ 1067.20 in 4 yr. What is the rate of interest

Accepted Answer

Given, T1 = 21/2 yr, T2 = 4 Yr According to the question. [P + (P x R x 4)/100] - [P + (P x R x 2.5)/100] = 1067.20 - 1012 = 55.2 ⇒ (1.5 x P x R)/100 = 55.2 ⇒ PR = (552 x 100) / 15 = 3680 For 4 yr. SI = PRT/100 = (3680 x 4)/100 = ₹ 147.2 ∴ Sum (P) = 1067.2 - 147.2 = ₹ 920 We have, PR = 3680 ∴ R = 3680/P = 3680/920 = 4%

Question 16

At simple interest, a sum becomes 3 times in 20 yr. Find the time, in which the sum will be double at the same rate of interest.

Accepted Answer

Here, n = 3, m = 2, T1 = 20 yr ∴ T2 = [(m -1) / (n - 1)] x T1 = (2 - 1) / (3 - 1) x 20 = 10 yr

Question 17

Simple interest for the sum of ₹ 1500 is ₹ 30 in 4 yr and ₹ 60 in 8 yr. Find the rate of simple interest.

Accepted Answer

Given, SI2 = 60, SI1 = 30, T1 = 4 yr, T2 = 8 yr According to the question, [(1500 x R x 8)/100] - [(1500 x R x 4)/100] = 60 - 30 ⇒ (6000 x R)/100 = 30 ∴ R = 30/60 = 1/2 = 0.5%

Question 18

At what rate per annum will the simple interest on a certain sum of money be 1/5 of the amount in 10 yr?

Accepted Answer

Let sum = P, then SI = P/5 time = 10 yr. ∴ Rate = (100 x P)/(P x 5 x 10) = 2%

Question 19

Mr. Deepak invested an amount of ₹ 21250 for 6 yr. At what rate of simple interest, will he obtain the total amount of ₹ 26350 at the end of 6 yr?

Accepted Answer

SI = 26350 - 21250 = ₹ 5100 ∴ Rate = (SI x 100) / (Principal x Time) = (5100 x 100) / (21250 x 6) = 4%

Question 20

The simple interest on ₹ 4000 in 3 yr at the rate of R % per annum equals equals to the simple interest on ₹ 5000 at the rate of 12% per annum in 2 yr The value of R is

Accepted Answer

Since, the two simple interest are equal. Then, (4000 x 3 x R)/100 = (5000 x 12 x 2)/100 ∴ R = 10%

Simple Interest Questions