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 = 84000 + 4200 = 88200
R = 5 and T = 1 year
Interest for the third year $=\dfrac{88200 \times 5 \times 1}{100}$
= 4410
Amount payable at the end of third year = 88200 + 4410 = 92610
Compound Interest for 3 years = 92610 -80000 = 12610
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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