RD Sharma Solutions For Class 8 Chapter 14 Compound Interest

RD Sharma Solutions For Class 8 Chapter 14 Compound Interest
1. Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2
years.
Solution:
Given details are,
e
Principal p = Rs 3000
Rate r = 5%
ut
Time = 2years
Interest for the first year = 3000×5×1/100 = 150

Amount at the end of first year = Rs 3000 + 300 = Rs 3150
tit
Principal interest for the second year = 3150×5×1/100 = 157.5
Amount at the end of second year = Rs 3150 + 157.5 = Rs 3307.5
∴ Compound Interest = Rs 3307.5 – Rs 3000 = Rs 307.5
s
2. What will be the compound interest on Rs. 4000 in two years when rate of interest is 5% per
annum?
In
Solution:
Given details are,
Rate r = 5%
sh
Time = 2years
By using the formula,
A = P 1+R/100 n
= 4000 1+15/1003
ka
= 4000 105/1003
= Rs 4410
∴ Compound Interest = A – P = Rs 4410 – Rs 4000 = Rs 410
3. Rohit deposited Rs. 8000 with a finance company for 3 years at an interest of 15% per
Aa
annum. What is the compound interest that Rohit gets after 3 years?
Solution:
Given details are,
Rate r = 15%
Time = 3years
A = P 1+R/100 n
= 8000 1+15/1003
= 8000 115/1003
= Rs 12167
4. Find the compound interest on Rs. 1000 at the rate of 8% per annum for 1 ½ years when
interest is compounded half yearly.
Solution:
Given details are,
te
Rate r = 8%
Time = 1 ½ years = 3/2 × 2 = 3 half years
itu
A = P 1+R/200 2n
= 1000 1+8/2003
= 1000 208/2003
st
= Rs 1124.86
∴ Compound Interest = A – P = Rs 1124.86 – Rs 1000 = Rs 124.86
5. Find the compound interest on Rs. 160000 for one year at the rate of 20% per annum, if the
In
interest is compounded quarterly.
Solution:
Given details are,
sh
Rate r = 20% = 20/4 = 5% forquarteryear

Time = 1year = 1 × 4 = 4 quarters
A = P 1+R/100 n
ka
= 160000 1+5/1004
= 160000 105/1004
= Rs 194481
Aa
6. Swati took a loan of Rs. 16000 against her insurance policy at the rate of 12 ½ % per
annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Given details are,
Rate r = 12 ½ % = 12.5%
Time = 3years
A = P 1+R/100 n
= 16000 1+12.5/1003
= 16000 112.5/1003
= Rs 22781.25
7. Roma borrowed Rs. 64000 from a bank for 1 ½ years at the rate of 10% per annum.
Compare the total compound interest payable by Roma after 1 ½ years, if the interest is
compounded half-yearly.
te
Solution:
Given details are,
itu
Rate r = 10 % = 10/2 % forhalfayear
Time = 1 ½ years = 3/2 × 2 = 3 halfyear
A = P 1+R/100 n
st
= 64000 1+10/2×1003
= 64000 210/2003
= Rs 74088
In
8. Mewa lal borrowed Rs. 20000 from his friend Rooplal at 18% per annum simple interest. He
lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
sh
Given details are,
Rate r = 18 %
Time = 2 years
ka

Interest amount Mewa lal has to pay,
Simple interest = P×T×R/100
Aa
= 20000×18×2/100 = 7200
Interest amount Rampal has to pay to Mewa lal,
A = P 1+R/100 n
= 20000 1+18/1002
= 20000 118/1002
= Rs 27848 – 20000 principalamount
= Rs 7848
∴ Mewa lal gain = Rs 7848–7200 = Rs 648
9. Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded
quarterly.
Solution:
Given details are,
Rate r = 20 % = 20/4 = 5% forquarterly
Time = 9 months = 9/3 = 3 forquarteryear
te
A = P 1+R/100 n
= 8000 1+5/1003
itu
= 8000 105/1003
= Rs 9261
10. Find the compound interest at the rate of 10% per annum for two years on that principal
st
which in two years at the rate of 10% per annum given Rs. 200 as simple interest.
Solution:
Given details are,
In
Simple interest SI = Rs 200
Rate r = 10 %
Time = 2 years
So, by using the formula,
sh
Simple interest = P×T×R/100
P = SI×100/ T×R
= 200×100 / 2 × 10
= 20000/20
ka
= Rs 1000
Now,
Rate of compound interest = 10%
Aa
Time = 2years
A = P 1+R/100 n
= 1000 1+10/1002
= 1000 110/1002
= Rs 1210
11. Find the compound interest on Rs. 64000 for 1 year at the rate of 10% per annum
compounded quarterly.
Solution:
Given details are,
Rate r = 10 % = 10/4 % forquarterly
Time = 1year = 1× 4 = 4 forquarterinayear
A = P 1+R/100 n
= 64000 1+10/4×1004
te
= 64000 410/4004
= Rs 70644.03
itu
12. Ramesh deposited Rs. 7500 in a bank which pays him 12% interest per annum
compounded quarterly. What is the amount which he receives after 9 months.
Solution:
Given details are,
st
Rate r = 12 % = 12/4 = 3 % forquarterly
Time = 9 months = 9/12years = 9/12 × 4 = 3 forquarterinayear
In
A = P 1+R/100 n
= 7500 1+3/1003
= 7500 103/1003
= Rs 8195.45
sh
∴ Required amount is Rs 8195.45

13. Anil borrowed a sum of Rs. 9600 to install a hand pump in his dairy. If the rate of interest
is 5 ½ % per annum compounded annually, determine the compound interest which Anil will
have to pay after 3 years.
ka
Solution:
Given details are,
Aa
Rate r = 5 ½ % = 11/2 %
Time = 3years
A = P 1+R/100 n
= 9600 1+11/2×1003
= 9600 211/2003
= Rs 11272.71
14. Surabhi borrowed a sum of Rs. 12000 from a finance company to purchase a refrigerator.
If the rate of interest is 5% per annum compounded annually, calculate the compound
interest that Surabhi has to pay to the company after 3 years.
Solution:
Given details are,
Rate r = 5 %
Time = 3years
te
A = P 1+R/100 n
= 12000 1+5/1003
itu
= 12000 105/1003
= Rs 13891.5
15. Daljit received a sum of Rs. 40000 as a loan from a finance company. If the rate of interest
st
is 7% per annum compounded annually, calculate the compound interest that Daljit pays after
2 years.
Solution:
In
Given details are,
Rate r = 7%
Time = 2years
sh
A = P 1+R/100 n
= 40000 1+7/1002
= 40000 107/1002
= Rs 45796
ka

