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\(\left(1+\frac{\mathrm{R}}{100}\right)^n \Rightarrow \frac{\mathrm{A}}{\mathrm{P}}=\left(1+\frac{\mathrm{R}}{100}\right)^n\)

⇒ \(\frac{24000}{12000}=\left(1+\frac{\mathrm{R}}{100}\right)^5 \Rightarrow 2=\left(1+\frac{\mathrm{R}}{100}\right)^5\)

∴ \(\left(1+\frac{R}{100}\right)^{20}\) = 2^{4} = 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} = 2^{4} [From (i)]

Comparing,

n = 4

∴ Period = 4 years