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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 =\dfrac{4000\times 12\times 1}{100}

= 480

Principal for the second year = 4000 + 480 = 4480

R = 12 and T = 1 year

Interest for second year =\dfrac{4480\times 12 \times 1}{100}

= 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 =\dfrac{6250\times 16\times 1}{100}

= 1000

Principal for the second year = 6250 + 1000 = 7250

R = 16 and T = 1 year

Interest for second year =\dfrac{7250\times 16 \times 1}{100}

= 1160

Principal for the third year = 7250 + 1160 = 8410

R= 16 and T = 1

Interest for third year =\dfrac{8410\times 16 \times 1}{100}

= 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 =\dfrac{80000\times 5 \times  1}{100}

= 4000

Principal for the second year = 80000 + 4000 = 84000

R = 5 and T = 1 year

Interest for the second year =\dfrac{84000\times 5 \times 1}{100}

= 4200

Principal for the third year = 80000 + 4200 = 84200

R = 5 and T = 1 year

Interest for the third year =\dfrac{84200\times 5 \times 1}{100}

= 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 =\dfrac{93750\times 9.6 \times 1}{100}

= 9000

Principal for the second year = 93750 + 9000 = 102750

R = 9.6 and T = 1 year

Interest for second year =\dfrac{102750 \times 9.6 \times 1}{100}

= 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 =\dfrac{112614\times 9.6 \times 1}{100}

= 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

\text{I}=\dfrac{\text{P}\times \times \text{R} \times \text{T}}{100}

Here2  , we have P = 7300 and R = 6 p.a and T = 2 years

\text{I}=\dfrac{7300\times 6 \times 2}{100}

= 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 =\dfrac{7300\times 6 \times 1}{100}

= 438

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

Interest for second year =\dfrac{7738 \times 6 \times 1}{100}

= 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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