Aa
EXERCISE 14.2 PAGE NO: 14.14

1. Compute the amount and the compound interest in each of the following by using the
formulae when :
i Principal = Rs 3000, Rate = 5%, Time = 2 years
ii Principal = Rs 3000, Rate = 18%, Time = 2 years
iii Principal = Rs 5000, Rate = 10 paise per rupee per annum, Time = 2 years
iv Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years
v Principal = Rs 12800, Rate = 7 ½ %, Time = 3 years
vi Principal = Rs 10000, Rate = 20% per annum compounded half-yearly, Time = 2 years
vii Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half yearly, Time
= 2 years.
Solution:
A = P 1+R/100 n
Let us solve
i Given, P = Rs 3000, rate = 5%, time = 2years

A = P 1+R/100 n
te
= 3000 1+5/1002
= 3000 105/1002
= Rs 3307.5
itu
Compound interest CI = A-P = Rs 3307.5 – 3000 = Rs 307.5
ii Given, P = Rs 3000, rate = 18%, time = 2years
A = P 1+R/100 n
= 3000 1+18/1002
= 3000 118/1002
st
= Rs 4177.2

iii Given, P = Rs 5000, rate = 10%, time = 2years
In
A = P 1+R/100 n
= 5000 1+10/1002
= 5000 110/1002
= Rs 6050
Compound interest CI = A-P = Rs 6050 – 5000 = Rs 1050

sh
iv Given, P = Rs 2000, rate = 4%, time = 3years

A = P 1+R/100 n
= 2000 1+4/1003
= 2000 104/1003
ka
= Rs 2249.72

v Given, P = Rs 12800, rate = 7 ½ % = 15/2% = 7.5%, time = 3years
A = P 1+R/100 n
= 12800 1+7.5/1003
Aa
= 12800 107.5/1003
= Rs 15901.4

vi Given, P = Rs 10000, rate = 20 % = 20/2 = 10% quarterly, time = 2years = 2 × 2 = 4years
A = P 1+R/100 n
= 10000 1+10/1004
= 10000 110/1004
= Rs 14641
vii Given, P = Rs 160000, rate = 10% = 10/2% = 5% halfyearly, time = 2years = 2×2 = 4 quarters
A = P 1+R/100 n
= 160000 1+5/1004
= 160000 105/1004
= Rs 194481

2. Find the amount of Rs. 2400 after 3 years, when the interest is compounded annually at the
rate of 20% per annum.
te
Solution:
Given details are,
itu
Rate r = 20% per annum
Time t = 3 years
A = P 1+R/100 n
st
= 2400 1+20/1003
= 2400 120/1003
= Rs 4147.2
In
∴ Amount is Rs 4147.2
3. Rahman lent Rs. 16000 to Rasheed at the rate of 12 ½ % per annum compound interest.
Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
sh
Given details are,
Rate r = 12 ½ % per annum = 12.5%
Time t = 3 years
ka
A = P 1+R/100 n
= 16000 1+12.5/1003
= 16000 112.5/1003
Aa
= Rs 22781.25
4. Meera borrowed a sum of Rs. 1000 from Sita for two years. If the rate of interest is 10%
compounded annually, find the amount that Meera has to pay back.
Solution:
Given details are,
Rate r = 10 % per annum
Time t = 2 years
A = P 1+R/100 n
= 1000 1+10/1002
= 1000 110/1002
= Rs 1210
∴ Amount is Rs 1210
5. Find the difference between the compound interest and simple interest. On a sum of Rs.
te
50,000 at 10% per annum for 2 years.
Solution:
Given details are,
itu
Rate r = 10 % per annum
Time t = 2 years
st
A = P 1+R/100 n
= 50000 1+10/1002
= 50000 110/1002
= Rs 60500
In
CI = Rs 60500 – 50000 = Rs 10500
We know that SI = PTR/100 = 50000×10×2/100 = Rs 10000

∴ Difference amount between CI and SI = 10500 – 10000 = Rs 500
sh
6. Amit borrowed Rs. 16000 at 17 ½ % per annum simple interest. On the same day, he lent it
to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years?
Solution:
Given details are,
ka
Rate r = 17 ½ % per annum = 35/2% or 17.5%
Time t = 2 years
Interest paid by Amit = PTR/100 = 16000×17.5×2/100 = Rs 5600
Amount gained by Amit:
Aa
A = P 1+R/100 n
= 16000 1+17.5/1002
= 16000 117.5/1002
= Rs 22090
CI = Rs 22090 – 16000 = Rs 6090
∴ Amit total gain is = Rs 6090 – 5600 = Rs 490
7. Find the amount of Rs. 4096 for 18 months at 12 ½ % per annum, the interest being
compounded semi-annually.
Solution:
Given details are,
Rate r = 12 ½ % per annum = 25/4% or 12.5/2%
Time t = 18 months = 18/12 × 2 = 3 half years
te
A = P 1+R/100 n
= 4096 1+12.5/2×1003
= 4096 212.5/2003
itu
= Rs 4913
8. Find the amount and the compound interest on Rs. 8000 for 1 ½ years at 10% per annum,
compounded half-yearly.
st
Solution:
Given details are,
In
Rate r = 10 % per annum = 10/2% = 5% halfyearly
Time t = 1 ½ years = 3/2 × 2 = 3 half years
A = P 1+R/100 n
sh
= 8000 1+5/1003
= 8000 105/1003
= Rs 9261
∴ CI = Rs 9261 – 8000 = Rs 1261
ka
9. Kamal borrowed Rs. 57600 from LIC against her policy at 12 ½ % per annum to build a
house. Find the amount that she pays to the LIC after 1 ½ years if the interest is calculated
half-yearly.
Solution:
Given details are,
Aa
Rate r = 12 ½ % per annum = 25/2×2% = 25/4% = 12.5/2% halfyearly
A = P 1+R/100 n
= 57600 1+12.5/2×1003
= 57600 212.5/2003
= Rs 69089.06
10. Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs.
64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid
by Abha after one year and a half.
Solution:
Given details are,
Rate r = 5 % per annum = 5/2% halfyearly
te
A = P 1+R/100 n
itu
= 64000 1+5/2×1003
= 64000 205/2003
= Rs 68921
∴ CI = Rs 68921 – 64000 = Rs 4921
st
11. Rakesh lent out Rs. 10000 for 2 years at 20% per annum, compounded annually. How
much more he could earn if the interest be compounded half-yearly?
Solution:
In
Given details are,
Time t = 2years
sh
A = P 1+R/100 n
= 10000 1+20/1002
= 10000 120/1002
= Rs 14400
ka
When the interest is compounded half yearly,

