Commercial mathematics
Compound Interest
12
Introduction
In the previous classes, you have learnt about simple interest and other related terms. You have
also solved many problems on simple interest. In this chapter, we shall learn about compound
interest, difference between simple and compound interest, computation of compound interest
as a repeated simple interest with a growing principal and also by use of formula.
2.1 Interest
It is the additional money besides the original money paid by the borrower to the moneylender (bank,
financial agency or individual) in lieu of the money used by him.
Principal. The money borrowed (or the money lent) is called principal.
Amount. The sum of the principal and the interest is called amount.
Thus, amount = principal + interest.
Rate. It is the interest paid on ` 100 for a specified period.
Time. It is the time for which the money is borrowed.
Simple Interest. It is the interest calculated on the original money (principal) for any given time
and rate.
Formula: Simple Interest =
Principal × Rate × Time
.
100
Compound Interest
At the end of the first year (or any other fixed period), if the interest accrued is not paid to the
moneylender but is added to the principal, then this amount becomes the principal for the
next year (or any other fixed period) and so on. This process is repeated until the amount for
the whole time is found.
The difference between the final amount and the (original) principal is called compound interest.
Remark
I n the case of simple interest, the principal remains constant for the whole time but in the
case of compound interest, the principal keeps on changing every year (or any other fixed
period).
If the interest is compounded annually, the principal changes after every year and if the
interest is compounded half-yearly (or any other fixed period), the principal changes after every
six months (or any other fixed period).
Illustrative Examples
Example 1. Find the amount and the compound interest on ` 15000 for 2 years at 8% per
annum.
Solution. Principal for the first year = ` 15000.
Interest for the first year = ` 15000 × 8 × 1
= ` 1200.
100
Amount after one year = ` 15000 + ` 1200 = ` 16200.
Principal for the second year = ` 16200.
Interest for the second year = ` 16200 × 8 × 1
= ` 1296.
100
Amount after 2 years = ` 16200 + ` 1296 = ` 17496.
Compound interest for 2 years = final amount – (original) principal
= ` 17496 – ` 15000 = ` 2496.
Note
he compound interest may also be obtained by adding together the interest of consecutive
T
years.
Thus, in the above example,
compound interest= interest of first year + interest of second year
= ` 1200 + ` 1296 = ` 2496.
Example 2. Find the amount and the compound interest on ` 25000 for 3 years at 12% per annum,
compounded annually.
Solution. Principal for the first year = ` 25000.
Interest for the first year = ` 25000 × 12 × 1
= ` 3000.
100
Amount after one year = ` 25000 + ` 3000 = ` 28000.
Principal for the second year = ` 28000.
Interest for the second year = ` 28000 × 12 × 1
= ` 3360.
100
Amount after 2 years = ` 28000 + ` 3360
= ` 31360.
Principal for the third year = ` 31360.
Interest for the third year = ` Amount after 3 years
Compound interest for 3 years
=
=
=
=
=
31360 × 12 × 1
= ` 3763·20
100
` 31360 + ` 3763·20
` 35123·20
final amount – (original) principal
` 35123·20 – ` 25000
` 10123·20.
Example 3. Find the compound interest to the nearest rupee on ` 7500 for 2 years
4 months at 12% per annum reckoned annually.
Solution. Principal for the first year = ` 7500.
316
Interest for the first year = ` 7500 × 12 × 1
= ` 900.
100
Understanding ICSE mathematics – Ix
Amount after one year = ` 7500 + ` 900 = ` 8400.
Principal for the second year = ` 8400.
Interest for the second year = ` 8400 × 12 × 1
= ` 1008.
100
Amount after 2 years = ` 8400 + ` 1008 = ` 9408.
1
Remaining time = 4 months = 4 year = year.
Principal for the next
1
year = ` 9408.
3
1
Interest for the next year = ` 3
9408 × 12 ×
100
3
12
1
3 = ` 376·32
Amount after 2 years 4 months = ` 9408 + ` 376·32
= ` 9784·32
∴ Compound interest for 2 years 4 months
= final amount – (original) principal
= ` 9784·32 – ` 7500 = ` 2284·32
= ` 2284 (to the nearest rupee).
Example 4. Find the amount and the compound interest on ` 16000 for 1 1 years at 10% per
2
annum, the interest being compounded half-yearly.
Solution. Since the rate of interest is 10% per annum, therefore, the rate of interest halfyearly = 1 of 10% = 5%.
2
Principal for the first half-year = ` 16000.
Interest for the first half-year = ` 16000 × 5 × 1
= ` 800.
100
Amount after the first half-year = ` 16000 + ` 800 = ` 16800.
Principal for the second half-year = ` 16800.
Interest for the second half-year = ` 16800 × 5 × 1
= ` 840.
100
Amount after one year = ` 16800 + ` 840 = ` 17640.
Principal for the third half-year = ` 17640.
Interest for the third half-year = ` 17640 × 5 × 1
= ` 882.
100
Amount after 1 1 years = ` 17640 + ` 882 = ` 18522.
