SChand CLASS 9 Chapter 2 Compound Interest Exercise 2(D)

 Exercise 2(D)

Question 1

The cost of a machine depreciates by 10% every year. If its present value is ₹ 18,000; what will be its value after three years.

Sol :

Principal=1800

Rate=10%

Time (n)=3 years

∴After 3 year value of machine

$=P\left(1-\frac{R}{100}\right)^{n}$

$=18000\left(1-\frac{10}{100}\right)^{3}$

$=18000\left(1-\frac{1}{10}\right)^{3}$

$=18000\left(\frac{9}{10}\right)^{3}$

$=18000\times \frac{9}{10}\times \frac{9}{10}\times \frac{9}{10}$

=18×9×9×9=13122

Question 2

The population of a town increased by 20% every year. If its present population is 2,16,000, find it population (i) after 2 years (ii) 2 years ago.

Sol :

Present population (P) =216000

Rate=20%

Time=2 year

(i) ∴After 2 year population

$=P\left(1+\frac{R}{100}\right)^{n}$

$=216000\left[1+\frac{20}{100}\right]^{2}$

$=216000 \times\left(1+\frac{1}{5}\right)^{2}$

$=216000 \times\left(\frac{6}{5}\right)^{2}$

$=216000 \times \frac{6}{5} \times \frac{6}{5}$

=8640×36=311040

(ii) 2 year ago population

$A \div\left(1+\frac{R}{100}\right)^{n}$

$=216000 \div\left(1+\frac{20}{100}\right)^{2}$

$=216000 \div\left(1+\frac{1}{5}\right)^{2}$

$=216000 \div\left(\frac{6}{5}\right)^{2}$

$=216000 \times\left(\frac{5}{6}\right)^{2}$

$= 216000 \times \frac{5}{6} \times \frac{5}{6}$

=150000

Question 3

The machinery of a certain factory is valued at ₹ 18,400 at the end of 1980. If it is supposed to depreciate each year at 8% of the value at the beginning of the year, calculate the value of the machine at the end of 1979 and 1981. 

Sol :

Value machinery at the end of 1980=18400

Rate=8%

(a) Time = 1 year ago  1979

∴$P=A\div \left[1-\fac{R}{100}\right]$

$=18400 \div\left(1-\frac{8}{100}\right)$

$= 10400 \div\left(\frac{23}{25}\right)$

$=\frac{18400 \times 25}{23}$

=20000

(b) Population after one year in 1981

$= P\left(1-\frac{R}{100}\right)$

$= 10400\left(1-\frac{8}{100}\right)$

$= 18400 \times \frac{23}{25}$

=16928

Question 4

The present value of a scooter is ₹ 15,360. If its value depreciates $12\frac{1}{2}/%$ every year, find its value after 3 years. 

Sol :

Principal=15360

Rate$=12 \frac{1}{2} \% \Rightarrow \frac{25}{2} \%$

Time (n) =3 year

After 3 year value 

$P\left(1-\frac{R}{100}\right)^{n}$

$= 1536 \times \left(1-\frac{25}{2 \times 1000}\right)^{3}$

$=15360\left(1-\frac{1}{8}\right)^{3}$

$= 15360\left[\frac{7}{8}\right]^{3}$

$=15360 \times \frac{7}{8} \times \frac{7}{8} \times \frac{7}{8}$

=10290

Question 5

A new car is purchased for ₹ 12,50,000. Its value depreciates at the rate of 10% in the first year, 8% in the 2nd year and then 6% every year. Find its value after 4 years.

Sol :

Principal=250000

Rate=10% First year, 8% Second year then

6% every year value of car after 4 year

$= P\left[1-\frac{R_{1}}{100} \right]\left[1-\frac{R_{2}}{100}\right)\left[1-\frac{R_{3}}{100}\right]\left[1-\frac{R_{4}}{100}\right]$

$=250000 \left(1-\frac{10}{100} \right) \left(1-\frac{8}{100}\right) \left(1-\frac{6}{100}\right) \left(1-\frac{6}{100}\right)$

$=250000 \left(1-\frac{1}{10} \right) \left(1-\frac{2}{25}\right) \left(1-\frac{3}{50}\right) \left(1-\frac{3}{50}\right)$

$=250000 \left(\frac{9}{10} \right) \left(\frac{23}{25}\right) \left(\frac{47}{50}\right) \left(\frac{47}{50}\right)$

$=\frac{10 \times 9 \times 23 \times 47 \times 47}{25}$

=182905.20

Question 6

The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. If the original count of the bacteria in a sample is 10000, find the bacteria count at the end of 3 hours.