Rate = 20/2 % = 10%
Time = 2×2 years = 4years
Aa
A = P 1+R/100 n
= 10000 1+10/1004
= 10000 110/1004
= Rs 14641
∴ Rakesh could earn Rs 14641–14400 = Rs 241 more

12. Romesh borrowed a sum of Rs. 245760 at 12.5% per annum, compounded annually. On
the same day, he lent out his money to Ramu at the same rate of interest, but compounded
semi-annually. Find his gain after 2 years.
Solution:
Given details are,
Rate r = 12.5% per annum
Time t = 2years
A = P 1+R/100 n
= 245760 1+12.5/1002
te
= 245760 112.5/1002
= Rs 311040
When compounded semi-annually,
itu
Rate = 12.5/2% = 6.25%
Time = 2×2 years = 4years
st
A = P 1+R/100 n
= 245760 1+6.25/1004
= 245760 106.25/1004
= Rs 313203.75
In
∴ Romesh gain is Rs 313203.75–311040 = Rs 2163.75
13. Find the amount that David would receive if he invests Rs. 8192 for 18 months at 12 ½ %
per annum, the interest being compounded half-yearly.
Solution:
sh
Given details are,
Rate r = 12 ½ % per annum = 25/2×2 = 25/4% = 12.5/2% halfyearly
Time t = 18 months = 18/12 = 1 ½ years = 3/2 ×2 = 3years
ka
A = P 1+R/100 n
= 8192 1+12.5/2×1003
= 8192 212.5/2003
Aa
= Rs 9826
14. Find the compound interest on Rs. 15625 for 9 months, at 16% per annum, compounded
quarterly.
Solution:
Given details are,
Rate r = 16% per annum = 16/4 = 4% quarterly
Time t = 9 months = 9/12 ×4 = 3quarters of a year
A = P 1+R/100 n
= 15625 1+4/1003
= 15625 104/1003
= Rs 17576
∴ CI = Rs 17576 – 15625 = Rs 1951
15. Rekha deposited Rs. 16000 in a foreign bank which pays interest at the rate of 20% per
te
annum compounded quarterly, find the interest received by Rekha after one year
Solution:
Given details are,
itu
Rate r = 20% per annum = 20/4 = 5% quarterly
Time t = 1 year = 4 quarters of a year
st
A = P 1+R/100 n
= 16000 1+5/1004
= 16000 105/1004
= Rs 19448.1
In
∴ CI = Rs 19448.1 – 16000 = Rs 3448.1
16. Find the amount of Rs. 12500 for 2 years compounded annually, the rate of interest being
15% for the first year and 16% for the second year.
sh
Solution:
Given details are,
Rate1 r = 15% and Rate2 = 16%
ka
Time t = 2 years
A = P (1 + R1/100 × 1 + R2/100)
= 12500 1+15/100×1+16/100
Aa
= 12500 1.15×1.16
= Rs 16675
∴ Amount after two years is Rs 16675
17. Ramu borrowed Rs. 15625 from a finance company to buy scooter. If the rate of interest
be 16% per annum compounded annually, what payment will he have to make after 2 ¼
years?
Solution:
Given details are,
Rate r = 16%
Time t = 2 ¼ years
A = P 1+R/100×1+R/100
= 15625 1+16/1002 × 1+(16/4/100)
= 15625 1+16/1002 × 1+4/100
= 15625 1.162 × 1.04
= Rs 21866
te
∴ Amount after 2 ¼ years is Rs 21866
18. What will Rs. 125000 amount to at the rate of 6%, if the interest is calculated after every
four months?
itu
Solution:
Given details are,
st
Time t = 1 year
Since interest is compounded after 4months, interest will be counted as 6/3 = 2% and
Time will be 12/4 = 3quarters
In
A = P 1+R/100 n
= 125000 1+2/1003
= 125000 102/1003
sh
= Rs 132651
19. Find the compound interest at the rate of 5% for three years on that principal which in
three years at the rate of 5% per annum gives Rs. 12000 as simple interest.
ka
Solution:
Given details are,
Simple interest SI = Rs 12000

Aa

Time t = 3 years
SI = PTR/100
P = SI×100 / T×R
= 12000×100 / 3×5
= 1200000/15
= 80000
A = P 1+R/100 n
= 80000 1+5/1003
= 80000 105/1003
= Rs 92610
∴ CI = Rs 92610 – 80000 = Rs 12610
20. A sum of money was lent for 2 years at 20% compounded annually. If the interest is
payable half-yearly instead of yearly, then the interest is Rs. 482 more. Find the sum.
Solution:
Given details are,
te
Rate r = 20% per annum = 20/2 = 10% halfyearly
Time t = 2 years = 2 × 2 = 4 half years
Principal be = Rs P
itu
P 1+R/100 n – P 1+R/100 n = 482
P 1+10/1004 – P 1+20/1002 = 482
P 110/1004 – P 120/1002 = 482
P 1.4641 – P 1.44 = 482
st
0.0241P = 482
P = 482/0.0241
= 20000
In
21. Simple interest on a sum of money for 2 years at 6 ½ % per annum is Rs. 5200. What will
be the compound interest on the sum at the same rate for the same period?
Solution:
sh
Given details are,

