RD Sharma Solutions Class 8 Compound Interest Exercise 14.1

RD Sharma Class 8 Solutions Chapter 14 Ex 14.1 PDF Free Download

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

 

 

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