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

Question 1

The cost of a machine is supposed to depreciate each year at 12% of its value at the beginning of the year. If the machine is valued at Rs. 44,000 at the beginning of 2008, find its value :
(i) at the end of 2009.
(ii) at the beginning of 2007.

Sol:

Cost of machine in 2008 = Rs. 44,000
Depreciation rate = 12%

(i)  ∴ Cost of machine at the end of 2009
= ${\displaystyle \text{P}{(1-\frac{r}{100})}^n}$

= ${\displaystyle 44,000{(1-\frac{12}{100})}^2}$

= ${\displaystyle 44,000\times {\left(\frac{88}{100}\right)}^2}$

= Rs. 34,073.60

(ii) Cost of machine at the beginning of 2007(P)

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

⇒ 44,000 = P${\displaystyle {(1-\frac{12}{100})}^1}$

⇒ 44,000 = P${\displaystyle {\left(\frac{88}{100}\right)}^1}$

⇒ P = ${\displaystyle \frac{44,000\times 100}{88}}$ = Rs. 50,000

Question 2

The value of an article decreases for two years at the rate of 10% per year and then in the third year it increases by 10%. Find the original value of the article, if its value at the end of 3 years is Rs. 40,095.

Sol:

Let X be the value of the article.

The value of an article decreases for two years at the rate of 10% per year.

The value of the article at the end of the 1^st year is
X - 10% of X = 0.90X

The value of the article at the end of the 2^nd year is
0.90X - 10% of (0.90X) = 0.81X

The value of the article increases in the 3^rd year by 10%.

The value of the article at the end of 3^rd  year is
0.81x + 10% of (0.81x) = 0.891x

The value of the article at the end of 3 years is Rs. 40,095.
0.891X = 40,095
⇒ X = 45,000
The original value of the article is Rs. 45,000.

Question 3

According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64,000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74,088 ?

Sol:

Population in 2009 (P) = 64,000
Let after n years its population be 74,088(A)
Growth rate= 5% per annum

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

⇒ ${\displaystyle \frac{74,088}{64,000}={\left(\frac{21}{20}\right)}^n}$

⇒ ${\displaystyle \frac{9,261}{8,000}={\left(\frac{21}{20}\right)}^n}$

⇒ ${\displaystyle {\left(\frac{21}{20}\right)}^3={\left(\frac{21}{20}\right)}^n}$

On comparing, we get,
n = 3 years

Question 4

The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 2,85,120.

Sol:

Let the population in the beginning of 1998 = P
The population at the end of 1999 = 2,85,120(A)
r₁ = - 12% and r₂ = + 8%

∴ A = P${\displaystyle (1-\frac{r_1}{100})(1+\frac{r_2}{100})}$

⇒ 2,85,120 = P${\displaystyle (1-\frac{12}{100})(1+\frac{8}{100})}$

⇒ 2,85,120 = P${\displaystyle \left(\frac{22}{25}\right)\left(\frac{27}{25}\right)}$

⇒ P = ${\displaystyle \frac{2,85,120\times 25\times 25}{22\times 27}}$ = 3,00,000. 

Question 5

A sum of money, invested at compound interest, amounts to Rs. 16,500 in 1 year and to Rs. 19,965 in 3 years. Find the rate per cent and the original sum of money invested.

Sol:

Let sum of money be Rs P and rate of interest = r %
Money after 1 year = Rs. 16,500
Money after 3 years = Rs. 19,965

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

⇒ 16,500 = P${\displaystyle {(1+\frac{r}{100})}^1}$             ...(1)

For 3 years
∴  A = P${\displaystyle {(1+\frac{r}{100})}^n}$
⇒ 19,965 = P${\displaystyle {(1+\frac{r}{100})}^3}$             ...(2)

Divide eqⁿ (2) by eqⁿ (1)

${\displaystyle \frac{19,965}{16,500}=\frac{P{(1+\frac{r}{100})}^3}{P{(1+\frac{r}{100})}^1}}$

⇒ ${\displaystyle \frac{121}{100}={(1+\frac{r}{100})}^2}$

⇒ ${\displaystyle {\left(\frac{11}{10}\right)}^2={(1+\frac{r}{100})}^2}$

On comparing, we get

⇒ ${\displaystyle \frac{11}{10}=1+\frac{r}{100}}$

⇒ r = 10%

Put value of r in eqⁿ(1)
16,500 = P${\displaystyle (1+\frac{10}{100})}$

⇒ P = ${\displaystyle \frac{16,500\times 10}{11}}$ = Rs. 15,000.

Question 6

The difference between C.I. and S.I. on Rs. 7,500 for two years is Rs. 12 at the same rate of interest per annum. Find the rate of interest.