Rate = 6 ½ % per annum = 13/2%
Simple Interest SI = Rs 5200

Time t = 2 years
ka
SI = PTR/100
P = SI×100 / T×R
= 5200×100 / 2×13/2
Aa
= 5200×100×2 / 2×13
= 1040000/26
= Rs 40000
Now,
P = Rs 40000
R = 13/2% = 6.5%
T = 2years
A = P 1+R/100 n
= 40000 1+6.5/1002
= 40000 106.5/1002
= Rs 45369
∴ CI = Rs 45369 – 40000 = Rs 5369
22. What will be the compound interest at the rate of 5% per annum for 3 years on that
principal which in 3 years at the rate of 5% per annum gives Rs. 1200 as simple interest.
te
Solution:
Given details are,
Rate = 5 % per annum
itu
Simple Interest SI = Rs 1200
Time t = 3 years
SI = PTR/100
st
P = SI×100 / T×R
= 1200×100 / 3×5
= 120000/15
In
= Rs 8000
Now,
P = Rs 8000
sh
R = 5%
T = 3years
A = P 1+R/100 n
ka
= 8000 1+5/1003
= 8000 105/1003
= Rs 9261
∴ CI = Rs 9261 – 8000 = Rs 1261
Aa

1. On what sum will the compound interest at 5% per annum for 2 years compounded
annually be Rs 164?
Solution:
Given details are,
Compound Interest CI = Rs 164
Time t = 2 years
Let P be ‘x’
CI = A – P
164 = P 1+R/100 n – P
= P [(1 + R/100)n – 1]
te
= x [(1 + 5/100)2 – 1]
= x [(105/100)2 – 1]
164 = x (1.052 – 1)
itu
x = 164 / (1.052 – 1)
= 164/0.1025
= Rs 1600
∴ The required sum is Rs 1600
st
2. Find the principal if the interest compounded annually at the rate of 10% for two years is
Rs. 210.
Solution:
In
Given details are,
Compound Interest CI = Rs 210

Time t = 2 years
sh

Let P be ‘x’
CI = A – P
210 = P 1+R/100 n – P
ka
= P [(1 + R/100)n – 1]
= x [(1 + 10/100)2 – 1]
= x [(110/100)2 – 1]
Aa
210 = x (1.12 – 1)
x = 210 / (1.12 – 1)
= 210/0.21
= Rs 1000
∴ The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded annually. Find the
sum.
Solution:
Given details are,
Amount = Rs 756.25
Time t = 2 years
A = P 1+R/100 n
756.25 = P 1+10/1002
te
P = 756.25 / 1+10/1002
= 756.25/1.21
= 625
itu
∴ The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ % per annum,
compounded half-yearly?
Solution:
st
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
In
Time t = 18months = 18/12years = 3/2 × 2 = 3 half years
A = P 1+R/100 n
4913 = P 1+25/4×1003
sh
P = 4913 / 1+25/4003
= 4913/1.19946
= 4096
∴ The principal amount is Rs 4096
ka
5. The difference between the compound interest and simple interest on a certain sum at 15%
per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Aa
Compound Interest CI – Simple Interest SI= Rs 283.50

Time t = 3 years
CI – SI = 283.50
P [(1 + R/100)n – 1] – PTR/100 = 283.50

P [(1 + 15/100)3 – 1] – P(315)/100 = 283.50
P
1.520–1
– 45P/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
∴ The sum is Rs 4000
te
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two
years Rs. 1290 as interest compounded annually, find the sum she borrowed.
Solution:
itu
Given details are,
Time = 2 years
st
CI = Rs 1290
CI = P [(1 + R/100)n – 1]
In
1290 = P [(1 + 15/100)2 – 1]
1290 = P
0.3225
sh
P = 1290/0.3225
= 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4% per
ka
annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Aa
CI = Rs 163.20
Principal P = Rs 2000
CI = P [(1 + R/100)n – 1]
163.20 = 2000[(1 + 4/100)n – 1]
163.20 = 2000 [(1.04)n -1]
163.20 = 2000 × 1.04n – 2000

163.20 + 2000 = 2000 × 1.04n
2163.2 = 2000 × 1.04n
1.04n = 2163.2/2000
1.04n = 1.0816
1.04n = 1.042
So on comparing both the sides, n = 2
∴ Time required is 2years
8. In how much time would Rs. 5000 amount to Rs. 6655 at 10% per annum compound
te
interest?
Solution:
Given details are,
itu
Rate = 10% per annum
A = Rs 6655
st
A = P 1+R/100n
6655 = 5000 1+10/100n
6655 = 5000 11/10n
In
11/10n = 6655/5000
11/10n = 1331/1000
11/10n = 11/103
So on comparing both the sides, n = 3
sh

9. In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest compounded half-
yearly?
Solution:
ka
Given details are,
Rate = 8% per annum = 8/2 = 4% halfyearly

A = Rs 4576
Aa
Let n be ‘2T’
A = P 1+R/100n
4576 = 4400 1+4/1002T
4576 = 4400 104/1002T
104/1002T = 4576/4400
104/1002T = 26/25
26/252T = 26/251
So on comparing both the sides, n = 2T = 1
∴ Time required is ½ year
10. The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per
annum is Rs. 20. Find the sum.
Solution:
Given details are,
Time = 2years
te
Compound Interest CI – Simple Interest SI= Rs 20
CI – SI = 20
itu
P [(1 + R/100)n – 1] – PTR/100 = 20
P [(1 + 4/100)2 – 1] – P(24)/100 = 20
P
51/625
st
– 2P/25 = 20
51/625P – 2/25P = 20
51P−50P/625 = 20
In
P = 20 × 625
P = 20/7.918
= 12500
sh

11. In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound interest?
Solution:
Given details are,
ka
Principal = Rs 1000
Amount = Rs 1331
Aa
Let time = T years

