Class 8 RD Sharma Solutions - Chapter 14 Compound Interest - Exercise 14.3 | Set 2

Chapter 14 of RD Sharma's Class 8 Mathematics book focuses on the Compound Interest. This chapter introduces the concept of the interest earned on an investment or loan where the interest itself earns interest over time. It builds on the principles of simple interest by incorporating the effects of compounding in which can significantly increase the total interest earned or paid.

Compound Interest

The Compound Interest is the interest on a loan or deposit calculated based on both the initial principal and the accumulated interest from the previous periods. Unlike simple interest in which is calculated only on the principal amount, compound interest is computed on the principal plus any interest that has already been added to it. This makes compound interest a powerful tool for growing investments or increasing the debt over time.

Chapter 14 Compound Interest - Exercise 14.3 | Set 1

Question 15. Find the rate percent per annum, if Rs. 2000 amounts to Rs. 2315.25 in a year and a half, interest being compounded six monthly.

Solution:

We have,

Principal = Rs 2000

Amount = Rs 2315.25

Time = 1 ½ years = 3/2 years

Let rate be = R % per annum

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

2315.25 = 2000 (1 + \frac{R}{100}  )^3/2

(1 + \frac{R}{100}  )^3/2 = 2315.25/2000

(1 + \frac{R}{100}  )^3/2 = (1.1576)

(1 + \frac{R}{100}  ) = 1.1025

\frac{R}{100}   = 1.1025 – 1

= 0.1025 × 100

= 10.25

Therefore,

Required Rate is 10.25% per annum.

Question 16. Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.

Solution:

We have,

Time = 3 years

Let rate be = R %

Also principal be = P

So, amount becomes = 2P

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

2P = P (1 + \frac{R}{100}  )³

(1 + \frac{R}{100}  )³ = 2

(1 + \frac{R}{100}  ) = 2^1/3

1 + \frac{R}{100}   = 1.2599

\frac{R}{100}   = 1.2599-1

= 0.2599

R = 0.2599 × 100

= 25.99

Therefore,

Required Rate is 25.99% per annum.

Question 17. Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly

Solution:

We have,

Time = 2 years = 2×2 = 4 half years

Let rate = R % per annum = R/2% half years

Let principal be = P

So, Amount becomes = 4P

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

4P = P (1 + \frac{R}{100 \times 2}  )⁴

(1 + \frac{R}{200}  )⁴ = 4

(1 + \frac{R}{200}  ) = 4^1/4

1 + \frac{R}{200}   = 1.4142

\frac{R}{200}   = 1.4142-1

= 0.4142

R = 0.4142 × 200

= 82.84%

Therefore,

Required Rate is 82.84% per annum.

Question 18. A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.

Solution:

We have,

Amount = Rs 5832

Time = 2 years

Rate = 8%

Let principal be = P

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

5832 = P (1 + \frac{8}{100}  )²

5832 = P (1.1664)

P = 5832/1.1664

= 5000

Therefore,

Required sum is Rs 5000.

Question 19. The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs. 360. Find the sum.

Solution:

We have,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + \frac{R}{100}  )ⁿ – 1] – (PTR)/100 = 360

Substituting the values, we have

P [(1 + \frac{7.5}{100}  )² – 1] – (P(2)(7.5))/100 = 360

P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

Therefore,

The sum is Rs 64000.

Question 20. The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years in Rs. 46. Determine the sum.

Solution:

We have,

Time = 3 years

Rate = 6\frac{2}{3}   % per annum = 20/3%

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 46

C.I – S.I = Rs 46

By using the formula,

P [(1 + \frac{R}{100}  )ⁿ – 1] – (PTR)/100 = 46

Substituting the values, we have

P [(1 + \frac{20}{3\times100}  )³ – 1] – (P(3)(20/3))/100 = 46

P[(1 + \frac{20}{300}  )³ – 1] – P/5 = 46

P[721/3375] – P/5 = 46

721/3375P – 1/5P = 46

(721P-675P)/3375 = 46

46P = 46 × 3375

46P = 46 × 3375/46

= 3375

Therefore, 

The sum is Rs 3375.

Question 21. Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n.

