Answer
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Hint: We will use the formula used for the compound interest because here, the depreciation will keep on happening on the new price.
Complete step-by-step answer:
Since, the cost of machine is given to be Rs. 7000
The depreciation is 8 % per annum.
Now, we will use the formula: $A = P{\left( {1 - \dfrac{R}{{100}}} \right)^n}$, where A is the final amount, P is the starting cost of the machine here, R is the rate of depreciation and n is the number of years.
Putting the given values in the formula mentioned above, we will then get the following equation:-
$ \Rightarrow A = 7000{\left( {1 - \dfrac{8}{{100}}} \right)^2}$
Simplifying the fraction in the parenthesis, we will then obtain the following equation:-
$ \Rightarrow A = 7000{\left( {1 - \dfrac{2}{{25}}} \right)^2}$
Simplifying by taking least common multiples on the right hand side, we will then obtain the following equation:-
$ \Rightarrow A = 7000{\left( {\dfrac{{25 - 2}}{{25}}} \right)^2}$
Simplifying the fraction in the parenthesis, we will then obtain the following equation:-
$ \Rightarrow A = 7000{\left( {\dfrac{{23}}{{25}}} \right)^2}$
We can write this as:-
$ \Rightarrow A = 7000 \times \dfrac{{23}}{{25}} \times \dfrac{{23}}{{25}}$
On simplifying the calculations, we will then obtain the following result:-
Amount = Rs. 5,924.80
The amount given is exactly the same.
Hence, the statement is true.
Hence, the correct option is (A).
Note:
The students must note that we have used here compound interest because every year the depreciation acts on the last depreciated price and the same thing happens in compound interest as well.
The students must commit to memory the following formula:-
$A = P{\left( {1 - \dfrac{R}{{100}}} \right)^n}$, where A is the final amount, P is the starting cost of the article, R is the rate of depreciation and n is the number of years.
The students must note that the concept of depreciation is useful in real life as well because it helps us to know the market value of our product after some years.
Complete step-by-step answer:
Since, the cost of machine is given to be Rs. 7000
The depreciation is 8 % per annum.
Now, we will use the formula: $A = P{\left( {1 - \dfrac{R}{{100}}} \right)^n}$, where A is the final amount, P is the starting cost of the machine here, R is the rate of depreciation and n is the number of years.
Putting the given values in the formula mentioned above, we will then get the following equation:-
$ \Rightarrow A = 7000{\left( {1 - \dfrac{8}{{100}}} \right)^2}$
Simplifying the fraction in the parenthesis, we will then obtain the following equation:-
$ \Rightarrow A = 7000{\left( {1 - \dfrac{2}{{25}}} \right)^2}$
Simplifying by taking least common multiples on the right hand side, we will then obtain the following equation:-
$ \Rightarrow A = 7000{\left( {\dfrac{{25 - 2}}{{25}}} \right)^2}$
Simplifying the fraction in the parenthesis, we will then obtain the following equation:-
$ \Rightarrow A = 7000{\left( {\dfrac{{23}}{{25}}} \right)^2}$
We can write this as:-
$ \Rightarrow A = 7000 \times \dfrac{{23}}{{25}} \times \dfrac{{23}}{{25}}$
On simplifying the calculations, we will then obtain the following result:-
Amount = Rs. 5,924.80
The amount given is exactly the same.
Hence, the statement is true.
Hence, the correct option is (A).
Note:
The students must note that we have used here compound interest because every year the depreciation acts on the last depreciated price and the same thing happens in compound interest as well.
The students must commit to memory the following formula:-
$A = P{\left( {1 - \dfrac{R}{{100}}} \right)^n}$, where A is the final amount, P is the starting cost of the article, R is the rate of depreciation and n is the number of years.
The students must note that the concept of depreciation is useful in real life as well because it helps us to know the market value of our product after some years.
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