SELINA Solution Class 9 Compound interest (without using formula) Chapter 2 Exercise 2C

Question 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.

Sol:

Rate of interest = ${\displaystyle \frac{\text{Difference in the interest of the two consecutive periods}\times 100}{\text{C.I. of preceeding year}\times \text{Time}}\%}$

= ${\displaystyle \frac{(7410-5700)\times 100}{5700\times 1}\%}$

= 30 %

Question 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.

Solution 1:

∵ 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. ${\displaystyle \frac{100\times \text{I}}{\text{P}\times \text{T}}\%}$
= ${\displaystyle 650\times \frac{1}{2}}$ % 
= 34 %

Solution 2:

Let sum of money = P, Rate = r % half-yearly.

Simple interest for I half-year on sum = Rs. 650.

Simple interest for II half-year on sum = P + 650 = Rs. 760.50

∵ 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.

Simple interest (S.I.) = ${\displaystyle \frac{P\times R\times T}{100}}$

∴ ${\displaystyle 110.50=\frac{650\times R\times 1}{100}}$

∴ 110.50 = 6.5 × R

∴ ${\displaystyle \frac{110.50}{6.5}}$ = R

∴ R = 17% Per half-yearly

∴ Rate of interest = 17 × 2 = 34% p.a.

Question 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.

Sol:

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. ${\displaystyle \frac{100\times \text{I}}{\text{P}\times \text{T}}\%}$
= ${\displaystyle \frac{100\times 264.60}{5,292\times 1}\%}$ 
= 5%

Question 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.

Sol:

Let the sum of money = Rs. 100
Interest on it for 1^st 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. ${\displaystyle \frac{100\times 5,292}{110.25}}$ = Rs. 4,800.

Question 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.

Sol:

(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. ${\displaystyle \frac{100\times \text{I}}{\text{P}\times \text{T}}}$ %

                             = ${\displaystyle \frac{100\times 108.90}{1089\times 1}}$ % = 10%

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

⇒ Amount in one year = Rs. 100 + Rs. 10 = Rs. 110
Similarly, C.I. for 2^nd year = 10% of Rs. 110 = Rs. 11
When C.I. for 2^nd year is Rs. 11, sum = Rs. 100
When C.I. for 2^nd year is Rs. 1089, sum = Rs. ${\displaystyle \frac{100\times 1089}{11}}$
= Rs. 9,900.

Question 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.

Sol:

For 1^st year
P = Rs. 8,000; A = 9,440 and T= 1 year
Interest = Rs. 9,440 - Rs. 8,000 = Rs. 1,440
Rate = ${\displaystyle \frac{\text{I}\times 100}{\text{P}\times \text{T}}\%}$

= ${\displaystyle \frac{1,440\times 100}{8,000\times 1}}$% = 18%

For 2^nd year
P= Rs. 9,440; R = 18% and T= 1 year

Interest = Rs ${\displaystyle \frac{9,440\times 18\times 1}{100}}$= Rs. 1,699.20

Amount = Rs. 9,440 + Rs. 1,699.20 = Rs. 11,139.20

For 3^rd year

P = Rs. 11,139.20; R = 18 % and T= 1year

Interest = Rs. ${\displaystyle \frac{11,139.20\times 18\times 1}{100}}$= Rs. 2,005.06

Question 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.

Sol:

For 1^st 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" xx 100 ]/["P" xx "T"] %

= [600 xx 100]/[15,000 xx 1/2]` % = 8% .

For 2^nd half - year : 
P = Rs. 15,600; R = 8% and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{15,600\times 8\times \frac{1}{2}}{100}}$ = Rs. 624

Amount = Rs. 15,600 + Rs. 624 = Rs. 16,224

For 3^rd half - year :

P = Rs. 16,224; R = 8 % and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{16,224\times 8\times \frac{1}{2}}{100}}$ = Rs. 648.96

Amount = Rs. 16,224 + Rs. 648.96 = Rs. 16,872.96.

