SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

v1 Measurement, STD1 M5 2021 HSC 26

The diagrams show two similar shapes. The dimensions of the small shape are enlarged by a scale factor of 1.5 to produce the large shape.
 

Calculate the area of the large shape.  (3 marks)

Show Answers Only

`94.5\ text(cm)^2`

Show Worked Solution

`text(Dimension of larger shape:)`

♦♦ Mean mark 32%.

`text(Width) = 6 xx 1.5 = 9\ text(cm)`

`text(Height) = 8 xx 1.5 = 12 \ text(cm)`

`text(Triangle height) = 2 xx 1.5 = 3\ text(cm)`

`:.\ text(Area)` `= 9 xx (12-3) + 1/2 xx 9 xx 3`
  `= 94.5\ text(cm)^2`

v1 Measurement, STD2 M7 2012 HSC 27c

A topographic map has a scale of 1 : 250 000.

  1. Two lookouts are 3.6 cm apart on the map.

     

    What is the actual distance between the two lookouts, in kilometres?   (1 mark)

  2. Two towns are 42.5 km apart. How far apart are the two towns on the map, in centimetres?   (1 mark)

Show Answers Only
  1. `text(9 km)`
  2. `text(17 cm)`
Show Worked Solution
♦ Mean mark 41%
MARKER’S COMMENT: Better responses converted between centimetres, metres and kilometres systematically using the map scale (1 unit on the map represents 250 000 of the same unit in reality).
i.    `text{Actual distance (3.6 cm)}` `= 3.6 xx 250\ 000`
  `= 900\ 000\ text(cm)`
  `= 9\ 000\ text(m)`
  `= 9\ text(km)`

 

`:.\ text(The 2 lookouts are 9 km apart.)`

♦ Mean mark 45%

ii.   `text(Towns are 42.5 km apart.)`

`text{From the scale, } 1\ \text{cm} = 250\ 000\ \text{cm} = 2\ 500\ \text{m} = 2.5\ \text{km}`

`=>\ \text{On the map, } 42.5\ \text{km} = 42.5/2.5 = 17\ \text{cm}`

`:.\ \text{Distance on the map is 17 cm.}`

v1 Algebra, STD2 A4 2017 HSC 28e

Sage brings 60 cartons of unpasteurised milk to the market each week. Each carton currently sells for $4 and at this price, all 60 cartons are sold each weekend.

Sage considers increasing the price to see if the total income can be increased.

It is assumed that for each $1 increase in price, 6 fewer cartons will be sold.

A graph showing the relationship between the increase in price per carton and the income is shown below.

 


 

  1. What price per carton should be charged to maximise the income?   (1 mark)

  2. What is the number of cartons sold when the income is maximised?   (1 mark)

  3. The cost of running the market stall is $40 plus $1.50 per carton sold.

    Calculate Sage's profit when the income earned from a day selling at the market is maximised.   (2 marks)

Show Answers Only

a.   `$7`

b.   `42`

c.   `$191`

Show Worked Solution

a.   `text(Graph is highest when increase = $3)`

`:.\ text(Carton price)\ = 4 + 3= $7`
 

b.   `text(Cartons sold)\ =60-(3 xx 6)=42`
  

c.   `text{Cost}\ = 42 xx 1.50 + 40 = $103`

`:.\ text(Profit when income is maximised)`

`= (42 xx 7)-103`

`= $191`

v1 Measurement, STD2 M7 2008 HSC 20 MC

A point `P` lies between a lamp post, 1.5 metres high, and a building, 7 metres high. `P` is 2.5 metres away from the base of the post.

From `P`, the angles of elevation to the top of the lamp post and to the top of the building are equal.
 

What is the distance, `x`, from `P` to the top of the tower?

  1. 10.60
  2. 12.50
  3. 13.60
  4. 14.55
Show Answers Only

`C`

Show Worked Solution

`text(Triangles are similar)\ \ text{(equiangular)}`

`text(In smaller triangle:)`

`h^2` `= 1.5^2 + 2.5^2`
  `= 8.5`
`h` `= sqrt 8.5`
   
`x/sqrt8.5` `= 7/1.5 \ \ \ text{(sides of similar Δs in same ratio)}`
`x` `= (7 sqrt 8.5)/1.5`
  `= 13.60…`

 
`=>  C`

v1 Algebra, STD2 A4 2009 HSC 28c

The brightness of a lamp \((L)\) is measured in lumens and varies directly with the square of the voltage \((V)\) applied, which is measured in volts.

When the lamp runs at 7 volts, it produces 735 lumens.

What voltage is required for the lamp to produce 1820 lumens? Give your answer correct to one decimal place.   (3 marks)

Show Answers Only

 `11.2\ \text(volts)`

Show Worked Solution
♦♦ Mean mark 22%
TIP: Establishing `L=k V^2` is the key part of solving this question.

`L prop V^2\ \ => \ \ L=kV^2`

`text(Find)\ k\ \text{given}\ L = 735\ \text{when}\ V = 7:`

`735` `= k xx 7^2`
`:. k` `= 735/49=15`

 
`text(Find)\ V\ text(when)\ L = 1820:`

`1820` `= 15 xx V^2`
`V^2` `= 1820/15=121.33…`
`V` `= sqrt{121.33} = 11.2\ text(volts)\ \ text{(to 1 d.p.)}`

v1 Algebra, STD2 A4 2021 HSC 24

A population of Tasmanian devils, `D`, is to be modelled using the function  `D = 650 (0.8)^t`, where `t` is the time in years.

  1. What is the initial population?   (1 mark)

  2. Find the population after 2 years.   (1 mark)

  3. On the axes below, draw the graph of the population against time, in the period  `t = 0`  to  `t = 6`.   (2 marks)
      

Show Answers Only

a.   `650`

b.   `416`

c.   `text{See Worked Solutions}`

Show Worked Solution

a.   `text{Initial population occurs when}\ \  t = 0:`

`D=650(0.8)^0=650 xx 1= 650`
 

b.    `text{Find} \ D \ text{when} \ \ t = 5: `

`D= 650 (0.8)^2= 416`

♦ Mean mark (c) 48%.

 
c.   `\text{At}\ t=6:`

`D=650(0.8)^6=170.39…`
 

v1 Algebra, STD2 A4 2021 HSC 10 MC

Which of the following best represents the graph of  \(y = 5 (0.4)^{x}\) ?
 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By elimination:}\)

♦ Mean mark 41%.

\(\text{When}\  x = 0, \ y = 5 \times (0.4)^0 = 5\)

\(\rightarrow\ \text{Eliminate B and C} \)

\(\text{As}\ \ x \rightarrow \infty, \ y \rightarrow 0 \)

\(\rightarrow\ \text{Eliminate A} \)

\(\Rightarrow D\)

v1 Measurement, STD2 M7 2016 HSC 16 MC

The width (`W`) of a road can be calculated using two similar triangles, as shown in the diagram.
  
  

What is the approximate width of the road?

  1. `12.8\ text(m)`
  2. `13.3\ text(m)`
  3. `14.6\ text(m)`
  4. `17.8\ text(m)`
Show Answers Only

`=> C`

Show Worked Solution

`text{Triangles are similar (equiangular)}`

`text(Using similar ratios:)`

`W/(6.5)` `= 18/8`
`:. W` `= (18 xx 6.5)/8`
  `= 14.62…`

 
`=> C`

v1 Measurement, STD2 M1 2020 HSC 5 MC

A pencil is measured to be 12.8 cm long, correct to one decimal place.

What is the percentage error in this measurement?

  1. 0.20%
  2. 0.39%
  3. 0.78%
  4. 1.56%
Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 44%.

`text{Absolute error} = 1/2 xx \text{precision} = 1/2 xx 0.1 = 0.05\ \text{cm}`

`%  \text{error}` `= \frac{0.05}{12.8} xx 100`
  `= 0.390625%`
  `= 0.39%`

`⇒ B`

v1 Measurement, STD2 M1 2019 HSC 8 MC

A newborn baby’s length is recorded as 52.4 cm.

What is the absolute error of this measurement?

  1. 0.25 cm
  2. 0.5 cm
  3. 1 cm
  4. 2 cm
Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 48%.

`text{Absolute error}` `= 1/2 xx\ text{precision}`
  `= 1/2 xx 1\ text{ cm}`
  `= 0.5\ text{ cm}`

`⇒ B`

v1 Measurement, STD2 M1 2015 HSC 12 MC

A sprinter’s reaction time at the start of a race was recorded as 0.25 seconds, correct to the nearest hundredth of a second.

What is the percentage error in this measurement, correct to one significant figure?

  1. 0.5%
  2. 1%
  3. 2%
  4. 3%
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 43%.

`text{Absolute error}\ =1/2 xx text{precision}\ = 1/2 xx 0.01 = 0.005\ text{s}`

`text{% error}` `=\ frac{text{absolute error}}{text{measurement}} xx 100%`  
  `=0.005/0.25 xx 100%`  
  `=2%`  

 
`⇒ C`

v1 Measurement, STD2 M1 2014 HSC 10 MC

The height of Mount Kosciuszko is measured to be 2228.1 m above sea level.

What is the percentage error in this measurement?

  1. 0.001%
  2. 0.002%
  3. 0.005%
  4. 0.011%
Show Answers Only

`B`

Show Worked Solution
 
♦ Mean mark 50%

`text{Absolute error}\ = 1/2 xx text{precision}\ = 1/2 xx 0.1 = 0.05\ text{m}`

`text{% error}` `= \frac{text{absolute error}}{text{measurement}} xx 100%`  
  `= 0.05 / 2228.1 xx 100%`  
  `= 0.0022%`  

 
`⇒  B`

v1 Measurement, STD2 M1 2016 HSC 1 MC

What is  0.04967  correct to two significant figures?

  1. 0.049
  2. 0.050
  3. 0.0496
  4. 0.0497
Show Answers Only

`B`

Show Worked Solution

`text(We are rounding 0.04967 to 2 significant figures.)`

• `text(First 2 significant digits: 4 and 9)`

• `text(Next digit is 6 → round **up**)`

• `text(0.04967 rounds to 0.050 (2 sig. fig.))`

`=> B`

v1 Measurement, STD2 M1 2015 HSC 1 MC

The distance from Earth to the Moon is approximately 384 400 km.

What is this distance in standard form correct to two significant figures?

  1. `3.84 × 10^5\ \text{km}`
  2. `3.8 × 10^5\ \text{km}`
  3. `3.9 × 10^5\ \text{km}`
  4. `3.84 × 10^6\ \text{km}`
Show Answers Only

`C`

Show Worked Solution

`384\ 400`

`= 3.844 × 10^5`

`\text(Rounded to 2 significant figures)  →  3.9 × 10^5\ \text{km}`

 
`⇒ C`

v1 Measurement, STD2 M1 2022 HSC 34

A composite solid is shown. The top section is a hemisphere with a diameter of 6 cm. The bottom section is a cylinder with a height of 3 cm and a diameter of 4 cm
 

Find the total volume of the composite solid in cm³, correct to 1 decimal place.  (4 marks)

Show Answers Only

`94.2 \ text{cm}^3`

Show Worked Solution
`text{Volume of Hemisphere}` `=2/3 pi r^3`  
  `=2/3 pi xx 3^3`  
  `=56.54\ text{cm}^3`  

 

`text{Volume of Cylinder}` `=pi r^2 h`  
  `=pi (2^2) xx 3`  
  `=37.69\ text{cm}^3`  

 

`text{Total Volume}` `=56.54+37.69`
  `=94.23`
  `=94.2 \ text{cm}^3`

v1 Measurement, STD2 M1 2014 HSC 27c

A swimming pool is in the shape of a rectangle with a semicircle at each end, as shown.

