# Class 8 RD Sharma Solutions – Chapter 14 Compound Interest – Exercise 14.2 | Set 1

• Last Updated : 08 Apr, 2021

### Question 1. Compute the amount and the compound interest in each of the following by using the formulae when:(i) Principal = Rs 3000, Rate = 5%, Time = 2 years(ii) Principal = Rs 3000, Rate = 18%, Time = 2 years(iii) Principal = Rs 5000, Rate = 10 paise per rupee per annum, Time = 2 years(iv) Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years(v) Principal = Rs 12800, Rate = 7 ½ %, Time = 3 years(vi) Principal = Rs 10000, Rate = 20% per annum compounded half-yearly, Time = 2 years(vii) Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half yearly, Time = 2 years.

Solution:

We have,

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A = P (1 + R/100)n

Let us solve

(i) Given, P = Rs 3000, rate = 5%, time = 2years
A = P (1 + R/100)n

Substituting the values we have,
= 3000 (1 + 5/100)2
= 3000 (105/100)2
= Rs 3307.5

Solving for Compound Interest, we get
Compound interest (CI) = A-P = Rs 3307.5 – 3000 = Rs 307.5

(ii) Given, P = Rs 3000, rate = 18%, time = 2years
A = P (1 + R/100)n

Substituting the values we have,
= 3000 (1 + 18/100)2
= 3000 (118/100)2
= Rs 4177.2

Solving for Compound Interest, we get
Compound interest (CI) = A-P = Rs 4177.2 – 3000 = Rs 1177.2

(iii) Given, P = Rs 5000, rate = 10%, time = 2years
A = P (1 + R/100)n

Substituting the values we have,
= 5000 (1 + 10/100)2
= 5000 (110/100)2
= Rs 6050

Solving for Compound Interest, we get
Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050

(iv) Given, P = Rs 2000, rate = 4%, time = 3years
A = P (1 + R/100)n

Substituting the values we have,
= 2000 (1 + 4/100)3
= 2000 (104/100)3
= Rs 2249.72

Solving for Compound Interest, we get
Compound interest (CI) = A-P = Rs 2249.72 – 2000 = Rs 249.72

(v) Given, P = Rs 12800, rate = 7 ½ % = 15/2% = 7.5%, time = 3years
A = P (1 + R/100)n
= 12800 (1 + 7.5/100)3
= 12800 (107.5/100)3
= Rs 15901.4
Compound interest (CI) = A-P = Rs 15901.4 – 12800 = Rs 3101.4

(vi) Given, P = Rs 10000, rate = 20 % = 20/2 = 10% (quarterly), time = 2years = 2 × 2 = 4years
A = P (1 + R/100)n
= 10000 (1 + 10/100)4
= 10000 (110/100)4
= Rs 14641

Solving for Compound Interest, we get
Compound interest (CI) = A-P = Rs 14641 – 10000 = Rs 4641

(vii) Given, P = Rs 160000, rate = 10% = 10/2% = 5% (half-yearly), time = 2years = 2×2 = 4 quarters
A = P (1 + R/100)n
= 160000 (1 + 5/100)4
= 160000 (105/100)4
= Rs 194481

Solving for Compound Interest, we get
Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481

### Question 2. Find the amount of Rs. 2400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.

Solution:

Given is the following set of values,
Principal (p) = Rs 2400
Rate (r) = 20% per annum
Time (t) = 3 years
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 2400 (1 + 20/100)3
= 2400 (120/100)3
= Rs 4147.2
∴ Amount is Rs 4147.2

### Question 3. Rahman lent Rs. 16000 to Rasheed at the rate of 12 ½ % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.

Solution:

Given :
Principal (p) = Rs 16000
Rate (r) = 12 ½ % per annum = 12.5%
Time (t) = 3 years
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 16000 (1 + 12.5/100)3
= 16000 (112.5/100)3
= Rs 22781.25
∴ Amount is Rs 22781.25

### Question 4. Meera borrowed a sum of Rs. 1000 from Sita for two years. If the rate of interest is 10% compounded annually, find the amount that Meera has to pay back.