A = P 1+R/100n
1331 = 1000 1+10/100T
1331 = 1000 110/100T
11/10T = 1331/1000
11/10T = 11/103
So on comparing both the sides, n = T = 3
12. At what rate percent compound interest per annum will Rs. 640 amount to Rs. 774.40 in 2
years?
Solution:
Given details are,
Principal = Rs 640
Amount = Rs 774.40
Time = 2 years
te
Let rate = R%
A = P 1+R/100n
itu
774.40 = 640 1+R/1002
1+R/1002 = 774.40/640
1+R/1002 = 484/400
1+R/1002 = 22/202
By cancelling the powers on both sides,
1+R/100 = 22/20
R/100 = 22/20 – 1
st
In
= 22−20/20
= 2/20
= 1/10
R = 100/10
sh
= 10%
∴ Required Rate is 10% per annum
13. Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest
being compounded half-yearly?
ka
Solution:
Given details are,
Principal = Rs 2000
Aa
Amount = Rs 2662
Time = 1 ½ years = 3/2 × 2 = 3 half years
Let rate be = R% per annum = R/2 % half yearly
A = P 1+R/100n
2662 = 2000 1+R/2×1003
1+R/2003 = 1331/1000
1+R/1003 = 11/103
By cancelling the powers on both sides,
1+R/200 = 11/10
R/200 = 11/10 – 1
= 11−10/10
= 1/10
R = 200/10
= 20%
te
14. Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest.
She lent this sum at the same rate to Hari for two years compound interest. At the end of two
years she received Rs. 210 as compound interest, but paid Rs. 200 only as simple interest.
Find the sum and the rate of interest.
itu
Solution:
Given details are,
C.I that Kamala receives = Rs 210
st
S.I that Kamala paid = Rs 200
Time = 2 years
So,
In
We know, SI = PTR/100
= P×2×R/100
P×R = 10000 ………….. Equation 1
CI = A – P
sh
CI = P [(1 + R/100)n – 1]
210 = P [(1 + R/100)2 – 1]
210 = P (12 + R2/1002 + 21R/100 – 1) byusingtheformula(a+b2)

ka
210 = P (1 + R2/10000 + R/50 – 1)

210 = P (R2/10000 + R/50)
210 = PR2/10000 + PR/50
We know PR = 10000 from Equation 1
Aa
210 = 10000R/10000 + 10000/50

210 = R + 200
R = 210 – 200
= 10%
In Equation 1, PR = 10000
P = 10000/R
= 10000/10
= 1000
∴ Required sum is Rs 1000
15. Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half,
interest being compounded six monthly.
Solution:
Given details are,
Principal = Rs 2000
te
Amount = Rs 2315.25
Time = 1 ½ years = 3/2 years
Let rate be = R % per annum
itu
A = P 1+R/100n
2315.25 = 2000 1+R/1003/2
1+R/1003/2 = 2315.25/2000
st
1+R/1003/2 = 1.1576
1+R/100 = 1.1025
R/100 = 1.1025 – 1
In
= 0.1025 × 100
= 10.25
∴ Required Rate is 10.25% per annum
16. Find the rate at which a sum of money will double itself in 3 years, if the interest is
sh
compounded annually.
Solution:
Given details are,
ka
Time = 3 years
Let rate be = R %
Also principal be = P
So, amount becomes = 2P
Aa
A = P 1+R/100n
2P = P 1+R/1003
1+R/1003 = 2
1+R/100 = 21/3
1 + R/100 = 1.2599
R/100 = 1.2599-1
= 0.2599
R = 0.2599 × 100
= 25.99
17. Find the rate at which a sum of money will become four times the original amount in 2
years, if the interest is compounded half-yearly.
Solution:
Given details are,
te
Time = 2 years = 2×2 = 4 half years
Let rate = R % per annum = R/2% half years
Let principal be = P
itu
So, Amount becomes = 4P
A = P 1+R/100n
st
4P = P 1+R/2×1004
1+R/2004 = 4
1+R/200 = 41/4
1 + R/200 = 1.4142
In
R/200 = 1.4142-1
= 0.4142
R = 0.4142 × 200
sh
= 82.84%
18. A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.
Solution:
ka
Given details are,

Amount = Rs 5832
Time = 2 years
Aa
Rate = 8%
Let principal be = P
A = P 1+R/100n
5832 = P 1+8/1002
5832 = P 1.1664
P = 5832/1.1664
= 5000
19. The difference between the compound interest and simple interest on a certain sum for 2
years at 7.5% per annum is Rs. 360. Find the sum.
Solution:
Given,
Time = 2 years
Rate = 7.5 % per annum
te
Let principal = Rs P
Compound Interest CI – Simple Interest SI = Rs 360

C.I – S.I = Rs 360
itu
P [(1 + R/100)n – 1] – PTR/100 = 360

P [(1 + 7.5/100)2 – 1] – P(27.5)/100 = 360
P
st
249/1600
– 3P/20 = 360
249/1600P – 3/20P = 360
In
249P−240P/1600 = 360
9P = 360 × 1600
P = 576000/9
= 64000
sh

20. The difference in simple interest and compound interest on a certain sum of money at
\(6\frac{2}{3}\) % per annum for 3 years in Rs. 46. Determine the sum.
ka
Solution:
Given,
Time = 3 years
Rate = \(6\frac{2}{3}\) % per annum = 20/3%
Aa
Let principal = Rs P
Compound Interest CI – Simple Interest SI = Rs 46

C.I – S.I = Rs 46
P [(1 + R/100)n – 1] – PTR/100 = 46

P [(1 + 20/3×100)3 – 1] – P(320/3)/100 = 46
P[(1 + 20/300)3 – 1] – P/5 = 46
P
721/3375
– P/5 = 46
721/3375P – 1/5P = 46
721P−675P/3375 = 46
46P = 46 × 3375
46P = 46 × 3375/46
= 3375
te
21. Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an
amount of Rs. 13230 after n years. Find the value of n.
itu
Solution:
Given details are,
Principal = Rs 12000
Amount = Rs 13230
Rate = 5% per annum
Let time = T years
st
In
A = P 1+R/100n
13230 = 12000 1+5/100T
13230 = 12000 105/100T
21/20T = 13230/12000
sh
21/20T = 441/400
21/20T = 21/202
ka
22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410
in 2 years?
Solution:
Given details are,
Aa
Principal = Rs 4000
Time = 2years
CI = Rs 410
Rate be = R% per annum
CI = P [(1 + R/100)n – 1]
410 = 4000 [(1 + R/100)2 – 1]
410 = 4000 1+R/1002 – 4000
410 + 4000 = 4000 1+R/1002
1+R/1002 = 4410/4000
1+R/1002 = 441/400
1+R/1002 = 21/202
By cancelling the powers on both the sides,
1 + R/100 = 21/20
R/100 = 21/20 – 1
te
= 21−20/20
= 1/20
R = 100/20
itu
=5
23. A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at
the end of 2 years. Find the sum deposited.
st
Solution:
Given details are,
Time = 2years
In
Amount = Rs 10404
Rate be = 2% per annum
Let principal be = Rs P
sh

