# ICSE Solutions for Selina Concise Chapter 2 Compound Interest (without using formula) Class 9 Maths

### Exercise 2(A)

1. Rs.16,000 is invested at 5% compound interest compounded per annum. Use the table, given below, to find the amount in 4 years.
 Year ↓ Initial amount(Rs.) Interest (Rs.) Final amount(Rs.) 1st 16,000 800 16,800 2nd 3rd 4th 5th
 Year ↓ Initial amount(Rs.) Interest (Rs.) Final amount(Rs.) 1st 16,000 800 16,800 2nd 16,800 840 17,640 3rd 17,640 882 18,522 4th 18,522 926.10 19,448.10 5th 19,448.10 972.405 20,420.505

Thus, the amount in 4 years is Rs. 19448.10.

2.1. Calculate the amount and the compound interest on :
Rs. 6,000 in 3 years at 5% per year.

For 1st year,
P = Rs. 6,000; R = 5%, and T = 1 year
∴ Interest = Rs. 6,000×5× 1/110 = Rs. 300.
And, amount = Rs. (6,000 + 300) = Rs. 6,300

For 2nd year,
P = Rs. 6,300; R = 5%, and T = 1 year
∴ Interest = Rs. 6,300×5× 1/100 = Rs. 315
And, amount = Rs. (6,300 + 315) = Rs. 6,615

For 3rd year,
P = Rs. 6,615; R = 5% and T = 1 year
∴ Interest = Rs. 6,615×5× 1/100 = Rs. 330.75
And, Amount = Rs. (6,615 + 330.75) = Rs. 6,945.75
∴ C.I. accrued = Final amount – Intitial Principal
= Rs. ( 6,945.75 – 6,000 )
= Rs. 945.75

2.2. Calculate the amount and the compound interest on :
Rs. 8,000 in 2.5 years at 15% per year.

For 1st year,
P = Rs. 8,000; R = 15%, and T = 1 year
∴ Interest = Rs. 8,000×15× 1/100 = Rs. 1200
And, amount = Rs. (8,000 + 1200) = Rs. 9,200

For 2nd year,
P = Rs. 9,200; R = 15%, and T = 1 year
∴ Interest = Rs. 9,200×15× 1/100= Rs. 1,380.
And, amount = Rs. (9,200 + 1,380) = Rs. 10,580
For the last 1212 year,
P = Rs. 10,580 ; R = 15% and T = 1212 year
∴ Interest = Rs. 10,580×15× 1/100 = Rs. 793.50
And, Amount = Rs. (10,580 + 793.50) = Rs. 11373.50
∴ C.I. accrued = Final amount – Intitial Principal
= Rs. ( 11,373.50 – 8,000 )
= Rs. 3373.50

3.1. Calculate the amount and the compound interest on :
₹ 4,600 in 2 years when the rates of interest of successive years are 10%and 12% respectively.

For 1st year
P = Rs. 4600
R = 10%
T = 1 year.
I = 4600×10× 1/100 = Rs. 460
A = 4600 + 460 = Rs. 5060

For 2nd year
P = Rs. 5060
R = 12%
T = 1 year
I = 5060×12× 1/100=607.20
A= 5060 + 607.20 = Rs. 5667.20
Compound interest = 5667.20 - 4600 = Rs. 1067.20
Amount after 2 years = Rs. 5667.20

3. Calculate the amount and the compound interest on :
Rs. 16,000 in 3 years, when the rates of the interest for successive years are 10%, 14% and 15% respectively.

For 1st year
P = Rs. 16000
R = 10%
T = 1 year
I = 16000×10× 1/100 = Rs. 1600
A = 16000 + 1600 = 17600

For 2nd year,
P = Rs. 17600
R = 14%
T = 1 year
I = 17600×14× 1/100 = Rs. 2464.
A = 1760 + 24654 = Rs. 20064

For 3rd year,
P = Rs. 20064
R = 15%
T = 1 year
I = 20064×15× 1/100 = 3009.60
Amount after 3 years = 20064 + 3009.60 = Rs. 23073.60
Compound interest = 23073.60 - 16000 = Rs. 7073.60

4. Find the compound interest, correct to the nearest rupee, on Rs. 2,400 for 2.5 years at 5 per cent per annum.

For 1st years
P = Rs. 2400
R = 5%
T = 1 year
I = 2400×5× 1/100 = 120
A = 2400 + 120 = Rs. 2520

For 2nd year
P = Rs. 2520
R = 5%
T = 1 year
I = 2520×5× 1/100 = Rs. 126.
A = 2520 + 126 = Rs. 2646

For final 1/2 year,
P = Rs. 2646
R = 5%
T = 1212 year
I = 2646×5× 1/100× 2 = Rs. 66.15
Amount after 1/2 years = 2646 + 66.15 = Rs. 2712.15
Compound interest = 2712.15 – 2400 = Rs. 312.15

