Class 8 RD Sharma Solutions - Chapter 14 Compound Interest - Exercise 14.2 | Set 2

Chapter 14 of RD Sharma's Class 8 textbook focuses on Compound Interest, a fundamental concept in financial mathematics. Exercise 14.2 | Set 2 builds upon the basic principles of compound interest introduced earlier, presenting more complex scenarios and applications. This set aims to enhance students' understanding of compound interest calculations, especially in situations involving different compounding periods, comparison of investment options, and real-world financial decision-making.

Question 11. Rakesh lent out Rs. 10000 for 2 years at 20% per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly?

Solution:

We have,

Principal (p) = Rs 10000

Rate (r) = 20% per annum

Time (t) = 2years

By using the formula,

A = P (1 + R/100)ⁿ

= 10000 (1 + 20/100)²

= 10000 (120/100)² = Rs 14400

When the interest is compounded half-yearly, 

New values are 

Rate of interest becomes= 20/2% = 10%

Time = 2×2 years = 4years

By using the formula, 

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 10000 (1 + 10/100)⁴

= 10000 (110/100)⁴ = Rs 14641

∴ Rakesh could earn Rs (14641 – 14400) = Rs 241 more

Question 12. Romesh borrowed a sum of Rs. 245760 at 12.5% per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after 2 years.

Solution:

Given: 

Principal (p) = Rs 245760

Rate (r) = 12.5% per annum

Time (t) = 2years

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 245760 (1 + 12.5/100)²

= 245760 (112.5/100)²

= Rs 311040

Now, When compounded semi-annually,

Rate = 12.5/2% = 6.25%

Time = 2×2 years = 4years

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 245760 (1 + 6.25/100)⁴

= 245760 (106.25/100)⁴ = Rs 313203.75

∴ Romesh gain is Rs (313203.75 – 311040) = Rs 2163.75

Question 13. Find the amount that David would receive if he invests Rs. 8192 for 18 months at 12 ½ % per annum, the interest being compounded half-yearly.

Solution:

Given, 

Principal (p) = Rs 8192

Rate (r) = 12 ½ % per annum = 25/2×2 = 25/4% = 12.5/2% (half-yearly)

Time (t) = 18 months = 18/12 = 1 ½ years = (3/2) × 2 = 3years

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 8192 (1 + 12.5/2×100)³

= 8192 (212.5/200)³

= Rs 9826

∴ Amount is Rs 9826

Question 14. Find the compound interest on Rs. 15625 for 9 months, at 16% per annum, compounded quarterly.

Solution:

Given,

Principal (p) = Rs 15625

Rate (r) = 16% per annum = 16/4 = 4% (quarterly)

Time (t) = 9 months = 9/12 × 4 = 3quarters of a year

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 15625 (1 + 4/100)³

= 15625 (104/100)³

= Rs 17576

∴ CI = Rs 17576 – 15625 = Rs 1951

Question 15. Rekha deposited Rs. 16000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year

Solution:

Given is,

Principal (p) = Rs 16000

Rate (r) = 20% per annum = 20/4 = 5% (quarterly)

Time (t) = 1 year = 4 quarters of a year

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 16000 (1 + 5/100)⁴

= 16000 (105/100)⁴= Rs 19448.1

∴ CI = Rs 19448.1 – 16000 = Rs 3448.1

Question 16. Find the amount of Rs. 12500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.

Solution:

We have the following set of values,

Principal (p) = Rs 12500

Rate1 (r) = 15% and Rate2 = 16%

Time (t) = 2 years

By using the formula,

A = P (1 + R1/100 × 1 + R2/100) = 12500 

Substituting the values we have, 

(1 + 15/100 × 1 + 16/100) = 12500 (1.15 × 1.16)

= Rs 16675

∴ Amount after two years is Rs 16675

Question 17. Ramu borrowed Rs. 15625 from a finance company to buy scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after 2 ¼ years?

