- 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%.
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