5. Calculate the compound interest for the second year on Rs. 8,000/- invested for 3 years at 10% per annum.

For 1st year
P = Rs. 8000
R = 10%
T = 1 year
I = 8000×10× 1/100 = 800
A = 8000 + 800 = Rs. 8800

For 2nd year
P = Rs. 8800
R = 10%
T = 1 year
I = 8800×10× 1/100
Compound interest for 2nd years = Rs. 880

6. A borrowed Rs. 2,500 from B at 12% per annum compound interest. After 2 years, A gave Rs. 2,936 and a watch to B to clear the account. Find the cost of the watch.

For 1st year,
P = Rs. 2500
R = 12%
T = 1 year
I = 2500×12× 1/100 = Rs. 300
Amount = 2500 + 300 = Rs. 2800

For 2nd year,
P = Rs. 2800
R = 12%
T = 1 year
I = 2800×12× 1/100 = Rs. 336
Amount = 2800 + 336 = Rs. 3136
Amount repaid by A to B = Rs. 2936
The amount of watch =Rs. 3136 – Rs. 2936 = Rs. 200

7. How much will Rs. 50,000 amount to in 3 years, compounded yearly, if the rates for the successive years are 6%, 8% and 10% respectively?

Interest for the first year = P×R×T/100
= 50,000×6×1/100
= Rs. 3,000
Amount for the first year = Rs. 50,000 + Rs. 3,000 = Rs. 53,000
Interest for the second year = P×R×T/100
= 53,000×8×1/100
= Rs. 4,240

Amount for the second year = Rs. 53,000 + Rs. 4,240 = Rs. 57,240
Interest for the third year = P×R×T/100
= 57,240×10×1/100
= Rs. 5,724
Amount for the third year = Rs. 57,240 + Rs. 5,724 = Rs. 62,964
Hence, the amount will be Rs. 62,964.

8. Meenal lends Rs. 75,000 at C.I. for 3 years. If the rate of interest for the first two years is 15% per year and for the third year it is 16%, calculate the sum Meenal will get at the end of the third year.

Interest for the first year = P×R×T/100
= 75,000×15×1/100
= Rs. 11,250

Amount for the first year = Rs. 75,000 + Rs. 3,000 = Rs. 86,250
Interest for the second year = P×R×T/100
= 86,250×15×1/100
= Rs. 12,937.5

Amount for the second year = Rs. 86,250 + Rs. 12,937.5 = Rs. 99,187.5
Interest for the third year = P×R×T/100
= 99,187.5×16×1/100
= Rs. 15,870

Amount for the third year = Rs. 99,187.5 + Rs. 15,870 = Rs. 1,15,057.5
Hence, the sum Meenal will get at the end of the third year is Rs. 1,15,057.5

9. Govind borrows Rs18,000 at 10% simple interest. He immediately invests the money borrowed at 10% compound interest compounded half-yearly. How much money does Govind gain in one year ?

To calculate S.I.
P = Rs. 18,000; R = 10% and T = 1 year
S.I. = Rs. 18,000×10×1/100 = Rs. 1,800.
To calculate C.I.

For 1st half- year :
P= Rs. 18,000; R = 10 % and T= 1212 year
Interest = Rs. 18,000×10×1/100×2 = Rs. 9000
Amount = Rs. 18,000 + Rs. 900 = Rs. 18,900

For 2nd year :
P= Rs. 18,900; R = 10% and T = 1212 year
Interest= Rs. 18,900×10×1/100×2 = Rs. 945.
Amount= Rs. 18,900 + Rs. 945 = Rs. 19,845
Compound interest = Rs. 19,845 – Rs. 18,000 = Rs. 1,845
His gain = Rs. 1,845 – Rs. 1,800= Rs. 45

10. Find the compound interest on Rs. 4,000 accrued in three years, when the rate of interest is 8% for the first year and 10% per year for the second and the third years.

Interest for the first year = P×R×T/100
= 4,000×8×1/100
= Rs. 3,20
Amount for the first year = Rs. 4,000 + Rs. 3,20 = Rs. 4,320

Interest for the second year = P×R×T/100
= 4,320×10×1/100
= Rs. 432
Amount for the second year = Rs. 4,320 + Rs. 432 = Rs. 4,752

Interest for the third year = P×R×T/100
= 4,752×10×1/100
= Rs. 475.20
Amount for the third year = Rs. 4,752 + Rs. 475.20 = Rs. 5,227.20

So, the compound interest = Rs. 5,227.20 – Rs. 4,000 = Rs. 1,227.20
Hence, the amount will get at the end of the third year is Rs. 1,227.50.

