# Selina Concise Solutions for Chapter 9 Simple and Compound Interest Class 8 ICSE Mathematics

### Exercise 9A

1. Find the interest and the amount on:
(i) ₹ 750 in 3 years 4 months at 10% per annum.
(ii) ₹ 5,000 at 8% per year from 23rd December 2011 to 29th July 2012.
(iii) ₹ 2,600 in 2 years 3 months at 1% per month.
(iv) ₹ 4,000 in 1 years at 2 paise per rupee per month.
Solution
(i) Given P = ₹750

∴ Amount (A) = P + I = ₹750 + ₹250 = ₹1000

(ii) Principal (P) = ₹5000
Rate (R) = 8% p.a.
Time (T) = 23 December 2011 to 29 July 2012

= 10 × 8 × 3
= ₹240
∴ Amount = P + I = ₹5000 + 240 = ₹5240

(iii) Here P = ₹2,600
Time (T) = 2 years 3 months  = 27 months
Rate (R) = 1 % per month
∴ Interest = (P × T × R)/100
= (2600 × 27 × 1)/100
= 26 × 27
= Rs. 702
∴ Amount = Rs.(2600 + 702) = Rs. 3302

(iv) Here  P = Rs. 4,000, Time (T) = 1 1/3 year
= 1 year + 12/3 months = 16months
Rate (R) = 2paise per rupee-per month = 2% per month
∴ Interest (I) = (P × R × T)/100
= (4,000 × 2 × 16)/100
= 40 × 32
= Rs.1280
∴ Amount(A) = P + I
= Rs. 4000 + Rs. 1280
= Rs. 5280

2. Rohit borrowed Rs. 24,000 at 7.5 percent per year. How much money will he pay at the end of 4th years to clear his debt?
Solution
Principal (P) = Rs. 24,000
Rate (R) = 7.5% P.A.
Time (T) = 4 years
S.I. = (P × R × T)/100
= Rs. (24,000 × 4 × 7.5)/100
= Rs. 240 × 4 × 7.5
= 240 × 30 = Rs. 7200
Amount needed to clear the debt at the end of 4th year = Rs. 24000 + Rs. 7200 = Rs. 3,1200

3. The interest on a certain sum of money is Rs. 1,480 in 2 years and at 10 per cent per year. Find the sum of money.
Solution
Let P = Rs x
Time (T) = 2years
Rate (R) = 0%
∴ Interest = (P × R × T)/100
= (x  × 10 × 2)/100 = x/5
x/5 = Rs. 1480  (Given)
∴ x = 1480 × 5 = Rs. 7400
Hence the money Rs. 7400

4. On what principal will the simple interest be Rs. 7,008 in 6 years 3 months at 5% per year?
Solution
Let principal = Rs. P
Time (T) = 6 years 3 months = 6 year + 3/12
year = 75/12 = 25/4 year = 6 ¼ years
Rate (R) = 5%
Simple Interest = Rs. 7,008
We know that
Simple Interest = (P × R × T)/100
⇒ 7,008 = (P × 25/4 × 5)/100
⇒ P = (7008 × 100 × 4)/(25 × 5)
= (7008 × 16)/5
= 112128/5
= Rs. 22425.60

5. Find the principal which will amount to Rs. 4,000 in 4 years at 6.25% per annum.
Solution
Let Principal = Rs. P, Time (T) = 4 years
Rate = 61/4 = 25/4%
Simple Interest = (P × R × T)/100
= (P × 25/4 × 4)/100
= P/4
∴ Amount = P + P/4 = 5P/4
⇒ 5P/4 = 4000
⇒ 5P = 4 × 4000
P = (4 × 4000)/5
= 4 × 800
⇒ P = Rs. 3200
Here, principal = Rs. 3200

6. (i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years?
(ii) At what rate per cent per year will a sum double itself in 6 years?
Solution
(i) P = Rs. 630, I = Rs. 126, T = 4 years
R = (100 × 1)/(P × T)
= (100 × 126)/(630 × 4)
= 100/20
= 5%

(ii) Let P = Rs. 100
∴ Amount = 2 × Rs. 100 = Rs. 200
Interest = A - P
= Rs. 200 - Rs. 100
= Rs. 100
T = 6 ¼ years = 25/4 years
R = (100 × 1)/(P × T)
= (100 × 100)/(100 × 25/4)%
= (100 × 100)/100 × 4/25
= 16%

7. (i) In how many years will Rs.950 produce Rs.399 as simple interest at 7% ?
(ii) Find the time in which Rs.1200 will amount to Rs.1536 at 3.5% per year.
Solution
(i) P = Rs. 950
S.I. = Rs. 399
R = 7%
We know that :
T = (100 × I)/(P × R)
= (100 × 399)/(950 × 7)
= (10 × 21)/(5 ×7)
= 2 × 3
= 6years

