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.