RD Sharma Solutions Class 8 Compound Interest Exercise 14.1

RD Sharma Solutions Class 8 Chapter 14 Exercise 14.1

RD Sharma Class 8 Solutions Chapter 14 Ex 14.1 PDF Download

Exercise 14.1

 

 Q1) Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.

 

Solution:

Principal for the first year = Rs 3,000

Interest for the first year=Rs3,000 x 5 x \(\frac{1}{100}\) = Rs 150

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

Amount at the end of the second year = Rs 3307.50

Compound interest=Rs 3,307.50 – 3,000 = Rs 307.50

 

Q2) 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 A = \(P(1+\frac{R}{100})^{n}\)

Giver P = Rs 4,000

R = 5% p.a

n = 2 years

Now,

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

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

= Rs 4410

And, CI = A – P = Rs 4,410 – Rs 4,000 = Rs 410

 

Q3) 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 = A = \(P(1+\frac{R}{100})^{n}\)

Given:

P = Rs 8,000

R = 15% p.a.

n = 3 years.

Now,

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

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

A = Rs. 12,167

And, CI = A – P = Rs 12,167 – Rs 8,000 = Rs 4,167

 

Q4) 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:

Given:

P = Rs 1,000

R = 8% p.a.

n = 1.5 years

We know that:

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

Now, CI = A – P = Rs 1,124.86 – Rs 1,000 = Rs 124.86

 

Q5) Find the compound interest on Rs 160000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.

 

Solution:

Given:

P = Rs 160,000

R = 20 % p. a.

n = 1 year

We know that:

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

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

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

A = Rs 19,4481

Now, CI = A – P = Rs 19,448.1 – Rs 16,000 = Rs 3,4481

 

Q6) 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:

Given:

P = 16,000

R = 12.5% p.a.

n = 3 years

We know that:

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

Now, CI = A – P = Rs 22,781.25 – Rs 16,000 = Rs 6,781.25

 

Q7) 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:

Given:

P = 64,000

R = 10% p.a.

n = 1.5 years

Amount after n years:

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

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

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

A = Rs 74,088

Now, CI = A – P = Rs 74,088 – Rs 64,000 = Rs 10,088

 

 Q8) 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:

SI for Mewa Lal = \(\frac{PRT}{100}\) = \(\frac{20000\times 18\times 21}{100}\) = Rs 7,200

Thus, he has to pay Rs 7,200 as interest after borrowing CI for Mewa Lal = A – P

= \(20000(1+\frac{18}{100})^{2}\) – 20,000

= \(20000(1.18)^{2}\) – 20,000

= 27,848- 20,000

= Rs 7,848

He gained Rs 7,848 as interest after lending. His gain in the whole transaction

= Rs 7,848 – Rs 7,200

= Rs 648

 

Q9) Find the compound interest on Rs 8000 for 9 months at 20% per annum compounded quarterly.

 

Solution:

P = Rs 8,000

T = 9 months = 3 quarters

R = 20% per annum = 5% per quarter

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

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

A = 9,261

The required amount is Rs 9,261.

Now,

CI = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

 

Q10) 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

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

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

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

A = Rs 1,210

Now,

CI = A – P = Rs 1,210 – Rs 1,000 = Rs 210

 

 Q11) Find the compound interest on Rs 64000 for 1 year at the rate of 10% per compounded quarterly.

 

Solution:

To calculate the interest compounded quarterly,

We have:

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 is Rs 70,644.03.

Now, CI = A – P = Rs 70,644.025 – Rs 64,000 = Rs 6,644.03

 

Q12) 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:

Given:

P = Rs 7,500

R = 12% p.a = 3% quarterly

T = 9 months = 3 quarters

We know that:

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 is Rs 8,195.45.

 

Q13) 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:

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

 Now,

CI = A – P = Rs 11,272.72 – Rs 9,600 = Rs 1,672.72

 

Q14) 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:

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 is Rs 13.891.50.

Now,

CI = A – P = Rs 13.891.50 – Rs 12,000 = Rs 1,891.50

 

Q15) 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:

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 is Rs 45,796.

Now,

CI = A – P = Rs 45.796 – Rs 40,000 = Rs 5,796