(ii) A = Rs. 1536
P = Rs. 1200
I = A – P
= Rs. 1536 – Rs. 1200
We know that :
T = (100 × I)/(P × R)
= (100 × 336)/(1200 × 3.5)
= (100 × 336 × 10)/(1200 × 35)  [∵ 1/3.5 = 10/35]
= (28 × 10)/35
= 8years

8. The simple interest on a certain sum of money is 3/8 of the sum in 6 1/4  years. Find the rate percent charged.
Solution
Let P = Rs. 8
S.I. = Rs. 3/8 × 8
= Rs. 3
T = 61/4 years = 25/4 years
We know that :
R = (100 × I)/(P × T)
= (100 × 3)/(8 × 25/4)
= (100 × 3)/8 × 4/25
= 2 × 3
= 6%

9. What sum of money borrowed on 24th May will amount to Rs. 10210.20 on 17th October of the same year at 5 percent per annum simple interest.
Solution
A = Rs. 10210.20
R = 5% P.A.
T = May + June + July + August + Sept.+ Oct.
= 7 + 30 + 31 + 31 + 30 + 17
= 146/365days = 2/5 year
We know that:
P + I = A

⇒ P = Rs. 10010
∴ Money to be borrowed = Rs. 10010

10. In what time will the interest on a certain sum of money at 6% be of itself?
Solution
Let P = Rs. 8
Interest = Rs. 8 × 5/8 = Rs. 5
R = 6%
T = (100 × I)/(P × R)
= (100 × 5)/(8 ×  6)
= 500/48
= 125/12 years
= 10 5/ 12 years
= 10 years 5 months
[∵ 5/12 year = 5/12 × 12 months = 5 months]
∴ Time = 10 years 5 months

11. Ashok lent out Rs.7000 at 6% and Rs.9500 at 5%. Find his total income from the interest in 3 years.
Solution
In Ist case:
P = Rs. 7000
R = 6%
T = 3 years
S.I. = (P × R × T)/100
= Rs. (7000 × 6 × 3)/100
= Rs. 1260
In IInd case:
P = Rs. 9500
R = 5%
T = 3 years
S.I. = (P × R × T)/100
= Rs. (9500 × 5 × 3)/100
= Rs. 1425
Total income from the interest = Rs. 1260 + Rs. 1425
= Rs. 2685

12. Raj borrows Rs.8,000; out of which Rs. 4500 at 5% and remainder at 6%. Find the total interest paid by him in 4 years.
Solution
Total sum borrowed by Raj = Rs. 8000
In the First case:
P = Rs. 4500
R = 5%
T = 4 years
S.I. = (P × R × T)/100
= Rs. (4500 × 5 × 4)/100
= Rs. 900
In the Second case:
P = Rs. 8000 - Rs. 4500
= 3500
R = 6%
T = 4 years
S.I. = (P × R × T)/100
= 35 × 6 × 4 = Rs. 840
= Rs. (3500 ×  6 × 4)/100
= 35 × 6 × 4 = Rs. 840
Total interest paid by Raj = Rs. 900 + Rs. 840
= Rs. 1740

13. Mohan lends Rs.4800 to John for 4 1/2 years and Rs.2500 to Shy am for 6 years and receives a total sum of Rs.2196 as interest. Find the rate percent per annum, it being the same in both the cases.
Solution
In the first case:
P = Rs. 4800
R = x% (Suppose)
T = 4 ½ years = 9/2 years
Interest = P × R × T)/100
= Rs. (4800 × x × 9)/(100 × 2)
= Rs. 24 × x × 9
= Rs. 216x
In the second case:
P = Rs. 2500
R = x %
T = 6 years
Interest = (P × R × T)/100
= Rs. (2500 × x × 6)/100
= Rs. 25 × x × 6
= Rs. 150x
According to statement,
Interest in first case + Interest in second case = Rs. 2196
∴  Rs. 216x + Rs. 150x = Rs. 2196
⇒ Rs. 366x = Rs. 2196
⇒ x = 2196/366
⇒ x = 6
∴ Rate = 6%

14. John lent Rs. 2550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old black and white television and Rs. 1422.50; find the price of the television.
Solution
P =  Rs.2550
R = 7.5%
T = 8 months = 8/12 years
= 2/3 years
S.I. = (P × R × T)/100
= Rs. 2550 × 7.5 × 2/3 × 1/100
= Rs. (2550 × 7.5 × 2)/(3 × 100)
= Rs. (2550 × 5)/100
= Rs. 12750/100
= Rs. 127.50
Amount = P + I
= Rs. 2550 + Rs. 127.50
= Rs. 2677.50
Mohan paid in cash = Rs. 1422.50
Price of the television = Amount – Paid in cash
= Rs. 2677.50 – Rs. 1422.50
= Rs. 1255