### Exercise 2(B)

1. Calculate the difference between the simple interest and the compound interest on Rs. 4,000 in 2 years at 8% per annum compounded yearly.

For 1st year
P = Rs. 4000
R = 8
T = 1 year
I = 4000×8×1/100= 320
A = 4000 + 320 = Rs. 4320

For 2nd year
P = Rs. 4320
R = 8%
T = 1 year
I = 4320×8×1/100 = Rs. 345.60
A = 4320 + 345.60 = 4665.60

Compound interest = Rs. 4665.60 – Rs. 4000 = Rs. 665.60
Simple interest for 2 years = 4000×8×2/100= Rs. 640
Difference of CI and SI = 665.60 – 640 = Rs 25.60.

2. A man lends Rs. 12,500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly ; find the difference between the C.I. fo the first year and the compound interest for the third year.

For 1st year
P = Rs. 12500
R = 12%
R = 1 year
I = 12500×12×1/100= Rs. 1500
A = 12500 + 1500 = Rs. 14000

For 2nd year
P = Rs. 1400
R = 15%
T = 1 year
I = 14000×15×1/100 = Rs. 2898
A = 1400 + 2100 = Rs. 16100

For 3rd year
P = Rs. 16100
R = 18%
T = 1 year
I = 16100×18×1/100 = Rs. 2898
A = 16100 + 2898 = Rs. 3998

Difference between the compound interest of the third year and first year
= Rs. 2893 – Rs. 1500
= Rs. 1398

3. A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by Rs. 96, find the sum of money.

Let money be Rs100

For 1st year
P = Rs. 100; R = 8% and T = 1 year.
Interest for the first year = Rs. 100×8×1/100 = Rs. 8
Amount = Rs. 100 + Rs. 8 = Rs. 108

For 2nd year
P = Rs.108; R = 8% and T= 1year.
Interest for the second year= Rs. 108×8×1/100 = Rs. 8.64

Difference between the interests for the second and first year = Rs. 8.64 - Rs. 8 = Rs. 0.64
Given that interest for the second year exceeds the first year by Rs. 96.
When the difference between the interests is Rs. 0.64, principal is Rs. 100

When the difference between the interests is Rs96, principal = Rs. 96×100/0.64 = Rs. 15,000.

4. A man borrows Rs. 6,000 at 5% C.I. per annum. If he repays Rs. 1,200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Given that the amount borrowed = Rs. 6,000.
Rate per annum = 5%

Interest on Rs. 6,000 = 5/100 x Rs. 6,000 = Rs. 300
So, amount at the end of the first year = Rs. 6,000+ Rs. 300 = Rs.6,300

Amount left to be paid = Rs. 6,300 – Rs. 1,200 = Rs. 5,100.

Interest on Rs. 5,100 = 5/100 x Rs. 5,100 = Rs. 255

So, amount at the end of the second year = Rs. 5,100 + Rs. 255 = Rs. 5,355.

Amount left to be paid = Rs. 5,355 – Rs. 1,200 = Rs. 4,155.
Hence, the amount of the loan outs tan ding at the beginning of the third year is Rs. 4,155.

5. A man borrows Rs. 5,000 at 12 percent compound interest payable every six months. He repays Rs. 1,800 at the end of every six months. Calculate the third payment he has to make at the end of 18 months in order to clear the entire loan.

For 1st six months :
P = Rs. 5,000, R = 12% and T = 6 months = 1212 year
∴ Interest = 5,000×1/2×100 = Rs. 300.
And, Amount = Rs. 5,000 + Rs. 300 = Rs. 5,300
Since, money repaid = Rs. 1,800
Balance = Rs. 5,300 – Rs. 1,800 = Rs. 3,500

For 2nd six months :
P = Rs. 3,500, R = 12% and T = 6 months = 1212 year
∴ Interest = 3,500×12×1/2×100 = Rs. 210.And, Amount = Rs. 3,500 + Rs. 210 = Rs. 3,710.

Again money repaid = Rs. 1,800
Balance = Rs. 3,710 – Rs. 1,800 = Rs. 1,910.

For 3rd six months :
P = Rs. 1,910, R = 12% and T = 6 months = 1212 year
∴ Interest = 1,910×12×1/2×100 = Rs. 114.60.

And, Amount = Rs. 1,910 + Rs. 114.60 = Rs. 2,024.60
Thus, the 3rd payment to be made to clear the entire loan is 2,024.60

6. On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is Rs. 180/- Find the sum lent out, if the rate of interest in both the cases is 10% per annum.Answer

Let principal p = Rs. 100.
R = 10%
T = 1 year
SI = 100×10× 1/100 = Rs. 10.

Compound interest payable half yearly
R = 5% half yearly
T = 1212 year = 1 half year
For first 1212 year
I = 100×5× 1/100= Rs. 5
A = 100 + 5 = Rs. 105

For second year
P = Rs. 105
I = 105×5× 1/100= Rs. 5.25

Total compound interest = 5 + 5.25 = Rs. 10.25
Difference of CI and SI = 10.25- 10 = Rs. 0.25
When difference in interest is Rs. 10.25, sum = Rs. 100.

If the difference is Rs. 1, sum = 100/0.25

If the difference is Rs. = 180, sum = 1000/0.25×180 = Rs. 72,000.

7. A manufacturer estimates that his machine depreciates by 15% of its value at the beginning of the year. Find the original value (cost) of the machine, if it depreciates by Rs. 5,355 during the second year.Answer

Let the original cost of the machine = Rs. 100.
∴ Depreciation during the 1st year = 15% of Rs. 100 = Rs, 15.
Value of the machine at the beginning of the 2nd year
= Rs.100 – Rs.15 = Rs. 85

∴ Depreciation during the 2nd year = 15% of Rs. 85 = Rs. 12.75

Now, when depreciation during 2nd year = Rs, 12.75,
Original cost = Rs. 100

∴ When depreciation during 2nd year = Rs. 5,355.
original cost = Rs. 100/12.75 ×5,355 = Rs. 42,000.
Hence, original cost of the machine is Rs. 42,000.

8.1. A man invest 5,600 at 14% per annum compound interest for 2 years. Calculate : The interest for the first year.

For 1st years
P = Rs. 5600
R = 14%
T = 1 year
I = 5600×14×1/100= Rs. 784

8.2. A man invest Rs. 5,600 at 14% per annum compound interest for 2 years. Calculate : The amount at the end of the first year.

Amount at the end of the first year = 5600 + 784 = Rs. 6384.

8.3. A man invest Rs. 5,600 at 14% per annum compound interest for 2 years. Calculate : The interest for the second year, correct to the nearest rupee.

For 2nd year
P = 6384
R = 14%
R = 1 year
I = 6,384×14× 1/100
= Rs. 803.76
= Rs. 894 (nearly)

9. A man saves Rs. 3,000 every year and invests it at the end of the year at 10% compound interest. Calculate the total amount of his savings at the end of the third years.

Savings at the end of every year = Rs. 3000

For 2nd year
P = Rs. 3000
R = 10%
T = 1 year
I = 3000×10× 1/100= 300
A = 3000 + 300 = Rs. 3300

For third year, savings = 3000
P = 3000 + 3300 = Rs. 6300
R = 10%
T = 1 year
I = 6300×10× 1/100 = Rs. 630.
A = 6300 + 630 = Rs. 6930

Amount at the end of 3rd year
= 6930 + 3000
= Rs. 9930

10. A man borrows Rs. 10,000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year and 42% of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt ?

The amount borrowed = Rs.10,000.
lnterest for the first year = P×R×T/100
= 10,000×5× 1/100
= Rs. 500

So, amount at the end of the first year
= Rs. 10,000 + Rs. 500
= Rs. 10,500

The man pays 35% of Rs.10,500 at the end of the first year
= 35/100 × 10,500 = Rs.3,675

So, amount left to be paid = Rs. 10,500 – Rs. 3,675 = Rs. 6,825.
lnterest for the second year = P×R×T/100
= 6,825×5×1/100
= Rs. 341.25

amount at the end of the second year
= Rs. 6825 + Rs. 341.25
= Rs. 7,166.25

The man pays 42% of Rs. 7166.25 at the end of the of the second year = 42100 × 7166.25 = Rs. 3,009.825

So, amount left to be paid = Rs. 7,166.25 – Rs. 3009.825
= Rs. 4156.425

lnterest for the third year = P×R×T/100

= 4,156.425×5×1/100
= Rs. 207.82125

So, amount at the end of the third year
= Rs. 4,156.425 + Rs. 207.82125 = Rs. 4,364.24625

Hence, he must pay Rs. 4,364.24625 at the end of the third year in order to dear the debt.

### Exercise 2(C)

1. A sum is invested at compound interest, compounded yearly. If the interest for two successive years is Rs. 5,700 and Rs. 7,410. calculate the rate of interest.

Rate of interest = Difference in the interest of the two consecutive periods×100/

C.I. of preceding year × Time %

= (7410 - 5700)× 100/5700× 1%

= 30%

2. A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are Rs. 650 and Rs. 760.50; find the rate of interest.

∵ Difference between the C.I. of two successive half-years
= Rs. 760.50 - Rs. 650 = Rs. 110.50
⇒ Rs. 110.50 is the interest of one half-year on Rs. 650
∴ Rate of interest = Rs. 100×I/P×T %
= (100×110.50)/(650× 1/2) %
= 34 %

3.1. A certain sum amounts to Rs. 5,292 in two years and Rs. 5,556.60 in three years, interest being compounded annually. Find : the rate of interest.

Amount in two years= Rs. 5,292
Amount in three years= Rs. 5,556.60
Difference between the amounts of two successive years
= Rs. 5,556.60 - Rs. 5,292 = Rs. 264.60
⇒ Rs. 264.60 is the interest of one year on Rs. 5,292
∴ Rate of interest = Rs. 100×I/P×T %
= (100×264.60)/(5,292×1) %
= 5%

3.2. A certain sum amounts to Rs. 5,292 in two years and Rs. 5,556.60 in three years, interest being compounded annually. Find: the original sum.

Let the sum of money = Rs. 100
Interest on it for 1st year= 5% of Rs. 100 = Rs. 5
⇒ Amount in one year= Rs. 100 + Rs. 5 = Rs. 105
Similarly, amount in two years = Rs. 105 + 5% of Rs. 105
= Rs. 105 + Rs. 5.25
= Rs. 110.25
When amount in two years is Rs. 110.25, sum = Rs. 100
⇒ When amount in two years is Rs. 5,292,
Sum = Rs. 100×5,292/110.25 = Rs. 4,800.

4. The compound interest, calculated yearly, on a certain sum of money for the second year is Rs. 1,089 and for the third year it is Rs. 1,197.90. Calculate the rate of interest and the sum of money.

(i) C.I. for second year = Rs. 1,089
C.I. for third year = Rs. 1,197.90
∵ Difference between the C.I. of two successive years
= Rs. 1,197.90 – Rs. 1089 = Rs. 108.90
⇒ Rs. 108.90 is the interest of one year on Rs.1089.

∴ Rate of interest = Rs. 100×I/P×T %
= 100×108.90/1089×1% = 10%

(ii) Let the sum of money = Rs.100
∴ Interest on it for 1st year = 10% of Rs.100= Rs.10

⇒ Amount in one year = Rs. 100 + Rs. 10 = Rs. 110
Similarly, C.I. for 2nd year = 10% of Rs. 110 = Rs. 11
When C.I. for 2nd year is Rs. 11, sum = Rs. 100
When C.I. for 2nd year is Rs. 1089, sum = Rs. 100×1089×11
= Rs. 9,900.

5. Mohit invests Rs. 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. 9,440. Calculate :

(i) the rate of interest per annum.

(ii) the amount at the end of the second year.

(iii) the interest accrued in the third year.

For 1st year
P = Rs. 8,000; A = 9,440 and T= 1 year
Interest = Rs. 9,440 – Rs. 8,000 = Rs. 1,440
Rate = I×100/P×T%
= 1,440×100/8,000×1% = 18%

For 2nd year
P= Rs. 9,440; R = 18% and T= 1 year
Interest = Rs 9,440×18×1×100= Rs. 1,699.20
Amount = Rs. 9,440 + Rs. 1,699.20 = Rs. 11,139.20

For 3rd year
P = Rs. 11,139.20; R = 18 % and T= 1year
Interest = Rs. 11,139.20×18×1/100= Rs. 2,005.06

6. Geeta borrowed Rs. 15,000 for 18 months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to Rs. 15,600; calculate :
(i) the rate of interest per annum.
(ii) the total amount of money that Geeta must pay at the end of 18 months in order to clear the account

For 1st half – year :
P= Rs. 15,000; A= Rs. 15,600 and T = ½ year
Interest = Rs. 15,600 – Rs. 15,000= Rs. 600
Rate = [I×100]/[P×T] %
= [600×100]/[15,000× 1/2]% = 8% .

For 2nd half – year :
P = Rs. 15,600; R = 8% and T = 1212 year
Interest = Rs. 15,600×8×0.5/100 = Rs. 624
Amount = Rs. 15,600 + Rs. 624 = Rs. 16,224

For 3rd half – year :
P = Rs. 16,224; R = 8 % and T = 1212 ye ar
Interest = Rs. 16,224×8×0.5/100= Rs. 648.96
Amount = Rs. 16,224 + Rs. 648.96 = Rs. 16,872.96.

7. Ramesh invests Rs. 12,800 for three years at the rate of 10% per annum compound interest. Find:
(i) the sum due to Ramesh at the end of the first year.
(ii) the interest he earns for the second year.
(iii) the total amount due to him at the end of the third year.

For 1st year :
P = Rs. 12,800; R = 10 % and T = 1 year
Interest = Rs. 12,800×10×1/100 = Rs. 1,280.
Amount = Rs. 12,800 + Rs. 1,280 = Rs. 14,080.

