Answer

Verified

421.8k+ views

Hint: Find the price after 1 year by calculating. Then find the price after two years. Subtract the price after two years from the cost price to get depreciation in the price of the machine after 2 years.

Complete step-by-step answer:

Let C be the cost price of the machine.

Let d be the rate of depreciation.

Given Data

C = 2,50,000 and d = 10%

Let P1 and P2 be the price of the machine after 1 year and 2 years respectively.

P1 = Cost price – rate loss due to depreciation in first year

Rate loss due to depreciation in 1 year can be determined as = $\dfrac{{{\text{C}} \times {\text{d}}}}{{100}}$

= $\dfrac{{2,50,000{\text{ }} \times {\text{ 10}}}}{{100}}$

= 25,000/-

Hence, P1 = C – $\dfrac{{{\text{C}} \times {\text{d}}}}{{100}}$

= 2,50,000 – 25,000

P1 = 2,25,000/-

Now, P2 = Price after one year – Rate loss due to depreciation in second year

Rate loss due to depreciation in second year =$\dfrac{{{\text{C}} \times {\text{d}}}}{{100}}$, for P2 Cost price C is nothing but P1, hence =$\dfrac{{{\text{P1}} \times {\text{d}}}}{{100}}$

= $\dfrac{{2,25,000{\text{ }} \times {\text{ 10}}}}{{100}}$

= 22,500/-

Hence, P2 = P1 – (P1 x d)/100

= 2,25,000 – 22,500

P2 = 2,02,500/-

The depreciation after two years can be computed as = Cost price – Price after two years

= C – P2

= 2,50,000 – 2,02,500

= 47,500/-

Note: In order to solve these types of questions, the key is to use the percentage formula correctly to find out the rate lost due to depreciation. Another thing to keep in mind is that while calculating the price after two years the original cost price should not be used but instead the price after one year should be used because there’s been a change already in that whole year.

Complete step-by-step answer:

Let C be the cost price of the machine.

Let d be the rate of depreciation.

Given Data

C = 2,50,000 and d = 10%

Let P1 and P2 be the price of the machine after 1 year and 2 years respectively.

P1 = Cost price – rate loss due to depreciation in first year

Rate loss due to depreciation in 1 year can be determined as = $\dfrac{{{\text{C}} \times {\text{d}}}}{{100}}$

= $\dfrac{{2,50,000{\text{ }} \times {\text{ 10}}}}{{100}}$

= 25,000/-

Hence, P1 = C – $\dfrac{{{\text{C}} \times {\text{d}}}}{{100}}$

= 2,50,000 – 25,000

P1 = 2,25,000/-

Now, P2 = Price after one year – Rate loss due to depreciation in second year

Rate loss due to depreciation in second year =$\dfrac{{{\text{C}} \times {\text{d}}}}{{100}}$, for P2 Cost price C is nothing but P1, hence =$\dfrac{{{\text{P1}} \times {\text{d}}}}{{100}}$

= $\dfrac{{2,25,000{\text{ }} \times {\text{ 10}}}}{{100}}$

= 22,500/-

Hence, P2 = P1 – (P1 x d)/100

= 2,25,000 – 22,500

P2 = 2,02,500/-

The depreciation after two years can be computed as = Cost price – Price after two years

= C – P2

= 2,50,000 – 2,02,500

= 47,500/-

Note: In order to solve these types of questions, the key is to use the percentage formula correctly to find out the rate lost due to depreciation. Another thing to keep in mind is that while calculating the price after two years the original cost price should not be used but instead the price after one year should be used because there’s been a change already in that whole year.

Recently Updated Pages

The branch of science which deals with nature and natural class 10 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

Assertion CNG is a better fuel than petrol Reason It class 11 chemistry CBSE

How does pressure exerted by solid and a fluid differ class 8 physics CBSE

Number of valence electrons in Chlorine ion are a 16 class 11 chemistry CBSE

What are agricultural practices? Define

What does CNG stand for and why is it considered to class 10 chemistry CBSE

The rate of evaporation depends on a Surface area b class 9 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

State whether the following statement is true or false class 11 physics CBSE

A night bird owl can see very well in the night but class 12 physics CBSE