### Exercise 9 B

1. The interest on a certain sum of money is 0.24 times of itself in 3 years. Find the rate of interest.
Solution
Let the sum borrowed = Rs. 100
Time = 3 years
Let rate of interest = r%
∴ Interest = (100 × 3 × r)/100  [∵ S.I. = (P × R × T)/100]
= 3r
= (0.24)(100)
= 24
(Given)
⇒ r = 24/3 = 8
Hence reqd. rate of interest = 8%

2. If ₹ 3,750 amount to ₹ 4,620 in 3 years at simple interest. Find:
(i) the rate of interest
(ii) the amount of Rs. 7,500 in 5  years at the same rate of interest
Solution
(i) In first case:
A = Rs. 4620
P = Rs. 3750
I = A – P = Rs. 4620 – Rs. 3750 = Rs. 870
T = 3 years
R = (100 × I)/(P × T)
= (100 × 870)/(3750 × 3)
= (100 × 290)/3750
= (4 × 29)/15
= 116/15
= 7 11/15%
In second case :
P  = Rs. 7500
R = 116/15%
T = 5 ½ years = 11/2 years
(ii) Interest = (P × R × T)/100
= Rs. (7500 × 11 × 116)/(2 × 15 × 100)
= (250 × 116 × 11)/100
= 10 × 29 × 11
= 290 × 11
= Rs. 3190
Amount = Rs. 7500 + 3190
= Rs. 10, 690

3. A sum of money, lent out at simple interest, doubles itself in 8 years. Find :
(i) the rate of interest
(ii) in how many years will the sum become triple (three times) of itself at the same rate per cent?
Solution
Let P = Rs. 100
A = Rs. 200
I = Rs. 200 – Rs. 100 = Rs. 100,
T = 8 years
R = (100 × I)/(P × T)
= (100 × 100)/(100 × 8)
= 100/8
= 25/2%
Now, again P = Rs. 100
A = Rs. 300
I = Rs. 300 – Rs. 100
= Rs. 200
R = 25/2%
T = (100 × I)/(P × R)
= (100 × 200)/(100 × 25/2)
= (100 × 200 × 2)/(100 × 25)
= 16 years
So, the given sum of money will become triple in 16 years.

4. Rupees 4000 amount to Rs.5000 in 8 years; in what time will Rs.2100 amount to Rs.2800 at the same rate?
Solution
In first case:
A = Rs. 5000
P = Rs. 4000
I = A – P
= Rs. 5000 – Rs. 4000
= Rs. 1000
T = 8 years
R = (100 × I)/(P × R)
= (100 × 1000)/(4000 × 8)
= 25/8%
In the second case:
A = Rs. 2800
P = Rs. 2100
I = Rs. 2800 - Rs. 2100
= Rs. 700
R = 25/8%
T = (100 × I)/(P × R)
= (100 × 700)/(2100 × 25/8)
= (100 × 700 × 8)/(2100 × 25)
= 32/3 years
= 10 2/3 years
10 2/3 × 12 months
= 10 24/3 months
= 10 years 8 months

5. What sum of money lent at 6.5% per annum will produce the same interest in 4 years as Rs.7500 produce in 6 years at 5% per annum?
Solution
In first case:
P = Rs. 7500
R = 5%
T = 6 years
Interest = (P × R × T)/100
= Rs. (7500 × 5 × 6)/100
= Rs. 75 × 5 × 6
= Rs. 2250
In second case :
According to the statement, interest = Rs. 2250
R = 6.5%  P.A.
T = 4 years
P = (100 × I)/(R × T)
= Rs. (100 × 2250)/(6.5 × 4)
= Rs. 225000/26
= Rs. 112500/13
= Rs. 8653.85
Required principal = Rs. 8653.85

6. A certain sum amounts to Rs.3825 in 4 years and to Rs.4050 in 6 years. Find the rate percent and the sum.
Solution
In 6 years sum amounts to = Rs. 4050
In 4 years sum amounts to = Rs. 3825
∴ Interest of 2 years = Rs. 4050 – Rs. 3825
= Rs. 225
Interest of 4 years = Rs. 225/2 × 4
= Rs. 450
(∵ Rs. 225 is interest for 2 years)
Now  P = A – I
= Rs. 3825 – Rs. 450
= Rs. 3375
I = Rs. 450
T = 4 years
R = (100 × I)/(P × T)
= (100 × I)/(P × T)
= (100 × 450)/(3375 × 4)
= 45000/13500%
= 450/135%
= 10/3%
= 3 1/3%
∴ R = 3 1/3 %
P = Rs. 3375

