Question

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}} \\\$

Answer 1

Hint – The depreciates rates every year can help forming a series, if the common difference of the series that is the difference between consecutive terms remains constant through than it forms an A.P, use the respective formula of series to get the answer at the end of 10 years.

Complete step-by-step 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 – We can easily eliminate some options directly after going through the question as the original price was 600000 and it depreciates in value every year hence eventually the cost after 10 years will be lesser than the original price so any option above 60000 or very close to 600000 is eventually wrong. Make the habit of option elimination by general common sense in examinations as it helps save a lot of time.