# RD Sharma Solutions Class 8 Compound Interest Exercise 14.1

## RD Sharma Solutions Class 8 Chapter 14 Exercise 14.1

#### Exercise 14.1

1. Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.

Solution:

According to the question,

Principal amount for the first year = Rs 3,000

Then, the interest for the first year=Rs3,000 x 5 x $\frac{1}{100}$ = Rs 150

i.e., The amount at the end of the first year = Rs 3,000 + Rs 150 = Rs 3,150

Principle Interest for the second year=Rs 3,150 x 5 x $\frac{1}{100}$ = Rs 157.50

i.e., The amount at the end of the second year = Rs 3,150+Rs 157.50= Rs 3307.50

Therefore,

The compound interest=The amount at the end of the second year – Principal amount for the first year

=Rs 3,307.50 – 3,000

= Rs 307.50

2. What will be the compound interest on Rs 4000 in two years when the rate of interest is 5% per annum?

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question,

P = Rs 4,000

R = 5% p.a

n = 2 years

Substituting in the equation,

A = $4000(1+\frac{5}{100})^{2}$

= $4000(\frac{105}{100})^{2}$

= Rs 4410

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

=A – P

= Rs 4,410 – Rs 4,000

= Rs 410

3. Rohit deposited Rs 8000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years?

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = Rs 8,000

R = 15% p.a.

n = 3 years.

Substituting in the equation,

A = $8000(1+\frac{15}{100})^{3}$

A = $8000(\frac{115}{100})^{3}$

A = Rs. 12,167

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

=A – P

= Rs 12,167 – Rs 8,000

= Rs 4,167

4. Find the compound interest on Rs 1000 at the rate of 8% per annum for $1\frac{1}{2}$ years when interest is compounded half-yearly.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = Rs 1,000

R = 8% p.a.

n = 1.5 years

Substituting in the equation,

A = $P(1+\frac{R}{200})^{2n}$

A = $1000(1+\frac{8}{200})^{3}$

A = $1000(\frac{208}{200})^{3}$

A = Rs 1,124.86

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 1,124.86 – Rs 1,000

= Rs 124.86

5. Find the compound interest on Rs 160000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = Rs 160,000

R = 20 % p. a.

n = 1 year

Substituting in the equation,

A = $P(1+\frac{R}{400})^{4n}$

A = $160000(1+\frac{20}{400})^{4}$

A = $160000(1.05)^{4}$

A = Rs 19,4481

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 19,448.1 – Rs 16,000

= Rs 3,4481

6. Swati took a loan of Rs 16000 against her insurance policy at the rate of $12\frac{1}{2}$% per annum. Calculate the total compound interest payable by Swati after 3 year.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = 16,000

R = 12.5% p.a.

n = 3 years

Substituting in the equation,

A = $P(1+\frac{R}{100})^{n}$

A = $16000(1+\frac{12.5}{100})^{3}$

A = $16000(\frac{112.5}{100})^{3}$

A = Rs 22,781.25

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 22,781.25 – Rs 16,000

= Rs 6,781.25

7. Roma borrowed Rs 64000 from a bank for $1\frac{1}{2}$ years at the rate of 10% per annum. Compute the total compound interest payable by Roma after $1\frac{1}{2}$ years, if the interest is compounded half-yearly.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = 64,000

R = 10% p.a.

n = 1.5 years

Amount after n years can be found out by substituting these values in the above equation:

A = $P(1+\frac{R}{200})^{2n}$

A = $64000(1+\frac{10}{200})^{3}$

A = $64000(\frac{210}{200})^{3}$

A = Rs 74,088

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 74,088 – Rs 64,000

= Rs 10,088

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:

Simple Interest for the amount Mewa Lal borrowed= $\frac{PRT}{100}$

= $\frac{20000\times 18\times 21}{100}$

= Rs 7,200

Hence, Mewa Lal has to pay Rs 7,200 as interest after borrowing.