7. At what rate percent of simple interest will the interest on Rs.3750 be one-fifth of itself in 4 years? To what will it amount in 15 years?
Solution
P = Rs. 3750
I = Rs. 3750  × 1/5
= Rs. 750
T = 4 years
R = (100 × I)/(P × T)
= (100 × 750)/(3750 × 4)
= (100 × 750)/(3750 × 4)
= 5%
Again,  P = Rs. 3750
Interest of 4 years = Rs. 750
Interest of 1 year = Rs. 750/4
Interest of 15 years = Rs.750/4 × 15
= Rs. (750 × 15)/4
= Rs. 5625/2
= Rs. 2812.50
Amount in 15 years will be = Rs. 3750  + Rs. 2812.50
= Rs. 6562.50
∴ Rate = 5%
Amount in 15 years will be = Rs. 6562.50

8. On what date will ₹ 1950 lent on 5th January, 2011 amount to ₹ 2125.50 at 5 percent per annum simple interest?
Solution
P = Rs. 1950
A = Rs. 2125.50
R = 5% p.a
I = A – P
= Rs. 2125.50 – Rs. 1950
= Rs. 175.50
T = (100 × I)/(P × R)
= (100 × 175.50)/(1950 × 5)
=  17550/9750
= 1755/975
= 117/65
= 9/5 years
= 4 1/5 years
= 1 years 292 days
∵ 4/5 years = 4/5 × 365 days = 292 days
Jan. +  Feb. + March + April + May  + June  + July  + Aug. + Sept. + Oct.
(31 – 5) + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 23
= 292 days
∴ Required date = 23rd October 2012

9. If the interest on Rs.2400 be more than the interest on Rs.2000 by Rs.60 in 3 years at the same rate percent; find the rate.
Solution
In first case:
P = Rs. 2400
R = x%  (Assume)
T = 3 years
Interest  = (P × R × T)/100
= Rs. (2400 × x × 3)/100
= Rs. 72x
In second case:
P = Rs. 2000
R = x%  (Rate same as in first case)
T = 3 years
Interest =  (P × R × T)/100
= Rs. (2000 × x × 3)/100
= Rs. 60x
According to the statement,
72x = 60x + 60
⇒ 72x – 60x = 60
⇒ 12x = 60
⇒ x = 60/12
⇒ x = 5
∴ Rate = 5%

10. Divide Rs. 15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at 4  per cent for 6 years.
Solution
Let one part = Rs. x
∴ Second part = Rs. (15,600 – x)
By the given condition = (x × 5 × 5)/100
= {(15,600 – x) × 9/2 × 6}/100
⇒ 25x = 27 × 15,600 – 27x
⇒ 25x + 27x = 27 × 15,600
⇒ 52x = 27 × 15, 6000
⇒ 52x = 27 × 15,600
⇒ x = (27 × 15,600)/52 = 27 × 300 = 8100
Hence one part = Rs. 8100 and second part = Rs. (15,600 – 8,100)
= Rs. 7,500

### Exercise 9 C

1. A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) final amount at the end of second year
(v) compound interest earned in 2 years.
Solution
(i) Here Principal (P) = Rs. 8,000
Rate of interest = 10%
Interest for the first year = (8,000 × 10 × 1)/100
= Rs. 800

(ii) ∴ Amount  = Rs.8,000 + Rs. 800
= Rs. 8,800
Thus principal for the second year = Rs, 8,800

(iii) Interest for the second year  = (8,800 × 10 × 1)/100
= Rs. 880

(iv) Amount at the end of second year = Rs. 8,800 + Rs. 880
= Rs 9,860

(v) Hence compound interest earned in 2 years = Rs. 9,680 – Rs. 8,000
= Rs. 1680

2. A man borrowed Rs. 20,000 for 2 years at 8% per year compound interest. Calculate :
(i) the interest of the first year.
(ii) the interest of the second year.
(iii) the final amount at the end of second year.
(iv) the compound interest of two years.
Solution
Here Principal (P) = Rs. 20,000, Time = 1 year
Rate = 8%

(i) ∴ Interest of the first year = (20,000 × 8 × 1)/100
= Rs.1600

(ii) ∴ Amount after one year
i.e., Principal for second year = Rs. 20,000 + Rs. 1,600
= Rs. 21,600
∴ Interest for second year = (21,600 × 8 × 1)/100
= 216 × 8
= Rs. 1728