Question 7

For 1^st 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" xx 100 ]/["P" xx "T"] %

= [600 xx 100]/[15,000 xx 1/2]` % = 8% .

For 2^nd half - year : 
P = Rs. 15,600; R = 8% and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{15,600\times 8\times \frac{1}{2}}{100}}$ = Rs. 624

Amount = Rs. 15,600 + Rs. 624 = Rs. 16,224

For 3^rd half - year :

P = Rs. 16,224; R = 8 % and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{16,224\times 8\times \frac{1}{2}}{100}}$ = Rs. 648.96

Amount = Rs. 16,224 + Rs. 648.96 = Rs. 16,872.96.

Sol:

For 1^st year :
P = Rs. 12,800; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{12,800\times 10\times 1}{100}}$ = Rs. 1,280.

Amount = Rs. 12,800 + Rs. 1,280 = Rs. 14,080.

For 2^nd year :
P = Rs. 14,080; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{14,080\times 10\times 1}{100}}$ = Rs. 1,408.

Amount = Rs. 14,080 + Rs. 1,408 = Rs. 15,488

For 3^rd year :
P = Rs. 15,488; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{15,488\times 10\times 1}{100}}$ = Rs. 1,548.80

Amount = Rs. 15,488 + Rs. 1,548.80 = Rs. 17,036.80

Question 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.

Sol:

(i) For 1^st year : 
P = Rs. 8,000; R = 7 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{8,000\times 7\times 1}{100}}$ = Rs. 560.

Amount = Rs. 8,000 + Rs. 560 = Rs. 8,560
Money returned = Rs. 3,560
Balance money for 2^nd year= Rs. 8,560 - Rs. 3,560 = Rs. 5,000

For 2^nd year :
P = Rs. 5,000; R = 7 % and T = 1 year.

Interest paid for the second year = Rs. ${\displaystyle \frac{5,000\times 7\times 1}{100}}$ 
= 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

Question 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.

Sol:

(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 = ${\displaystyle \frac{400}{4000} \times 100\%}$ = 10%

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

Original Cost = ${\displaystyle \frac{100}{10}\times 4000}$ = 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(1^st year + 2^nd year + 3^rd 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

Question 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.

Sol:

Let the sum of money be Rs.100.
Rate of interest = 10% p.a.
Interest at the end of 1^st year = 10% of Rs. 100 = Rs. 10
Amount at the end of 1^st year = Rs. 100 + Rs. 10 = Rs. 110
Interest at the end of 2^nd year = 10% of Rs. 110 = Rs. 11
Amount at the end of 2^nd year = Rs. 110 + Rs. 11 = Rs.121
Interest at the end of 3^rd year =10% of Rs. 121= Rs. 12.10
Difference between interest of 3^rd year and 1^st 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 = ${\displaystyle \frac{100\times 252}{2.10}}$ = Rs.12,000.

Question 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 ?

Sol:

For 1^st year :
P = Rs. 10,000; R = 10% and T = 1 year

Interest = Rs. ${\displaystyle \frac{10,000\times 10\times 1}{100}}$= Rs.1,000

Amount at the end of 1^st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000

Money paid at the end of 1^st year = 30% of Rs. 10,000 = Rs. 3,000

∴ Principal for 2^nd year = Rs. 11,000 - Rs. 3,000 = Rs. 8,000

For 2^nd year :

P = Rs. 8,000; R = 10% and T = 1 year

Interest = Rs. ${\displaystyle \frac{8,000\times 10\times 1}{100}}$ = Rs. 800

Amount at the end of 2^nd year = Rs. 8,000 + Rs. 800 = Rs. 8,800

Money paid at the end of 2^nd year = 30% of Rs. 10,000 = Rs. 3,000

∴ Principal for 3^rd year = Rs. 8,800 - Rs. 3,000 =Rs. 5,800.

Question 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 ?

