Compound Interest For Class 9

 

Compound Interest For Class 9

Exercise

Question #1

Ramesh invest ₹ 12,800 for three years at the rate of 10% per annum compound interest. Find

i)                 The sum due to Ramesh at the end of the first year.

ii)                The interest he earned for the second year.

iii)              The total amount due to him at the end of the third year.

Solution :

i)                 P = ₹ 12,800

R = 10%

T = 3 years

 

CI for first year  = prt/100

                             = 12800 * 10* 1/100

                             = 1280

 

Sum = P + I = 12800 + 1280 =  ₹ 14080 ( P for second year )

 

ii)                CI for second year = prt/100

                                                = 14080 * 10 * 1/100

                                                = ₹ 1408

Sum = P  + I = 14080 + 1408 = ₹ 15488 ( P for second year)

 

iii)              CI for third year = prt/100

                                            = 15488 * 10 * 1/100

                                            = ₹ 1548.8

Amount at the end of third year = P  + I = 15488 + 1548.8 = ₹ 17036.8 ( P for second year)

 

 

Question 2 :

Aparna borrows a sum of ₹ 2500 for two years three months at 8% p.a. compounded annually.

Find i) The CI for two years.

        ii) The amount at the end of 2 years 3 months.

Solution :

i)                 A = P(1 + r/100)ⁿ

                      A  = 2500 (1 + 8/100)²

                                   A = 2500 (108/100)  (108/100) 

                       A =  ₹ 2916

                  CI for 2 years = 2916 -2500 = 416

                  Now, Principal for next 4 month =  ₹ 2916

                CI for 3 month = 2916 * 3/12 * 8/100 = 58.32

ii) The amount at the end of 2 years 3 months = 2916 + 58.32 = ₹ 2974.32

Question 3 :

How much will ₹ 20,000 amount to in 2 years at CI if the rate for successive years are 8 % and 10% p.a. respectively?

Solution :

P = 20000 , t = 2 years , r1 = 8% , r2 = 10%

A = P ( 1 + r1/100) (1 + r2/100)

   = 20000 * (1 + 8/100 ) * ( 1 + 10/100)

   = 20000 * 108/100 * 110/100

  = 2 * 108 * 110

  = ₹ 23,760

Question 4

Find the CI on ₹ 64000 for 1 ½ years at 5%p.a., the interest being compounded half-yearly.

Solution

Solution ;

P = ₹ 64000

R = 5%

N = 1 ½ = 3/2

A = P( 1 + r/200)²ⁿ

    = 64000 ( 1 + 5/200 )^2*3/2

     = 64000 ( 1 + 1/40)³

    = 64000 ( 41/40)³

    = 64000 ( 41/40) ( 41/40) ( 41/40)

     = 64000 * 68921/64000

    = ₹ 68921

 

CI = A – P = 68921 – 64000 =  ₹ 4,921

Question 5.

5. Mr. Britto borrowed ₹ 28000 for 2 years. The rate of interest for two successive years are 8% and 10% respectively. If the repays ₹ 5240 at the end of the first year, find the outstanding amount at the end of the second year.

Solution :

P: ₹ 28000

R1: 8% and R2 = 10%

Amount after one year = P (1 + R1/100)

                                          = 28000 (1 + 8/100)

                                         = 28000 * 108/100

                                         = 280 *108

                                         = 30240

Money returned = ₹ 5240

Principal for 2^nd year = 30240 – 5240

                                      = ₹ 25000

Amount after 2^nd year = 25000 (1 + 10/100)

                                        = 25000 (110/100)

                                        = ₹ 27500

Amount at the end of second year = ₹ 27500

6. In what time will ₹ 15000 yield ₹ 4965 as compound interest at 10% p.a. compounded annually ?

Solution: P = ₹15000; CI = ₹ 4965; r = 10%

A= P + Cl = 15000 + 4965 = ₹ 19965

A = P (1 + r/100)ⁿ

19965 = 15000 (1 + 10/100) ⁿ

19965 = 15000 (1 + 1/10) ⁿ

19965 = 15000 (11/10) ⁿ

19965/15000 =  (1 + 1/10) ⁿ

1331/1000 =  (11/10) ⁿ

(11/10) ³ = (11/10) ⁿ

Base are same , so n = 3

Question 7.

7. What sum of money will amount to ₹11,025 in 2 years at 5% p.a. compounded annually?

Solution: A = ₹11025; r = 5 %; n = 2 years

A = P (1 + r/100) ⁿ

11025 = P (1 + 5/100) ²

11025 = P (1 + 1/20)²

11025 = P (21/20) ²

11025 = P(21/20) (21/20)

P = 11025 * 20*20 /21*21

P = ₹ 10000

Question 8.

8. Arun borrows ₹ 24000 from Bryan at SI for 2 years at 5% p.a. and immediately lends out this money to Chand for 2 years at 5% p.a. compounded annually. What is Arun’s gain in this transaction?

Solution: P = ₹ 24000; n = 2 years; r = 5%

SI = Prt /100

SI = 24000 * 5 * 2 /100

SI = ₹ 2400

Amount when compounded annually,

A = P (1 + r/100) ⁿ

A= 24000 (1 + 5/100)²

A = 24000 (1 + 1/20)²

A = 24000 (21/20) (21/20)

A = 60 * 21 *21

A = 26460

CI = ₹ (26460 – 24000) = ₹ 2460

Arun Gain = CI – SI = 2460 – 2400 = ₹ 60 

 

Question 9.

9. A certain sum amounts to ₹ 17,640 in 2 years and to ₹ 18,522 in 3 years at compound interest. Find the rate and the sum.

Solution:

Here, we subtract the amounts to get the interest for 1 year.

Amount in 3 years = P + CI(3^rd year) = ₹ 18522

Amount in 2 years = P + CI(2^nd year) = ₹ 17640

CI for 1 year = 18522 – 17640 = ₹ 882

₹ 882 is the interest on 17640 for 1 year.

R = 100 * I/P*T

   = 100 * 882 /17640 * 1 = 5%

R = 5%

A = P (1 + r/100) ⁿ

17640= P (1 + 5/100)²

17640 = P (1 + 1/20)²

17640 = P(21/20) (21/20)

P = 17640 * 20/21 * 20/21

A = ₹ 16000

Question 10.

10. There were 7,20,000 people in a town. If the population of this town increases at the rate of 5% p.a. Find the population at the end of 3 years.

Solution :

Population after n years = Present Population (1 + r/100) ⁿ

                                           = 720000 (1 + 5/100) ³

                                           = 720000 (1 + 1/20) ³

                                            = 720000 (21/20) ³

                                            = 720000 (21/20) (21/20) (21/20)

                                           = 720000 (21/20) (21/20) (21/20)

                                           = 720 (21/2) (21/2) (21/2)

                                           = 90 * 21 *21 *21 = 8,33,490

The population at the end of 3 years = 8,33,490