Solution:

Given details are,

Principal (p) = Rs 15625

Rate (r) = 16%

Time (t) = 2 ¼ years

By using the formula,

A = P (1 + R/100 × 1 + R/100)= 15625 

Substituting the values we have, 

(1 + 16/100)² × (1 + (16/4)/100)= 15625 

(1 + 16/100)² × (1 + 4/100)= 15625 

(1.16)² × (1.04)= Rs 21866

∴ Amount after 2 ¼ years is Rs 21866

Question 18. What will Rs. 125000 amount to at the rate of 6%, if the interest is calculated after every four months?

Solution:

Given, 

Principal (p) = Rs 125000

Rate (r) = 6% per annum

Time (t) = 1 year

Since interest is compounded after 4months, interest will be counted as 6/3 = 2% and,

Time will be 12/4 = 3quarters

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 125000 (1 + 2/100)³

= 125000 (102/100)³

= Rs 132651

∴ Amount is Rs 132651

Question 19. Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs. 12000 as simple interest.

Solution:

Given,

Simple interest (SI) = Rs 12000

Rate (r) = 5% per annum

Time (t) = 3 years

SI = (PTR)/100P 

= (SI×100)/(T×R)

Solving the equations, 

= (12000×100) / (3×5)

= 1200000/15= 80000

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 80000 (1 + 5/100)³

= 80000 (105/100)³

= Rs 92610

∴ CI = Rs 92610 – 80000 = Rs 12610

Question 20. A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs. 482 more. Find the sum.

Solution:

We have,

Rate (r) = 20% per annum = 20/2 = 10% (half yearly)

Time (t) = 2 years = 2 × 2 = 4 half years

Principal be = Rs P

P (1 + R/100)ⁿ – P (1 + R/100)ⁿ 

= 482P (1 + 10/100)⁴ – P (1 + 20/100)² 

= 482P (110/100)⁴ – P (120/100)²

 = 482P (1.4641) – P (1.44) 

= 4820.0241P

 = 482P = 482/0.0241

= 20000

∴ Amount is Rs 20000

Question 21. Simple interest on a sum of money for 2 years at 6 ½ % per annum is Rs. 5200. What will be the compound interest on the sum at the same rate for the same period?

Solution:

Given is,

Rate = 6 ½ % per annum = 13/2%

Simple Interest (SI) = Rs 5200

Time (t) = 2 years

By using the formula,

SI = (PTR)/100P = (SI×100) / (T×R)

= (5200×100) / (2×13/2)

= (5200×100×2) / (2×13)

= 1040000/26

= Rs 40000

Now, P = Rs 40000R 

= 13/2% = 6.5%T = 2years

By using the formula,

A = P (1 + R/100)ⁿ

Substituting the values we have, 

= 40000 (1 + 6.5/100)²

= 40000 (106.5/100)²

= Rs 45369

∴ CI = Rs 45369 – 40000 = Rs 5369

Question 22. What will be the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs. 1200 as simple interest.

Solution:

Rate = 5 % per annum

Simple Interest (SI) = Rs 1200

Time (t) = 3 years

By using the formula,

SI = (PTR)/100P = (SI×100) / (T×R)

= (1200×100) / (3×5)

= 120000/15

= Rs 8000

Now, P = Rs 8000R 

= 5%T = 3years

By using the formula,

A = P (1 + R/100)ⁿ

= 8000 (1 + 5/100)³

= 8000 (105/100)³

= Rs 9261

∴ CI = Rs 9261 – 8000 = Rs 1261

Summary

Exercise 14.2 | Set 2 of Chapter 14 delves deeper into compound interest calculations, presenting a variety of scenarios that challenge students to apply their understanding in more complex situations. This set covers topics such as calculating compound interest for fractional time periods, determining interest rates and time periods given other parameters, comparing different compounding frequencies, and analyzing the growth of investments over time. The problems are designed to enhance students' analytical and problem-solving skills, preparing them for real-world financial calculations and decision-making. By working through these exercises, students gain a more nuanced understanding of how compound interest works in various contexts, setting a foundation for more advanced financial mathematics in future studies.