2
∴ Compound interest for 1 1 years = final amount – (original) principal
2
= ` 18522 – ` 16000 = ` 2522.
Example 5. Nikita invests ` 6000 for two years at a certain rate of interest compounded annually.
At the end of first year it amounts to ` 6720. Calculate :
(i) the rate of interest.
(ii) the amount at the end of the second year.
Solution. Given, principal = ` 6000, amount after one year = ` 6720.
(i) Interest for the first year = ` 6720 – ` 6000 = ` 720.
Let the rate of interest be R% p.a., then
S.I. =
P×R×T
6000 × R × 1
⇒ 720 =
⇒ R = 12.
100
100
Hence, the rate of interest = 12% p.a.
Compound interest
137
(ii) Principal for the second year = ` 6720.
Interest for the second year = `
6720 × 12 × 1
= ` 806·40
100
∴ The amount at the end of second year= ` 6720 + ` 806·40
= ` 7526·40.
Example 6. Calculate the amount due and the compound interest on ` 7500 in 2 years when the
rate of interest on successive years is 8% and 10% respectively.
Solution.
Principal for the first year = ` 7500, rate = 8%.
Interest for the first year = `
7500 × 8 × 1
= ` 600.
100
Amount after the first year = ` 7500 + ` 600 = ` 8100.
Principal for the second year = ` 8100, rate = 10%.
Interest for the second year = ` 8100 × 10 × 1
= ` 810.
100
∴Amount due after 2 years = ` 8100 + ` 810 = ` 8910.
Compound interest for 2 years = amount – principal
= ` 8910 – ` 7500 = ` 1410.
Example 7. Calculate the difference between the compound interest and the simple interest on
` 12000 at 9% per annum in 2 years.
Solution. Given principal = ` 12000, rate = 9% p.a. and time = 2 years
12000 × 9 × 2
∴S.I. = ` = ` 2160.
100
For C.I.
Principal for the first year = ` 12000.
Interest for the first year = ` 12000 × 9 × 1
= ` 1080.
100
Amount after one year = ` 12000 + ` 1080 = ` 13080.
Principal for the second year = ` 13080.
Interest for the 2nd year = `
13080 × 9 × 1
100
= ` 1177·20
∴
C.I. of 2 years = ` 1080 + ` 1177·20 = ` 2257·20.
∴ Difference between compound interest and simple interest in 2 years
= ` 2257·20 – ` 2160
= ` 97·20.
Example 8. The simple interest on a sum of money for 2 years at 4% per annum is ` 340.
Find
(i) the sum of money
(ii) the compound interest on this sum for one year payable half-yearly at the same rate.
Solution. Given, S.I. = ` 340, rate = 4% p.a. and time = 2 years
(i) Let the sum of money be P, then
S.I. =
⇒
318
P×R×T
100
340 × 100
P = ` 4×2
⇒ ` 340 =
= ` 4250.
Understanding ICSE mathematics – Ix
P×4×2
100
(ii)Since the rate of interest is 4% per annum, therefore, the rate of interest halfyearly = 2%.
Principal for the first half-year = ` 4250.
Interest for the first half-year = `
4250 × 2 × 1
= ` 85.
100
∴Amount after the first half-year = ` 4250 + ` 85 = ` 4335.
Principal for the 2nd half-year = ` 4335.
Interest for the 2nd half-year = `
4335 × 2 × 1
= ` 86·70
100
∴ Compound interest on the above sum for one year payable half-yearly
= ` 85 + ` 86·70 = ` 171·70.
Example 9. The simple interest on a certain sum of money for 3 years at 5% per annum is ` 1200.
Find the amount due and the compound interest on this sum of money at the same rate after 3 years,
interest is reckoned annually.
Solution. Given simple interest for 3 years = ` 1200.
1
of ` 1200 = ` 400
3
P×5×1
⇒ ` 400 =
100
∴Simple interest for one year =
S.I. =
P×R×T
100
⇒
400 × 100
= ` 8000.
5×1
P = ` ∴Amount after one year = ` 8000 + ` 400 = ` 8400.
Principal for the second year = ` 8400.
Interest for the second year = ` 8400 × 5 × 1
= ` 420.
100
Amount after 2 years = ` 8400 + ` 420 = ` 8820.
Interest for the third year = `
8820 × 5 × 1
= ` 441.
100
∴Amount due after 3 years = ` 8820 + ` 441 = ` 9261.
Compound interest for 3 years = ` 9261 – ` 8000 = ` 1261.
Example 10. Ranbir borrows ` 20000 at 12% per annum compound interest. If he repays
` 8400 at the end of the first year and ` 9680 at the end of the second year, find the amount of the loan
outstanding at the beginning of the third year.
Solution.
Principal for the first year = ` 20000, rate = 12%.
Interest for the first year = ` 20000 × 12 × 1
= ` 2400.
100
Amount after the first year = ` 20000 + ` 2400 = ` 22400.
Money refunded at the end of first year = ` 8400.
∴
Principal for the second year = ` 22400 – ` 8400 = ` 14000.
