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Steve owns gardening equipment that had an initial value of $12 000.
He will use the unit cost method to depreciate the value of this equipment per hour used.
Steve uses the gardening equipment for 960 hours per year.
The value of the gardening equipment is $7680 after two years.
By what amount does the gardening equipment depreciate per hour used?
\(A\)
\(\text{Value loss }=12\,000-7680=$4320\)
\(\text{Hours of use }=2 \times 960=1920\ \text{hours} \)
\(\text{Depreciation per hour }=\dfrac{4320}{1920}=\$2.25\)
\(\Rightarrow A\)
Gus purchases a coffee machine for $15 000 and depreciates its value using the unit cost method.
The rate of depreciation is $0.04 per cup of coffee made.
A recurrence relation that models the year-to-year value \(G_n\), in dollars, of the machine is
Question 18
A rule for \(G_n\), the value of the machine after \(n\) years is
Question 19
The number of cups made by the machine per year is
\(\text{Question 18:}\ C\)
\(\text{Question 19:}\ E\)
\(\text{Question 18}\)
\(G_n =\ \text{cost}-(\text{cups of coffee}\ \times 0.04) \times n \)
\(G_n =\ 15\ 000-(\text{cups of coffee}\ \times 0.04) \times n \)
\(\text{Option C is the only possible correct answer.}\)
\(\Rightarrow C\)
\(\text{Question 19}\)
\(\text{Using the}\ G_n\ \text{rule from question 18:}\)
| \(\text{cups of coffee}\ \times 0.04\) | \(=1314\) | |
| \(\text{cups of coffee}\) | \(=\dfrac{1314}{0.04}\) | |
| \(=32\ 850\) |
\(\Rightarrow E\)
Sienna owns a coffee shop. A coffee machine, purchased for $12 000, is depreciated in value using the unit cost method. The rate of depreciation is $0.05 per cup of coffee made. The recurrence relation that models the year-to-year value, in dollars, of the coffee machine is `M_0 = 12 \ 000,` `M_{n+1} = M_n - 1440` --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) ---
`M_n =`
`+`
`xx n`
Show Answers Only
Show Worked Solution
a.
`text{Cups of coffee}`
`= 1440/0.05`
`= 28 \ 800`
b.
`text{Flat rate (%)}`
`= {1440}/{12 \000}`
`= 12text{% p.a.}`
c. `M_n = 12 \ 000 + (-0.05) xx n`
The value of a van purchased for $45 000 is depreciated by `k`% per annum using the reducing balance method.
After three years of this depreciation, it is then depreciated in the fourth year under the unit cost method at the rate of 15 cents per kilometre.
The value of the van after it travels 30 000 km in this fourth year is $26 166.24
The value of `k` is
`B`
`text(Depreciation rate) = 1 – k/100`
`S = 45\ 000 (1 – k/100)^3 – 30\ 000 xx 0.15`
| `26\ 166.24` | `= 45\ 000 (1 – k/100)^3 + 4500` |
| `30\ 666.24` | `= 45\ 000 (1 – k/100)^3` |
`text(Testing each option:)`
`k = 12`
`=> B`
Marlon plays guitar in a band.
He paid $3264 for a new guitar.
The value of Marlon's guitar will be depreciated by a fixed amount for each concert that he plays.
After 25 concerts, the value of the guitar will have decreased by $200.
--- 2 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
Complete the rule below by writing the appropriate numbers in the boxes provided. (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
`G_n =` – × `n`
After how many concerts will the value of Marlon's guitar first fall below $2500? (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
| a. `text(Value)` | `= 3264-200` |
| `= $3064` |
| b. `text(Depreciation per concert)` | `= (200)/(25)` |
| `= $8 \ text(per concert)` |
c. `G_n = 3264 – 8 xx n`
d. `text(Find) \ n \ text(when) \ \ G_n = 2500:`
| `2500` | `= 3264 – 8n` |
| `n` | `= (3264-2500)/8` |
| `= 95.5` |
`:. \ text(After) \ 96 \ text(concerts, value first falls below $2500)`
A printer was purchased for $680.
After four years the printer has a value of $125.
On average, 1920 pages were printed every year during those four years.
The value of the printer was depreciated using a unit cost method of depreciation.
The depreciation in the value of the printer, per page printed, is closest to
`E`
`text(Value lost each year)`
`= (680 – 125)/4`
`= $138.75`
`:.\ text(Depreciation per page)`
`= 138.75/1920`
`= 0.072…`
`~~ 7\ text(cents per page.)`
`=> E`
A delivery truck when new was valued at $65 000.
The truck’s value depreciates at a rate of 22 cents per kilometre travelled.
After it has travelled a total distance of 132 600 km, the value of the truck will be
A. `$14\ 300`
B. `$22\ 100`
C. `$22\ 516`
D. `$29\ 172`
E. `$35\ 828`
`E`
| `text(Value of truck)` | `= $65\ 000 − 132\ 600 xx 0.22` |
| `= $35\ 828` |
`=> E`
A car is valued at $30 000 when new.
Its value is depreciated by 25 cents for each kilometre it travels.
The number of kilometres the car travels before its value depreciates to $8000 is
A. 32 000
B. 55 000
C. 88 000
D. 120 000
E. 550 000
`C`
`text(New car = 30 000)`
`text(Depreciation) = 0.25x \ \ (x = text(kilometres travelled) )`
`:.\ text(Car value = 30 000) − 0.25x`
`text(Find)\ x\ text(when car value = $8000)`
| `8000` | `= 30 \ 000 − 0.25x` |
| `0.25x` | `= 22 \ 000` |
| `x` | `= 88 \ 000` |
`=> C`
A photocopier is depreciated by $0.04 for each copy it makes.
Three years ago the photocopier was purchased for $48 000.
Its depreciated value now is $21 000.
The total number of copies made by the photocopier in the three years is
A. `108\ 000`
B. `192\ 000`
C. `276\ 000`
D. `525\ 000`
E. `675\ 000`
`E`
| `text(Depreciation)` | `= 48\ 000 − 21\ 000` |
| `= 27\ 000` | |
| `:.\ text(No. copies)` | `= (27\ 000)/0.04` |
| `= 675\ 000\ text(copies)` |
`=> E`
A van is purchased for $56 000.
Its value depreciates at a rate of 42 cents for each kilometre that it travels.
The value of the van after it has travelled 32 000 km is
A. `$13\ 440`
B. `$26\ 880`
C. `$29\ 120`
D. `$32\ 480`
E. `$42\ 560`
`E`
| `text(Depreciation)` | `= 0.42 xx 32\ 000` |
| `= 13\ 440` |
| `:.\ text(Value of car)` | `= 56\ 000 – 13\ 440` |
| `= $42\ 560` |
`=> E`