(iii) Final amount at the end of second year = Rs.(21,600 + 1728)
= Rs. 23,328

(iv) Interest of two years = Rs. 23,328 – Rs. 20,000
= Rs. 3,328

3. Calculate the amount and the compound interest on Rs. 12,000 in 2 years and at 10% per year.
Solution
For 1st year
Principal (P) = Rs. 12,000
Rate (R) = 10%
Time (T) = 1 year
I = Interest = (12,000 × 10 × 1)/100
= 120 × 10
= Rs. 1200
Amount = P + I
=  Rs. 12,000 + Rs. 1200
= Rs. 13,200
For IInd year
P = Rs. 13,200,  R = 10%, Time(T) = 1 year
∴ Interest  = (13,200 × 10 × 1)/100
= 132 × 10
= Rs. 1320
∴ Amount in 2 years = Rs.(13,200) + (1320)
= Rs. 14520
Compound interest in 2 years = Rs. 1200 + Rs. 1320
= Rs. 2520
[or directly = Rs. 14520 – Rs. 12000 = Rs. 2520]

4. Calculate the amount and the compound interest on Rs. 10,000 in 3 years at 8% per annum.
Solution
For 1st year
Principal (P) = Rs. 10,000, Rate (R) = 8%
Time (T) = 1 year
∴ Interest = (10,000 × 8 × 1)/100
= 100 × 8
= Rs. 800
For 2nd year
P = Rs. 10,000 + Rs. 800 = Rs. 10,800
Rate (R) = 8%   Time (T)  = 1 year
∴ Interest = (10,800 × 8 × 1)/100
= 108 × 8
= Rs. 864
For 3rd year
∴ P = Rs. 10,800 + Rs. 864 = Rs. 11664
R = 8%
T = 1 year
∴ Interest = (11664 × 8 × 1)/100
= (11664 × 2)/25
= Rs. 933.12
∴ Amount = Rs.11664 + 933.12 = Rs. 12597.12
Hence, required amount = Rs. 12597.12
∴ Compound Interest = Rs. 12597.12 – 10000
= Rs. 2597.12

5. Calculate the compound interest on Rs. 5,000 in 2 years; if the rates of interest for successive years be 10% and 12% respectively.
Solution
For 1st year
Principal (P) = Rs. 5,000, Rate (R) =  10%
Time (T) = 1 year
∴ Interest  = (5,000 × 10 × 1)/100
= 50 × 10
= Rs. 500
∴ Amount at the end of 1st year = Rs. (5000 + 500) = Rs. 5500
For 2nd year
P = Rs. 5500, rate = 12%, T = 1 year
∴ Interest  = (5500 × 12 × 1)/100
= 55 × 12
= Rs. 660
∴ Amount at the end of 2nd year = Rs. 5500 + Rs. 660 = Rs. 6160
Hence compound Interest = Rs. 6160 – Rs. 5000
= Rs. 1160

6. Calculate the compound interest on Rs. 15,000 in 3 years; if the rates of interest for successive years be 6%, 8% and 10% respectively.
Solution
For 1st year
Principal (P) = Rs. 15,000 ,   Rate (R) = 6%
Time (T) = 1 year
∴ Interest = (15,000 × 6 × 1)/100
= 150 × 6
= Rs. 900
∴ Amount at the end of 1st year = Rs. 15,000 + Rs. 900
= Rs.15900
For 2nd year
P = Rs. 15900, R = 8%,  T = 1 year
∴ Interest = (15,900 × 8 × 1)/100
= 159 × 8
= Rs. 1272
∴ Amount at the end of 2nd year = Rs. (15,900 + 1272)
= Rs. 17172
For 3rd year
P = Rs. 17172, R = 10% ,  T = 1 year
∴ Interest = (17172 × 10 × 1)/100
= Rs. 1717.20
∴ Amount at the end of 3rd year = Rs. (17172 + 1717.20)
= Rs. 18889.20
∴ Compound interest = 18889.20 – 15,000
= Rs. 3889.20

7. Mohan borrowed Rs. 16,000 for 3 years at 5% per annum compound interest. Calculate the amount that Mohan will pay at the end of 3 years.
Solution
For 1st year
Principal (P) = Rs. 16,000.
Rate (R) = 5%
Time (T) = 1 year
∴ Interest = (16,000 × 5 × 1)/100
= 160 × 5
= Rs. 800
∴ Amount at the end of 1st year = Rs. (16,000 + 800)
= Rs. 16,800
For 2nd year
P = Rs. 16,800
R = 5%
T = 1 year
∴  Interest = (16,800 × 5 × 1)/100
= 168 × 5
= Rs. 840
∴  Amount at the end of 2nd year = Rs.(16,800 + 840)
= Rs. 17640
For 3rd year
P = 17640,
R = 5%,
T  = 1 year
∴ Interest = (17640 × 5 × 1)/100
= 1764/2
= Rs. 882
∴ Amount at the end of 3rd year = Rs. (17640 + 882)
= Rs. 18522
Hence reqd. amount = Rs. 18522

