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Financial Maths, STD1 F1 2021 HSC 12

A dance school runs a holiday program and charges $150 per student.

The costs of running the program are:

Dance teacher: $110 per hour

Room hire: $52 per hour plus 10% GST.

The program goes for two hours a day for five days.

What profit does the dance school make if 20 students pay for the program?  (3 marks)

Show Answers Only

`$1328`

Show Worked Solution

`text(Revenue) = 20 xx 150 = $3000`

`text(C)text(ost per hour)` `= 110 + (52 + \text{GST})`
  `= 110 + 52 xx 1.1`
  `= $167.20`
`text(Total cost)` `= 2 xx 5 xx 167.20`
  `= $1672`

 

`:.\ text(Profit)` `= 3000 – 1672`
  `= $1328`

Measurement, STD1 M4 2021 HSC 11

Chocolate of a particular brand can be bought in three different sizes.

Option 1: 100 grams for $1.50

Option 2: 300 grams for $4.20

Option 3: 500 grams for $7.25

Which option gives the lowest price per 100 grams? Justify your answer with calculations.  (2 marks)

Show Answers Only

`text(Option 2 is the cheapest.)`

Show Worked Solution

`text(Compare prices per 100 grams:)`

`text(Option 1) = $1.50`

`text(Option 2) = 4.20 ÷ 3 = $1.40`

`text(Option 3) = 7.25 ÷ 5 = $1.45`

`:.\ text(Option 2 is the cheapest.)`

Calculus, 2ADV C4 2021 HSC 28

The region bounded by the graph of the function  `f(x) = 8 - 2^x`  and the coordinate axes is shown
 

  1. Show that the exact area of the shaded region is given by  `24 - 7/ln2`.  (3 marks)

  2. A new function  `g(x)`  is found by taking the graph of   `y = -f(-x)`  and translating it by 5 units to the right.
  3. Sketch the graph of  `y = g(x)`  showing the `x`-intercept and the asymptote.  (2 marks)

  4. Hence, find the exact value of  `int_2^5 g(x)\ dx`.  (1 mark)

Show Answers Only
  1. `text(See Worked Solution)`
  2.  
  3. `7/(ln2) – 24`
Show Worked Solution

a.   `xtext(-intercept occurs when)`

`8 – 2^x = 0 \ => \ x = 3`

`text(Area)` `= int_0^3 8 – 2^x\ dx`
  `= [8x – (2^x)/(ln2)]_0^3`
  `= 24 – 8/(ln 2) – (0 – 1/(ln2))`
  `= 24 – 8/(ln2) + 1/(ln2)`
  `= 24 – 7/(ln2)\ \ text(u²)`

♦ Mean mark part (b) 48%.

b.

`y = f(–x) -> text(reflect)\ \ y = f(x)\ \ text(in the)\ ytext(-axis)`

`y = -f(–x) -> text(reflect)\ \ y = f(–x)\ \ text(in the)\ xtext(-axis)`
 

♦♦♦ Mean mark part (c) 13%.

c.   `int_2^5 g(x)\ dx\ \ text{is the same area as found in part (a)}`

`text(except it is below the)\ xtext(-axis.)`

`:. int_2^5 g(x)\ dx = 7/(ln2) – 24`

Financial Maths, STD1 F2 2021 HSC 9 MC

James invests $5000 at 3% per annum simple interest for 2 years, while Sally invests $5000 at 3% per annum compounded annually for 2 years.

For the given investments, which of the following statements is TRUE?

  1. James earns more interest than Sally.
  2. Sally earns more interest than James.
  3. James and Sally earn the same amount of interest.
  4. There is not enough information to compare the interest earned by James and Sally.
Show Answers Only

`B`

Show Worked Solution

`text{Investments earn more when interest is compounded (vs simple}`

`text{interest) because interest is paid on both the principal and earlier}`

`text{interest.}`

`:.\ text(Sally earns more interest than James.)`

`=> B`

Measurement, STD1 M4 2021 HSC 6 MC

Blood pressure is recorded as systolic pressure/diastolic pressure and measured in mmHg.

When standing upright, Mary’s blood pressure was 139/85. Three minutes after sitting down, her blood pressure was 118/74.

What was the change in Mary’s diastolic blood pressure?

  1. There was an increase of 11 mmHg.
  2. There was an increase of 21 mmHg.
  3. There was a decrease of 11 mmHg.
  4. There was a decrease of 21 mmH.
Show Answers Only

`C`

Show Worked Solution

`text(Diastolic pressure went from 85 to 74,)`

♦ Mean mark 45%.

