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v1 Financial Maths, STD2 F1 SM-Bank 6

Jasmine intends to keep a car purchased for $11 200 for 10 years. At the end of this time its value will be $2400.

  1. By what amount, in dollars, would the car’s value depreciate annually if Jasmine used the flat rate method of depreciation?  (1 mark)
  2. Determine the annual flat rate of depreciation correct to one decimal place.  (1 mark)
Show Answers Only

a.    `$880`

b.    `7.9\text{% (1 d.p.)}`

Show Worked Solution
a.     `\text{Depreciation}` `= 11\,200-2400`
    `=  $8800`

`:.\ \text{Annual depreciation}`

`= 8800 / 10`

`= $880`

b.    `:.\ \text{Flat rate of depreciation}`

`= 880 / (11\ 200) \times 100\text{%}`

`= 7.857…`

`= 7.9\text{% (1 d.p.)}`

v1 Financial Maths, STD2 F1 SM-Bank 3

A business purchased a delivery van for $48 000.

For accounting purposes, the van is depreciated using the straight-line depreciation method.

The van is depreciated at a flat rate of 12% of the purchase price each year.

  1. By how many dollars will the van depreciate annually? (1 mark)
  2. Calculate the value of the van after two years. (1 mark)
  3. After how many years will the van be worth $14 400? (1 mark)
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a.    `$5760`

b.    `$36\ 480`

c.    `6\ text(years)`

Show Worked Solution
a.     `text(Annual depreciation)` `= 12\% xx 48\ 000`
    `= $5760`

b.    `text(After 2 years,)`

`text(Value)` `= V_0-Dn`
  `= 48\ 000-(2 xx 5760)`
  `= $36\ 480`

c.    `text(Find)\ n\ text(when value = $14 400)`

`14\ 400` `= 48\ 000-5760 xx n`
`5760n` `= 33\ 600`
`:.n` `= (33\ 600) / 5760 = 6`
  `= 6\ \text{years}`

v1 Financial Maths, STD2 F1 2021 HSC 19

Sophie bought a set of gym equipment four years ago. It depreciated by $1800 each year using the straight-line method of depreciation. The equipment is now valued at $6200.

Find the initial value of the gym equipment.  (2 marks)

Show Answers Only

`$13\ 400`

Show Worked Solution

`text{Find initial value}\ (V_0):`

`S` `=V_0-Dn`  
`6200` `=V_0-1800 xx 4`  
`V_0` `=6200 + 7200`  
  `=$13\ 400`  

v1 Financial Maths, STD2 F1 2005 HSC 25a

Reece is preparing his annual budget for 2006.

His expected income is:

• $90 every week as a swimming coach
• Interest earned from an investment of $4500 at a rate of 4% per annum.

His planned expenses are:

• $30 every week on transport
• $12 every week on lunches
• $48 every month on entertainment.

Reece will save his remaining income. He uses the spreadsheet below for his budget.
 

2UG-2005-25a
 

  1. Determine the values of `X`, `Y` and `Z`. (Assume there are exactly 52 weeks in a year.)  (3 marks)

At the beginning of 2006, Reece starts saving.

  1. Will Reece have saved enough money during 2006 for a deposit of $2500 on a car if he keeps to his budget? Justify your answer with suitable calculations.  (2 marks)

Show Answers Only

a.    `X =$180, Y = $1560\ text(and) \ Z = $576`

b.    `text(Reece will NOT have saved enough for a $2500 deposit.)`

Show Worked Solution

a.    `text(Interest on Investment) = X`

`X` `= 4500 xx 4 text(%)`
  `= $180`

 
`text(Transport =)\ Y`

`Y` `= 52 × 30`
  `= $1560`

 
`text(Entertainment =)\ Z`

`Z` `= 48 × 12`
  `= $576`

 

b.     `text(Total Income)` `= 4680 + 180`
    `= $4860`
`text(Total Expenses)` `= 1560 + 624 + 576`
  `= $2760`
`text(Savings)` `= 4860 − 2760`
  `= $2100`

 
`:.\ text(Reece will NOT have saved enough for)`

`text(a $2500 deposit.)` 

v1 Financial Maths, STD2 F1 2019 HSC 9 MC

What is the interest earned, in dollars, if $1200 is invested for `x` months at a simple interest rate of 4% per annum?

  1. `4x`
  2. `40x`
  3. `400x`
  4. `4800x`
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`A`

Show Worked Solution

♦♦♦ Mean mark 20%!

`text(Interest)` `= 1200 xx x/12 xx 4/100`
  `= 4x`

`=> A`

v1 Financial Maths, STD2 F1 2024 HSC 13 MC

Alicia takes out a loan of $7500. Simple interest is charged on the loan.

The loan and the interest charged will be repaid by making monthly repayments of $250 over 3 years.

What simple interest rate per annum, to the nearest percent, is charged on the loan?

  1. 7%
  2. 9%
  3. 12%
  4. 15%
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Number of repayments} = 3 \times 12 = 36\)

\(\text{Total repayments} = 36 \times 250 = \$9000\)

\(\text{Interest paid} = 9000 – 7500 = \$1500\)

  \(I\) \(= Prn\)
  \(1500\) \(= 7500 \times r \times 3\)
  \(r\) \(= \dfrac{1500}{7500 \times 3} = 0.0666…\)
    \(= 6.67\% \approx 7\%\)

 
\(\Rightarrow A\)

♦ Mean mark 39%.

v1 Financial Maths, STD2 F1 2017 HSC 11 MC

A car was bought for $22 500 and one year later its value had depreciated to $18 450.

What is the approximate depreciation, expressed as a percentage of the purchase price?

  1. 18%
  2. 22%
  3. 78%
  4. 82%
Show Answers Only

`A`

Show Worked Solution
`text(Net Depreciation)` `= 22\ 500-18\ 450`
  `= $4050`

 

`:. %\ text(Depreciation)` `= 4050 / (22\ 500) xx 100`
  `= 18%`

`=> A`

v1 Financial Maths, STD2 F1 2013 HSC 13 MC

Jared borrowed $9 000. He repaid the loan in full at the end of three years with a lump sum of $10 350.

What annual simple interest rate was he charged?

  1. 6%
  2. 5.25%
  3. 5%
  4. 4.5%
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`C`

Show Worked Solution

`text(Total interest paid) = 10\ 350-9000 = $1350`

`I` `=Prn`
`1350` `=9000 xx r xx 3`
`r` `=1350/(27\ 000)`
  `=5 text(%)`

 
`=> C`

v1 Financial Maths, STD2 F1 2010 HSC 5 MC

Lina invests $2500 for 1 year and 2 months. The simple interest is calculated at a rate of 5% per annum.

What is the total value of the investment at the end of this period?

  1. $2700
  2. $2650
  3. $2630
  4. $2800 
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`C`

Show Worked Solution
`text(Interest)` `=Prn`
  `=2500 xx \text(5%)\ xx 14/12`
  `=$130`

 

`:.\ text(Value of Investment)` `=2500+130`
  `=$2630`

`=>  C`

v1 Financial Maths, STD2 F1 2023 HSC 4 MC

A company car was valued at $72 000 when new. The value of the car depreciates at a rate of 18 cents per kilometre travelled.

What is the value of the car after it has travelled a total distance of 145 000 km?

  1. $45 900
  2. $31 920
  3. $25 920
  4. $16 100
Show Answers Only

`A`

Show Worked Solution
`text{Depreciation}` `=0.18 xx 145\ 000`  
  `=$26\ 100`  

 

`text{Car value}` `=72\ 000-26\ 100`  
  `=$45\ 900`  

 
`=>A`