ML Aggarwal Solution Class 9 Chapter 2 Compound Interest Exercise 2.3

 Exercise 2.3

Question 1

Sol :

Present Population=200000

Population after one year=Present Population$(1+\frac{10}{100})$

=200000×1.1

=220000

Population after second year$=220000\times \left(1+\frac{15}{100}\right)$

=220000(1+0.15)

=220000(1.15)

=253000

∴Population of the town at the end of the two years=253000

Question 2

Sol :

Present population=15625

Population after 3 years=Present population $\times \left(1+\frac{4}{100}\right)^3$

$=15625 \left(1+\frac{1}{25}\right)^3$

$=15625 \left(\frac{25+1}{25}\right)^3$

$=15625 \left(\frac{26}{25}\right)^3$

$=15625 \times \frac{26\times 26\times 26}{25\times 25\times 25}$

=26×26×26

=17576

∴Increase in population

=17576-15625

=1951

Question 3

Sol :

Present Population=6760000

(i) Its population 2 years hence

$=6760000\left(1+\frac{4}{100}\right)^2$

$=6760000\left(1+\frac{1}{25}\right)^2$

$=6760000\left(\frac{26}{25}\right)^2$

$=\frac{6760000\times 26\times 26}{25\times 25}$

=6760000×1.04×1.04

=7311616

(ii) Its population 2 years ago

$=6760000\left(1-\frac{4}{100}\right)^2$

$=6760000\left(1-\frac{1}{25}\right)^2$

$=67600000\left(\frac{25-1}{25}\right)^2$

$=67600000\left(\frac{24}{25}\right)^2$

=6760000×0.96×0.96

=6760000×0.9246

=6250000

Question 4

Sol :

Refrigerator value after two years$=9000\left(1-\frac{5}{100}\right)^2$

$=9000 \left(\frac{19}{20}\right)^2$

$=9000 \times \frac{19}{20}\times \frac{19}{20}$

$=\frac{16254}{2}$

=8122.5

∴Total depreciation=9000-8122.5=887.5

Question 5

Sol :

Purchased cost of scooter=24000

Rate of depreciation=5%

Scooter value after 3 years

$=24000\left(1-\frac{5}{100}\right)^3$

$=24000\left(1-\frac{1}{20}\right)^3$

$=24000\left(\frac{20-1}{20}\right)^3$

$=24000\left(\frac{19}{20}\right)^3$

$=\frac{24000\times 19\times 19 \times 19}{20\times 20\times 20}$

=3×361×19

=57×361

=20577

∴Scooter Value after three years=20577

Question 6

Sol :

Present Production=2187 quintals 

Wheat Production 2 years ago$=2187\left(1-\frac{8}{100}\right)^2$

$=2187\left(\frac{12.5-1}{12.5}\right)^2$

$=2187\left(\frac{11.5}{12.5}\right)^2$

$=2187 (0.92)^2$

=2187×0.8464

=1851

∴Wheat Production two years ago=1851

Question 7

Sol :

Rate of depreciation=5%

Present Value of property=411540

Property value 3 years ago=V_o

$411540=V_o \left(1-\frac{5}{100}\right)^3$

$411540=V_o\left(1-\frac{1}{20}\right)^3$

$\frac{411540\times 8000}{19\times 361}=V_o$

$\frac{411540 \times 8000}{6859}=V_o$

$V_o=480000$

∴Property Value 3 years ago=480000

Question 8

Sol :

Purchased cost=16000

Rate of scooter after 2 years=14400

$16000=14400\left(1+\frac{r}{100}\right)^2$

$\frac{16000}{14400}=\left(1+\frac{r}{100}\right)^2$

$\sqrt{\frac{10}{9}}=\frac{100+r}{100}$

$\sqrt{\frac{10}{9}}=\frac{100+r}{100}$

300-3r=315

3r=15

r=5

Question 9

Sol :

Here , Production of cars in year 2011-2012 =80000

Production of cars in year 2014-2015 =92160

$V=V_o \left(1+\frac{r}{100}\right)^n$

$92160=80000 \left(1+\frac{r}{100}\right)^3$

$\frac{92160}{80000}=\left(1+\frac{r}{100}\right)^3$

$\left(\frac{21}{20}\right)^3=\left(1+\frac{r}{100}\right)^3$

$\frac{21}{20}=1+\frac{r}{100}$

100+r=105

r=5% p.a

Question 10

Sol :

Here, Present Value of Machine=500000

n years after machine cost=364500

$500000=364500\left(1-\frac{10}{100}\right)^n$

$\frac{5000}{3645}=(0.9)^n$

364500=500000(1-0.1)ⁿ

$\frac{364500}{500000}=(0.9)^n$

0.729=(0.9)ⁿ

(0.9)³=(0.9)ⁿ

∴Number of years=3