Question Bank
Compound Interest
1. The simple interest on a certain sum of money for 3 years at 5% p.a.
is Rs 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 is Rs 1200.
1
Simple interest for one year = of Rs 1200 = Rs 400.
3
P×r ×T P×5×1
We know that, S.I. =
=
100
100
= Rs 400
⇒
P = Rs 8000.
Now, P = Rs 8000, r = 5% p.a., n = 3 years
n
r ⎞
⎛
∴ A = P ⎜1 +
⎟
100 ⎠
⎝
3
5 ⎞
⎛
= Rs 8000 × ⎜1 +
⎟
100 ⎠
⎝
3
⎛ 21 ⎞
= Rs 8000 × ⎜ ⎟
⎝ 20 ⎠
21
21
21
= Rs 8000 ×
×
×
20
20
20
= Rs 9261
∴ Amount after 3 years = Rs 9261
Compound interest for 3 years
=A–P
= Rs 9261 – Rs 8000
= Rs 1261.
2. A sum of Rs 9,600 is invested for 3 years at 10% p.a. compound
interest :
Math Class X
1
Question Bank
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in 2 years.
(iv) Find the difference between the answers in (ii) and (i) and
find the interest on this sum for one year.
(v) Hence, write down the compound interest for the third year.
Solution.
(i) P = Rs 9,600, r = 10% p.a., n = 1 year
n
r ⎤
⎡
∴ A = P ⎢1 +
100 ⎥⎦
⎣
1
10 ⎤
⎡
⇒ A = Rs 9,600 ⎢1 +
100 ⎥⎦
⎣
11
= Rs 9,600 ×
10
= Rs 10,560
∴ The sum due at the end of the first year = Rs 10,560
(ii) Amount at the end of first year = The principal for the
second year.
P = Rs 10,560, r = 10% p.a., n = 1 year.
1
10 ⎤
⎡
∴ A = Rs 10,560 ⎢1 +
100 ⎥⎦
⎣
11
⇒ A = Rs 10,560 ×
10
= Rs 11,616.
∴ The sum due at the end of the second year = Rs 11,616.
(iii) Compound interest earned in 2 years
= Amount at the end of 2nd year – Principal
= Rs 11,616 – Rs 9,600 = Rs 2,016.
(iv) Difference between the answers in (ii) and (i).
= Rs 11,616 – Rs 10,560 = Rs 1,056.
The interest on this sum for 1 year
Math Class X
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Question Bank
Rs 1056 × 1 × 10
= Rs 105.60
100
(v) Principal for the third year = Rs 11,616
Rate = 10%
Rs 11,616 × 1 × 10
∴ Compound interest for third year =
100
= Rs 1161.60
3. A sum of money is invested at a constant rate of compound interest
payable annually. The interest in two successive years are Rs 225
and Rs 240. Find the interest in the third year.
Solution.
1
r ⎤
⎡
For 1st year, A1 = P ⎢1 +
100 ⎥⎦
⎣
r ⎤
⎡
∴ CI for 1st year = A1 – P = P ⎢1 +
⎥– P
100
⎣
⎦
Pr
⇒
225 =
... (i)
100
2
r ⎤
⎡
For 2nd year A2 = P ⎢1 +
100 ⎥⎦
⎣
∴ CI for 2nd year
2
r ⎞
r ⎞
r ⎞
⎛
⎛
⎛
= A 2 – P ⎜1 +
=
P
1
+
–
P
1
+
⎟
⎜
⎟
⎜
⎟
100 ⎠
100 ⎠
100 ⎠
⎝
⎝
⎝
r ⎞ ⎡
r
Pr ⎛
r ⎞
⎛
⎤
= P ⎜1 +
– 1⎥ =
⎟ ⎢1 +
⎜1 +
⎟
100 ⎠ ⎣
100
100 ⎠
⎝
⎦ 100 ⎝
Pr ⎛
r ⎞
⇒ 240 =
... (ii)
⎜1 +
⎟
100 ⎝
100 ⎠
240
r
Dividing (ii) by (i), we get
=1+
225
100
16
r
r
r
100
2
=1+
=1+
=6
⇒
⇒
⇒ r
15
100
100
100
15
3
=
Math Class X
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Question Bank
2
in (i), we get
3
P × 20
225 × 3 × 100
225 =
= 3375
⇒ P=
3 × 100
20
3
r ⎞
⎛
For 3rd year A3 = P ⎜1 +
⎟
100 ⎠
⎝
3
2
r ⎞
r ⎞
⎛
⎛
∴
CI for 3rd year = P ⎜1 +
⎟ – P ⎜1 +
⎟
100 ⎠
100 ⎠
⎝
⎝
2
r ⎞
r
⎛
= P ⎜1 +
×
⎟
100 ⎠ 100
⎝
Substituting r = 6
2
3375 × 20 ⎛
20 ⎞
=
⎜1 +
⎟
3 × 100 ⎝
3 × 100 ⎠
3375 × 20 16 16
=
×
×
= 256
3 × 100
15 15
Hence, interest in the third year = Rs 256
4. A sum of Rs 24000 is lent out for 2 years at compound interest, the
rate of interest being 10% for the first year and 12% for the second
year. The borrower returns some money at the end of the first year
and on paying Rs 12768 at the end of the second year, the total debt
is cleared. Calculate the amount of money returned at the end of the
first year.
