SELINA Solution Class 9 Compound interest (Using formula) Chapter 3 Exercise 3E

Question 1

Simple interest on a sum of money for 2 years at 4% is Rs .450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half yearly.

Sol:

1^st case
Given :  S.I. = Rs 450 ; Time = 2 years and Rate = 4%
∴ Principle = ${\displaystyle \frac{\text{I}\times 100}{\text{R}\times \text{T}}=\frac{450\times 100}{4\times 2}}$ = Rs. 5625.

2^nd case ( compounded half-yearly )
P = Rs. 5,625 ; n = 1 year and r = 4%
∴ A = P${\displaystyle {(1+\frac{r}{2\times 100})}^{n\times 2}=5,625{(1+\frac{4}{2\times 100})}^{1\times 2}}$

=${\displaystyle 5625{\left(\frac{51}{50}\right)}^2}$

= Rs. 5852.25

∴ C.I. = 5,852.25 - 5,625 = Rs. 227.25

Question 2

Find the compound interest to the nearest rupee on Rs. 10,800 for ${\displaystyle 2\frac{1}{2}}$ years at 10% per annum.

Sol:

Given : P = Rs. 10,800 ; Time = ${\displaystyle 2\frac{1}{2}}$ years and Rate = 10% p.a.

For 2 years
A = P${\displaystyle {(1+\frac{r}{100})}^n}$ = 10,800( 1 + 10/100 )^2 = Rs. 13,068

For ${\displaystyle \frac{1}{2}}$ year
∴ A = P${\displaystyle {(1+\frac{r}{2\times 100})}^{n\times 2} =13068{(1+\frac{10}{2\times 100})}^{\frac{1}{2}\times 2}}$

= 13068 x ${\displaystyle \frac{21}{20}}$ = Rs. 13721.40 = Rs. 13721 ( nearest rupee)
∴ Rs.13,721 - Rs.10,800 = Rs.2,921

Question 3

The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is Rs.97,200, find:

1.   Its value after 2 years.
2.   Its value when it was purchased.

Sol:

(i) Present value of machine(P) =  Rs.97,200 
Depreciation rate = 10%

∴ Value of machine after 2 years = P${\displaystyle {(1-\frac{r}{100})}^n}$

                                                     = 97,200${\displaystyle {(1-\frac{10}{100})}^2}$

                                                     = 97,200${\displaystyle {\left(\frac{9}{10}\right)}^2}$
                                                     = Rs. 78732.

(ii) Present value of machine(A) = Rs.97,200
Depreciation rate = 10% and time = 2 years
To calculate the cost 2 years ago

∴ A = P${\displaystyle {(1-\frac{r}{100})}^n}$

⇒ 97,200 = P${\displaystyle {(1-\frac{10}{100})}^2}$

⇒ 97,200 = P${\displaystyle {\left(\frac{9}{10}\right)}^2}$

⇒ P = Rs. 97,200 x ${\displaystyle {\left(\frac{10}{9}\right)}^2}$  = 1,20,000

Question 4

Anuj and Rajesh each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. Rajesh received Rs. 64 more than Anuj. Find the money lent by each and interest received.

Sol:

Let the sum of money lent by both Rs. y
For Anuj
P = Rs.y ; rate = 8% and time = 2 years

∴ S.I. = ${\displaystyle \frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{y\times 8\times 2}{100}=\frac{4y}{25}}$

 For Rajesh
P = Rs. y ; rate = 8% and time = 2 years

∴ C.I. = P${\displaystyle [{(1+\frac{r}{100})}^n-1]=y[{(1+\frac{8}{100})}^2-1]=\frac{104y}{625}}$

Given : C.I. - Rs. 64

⇒ ${\displaystyle \frac{104y}{625}-\frac{4y}{25}=64}$

⇒ ${\displaystyle \frac{4y}{625}=64}$

⇒ ${\displaystyle y=\frac{64\times 625}{4}}$  = Rs. 10,000

Interest received by Anuj = ${\displaystyle \frac{4\times 10,000}{25}}$ = Rs. 1600

Interest received by Rajesh = ${\displaystyle \frac{104\times 10,000}{625}}$ = Rs. 1664.

Question 5

Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest on Rs. 4,715 for 5 years, both at the rate of 5% per annum.

Sol:

Given : Principal = Rs.4,715; time = 5 years and rate= 5% p.a.

∴ S.I. = ${\displaystyle \frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{4715\times 5\times 5}{100}=Rs.1,178.75}$

Then C.I. = Rs.1,178.75 x 4 = Rs. 4,715
Time = 2 years and rate = 5%

∴ C.I. = P${\displaystyle [{(1+\frac{r}{100})}^n-1]}$

⇒ 4,715 = P${\displaystyle [{(1+\frac{5}{100})}^2-1]}$

⇒ 4,715 = P${\displaystyle \left[\frac{41}{400}\right]}$

⇒ P = Rs. ${\displaystyle \frac{4,715\times 400}{41}}$ = Rs. 46,000.

Question 6

A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to Rs. 4,950, find the sum invested.

Sol:

Given : C.I. for the 2^nd year = Rs. 4,950 and rate = 15%

Then, C.I. = P${\displaystyle [{(1+\frac{r}{100})}^n-1]}$

⇒ 4,950 = P${\displaystyle [{(1+\frac{15}{100})}^1-1]}$

⇒ 4,950 = P${\displaystyle \left[\frac{3}{20}\right]}$

⇒ P = ${\displaystyle \frac{4,950\times 20}{3}}$

⇒ P = Rs. 33,000.

Then amount at the end of 2^nd year= Rs. 33,000
For first 2 years
 A = Rs. 33,000 ; r₁ = 10%

∴ A = P${\displaystyle (1+\frac{r_1}{100})}$

⇒ 33,000 = P${\displaystyle (1+\frac{10}{100})}$

⇒ 33,000 = P${\displaystyle \left(\frac{11}{10}\right)}$

⇒ P = ${\displaystyle \frac{33,000\times 10}{11}}$ = Rs. 30,000
The sum invested is Rs.30,000.

Question 7

A sum of money is invested at 10% per annum compounded half yearly. If the difference of amounts at the end of 6 months and 12 months is Rs.189, find the sum of money invested.

Sol:

Let the sum of money be Rs. y
and rate = 10% p.a. compounded half yearly

 For first 6 months
∴ A = P${\displaystyle {(1+\frac{r}{2\times 100})}^{n\times 2}=y{(1+\frac{10}{2}\times 100)}^{\frac{1}{2}\times 2}=\left(\frac{21}{20}\right)y}$

 For first 12 months
∴ A = P${\displaystyle {(1+\frac{r}{2\times 100})}^{n\times 2}=y{(1+\frac{10}{2}\times 100)}^{1\times 2}=\left(\frac{441}{400}\right)y}$

Given: The difference between the above amounts = Rs.189
⇒ ${\displaystyle \left(\frac{441}{400}\right)y -\left(\frac{21}{20}\right)y=189}$

⇒ ${\displaystyle \left(\frac{21}{400}\right)y=189}$

⇒ ${\displaystyle y=\frac{189\times 400}{21}}$

y = 3600.

Question 8

Rohit borrows Rs. 86,000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit's profit in the transaction at the end of two years.

Sol:

P = Rs. 86,000; time = 2 years and rate = 5% p.a.
To calculate S.I.

∴ S.I. = ${\displaystyle \frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{86,000\times 5\times 2}{100}=\text{Rs.}8,600}$

 To calculate C.I.
∴  C.I. = P${\displaystyle [{(1+\frac{r}{100})}^n-1]}$

= ${\displaystyle 86,000[{(1+\frac{5}{100})}^2-1]}$

= ${\displaystyle 86,000\left(\frac{41}{400}\right)}$ = Rs. 8,815

Profit = C.I. - S.I. = Rs.8,815 - Rs.8,600 = Rs.215

Question 9

The simple interest on a certain sum of money for 3 years at 5% per annum is Rs.1,200. Find the amount and the compound interest due on this sum of money at the same rate and after 2 years. Interest is reckoned annually.

Sol:

Let Rs.X be the sum of money.
Rate = 5 % p.a. Simple interest = Rs.1,200, n = 3 years.

${\displaystyle 1,200=\frac{\text{X}\times 5\times 3}{100}}$

⇒ X = ${\displaystyle \frac{12,00,000}{15}}$

⇒ X = 8,000

The amount due and the compound interest on this sum of money at the same rate and after 2 years.
P = Rs. 8,000 ; rate = 5% p.a., n = 3 years

∴ A = P${\displaystyle {(1+\frac{r}{100})}^n}$

⇒ A = 8,000${\displaystyle {(1+\frac{5}{100})}^2}$

⇒ A = 8,000( 1.1025 )

⇒ A = 8,820

C.I. = A - P
⇒ C.I. = 8,820 - 8,000
⇒ C.I. = 820.
The amount due after 2 years is Rs. 8,820 and the compound interest is Rs. 820.

Question 10

Nikita invests Rs.6,000 for two years at a certain rate of interest compounded annually. At the end of first year it amounts to Rs.6,720. Calculate:
(a) The rate of interest.
(b) The amount at the end of the second year.

Sol:

Let X % be the rate of interest.
P = Rs. 6,000, n = 2 years, A = Rs.6,720
For the first year
A = P${\displaystyle {(1+\frac{r}{100})}^n}$

⇒ 6,720 = 6,000${\displaystyle {(1+\frac{\text{X}}{100})}^1}$

⇒ 6,720 - 6,000 = 60X
⇒ X = 12

The rate of interest is X % = 12 %.
The amount at the end of the second year.

A = P${\displaystyle {(1+\frac{r}{100})}^n}$

⇒ A = 6,000${\displaystyle {(1+\frac{12}{100})}^2}$

⇒ A = 6,000${\displaystyle {\left(\frac{112}{100}\right)}^2}$

⇒ A = 7,526.40
The amount at the end of the second year = Rs. 7,526.40