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Page 1
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.2 PAGE NO: 14.14
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:
By using the formula,
A = P (1 + R/100)
n
Let us solve
(i) Given, P = Rs 3000, rate = 5%, time = 2years
A = P (1 + R/100)
n
= 3000 (1 + 5/100)
2
= 3000 (105/100)
2
= Rs 3307.5
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
= 3000 (1 + 18/100)
2
= 3000 (118/100)
2
= Rs 4177.2
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
= 5000 (1 + 10/100)
2
= 5000 (110/100)
2
= Rs 6050
Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050
(iv) Given, P = Rs 2000, rate = 4%, time = 3years
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.2 PAGE NO: 14.14
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:
By using the formula,
A = P (1 + R/100)
n
Let us solve
(i) Given, P = Rs 3000, rate = 5%, time = 2years
A = P (1 + R/100)
n
= 3000 (1 + 5/100)
2
= 3000 (105/100)
2
= Rs 3307.5
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
= 3000 (1 + 18/100)
2
= 3000 (118/100)
2
= Rs 4177.2
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
= 5000 (1 + 10/100)
2
= 5000 (110/100)
2
= Rs 6050
Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050
(iv) Given, P = Rs 2000, rate = 4%, time = 3years
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
A = P (1 + R/100)
n
= 2000 (1 + 4/100)
3
= 2000 (104/100)
3
= Rs 2249.72
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
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
Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481
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 details are,
Principal (p) = Rs 2400
Rate (r) = 20% per annum
Time (t) = 3 years
By using the formula,
A = P (1 + R/100)
n
= 2400 (1 + 20/100)
3
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.2 PAGE NO: 14.14
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:
By using the formula,
A = P (1 + R/100)
n
Let us solve
(i) Given, P = Rs 3000, rate = 5%, time = 2years
A = P (1 + R/100)
n
= 3000 (1 + 5/100)
2
= 3000 (105/100)
2
= Rs 3307.5
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
= 3000 (1 + 18/100)
2
= 3000 (118/100)
2
= Rs 4177.2
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
= 5000 (1 + 10/100)
2
= 5000 (110/100)
2
= Rs 6050
Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050
(iv) Given, P = Rs 2000, rate = 4%, time = 3years
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
A = P (1 + R/100)
n
= 2000 (1 + 4/100)
3
= 2000 (104/100)
3
= Rs 2249.72
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
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
Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481
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 details are,
Principal (p) = Rs 2400
Rate (r) = 20% per annum
Time (t) = 3 years
By using the formula,
A = P (1 + R/100)
n
= 2400 (1 + 20/100)
3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= 2400 (120/100)
3
= Rs 4147.2
? Amount is Rs 4147.2
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 details are,
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
= 16000 (1 + 12.5/100)
3
= 16000 (112.5/100)
3
= Rs 22781.25
? Amount is Rs 22781.25
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:
Given details are,
Principal (p) = Rs 1000
Rate (r) = 10 % per annum
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
= 1000 (1 + 10/100)
2
= 1000 (110/100)
2
= Rs 1210
? Amount is Rs 1210
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
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.2 PAGE NO: 14.14
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:
By using the formula,
A = P (1 + R/100)
n
Let us solve
(i) Given, P = Rs 3000, rate = 5%, time = 2years
A = P (1 + R/100)
n
= 3000 (1 + 5/100)
2
= 3000 (105/100)
2
= Rs 3307.5
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
= 3000 (1 + 18/100)
2
= 3000 (118/100)
2
= Rs 4177.2
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
= 5000 (1 + 10/100)
2
= 5000 (110/100)
2
= Rs 6050
Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050
(iv) Given, P = Rs 2000, rate = 4%, time = 3years
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
A = P (1 + R/100)
n
= 2000 (1 + 4/100)
3
= 2000 (104/100)
3
= Rs 2249.72
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
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
Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481
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 details are,
Principal (p) = Rs 2400
Rate (r) = 20% per annum
Time (t) = 3 years
By using the formula,
A = P (1 + R/100)
n
= 2400 (1 + 20/100)
3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= 2400 (120/100)
3
= Rs 4147.2
? Amount is Rs 4147.2
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 details are,
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
= 16000 (1 + 12.5/100)
3
= 16000 (112.5/100)
3
= Rs 22781.25
? Amount is Rs 22781.25
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:
Given details are,
Principal (p) = Rs 1000
Rate (r) = 10 % per annum
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
= 1000 (1 + 10/100)
2
= 1000 (110/100)
2
= Rs 1210
? Amount is Rs 1210
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
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
By using the formula,
A = P (1 + R/100)
n
= 50000 (1 + 10/100)
2
= 50000 (110/100)
2
= Rs 60500
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
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
= 16000 (1 + 17.5/100)
2
= 16000 (117.5/100)
2
= Rs 22090
CI = Rs 22090 – 16000 = Rs 6090
? Amit total gain is = Rs 6090 – 5600 = Rs 490
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
= 4096 (1 + 12.5/2×100)
3
= 4096 (212.5/200)
3
= Rs 4913
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.2 PAGE NO: 14.14
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:
By using the formula,
A = P (1 + R/100)
n
Let us solve
(i) Given, P = Rs 3000, rate = 5%, time = 2years
A = P (1 + R/100)
n
= 3000 (1 + 5/100)
2
= 3000 (105/100)
2
= Rs 3307.5
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
= 3000 (1 + 18/100)
2
= 3000 (118/100)
2
= Rs 4177.2
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
= 5000 (1 + 10/100)
2
= 5000 (110/100)
2
= Rs 6050
Compound interest (CI) = A-P = Rs 6050 – 5000 = Rs 1050
(iv) Given, P = Rs 2000, rate = 4%, time = 3years
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
A = P (1 + R/100)
n
= 2000 (1 + 4/100)
3
= 2000 (104/100)
3
= Rs 2249.72
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
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
Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481
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 details are,
Principal (p) = Rs 2400
Rate (r) = 20% per annum
Time (t) = 3 years
By using the formula,
A = P (1 + R/100)
n
= 2400 (1 + 20/100)
3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= 2400 (120/100)
3
= Rs 4147.2
? Amount is Rs 4147.2
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 details are,
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
= 16000 (1 + 12.5/100)
3
= 16000 (112.5/100)
3
= Rs 22781.25
? Amount is Rs 22781.25
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:
Given details are,
Principal (p) = Rs 1000
Rate (r) = 10 % per annum
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
= 1000 (1 + 10/100)
2
= 1000 (110/100)
2
= Rs 1210
? Amount is Rs 1210
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
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
By using the formula,
A = P (1 + R/100)
n
= 50000 (1 + 10/100)
2
= 50000 (110/100)
2
= Rs 60500
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
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
= 16000 (1 + 17.5/100)
2
= 16000 (117.5/100)
2
= Rs 22090
CI = Rs 22090 – 16000 = Rs 6090
? Amit total gain is = Rs 6090 – 5600 = Rs 490
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
= 4096 (1 + 12.5/2×100)
3
= 4096 (212.5/200)
3
= Rs 4913
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
? Amount is Rs 4913
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,
A = P (1 + R/100)
n
= 8000 (1 + 5/100)
3
= 8000 (105/100)
3
= Rs 9261
? CI = Rs 9261 – 8000 = Rs 1261
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
= 57600 (1 + 12.5/2×100)
3
= 57600 (212.5/200)
3
= Rs 69089.06
? Amount is Rs 69089.06
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
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