Sol :

Principal=10000

First hour (R₁)=10%

Second Hour (R₂)=10%

Third Hour (R₃)=10%

After three hour number of bacteria

$=P\left(1+\frac{R_{1}}{100}\right)\left(1-\frac{R_{2}}{100}\right)\left(1+\frac{R_{3}}{100}\right)$

$=10000\left(1+\frac{10}{104}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{1}{100}\right)$

$=10000 \times\left(1+\frac{1}{10}\right)\left(1-\frac{1}{10}\right) \left(1+\frac{1}{10}\right)$

$=10000 \times \frac{11}{10} \times \frac{9}{10} \times \frac{11}{10}$

=10890

Question 7

The production of refrigerators in factory rose from 40000 to 48400 in 2 years. Find the rate of growth p.a.

Sol :

Principal=40000

Amount=40400

Time (n)=2 year

Rate=R%

∴$A=P\left[1+\frac{R}{100}\right]^n$

$\frac{A}{P}=\left[1+\frac{R}{100}\right]^n$

$\frac{48400}{40000}=\left[1+\frac{R}{100}\right]^2$

$\frac{4}{4} \times \frac{121}{100}=\left[1+\frac{R}{100}\right]^2$

$\frac{121}{100}=\left[1+\frac{R}{100}\right]^2$

$\left(\frac{11}{10}\right)^2=\left[1+\frac{R}{100}\right]^2$

$\frac{11}{10}=\left[1+\frac{R}{100}\right]^2$

$\frac{R}{100}=\frac{11}{10}-1$

$\frac{R}{100}=\frac{1}{10}$

$\frac{R}{100}=\frac{1}{10}$

$R=\frac{1}{10}\times 100$

R=10%

Question 8

The value of a flat worth ₹ 5,00,000 is depreciating at the rate of 10% p.a. In how many years will its value be reduced to ₹ 364500?

Sol :

Principal=500000

Amount=364500

Rate=10%

Time=n year

$A \Rightarrow P\left(1+\frac{R}{100}\right)^{n}$

$\frac{A}{P}=\left(1+\frac{R}{100}\right)^{n}$

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

$\frac{5}{5}\left(\frac{729}{1000}\right)=\left(1-\frac{1}{10}\right)^{n}$

$\frac{729}{1000}=\left(\frac{9}{10}\right)^{n}$

$\left(\frac{9}{10}\right)^{3}=\left(\frac{9}{10}\right)^{n}$

n=3

Question 9

Rachit bought a flat for ₹ 10 lakh and a car for ₹ 3,20,000 at the same time. The price of the flat appreciates uniformly at the rate of 20% p.a.; while the price of the car depreciates at the rate of 15% p.a. If Rachit sells the flat and car after 3 years, what will be his profit or loss?

Sol :

Price of flat=1000000

Price of car=320000

Rate of deprecation of flat (R₁)=20%

Rate of deprecation of flat (R₂)=15%

Time =3 year

(a) After 3 year value of flat $=P\left(1+\frac{R}{100}\right)^{n}$

$=1000000\left(1+\frac{20}{100}\right)^{3}$

$=1000000\left(1+\frac{1}{5}\right)^{3}$

$=1000000\left(\frac{6}{5}\right)^{3}$

$=1000000 \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}$

=1728000

(b) After 3 year value of car$=P\left(1-\frac{R_{2}}{100}\right)^{n}$

$=320000\left(1-\frac{15}{100}\right)^{3}$

$=320000\left(1-\frac{3}{20}\right)^{3}$

$=320000\left(\frac{17}{20}\right)^{3}$

$=320000 \times \frac{17}{20} \times \frac{17}{20} \times \frac{17}{20}$

$=320 \times \frac{17 \times 17 \times 17}{8}$

=40×17×17×17=196520

Total cost of flat and car

=1000000+320000=1320000

Selling price=1720000+196520

=1924520

Gain=1924520-1320000

=604520

Question 10

8000 workers were employed by a company to complete a job in 4 years. At the end of first year, 5% of the workers were retrenched. At the end of second year, 5% of those working at that time were retrenched. However to complete the job in time, the number of workers was increased by 10% of those working at the end of third year. How many workers were working during the fouth year ?

Sol :

Number of workers in the beginning=8000

After first year rate=5%

After Second year rate=5%

After Third year rate=10%

∴After third year number of workers in the beginning of fourth year

$=P\left(1-\frac{R_{1}}{100}\right)\left(1-\frac{R_{2}}{100}\right)\left(1+\frac{R_{2}}{100}\right)$

$=8000\left(1-\frac{5}{100}\right)\left(1-\frac{5}{100}\right)\left(1+\frac{10}{100}\right)$

$=8000\left(1-\frac{1}{20}\right)\left(1-\frac{1}{20}\right)\left(1+\frac{1}{10}\right)$

$=8000\left(\frac{19}{20}\right)\left(1-\frac{19}{20}\right)\left(\frac{11}{10}\right)$

=19×19×11=7942