# A piece of equipment cost a certain factory Rs. 600,000. If it depreciates in value, 15 percent the first year, 13.5 percent the next year, 12 percent the third year, and so on, what will be its value at the end of 10 years, all percentages applying to the original cost?

$

{\text{A}}{\text{. Rs}}{\text{. 200,000}} \\

{\text{B}}{\text{. Rs}}{\text{. 105,000}} \\

{\text{C}}{\text{. Rs}}{\text{. 405,000}} \\

{\text{D}}{\text{. Rs}}{\text{. 650,000}} \\

$

Last updated date: 23rd Mar 2023

•

Total views: 307.2k

•

Views today: 6.86k

Answer

Verified

307.2k+ views

Hint – In this question apply the formula of sum of an A.P which is given as ${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, so use this property of A.P to reach the answer.

The cost of the piece of equipment $ = 600,000{\text{ Rs}}$.

It is given that the cost of a piece of equipment depreciates in value, 15 percent the first year, 13.5 percent the next year, 12 percent the third year, and so on and all the percentages applying to the original cost.

Now we have to find out its depreciation value at the end of 10 years.

Now as we see that $\left( {15,{\text{ 13}}{\text{.5, 12,}}........} \right)$ forms an A.P.

With first term $\left( {{a_1}} \right) = 15$, common difference (d) $\left( {13.5 - 15} \right) = \left( {12 - 13.5} \right) = - 1.5$

And the number of terms (n) = 10.

Therefore the depreciation value (D.V) of the piece of equipment is original cost minus original cost multiplied by sum of all depreciation percentages.

$D.V = 600,000 - 600,000\left( {\dfrac{{15}}{{100}} + \dfrac{{13.5}}{{100}} + \dfrac{{12}}{{100}} + ............} \right)$

$D.V = 600,000 - \dfrac{{600,000}}{{100}}\left( {15 + 13.5 + 12 + .....} \right)$

Now apply the formula of sum of an A.P which is given as ${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$

Therefore the value of $\left( {15 + 13.5 + 12 + .....} \right)$

$

{S_n} = \dfrac{{10}}{2}\left( {2 \times 15 + \left( {10 - 1} \right)\left( { - 1.5} \right)} \right) \\

\Rightarrow {S_n} = 5\left( {30 - 13.5} \right) = 82.5 \\

$

Therefore depreciation value (D.V) of the piece of equipment

$

D.V = 600,000 - \dfrac{{600,000}}{{100}}\left( {82.5} \right) \\

\Rightarrow D.V = 600,000 - 495,000 = 105,000 \\

$

Hence option (B) is correct.

Note – In such types of questions first check that whether the percentage of depreciation forms any series or not if yes then check which series after that first find out the sum of this series which is the total percentage of depreciation then calculate this percentage value and subtract this value from the original cost which is the required depreciation value of the equipment after 10 years.

The cost of the piece of equipment $ = 600,000{\text{ Rs}}$.

It is given that the cost of a piece of equipment depreciates in value, 15 percent the first year, 13.5 percent the next year, 12 percent the third year, and so on and all the percentages applying to the original cost.

Now we have to find out its depreciation value at the end of 10 years.

Now as we see that $\left( {15,{\text{ 13}}{\text{.5, 12,}}........} \right)$ forms an A.P.

With first term $\left( {{a_1}} \right) = 15$, common difference (d) $\left( {13.5 - 15} \right) = \left( {12 - 13.5} \right) = - 1.5$

And the number of terms (n) = 10.

Therefore the depreciation value (D.V) of the piece of equipment is original cost minus original cost multiplied by sum of all depreciation percentages.

$D.V = 600,000 - 600,000\left( {\dfrac{{15}}{{100}} + \dfrac{{13.5}}{{100}} + \dfrac{{12}}{{100}} + ............} \right)$

$D.V = 600,000 - \dfrac{{600,000}}{{100}}\left( {15 + 13.5 + 12 + .....} \right)$

Now apply the formula of sum of an A.P which is given as ${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$

Therefore the value of $\left( {15 + 13.5 + 12 + .....} \right)$

$

{S_n} = \dfrac{{10}}{2}\left( {2 \times 15 + \left( {10 - 1} \right)\left( { - 1.5} \right)} \right) \\

\Rightarrow {S_n} = 5\left( {30 - 13.5} \right) = 82.5 \\

$

Therefore depreciation value (D.V) of the piece of equipment

$

D.V = 600,000 - \dfrac{{600,000}}{{100}}\left( {82.5} \right) \\

\Rightarrow D.V = 600,000 - 495,000 = 105,000 \\

$

Hence option (B) is correct.

Note – In such types of questions first check that whether the percentage of depreciation forms any series or not if yes then check which series after that first find out the sum of this series which is the total percentage of depreciation then calculate this percentage value and subtract this value from the original cost which is the required depreciation value of the equipment after 10 years.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE