After studying the chapter Compound Interest you will learn to define compound interest; to distinguish simple interest and compound interest; to derive the formula for amount when interest is compounded; to calculate compound interest using ready rockers; to solve problems using the formula of compound interest.
2.2.1 Introduction – Compound Interest:
In banks, insurance corporations and the other financial institutions, when they issue loans or accept deposits, the interest is calculated not on the original principal throughout the period. They calculate interest at regular intervals like quarterly, half yearly or yearly. The interest is added to the principal and further interest is calculated on the principal and the accrued interest. In these cases, the principal increases. We say that the interest is compounded in all these situations. The concept of compounding is very useful in predicting the population of a country,, growth in the food production, decay of a radioactive substance etc.
In the unit compound interest we study the calculation of compound interest and derive the formula to find the compound interest.
2.2.2 Compound Interest – Compound Interest:
Simple interest is calculated using the formula,
I = P x T x R/100
Where I = simple interest;
P = principal amount;
T = Time;
R = Rate of interest.
Example: If the principal Rs. 2200 and rate 10%, then the interst after 1 year, I = P x T x R/100
= 2200 x 1 x 10/100 = 200.
2.2.3 Difference between simple interest and compound interest
Let us consider some examples for calculating compound interest and compare it with simple interest.
Example 1:
Rs. 20000 is lent for 2 years at 5% compound interest. Calculate the compound interest and the amount after 2years if the interest is compounded annually. What would be have been the interest earned if the same amount was lent for simple interest at the same rate for the same duration?
Solution:
Principal for the first year = 20,000 x 1 x 5/100 = 1000(Using simple interest formula)
Principal for the second year = 20,000 + 1000 = 21000
Interest for the second year = 21000 x 5 /100 = 1050.
Amount after 2 years = 21000 + 1050 = 22050
Compound interest = amount – principal = 22050 – 20000 = 2050
Hence compound interest ater 2 years is 2050 Rs.
Simple interest on 20000 for 2 years at the rate of 5% is, I = P x T x R/100 = 20,000 x 2 x 5/100 = 2,000.
Major differences between Simple interest and Compound interest:
| SIMPLE INTEREST | COMPOUND INTEREST |
| 1. It is the interest calculated on the principal alone. | 1. It is the interest calculated on the principal and accrued interest. |
| 2. Principal remains the same throughout the period. | 2. Principal goes on increasing for every period of time. |
2.2.4 Derivation of Compound Interest – Compound Interest:
Let P be the principal R be the rate of interest per annum and n be the period or time or number of years. We derive the formula in three steps:
Example: Find the compound interest on Rs. 12000 for 2 years at 4% per annum
Solution:
Given P = Rs. 12000;
R = 4%
n = 2years
Using the formula,
Therefore, Compound interest = A – P = 12979.2 – 12000 = Rs. 979.2
Example: Find the amount on Rs. 2000 after 2years if the rates of interest are 2% and 3% for successive years.
Solution:
We have P = 20000 , R1 = 2% and R2 = 3% and n = 2 years
Here, we can directly use the formula,
= 20000 (1.02)(1.03)
= 21, 012
Hence the amount after 2years is Rs. 21,012.
2.2.5 Calculation of Compound Interest Using the ready Reckoners – Compound Interest:
Compound interest can be easily calculated using ready reckoners. A ready reckoner is a pre-calculated table for interest for different periods and different times.
Compound Interest – Exercise 2.2.5
- Calculate the amount and compound interest for the following:
(a) Rs12,000 for 2 years at 10% compounded annually.
(b) Rs 20,000 for 3 years at 8% compounded annually.
(c) Rs 5,000 for 1 year at 4% compounded semi-annually.
(d) Rs 10,000 for 11/2 years at 5% compounded half – yearly.
(e) Rs 500 for 1year at 2% compounded quarterly.
Solution:
(a) Rs12,000 for 2 years at 10% compounded annually.
P = 12,000;
R = 10%
n = 2
A = P [1 + R/100]n
=12,000[1 + 10/100]2
=12,000[1+0.1]2
=12,000[1.1]2
= 12,000[1.21]
=14,520.
Compound interest = Amount – Principal
= 14,520 – 12,000
= 2520
(b) Rs 20,000 for 3 years at 8% compounded annually.
P = 20,000;
R = 8%
n = 3
A = P [1 + R/100]n
= 20000[1 + 8/100]3
= 20,000 [1.08]3
= 20,000[1.259712]
=25,194.24
Compound interest = Amount – Principal
= 25,194.24 – 20,000
= 5194.24
(c) Rs 5,000 for 1 year at 4% compounded semi-annually.
P = 5,000
n = 1
R = 4
A = P(1 + R/2×100)2n
A = 5,000x(1 + 4/2×100)2×1
=5000x(1.02)2
= 5,202
Compound interest = Amount – Principal
=5,202 – 5,000
= 202
(d) Rs 10,000 for 11/2 years at 5% compounded half – yearly.
P = 10,000
n = 11/2
R = 5
A = P(1 + R/2×100)2n
= 10,000(1 + 5/2×100)2×1.5
= 10,000(41/40)3
= 10,000(1.02)3
= 10,768.91
Compound interest = Amount – Principal
=10,768.91 – 10,000
= 768.91
=769.00
(e) Rs 500 for 1year at 2% compounded quarterly.
P = 500
n = 1
R = 2
A = P(1 + R/4×100)4n
=500(1 + 2/4×100)4×1
= 500(1 + 1/200)4
= 500(201/200)4
=510.07
Compound interest = Amount – Principal
= 510.07 – 500
= 10.07
- A man invests Rs 5000 for 2 years at compound interest. After one year his money amount to Rs 5150. Find the interest for second year.
Solution:
Principal for the first year = P1 = 5000.00
Amount at the end of first year = A1 = 5150.00
Interest = A1 – P1 = 5150 – 5000 =150
Rate of interest = R = 100xI/PT = 100×150/5000 = 15000/5000 = 3
P = 5000 + 150 = 5150
Interest in the second year = PTR/100 = 5150x1x3/100 = 154.5
- Find the amount on 10000 after 2 years if the rate of interest are 3% and 4% doe successive years.
Solution:
P1 = 10000.00
R1 = 3%
T = 1
I1 = PTR/100 = 10000x1x3/100 = 300
Principal for the second year P2 = 10000 + 300 = 10300
R2 = 4%
I2 = PTR/100 = 10300x1x4/100 = 412
Amount at the end of second year P2 = 10300 + 412 = 10712
- Pralhad invests a sum of money in a bank and gets Rs 3307.5 Rs 3472.87 in 2nd and 3rd year respectively. Find the sum he invested.
Solution:
Amount at the end of second year A2 = RS 3307.50
Amount at the end of third year A3 = Rs 3472.87
Interest on Rs 3307.50 for a year = 3472.87 – 3307.50 = 165.37
Rate of interest = 100xI/PT = 100*165.37/3307.50*1 = 5%
Let P be the initial investment
Amount after 2 years A = P(1+ R/100)n
3307.50 = P(1+5/100)2
3307.50 = P(21/20)2
P = 3020.00
- On what sum of money will be difference between the simple interest and compound interest for 2 years at 4% per annum will be equal to RS 100? (Hint : Assuming the principal to be RS 100 first calculate the SI and CI, then proceed)
Solution:
Let the principal be Rs. 100.00
R = 4
T = 2 years
S.I = PTR/100 = 100*2*4/100 = Rs. 8.00
Amount at the end of 2 years if C.
I is calculated = 100(1 + 4/100)2
= 100*26/25*26/25=108.16
C.I = 108.16 – 100.00 = 8.16
Difference between SI and CI = 8.16 – 8.00 = 0.16
When the difference is Rs. 0.16
The amount invested = Rs. 100.00
The amount invested when the difference between SI and CI is Rs. 100.00 = 100*100/0.16 = Rs. 62500.00
- A sum of money is invested at compound interest payable annually. The interest in two successive year are RS 275.00 and Rs 300.00. Find the rate of interest.
Solution:
Interest at the end of 1 year = 275
Interest at the end of II year = 300
Interest per year = 300 -275 = RS. 25
Interest on RS 100 per year = 25
I = PTR/100
25 = 275*1*R/100 = 25*100/275 = 2500/275 = 9.09
Rate of interest = 9.09
- The difference between compound interest and simple interest on a certain sum for 2 year at 7 ½% per annum is RS. 360. Find the sum and verify answer.
Solution:
To find the simple interest and compound interest let us assume that the principal be Rs. 100
Time = 2 years
Rate =7 ½ % = 15/2%
SI = PTR/100 = 100*2*15/100*2 = 15
CI = P[(1 + R/100)n – 1]
= 100[(1 + 15/2×100)2 – 1]
= 100[(215/200)2 – 1]
= 100[46225/40000 – 1 ]
= 100[46225-40000/40000]
= 100*6225/40000 = 6225/400
= 15.5625
CI – SI = 15.5625 – 15 = 0.5625
When P = 100 ; Difference in CI and SI = 0.5625
If the difference is Rs. 360 then principal = 100/0.5625 x 360 = 64000
Thus, the principal is Rs. 64000
Verification:
S.I = PTR/100 = 64000x2x15/2×100 = 9600
CI = P[(1 + R/100)n – 1] = 64000x[(1+15/2×100)2 – 1] = 9960
SI – CI = 9960 – 9600 = 360
- Teju invests Rs. 12,000 at 5% interest compounded annually. IF he receives an amount Rs. 13320 at the end find the period.