8. Rekha borrowed Rs. 40,000 for 3 years at 10% per annum compound interest. Calculate the interest paid by her for the second year.
Solution
For 1st year
Principal = Rs. 40,000,
Rate = 10%,
Time = 1 year
∴ Interest = (40,000 × 10 × 1)/100
= 400 × 10
= Rs. 4000
∴ Amount at the end of 1st year = Rs. (40,000 + 4000)
= Rs.44,000
For 2nd year
P = Rs. 44,000,
R = 10%,
T = 1 year
∴ Interest = Rs.(44,000 × 10 × 1)/100
= 440 × 10
= Rs. 4400
Thus interest earned in the second year = Rs. 4400

9. Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.
Solution
Principal (P) = Rs. 15000
Rate (R) = 6% p.a.
Period (n) = 5 years
Interest for the first year  = PRT/100
= (15000 × 6 × 1)/100
= Rs. 900
∴ Amount for the first year = Rs. 15000 + 900
= Rs. 15900
Principal for the second year = Rs. 15900
Interest for the second year = (15900 × 6 × 1)/100
= 159 × 6 = Rs. 954

10. A man invests Rs. 9600 at 10% per annum compound interest for 3 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of the first year.
(iii) the interest for the second year.
(iv) the interest for the third year.
Solution
Principal (P) = Rs. 9600
Rate (R) = 10% p.a.
Period (P) = 3 years
(i) ∴ Interest for the first year = PRT/100
= (9600 × 10 × 1)/100
= Rs. 960

(ii) Amount at the end of first year = P + S.I. = Rs. 9600 + 960
= Rs. 10560

(iii) Principal for the second year = Rs. 10560
Interest for the second year = (10560 × 10 × 1)/100
= Rs. 1056
∴ Amount after second year = Rs. 10560 + 1056
= Rs. 11616

(iv) Principal for the third year = Rs. 11616
Interest for the third year = (11616 × 10 × 1)/100
= 116.16 × 10
= Rs. 1161.60

11. A person invests Rs. 5,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,600. Calculate :
(i) the rate of interest per year.
(ii) the amount at the end of the second year.
Solution
Principal (P) = Rs. 5000
Period (T) = 2 years
Amount at the end of one year = Rs. 5600
∴ Interest for the first year = A – P
= Rs. 5600 – 5000
= Rs. 600
(i) ∴ Rate of interest = (S.I. × 100)/(P × T)
= (600 × 100)/(5000 × 1)
= 12% p.a.

(ii) Principal for the second year = Rs. 5600
Interest for the second year = (5600 × 12 × 1)/100
= ₹672
∴ Amount at the end of second year = P + S.I.
= 5600 + 672
= ₹6272

12. Calculate the difference between the compound interest and the simple interest on ₹ 7,500 in two years and at 8% per annum.
Solution
Principal (P) = ₹7500
Rate (R) = 8% p.a.
Period (T) = 2 years
∴ Simple Interest = PRT/100
= (7500 × 8 × 2)/100
= ₹ 1200
Interest for the second year = (7500 × 8 × 1)/100
= ₹ 600
∴ Amount at the end of first year = P + S.I.
= ₹ 7500 + ₹ 600
= ₹ 8100
∴ Interest for the second year = (8100 × 8 × 1)/100
= ₹ 648
∴ Total C.I. for 2 years  = ₹ 600 + ₹ 648
= ₹1248
∴ Difference between C.I. and S.I. for 2 years =  ₹ 1248 - ₹1200
= ₹ 48

13. Calculate the difference between the compound interest and the simple interest on ₹ 8,000 in three years and at 10% per annum.
Solution
Principal (P) = ₹8000
Rate (R) = 10% p.a.
Period (T) = 3 years
∴ S.I. for 3 years = PRT/100
= (8000 × 10 × 3)/100
= ₹ 2400
Now, S.I. for 1st year = ₹ (8000 × 10 × 1)/100
= 80 × 10 × 1
= ₹ 800
Amount for the first year = P + S.I.
= ₹ 8000 + ₹ 800
= ₹ 8800
Principal for the second year = ₹8800
Interest for the second year = (8800 × 10 × 1)/100
₹880
∴ Amount after second year = ₹ 8800 + ₹ 880
= ₹ 9680
Principal for the third year = ₹ 9680
Interest  for the third year  = ₹ (9680 × 10 × 1)/100
= ₹ 968
∴ C.I.  for 3 years = ₹800 + ₹880 + ₹968
= ₹ 2648
∴ Difference between C.I. and S.I.  for 3 years = ₹2648 - ₹2400
= ₹248

