# S.chand mathematics solution class 8 chapter 9 simple interest and compound interest

## Exercise 9 (B)

QUESTION 1

Find the amount and compound interest on 4000 at 12% p.a. for 2 years , compounded annually .

Sol :

Principal  for first year = 4000 and R = 12 , T = 1 year

Interest for the first year = 480

Principal for the second year = 4000 + 480 = 4480

R = 12 and T = 1 year

Interest for second year = 537.6

Amount payable at the end = 4480 + 537.6 = 5017.6

Compound interest for 2 years = 5017.6 – 4000 = 1017.6

QUESTION 2

Find the compound interest on 6250 at 16% p.a. for 3 years , compounded annually .

Sol :

Principal  for first year = 6250 and R = 16 , T = 1 year

Interest for the first year = 1000

Principal for the second year = 6250 + 1000 = 7250

R = 16 and T = 1 year

Interest for second year = 1160

Principal for the third year = 7250 + 1160 = 8410

R= 16 and T = 1

Interest for third year = 1345.6

Amount payable at the end = 8410 + 1345.6 = 9755.6

Compound interest for 3 years = 9755.6 – 6250 = 3505.6

QUESTION 3

To renovate his ice cream parlour , Pratham took a loan of 80000 from a bank . If the bank charges interest at the rate of 5% p.a , compounded annually , find the compound interest paid by Pratham at the end of 3 years .

Sol :

Principal for the first year = 80000

R = 5 and T = 1 year

Interest for the first year = 4000

Principal for the second year = 80000 + 4000 = 84000

R = 5 and T = 1 year

Interest for the second year = 4200

Principal for the third year = 80000 + 4200 = 84200

R = 5 and T = 1 year

Interest for the third year = 4210

Amount payable at the end of third year = 84200 + 4210 = 88410

Compound Interest for 3 years = 80000 – 88410 = 8410

QUESTION 4

Maria invest 93,750 at 9.6% per annum for 3 years and the interest is compounded annually .
Calculate :
(i) The amount standing to her credit at the end of the second year.
(ii) The interest for the third year .
(iii) The compound interest for the three years .

Sol : (i)

Principal for the first year = 93750

R = 9.6 and T = 1 year

Interest for the first year = 9000

Principal for the second year = 93750 + 9000 = 102750

R = 9.6 and T = 1 year

Interest for second year = 9864

Amount credit at the end of second year = 102750 + 9864 = 112614

Sol : (ii)

Principal for third year = 102750 + 9864 = 112614

R = 9.6 p.a and T = 1 year

Interest for third year = 10810.944

Amount payable at the end of third year = 112614 + 10810.94 = 123424.944

Sol : (iii)

Compound interest for 3 years = 123424.944 – 93750

= 29674.944

QUESTION 5

What is the difference between the simple and compound interest on 7300 at the rate of 6% per annum in 2 years .

Sol :

Lets find simple interest first Here2  , we have P = 7300 and R = 6 p.a and T = 2 years = 876

And now we have to find compound interest

Principal for the first year = 7300 , R = 6 p.a. and T = 1

Interest for the first year = 438

Principal for second year = 7300 + 438 = 7738 , R = 6 and T = 1 year

Interest for second year = 464.28

Amount payable at the end of 3 years = 7738 + 464.28 = 8202.28

Compound interest for 2 years = 8202.28 – 7300 = 902.28

Difference between Simple interest and Compound interest for 2 years

= Compound Interest – Simple interest

= 902.28 – 876

= 26.28

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