Answer
Verified
424.5k+ 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
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE