Students can cross-reference their work with OP Malhotra Class 9 Solutions Chapter 2 Compound Interest Chapter Test to ensure accuracy.

## S Chand Class 9 ICSE Maths Solutions Chapter 2 Compound Interest Chapter Test

Question 1.
Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At end of first year it amounts to ₹ 6,720. Calculate:
(i) the rate of interest
(ii) the amount at the end of the second year.
Solution:
Nikita invests (P) = ₹ 6000
Period (T) = two years
and after 1 year she got back ₹ 6720
∴ Interest for 1 year = ₹ (6720 – 6000) = ₹ 720
(i) ∴ Rate of interest compounded annually
= $$\frac{720 \times 100}{6000}$$ = 12%

(ii) Interest at end of second year
$$\frac{6720 \times 12 \times 1}{100}$$
∴ Amount at end of second year = ₹ (6720 + 806.40) = ₹ 7,526.40

Question 2.
Rohit borrows ₹ 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit’s profit in the transaction at the end of two years.
Solution:
Rohit borrows ₹ 86,000 for 2 years at simple interest at 5% per annum
∴ Interest = $$\frac{86,000 \times 2 \times 5}{100}$$ = ₹ 8600
Rohit gets ₹ 8600 in 2 years as simple interest Rohit lends money to Akshay at 5% p.a. compounded annually for 2 years
Amount he gets back after 2 years is
A = P $$\left(1+\frac{r}{100}\right)^n=86,000\left(1+\frac{5}{100}\right)^2$$
= 86,000 $$\left(1+\frac{1}{20}\right)^2$$ [t = 2, r = 5%]
A = 86,000 x $$\frac { 21 }{ 20 }$$ x $$\frac { 21 }{ 20 }$$
A = ₹ 94,815
C.I. = A – P
= 94,815 – 86,000
C.I. = ₹ 8815
Rohit’s profit in the transaction at the end of two years = ₹(8815 – 8600) = ₹ 215 Question 3.
Mr. Kumar borrowed ₹ 15,000 for two years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays ₹ 6,200 at the end of the first year, find the outstanding amount at the end of the second year.
Solution:
For 1st year :
S.I = $$\frac{15000 \times 8 \times 1}{100}$$ = ₹ 1200
Amount = ₹ (15000 + 1200) = ₹ 16200
Remaining amount after repayment
= ₹ (16200 – 6200) = ₹ 10000
For 2nd year :
P = ₹ 10000
S.I.= $$\frac{10000 \times 10 \times 1}{100}$$ = ₹ 1000
Amount at the end of 2nd year = ₹ 10000 + ₹ 1000 = ₹ 11000

Question 4.
In what period of time will ₹ 12,000 yield ₹ 3,972 as compound interest at 10% per annum, if compounded on an yearly basis?
Solution:
P = ₹ 12000
A = ₹ 12000 + ₹ 3972 = ₹ 15972
R = 10% p.a., n = ? Comparing,
n = 3
∴ Time = 3 years

Question 5.
On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹ 25 if the rate of interest charged for both is 5%?
Solution:
Difference between C.I. and S.l. = ₹25
Let principal (P) = ₹ 100
Rate (R) = 5% p.a. and period (n) = 2 years
Then S.I. = $$\frac{\mathrm{PRT}}{100}=\frac{100 \times 5 \times 2}{100}$$ = ₹ 10
and when interest is compounded annually
A = P$$\left(1+\frac{\mathrm{R}}{100}\right)^n=100\left(1+\frac{5}{100}\right)^2$$
= ₹100 x $$\frac{21}{20} \times \frac{21}{20}=₹ \frac{441}{4}$$
and C.I. = A – P = ₹ $$\frac { 441 }{ 4 }$$ – 100
= ₹$$\frac{441-400}{4}=₹ \frac{41}{4}$$
Now difference between C.I. and S.l.
= ₹$$\frac { 41 }{ 4 }$$ – 10
= ₹ $$\frac { 41-40 }{ 4 }$$ = ₹$$\frac { 1 }{ 4 }$$
If difference is ₹$$\frac { 1 }{ 4 }$$, then principal = ₹ 100
and if difference is ₹ 1, then principal
= ₹$$\frac { 100×4 }{ 1 }$$
and if difference is ₹ 25, the principal
= $$\frac{100 \times 4 \times 25}{1}$$ = ₹ 10000

Multiple Choice Questions (MCQs)