Solution.
Principal for the first year = Rs 24000,
Rate = 10% p.a.
10 ⎞
⎛
Amount at the end of first year = Rs 24000 ⎜1 +
⎟
100 ⎠
⎝
= Rs 26400
Let the amount of money returned at the end of first year be
Rs x.
Then, principal for the second year = Rs (26400 – x), Rate = 12% p.a.
Math Class X
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Question Bank
12 ⎞
⎛
Amount at the end of second year = Rs (26400 – x) ⎜1 +
⎟
100 ⎠
⎝
28
= Rs (26400 – x) ×
25
28
= 12768
But
(26400 – x) ×
25
1268 × 25
⇒
26400 – x =
= 11400
28
⇒
x = 26400 – 11400 = 15000
Hence, amount of money returned at the end of 1st year = Rs 15000
5. The cost of a car, purchased 2 years ago, depreciates at the rate of
20% p.a. If its present value is Rs 315600, find :
(i) its value after 2 years
(ii) its value, when it was purchased 2 years ago.
Solution.
(i) Here, r = 20% p.a. and present value = Rs 315600
Value of the car after 2 years
2
r ⎤
⎡
= (Its present value) × ⎢1 +
100 ⎥⎦
⎣
2
⎡
20 ⎞ ⎤
⎛
= Rs ⎢315600 × ⎜ 1 –
⎟ ⎥
100
⎝
⎠ ⎥⎦
⎢⎣
4
4⎤
⎡
= Rs ⎢315600 × × ⎥
5
5⎦
⎣
= Rs 201984
∴ Value of the car after 2 years = Rs 201984
(ii) The car was purchased 2 years ago.
2
r ⎤
⎡
(Value of the car 2 years ago) × ⎢1 +
100 ⎥⎦
⎣
= Its present value.
Math Class X
5
Question Bank
⇒ Value of the car 2 years ago
⎡
⎤
⎢
⎥
Its present value ⎢ 315600 ⎥
=
=
2
2
⎢⎛
20 ⎞ ⎥
r
⎛
⎞
⎢ ⎜1 –
⎜1 –
⎟
⎟ ⎥
100 ⎠
100 ⎠ ⎦
⎝
⎣⎝
5
5⎤
⎡
= Rs ⎢315600 × × ⎥ = 493125
4
4⎦
⎣
∴ Value of the car 2 years ago = Rs 493125
6. The population of a town increased at 5% of its population every
year. The present population of the town is 21000.
(i) What was the population of the town a year ago?
(ii) What would be the population of the town after 4 years?
Solution. Here, r = 5% and present population = 21000
r ⎤
⎡
(i) (Population 1 year ago) × ⎢1 +
100 ⎥⎦
⎣
= Present population
Pr esent population
⇒
Population 1 year ago =
r ⎞
⎛
⎜1 +
⎟
100 ⎠
⎝
21000
20
=
= 21000 ×
5 ⎞
21
⎛
1
+
⎜
⎟
100 ⎠
⎝
= 20000
r ⎤
⎡
(ii) (Present population) × ⎢1 +
100 ⎥⎦
⎣
= Population after 1 year
Math Class X
6
Question Bank
5 ⎤
⎡
Population after 1 year = 21000 × ⎢1 +
100 ⎥⎦
⎣
21
= 21000 ×
20
= 22050
7. A man bought a plot of land for Rs 160000 and a car for Rs 180000
at the same time. The price of the plot of land grows uniformly at
the rate of 25% p.a. while the prices of the car depreciates by 20%
p.a. If the man sells the plot of land as well the car after 3 years,
what were be his profit or loss?
Solution.
Total money invested = Rs (160000 + 180000)
= Rs 340000
3
25 ⎤
⎡
Price of the plot of land after 3 years = Rs 160000 ⎢1 +
100 ⎥⎦
⎣
5
5
5
= Rs 160000 × × ×
4
4
4
= Rs 312500
3
20 ⎤
⎡
Price of the car after 3 years = Rs 180000 ⎢1 –
100 ⎥⎦
⎣
4
4
4
= Rs 180000 × × ×
5
5
5
= Rs 92160
Selling prices of both the items = Rs (312500 + 92160)
= Rs 404660
∴
Profit earnerd = Rs (404660 – 340000)
= Rs 64660
⇒
Math Class X
7
Question Bank