`text(or decreased by 11 mmHg.)`

`=> C`

Calculus, 2ADV C4 2021 HSC 27

Kenzo has a solar powered phone charger. Its power, `P`, can be modelled by the function

`P(t) = 400 sin(pi/12 t),\ \ 0 <= t <= 12`,

where  `t`  is the number of hours after sunrise.

  1. Sketch the graph of  `P` for  `0 ≤ t ≤ 12`.  (2 marks)

Power is the rate of change of energy. Hence the amount of energy, `E` units, generated by the solar powered phone charger from  `t = a`  to  `t = b`,  where  `0 ≤ a ≤ b ≤ 12` is given by

`E = int_a^b P(t)\ dt`.

  1. Show that  `E = 4800/pi (cos\ (api)/12 - cos\ (bpi)/12)`.  (2 marks)

  2. To make a phone call, a phone battery needs at least 300 units of energy. Kenzo woke up 3 hours after sunrise and found that his phone battery had no units of energy. He immediately began to use his solar powered charger to charge his phone battery.
  3. Find the least amount of time he needed to wait before he could make a phone call. Give your answer correct to the nearest minute.  (3 marks)

  4. The next day, Kenzo woke up 6 hours after sunrise and again found that his phone battery had no units of energy. He immediately began to use his solar powered charger to charge his phone battery.
  5. Would it take more time or less time or the same amount of time, compared to the answer in part (c), to charge his phone battery in order to make a phone call? Explain your answer by referring to the graph drawn in part (a).  (1 mark)

Show Answers Only
  1.  
  2. `text(See Worked Solution)`
  3. `text(57 minutes)`
  4. `text(The power produced is at its peak when)\ t = 6`
  5. `:.\ text(It will charge in less time.)`
Show Worked Solution

a.

b.    `E` `= int_a^b 400 sin(pi/12 t)\ dt`
    `= [-400 · 12/pi cos(pi/12 t)]_a^b`
    `= -4800/pi cos(pi/12 b) – (-4800/pi cos(pi/12 a))`
    `= 4800/pi(cos\ (api)/12 – cos\ (bpi)/12)`

 

♦♦ Mean mark part (c) 29%
COMMENT: It is arguable that the nearest minute may also be 3h 58 m as the phone is not adequately charged at 3h 57 m.

c.   `text(Find)\ \ b\ \ text(given)\ \ E = 300\ \ text(and)\ \ a = 3:`

`300` `= 4800/pi (cos\ pi/4 – cos\ (bpi)/12)`
`(300pi)/4800` `= 1/sqrt2 – cos\ (bpi)/12`
`cos\ (bpi)/12` `= 1/sqrt2 – pi/16`
`(bpi)/12` `= cos^(-1) (1/sqrt2 – pi/16)`
`b` `= 12/pi cos^(-1)(1/sqrt2 – pi/16)`
  `= 3.952…`
  `= 3\ text{h 57 m  (nearest minute)}`

 
`:.\ text(Least time before phone is charged = 57 minutes)`

 

♦♦ Mean mark part (d) 22%

d.   `text(The power produced is at its peak when)\ \ t = 6.`

`:.\ text(It will charge in less time.)`

Statistics, 2ADV S3 2021 HSC 33

People are given a maximum of six hours to complete a puzzle. The time spent on the puzzle, in hours, can be modelled using the continuous random variable \(X\) which has probability density function

\(f(x)= \begin{cases}
\dfrac{A x}{x^2+4} & \text{for } 0 \leq x \leq 6,(\text { where } A>0) \\
\ \\
0 & \text {for all other values of } x
\end{cases}\)

The graph of the probability density function is shown below. The graph has a local maximum.
 

  1. Show that  \(A=\dfrac{2}{\ln 10}\).  (2 marks)

  2. Show that the mode of \(X\) is two hours.  (2 marks)

  3. Show that  \(P(X<2)=\log _{10} 2\).  (2 marks)

  4. The Intelligence Quotient (IQ) scores of people are normally distributed with a mean of 100 and standard deviation of 15.
  5. It has been observed that the puzzle is generally completed more quickly by people with a high IQ.
  6. It is known that 80% of people with an IQ greater than 130 can complete the puzzle in less than two hours.
  7. A person chosen at random can complete the puzzle in less than two hours.
  8. What is the probability that this person has an IQ greater than 130? Give your answer correct to three decimal places.  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solution}\)
  2. \(\text{See Worked Solution}\)
  3. \(\text{See Worked Solution}\)
  4. \(0.066\)
Show Worked Solution

a.  \(\displaystyle\int_0^6 \dfrac{A x}{x^2+4} \, d x=1\)

