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 =4000×12×1100
= 480
Principal for the second year = 4000 + 480 = 4480
R = 12 and T = 1 year
Interest for second year =4480×12×1100
= 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 =6250×16×1100
= 1000
Principal for the second year = 6250 + 1000 = 7250
R = 16 and T = 1 year
Interest for second year =7250×16×1100
= 1160
Principal for the third year = 7250 + 1160 = 8410
R= 16 and T = 1
Interest for third year =8410×16×1100
= 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 =80000×5×1100
= 4000
Principal for the second year = 80000 + 4000 = 84000
R = 5 and T = 1 year
Interest for the second year =84000×5×1100
= 4200
Principal for the third year = 84000 + 4200 = 88200
R = 5 and T = 1 year
Interest for the third year =88200×5×1100
= 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 =93750×9.6×1100
= 9000
Principal for the second year = 93750 + 9000 = 102750
R = 9.6 and T = 1 year
Interest for second year =102750×9.6×1100
= 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 =112614×9.6×1100
= 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
I=P××R×T100
Here2 , we have P = 7300 and R = 6 p.a and T = 2 years
I=7300×6×2100
= 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 =7300×6×1100
= 438
Principal for second year = 7300 + 438 = 7738 , R = 6 and T = 1 year
Interest for second year =7738×6×1100
= 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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