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Page 1
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)
n
- 1] – (PTR)/100 = 283.50
P [(1 + 15/100)
3
- 1] – (P(3)(15))/100 = 283.50
P[1.520 - 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
? The sum is Rs 4000
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the
end of two years Rs. 1290 as interest compounded annually, find the sum she
borrowed.
Solution:
Given details are,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)
n
- 1]
1290 = P [(1 + 15/100)
2
- 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
? The sum is Rs 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4%
per annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Rate = 4 % per annum
CI = Rs 163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)
n
- 1]
163.20 = 2000[(1 + 4/100)
n
- 1]
163.20 = 2000 [(1.04)
n
-1]
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)
n
- 1] – (PTR)/100 = 283.50
P [(1 + 15/100)
3
- 1] – (P(3)(15))/100 = 283.50
P[1.520 - 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
? The sum is Rs 4000
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the
end of two years Rs. 1290 as interest compounded annually, find the sum she
borrowed.
Solution:
Given details are,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)
n
- 1]
1290 = P [(1 + 15/100)
2
- 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
? The sum is Rs 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4%
per annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Rate = 4 % per annum
CI = Rs 163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)
n
- 1]
163.20 = 2000[(1 + 4/100)
n
- 1]
163.20 = 2000 [(1.04)
n
-1]
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
163.20 = 2000 × (1.04)
n
– 2000
163.20 + 2000 = 2000 × (1.04)
n
2163.2 = 2000 × (1.04)
n
(1.04)
n
= 2163.2/2000
(1.04)
n
= 1.0816
(1.04)
n
= (1.04)
2
So on comparing both the sides, n = 2
? Time required is 2years
8. In how much time would Rs. 5000 amount to Rs. 6655 at 10% per annum
compound interest?
Solution:
Given details are,
Rate = 10% per annum
A = Rs 6655
Principal (P) = Rs 5000
By using the formula,
A = P (1 + R/100)
n
6655 = 5000 (1 + 10/100)
n
6655 = 5000 (11/10)
n
(11/10)
n
= 6655/5000
(11/10)
n
= 1331/1000
(11/10)
n
= (11/10)
3
So on comparing both the sides, n = 3
? Time required is 3years
9. In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest
compounded half-yearly?
Solution:
Given details are,
Rate = 8% per annum = 8/2 = 4% (half yearly)
A = Rs 4576
Principal (P) = Rs 4400
Let n be ‘2T’
By using the formula,
A = P (1 + R/100)
n
4576 = 4400 (1 + 4/100)
2T
4576 = 4400 (104/100)
2T
(104/100)
2T
= 4576/4400
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
EXERCISE 14.3 PAGE NO: 14.20
1. On what sum will the compound interest at 5% per annum for 2 years
compounded annually be Rs 164?
Solution:
Given details are,
Rate = 5 % per annum
Compound Interest (CI) = Rs 164
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
164 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 5/100)
2
– 1]
= x [(105/100)
2
– 1]
164 = x ((1.05)
2
– 1)
x = 164 / ((1.05)
2
– 1)
= 164/0.1025
= Rs 1600
? The required sum is Rs 1600
2. Find the principal if the interest compounded annually at the rate of 10% for two
years is Rs. 210.
Solution:
Given details are,
Rate = 10 % per annum
Compound Interest (CI) = Rs 210
Time (t) = 2 years
By using the formula,
Let P be ‘x’
CI = A – P
210 = P (1 + R/100)
n
– P
= P [(1 + R/100)
n
- 1]
= x [(1 + 10/100)
2
– 1]
= x [(110/100)
2
– 1]
210 = x ((1.1)
2
– 1)
x = 210 / ((1.1)
2
– 1)
= 210/0.21
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
= Rs 1000
? The required sum is Rs 1000
3. A sum amounts to Rs. 756.25 at 10% per annum in 2 years, compounded
annually. Find the sum.
Solution:
Given details are,
Rate = 10 % per annum
Amount = Rs 756.25
Time (t) = 2 years
By using the formula,
A = P (1 + R/100)
n
756.25 = P (1 + 10/100)
2
P = 756.25 / (1 + 10/100)
2
= 756.25/1.21
= 625
? The principal amount is Rs 625
4. What sum will amount to Rs. 4913 in 18 months, if the rate of interest is 12 ½ %
per annum, compounded half-yearly?
Solution:
Given details are,
Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly
Amount = Rs 4913
Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years
By using the formula,
A = P (1 + R/100)
n
4913 = P (1 + 25/4 ×100)
3
P = 4913 / (1 + 25/400)
3
= 4913/1.19946
= 4096
? The principal amount is Rs 4096
5. The difference between the compound interest and simple interest on a certain
sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.