Solution:

We have,
Principal (p) = Rs 1000
Rate (r) = 10 % per annum
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 1000 (1 + 10/100)2
= 1000 (110/100)2
= Rs 1210
∴ Amount is Rs 1210

### Question 5. Find the difference between the compound interest and simple interest. On a sum of Rs. 50,000 at 10% per annum for 2 years.

Solution:

Given details are,
Principal (p) = Rs 50000
Rate (r) = 10 % per annum
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 50000 (1 + 10/100)2
= 50000 (110/100)2
= Rs 60500

Calculating for Compound Interest, we have
CI = Rs 60500 – 50000 = Rs 10500
We know that SI = (PTR)/100 = (50000 × 10 × 2)/100 = Rs 10000
∴ Difference amount between CI and SI = 10500 – 10000 = Rs 500

### Question 6. Amit borrowed Rs. 16000 at 17 ½ % per annum simple interest. On the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years?

Solution:

Given details are,
Principal (p) = Rs 16000
Rate (r) = 17 ½ % per annum = 35/2% or 17.5%
Time (t) = 2 years
Interest paid by Amit = (PTR)/100 = (16000×17.5×2)/100 = Rs 5600
Amount gained by Amit:
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 16000 (1 + 17.5/100)2
= 16000 (117.5/100)2
= Rs 22090

Calculating for Compound Interest, we have
CI = Rs 22090 – 16000 = Rs 6090
∴ Amit’s total gain is = Rs 6090 – 5600 = Rs 490

### Question 7. Find the amount of Rs. 4096 for 18 months at 12 ½ % per annum, the interest being compounded semi-annually.

Solution:

Given details are,
Principal (p) = Rs 4096
Rate (r) = 12 ½ % per annum = 25/4% or 12.5/2%
Time (t) = 18 months = (18/12) × 2 = 3 half years
By using the formula,
A = P (1 + R/100)n

Substituting the given values we have,
= 4096 (1 + 12.5/2×100)3
= 4096 (212.5/200)3
= Rs 4913
∴ Amount is Rs 4913

### Question 8. Find the amount and the compound interest on Rs. 8000 for 1 ½ years at 10% per annum, compounded half-yearly.

Solution:

Given details are,
Principal (p) = Rs 8000
Rate (r) = 10 % per annum = 10/2% = 5% (half yearly)
Time (t) = 1 ½ years = (3/2) × 2 = 3 half years
By using the formula,

Substituting the values we have,
A = P (1 + R/100)n
= 8000 (1 + 5/100)3
= 8000 (105/100)3
= Rs 9261

Calculating for Compound Interest, we have
∴ CI = Rs 9261 – 8000 = Rs 1261

### Question 9. Kamal borrowed Rs. 57600 from LIC against her policy at 12 ½ % per annum to build a house. Find the amount that she pays to the LIC after 1 ½ years if the interest is calculated half-yearly.

Solution:

Given details are,
Principal (p) = Rs 57600
Rate (r) = 12 ½ % per annum = 25/2×2% = 25/4% = 12.5/2% (half yearly)
Time (t) = 1 ½ years = (3/2) × 2 = 3 half years
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 57600 (1 + 12.5/2×100)3
= 57600 (212.5/200)3
= Rs 69089.06
∴ Amount is Rs 69089.06

### Question 10. Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs. 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.

Solution:

Given details are,
Principal (p) = Rs 64000
Rate (r) = 5 % per annum = 5/2% (half yearly)
Time (t) = 1 ½ years = (3/2) × 2 = 3 half years
By using the formula,
A = P (1 + R/100)n

Substituting the values we have,
= 64000 (1 + 5/2×100)3
= 64000 (205/200)3
= Rs 68921

Calculating for Compound Interest, we have
∴ CI = Rs 68921 – 64000 = Rs 4921

### Chapter 14 Compound Interest – Exercise 14.2 | Set 2

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