14. Rohit borrowed ₹ 40,000 for 2 years at 10% per annum C.I. and Manish borrowed the same sum for the same time at 10.5% per annum simple interest. Which of these two gets less interest and by how much?
Solution
In first case,
Sum borrowed (P) = ₹40000
Rate (R) = 10 p.a. compounded annually
Time (T) = 2 years
∴ Interest for first year = PRT/100
= ₹ (4000 × 10 × 1)/100
= ₹ 4000
Amount after one year  = ₹ 40000 + 4000
= ₹ 44000
Principal for the second year = ₹ 44000
∴ Interest for the second year = (44000 × 10 × 1)/100
= ₹4400
∴ Compound Interest for 2 years = ₹4000 + 4400
= ₹ 8400
In second case,
Principal (P) = ₹40000
Rate (R) = 10.5% p.a.
Time (T)  = 2 years
∴ Simple interest = PRT/100
= (40000 × 10.5 × 2)/100
= ₹ (40000 × 105 × 2)/(100 × 10)
= ₹ 8400
In both the cases, interest is same.

15. Mr. Sharma borrowed ₹ 24,000 at 13% p.a. simple interest and an equal sum at 12% p.a. compound interest. Find the total interest earned by Mr. Sharma in 2 years.
Solution
Sum borrowed (P) = ₹24000
Rate (R) = 13% p.a.
Time (T) = 2 years
In case of simple interest,
Simple interest for 2 years = PRT/100
= ₹(24000 × 13 × 2)/100
= ₹6240
In case of compound interest,
Interest for the first year = (24000 × 12 × 1)/100
= ₹ 2880
Amount after first year = ₹24000 + 2880
= ₹26880
Interest for the first year = (26880 × 12 × 1)/100
= ₹322560/100
= ₹ 3225. 60
∴ C.I for 2 years = ₹ 2880 + ₹ 3225.60
= 6105.60
Total interest = ₹ 6240 + ₹6105.60
= ₹ 12345.60

16. Peter borrows ₹ 12,000 for 2 years at 10% p.a. compound interest. He repays ₹ 8,000 at the end of first year. Find:
(i) the amount at the end of first year, before making the repayment.
(ii) the amount at the end of first year, after making the repayment.
(iii) the principal for the second year.
(iv) the amount to be paid at the end of second year, to clear the account.
Solution
Sum borrowed = ₹12000
Rate (R) = 10% p.a.  compound annually
Time (T) = 2 years
Interest for the first year = PRT/100
= (12000 × 100 × 1)/100
= ₹1200

(i) Amount = ₹12000 + 1200 = ₹13200
Amount paid = ₹ 8000

(ii) Balance amount = ₹13200 – 8000
= ₹5200

(iii) ∴ Principal for the second  year = ₹ 5200

(iv) Interest for the second year = (5200 × 10 × 1)/100
= ₹ 520
∴ Amount = ₹ 5200 + ₹ 520
= ₹ 5720

17.  Gautam takes a loan of ₹ 16,000 for 2 years at 15% p.a. compound interest. He repays ₹ 9,000 at the end of first year. How mucH must he pay at the end of second year to clear the debt?
Solution
Loan taken (P) = ₹ 16000
Rate (R) = 15% p.a.
Time (T) = 2 years
∴ Interest for the first year = PRT/100
= (16000 × 15 × 1)/100
= ₹ 2400
Amount after one year = ₹ 16000 + ₹ 2400
= ₹ 18400
At the end of one year aount paid back = ₹ 9000
Balance amount = ₹18400 – 9000
= ₹9400
Interest for the second year = (9400 × 15 × 1)/100
= ₹ 1410
Amount after second year = ₹ 9400 + ₹ 1410
= ₹ 10810

18. A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹ 12,000. Find:
(i) the sum of money.
(ii) the compound interest earned by this money in two years and at 10% p.a. compound interest.
Solution
Rate (R) = 8% p.a.
Period (T) = 5 years
Interest (I) = ₹ 12000
(i) ∴ Sum = (I × 100)/(R × T)
=  ₹(12000 × 100)/(8 × 5)
= ₹30000

(ii) Rate (R) = 10% p.a.
Time (T) = 2 years
Principal (P) = ₹  30000
Interest for the first year = PRT/100
= ₹(30000 × 10 × 1)/100
= ₹3000
∴ Amount after one year =  ₹30000 + 3000
= 33000
Principal for the second year = ₹ 33000
Interest for the second year = (33000 × 10 × 1)/100
= ₹ 3300
∴ Compound  Interest for two years = ₹3000 + 3300
= ₹6300

19. Find the amount and the C.I. on ₹ 12,000 at 10% per annum compounded half-yearly.
Solution
Principal (P) = ₹12,000
Rate (R) = 10%
Time (T) = 1 years
Amount = P × {1 + r/(2 × 100)}nx2
= ₹12,000 × (1 + 10/200)2
= ₹12,000 × (210/200)2
= ₹12,000 × 21/20 × 21/20
= ₹13,230
C.I. = Amount – Principal
= ₹13230 - ₹ 12000
= ₹1230

20. Find the amount and the C.I. on ₹ 8,000 in 1 1/2 year at 20% per year compounded half-yearly.
Solution
Principal (P) = ₹8000
Rate (R) = 20%
Time = 1 ½ years = 3/2 years
Amount = Principal × {1 + r/(2 × 100)}nx2
= ₹8000 × (1 + 20/200)3/2x2
= ₹8000 × (220/200)3
= ₹ 8000 × 11/10 × 11/10 × 11/10
= ₹ 10648
C.I. = Amount – Principal
= ₹10648 - ₹8000
= ₹2648

21. Find the amount and the compound interest on ₹ 24,000 for 2 years at 10% per annum compounded yearly.
Solution
Principal (P) = ₹24,000
Time (T) = 2 years
Rate (R) = 10%
Amount = Principal – {1 + r/(2 × 100)}nx2
= ₹24,000 × (1 + 10/200)2x2
= ₹24,000 × (210/200)4
= ₹24,000 × 21/20 × 21/20 × 21/10 × 21/20
= ₹29, 172
C.I. = Amount – Principal
= ₹29,172 - ₹24,000
= ₹5,172

22. Find the amount and the compound interest on ₹ 16,000 for 3 years at 5% per annum compounded annually.
Solution
Principal (P) = ₹16,000
Time (T) = 3 years
Rate (R) = 5%
Amount = Principal × {1 + r/(2 × 100)}nx2
= ₹16,000 × (1 + 5/200)3x2
= ₹16,000 × (205/200)6
= ₹ 16,000 × 41/40 × 41/40 × 41/40 × 41/40 × 41/40 × 41/40
= ₹18,555
C.I. = Amount – Principal
= ₹18,555 - ₹16,000
= ₹2555

23. Find the amount and the compound interest on ₹ 20,000 for 1 ½ years at 10% per annum compounded half-yearly.
Solution
Principal (P) = ₹20,000
Time (T) = 1 ½ years = 3/2 years
Rate (R) = 10%
Amount = P ×{1 + r/(2 × 100)}nx2
= ₹20,000 × (1 + 10/200)3/2x2
= ₹20,000 × (210/200)3
= ₹20,000 × 21/20 × 21/20 × 21/20
= ₹23,152.50
C.I. = Amount – Principal
= ₹23,152.50 - ₹20,000
= ₹3, 152.50

24. Find the amount and the compound interest on ₹ 32,000 for 1 year at 20% per annum compounded half-yearly.
Solution
Principal (P) = ₹32,000
Time (T) = 1 year
Rate (R) = 20%
Amount = Principal × {1 + r/(2 × 100)}nx2
= ₹32,000 × (1 + 20/200)1x2
= ₹32,000 × (11/10)2
= ₹32,000 × 11/10 × 11/10
= ₹38,720
C.I. = Amount – Principal
= ₹38,720 - ₹32,000
= ₹6,720
C.I = Amount – Principal
= ₹38,720 - ₹ 32,000
= ₹6,720

25. Find the amount and the compound interest on ₹ 4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.
Solution
Principal (P) = ₹4,000
Time (T) = 2 years
Rate (R1) = 10% and rate (R2) = 15%
Amount = P(1 + R1/100) (1 + R2/100)
= ₹4,000 (1 + 10/100)(1 + 15/100)
= ₹4,000 × 11/10 × 23/20
= ₹5060
C.I. = Amount - Principal
= ₹5060 -  ₹4000
= ₹1060

26. Find the amount and the compound interest on ₹ 10,000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.
Solution
Principal (P) = ₹10,000
Time (t) = 3 years
Rate (R1) = 10%
Rate (R2) = 15%
Rate (R3) = 20%
Amount = P(1 + R1/100) (1 + R2/100)(1 + R3/100)
= ₹10,000 × (1 + 10/100)(1 + 15/100)(1 + 20/100)
= ₹10,000 × 11/10 × 23/20 × 6/5
= ₹15,180
C.I. = Amount – Principal
= ₹15,180 - ₹10,000
= ₹5180