In Exercise 14.3 of Chapter 14, we shall discuss how to compute each of the three quantities, i.e., P, T, R, when sufficient data is given. RD Sharma Solutions are the best study materials, developed in accordance with the latest CBSE guidelines. Solutions given here are solved and prepared by our subject experts in a step-by-step format to help students understand the concept easily. Moreover, regular practice using these solutions helps in scoring good marks in the final exam. Students can download the PDF material from the links provided below.
RD Sharma Solutions for Class 8 Maths Exercise 14.3 Chapter 14 Compound Interest
Access answers to Maths RD Sharma Solutions for Class 8 Exercise 14.3 Chapter 14 Compound Interest
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
= 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
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]
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
(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
∴ Time required is 3years
12. At what rate percent compound interest per annum will Rs. 640 amount to Rs. 774.40 in 2 years?
Solution:
Given details are,
Principal = Rs 640
Amount = Rs 774.40
Time = 2 years
Let rate = R%
By using the formula,
A = P (1 + R/100)n
774.40 = 640 (1 + R/100)2
(1 + R/100)2 = 774.40/640
(1 + R/100)2 = 484/400
(1 + R/100)2 = (22/20)2
By cancelling the powers on both sides,
(1 + R/100) = (22/20)
R/100 = 22/20 – 1
= (22-20)/20
= 2/20
= 1/10
R = 100/10
= 10%
∴ Required Rate is 10% per annum
13. Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest being compounded half-yearly?
Solution:
Given details are,
Principal = Rs 2000
Amount = Rs 2662
Time = 1 ½ years = 3/2 × 2 = 3 half years
Let rate be = R% per annum = R/2 % half yearly
By using the formula,
A = P (1 + R/100)n
2662 = 2000 (1 + R/2×100)3
(1 + R/200)3 = 1331/1000
(1 + R/100)3 = (11/10)3
By cancelling the powers on both sides,
(1 + R/200) = (11/10)
R/200 = 11/10 – 1
= (11-10)/10
= 1/10
R = 200/10
= 20%
∴ Required Rate is 20% per annum
14. Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received Rs. 210 as compound interest, but paid Rs. 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Given details are,
C.I that Kamala receives = Rs 210
S.I that Kamala paid = Rs 200
Time = 2 years
So,
We know, SI = PTR/100
= P×2×R/100
P×R = 10000 ………….. Equation 1
CI = A – P
CI = P [(1 + R/100)n – 1]
210 = P [(1 + R/100)2 – 1]
210 = P (12 + R2/1002 + 2(1)(R/100) – 1) (by using the formula (a+b)2)
210 = P (1 + R2/10000 + R/50 – 1)
210 = P (R2/10000 + R/50)
210 = PR2/10000 + PR/50
We know PR = 10000 from Equation 1
210 = 10000R/10000 + 10000/50
210 = R + 200
R = 210 – 200
= 10%
In Equation 1, PR = 10000
P = 10000/R
= 10000/10
= 1000
∴ Required sum is Rs 1000
15. Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half, interest being compounded six monthly.
Solution:
Given details are,
Principal = Rs 2000
Amount = Rs 2315.25
Time = 1 ½ years = 3/2 years
Let rate be = R % per annum
By using the formula,
A = P (1 + R/100)n
2315.25 = 2000 (1 + R/100)3/2
(1 + R/100)3/2 = 2315.25/2000
(1 + R/100)3/2 = (1.1576)
(1 + R/100) = 1.1025
R/100 = 1.1025 – 1
= 0.1025 × 100
= 10.25
∴ Required Rate is 10.25% per annum
16. Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Given details are,
Time = 3 years
Let rate be = R %
Also principal be = P
So, amount becomes = 2P
By using the formula,
A = P (1 + R/100)n
2P = P (1 + R/100)3
(1 + R/100)3 = 2
(1 + R/100) = 21/3
1 + R/100 = 1.2599
R/100 = 1.2599-1
= 0.2599
R = 0.2599 × 100
= 25.99
∴ Required Rate is 25.99% per annum
17. Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly.
Solution:
Given details are,
Time = 2 years = 2×2 = 4 half years
Let rate = R % per annum = R/2% half years
Let principal be = P
So, Amount becomes = 4P
By using the formula,
A = P (1 + R/100)n
4P = P (1 + R/2×100)4
(1 + R/200)4 = 4
(1 + R/200) = 41/4
1 + R/200 = 1.4142
R/200 = 1.4142-1
= 0.4142
R = 0.4142 × 200
= 82.84%
∴ Required Rate is 82.84% per annum
18. A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Given details are,
Amount = Rs 5832
Time = 2 years
Rate = 8%
Let principal be = P
By using the formula,
A = P (1 + R/100)n
5832 = P (1 + 8/100)2
5832 = P (1.1664)
