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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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