The pool is 7000 mm long, 4000 mm wide, and has a depth of 2100 mm.  
  

How much water is needed to fill the pool, to the nearest litre?   (4 marks) 

Show Answers Only

`51 \ 589 \ text(L)`

Show Worked Solution

`V = Ah` 

♦ Mean mark 41%
STRATEGY: Adjusting measurements to metres makes the final conversion to litres simple.

`text(Finding Area of base)`

`text(Semi-circles have radius 2000 mm) = 2 \ text(m)`

`:.\ text(Area of 2 semicircles)`

`=2 xx 1/2 xx pi r^2`

`= pi xx 2^2`

`= 12.56 \ text(m)^2`
 

`text(Area of rectangle)`

`= l xx b`

`= (7-2 xx 2) xx 4`

`= 12\ text(m)^2`

 

`:.\ text(Volume)` `= Ah`
  `= (12.56… + 12) xx 2.1`
  `= 51.589…\ text(m)^3`
  `= 51 \ 589 \ text(L)\ \ text{(using 1m³} = 1000\ text{L)}`
   

v1 Measurement, STD2 M1 2013 HSC 12 MC

A hemisphere sits perfectly on top of a cylinder to form a solid. 

What is the volume of the solid?

  1. 1750 cm³
  2. 1950 cm³
  3. 2150 cm³
  4. 2350 cm³
Show Answers Only

`C`

Show Worked Solution
`text(Volume )` `=text{Vol (cylinder)} +text{Vol (hemisphere)}`
  `= pi r^2h+2/3pi r^3`
  `= pi xx 6^2 xx 15 + 2/3pi xx 6^3`
  `=2149.84\ text(cm)^3`

 
`=>\ C`

v1 Measurement, STD2 M1 2018 HSC 30a

A cylindrical oil tank has a height of 7 metres and a capacity of 1.5 megalitres.
 

What is the diameter of the oil tank? Give your answer in metres, correct to two decimal places.  (3 marks)

Show Answers Only

`16.50\ text{m}`

Show Worked Solution

`text{Converting megalitres to m³  (using 1 m³ = 1000 L):}`

♦ Mean mark 48%.

`1.5\ text(ML)` `= (1.5 xx 10^6)/(10^3)`
  `= 1.5 xx 10^3\ text(m)^3`
  `= 1500\ text(m)^3`

 

`V` `= pir^2h`
`1500` `= pi xx r^2 xx 7`
`r^2` `= 1500/(pi xx 7)`
`sqrt(r^2)` `= sqrt(68.21)`
  `= 8.26\ text{m}`

 

`text{Diameter}` `=2r`  
  `=2xx8.26 \ text{m}`  
  `=16.52\ text{m}`  

 

v1 Measurement, STD2 M1 2009 HSC 11 MC

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  

  1. 68 m²
  2. 73 m²
  3. 95 m²
  4. 193 m²
Show Answers Only

`C`

Show Worked Solution
`text(Area)` `=\ text(Area of Sector – Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx bh)`
  `= (90/360 xx pi xx 12^2)-(1/2 xx 6 xx 6)`
  `= 113.097…-18`
  `= 95.097…\ text(m²)`

`=> C`

v1 Measurement, STD2 M1 2021 HSC 16

The surface area, `A`, of a sphere is given by the formula

`A = 4 pi r^2,`

where `r` is the radius of the sphere.

A satellite dish resembles the inner surface of the lower half of a sphere with a radius of 1.5 meters.

 

Find the surface area of the satellite dish in square metres, correct to one decimal place.   (2 marks)

Show Answers Only

`14.1\ text{m}^2`

Show Worked Solution
`A` `= frac{1}{2} times 4 pi r^2`
  `= 2 pi r^2`
  `= 2 pi times (1.5)^2`
  `= 2 pi times 2.25`
  `= 14.137…`
  `= 14.1\ text{m}^2\ \text{(1 d.p.)}`

v1 Measurement, STD2 M1 2019 HSC 16

A decorative light fixture is in the shape of a hollow hemisphere with a diameter of 24 cm.
 

The inside of the fixture is to be coated with reflective paint.

What is the area to be painted on the inside surface? Give your answer correct to the nearest square centimetre.   (2 marks)

Show Answers Only

`905\ \text{cm}^2`

Show Worked Solution
`A` `= 2 pi r^2`
  `= 2 × pi × 12^2`
  `= 2 × pi × 144 = 905.0…`
  `≈ 905\ \text{cm}^2`

v1 Measurement, STD2 M1 2016 HSC 30c

A landscape artist was commissioned to design a garden consisting of part of a circle, with centre `O`, and a rectangle, as shown in the diagram. The radius `OC` of the circle is 20 m, the width `BC` of the rectangle is 10 m, and `DOC` is 100°.
 