P = 5832/1.1664
= 5000
∴ Required sum is Rs 5000
19. The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs. 360. Find the sum.
Solution:
Given,
Time = 2 years
Rate = 7.5 % per annum
Let principal = Rs P
Compound Interest (CI) – Simple Interest (SI) = Rs 360
C.I – S.I = Rs 360
By using the formula,
P [(1 + R/100)n – 1] – (PTR)/100 = 360
P [(1 + 7.5/100)2 – 1] – (P(2)(7.5))/100 = 360
P[249/1600] – (3P)/20 = 360
249/1600P – 3/20P = 360
(249P-240P)/1600 = 360
9P = 360 × 1600
P = 576000/9
= 64000
∴ The sum is Rs 64000
20. The difference in simple interest and compound interest on a certain sum of money at
Solution:
Given,
Time = 3 years
Rate =
Let principal = Rs P
Compound Interest (CI) – Simple Interest (SI) = Rs 46
C.I – S.I = Rs 46
By using the formula,
P [(1 + R/100)n – 1] – (PTR)/100 = 46
P [(1 + 20/3×100)3 – 1] – (P(3)(20/3))/100 = 46
P[(1 + 20/300)3 – 1] – P/5 = 46
P[721/3375] – P/5 = 46
721/3375P – 1/5P = 46
(721P-675P)/3375 = 46
46P = 46 × 3375
46P = 46 × 3375/46
= 3375
∴ The sum is Rs 3375
21. Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n.
Solution:
Given details are,
Principal = Rs 12000
Amount = Rs 13230
Rate = 5% per annum
Let time = T years
By using the formula,
A = P (1 + R/100)n
13230 = 12000 (1 + 5/100)T
13230 = 12000 (105/100)T
(21/20)T = 13230/12000
(21/20)T = 441/400
(21/20)T = (21/20)2
So on comparing both the sides, n = T = 2
∴ Time required is 2years
22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?
Solution:
Given details are,
Principal = Rs 4000
Time = 2years
CI = Rs 410
Rate be = R% per annum
By using the formula,
CI = P [(1 + R/100)n – 1]
410 = 4000 [(1 + R/100)2 – 1]
410 = 4000 (1 + R/100)2 – 4000
410 + 4000 = 4000 (1 + R/100)2
(1 + R/100)2 = 4410/4000
(1 + R/100)2 = 441/400
(1 + R/100)2 = (21/20)2
By cancelling the powers on both the sides,
1 + R/100 = 21/20
R/100 = 21/20 – 1
= (21-20)/20
= 1/20
R = 100/20
= 5
∴ Required Rate is 5% per annum
23. A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.
Solution:
Given details are,
Time = 2years
Amount = Rs 10404
Rate be = 2% per annum
Let principal be = Rs P
By using the formula,
A = P [(1 + R/100)n
10404 = P [(1 + 2/100)2]
10404 = P [1.0404]
P = 10404/1.0404
= 10000
∴ Required sum is Rs 10000
24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum compound interest?
Solution:
Given details are,
Principal = Rs 1600
Amount = Rs 1852.20
Rate = 5% per annum
Let time = T years
By using the formula,
A = P (1 + R/100)n
1852.20 = 1600 (1 + 5/100)T
1852.20 = 1600 (105/100)T
(21/20)T = 1852.20/1600
(21/20)T = 9261/8000
(21/20)T = (21/20)3
So on comparing both the sides, n = T = 3
∴ Time required is 3years
25. At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?
Solution:
Given details are,
Principal = Rs 1000
Amount = Rs 1102.50
Rate = R% per annum
Let time = 2 years
By using the formula,
A = P (1 + R/100)n
1102.50 = 1000 (1 + R/100)2
(1 + R/100)2 = 1102.50/1000
(1 + R/100)2 = 4410/4000
(1 + R/100)2 = (21/20)2
1 + R/100 = 21/20
R/100 = 21/20 – 1
= (21-20)/20
= 1/20
R = 100/20
= 5
∴ Required Rate is 5%
26. The compound interest on Rs. 1800 at 10% per annum for a certain period of time is Rs. 378. Find the time in years.
Solution:
Given details are,
Principal = Rs 1800
CI = Rs 378
Rate = 10% per annum
Let time = T years
By using the formula,
CI = P [(1 + R/100)n – 1]
378 = 1800 [(1 + 10/100)T – 1]
378 = 1800 [(110/100)T – 1]
378 = 1800 [(11/10)T – 1800
378 + 1800 = 1800 [(11/10)T
(11/10)T = 2178/1800
(11/10)T = 726/600
(11/10)T = 121/100
(11/10)T = (11/10)2
So on comparing both the sides, n = T = 2
∴ Time required is 2years
27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually?
Solution:
Given details are,
Time = 2years
Amount = Rs 45582.25
Rate be = 6 ¾ % per annum = 27/4%
Let principal be = Rs P
By using the formula,
A = P [(1 + R/100)n
45582.25 = P [(1 + 27/4×100)2]
45582.25 = P (1 + 27/400)2
45582.25 = P (427/400)2
45582.25 = P × 427/400 × 427/400
P = (45582.25 × 400 × 400) / (427×427)
P = 7293160000/182329
= 40000
∴ Required sum is Rs 40000
28. Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Given details are,
Time = 2years
Amount = Rs 453690
Rate be = 6.5 % per annum
Let principal be = Rs P
By using the formula,
A = P [(1 + R/100)n
453690 = P [(1 + 6.5/100)2]
453690 = P (106.5/100)2
453690 = P × 106.5/100 × 106.5/100
P = (453690 × 100 × 100) / (106.5×106.5)
P = 4536900000/11342.25
= 400000
∴ Required sum is Rs 400000
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