Solution:
Given details are,
Rate = 15 % per annum
Compound Interest (CI) – Simple Interest (SI)= Rs 283.50
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
Time (t) = 3 years
By using the formula,
CI – SI = 283.50
P [(1 + R/100)
n
- 1] – (PTR)/100 = 283.50
P [(1 + 15/100)
3
- 1] – (P(3)(15))/100 = 283.50
P[1.520 - 1] – (45P)/100 = 283.50
0.52P – 0.45P = 283.50
0.07P = 283.50
P = 283.50/0.07
= 4000
? The sum is Rs 4000
6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the
end of two years Rs. 1290 as interest compounded annually, find the sum she
borrowed.
Solution:
Given details are,
Rate = 15 % per annum
Time = 2 years
CI = Rs 1290
By using the formula,
CI = P [(1 + R/100)
n
- 1]
1290 = P [(1 + 15/100)
2
- 1]
1290 = P [0.3225]
P = 1290/0.3225
= 4000
? The sum is Rs 4000
7. The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4%
per annum. Find the period for which the compound interest is Rs. 163.20.
Solution:
Given details are,
Rate = 4 % per annum
CI = Rs 163.20
Principal (P) = Rs 2000
By using the formula,
CI = P [(1 + R/100)
n
- 1]
163.20 = 2000[(1 + 4/100)
n
- 1]
163.20 = 2000 [(1.04)
n
-1]
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
163.20 = 2000 × (1.04)
n
– 2000
163.20 + 2000 = 2000 × (1.04)
n
2163.2 = 2000 × (1.04)
n
(1.04)
n
= 2163.2/2000
(1.04)
n
= 1.0816
(1.04)
n
= (1.04)
2
So on comparing both the sides, n = 2
? Time required is 2years
8. In how much time would Rs. 5000 amount to Rs. 6655 at 10% per annum
compound interest?
Solution:
Given details are,
Rate = 10% per annum
A = Rs 6655
Principal (P) = Rs 5000
By using the formula,
A = P (1 + R/100)
n
6655 = 5000 (1 + 10/100)
n
6655 = 5000 (11/10)
n
(11/10)
n
= 6655/5000
(11/10)
n
= 1331/1000
(11/10)
n
= (11/10)
3
So on comparing both the sides, n = 3
? Time required is 3years
9. In what time will Rs. 4400 become Rs. 4576 at 8% per annum interest
compounded half-yearly?
Solution:
Given details are,
Rate = 8% per annum = 8/2 = 4% (half yearly)
A = Rs 4576
Principal (P) = Rs 4400
Let n be ‘2T’
By using the formula,
A = P (1 + R/100)
n
4576 = 4400 (1 + 4/100)
2T
4576 = 4400 (104/100)
2T
(104/100)
2T
= 4576/4400
RD Sharma Solutions for Class 8 Maths Chapter 14 –
Compound Interest
(104/100)
2T
= 26/25
(26/25)
2T
= (26/25)
1
So on comparing both the sides, n = 2T = 1
? Time required is ½ year
10. The difference between the S.I. and C.I. on a certain sum of money for 2 years at
4% per annum is Rs. 20. Find the sum.
Solution:
Given details are,
Rate = 4 % per annum
Time = 2years
Compound Interest (CI) – Simple Interest (SI)= Rs 20
By using the formula,
CI – SI = 20
P [(1 + R/100)
n
- 1] – (PTR)/100 = 20
P [(1 + 4/100)
2
- 1] – (P(2)(4))/100 = 20
P[51/625] – (2P)/25 = 20
51/625P – 2/25P = 20
(51P-50P)/625 = 20
P = 20 × 625
P = 20/7.918
= 12500
? The sum is Rs 12500
11. In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound
interest?
Solution:
Given details are,
Principal = Rs 1000
Amount = Rs 1331
Rate = 10% per annum
Let time = T years
By using the formula,
A = P (1 + R/100)
n
1331 = 1000 (1 + 10/100)
T
1331 = 1000 (110/100)
T
(11/10)
T
= 1331/1000
(11/10)
T
= (11/10)
3
So on comparing both the sides, n = T = 3
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Compound Interest - EXERCISE 14.3 Notes
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Compound Interest - EXERCISE 14.3 Class 8 Questions
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The content of Compound Interest - EXERCISE 14.3 is prepared as per the latest Class 8 syllabus.
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