Question 1

A father divided his property between his two sons A and B. A invests the amount at compound interest of 8% per annum and B invests the amount at 10% per annum simple interest. At the end of 2 yr, the interest received by B is ₹ 1336 more than the i…

Accepted Answer

Suppose A gets ₹ P and B gets ₹ (25000 - P). Interest received by A at the rate of 8% per annum CI = P(1 + 8/100)2 - P = P(27/25)2 - P = 104P/625 Interest received by B at the rate of 10% per annum SI = [(25000 - P) x 10 x 2]/100 = (25000 - P)/5 According to the question, (25000 - P)/5 = 104P/(625 + 1336) ∴ P = 10000

Question 2

What is the difference between the compound interest and simple interest calculated on an amount of ₹ 16200 at the end of 3 yr at 25% pa? (Rounded off to two digits after decimal)

Accepted Answer

Let required difference be ₹ D. By formula, D = P x (R/100)2 x [(3 + R)/100] = 16200 x (25/100)2 x [(3 + 25)/100] = 16200 x 625/10000 x 13/4 = 162 x 625 x 13/4 x 100 = ₹ 3290.63

Question 3

Divide ₹ 2602 between M and N, so that the amount of M after 7 yr is equal to the amount of N after 9 yr, the interest being compounded at 4% pa.

Accepted Answer

Let the first part be ₹ M . Then , second part = (2602 - M) According to the question, M(1 + 4/100)4 = (2602 - M) (1+4/100)9 ⇒ M/(2602 - M) = [(1 + 4/100)9]/[(1 + 4/100)7] ⇒ M/(2602 - M) = (1 + 4/100)2 = (26/25) x (26/25) = 676/625 ⇒ 625M = 676 x 2602 - 676M ⇒ 1301M = 676 x 2602 ∴ M = (676 x 2602)/1301 = 676 x 2 = 1352 ∴ The two parts are ₹ 1352 and ₹ (2602 - 1352) = ₹ 1250

Question 4

SBI lent ₹ 1331 lakh to the TATA group at compound interest and got ₹1728 lakh after 3 yr. What is the rate of interest charged, if compounded annually?

Accepted Answer

According to the question, 1728 = 1331(1 + R/100)3 1728/1331 = (1 + R/100)3 ⇒ (12/11)3 = (1 + R/100)3 ⇒ 1 + R/100 = 12/11 ⇒ R/100 = (12/11) - 1 = 1/11 ∴ R = 100/11= 9.09%

Question 5

A sum of money lent at compound interest for 2 yr at 20 % pa would fetch ₹ 964 more, if the interest was payable half-yearly than if it was payable annually. What is the sum?

Accepted Answer

Let the sum be ₹ P . Then, CI when compounded half - yearly = [P x (1 + 10/100)4 - P] = 4641P/10000 CI when compounded annually = [ P x (1 + 20/100)2 - P] = 11P/25 According to the question, 4641P/10000 - 11P/25 = 964 ⇒ [(4641 - 4400)/10000] x P = 964 ∴ P = (964 x 10000)/241 = ₹ 40000

Question 6

The compound interest on a sum of ₹ 4000 becomes ₹ 630.50 in 9 months. Find the rate of interest, if interest is compounded quarterly.

Accepted Answer

Given, P =₹ 4000, n = 9 months = 3/4 yr and CI = ₹ 630.50 Amount = P + CI = 4000 + 630.50 = ₹ 4630.50 According to the formula, ⇒ Amount = P[(1+R/(100 x 4)]4n ⇒ 4630.50 = 4000(1+R/400)4 x 3/4 ⇒ 4630.50 = 4000[(400 + R)/400]3 ⇒ 4630.50/4000 = [(400 + R)/400]3 ⇒ 9261/8000 = [(400 + R)/400]3 ⇒ (21/20)3 = [(400 + R)/400]3 ⇒ [(400 + R)/400]/400 = 21/20 ⇒ 400 + R = 21 x 20 = 420 ∴ R = 420 - 400 = 20%

Question 7

Income of Shantanu was ₹ 4000. In the first 2 yr. his income decreased by 10% and 5% respectively but in the third year, the income increased by 15%. What was his income at the end of third year?

Accepted Answer

Given, P = ₹ 4000 R1= 10% (decreased), R2 = 5% (decreases) and R3 = 15% (growth) ∴ According to the formula, Income at the end of third year = P(1 - R1/100)(1 - R2/100)(1 + R3/100) = 4000(1- 10/100) (1 - 5/100) (1 + 15/100) = 4000 x (9/10) x (19/20) x (23/20) = 9 x 19 x 23 = ₹ 3933

Question 8

The simple interest for certain sum in 2 yr at 4% pa is ₹ 80. What will be the compound interest for the same sum, if conditions of rate and time period are same?

Accepted Answer

Given, n = 2 yr, R = 4% and SI = 80 According to the formula, CI = SI(1 + 4/200) = 80 x 51/50 = ₹ 81.60

Question 9

A sum invested for 3 yr compounded at 5%, 10% and 20% respectively. In 3 yr, if the sum amounts to ₹ 16632, then find the sum.

Accepted Answer

Let the required sum be P Then, P (1 + 5/100) (1 + 10/100) (1 + 20/100) = 16632 ⇒ P x (105/100) x (110/100) x (120/100) = 16632 ⇒ P = 16632 x [(100 x 100 x 100) / (105 x 110 x 120)] ∴ P = ₹ 12000

Question 10

A man borrows ₹ 5100 to be paid back with compound interest at the rate of 4 % pa by the end of 2 yr in two equal yearly installments. How much will each installment be?

Accepted Answer

Let each installment be ₹ N . Then , according to the question , N/(1 + 4/100 ) + N/(1 + 4/100 )2 = 5100 ⇒ 25N/26 + 625N/676 = 5100 ⇒ 1275N = 5100 x 676 ∴ N = (5100 x 676)/1275 = ₹ 2704

Question 11

An amount is invested in a bank at compound rate of interest . The total amount, including interest, after first and third year is ₹ 1200 and ₹ 1587, respectively . What is the rate of interest?

Accepted Answer

Let amount be ₹ P and rate of interest be R % annually . According to the question, Amount after 1st yr = ₹ 1200 ⇒ P(1 + R/100) = 1200 ...(i) Amount after 3rd yr = 1587 ⇒ P(1 + R/100)3 = 1587 ...(ii) On dividing Eq. (ii) from Eq. (i), we get (1 + R/100)2 = 1587/1200 = 529/400 ⇒ 1 + R/100 = 23/20 ⇒ R/100 = 3/20 ∴ R = 15 %

Question 12

A sum of ₹ 11000 was taken as loan. This is to be paid in two equal annual installments. If the rate of interest be 20% compounded annually, then the value of each installment is

Accepted Answer

Let value of each installment be ₹ N . Then, N/(1 + 20/100) + N/(1 + 20/100)2 = 11000 ⇒ N(5/6 + 25/36) = 11000 ⇒ N(56/36) = 11000 ∴ N = ₹ 7200

Question 13

A sum of ₹ 8448 is to be divided between A and B who are respectively 18 and 19 yr old, in such a way that if their shares be invested at 6.25 % per annum at compound interest, they will receive equal amounts on attaining the age of 21 yr. The prese…

Accepted Answer

Let shares of A and B be ₹ N and ₹ (8448 - N), respectively, Amount got by A after 3 yr = Amount got by B after 2 yr N(1 + 6.25/100)3 = (8448 - N) x (1 + 6.25/100)2 ⇒ 1 + 6.25/100 = (8448 - N)/N ⇒ 1 + 1/16 = (8448 - N)/N ⇒ 17/16 = (8448 - N)/N ⇒ 17N = 135168 - 16N ⇒ N = 4096

Question 14

The simple interest on a certain sum of money for 3 yr at 8 % pa is half the compound interest on ₹ 8000 for 2 yr at 10 % pa. Find the sum on which simple interest is calculated

Accepted Answer

CI = 8000 x (1 + 10/100)2 - 8000 = 8000 x (11/10) x (11/10) - 8000 = ₹ 9680 - 8000 = ₹ 1680 ∴ Sum = (840 x 100)/(3 x 8) = ₹ 3500 [ ∵ SI is half of CI ] ∴ SI = 1680/2 = ₹ 840

Question 15

During the first year, the population of a village is increased by 5% and the second year, it is diminished by 5%. At the end of the second year, its population was 47880. What was the population at the beginning of the first year?

Accepted Answer

Let the population at the beginning of the first year be N . Then, according to the question, N x [1+ 5/100] x [1 - 5/100] = 47880 ⇒ N x (105/100) x (95/100) = 47880 ∴ N = 47880 x (100/105) x (100/95) = 48000

Question 16

What amount will be received on a sum of ₹ 1750 in 2 1/ 2 yr, if the interest is compounded at the rate of 8% pa?

Accepted Answer

Given, P = ₹ 1750, R = 8%, n = 2 and a/b = 1/2 According to the formula, Amount = P(1 + R/100)n x [1 + {(a/b) x R}/100] =1750 (1 + 8/100 )2 [1 + {(1/2) x 8} /100] =1750 (27/25)2 x 26/25 = 1750 x (27/25) x (27/25) x (26/25) =₹ 2122.848 = ₹ 2122.85

Question 17

Find what is that first years in which a sum of money will become more than double in amount if put out at compound interest at the rate of 10% per annum?

Accepted Answer

Here, P(1 + 10/100)t > 2P ⇒ (11/10)t > 2 When t = 8 → (11/10)8 = 2.14358 t =7 → (11/10)7 = 1.9487 By trial, [11 x 11 x 11 x 11 x 11 x 11 x 11 x 11] / [10 x 10 x 10 x 10 x 10 x 10 x 10 x 10] >2 Hence, the first years in which sum of money will become more than double in amount is 8th year

Question 18

Sharon Stone deposits $2000 at the end of each year in an account earning 10% compounded annually. Determine how much money she has after 25 years. How much interest did she earn?

Accepted Answer

S=R[(1+i)^n-1]/i

Question 19

Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually

Accepted Answer

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years. Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)] =Rs. [8000 * (23/20) * (23/20) * (21/20)] = Rs. 11109. . :. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Question 20

A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is

Accepted Answer

Let sum=Rs.x C.I. when compounded half yearly = x 1 + 10 100 4 - x = 4641 10000 C.I. when compounded annually = x 20 100 2 - x = 11 25 4641 10000 x - 11 25 x = 482 => x=20000

Compound Interest Questions