What is the area of the whole garden, correct to the nearest square metre?  (5 marks)

Show Answers Only

`6281\ text{m²  (nearest m²)}`

Show Worked Solution

`text(In)\ \triangle ODC,`

`sin50^@` `= (ED)/20`
`ED` `= 20 \times \sin50^@`
  `= 34.472`
`:. DC` `= 2 \times 34.472 = 68.944\ \text{m}`

 

`cos50^@` `= (OE)/20`
`:. OE` `= 20 \times \cos50^@ = 28.925`

 

`text(Area of)\ \triangle ODC`

`= \frac{1}{2} \times 68.944 \times 28.925 = 997.12\ \text{m}^2`

 

`text(Area of rectangle ABCD)` `= 10 \times 68.944 = 689.44\ \text{m}^2`

 

`text(Area of major sector DOAC)`

`= \pi \times 20^2 \times \frac{260}{360} = 4594.58\ \text{m}^2`

 

`:.\ \text{Area of garden}`

`= 997.12 + 689.44 + 4594.58 = 6281.14`

`= 6281\ \text{m² (nearest m²)}`

v1 Measurement, STD2 M1 2018 HSC 27c

A farmer is designing a chicken coop with a roof shaped like half a cylinder, open at both ends. The structure has a diameter of 4 metres and a length of 12 metres.
 

 
The curved roof is to be made of aluminum sheets.

What area of aluminum sheets is required, to the nearest m²?  (2 marks)

Show Answers Only

`75\ text(m²  (nearest m²))`

Show Worked Solution

`text(Flatten out the half cylinder,)`

`text(Width)` `= 1/2 xx text(circumference)`
  `= 1/2 xx pi xx 4`
  `= 6.283…`

 

`:.\ text(Sheeting required)` `= 12 xx 6.283…`
  `= 75.39…`
  `= 75\ text(m²  (nearest m²))`

v1 Measurement, STD2 M1 2009 HSC 23c

The diagram shows the shape and dimensions of a floor which is to be tiled.
 

  1. Find the area of the floor.   (2 marks)

  2. Tiles are sold in boxes. Each box holds one square metre of tiles and costs $60. When buying the tiles, 10% more tiles are needed, due to cutting and wastage.

     

    Find the total cost of the boxes of tiles required for the floor.   (2 marks)

Show Answers Only

i.    `17\ text(m²)`

ii.   `$1140`

Show Worked Solution
i.
`text(Area)` `=\ text(Area of big square – Area of 2 cut-out squares`
  `= (3 + 2) xx (3 + 2)\-2 xx (2 xx 2)`
  `= 25\-8`
  `= 17\ text(m²)`

 

ii. `text(Tiles required)` `= (17 +10 text{%}) xx 17`
    `= 18.7\ text(m²)`

 

 `=>\ text(19 boxes are needed)`

`:.\ text(Total cost of boxes)` `=19 xx $60`
  `= $1140`

v1 Financial Maths, STD2 F4 2008 HSC 27c

A laptop depreciated in value by 20% per annum. Three years after it was purchased, it had depreciated to a value of $2048, using the declining balance method.

What was the purchase price of the laptop?   (2 marks)

Show Answers Only

`$4000`

Show Worked Solution

`S = V_0 (1 – r)^n`

`2048` `= V_0 (1-0.20)^3`
`2048` `= V_0 (0.80)^3`
`V_0` `= 2048 / 0.512`
  `= 4000`

 

`:.\ text(The purchase price) = $4000`

v1 Financial Maths, STD2 F4 2005 HSC 26a

A new high-end coffee machine is purchased for $25 000 in January 2020.

At the end of each year, starting in 2021, the machine depreciates in value by 15% per annum, using the declining balance method of depreciation.

In which year will the value of the machine first fall below $15 000? (2 marks)

Show Answers Only

`text(The value falls below $15 000 in the fourth year)`

`text{which will be during 2024.}`

Show Worked Solution

`text(Using the formula:)\ \ S = V_0(1-r)^n`

`text(where) \ \ V_0 = 25\ 000,\ \ r = 0.15`

`text(If)\ n = 1\ \text{(2021)}`

`S` `= 25\ 000(0.85)^1`
  `=21\ 250`

 

`text(If)\ n = 2\ \text{(2022)}`

`S` `=25\ 000(0.85)^2`
  `=25\ 000(0.7225)`
  `=18\ 062.50`

 

`text(If)\ n = 3\ \text{(2023)}`

`S` `=25\ 000(0.85)^3`
  `=25\ 000(0.614125)`
  `=15\ 353.13`

 

`text(If)\ n = 4\ \text{(2024)}`

`S` `=25\ 000(0.85)^4`
  `=25\ 000(0.52200625)`
  `=13\ 050.16`

 

`:.\ \text{The value first falls below $15 000 in the fourth year}`

`text{which will be during 2024.}`

v1 Financial Maths, STD2 F4 2019 HSC 37

A machine is purchased for $32 800. Each year the value of the machine is depreciated by the same percentage.

The table shows the value of the machine, based on the declining-balance method of depreciation, for the first three years.

\[ \begin{array} {|c|c|} \hline \textit{End of year} & \textit{Value} \\ \hline 1 & \$27\,056.00 \\ \hline 2 & \$22\,888.16 \\ \hline 3 & \$19\,377.82 \\ \hline \end{array} \]

What is the value of the machine at the end of 10 years?  (3 marks)

Show Answers Only

`$4783.78`

Show Worked Solution

`text(Find the depreciation rate:)`

`S` `= V_0(1-r)^n`
`27\ 056` `= 32\ 800(1-r)^1`
`1-r` `= \frac{27\ 056}{32\ 800} = 0.82488`
`r` `= 0.17512`

 

`:.\ \text(Value after 10 years)`

`= 32\ 800(1-0.17512)^{10}`

`= 32\ 800(0.82488)^{10}`

`= 32\ 800 × 0.1458`

`= $4783.78`

v1 Financial Maths, STD2 F4 2018 HSC 26h

A piece of machinery is purchased for $18,500.

The value of the machine depreciates by 14% each year using the declining-balance method.