A = P [(1 + R/100)n
10404 = P [(1 + 2/100)2]
ka
10404 = P
1.0404
P = 10404/1.0404
= 10000
Aa

24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum
compound interest?
Solution:
Given details are,
Principal = Rs 1600
Amount = Rs 1852.20
Rate = 5% per annum
Let time = T years
A = P 1+R/100n
1852.20 = 1600 1+5/100T
1852.20 = 1600 105/100T
21/20T = 1852.20/1600
21/20T = 9261/8000
te
21/20T = 21/203
itu
25. At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound
interest?
Solution:
Given details are,
st
Principal = Rs 1000
Amount = Rs 1102.50
Rate = R% per annum
In
Let time = 2 years
A = P 1+R/100n
sh
1102.50 = 1000 1+R/1002

1+R/1002 = 1102.50/1000
1+R/1002 = 4410/4000
1+R/1002 = 21/202
1 + R/100 = 21/20
ka
R/100 = 21/20 – 1
= 21−20/20
= 1/20
Aa
R = 100/20
=5
∴ Required Rate is 5%
26. The compound interest on Rs. 1800 at 10% per annum for a certain period of time is Rs.
378. Find the time in years.
Solution:
Given details are,
Principal = Rs 1800
CI = Rs 378
Let time = T years
CI = P [(1 + R/100)n – 1]
378 = 1800 [(1 + 10/100)T – 1]
378 = 1800 [(110/100)T – 1]
te
378 = 1800 [(11/10)T – 1800
378 + 1800 = 1800 [(11/10)T
itu
11/10T = 2178/1800
11/10T = 726/600
11/10T = 121/100
11/10T = 11/102
st
27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest
being compounded annually?
In
Solution:
Given details are,
Time = 2years
sh
Amount = Rs 45582.25
Rate be = 6 ¾ % per annum = 27/4%
ka
A = P [(1 + R/100)n
45582.25 = P [(1 + 27/4×100)2]
45582.25 = P 1+27/4002
45582.25 = P 427/4002
Aa
45582.25 = P × 427/400 × 427/400
P = 45582.25×400×400 / 427×427
P = 7293160000/182329
= 40000
28. Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded
annually. Find the sum.
Solution:
Given details are,
Time = 2years
Amount = Rs 453690
Rate be = 6.5 % per annum
A = P [(1 + R/100)n
te
453690 = P [(1 + 6.5/100)2]
453690 = P 106.5/1002
453690 = P × 106.5/100 × 106.5/100
itu
P = 453690×100×100 / 106.5×106.5
P = 4536900000/11342.25
= 400000
st
∴ Required sum is Rs 400000 In
1. The present population of a town is 28000. If it increases at the rate of 5% per annum, what
will be its population after 2 years?
Solution:
sh
Given details are,

Present population of town is = 28000
Rate of increase in population is = 5% per annum
ka
Number of years = 2
A = P 1+R/100n
Population of town after 2 years = 28000 1+5/1002
Aa
= 28000 1.052
= 30870
∴ Population of town after 2 years will be 30870
2. The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5%
respectively, calculate the population of city after 3 years.
Solution:
Given details are,
Population of city P = 125000

Annual birth rate R1= 5.5 %
Annual death rate R2 = 3.5 %
Annual increasing rate R = (R1 – R2) = 5.5 – 3.5 = 2 %
Number of years = 3
A = P 1+R/100n
So, population after two years is = 125000 1+2/1003
te
= 125000 1.023
= 132651
∴ Population after 3 years will be 132651
itu
3. The present population of a town is 25000. It grows at 4%, 5% and 8% during first year,
second year and third year, respectively. Find its population after 3 years.
Solution:
Given details are,
st
Present population is = 25000
First year growth R1 = 4%
Second year growth R2 = 5%
In
Third year growth R3 = 8%
Number of years = 3
sh
A = P 1+R/100n
So, population after three years = P (1 + R1/100) (1 + R2/100) (1 + R3/100)
= 250001+4/100 1+5/100 1+8/100

= 25000 1.04 1.05 1.08
ka
= 29484
∴ Population after 3 years will be 29484
4. Three years ago, the population of a town was 50000. If the annual increase during three
successive years be at the rate of 4%, 5% and 3%, respectively, find the present population.
Aa
Solution:
Given details are,
Three years ago population of town was = 50000
Annual increasing in 3 years = 4%,5%, 3%, respectively
So, let present population be = x
A = P 1+R/100n
x = 50000 1+4/100 1+5/100 1+3/100
= 50000 1.04 1.05 1.03
= 56238
∴ Present population of the town is 56238
5. There is a continuous growth in population of a village at the rate of 5% per annum. If its
present population is 9261, what it was 3 years ago?
Solution:
Given details are,
te
Present population of town is = 9261
Continuous growth of population is = 5%
itu
So, let population three years ago be = x
A = P 1+R/100n
9261 = x 1+5/100 1+5/100 1+5/100
st
9261 = x 1.05 1.05 1.05
= 8000
In
6. In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual
rate of growth of the production of scooters.
Solution:
Given details are,
sh
Initial production of scooters is = 40000

Final production of scooters is = 46305
Time = 3 years
ka
Let annual growth rate be = R%

A = P 1+R/100n
46305 = 40000 1+R/100 1+R/100 1+R/100
Aa
46305 = 40000 1+R/1003

1+R/1003 = 46305/40000
= 9261/8000
= 21/203
1 + R/100 = 21/20
R/100 = 21/20 – 1
R/100 = 21−20/20
= 1/20
R = 100/20
=5
∴ Annual rate of growth of the production of scooters is 5%
7. The annual rate of growth in population of a certain city is 8%. If its present population is
196830, what it was 3 years ago?
Solution:
Given details are,
Annual growth rate of population of city is = 8%
te
Present population of city is = 196830
Let population of city 3 years ago be = x
itu
A = P 1+R/100
196830 = x 1+8/100 1+8/100 1+8/100
196830 = x 27/25 27/25 27/25
196830 = x 1.08 1.08 1.08
st
196830 = 1.259712x
x = 196830/1.259712
= 156250
In
∴ Population 3 years ago was 156250
8. The population of a town increases at the rate of 50 per thousand. Its population after 2
years will be 22050. Find its present population.
Solution:
sh
Given details are,

