Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Compound Interest - EXERCISE 14.2
Compound Interest - EXERCISE 14.2 | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.93/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Compound Interest - EXERCISE 14.2 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for Compound Interest - EXERCISE 14.2 have been prepared according to the Class 8 exam syllabus. Information about Compound Interest - EXERCISE 14.2 covers topics
like and Compound Interest - EXERCISE 14.2 Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Compound Interest - EXERCISE 14.2.
Introduction of Compound Interest - EXERCISE 14.2 in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & Compound Interest - EXERCISE 14.2 in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: Compound Interest - EXERCISE 14.2 | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for Compound Interest - EXERCISE 14.2 | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about Compound Interest - EXERCISE 14.2
In this doc you can find the meaning of Compound Interest - EXERCISE 14.2 defined & explained in the simplest way possible. Besides explaining types of
Compound Interest - EXERCISE 14.2 theory, EduRev gives you an ample number of questions to practice Compound Interest - EXERCISE 14.2 tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
shortcuts and tricks
,Objective type Questions
,past year papers
,Compound Interest - EXERCISE 14.2 | Mathematics (Maths) Class 8
,mock tests for examination
,Exam
,practice quizzes
,Compound Interest - EXERCISE 14.2 | Mathematics (Maths) Class 8
,Viva Questions
,ppt
,Compound Interest - EXERCISE 14.2 | Mathematics (Maths) Class 8
,video lectures
,study material
,MCQs
,Semester Notes
,Previous Year Questions with Solutions
,Free
,Sample Paper
,Summary
,Important questions
,Extra Questions
;
Additional Information about Compound Interest - EXERCISE 14.2 for Class 8 Preparation
Compound Interest - EXERCISE 14.2 Free PDF Download
The Compound Interest - EXERCISE 14.2 is an invaluable resource that delves deep into the core of the Class 8 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Compound Interest - EXERCISE 14.2 now and kickstart your journey towards success in the Class 8 exam.
Importance of Compound Interest - EXERCISE 14.2
The importance of Compound Interest - EXERCISE 14.2 cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Compound Interest - EXERCISE 14.2 Notes
Compound Interest - EXERCISE 14.2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Compound Interest - EXERCISE 14.2.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Compound Interest - EXERCISE 14.2 Notes on EduRev are your ultimate resource for success.
Compound Interest - EXERCISE 14.2 Class 8 Questions
The "Compound Interest - EXERCISE 14.2 Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Compound Interest - EXERCISE 14.2 on the App
Students of Class 8 can study Compound Interest - EXERCISE 14.2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Compound Interest - EXERCISE 14.2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Compound Interest - EXERCISE 14.2 is prepared as per the latest Class 8 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup