Depreciation
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Question 1 of 4
1. Question
A television set with an original price of `$1750` depreciates in value by `9%` per year over `8` years. Solve for the following values:Round your answer to two decimal places
`\text(New price (A)):$` (822.94)` \text(Depreciation Amount (Dep)):$` (927.06)
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Depreciation Formula
$$\mathsf{\color{#00880A}{A}}=\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$Depreciation Amount
$$\mathsf{Dep=\color{#9a00c7}{P}\color{#00880A}{A}}$$First, summarise the data and draw a diagram for easier understanding of the problem`\text(Original amount (P))=$1750``\text(Depreciation rate (R))=9% (0.09)``\text(Depreciation time (n))=8 \text(years)``\text(New amount (A))=?`Substitute the known values into the depreciation formula to solve for the new amount.Use the decimal value of the percentage.$$\mathsf{\color{#00880A}{A}}$$ `=` $$\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$ $$\mathsf{\color{#00880A}{A}}$$ `=` $$\color{#9a00c7}{1750}(1\color{#e85e00}{0.09})^{\color{#007DDC}{8}}$$ Substitute known values `=` `1750times0.4702525` `=` `$822.94` Rounded to two decimal places Finally, subtract the new amount from the original amount to get the depreciation amount.$$\mathsf{Dep}$$ `=` $$\mathsf{\color{#9a00c7}{P}\color{#00880A}{A}}$$ $$\mathsf{Dep}$$ `=` $$\color{#9a00c7}{1750}\color{#00880A}{822.94}$$ `=` `$927.06` `A=$822.94``\text(Dep)=$927.06` 

Question 2 of 4
2. Question
A truck with an original price of $$$190{,}000$$ depreciates in value by `15%` per year over `5` years. Solve for the following values:Round the new amount to two decimal placesRound the depreciation amount to a whole number
`\text(New price (A)):$` (84304.01)` \text(Depreciation Amount (Dep)):$` (105696)
Hint
Help VideoCorrect
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Incorrect
Depreciation Formula
$$\mathsf{\color{#00880A}{A}}=\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$Depreciation Amount
$$\mathsf{Dep=\color{#9a00c7}{P}\color{#00880A}{A}}$$First, summarise the data and draw a diagram for easier understanding of the problem$$\text{Original amount (P)}=$190{,}000$$`\text(Depreciation rate (R))=15% (0.15)``\text(Depreciation time (n))=5 \text(years)``\text(New amount (A))=?`Substitute the known values into the depreciation formula to solve for the new amount.Use the decimal value of the percentage.$$\mathsf{\color{#00880A}{A}}$$ `=` $$\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$ $$\mathsf{\color{#00880A}{A}}$$ `=` $$\color{#9a00c7}{190{,}000}(1\color{#e85e00}{0.15})^{\color{#007DDC}{5}}$$ Substitute known values `=` $$190{,}000\times(0.85)^{5}$$ `=` $$190{,}000\times0.4437053$$ `=` $$$84{,}304.01$$ Rounded to two decimal places Finally, subtract the new amount from the original amount to get the depreciation amount.$$\mathsf{Dep}$$ `=` $$\mathsf{\color{#9a00c7}{P}\color{#00880A}{A}}$$ $$\mathsf{Dep}$$ `=` $$\color{#9a00c7}{190{,}000}\color{#00880A}{84{,}304.01}$$ `=` $$$105{,}696$$ Rounded to a whole number $$A=$84{,}304.01$$$$\text{Dep}=$105{,}696$$ 

Question 3 of 4
3. Question
An offroad vehicle depreciates in value by `14%` per year. After `3` years, its new price is now $$$27{,}300$$. What was its original price?Round your answer to a whole number `\text(Original price )=` (42921)
Hint
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Depreciation Formula
$$\mathsf{\color{#00880A}{A}}=\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$Depreciation Amount
$$\mathsf{Dep=\color{#9a00c7}{P}\color{#00880A}{A}}$$First, summarise the data and draw a diagram for easier understanding of the problem$$\text{New amount (A)}=$27{,}300$$`\text(Depreciation rate (R))=14% (0.14)``\text(Depreciation time (n))=3 \text(years)``\text(Original price (P))=?`Substitute the known values into the depreciation formula to solve for the new amount.Use the decimal value of the percentage.$$\mathsf{\color{#00880A}{A}}$$ `=` $$\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$ $$\color{#00880A}{27{,}300}$$ `=` $$\color{#9a00c7}{P}(1\color{#e85e00}{0.14})^{\color{#007DDC}{3}}$$ Substitute known values $$27{,}300$$ `=` $$P\times(0.86)^{3}$$ $$27{,}300\color{#CC0000}{\div(0.86)^{3}}$$ `=` `Ptimes(0.86)^3``divide(0.86)^3` Divide both sides by `(0.86)^3` $$\frac{27{,}300}{0.636056}$$ `=` `P` $$42{,}920.75$$ `=` `P` `P` `=` $$$42{,}921$$ Rounded to a whole number `\text(Original price)=$42,921` 
Question 4 of 4
4. Question
If a computer with a value of `$1550` depreciates at `8%` per year, how many years will it take for its price to drop to `$865?` (7) `\text(years)`
Hint
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Correct!
Incorrect
Depreciation Formula
$$\mathsf{\color{#00880A}{A}}=\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$Depreciation Amount
$$\mathsf{Dep=\color{#9a00c7}{P}\color{#00880A}{A}}$$First, summarise the data and draw a diagram for easier understanding of the problem`\text(Original amount (P))=$1550``\text(Depreciation rate (R))=8% (0.08)``\text(New amount (A))=$865``\text(Depreciation time (n))=?`Substitute the known values into the depreciation formula to solve for the depreciation time.Use the decimal value of the percentage.$$\mathsf{\color{#00880A}{A}}$$ `=` $$\mathsf{\color{#9a00c7}{P}(1\color{#e85e00}{R})^{\color{#007DDC}{n}}}$$ $$\color{#00880A}{865}$$ `=` $$\color{#9a00c7}{1550}(1\color{#e85e00}{0.08})^{\color{#007DDC}{n}}$$ Substitute known values `865` `=` `1550times(0.92)^n` `865``divide1550` `=` `1550times(0.92)^n``divide1550` Divide both sides by `1550` $$\frac{865}{1550}$$ `=` `(0.92)^n` `0.558` `=` `(0.92)^n` Rounded to three decimal places Finally, use trial and error by substituting values to `n` to which will make the left and right sides equal.Start by substituting `n=6` and `n=7`.`n=6``0.558` `=` `(0.92)^n` `0.558` `=` `(0.92)^6` Substitute `n=6` `0.558` `=` `0.606355001344` `0.558` `=` `0.606` Rounded to three decimal places `n=7``0.558` `=` `(0.92)^n` `0.558` `=` `(0.92)^7` Substitute `n=7` `0.558` `=` `0.55784660123648` `0.558` `=` `0.558` Rounded to three decimal places Substituting `n=7` makes the left and right sides of the equation equal.Therefore, it will take `7` years for the computer’s price to drop to `$865``7 \text(years)`
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