Growth rate of population of town is = 50/1000×100 = 5%
Population after 2 years is = 22050
ka
So, let present population of town be = x

A = P 1+R/100
22050 = x 1+5/100 1+5/100
Aa
22050 = x 105/100 105/100

22050 = x 1.05 1.05
22050 = 1.1025x
x = 22050/1.1025
= 20000
9. The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the
second hour and again increases by 12% in the third hour. If the count of bacteria in the
sample is 13125000, what will be the count of bacteria after 3 hours?
Solution:
Given details are,
Count of bacteria in sample is = 13125000
The increase and decrease of growth rates are = 10%, -8%, 12%
So, let the count of bacteria after 3 hours be = x
A = P 1+R/100
te
x = 13125000 1+10/100 1–8/100 1+12/100
x = 13125000 110/100 92/100 112/100
x = 13125000 1.1 0.92 1.12
itu
= 14876400
∴ Count of bacteria after three hours will be 14876400
10. The population of a certain city was 72000 on the last day of the year 1998. During next
year it increased by 7% but due to an epidemic it decreased by 10% in the following year.
What was its population at the end of the year 2000?
Solution:
Given details are,
st
In
Population of city on last day of year 1998 = 72000
Increasing rate R in 1999 = 7%

Decreasing rate R in 2000 = 10 %
sh
A = P 1+R/100
x = 72000 1+7/100 1–10/100
= 72000 107/100 90/100
= 72000 1.07 0.9
= 69336
ka
∴ Population at the end of the year 2000 will be 69336

11. 6400 workers were employed to construct a river bridge in four years. At the end of the
first year, 25% workers were retrenched. At the end of the second year, 25% of those working
at that time were retrenched. However, to complete the project in time, the number of workers
Aa
was increased by 25% at the end of the third year. How many workers were working during
the fourth year?
Solution:
Given details are,
Initial number of workers are = 6400
At the end of first year = 25% retrenched
At the end of second year = 25% retrenched
At the end of third year = 25% increased
A = P 1+R/100
x = 6400 1–25/100 1–25/100 1+25/100
= 6400 75/100 75/100 125/100
= 6400 0.75 0.75 1.25
= 4500
∴ Number of workers working during the fourth year is 4500
12. Aman started a factory with an initial investment of Rs 100000. In the first year, he
te
incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in
the third year rose to 12%. Calculate his net profit for the entire period of three years.
Solution:
itu
Given details are,
Initial investment by Aman = Rs.100000
In first year = incurred a loss of 5%
In second year = earned a profit of 10%
In third year = earned a profit of 12 %
st
In
A = P 1+R/100
x = 100000 1–5/100 1+10/100 1+12/100
= 100000 95/100 110/100 112/100
= 100000 0.95 1.1 1.12
= 117040
sh
∴ Aman’s net profit for entire three years is 117040 – 100000 = Rs 17040
13. The population of a town increases at the rate of 40 per thousand annually. If the present
population be 175760, what was the population three years ago.
Solution:
ka
Given,
Annul increase rate of population of town = 40/1000× 100 = 4%
Present population of town = 175760
Aa
So, let the population of town 3 years ago be = x

A = P 1+R/100
175760 = x 1+4/100 1+4/100 1+4/100
175760 = x 104/100 104/100 104/100
175760 = x 1.04 1.04 1.04
175760 = 1.124864x
x = 175760/1.124864
= 156250
∴ Population 3 years ago was 156250
14. The population of a mixi company in 1996 was 8000 mixies. Due to increase in demand it
increases its production by 15% in the next two years and after two years its demand
decreases by 5%. What will its production after 3 years?
Solution:
Given,
te
Population of mixi company in 1996 was = 8000 mixies
Production growth rate in next 2 years is = 15 %
Decrease rate in 3rd year is = 5%
itu
A = P 1+R/100
x = 8000 1+15/100 1+15/100 1–5/100
= 8000 115/100 115/100 95/100
st
= 8000 1.15 1.15 0.95
= 10051
∴ Production after three years will be 10051 mixies
In
15. The population of a city increases each year by 4% of what it had been at the beginning of
each year. If the population in 1999 had been 6760000, find the population of the city
in 1 2001 ii 1997.
Solution:
Given details are,
sh
Annual increase rate of population of city is = 4%

Population in 1999 was = 6760000
i Population of the city in 2001 2yearsafter

ka
A = P 1+R/100
x = 6760000 1+4/100 1+4/100
= 6760000 104/100 104/100
= 6760000 1.04 1.04
Aa
= 7311616
∴ Population in the year 2001 is 7311616
ii Population of city in 1997 2yearsago

A = P 1+R/100
x = 6760000 1–4/100 1–4/100
= 6760000 96/100 96/100
= 6760000 0.96 0.96
= 6230016
∴ Population in the year 1997 was 6230016
16. Jitendra set up a factory by investing Rs. 2500000. During the first two successive years
his profits were 5% and 10%, respectively. If each year the profit was on previous year’s
capital, compute his total profit.
Solution:
Given details are,
te
Initial investment by Jitendra was = Rs 2500000
Profit in first 2 successive years were = 5% and 10%
Final investment after two successive profits = 2500000 1+5/100 1+10/100
itu
= 2500000 105/100 110/100
= 2500000 1.05 1.1
= 2887500
∴ Jitendra total profit is = 2887500 – 2500000 = Rs 387500
st
In
1. Ms. Cherian purchases a boat for Rs. 16000. If the total cost of the boat is depreciating at
the rate of 5% per annum, calculate its value after 2 years.
Solution:
Given details are,
sh
Price of a boat is = Rs 16000

Depreciation rate = 5% per annum
ka
A = P 1+R/100
= P 1+R/1002
Since it is depreciation we use P 1–R/100n
= 16000 1–5/100 1–5/100
= 16000 95/100 95/100
Aa
= 16000 0.95 0.95

= 14440
∴ Value of the boat after two years is Rs 14440
2. The value of a machine depreciates at the rate of 10% per annum. What will be its value 2
years hence, if the present value is Rs 100000? Also, find the total depreciation during this
period.
Solution:
Given details are,
Present value of machine is = Rs 100000
Rate of depreciation = 10% per annum
A = P 1+R/100
= 100000 1–10/100 1–10/100
= 100000 90/100 90/100
= 100000 0.9 0.9
= 81000
te
Value of machine after two years will be Rs 81000
∴ Total depreciation during this period is Rs 100000–81000 = Rs 19000

