Answer
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Hint:
Depreciation refers to when the value of something goes down over time. The value of a machine or any other article subject to wear and tear decreases with time. Relative decrease in the value of a machine is called its depreciation. The decrease in the value per year is called the rate of depreciation. Thus, if P is the principal amount of a machine and r% is the rate of depreciation per annum, the value of machine (A) after n years.
$A = P{\left( {1 - \dfrac{r}{{100}}} \right)^n}$
Here, we have to apply the above formula to calculate the value of the lamp after 3 years.
Complete step by step solution:
According to the question:
Initial cost of emergency lamp = P = Rs.1100
Rate of depreciation = r = 10%
And time = n = 3years
∴Cost of the emergency lamp (A) after 3 years$ = P{\left( {1 - \dfrac{r}{{100}}} \right)^n}$
$\begin{gathered}
= 1100{\left( {1 - \dfrac{r}{{100}}} \right)^n} \\
= 1100{\left( {1 - \dfrac{{10}}{{100}}} \right)^3} \\
= 1100{\left( {1 - \dfrac{1}{{10}}} \right)^3} \\
= 1100{\left( {\dfrac{9}{{10}}} \right)^3} \\
\end{gathered} $
$\begin{gathered}
= 1100 \times {\left( {0.9} \right)^3} \\
= 1100 \times 0.729 \\
= Rs.801.90 \\
\end{gathered} $
Therefore, the value of the emergency lamp at the end of 3 years is Rs.801.90.
∴Option (C) is correct
Note:
For such types of questions we use the formula $A = P{\left( {1 - \dfrac{r}{{100}}} \right)^n}$and after simplifying the expression of power, we multiply it.
An equal amount of depreciation is charged every year throughout the useful life of an asset. After the useful life of the asset, it’s value becomes equal to its residual value.
When the amount of depreciation and the corresponding period are plotted on a graph, it results in a straight line. The number of years a machine can be effectively used is called its life span.
If the scrap value and useful life is given then to calculate the amount of depreciation we use the given formula:
Amount of Depreciation = $\dfrac{{\left( {{\text{Cost of Asset}} - {\text{Net Residual Value}}} \right)}}{{{\text{Useful Life}}}}$
Depreciation refers to when the value of something goes down over time. The value of a machine or any other article subject to wear and tear decreases with time. Relative decrease in the value of a machine is called its depreciation. The decrease in the value per year is called the rate of depreciation. Thus, if P is the principal amount of a machine and r% is the rate of depreciation per annum, the value of machine (A) after n years.
$A = P{\left( {1 - \dfrac{r}{{100}}} \right)^n}$
Here, we have to apply the above formula to calculate the value of the lamp after 3 years.
Complete step by step solution:
According to the question:
Initial cost of emergency lamp = P = Rs.1100
Rate of depreciation = r = 10%
And time = n = 3years
∴Cost of the emergency lamp (A) after 3 years$ = P{\left( {1 - \dfrac{r}{{100}}} \right)^n}$
$\begin{gathered}
= 1100{\left( {1 - \dfrac{r}{{100}}} \right)^n} \\
= 1100{\left( {1 - \dfrac{{10}}{{100}}} \right)^3} \\
= 1100{\left( {1 - \dfrac{1}{{10}}} \right)^3} \\
= 1100{\left( {\dfrac{9}{{10}}} \right)^3} \\
\end{gathered} $
$\begin{gathered}
= 1100 \times {\left( {0.9} \right)^3} \\
= 1100 \times 0.729 \\
= Rs.801.90 \\
\end{gathered} $
Therefore, the value of the emergency lamp at the end of 3 years is Rs.801.90.
∴Option (C) is correct
Note:
For such types of questions we use the formula $A = P{\left( {1 - \dfrac{r}{{100}}} \right)^n}$and after simplifying the expression of power, we multiply it.
An equal amount of depreciation is charged every year throughout the useful life of an asset. After the useful life of the asset, it’s value becomes equal to its residual value.
When the amount of depreciation and the corresponding period are plotted on a graph, it results in a straight line. The number of years a machine can be effectively used is called its life span.
If the scrap value and useful life is given then to calculate the amount of depreciation we use the given formula:
Amount of Depreciation = $\dfrac{{\left( {{\text{Cost of Asset}} - {\text{Net Residual Value}}} \right)}}{{{\text{Useful Life}}}}$
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