Sol:

For 1^st year :

P = Rs. 10,000; R = 10% and T = 1 year

Interest = Rs. ${\displaystyle \frac{10,000\times 10\times 1}{100}}$ = Rs. 1,000

Amount at the end of 1^st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000

Money paid at the end of 1^st year = 20% of Rs. 11,000 = Rs. 2,200

∴ Principal for 2^nd year = Rs. 11,000 - Rs. 2,200 = Rs. 8,800

For 2^nd year :

P = Rs. 8,800; R = 10% and T= 1 year

Interest = Rs. ${\displaystyle \frac{8,800\times 10\times 1}{100}}$= Rs. 880

Amount at the end of 2^nd year = Rs. 8,800 + Rs. 880 = Rs. 9,680

Money paid at the end of 2^nd year = 20% of Rs. 9,680 = Rs.1,936

∴ Principal for 3^rd year =Rs. 9,680 - Rs. 1,936 = Rs. 7,744.

SELINA Solution Class 9 Compound interest (without using formula) Chapter 2 Exercise 2B

Question 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.

Sol:

For 1^st year
P = Rs. 4000
R = 8
T = 1 year

I = ${\displaystyle \frac{4000\times 8\times 1}{100}}$ = 320

A = 4000 + 320 = Rs. 4320

For 2^nd year
P = Rs. 4320
R = 8%
T = 1 year

I = ${\displaystyle \frac{4320\times 8\times 1}{100}}$ = Rs. 345.60

A = 4320 + 345.60 = 4665.60

Compound interest = Rs. 4665.60 - Rs. 4000 = Rs. 665.60

Simple interest for 2 years = ${\displaystyle \frac{4000\times 8\times 2}{100}}$ = Rs. 640

Difference of CI and SI = 665.60 - 640 = Rs 25.60.

Question 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.

Sol:

For 1^st year
P = Rs. 12500
R = 12%
R = 1 year

I = ${\displaystyle \frac{12500\times 12\times 1}{100}}$ = Rs. 1500

A = 12500 + 1500 = Rs. 14000

For 2^nd year
P = Rs. 1400
R = 15%
T = 1 year

I = ${\displaystyle \frac{14000\times 15\times 1}{100}}$ = Rs. 2100

A = 14000 + 2100 = Rs. 16100

For 3^rd year
P = Rs. 16100
R = 18%
T = 1 year

I = ${\displaystyle \frac{16100\times 18\times 1}{100}}$ = Rs. 2898

A = 16100 + 2898 = Rs. 18,998

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

Question 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.

Sol:

Let money be Rs100
For 1^st year
P = Rs. 100; R = 8% and T = 1 year.
Interest for the first year = Rs. ${\displaystyle \frac{100\times 8\times 1}{100}}$ = Rs. 8
Amount = Rs. 100 + Rs. 8 = Rs. 108

For 2^nd year
P = Rs.108; R = 8% and T= 1year.
Interest for the second year= Rs. ${\displaystyle \frac{108\times 8\times 1}{100}}$ = 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. ${\displaystyle \frac{96\times 100}{0.64}}$ = Rs. 15,000.

Question 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.

Sol:

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

Interest on Rs. 6,000 = ${\displaystyle \frac{5}{100}}$ 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 = ${\displaystyle \frac{5}{100}}$ 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.

Question 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.

Sol:

For 1^st six months :
P = Rs. 5,000, R = 12% and T = 6 months = ${\displaystyle \frac{1}{2}}$ year
∴ Interest = ${\displaystyle \frac{5,000\times 12\times 1}{2\times 100}}$ = 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 2^nd six months :
P = Rs. 3,500, R = 12% and T = 6 months = ${\displaystyle \frac{1}{2}}$ year

∴ Interest = ${\displaystyle \frac{3,500\times 12\times 1}{2\times 100}}$ = 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 3^rd six months :
P = Rs. 1,910, R = 12% and T = 6 months = ${\displaystyle \frac{1}{2}}$ year
∴ Interest = ${\displaystyle \frac{1,910\times 12\times 1}{2\times 100}}$ = Rs. 114.60.