What is the value of the machine after three years?  (2 marks)

Show Answers Only

`$11\ 767\ \ (text(nearest dollar))`

Show Worked Solution
`S` `= V_0(1 – r)^n`
  `= 18\ 500(1 – 0.14)^3`
  `= 18\ 500(0.86)^3`
  `= 18\ 500 × 0.636056`
  `= 11\ 767.04`
  `= $11\ 767\ \ (text(nearest dollar))`

v1 Financial Maths, STD2 F4 2014 HSC 9 MC

A laptop is purchased for $2500. It depreciates at a rate of 25% per annum using the declining balance method.

What will be the salvage value of the laptop after 2 years, to the nearest dollar?

  1. $1406
  2. $1250
  3. $1681
  4. $1875
Show Answers Only

`A`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 2500 (1-25/100)^2`
  `= 2500 (0.75)^2`
  `= 2500 × 0.5625`
  `= $1406.25`

 

`Rightarrow \ text(To the nearest dollar, the salvage value is **$1406**)`

`Rightarrow A`

v1 Financial Maths, STD2 F4 2021 HSC 4 MC

A machine was purchased for $3200 four years ago. It depreciates at a rate of 12% per year using the declining-balance method.

What is the machine’s current value, to the nearest dollar?

  1. $1850
  2. $1919
  3. $1945
  4. $2010
Show Answers Only

`B`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 3200 (1-0.12)^4`
  `= 3200 (0.88)^4`
  `= 3200 × 0.5997`
  `= 1919`

⇒ `B`

v1 Financial Maths, STD2 F4 2021 HSC 26

Mila plans to invest $42 000 for 1.5 years. She is offered two different investment options.

Option A:  Interest is paid at 5% per annum compounded monthly.

Option B:  Interest is paid at `r` % per annum simple interest.

  1. Calculate the future value of Mila's investment after 1.5 years if she chooses Option A. (2 marks)
  2. Find the value of `r` in Option B that would give Mila the same future value after 1.5 years as for Option A. Give your answer correct to two decimal places. (2 marks)
Show Answers Only
  1. `$45\ 264.08`
  2. `5.18text(%)`
Show Worked Solution
a.   `r` `= text(5%)/12 = text(0.4167%) = 0.004167\ \text(per month)`
  `n` `= 12 × 1.5 = 18`
`FV` `= PV(1 + r)^n`
  `= 42\ 000(1 + 0.004167)^{18}`
  `= $45\ 264.08`

 

b.   `I` `= Prn`
  `3\ 264.08` `= 42\ 000 × r × 1.5`
  `r` `= 3\ 264.08 / (42\ 000 × 1.5)`
    `= 0.0518…`
    `= 5.18\ \text{% (to 2 d.p.)}`

v1 Financial Maths, STD2 F4 2015 HSC 26d

A laptop currently costs $850.

Assuming a constant annual inflation rate of 3.2%, calculate the cost of the same laptop in 4 years’ time.  (2 marks)

Show Answers Only

`$962.38\ \text{(nearest cent)}`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 850(1.032)^4`
  `= 850(1.132216)`
  `= 962.3836…`
  `= $962.38\ \text{(nearest cent)}`

v1 Financial Maths, STD2 F4 2022 HSC 11 MC

In eight years, the future value of an investment will be $120 000. The interest rate is 6% per annum, compounded half-yearly.

Which equation will give the present value `(PV)` of the investment?

  1. `PV=(120\ 000)/((1+0.06)^(8))`
  2. `PV=(120\ 000)/((1+0.03)^(8))`
  3. `PV=(120\ 000)/((1+0.03)^(16))`
  4. `PV=(120\ 000)/((1+0.06)^(16))`
Show Answers Only

`C`

Show Worked Solution

`text{Compounding periods} = 8 xx 2 = 16`

`text{Compounding rate} = (6text{%}) / 2 = 3text{%} = 0.03`

`PV = (120\ 000) / ((1 + 0.03) ^{16})`

`=> C`

v1 Financial Maths, STD2 F1 2004 HSC 27b

Jack is paid at these rates:

\begin{array}{|l|} \hline \text{Weekday rate} & \text{\(\$\)15 per hour}\\ \text{Saturday rate} & \text{Time-and-a-half}\\ \text{Sunday rate} & \text{Double time}\\ \hline \end{array}

His time sheet for last week is:

\begin{array}{|l|l|l|l|} \hline & \textit{Start} & \textit{Finish} & \textit{Unpaid break}\\ \hline \text{Friday} & \text{9.00 am} & \text{4.00 pm} & \text{1 hour} \\ \text{Saturday} & \text{9.00 am} & \text{4.00 pm} & \text{1 hour} \\ \text{Sunday} & \text{9.00 am} & \text{1.00 pm} & \text{30 minutes} \\ \hline \end{array}

  1. Calculate Jack's gross pay for last week.  (3 marks)
  2. Jack decides not to work on Sundays. He wants to keep his weekly gross pay the same. How many extra hours at the weekday rate must he work?  (1 mark)
Show Answers Only
  1. \(\$330\)
  2. \(\text{7 extra hours}\)
Show Worked Solution
i.     \(\text{Pay(Fri)}\) \(=6 \ \text {hours } \times 15.00\)
    \(=\$90.00\)
  \(\text{Pay(Sat)}\) \(=6 \ \text {hours } \times 1.5 \times 15.00\)
    \(=\$135.00\)
  \(\text{Pay(Sun)}\) \(=3.5 \ \text {hours } \times 2 \times 15.00 \)
    \(=\$105.00\)

\(\therefore \text{Gross pay }\) \(=90+135+105\)
  \(=\$330\)

 

ii.     \(\text{Pay on Sun}=\$105\)

\(\text{Weekly equivalent hours}\) \(=\dfrac{105}{15}\)
  \(=7 \ \text{hours}\)

 

\(\therefore \ \text{He will have to work 7 extra hours on a weekday}\)

v1 Financial Maths, STD2 F1 2022 HSC 21

A real estate agent's commission for selling houses is 2% for the first $900 000 of the sale price and 1.5% for any amount over $900 000.

Calculate the commission earned in selling a house for $1 200 000.  (2 marks)

Show Answers Only

`$22\ 500`

Show Worked Solution
`text{Commission}` `=900\ 000 xx 2text{%} + (1\ 200\ 000-900\ 000) xx 1.5text{%}`
  `=900\ 000 xx 0.02 + 300\ 000 xx 0.015`
  `=18\ 000 + 4\ 500`
  `=$22\ 500`

v1 Financial Maths, STD2 F1 2013 HSC 11 MC

An enterprise agreement has the following annual salary arrangements:
 

Rowan's employer pays 5% more than the enterprise agreement. He is on Step 2 and receives an allowance for Leader 1.

What is Rowan's gross monthly pay?

  1. $3575.50
  2. $3730.00
  3. $3937.50
  4. $4150.00
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 46%
`text(Gross annual pay)` `=(40\ 000+5000)xx1.05`
  `=$47\ 250`

 

`:.\ text(Gross monthly pay)` `=(47\ 250)/12`
  `=$3937.50`

 
`=>\ C`

v1 Financial Maths, STD2 F1 2007 HSC 3 MC

Dan is about to go on holidays for four weeks. His weekly salary is $375  and his holiday loading is 20% of four weeks pay.

What is Dan's total pay for the four weeks holiday?

  1. $400
  2. $1250
  3. $1800
  4. $2150
Show Answers Only

`C`

Show Worked Solution
`text(Salary)\ text{(4 weeks)}` `= 4 xx 375`
  `= $1500`

 

`text(Holiday loading)` `= 1500 xx 20%`
  `= $300`

 

`:.\ text(Total pay)` `= 1500+300`
  `= $1800`

 
`=>  C`

v1 Financial Maths, STD2 F1 2008 HSC 7 MC

Maya’s regular wage is $20 per hour. One week, she worked 9 hours at double time.

How many hours would Maya need to work at time-and-a-half to earn the same amount?

  1. 10
  2. 11
  3. 12
  4. 13
Show Answers Only

`C`

Show Worked Solution

`text(Amount earned at double time)`

`= 9 xx 2 xx 20`

`= $360`

`text(Time-and-a-half rate)` `= 1.5 xx 20`
  `= $30 text(/hr)`

 
`:.\ text(# Hours at time-and-a-half)`

`= 360 / 30`

`= 12\ text(hrs)`

`=>  C`

v1 Financial Maths, STD2 F1 2014 HSC 13 MC

Mark works as a car salesperson. His commission is based on a sliding scale of 5% on the first $3000 of his sales, 3% on the next $1500, and 1.5% thereafter.

What is Mark’s commission when his total sales are $6,250? 

  1. $202.75
  2. $210.50
  3. $216.25
  4. $221.25
Show Answers Only

`D`

Show Worked Solution

`text(Commission)`

`= (3000 xx text(5%)) + (1500 xx text(3%)) + (6250 – 4500) xx text(1.5%)`

`= (3000 xx 0.05) + (1500 xx 0.03) + (1750 xx 0.015)`

`= 150 + 45 + 26.25`

`= 221.25`

`=> D`

v1 Financial Maths, STD2 F1 2006 HSC 22 MC

This income tax table is used to calculate Daniel’s tax payable.

Taxable income Tax payable
$0 − $11 000 Nil
$11 001 − $42 400 20 cents for each $1 over $11 000
$42 401 − $78 800 $6280 plus 33 cents for each $1 over $42 400
$78 801 − $108 400 $18 292 plus 37 cents for each $1 over $78 800
$108 401 and over $31 316 plus 48 cents for each $1 over $108 400

Daniel’s taxable income increases from $45 000 to $80 000.

What percentage of his increase will he pay in additional tax?

  1. `text(22.7%)`
  2. `text(25.5%)`
  3. `text(33.1%)`
  4. `text(40.6%)`
Show Answers Only

`C`

Show Worked Solution
`text(Tax on $45 000)` `= 6280 + 0.33 xx (45\ 000-42\ 400)`
  `= 6280 + 0.33 xx 2600`
  `= 6280 + 858`
  `= $7138`
`text(Tax on $80 000)` `= 18\ 292 + 0.37 xx (80\ 000-78\ 800)`
  `= 18\ 292 + 0.37 xx 1200`
  `= 18\ 292 + 444`
  `= $18\ 736`
`:.\ text(Extra tax)` `= 18\ 7360-7138`
  `= $11\ 598`

`:.\ \text{% of income increase paid in tax}`

`= (11\ 598) / (35\ 000) xx 100`

`=33.14%`

`=> C`

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 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%
Show Answers Only

`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 
Show Answers Only

`C`

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

 

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

`=>  C`

Advanced Trigonometry, 2ADV T2 2019 HSC 13a

Solve  \(2\sin\,\theta\,\cos\,\theta=\sin\,\theta\)  for  \(0^{\circ} \leq \theta \leq 360^{\circ}\).   (3 marks)

Show Answers Only

\(\theta=0^{\circ}, 60^{\circ}, 180^{\circ}, 300^{\circ}, 360^{\circ}\)

Show Worked Solution

♦ Mean mark 49%.

\(2\sin\,\theta\,\cos\,\theta\) \(=\sin\,\theta\)
\(2\sin\,\theta\,\cos\,\theta-\sin\,\theta\) \(= 0\)
\(\sin\,\theta(2\cos\,\theta-1)\) \(=0\)

 
\(\sin\,\theta=0:\)

\(\Rightarrow \theta=0^{\circ}, 180^{\circ}, 360^{\circ}\)
 

\(\cos\,\theta=\dfrac{1}{2}:\)

\(\text{Reference angle}\ =60^{\circ}\)

\(\text{cos is positive in 1st/4th quadrants.}\)

\(\Rightarrow \theta=60^{\circ}, (360-60)^{\circ} = 60^{\circ}, 300^{\circ}\) 

Advanced Trigonometry, 2ADV T2 2004 HSC 8a

  1. Show that  \(\cos\,\theta\,\tan\,\theta = \sin\,\theta\).   (1 mark)

  2. Hence solve  \(8\sin\,\theta\,\cos\,\theta\,\tan\,\theta = \dfrac{1}{\sin\,\theta}\)  for  \(0^{\circ} \leq \theta \leq 360^{\circ}\).   (2 marks)

Show Answers Only
  1. \(\text{Proof (See Worked Solutions)}\)
  2. \(30^{\circ}, 150^{\circ}\)
Show Worked Solution

a.    \(\text{Prove}\ \cos\,\theta\,\tan\,\theta = \sin\,\theta\)

\(\text{LHS}\) \(= \cos\,\theta\,\tan\,\theta\)
  \(=\cos\,\theta \times \dfrac{\sin\,\theta}{\cos\,\theta}\)
  \(= \sin\,\theta\)
  \(=\ \text{RHS}\)

 

b.    \(8\sin\,\theta\,\cos\,\theta\,\tan\,\theta\) \(= \dfrac{1}{\sin\,\theta}\)
  \(8\sin^{2}\theta\) \(= \dfrac{1}{\sin\,\theta}\)
  \(8\sin^{3}\theta\) \(= 1\) 
  \(\sin^{3}\theta\) \(=\dfrac{1}{8}\)
  \(\sin\,\theta\)  \(=\dfrac{1}{2}\) 

 
\(\text{Reference angle:}\ \theta=30^{\circ}\)

\(\text{sin is positive in 1st/2nd quadrants.}\)

\(\therefore \ \theta=30^{\circ}, (180-30)^{\circ}=30^{\circ}, 150^{\circ}\)

Advanced Trigonometry, 2ADV T2 2014 HSC 7 MC

How many solutions of the equation  `(sin theta-1)(tan theta + 2) = 0`  lie between `0°` and `360°`?

  1. `1`
  2. `2`
  3. `3`
  4. `4`
Show Answers Only

`B`

Show Worked Solution
♦♦♦ Mean mark 25%, making it the toughest MC question in the 2014 exam.

`text(When)\ (sin theta-1)(tan theta + 2) = 0`

`(sin theta-1) = 0\ \ text(or)\ \ tan theta + 2 = 0`

`text(If)\ \ sin theta-1= 0:`

`sin theta= 1\ \ =>\ \ theta= 90°,\ \ \ 0° < theta < 360°`
 

`text(If)\ \ tan theta + 2= 0:`

`tan theta= -2`

`text{Since}\ tan\ 90°\ text{is undefined, there are only 2 solutions when}`

`tan theta = -2\ \text{(which occurs in the 1st and 4th quadrants).}`
  

`:.\ 2\ text(solutions)`

`=>  B`

Advanced Trigonometry, 2ADV T2 2008 HSC 6a

Solve  \(2\sin^{2}\left( \dfrac{\theta}{3}\right) = 1\)  for  \(-180^{\circ} \leq \theta \leq 180^{\circ}\).   (3 marks)

Show Answers Only

\(\theta=-135^{\circ}, 135^{\circ}\)

Show Worked Solution
\(2\sin^{2}\left( \dfrac{\theta}{3}\right)\) \(= 1\)
\(\sin^{2}\left( \dfrac{\theta}{3}\right)\) \(= \dfrac{1}{2}\)
\(\sin\left( \dfrac{\theta}{3}\right)\) \(= \pm \dfrac{1}{\sqrt{2}}\)

 

MARKER’S COMMENT: Many students had problems adjusting their answer to the given domain, especially when dealing with negative angles.

\(\text{When}\ \ \sin\left( \dfrac{\theta}{3}\right)= \dfrac{1}{\sqrt{2}}:\)

\(\dfrac{\theta}{3}\) \(= 45, 180-45=45^{\circ}, 135^{\circ}\)
\(\theta\) \(= 135^{\circ}, 405^{\circ}\)

 

\(\text{When}\ \ \sin\left( \dfrac{\theta}{3}\right)= -\dfrac{1}{\sqrt{2}}:\)

\(\dfrac{\theta}{3}\) \(= -45^{\circ}, -135^{\circ}\)
\(\theta\) \(= -135^{\circ}, -405^{\circ}\)

 

\(\therefore \theta=-135^{\circ}, 135^{\circ}\ \ \text{for}\ \ -180^{\circ} \leq \theta \leq 180^{\circ}\)

Advanced Trigonometry, 2ADV T2 2023 HSC 20

Find all the values of `theta`, where  `0^@ <=theta <= 360^@`, such that

`sin(theta-60^@)=-sqrt3/2`   (3 marks)

Show Answers Only

`theta=0^@, 300^@ and 360^@`

Show Worked Solution

`sin60^@=sqrt3/2\ \ =>\ \ text{Reference angle}\ =60^@`

`=>\ text{sin is negative in 3rd and 4th quadrants}`

`sin(theta-60^@)` `=180+60, 360-60`  
  `=240^@, 300^@`  

 
`theta-60^@=240^@\ \ =>\ \ theta=300^@`

`theta-60^@=300^@\ \ =>\ \ theta=360^@`

 
`text{Consider}\ theta = 0^@`

`sin(0-60^@)=sin(-60^@)=-sqrt3/2`

`:.theta=0^@, 300^@ and 360^@`

Advanced Trigonometry, 2ADV T2 2016 HSC 11g

Solve  \(\sin\left(\dfrac{x}{2}\right)= \dfrac{1}{2}\)  for  \(0^{\circ} \leq x \leq 360^{\circ}\).   (2 marks)

