Ramu purchased a boat for ₹ 16000. If the cost of the boat is depreciating at the rate of \[5\% \] p.a., calculate its value after 2 years.
Answer
Verified
466.5k+ views
Hint: Here, we need to find the value of the boat after 2 years. We will find the amount of depreciation during the first year and use it to find the value of the boat at the end of the first year. Similarly, we will find the amount of depreciation during the second year and use it to find the value of the boat at the end of 2 years.
Complete step-by-step answer:
Depreciation is the loss in value of something due to wear and tear, use, or becoming obsolete. It is usually calculated on the value of the asset at the beginning of the year.
It is given that the rate of depreciation is \[5\% \] p.a.
This means that the value of the boat at the end of the first year is \[5\% \] less than its value at the beginning of the year.
Similarly, the value of the boat at the end of the second year will be \[5\% \] less than its value at the beginning of the second year.
First, we will calculate the amount of depreciation during the first year.
The value lost is \[5\% \] of the cost of the boat.
Therefore, we get
Amount of depreciation during the first year \[ = 5\% \times 16000 = \dfrac{5}{{100}} \times 16000 = 800\]
Therefore, the value of the boat lost due to depreciation during the first year is ₹ 800.
The value of the boat at the end of the first year is the difference between the value of the boat at the beginning of the first year, and the amount of depreciation.
Therefore, we get
Value of the boat at end of first year \[ = 16000 - 800 = 15200\]
Now, we will calculate the value of the boat at the end of 2 years.
The value of the boat at beginning of second year \[ = \] value of boat at end of first year \[ = 15200\]
The value lost during the second year is \[5\% \] of the value of the boat at the beginning of the second year.
Therefore, we get
Amount of depreciation during the second year \[ = 5\% \times 15200 = \dfrac{5}{{100}} \times 15200 = 760\]
Therefore, the value of the boat lost due to depreciation during the second year is ₹ 760.
The value of the boat at the end of the second year is the difference between the value of the boat at the beginning of the second year, and the amount of depreciation.
Therefore, we get
Value of the boat at end of 2 years \[ = 15200 - 760 = 14440\]
Therefore, the value of the boat after 2 years is ₹ 14440.
Note: We can also use the formula \[A = P{\left( {1 - \dfrac{R}{{100}}} \right)^n}\] for calculating the value of an asset after \[n\] years, where the rate of depreciation is \[R\]. Here, \[P\] is the original cost of the object or the asset.
Substituting \[P = 16000\], \[R = 5\% \], and \[n = 2\] in the formula, we get
\[A = 16000{\left( {1 - \dfrac{5}{{100}}} \right)^2}\]
Simplifying the expression, we get
\[\Rightarrow A = 16000{\left( {\dfrac{{95}}{{100}}} \right)^2} \\
\Rightarrow A = 16000{\left( {\dfrac{{19}}{{20}}} \right)^2} \\
\Rightarrow A = 16000 \times \dfrac{{19}}{{20}} \times \dfrac{{19}}{{20}} \\\]
Multiplying the terms, we get
\[ \Rightarrow A = 14440\]
Therefore, the value of the boat after 2 years is ₹ 14440.
Complete step-by-step answer:
Depreciation is the loss in value of something due to wear and tear, use, or becoming obsolete. It is usually calculated on the value of the asset at the beginning of the year.
It is given that the rate of depreciation is \[5\% \] p.a.
This means that the value of the boat at the end of the first year is \[5\% \] less than its value at the beginning of the year.
Similarly, the value of the boat at the end of the second year will be \[5\% \] less than its value at the beginning of the second year.
First, we will calculate the amount of depreciation during the first year.
The value lost is \[5\% \] of the cost of the boat.
Therefore, we get
Amount of depreciation during the first year \[ = 5\% \times 16000 = \dfrac{5}{{100}} \times 16000 = 800\]
Therefore, the value of the boat lost due to depreciation during the first year is ₹ 800.
The value of the boat at the end of the first year is the difference between the value of the boat at the beginning of the first year, and the amount of depreciation.
Therefore, we get
Value of the boat at end of first year \[ = 16000 - 800 = 15200\]
Now, we will calculate the value of the boat at the end of 2 years.
The value of the boat at beginning of second year \[ = \] value of boat at end of first year \[ = 15200\]
The value lost during the second year is \[5\% \] of the value of the boat at the beginning of the second year.
Therefore, we get
Amount of depreciation during the second year \[ = 5\% \times 15200 = \dfrac{5}{{100}} \times 15200 = 760\]
Therefore, the value of the boat lost due to depreciation during the second year is ₹ 760.
The value of the boat at the end of the second year is the difference between the value of the boat at the beginning of the second year, and the amount of depreciation.
Therefore, we get
Value of the boat at end of 2 years \[ = 15200 - 760 = 14440\]
Therefore, the value of the boat after 2 years is ₹ 14440.
Note: We can also use the formula \[A = P{\left( {1 - \dfrac{R}{{100}}} \right)^n}\] for calculating the value of an asset after \[n\] years, where the rate of depreciation is \[R\]. Here, \[P\] is the original cost of the object or the asset.
Substituting \[P = 16000\], \[R = 5\% \], and \[n = 2\] in the formula, we get
\[A = 16000{\left( {1 - \dfrac{5}{{100}}} \right)^2}\]
Simplifying the expression, we get
\[\Rightarrow A = 16000{\left( {\dfrac{{95}}{{100}}} \right)^2} \\
\Rightarrow A = 16000{\left( {\dfrac{{19}}{{20}}} \right)^2} \\
\Rightarrow A = 16000 \times \dfrac{{19}}{{20}} \times \dfrac{{19}}{{20}} \\\]
Multiplying the terms, we get
\[ \Rightarrow A = 14440\]
Therefore, the value of the boat after 2 years is ₹ 14440.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success
Master Class 9 English: Engaging Questions & Answers for Success
Master Class 9 Science: Engaging Questions & Answers for Success
Master Class 9 Social Science: Engaging Questions & Answers for Success
Master Class 9 Maths: Engaging Questions & Answers for Success
Class 9 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
AIM To prepare stained temporary mount of onion peel class 7 biology CBSE
The domestic sewage in large cities a Has a high BOD class 7 biology CBSE
Aeroplanes fly in which of the following layers of class 7 social science CBSE
List five rational numbers between 2 and 1 class 7 maths CBSE
Find the following percentage value 115 of 250-class-7-maths-CBSE
The product of two coprimes is 117 Their LCM should class 7 maths CBSE