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

Question 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.

Sol:

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

SI = ${\displaystyle \frac{100\times 10\times 1}{100}}$ = Rs. 10.

Compound interest payable half yearly
R = 5% half yearly
T = ${\displaystyle \frac{1}{2}}$ year = 1 half year
For first ${\displaystyle \frac{1}{2}}$ year
I = ${\displaystyle \frac{100\times 5\times 1}{100}}$= Rs. 5

A = 100 + 5 = Rs. 105
For second year
P = Rs. 105  

I = ${\displaystyle \frac{105\times 5\times 1}{100}}$ = 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 = ${\displaystyle \frac{100}{0.25}}$

If the difference is Rs. = 180, sum = ${\displaystyle \frac{100}{0.25}\times 180}$ = Rs. 72,000.

Question 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.

Sol:

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

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

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

∴ when depreciation during 2nd year = Rs. 5,355.
original cost = Rs. ${\displaystyle \frac{100}{12.75}\times 5,355}$ = Rs. 42,000.
Hence, original cost of the machine is Rs. 42,000.

Question 8.1

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

Sol:

For 1^st years
P = Rs. 5600
R = 14%
T = 1 year

I = ${\displaystyle \frac{5600\times 14\times 1}{100}}$ = Rs. 784

Question 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.

Sol:

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

Question 8.3

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

Sol:

For the first year:
P = Rs. 5,600, N = 1 year and R = 14%
We have,
S.I. = ${\displaystyle \frac{\text{PNR}}{100}=\frac{5,600\times 1\times 14}{100}=Rs.784}$.

And Amount at the end of first year P + S.I. = Rs. 5600 + Rs. 784 = Rs. 6,384.

Now for the second year:

For 2^nd year
P = 6384, R = 14%, N = 1 year

S.I. = ${\displaystyle \frac{\text{PNR}}{100}=\frac{6,384\times 14\times 1}{100}=Rs.893.76}$
To the nearest rupee, it is Rs. 894 (nearly).

Question 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.

Sol:

Savings at the end of every year = Rs. 3000
For 2^nd year
P = Rs. 3000
R = 10%
T = 1 year

I = ${\displaystyle \frac{3000\times 10\times 1}{100}}$ = 300

A = 3000 + 300 = Rs. 3300

For third year, savings = 3000
P = 3000 + 3300 = Rs. 6300
R = 10%
T = 1 year

I = ${\displaystyle \frac{6300\times 10\times 1}{100}}$ = Rs. 630.
A = 6300 + 630 = Rs. 6930

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

Question 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 ?

Sol:

For the first year,
P₁ = 10,000, R = 5%
A₁ = 10,000(1 + ${\displaystyle \frac{5}{100}}$) 

= 10,000 × ${\displaystyle \frac{105}{100}}$

= Rs. 10,500.

At the end of the first year, he repays 35% of the sum borrowed so he repays the amount = 10,500 × ${\displaystyle \frac{35}{100}}$ = Rs. 3,500.

Left amount = 10,500 - 3,500 = Rs. 7,000.

For the second year,
P₂ = Rs. 7000, R = 5%

A₂ = 7000(1 + ${\displaystyle \frac{5}{100}}$)

= 7000 × ${\displaystyle \frac{105}{100}}$

= Rs. 7,350.

At the end of the second year, he repays 42% of the sum borrowed so he repays the amount =

 

10000 × ${\displaystyle \frac{42}{100}}$ = Rs. 4200

 

Left amount = 7350 − 4200 = Rs. 3150

 

For the Third year

 

P₃ ​= Rs. 3150, R = 5%

 

A₃ ​= 3150( 1+ ${\displaystyle \frac{5}{100}}$ ​)

 

= 3150 ×${\displaystyle \frac{105}{100}}$

 

= Rs. 3307.50

 

Hence he pays Rs. 3307.50 at the end of the third year in order to clear the debt.