Solution:

We have,

Principal = Rs 12000

Amount = Rs 13230

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

13230 = 12000 (1 + \frac{5}{100}  )^T

13230 = 12000 (\frac{105}{100}  )^T

(21/20)^T = 13230/12000

(21/20)^T = 441/400

(21/20)^T = (21/20)²

So on comparing both the sides, n = T = 2

Therefore, 

Time required is 2 years.

Question 22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?

Solution:

We have,

Principal = Rs 4000

Time = 2 years

CI = Rs 410

Rate be = R% per annum

By using the formula,

CI = P [(1 + \frac{R}{100}  )ⁿ – 1]

Substituting the values, we have

410 = 4000 [(1 + \frac{R}{100}  )² – 1]

410 = 4000 (1 + \frac{R}{100}  )² – 4000

410 + 4000 = 4000 (1 + \frac{R}{100}  )²

(1 + \frac{R}{100}  )² = 4410/4000

(1 + \frac{R}{100}  )² = 441/400

(1 + \frac{R}{100}  )² = (21/20)²

By canceling the powers on both the sides,

1 + \frac{R}{100}   = 21/20

\frac{R}{100}   = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5% per annum.

Question 23. A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.

Solution:

We have,

Time = 2years

Amount = Rs 10404

Rate be = 2% per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

10404 = P [(1 + \frac{2}{100}  )²]

10404 = P [1.0404]

P = 10404/1.0404

= 10000

Therefore, 

Required sum is Rs 10000.

Question 24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum compound interest?

Solution:

We have,

Principal = Rs 1600

Amount = Rs 1852.20

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

1852.20 = 1600 (1 + \frac{5}{100}  )^T

1852.20 = 1600 (\frac{105}{100}  )^T

(21/20)^T = 1852.20/1600

(21/20)^T = 9261/8000

(21/20)^T = (21/20)³

So on comparing both the sides, n = T = 3

Therefore,

Time required is 3 years.

Question 25. At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?

Solution:

We have,

Principal = Rs 1000

Amount = Rs 1102.50

Rate = R% per annum

Let time = 2 years

By using the formula,

A = P (1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

1102.50 = 1000 (1 + \frac{R}{100}  )²

(1 + \frac{R}{100}  )² = 1102.50/1000

(1 + \frac{R}{100}  )² = 4410/4000

(1 +\frac{R}{100}  )² = (21/20)²

1 + \frac{R}{100}   = 21/20

\frac{R}{100}  = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5%.

Question 26. The compound interest on Rs. 1800 at 10% per annum for a certain period of time is Rs. 378. Find the time in years.

Solution:

We have,

Principal = Rs 1800

CI = Rs 378

Rate = 10% per annum

Let time = T years

By using the formula,

CI = P [(1 + \frac{R}{100}  )ⁿ – 1]

Substituting the values, we have

378 = 1800 [(1 + \frac{10}{100}  )^T – 1]

378 = 1800 [(\frac{110}{100}  )^T – 1]

378 = 1800 [(\frac{11}{10}  )^T – 1800

378 + 1800 = 1800 [(\frac{11}{10}  )^T

(11/10)^T = 2178/1800

(11/10)^T = 726/600

(11/10)^T = 121/100

(11/10)^T = (11/10)²

So on comparing both the sides, n = T = 2

Therefore,

Time required is 2 years.

Question 27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually

Solution:

We have,

Time = 2years

Amount = Rs 45582.25

Rate be = 6 ¾ % per annum = 27/4%

Let principal be = Rs P

By using the formula,

A = P [(1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

45582.25 = P [(1 + 27/4×100)²]

45582.25 = P (1 + \frac{27}{100}  )²

45582.25 = P (\frac{427}{400}  )²

45582.25 = P × 427/400 × 427/400

P = (45582.25 × 400 × 400) / (427×427)

P = 7293160000/182329

= 40000

Therefore,

Required sum is Rs 40000.

Question 28. Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.

Solution:

We have,

Time = 2years

Amount = Rs 453690

Rate be = 6.5 % per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + \frac{R}{100}  )ⁿ

Substituting the values, we have

453690 = P [(1 + \frac{6.5}{100}  )²]

453690 = P (\frac{106.5}{100}  )²

453690 = P × 106.5/100 × 106.5/100

P = (453690 × 100 × 100) / (106.5×106.5)

P = 4536900000/11342.25

= 400000

Therefore,

Required sum is Rs 400000.

Read More:

• Compound Interest Formula
• Compound Interest – Aptitude Questions and Answers