Solution:
P = 12,000
R = 5%
A = 13230
A = P(1+R/100)n
13230 = 12000(1+ 5/100)n
13230/12000 = (1+5/100)n
21/20 = (21/20)n
n = 2
- The present population of a village is Rs . 18000. It is estimated that the population of the village grows by 3% per year . Find the population of the village after 4 years.
Solution:
P = 18000
R = 3%
n = 4
A = P(1+R/100)n
= 18000(1 + 3/100)4
= 18000x103x103x103x103/100x100x100x100
= 20256.15858
= 20259
Population after 4 years = 20259
- Jeshu purchased a bike by paying Rs. 52000. If the value depreciates by 2% every year. Find the value of the bike after 3 years .
Solution:
Original price of the bike =rs.52,000
Rate of depreciation =2%
Price after 3 years = P(1 – R/100) n
= 52000 x (1 – 2/100)3
= 52000 x (98/100)3
= 48941.98 ≈ 48942
- Using the ready reckoner find the compound interest in the following:
(a). Principal Rs 15000 for 4 years at 6.5% p.a.
(b). Principal Rs 22000 for 5 years at 12% p.a.
Solution:
(a) Principal Rs 15000 for 4 years at 6.5% p.a.
P = 15000
T = 4
R = 6.5%
From the table the interest on Rs. 1 t 6.5% for 4 years is Rs. 0.2865
CI on Rs. 15000 = 0.2865*15000 = Rs. 4297.5
(b)P = Rs. 22000
R = 12%
n = 5years
From the table the interest on Rs. 1 t 12% for 5 years is Rs. 0.7623
CI on Rs. 22000 for 5 years
= 0.7623*22000 = Rs. 16770.60 ≈ 16771
- Using the ready reckoner find the period of interest in the following :
(a) Compound interest Rs 4026 at 7% p.a. principal Rs 10000
(b) Amount RS 16939.2 principal RS 12000 at 9% p.a.
Solution:
(a) P = 30000 ; R = 7% ; CI = Rs. 4026
Thus, Rs. 4026 is the compound interest for the principal of Rs. 10000 at 7%.
Now, we have to find compound interest for Re. 1. For Rs. 10000, th compound interest is Rs. 4026
For the principal Re. 1, the compound interest is = 4026/10000 = 0.4026
From the ready recknors 0.4026 corresponds to 5 years.
Therefore, n = 5.
(b) Amount RS 16939.2 principal RS 12000 at 9% p.a.
We have A = Rs. 16939.2
P = 12000
R = 9
Compound interest = A – P = 16939.2 – 12000 = 4939.2
Thus, Rs. 4939.2 is the compound interest for the principal of Rs. 12000 at 9%.
Now, we have to find compound interest for Re. 1. For Rs. 12000, the compound interest is Rs. 4939.2
For the principal Re. 1, the compound interest is = 4939.2/12000 =0.4116
From the ready recknors 0.4116 corresponds to 4 years.
Therefore, n = 4.
- Using the ready recknors ,Find the rate of interest in the following :
(a) Compound interest RS 1733.6 for 3 years on principal of RS 11000
(b) Amount RS 35246 principal SR 20000 for 5 years .
Solution :
(a) Compound interest = Rs. 1733.6
n = 3
P = Rs. 11000
Thus, Rs. 4939.2 is the compound interest for the principal of Rs. 12000 at 9%.
Now, we have to find compound interest for Re. 1. For Rs. 11000, the compound interest is Rs. 1733.6
For the principal Re. 1, the compound interest is = 1733.6 /11000 = 0.1576
From the ready recknors 0.1576 corresponds to 5%
Therefore the rate of interest is 5%.
(b)
Amount = Rs. 35246
P = Rs. 20000
n = 5
Compound interest = A – P = 35246 – 20000 = 15246
Now, we have to find compound interest for Re. 1. For Rs. 20000, the compound interest is Rs. 15246
For the principal Re. 1, the compound interest is = 15246 /20000 = 0.7623
From the ready recknors 0.7623 corresponds to 12%
Therefore the rate of interest is 12%.
