Answer
454.2k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)