Question 6.
A sum of ₹ 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become
(a) ₹ 48,000
(b) ₹ 96,000
(c) ₹ 1,90,000
(d) ₹ 1,92,000
Solution:
(d) ₹ 1,92,000
Sum (P) = ₹ 12000
∴ A = ₹ 12000 x 2 = ₹ 24000
Period (n) = 5 years
∴ A = P$$\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n$$
⇒ $$\frac{24000}{12000}=\left(1+\frac{\mathrm{R}}{100}\right)^5 \Rightarrow 2=\left(1+\frac{\mathrm{R}}{100}\right)^5$$
∴ $$\left(1+\frac{R}{100}\right)^{20}$$ = 24 = 16
Now, amount after 20 years
= P$$\left(1+\frac{\mathrm{R}}{100}\right)^{20}=12000\left[\left(1+\frac{\mathrm{R}}{100}\right)^5\right]^4$$
= ₹ 12000 x (2)4 = ₹ 12000 x 16
= ₹ 1,92,000 Question 7.
The difference between C.I. (compound interest) and S.I. (simple interest) on a sum of ₹ 4,000 for 2 years at 5% p.a. payable yearly is
(a) ₹ 20
(b) ₹ 10
(c) ₹ 50
(d) ₹ 60
Solution:
(b) ₹ 10
Sum (P) = ₹ 4000
Rate (R) = 5% p.a.
Period (n) = 2 years
∴ S.I = $$\frac{\mathrm{PRT}}{100}=\frac{4000 \times 5 \times 2}{100}$$ = ₹ 400
and when interest is compounded annually,
then A = $$\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow 4000\left(1+\frac{5}{100}\right)^2$$
= ₹ 4000 x $$\frac { 21 }{ 20 }$$ x $$\frac { 21 }{ 20 }$$ = ₹ 4410
and C.I. = A – P = ₹ 4410 – 4000 = ₹ 410
and difference between C.I. and S.I.
= ₹ 410 – 400 = ₹ 10

Question 8.
If the difference between simple interest and compound interest on a certain sum of money for 3 years at 10% per annum is ₹ 31, the sum is
(a) ₹ 500
(b) ₹ 50
(c) ₹ 1000
(d) ₹ 1250
Solution:
(c) ₹ 1000
Difference in C.I. and S.I. = ₹ 31
Let principal (P) = ₹ 100
Rate (R) = 10% p.a.
Period (n) = 3 years
∴ S.I. = $$\frac{\text { PRT }}{100}=₹ \frac{100 \times 10 \times 3}{100}$$ = ₹ 30
When interest is compounded annually then
A = P$$\left(1+\frac{\mathrm{R}}{100}\right)^n$$
= 100$$\left(1+\frac{10}{100}\right)^3=₹ 100 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$$
= ₹ $$\frac { 1331 }{ 10 }$$
∴ C.I. = A – P = ₹ $$\frac{1331-100}{10}=₹ \frac{331}{10}$$
and difference between C.I. and S.I.
= ₹$$\frac{1331-100}{10}=₹ \frac{331}{10}$$
If difference is ₹$$\frac { 31 }{ 10 }$$ then principal = ₹ 100
If difference is ₹ 1 then principal = ₹$$\frac { 100×10 }{ 31 }$$
and if difference is ₹31, then principal
= $$\frac { 100×10×31 }{ 31 }$$ = ₹1000 Question 9.
On what sum of money will compound interest for 2 years at 5% per annum amount to ₹164?
(a) ₹ 1600
(b) ₹ 1500
(c) ₹ 1400
(d) ₹ 1700
Solution:
(a) ₹ 1600
C.I. = ₹ 164
Rate = 5% p.a.
and period (n) = 2 years
Let principal (P) = ₹ 100
Then A = P$$\left(1+\frac{\mathrm{R}}{100}\right)^n$$
= ₹$$\left(1+\frac{5}{100}\right)^2=₹ 100 \times \frac{21}{20} \times \frac{21}{20}$$
= ₹ $$\frac { 441 }{ 4 }$$
and C.I. = A – P
= ₹ $$\frac { 441 }{ 4 }$$ – 100 = ₹ $$\frac { 41 }{ 4 }$$
If C.I. is ₹ $$\frac { 41 }{ 4 }$$then principal = ₹ 100
If C.I. is ₹ 1 then principal = ₹ $$\frac { 100×4 }{ 41 }$$
and if C.L is ₹ 164, then principal
= ₹ $$\frac { 100×4×164 }{ 41 }$$
= ₹ 1600

Question 10.
A sum of money becomes eight times in 3 years. If the rate is compounded annually, in how much time will the same amount at the same compound rate becomes sixteen times?
(a) 6 years
(b) 4 years
(c) 8 years
(d) 5 years
Solution:
(b) 4 years
Let principal (P) = ₹ 100
Then amount (A) = ₹ 100 x 8 = ₹ 800
Period (n) = 3 years
∴ A = P$$\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n$$
⇒ $$\frac{800}{100}=\left(1+\frac{\mathrm{R}}{100}\right)^3$$
⇒ $$\left(1+\frac{\mathrm{R}}{100}\right)^3=8=2^3 \Rightarrow\left(1+\frac{\mathrm{R}}{100}\right)$$ = 2 … (i)
In second case,
P = ₹ 100
Then A = ₹ 100 x 16 = 1600
∴ $$\frac { A }{ P }$$ = $$\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{1600}{100}=\left(1+\frac{\mathrm{R}}{100}\right)^n$$
⇒ $$\left(1+\frac{\mathrm{R}}{100}\right)^n$$ = 16
⇒ (2)n = 24                 [From (i)]
Comparing,
n = 4
∴ Period = 4 years