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=Vo
$411540=V_o \left(1-\frac{5}{100}\right)^3$
$411540=V_o\left(1-\frac{1}{20}\right)^3$
$\frac{411540\times 8000}{19\times 361}=V_o$
$\frac{411540 \times 8000}{6859}=V_o$
$V_o=480000$

∴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

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