For 2nd year :
P = Rs. 14,080; R = 10 % and T = 1 year
Interest = Rs. 14,080×10×1/100 = Rs. 1,408.
Amount = Rs. 14,080 + Rs. 1,408 = Rs. 15,488

For 3rd year :
P = Rs. 15,488; R = 10 % and T = 1 year
Interest = Rs. 15,488×10×1/100 = Rs. 1,548.80
Amount = Rs. 15,488 + Rs. 1,548.80 = Rs. 17,036.80

8. Rs. 8,000 is lent out at 7% compound interest for 2 years. At the end of the first year Rs. 3,560 are returned. Calculate :
(i) the interest paid for the second year.
(ii) the total interest paid in two years.
(iii) the total amount of money paid in two years to clear the debt.

(i) For 1st year :
P = Rs. 8,000; R = 7 % and T = 1 year
Interest = Rs. 8,000×7×1/100 = Rs. 560.
Amount = Rs. 8,000 + Rs. 560 = Rs. 8,560
Money returned = Rs. 3,560
Balance money for 2nd year= Rs. 8,560 – Rs. 3,560 = Rs. 5,000

For 2nd year :
P = Rs. 5,000; R = 7 % and T = 1 year.
Interest paid for the second year = Rs. 5,000×7×1/100 = Rs. 350

(ii) The total interest paid in two years= Rs. 350 + Rs. 560 = Rs. 910

(iii) The total amount of money paid in two years to clear the debt = Rs. 8,000+ Rs. 910 = Rs. 8,910

9. The cost of a machine depreciated by Rs. 4,000 during the first year and by Rs. 3,600 during the second year. Calculate :
1. The rate of depreciation.
2. The original cost of the machine.
3. Its cost at the end of the third year.

(i) Difference between depreciation in value between the first and second years Rs. 4,000 – Rs. 3,600 = Rs. 400.
⇒ Depreciation of one year on Rs. 4,000 = Rs. 400
⇒ Rate of depreciation = 40/4000 ×100% = 10%

(ii) Let Rs.100 be the original cost of the machine.
Depreciation during the 1st year = 10% of Rs.100 = Rs.10
When the values depreciates by Rs.10 during the 1st year, Original cost = Rs.100
⇒ When the depreciation during 1st year = Rs. 4,000

Original Cost = 100/10 ×4000 = Rs. 40,000

The original cost of the machine is Rs. 40,000.

(iii) Total depreciation during all the three years
= Depreciation in value during(1st year + 2nd year + 3rd year)
= Rs. 4,000 + Rs. 3,600 + 10% of (Rs. 40,000 – Rs. 7,600)
= Rs. 4,000 + Rs. 3,600 + Rs. 3,240
= Rs.10,840

The cost of the machine at the end of the third year
= Rs. 40,000 – Rs.10,840 = Rs. 29,160

10. Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs. 252.

Let the sum of money be Rs.100.
Rate of interest = 10% p.a.
Interest at the end of 1st year = 10% of Rs. 100 = Rs. 10
Amount at the end of 1st year = Rs. 100 + Rs. 10 = Rs. 110
Interest at the end of 2nd year = 10% of Rs. 110 = Rs. 11
Amount at the end of 2nd year = Rs. 110 + Rs. 11 = Rs.121
Interest at the end of 3rd year =10% of Rs. 121= Rs. 12.10
Difference between interest of 3rd year and 1st year
= Rs. 12.10 – Rs. 10 = Rs. 2.10
When difference is Rs. 2.10, principal is Rs. 100
When difference is Rs. 252, principal = 100× 252/2.10 = Rs.12,000.

11. A man borrows Rs.10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 30% of the sum borrowed. How much money is left unpaid just after the second year ?

For 1st year :
P = Rs. 10,000; R = 10% and T = 1 year
Interest = Rs. 10,000×10×1/100= Rs.1,000
Amount at the end of 1st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000
Money paid at the end of 1st year = 30% of Rs. 10,000 = Rs. 3,000
∴ Principal for 2nd year = Rs. 11,000 – Rs. 3,000 = Rs. 8,000

For 2nd year :

P = Rs. 8,000; R = 10% and T = 1 year
Interest = Rs. 8,000×10×1/100 = Rs. 800
Amount at the end of 2nd year = Rs. 8,000 + Rs. 800 = Rs. 8,800
Money paid at the end of 2nd year = 30% of Rs. 10,000 = Rs. 3,000
∴ Principal for 3rd year = Rs. 8,800 – Rs. 3,000 =Rs. 5,800.