Interest for the second year = ` 14000 × 12 × 1
= ` 1680.
100
Amount after the second year = ` 14000 + ` 1680 = ` 15680.
Money refunded at the end of 2nd year = ` 9680.
∴ The loan outstanding at the beginning of the third year
= ` 15680 – ` 9680 = ` 6000.
Compound interest
139
Example 11. Mr. Kumar borrowed ` 15000 for two years. The rate of interest for the two successive
years are 8% and 10% respectively. If he repays ` 6200 at the end of first year, find the outstanding
amount at the end of the second year.
Solution.
Principal for the first year = ` 15000, rate = 8% p.a.
Interest for the first year = `
Amount after one
Money repaid at the end of first
∴
Principal for the second
rate of interest for second
year
year
year
year
=
=
=
=
15000 × 8 × 1
= ` 1200.
100
` 15000 + ` 1200 = ` 16200.
` 6200.
` 16200 – ` 6200 = ` 10000;
10% p.a.
Interest for the second year = ` 10000 × 10 × 1
= ` 1000.
100
Amount after second year = ` 10000 + ` 1000 = ` 11000.
∴ The amount outstanding at the end of second year = ` 11000.
Example 12. Sulekha deposits ` 8000 in a bank every year in the beginning of the year, at 10% per
annum compound interest. Calculate the amount due to her at the end of three years. Also find her gain
in three years.
Solution.
Principal for the first year = ` 8000, rate = 10% p.a.
Interest for the first year = ` 8000 × 10 × 1
= ` 800.
100
Amount after one year = ` 8000 + ` 800 = ` 8800.
Money deposited at the beginning of 2nd year = ` 8000.
Principal for the 2nd year = ` 8800 + ` 8000 = ` 16800.
Interest for the 2nd year = ` 16800 × 10 × 1 = ` 1680.
100
Amount after 2 years = ` 16800 + ` 1680 = ` 18480.
Money deposited at the beginning of 3rd year = ` 8000.
Principal for the 3rd year = ` 18480 + ` 8000 = ` 26480.
Interest for the 3rd year = ` Amount after 3 years =
∴ The amount due to Sulekha at the end of 3 years =
Money deposited by Sulekha in 3 years =
=
∴Gain of Sulekha in 3 years =
=
26480 × 10 × 1
= ` 2648.
100
` 26480 + ` 2648 = ` 29128.
` 29128.
` 8000 + ` 8000 + ` 8000
` 24000.
` 29128 – ` 24000
` 5128.
Example 13. A man borrows ` 5000 at 12% compound interest per annum, interest payable after
six months. He pays back ` 1800 at the end of every six months. Calculate the third payment he has to
make at the end of 18 months in order to clear the entire loan.
Solution. Since the rate of interst is 12% per annum, therefore, rate of interest halfyearly = 6%.
Principal for the first six months = ` 5000.
410
Interest for the first six months = ` 5000 × 6 × 1
= ` 300.
100
Amount after first six months = ` 5000 + ` 300 = ` 5300.
Money refunded at the end of first six months = ` 1800.
Understanding ICSE mathematics – Ix
∴
Principal for the second six months = ` 5300 – ` 1800
= ` 3500.
Interest for the second six months = `
Amount after second six months
Money refunded at the end of second six months
∴
Principal for the third six months
=
=
=
=
=
3500 × 6 × 1
= ` 210.
100
` 3500 + ` 210
` 3710.
` 1800.
` 3710 – ` 1800
` 1910.
Interest for the third six months = ` 1910 × 6 × 1
= ` 114·60.
100
∴ The payment to be made at the end of 18 months to clear the entire loan
= ` 1910 + ` 114·60
= ` 2024·60.
Exercise 2.1
1. Find the amount and the compound interest on ` 8000 at 5% per annum for 2 years.
2. A man invests ` 46875 at 4% per annum compound interest for 3 years. Calculate :
(i) the interest for the first year.
(ii) the amount standing to his credit at the end of the second year.
(iii) the interest for the third year.
3. Calculate the compound interest for the second year on ` 8000 invested for
3 years at 10% p.a.
Also find the sum due at the end of third year.
4. Ramesh invests ` 12800 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 earns for the second year.
(iii) the total amount due to him at the end of three years.
5. The simple interest on a sum of money for 2 years at 12% per annum is ` 1380. Find:
(i) the sum of money.
(ii) the compound interest on this sum for one year payable half-yearly at the same
rate.
6.A person invests ` 10000 for two years at a certain rate of interest, compounded
annually. At the end of one year this sum amounts to ` 11200. Calculate :
(i) the rate of interest per annum.
(ii) the amount at the end of second year.
7. A man invests ` 5000 for three years at a certain rate of interest, compounded
annually. At the end of one year it amounts to ` 5600. Calculate :
(i) the rate of interest per annum.
(ii) the interest accrued in the second year.
(iii) the amount at the end of the third year.
8. Find the amount and the compound interest on ` 2000 at 10% p.a. for 2 1 years,
2
compounded annually.
Compound interest
141