3. Pritam bought a plot of land for Rs. 640000. Its value is increasing by 5% of its previous
itu
value after every six months. What will be the value of the plot after 2 years?
Solution:
Given details are,
Price of land is = Rs 640000
Rate of increase = 5% in every six month
st
In
A = P 1+R/100n
= 640000 1+5/100 1+5/100 1+5/100 1+5/100
= 640000 105/100 105/100 105/100 105/100
= 640000 1.025 1.025 1.025 1.025
= 706440.25
sh
∴ The value of the plot after two years will be Rs 706440.25

4. Mohan purchased a house for Rs. 30000 and its value is depreciating at the rate of 25% per
year. Find the value of the house after 3 years.
Solution:
ka
Given details are,

Price of house is = Rs 30000
Depreciation rate is = 25% per year
Aa
A = P 1+R/100n
= 30000 1–25/100 1−25/100 1–25/100
= 30000 75/100 75/100 75/100
= 30000 0.75 0.75 0.75
= 12656.25
∴ The value of the house after 3 years is Rs 12656.25
5. The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years
ago. If its present value is Rs. 43740, find its purchase price.
Solution:
Given details are,
Present value of machine is = Rs 43740
Depreciation rate of machine is = 10% per annum
Let the purchase price 3 years ago be = Rs x
A = P 1+R/100n
te
43740 = x 1–10/100 1–10/100 1–10/100
43740 = x 90/100 90/100 90/100
43740 = x 0.9 0.9 0.9
itu
43740 = 0.729x
x = 43740/0.729
= 60000
∴ The purchase price is Rs 60000
st
6. The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per
annum. If its present value is Rs. 9680, for how much was it purchased?
Solution:
In
Given details are,
Present value of refrigerator is = Rs 9680
Depreciation rate is = 12%
sh
Let the price of refrigerator 2 years ago be = Rs x

A = P 1+R/100n
9680 = x 1–12/100 1–12/100
ka
9680 = x 88/100 88/100

9680 = x 0.88 0.88
9680 = 0.7744x
x = 9680/0.7744
Aa
= 12500
∴ The refrigerator was purchased for Rs 12500
7. The cost of a T.V. set was quoted Rs. 17000 at the beginning of 1999. In the beginning of
2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4%
in the beginning of 2001. What was the cost of the T.V. set in 2001?
Solution:
Given details are,
Cost of T.V at beginning of 1999 is = Rs 17000
Hiked in price in the year 2000 is = 5%
Depreciation rate in the year 2001 is = 4%
A = P 1+R/100n
= 17000 1+5/100 1–4/100
= 17000 105/100 96/100
= 17000 1.05 0.96
= 17136
te
∴ The cost of TV set in the year 2001 is Rs 17136
8. Ashish started the business with an initial investment of Rs. 500000. In the first year he
incurred a loss of 4%. However during the second year he earned a profit of 5% which in third
itu
year rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Given,
Initial investment by Ashish is = Rs 500000
Incurred loss in the first year is = 4%
Profit in 2nd year is = 5 %
st
In
Profit in 3rd year is = 10%
A = P 1+R/100n
= 500000 1–4/100 1+5/100 1+10/100
sh
= 500000 96/100 105/100 110/100

= 500000 0.96 1.05 1.1
= 554400
∴ The net profit for the entire period of 3 years is Rs 554400
ka
Aa

RD Sharma Solutions For Class 8 Chapter 14 Compound Interest

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RD Sharma Solutions For Class 8 Chapter 14 Compound Interest

Interest for the first year = 3000×5×1/100 = 150

Rate r = 20% = 20/4 = 5% forquarteryear

Given details are,

By using the formula,

Simple interest = P×T×R/100

∴ Required amount is Rs 8195.45

By using the formula,

∴ Compound Interest = A – P = Rs 45796 – Rs 40000 = Rs 5796

EXERCISE 14.2 PAGE NO: 14.14

i Given, P = Rs 3000, rate = 5%, time = 2years

Compound interest CI = A-P = Rs 4177.2 – 3000 = Rs 1177.2

Compound interest CI = A-P = Rs 6050 – 5000 = Rs 1050

iv Given, P = Rs 2000, rate = 4%, time = 3years

Compound interest CI = A-P = Rs 2249.72 – 2000 = Rs 249.72

Compound interest CI = A-P = Rs 15901.4 – 12800 = Rs 3101.4

Compound interest CI = A-P = Rs 194481 – 160000 = Rs 34481

Given details are,

By using the formula,

We know that SI = PTR/100 = 50000×10×2/100 = Rs 10000

By using the formula,

By using the formula,

When the interest is compounded half yearly,

By using the formula,

∴ Rakesh could earn Rs 14641–14400 = Rs 241 more

Given details are,

By using the formula,

Simple interest SI = Rs 12000

Rate r = 5% per annum

Given details are,

Simple Interest SI = Rs 5200

By using the formula,

EXERCISE 14.3 PAGE NO: 14.20

Compound Interest CI = Rs 210

By using the formula,

Rate = 15 % per annum

Compound Interest CI – Simple Interest SI= Rs 283.50

P [(1 + R/100)n – 1] – PTR/100 = 283.50

163.20 = 2000 × 1.04n – 2000

∴ Time required is 3years

Given details are,

Rate = 8% per annum = 8/2 = 4% halfyearly

∴ The sum is Rs 12500

Let time = T years

210 = P (12 + R2/1002 + 21R/100 – 1) byusingtheformula(a+b2)

210 = P (1 + R2/10000 + R/50 – 1)

210 = 10000R/10000 + 10000/50

By using the formula,

Given details are,

Compound Interest CI – Simple Interest SI = Rs 360

P [(1 + R/100)n – 1] – PTR/100 = 360

∴ The sum is Rs 64000

Compound Interest CI – Simple Interest SI = Rs 46

P [(1 + R/100)n – 1] – PTR/100 = 46

By using the formula,

∴ Required sum is Rs 10000

1102.50 = 1000 1+R/1002

45582.25 = P × 427/400 × 427/400

Given details are,

Population of city P = 125000