\(\begin{aligned} \dfrac{A}{2} \int_0^6 \dfrac{2 x}{x^2+4} d x & =1 \\
\dfrac{A}{2}\left[\ln \left(x^2+4\right)\right]_0^6 & =1 \\
\dfrac{A}{2}(\ln 40-\ln 4) & =1 \\
\dfrac{A}{2} \ln \left(\dfrac{40}{4}\right) & =1 \\
\dfrac{A}{2} \ln 10 & =1 \\
A & =\dfrac{2}{\ln 10}\end{aligned}\)

b.  \(\text{Mode \(\rightarrow f(x)\) is a MAX}\)

♦♦♦ Mean mark part (b) 24%.

\(\begin{aligned}
f(x) & =\dfrac{A x}{x^2+4} \\
f^{\prime}(x) & =\dfrac{A\left(x^2+4\right)-A x(2 x)}{\left(x^2+4\right)^2} \\
& =\dfrac{A x^2+4 A-2 A x^2}{\left(x^2+4\right)^2} \\
& =\dfrac{A\left(4-x^2\right)}{\left(x^2+4\right)^2}
\end{aligned}\)

\(\text{\(f(x)\) max occurs when \(f^{\prime}(x)=0\) :}\)

\(\begin{aligned}
4-x^2 & =0 \\
x & =2 \quad(x>0)
\end{aligned}\)

♦♦ Mean mark part (c) 30%.
c.    \(P(X<2)\) \(=\displaystyle \int_0^2 \dfrac{A x}{x^2+4} d x\)
    \(=\dfrac{A}{2}\left[\ln \left(x^2+4\right)\right]_0^2\)
    \(=\dfrac{1}{\ln 10}(\ln 8-\ln 4)\)
    \(=\dfrac{1}{\ln 10}\left(\ln \dfrac{8}{4}\right)\)
    \(=\dfrac{1}{\ln 10} \cdot \ln 2\)
    \(=\log _{10} 2\)

 

d.   \(z \text{-score}(130)=\dfrac{x-\mu}{\sigma}=\dfrac{130-100}{15}=2\)

♦♦ Mean mark part (d) 25%.

\(P(z>2)=2.5 \%\)

\begin{aligned}
P(\text { IQ }>130 \mid x<2) & =\dfrac{P(\text { IQ }>130 \cap X<2)}{P(X<2)} \\
& =\dfrac{0.8 \times 0.025}{\log _{10} 2} \\
& =0.0664 \ldots \\
& =0.066
\end{aligned}

Statistics, 2ADV S2 2021 HSC 17

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2 − 0.011x`.

The correlation coefficient is  `r = –0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.  (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.  (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6 − 0.683x`.
  4. The correlation coefficient is  `r = − 0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.  (1 mark)

Show Answers Only
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `text(Latitude. Correlation coefficient shows a stronger relationship.)`
Show Worked Solution
a.i.    `y` `=29.2 – 0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
 

a.ii.  `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

b.  `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Statistics, STD2 S4 2021 HSC 33

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2 − 0.011x`.

The correlation coefficient is  `r = –0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.  (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.  (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6 − 0.683x`.
  4. The correlation coefficient is  `r = − 0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.  (1 mark)

Show Answers Only
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `text(Latitude. Correlation coefficient shows a stronger relationship.)`
Show Worked Solution
a.i.    `y` `=29.2 – 0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
♦♦ Mean mark part a.ii. 28%.

 
a.ii.  `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

♦♦♦ Mean mark part b 18%.

b.  `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Financial Maths, 2ADV M1 2021 HSC 29

  1. On the day that Megan was born, her grandfather deposited $5000 into an account earning 3% per annum compounded annually. On each birthday after this, her grandfather deposited $1000 into the same account, making his final deposit on Megan’s 17th birthday. That is, a total of 18 deposits were made.
  2. Let `A_n` be the amount in the account on Megan’s `n`th birthday, after the deposit is made.
  3. Show that  `A_3 = $8554.54`.  (2 marks)

