Question

The cost of a machine depreciated by Rs 4000/- during the 1st year and by Rs 3600/- during the 2nd year. Calculate :
i) The rate of depreciation
ii) The original cost of the machine
iii) Its cost at the end of the third year

Answer 1

4000-3600=400/4000×100=10%
rate of depreciation is 10%
original cost = x
x × 9/10×9/10=3600
x=4444.44
cost at the end of 3rd year is
3600×100/81×9/10×9/10×9/10=3240

Answer 2

i) Rate of depreciation is 10%

ii) Original cost of machine is Rs.40000  

iii) The price of the machine at the end of the third year is Rs.29160.

GIVEN:

Cost of machine depreciated during 1st year = Rs.4000 and during 2nd year = Rs.3600

To find:

i) Rate of depreciation  

ii) The original cost of the machine  

iii) Its cost at the end of the third year

Solution :

i) Rate of depreciation:

Let the original price of the machine by X and the rate of the depreciation is R.

Now the cost of the machine calculated after 1 year is Y, therefore according to the formula

$Y=X\left(1-\frac{R}{100}\right)^{1}$

Now as the question says the price of machine depreciated during the first year is 4000, that means    

X–Y = 4000, hence keeping the value of  

$X-X\left(1-\frac{R}{100}\right)^{1}=4000 ; \frac{X R}{100}=4000 ; X=\frac{400000}{R}$

Now let us take Z as the cost in the second year of the machine

$\begin{array}{l}{Z=Y\left(1-\frac{R}{100}\right)^{1} \text { putting the value of } Y=X\left(1-\frac{R}{100}\right)^{1} \text { in } Z=Y\left(1-\frac{R}{100}\right)^{1} \text { we get, }} \\ {Z=X\left(1-\frac{R}{100}\right)^{1}\left(1-\frac{R}{100}\right)^{1}=X\left(1-\frac{R}{100}\right)^{2}}\end{array}$

$Z=X\left(1-\frac{R}{100}\right)^{2}$ now we know that Z – Y = 3600 so subtracting Y from z we get

$X\left(1-\frac{R}{100}\right)^{2}-X\left(1-\frac{R}{100}\right)^{1}=X\left(1-\frac{R}{100}\right)^{1}\left(X\left(1-\frac{R}{100}\right)^{1}+1\right)$

$X\left(1-\frac{R}{100}\right)^{1}\left(X\left(1-\frac{R}{100}\right)^{1}+1\right)=3600$ and in place of X put the value of X as $X=\frac{400000}{R}$, with this we find the value of R

$\frac{400000}{R}\left(1-\frac{R}{100}\right)^{1}\left(\frac{400000}{R}\left(1-\frac{R}{100}\right)^{1}+1\right)=3600$, the value of R is equal to 10%

ii) Original cost of the machine:

Now, we know that rate of depreciation = 10% then the value of the machine is  $X=\frac{400000}{R}$;  $X=\frac{400000}{10}=40000$Rupees.

iii) Cost of machine at the end of the 3rd year:

Now, Cost of machine at the end of the 3rd year is

The price of the machine depreciated after the 3rd year is H

$H=P\left(1-\frac{R}{100}\right)^{3}$

$H=40000\left(1-\frac{10}{100}\right)^{3}=29160\ \text{rupees}$