Compound Interest for the amount Mewa Lal lend= $20000(1+\frac{18}{100})^{2}$ – 20,000

= $20000(1.18)^{2}$ – 20,000

= 27,848- 20,000

= Rs 7,848

i.e., Mewa Lal gained Rs 7,848 as interest after lending the money.

Therefore, his total gain= Compound Interest for the amount he lend-Simple Interest for the amount he borrowed

= Rs 7,848 – Rs 7,200

= Rs 648

9. Find the compound interest on Rs 8000 for 9 months at 20% per annum compounded quarterly.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = Rs 8,000

T = 9 months = 3 quarters

R = 20% per annum = 5% per quarter

Substituting in the equation,

A = $8000(1+\frac{5}{100})^{3}$

A = $8000(1.05)^{3}$

A = 9,261

Thus, the required amount = Rs 9,261.

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

=A – P

= Rs 9,261 – Rs 8,000

= Rs 1,261

10. Find the compound interest at the rate of 10% per annum for two years on that principle which in two years at the rate of 10% per annum gives Rs 200 as simple interest.

Solution:

SI = $\frac{PRT}{100}$

:. P = $\frac{SI\times 100}{RT}$

P = $\frac{200\times 100}{10\times 2}$

P = Rs 1,000

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Substituting the values in the equation,

A = $P(1+\frac{R}{100})^{n}$

A = $1000(1+\frac{10}{100})^{2}$

A = $1000(1.1)^{2}$

A = Rs 1,210

Thus, the required amount =Rs 1,210.

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 1,210 – Rs 1,000

= Rs 210

11. Find the compound interest on Rs 64000 for 1 year at the rate of 10% per compounded quarterly.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Substituting the values in the equation as per the question,

A = $P(1+\frac{R}{400})^{4n}$

A = $64000(1+\frac{10}{400})^{4\times 1}$

A = $64000(1.025)^{4}$

A = 70,644.03

Thus, the required amount =Rs 70,644.03.

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 70,644.025 – Rs 64,000

= Rs 6,644.03

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:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Given in the question:

P = Rs 7,500

R = 12% p.a = 3% quarterly

T = 9 months = 3 quarters

Substituting the values in the equation,

A = $P(1+\frac{R}{100})^{n}$

A = $7500(1+\frac{3}{100})^{3}$

A = $7500(1.03)^{3}$

A = 8,195.45

Thus, the required amount = Rs 8,195.45.

13. Anil borrowed a sum of Rs 9600 to install a hand pump in his dairy. If the rate of 1 interest is $5\frac{1}{2}$% per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Substituting the values in the equation as per the question,

A = $P(1+\frac{R}{100})^{n}$

A = $9600(1+\frac{5.5}{100})^{3}$

A = $9600(1.055)^{3}$

A = Rs 11,272.72

Thus, the required amount =Rs 11,272.72.

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

=A – P

= Rs 11,272.72 – Rs 9,600

= Rs 1,672.72

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:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Substituting the values in the equation as per the question,

A = $P(1+\frac{R}{100})^{n}$

A = $12000(1+\frac{5}{100})^{3}$

A = $12000(1.05)^{3}$

A = Rs 13,891.50

Thus, the required amount =Rs 13.891.50.

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

= A – P

= Rs 13.891.50 – Rs 12,000

= Rs 1,891.50

15. Daljit received a sum of Rs. 40000 as a loan from a finance company. If the rate of interest is 7% per annum corn compounded annually, calculate the compound interest that Daljit pays after 2 years.

Solution:

We know that,

Amount A at the end of n years at the rate of R% per annum is given by the formula,

A = $P(1+\frac{R}{100})^{n}$

Substituting the values in the equation as per the question,

A = $P(1+\frac{R}{100})^{n}$

A = $40000(1+\frac{7}{100})^{2}$

A = $40000(1.07)^{2}$

A = Rs 45,796

Thus, the required amount =Rs 45,796.

Therefore, the compound interest, CI = Amount A at the end of n years- Principal amount P

=A – P

= Rs 45.796 – Rs 40,000

= Rs 5,796