Question

A scooter was bought at Rs. 42,000. Its value depreciated at the rate of \[8\% \] per annum. Find its value after one year.

Answer 1

Hint: Here, we will find the depreciated value of a scooter. We will substitute the given values in the formula of the compound interest and simplify it further to find the value of depreciation on the scooter. Depreciation is defined as a reduction in the cost of an asset until the value of the asset becomes zero or negligible.

Formula Used:
If the amount is compounded annually, then the amount is given by \[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t}\], where \[A\] is the amount, \[P\] is the principal, \[R\] is the rate of Interest and \[t\] is the number of years.

Complete step-by-step answer:
We are given that a scooter was bought at Rs 42,000.
Since the rate of interest is depreciated, the rate of interest is \[ - 8\% \] per annum.
Now, we will use the amount formula.
Substituting \[P = 42000\], \[R = - 8\% \] and \[t = 1\] in the formula \[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t}\], we get
\[ \Rightarrow A = 42000{\left( {1 + \dfrac{{ - 8}}{{100}}} \right)^1}\]
By taking LCM, we get
\[ \Rightarrow A = 42000{\left( {1 \times \dfrac{{100}}{{100}} - \dfrac{8}{{100}}} \right)^1}\]
\[ \Rightarrow A = 42000{\left( {\dfrac{{100}}{{100}} - \dfrac{8}{{100}}} \right)^1}\]
Subtracting the terms, we get
\[ \Rightarrow A = 42000{\left( {\dfrac{{92}}{{100}}} \right)^1}\]
Dividing 42000 by 100, we get
\[ \Rightarrow A = 420 \times 92\]
Now multiplying the terms, we get
\[ \Rightarrow A = 38,640\]
Thus, the amount of a scooter is Rs 38,640.
Therefore, the value after 1 year after depreciation is Rs 38,640.

Note: Here, we have to note that if the value is depreciated then the rate of interest will be negative. Thus depreciation is the decrease in the value of the assets. The depreciation amount can be calculated by using the compound interest formula. Compound interest is defined as the interest calculated for the principal and the interest accumulated over a period of years before.