12. A man borrows Rs.10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year ?

For 1st year:

P = Rs. 10,000; R = 10% and T = 1 year
Interest = Rs. 10,000×10×1/100 = Rs. 1,000
Amount at the end of 1st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000
Money paid at the end of 1st year = 20% of Rs. 11,000 = Rs. 2,200
∴ Principal for 2nd year = Rs. 11,000 – Rs. 2,200 = Rs. 8,800

For 2nd year :

P = Rs. 8,800; R = 10% and T= 1 year
Interest = Rs. 8,800×10×1/100= Rs. 880
Amount at the end of 2nd year = Rs. 8,800 + Rs. 880 = Rs. 9,680
Money paid at the end of 2nd year = 20% of Rs. 9,680 = Rs.1,936
∴ Principal for 3rd year =Rs. 9,680 – Rs. 1,936 = Rs. 7,744.

### Exercise 2(D)

1. What sum will amount of Rs. 6,593.40 in 2 years at C.I. , if the rates are 10 per cent and 11 per cent for the two successive years ?

Let principal (p) = Rs. 100

For 1st year :
P = Rs. 100
R = 10%
T = 1 year
I = 100×10×1/100 = Rs. 10.
A = 100 + 10 = Rs. 110

For 2nd year :
P = Rs. 110
R = 11%
T = 1 year
I = 110×11×1/100 = Rs. 12.10.
A = 110 + 12.10 = Rs. 122.10

If Amount is Rs. 122.10 on a sum of Rs. = 100

If amount is Rs. 1, sum = 100122.10

If amount is Rs. 6593.40, sum = 100/122.10×6593.40 = Rs. 5400

2. The value of a machine depreciated by 10% per year during the first two years and 15% per year during the third year. Express the total depreciation of the machine, as per cent, during the three years.

Let the value of machine in the beginning = Rs. 100

For 1st year depreciation = 10% of Rs. 100 = Rs. 100
Value of machine for second year = 100 – 10 = Rs. 90

For 2nd year depreciation = 10% of 90 = Rs. 9
Value of machine for third year = 90 – 9 = Rs. 81

For 3rd year depreciation = 15% of 81 = Rs. 12.15
Value of machine at the end of third year = 81 - 12.15 = Rs. 68.85

Net depreciation = Rs. 100 - Rs. 68.85 = Rs. 31.15 Or 31.15%

3. Rachna borrows Rs. 12,000 at 10 percent per annum interest compounded half-yearly. She repays Rs. 4,000 at the end of every six months. Calculate the third payment she has to make at end of 18 months in order to clear the entire loan. Answer

For 1st half – year :
P = Rs. 12,000; R = 10 % and T = 1/2 year
Interest = Rs. [12,000×10×1]/[100 ×2] = Rs. 600
Amount = Rs. 12,000 + Rs. 600 = Rs. 12,600

Money paid at the end of 1st half year = Rs. 4,000
Balance money for 2nd half-year = Rs. 12,600 – Rs. 4,000 = Rs. 8,600

For 2nd half – year :

P = Rs. 8,600; R = 10% and T = 1/2 year
Interest = Rs. [8,600×10 ×1]/[100×2] =Rs. 430
Amount = Rs. 8,600 + Rs. 430 = Rs. 9,030

Money paid at the end of 2nd half-year = Rs. 4,000
Balance money for 3rd half – year = Rs. 9,030 – Rs. 4,000 = Rs. 5,030

For 3rd half-year

P = Rs. 5,030; R = 10% and T = 1/2 year
Interest = Rs. [5,030×10×1]/[100×2] = Rs. 251.50

Amount = Rs. 5,030 + Rs. 251.50 = Rs. 5,281.50

4. On a certain sum of money, invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is Rs. 2,652. Find the sum. Answer

Let Principal = Rs.100

For 1st year
P = Rs. 100; R = 10 % and T = 1 year
Interest = Rs.[100×10×1]/100 = Rs. 10
Amount = Rs. 100 + Rs. 10 = Rs. 110

For 2nd year

P = Rs. 110; R = 10 % and T = 1 year
Interest = Rs. [110×10×1]/ = Rs. 11
Amount = Rs. 110 + Rs. 11 = Rs. 121

For 3rd year

P = Rs. 121; R = 10 % and T = 1 year
Interest = Rs [121×10×1]/= Rs. 12.10

Sum of C.I. for 1st year and 3rd year = Rs. 10 + Rs. 12.10 = Rs. 22.10

When sum is Rs. 22.10, principal is Rs. 100

When sum is Rs. 2,652, principal = Rs. [100×2652]/[22.10] = Rs. 12,000.

5. During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates by Rs. 2,640 during the second financial year of its purchase.

Let original value of machine = Rs. 100

For 1st year
P = Rs. 100; R = 12% and T = 1 year
Depreciation in 1st year = Rs [100×12×1]/ = Rs.12
Value at the end of 1st year = Rs. 100 – Rs. 12 = Rs. 88

For 2nd year
P = Rs. 88; R = 12% and T = 1 year
Depreciation in 2nd year = Rs.[88×12×1]/= Rs. 10.56
When depreciation in 2nd year is Rs.10.56, original cost is Rs.100
When depreciation in 2nd year is Rs.2,640, original cost = [100×2640 ]/[10.56] = Rs. 25,000

6. Find the sum on which the difference between the simple interest and compound interest at the rate of 8% per annum compounded annually would be Rs. 64 in 2 years.

Let Rs. X be the sum.
Simple Interest (I) = [X×8×1]/100 = 0.08X

Compound interest

For 1st year :
P = Rs. X, R = 8% and T = 1
⇒ Interest (I) = [X×8×1]/100 = 0.08X

For 2nd year :
P = Rs. X + Rs. 0.08X = Rs.1.08X
⇒ Interest (I) = [1.08X×8×1]/100 = 0.0864X

The difference between the simple interest and compound interest at the rate of 8% per annum compounded annually should be Rs. 64 in 2 years.
⇒ Rs. 0.08X – Rs. 0.0864X = Rs. 64
⇒ Rs. 0.0064X = Rs. 64
⇒ x = Rs.10,000
Hence the sum is Rs.10,000

7. A sum of Rs. 13,500 is invested at 16% per annum compound interest for 5 years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of first year.
(iii) the interest for the second year, correct to the nearest rupee.

For 1st year :
P = Rs. 13,500; R = 16% and T = 1 year
Interest = Rs. [13,500×16×1]/ = Rs. 2,160
Amount = Rs. 13,500 + Rs. 2,160= Rs. 15,660

For 2nd year :
P = Rs. 15,660; R = 16% and T = 1 year
Interest = Rs. [15,660×16×1]/ = Rs. 2,505.60 = Rs. 2,506

8.Saurabh invests Rs. 48,000 for 7 years at 10% per annum compound interest. Calculate:
(i) the interest for the first year.
(ii) the amount at the end of second year.
(iii) the interest for the third year.

For 1st year :
P = Rs. 48,000; R = 10 % and T = 1 year
Interest = Rs. [48,000×10×1]/ = Rs. 4,800
Amount = Rs. 48,000 + Rs. 4,800 = Rs. 52,800

For 2nd year :

P = Rs. 52,800; R = 10 % and T = 1 year
Interest = Rs. [52,800×10 ×1]/ = Rs. 5,280
Amount = Rs. 52,800 + Rs. 5,280 = Rs. 58,080

For 3rd year :
P = Rs. 58,080; R = 10% and T = 1 year
Interest = Rs. [58,080×10×1]/` = Rs. 5,808

9. Ashok borrowed Rs. 12,000 at some rate on compound interest. After a year, he paid back Rs.4,000. If the compound interest for the second year is Rs. 920, find:
1. The rate of interest charged
2. The amount of debt at the end of the second year

(i) Let X% be the rate of interest charged.

For 1st year :
P = Rs.12,000, R = X% and T = 1
⇒ Interest (I) = [12,000×X×1]/ = 120X

For 2nd year:
After a year, Ashok paid back Rs. 4,000.
P = Rs.12,000 + Rs. 120X – Rs. 4,000 = Rs. 8,000 + Rs.120X
⇒ Interest (I) = [( 8000 + 120X)×X]/100 = ( 80X + 1.20X2 )

The compound interest for the second year is Rs. 920.
Rs. ( 80X + 1.20X) = Rs. 920
⇒ 1.20X2 + 80X – 920 = 0
⇒ 3X2 + 200X – 2300 = 0
⇒ 3X2 + 230X – 30X – 2300 = 0
⇒ X(3X + 230) -10(3X + 230) = 0
⇒ (3X + 230)(X – 10) = 0
⇒ X = -230/3 or X = 10
As rate of interest cannot be negative so x = 10.
Therefore, the rate of interest charged is 10%.

(ii) For 1st year :
Interest = Rs.120X = Rs.1200

For 2nd year :
Interest = Rs.( 80X + 1.20X) = Rs.920

The amount of debt at the end of the second year is equal to the addition of principal of the second year and interest for the two years.
Debt = Rs. 8,000 + Rs. 1200 + Rs. 920 = Rs. 10,120

10. On a certain sum of money, lent out at C.I., interests for first, second and third years are Rs. 1,500; Rs. 1,725 and Rs. 2,070 respectively. Find the rate of interest for the:(i) second year(ii) third year. Answer

Total interest obtained in the first year = Rs, 1500
lnterest for the second year – Total interest obtained in the first year
= Rs. 1725 – Rs. 1500
= Rs. 225

Rate of interest for the second year = 225/1500 ×100 = 15%

Interest for the third year – Interest for the second year
= Rs. 2,070 – Rs. 1,725
= Rs. 345

Rate of interest for the third year
= 345/[1,725] ×100 = 20%

So, rate of interest for the second year and third year are 15% and 20% respectively.