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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.
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