Sol:

Given : P = Rs. 7,500 and Time(n) = 2 years
Let rate of interest = y%

∴ S.I. = ${\displaystyle \frac{\text{P}\times \text{R}\times \text{T}}{100} =\frac{7500\times y\times 2}{100} =Rs.150y}$ 

∴ C.I. = P${\displaystyle {(1+\frac{r}{100})}^n-\text{P} =\text{Rs}.7,500{(1+\frac{y}{100})}^2-\text{Rs.}7,500}$

Given : C.I. : S.I. = Rs. 12

⇒ ${\displaystyle 7,500{[1+\frac{y}{100}]}^2-7,500-15y=12}$

⇒ ${\displaystyle 7,500[1+\frac{y^2}{10000}+\frac{2y}{100}]-7,500-150y=12}$

⇒ ${\displaystyle 7,500+\frac{7500y^2}{10000}+150y-7,500-150y=12}$

⇒ ${\displaystyle \frac{3y^2}{4}=12}$

⇒ y² = 16          
⇒ y = 4 %

Question 7

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

Sol:

Let Principal be Rs y and rate= r%
According to 1^st condition
Amount in 10 years = Rs 3y

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

⇒ 3y = y ${\displaystyle {(1+\frac{r}{100})}^{10}}$

⇒ 3 =${\displaystyle {(1+\frac{r}{100})}^{10}}$               ...(1)

According to 2^nd condition
Let after n years amount will be Rs 27y

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

⇒ 27y = y ${\displaystyle {(1+\frac{r}{100})}^n}$

⇒ (3)^3 =${\displaystyle {(1+\frac{r}{100})}^n}$

Put value from first equation

⇒ ${\displaystyle {\left[{(1+\frac{r}{100})}^{10}\right]}^3={(1+\frac{r}{100})}^n}$

On comparing, we get
n = 10 x 3 = 30 years

Question 8

Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying Rs.19,360 at the end of the second year and Rs. 31,944 at the end of the third year he clears the debt; find the sum borrowed by him.

Sol:

At the end of the two years the amount is
${\displaystyle {\text{A}}_1=\text{P}{(1+\frac{r}{100})}^n}$

⇒ ${\displaystyle {\text{A}}_1=\text{P}{(1+\frac{10}{100})}^2}$

Mr. Sharma paid Rs.19,360 at the end of the second year.
So for the third year the principal is A₁ - 19,360.
Also he cleared the debt by paying Rs.31,944 at the end of the third year.

${\displaystyle {\text{A}}_2=\text{P}{(1+\frac{r}{100})}^n}$

⇒  31,944 = ${\displaystyle (\text{P}{(1+\frac{10}{100})}^2-19,360){(1+\frac{10}{100})}^1}$

⇒  29040 = ${\displaystyle (\text{P}{(1+\frac{10}{100})}^2-19,360)}$

⇒  ${\displaystyle (\text{P}{(1+\frac{10}{100})}^2=48,400}$

⇒  P = Rs. 40,000
Mr. Sharma borrowed Rs. 40,000.

Question 9

The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is Rs. 15. Find the sum of money lent out.

Sol:

Let sum of money be Rs. y
To calculate S.I.
S.I. = ${\displaystyle \frac{\text{P}\times \text{R}\times \text{T}}{100}=\frac{y\times 10\times 1}{100}=Rs.\frac{y}{10}}$

To calculate C.I.(compounded half-yearly)
∴ C.I. = P${\displaystyle [{(1+\frac{r}{2\times 100})}^{n\times 2}-1]=y[{(1+\frac{10}{2x100})}^{1\times 2}-1]}$

= ${\displaystyle y[{\left(\frac{21}{20}\right)}^2-1]=\left(\frac{41}{400}\right)y}$

Given : C.I. - S.I. = Rs. 15

⇒ ${\displaystyle \left(\frac{41}{400}\right)y-\frac{y}{10}=15}$

⇒ `y/400 = 15 ⇒ y = Rs. 6,000.

Question 10

The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?

Sol:

Let Rs.X and Rs.Y be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.
Pramod will attain the age of 25 years  after 25 - 16 = 9 years
Rohit will attain the age of 25 years  after 25 -18 = 7 years

X${\displaystyle {(1+\frac{5}{100})}^9=y{(1+\frac{5}{100})}^7}$

⇒ ${\displaystyle \frac{\text{X}}{\text{Y}}=\frac{1}{{(1+\frac{5}{100})}^2}}$

⇒ ${\displaystyle \frac{\text{X}}{\text{Y}}=\frac{400}{441}}$

Pramod and Rohit should invest in 400 : 441 ratio respectively such that they will get the same sum on attaining the age of 25 years.