Show Answers Only

\(x = 60^{\circ}, 300^{\circ}\)

Show Worked Solution

\(\sin\left(\dfrac{x}{2}\right)= \dfrac{1}{2}\ \ \text{for}\ \ 0^{\circ} \leq x \leq 360^{\circ}\)

\(\Rightarrow\ \text{Reference angle}\ = 30^{\circ}\)

\(\dfrac{x}{2}\) \(= 30^{\circ}, 180-30, 360+30, …\)
  \(=30^{\circ}, 150^{\circ}, 390^{\circ},\ …\)

 

\(\therefore x = 60^{\circ}, 300^{\circ}\ \ \text{for}\ \ 0^{\circ} \leq x \leq 360^{\circ}\)

Advanced Trigonometry, 2ADV T2 2011 HSC 2b

Find the exact values of \(x\) such that  \(2\sin\,x =-\sqrt{3}\), where  \(0^{\circ} \leq x \leq 360^{\circ}\).   (2 marks)

Show Answers Only

\(240^{\circ}, 300^{\circ}\)

Show Worked Solution
MARKER’S COMMENT: Better responses found the reference angle and then identified the correct quadrants, as shown.
\(2\sin\,x\) \(=-\sqrt{3}\ \ \text{where}\ \ 0^{\circ} \leq x \leq 360^{\circ}\)
\(\sin\,x\) \(= -\dfrac{\sqrt{3}}{2}\)
\(\sin\,60^{\circ}\) \(= \dfrac{\sqrt{3}}{2}\)

 

\(\text{Since}\ \sin\,x\ \text{is negative in 3rd/4th quadrants:}\)

\(x\) \(= 180+60,\ 360-60\)
  \(= 240^{\circ}, 300^{\circ}\)

Advanced Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of  \(2\sin\,\theta = 1\)  for  \(0^{\circ} \leq \theta \leq 360^{\circ}\).   (2 marks)

Show Answers Only

\(30^{\circ}, 150^{\circ}\)

Show Worked Solution

\(2\sin\,\theta=1\ \ \Rightarrow\ \ \sin\,\theta=\dfrac{1}{2}\)

\(\text{Reference angle:}\ \sin\,30^{\circ}=\dfrac{1}{2}\)

\(\text{Since sin is positive in the 1st/2nd quadrants:}\)

\(\therefore\ \theta\) \(= 30^{\circ}, 180-30^{\circ}\)
  \(= 30^{\circ}, 150^{\circ}\)

Advanced Trigonometry, 2ADV T2 2009 HSC 1e

Find the exact value of \(\theta\) such that  \(2\cos\,\theta = 1\), where  \(0^{\circ} \leq \theta \leq 90^{\circ}\).   (2 marks)

Show Answers Only

 \(\theta = 60^{\circ}\)

Show Worked Solution
\(2 \cos\,\theta\) \(= 1\)
\(\cos\,\theta\) \(= \dfrac{1}{2}\)
\(\therefore \theta\) \(= 60^{\circ},\ \ \ \ 0^{\circ} \leq \theta \leq 90^{\circ}\)

Advanced Trigonometry, 2ADV T2 2007 HSC 4a

Solve  \(\sqrt{2}\,\sin\,x = 1\)  for  \(0^{\circ} \leq x \leq 360^{\circ}.\)   (2 marks)

Show Answers Only

\(45^{\circ}, 135^{\circ}\)

Show Worked Solution

\(\sqrt{2}\,\sin\,x = 1\ \ \Rightarrow \ \ \sin\,x= \dfrac{1}{\sqrt{2}} \)

\(\text{Reference angle:}\ \sin\,45^{\circ} = \dfrac{1}{\sqrt{2}}\)

\(\text{Since sin is positive in 1st/2nd quadrants:}\)

\(\therefore x\) \(= 45^{\circ}, (180-45)^{\circ}\)
  \(= 45^{\circ}, 135^{\circ}\)

Advanced Trigonometry, 2ADV T2 2005 HSC 2a

Solve  \(\cos\,\theta = \dfrac{1}{\sqrt{2}}\)  for  \(0^{\circ} ≤ \theta ≤ 360^{\circ}\).   (2 marks)

Show Answers Only

\(45^{\circ}, 315^{\circ}`

Show Worked Solution

\(\cos\,\theta = \dfrac{1}{\sqrt{2}}\)  for  \(0^{\circ} ≤ \theta ≤ 360^{\circ}\)

\(\text{Reference angle:}\ \cos\,45^{\circ} = \dfrac{1}{\sqrt{2}}\)

\(\text{Since cos is positive in 1st/4th quadrants:}\)

\(\theta\) \(= 45^{\circ}, 360-45`
  \(= 45^{\circ}, 315^{\circ}\)

 

Advanced Trigonometry, 2ADV T2 2012 HSC 6 MC

What are the solutions of  `sqrt3 tanx = -1`  for  `0^@<=x<=360^@`? 

  1. `120^@\ text(and)\ 240^@` 
  2. `120^@\ text(and)\ 300^@`
  3. `150^@\ text(and)\ 210^@` 
  4. `150^@\ text(and)\ 330^@` 
Show Answers Only

`D`

Show Worked Solution
`sqrt3 tanx` `= -1`
`tanx` `= -1/sqrt3`

 
`text(When)\ tanx = 1/sqrt3,\ \ x=30^@`

`text(S)text(ince)\ tanx\ text{is negative in 2nd/4th quadrant:}`

`:. x` ` = 180-30,360-30`
  `= 150^@,\ 330^@`

 
`=>  D`