  4. On her 17th birthday, just after the final deposit is made, Megan has $30 025.83 in her account. You are NOT required to show this.
  5. Megan then decides to leave all the money in the same account continuing to earn interest at 3% per annum compounded annually. On her 18th birthday, and on each birthday after this, Megan withdraws $2000 from the account.
  6. How many withdrawals of $2000 will Megan be able to make?  (3 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `20`
Show Worked Solution

a.   `A_1 = 5000(1.03) + 1000`

`A_2` `= [5000(1.03) + 1000](1.03) + 1000`
  `= 5000(1.03)^2 + 1000(1.03) + 1000`
`A_3` `= 5000(1.03)^3 + 1000(1.03)^2 + 1000(1.03) + 1000`
  `= 8554.535`
  `= $8554.54`

 

b.   `text(Megan’s 18th birthday onwards:)`

`A_1 = 30\ 025.83(1.03) – 2000`

`A_2` `= [30\ 025.83(1.03) – 2000](1.03) – 2000`
  `= 30\ 025.83(1.03)^2 – 2000(1.03) – 2000`
  `vdots`
`A_n` `= 30\ 025.83(1.03)^n – 2000(1 + 1.03 + 1.03^2 + … + 1.03^(n – 1))`

 

`text(Find)\ n\ text(when)\ \ A_n = 0:`

`30\ 025.83(1.03)^n` `= 2000[(a(r^n – 1))/(r – 1)]`
  `= 2000((1.03^n – 1)/(1.03 – 1))`
`30\ 025.83(1.03)^n` `= 66\ 666.67(1.03)^n – 66\ 666.67`
`36\ 640.84(1.03)^n` `= 66\ 666.67`
`1.03^n` `= (66\ 666.67)/(36\ 640.84)`
`n ln 1.03` `= ln ((66\ 666.67)/(36\ 640.84))`
`n` `= (ln ((66\ 666.67)/(36\ 640.84)))/(ln 1.03)`
  `= 20.24…`

 
`:.\ text(Megan can make 20 withdrawals of $2000.)`

Measurement, STD2 M6 2021 HSC 32

A right-angled triangle  `XYZ`  is cut out from a semicircle with centre `O`. The length of the diameter  `XZ`  is 16 cm and  `angle YXZ`  = 30°, as shown on the diagram.
 


 

  1. Find the length of  `XY`  in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of the shaded region in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86 \ text{cm}`
  2. `45.1 \ text{cm}^2`
Show Worked Solution

 

a.    `cos 30^@` `=(XY)/16`
  `XY` `= 16 \ cos 30^@`
    `= 13.8564`
    `= 13.86 \ text{cm (2 d.p.)}`

 

b.    `text{Area of semi-circle}` `= 1/2 times pi r^2`
    `= 1/2 pi times 8^2`
    `= 100.531 \ text{cm}^2`

♦ Mean mark part (b) 36%.
`text{Area of} \ Δ XYZ` `= 1/2 ab\ sin C`  
  `= 1/2 xx 16 xx 13.856 xx sin 30^@`  
  `= 55.42 \ text{cm}^2`  

 

`:. \ text{Shaded Area}` `= 100.531-55.42`  
  `= 45.111`  
  `= 45.1 \ text{cm}^2 \ \ text{(1 d.p.)}`  

Trigonometry, 2ADV T1 2021 HSC 18

The diagram shows a triangle `ABC` where `AC` = 25 cm, `BC` = 16 cm, `angle BAC` = 28° and angle `ABC` is obtuse.
 


 

Find the size of the obtuse angle `ABC` correct to the nearest degree.  (3 marks)

Show Answers Only

`133°`

Show Worked Solution

`text(Using the sine rule:)`

`sin theta/25` `= (sin 28°)/16`
`sin theta` `= (25 xx sin 28°)/16`
`sin theta` `= 0.73355`
`theta` `= 47°`
 
`:. angleABC` `= 180-47`
  `= 133°`

Financial Maths, 2ADV M1 2021 HSC 25

A table of future value interest factors for an annuity of $1 is shown.
 

   

Simone deposits $1000 into a savings account at the end of each year for 8 years. The interest rate for these 8 years is 0.75% per annum, compounded annually.

After the 8th deposit, Simone stops making deposits but leaves the money in the savings account. The money in her savings account then earns interest at 1.25% per annum, compounded annually, for a further two years.

Find the amount of money in Simone's savings account at the end of ten years.  (3 marks)

Show Answers Only

`$8419.81`

Show Worked Solution

`text(In 1st 8 years:)`

Mean mark 52%.

`text(Future value factor = 8.2132)`

`text(Value of annuity)` `= 8.2132 xx 1000`
  `= $8213.20`
 
`text(After 10 years:)` 
`text(Value of investment)` `= 8213.2 xx (1.0125)^2`
  `= $$8419.81`

Statistics, STD2 S5 2021 HSC 38

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable lies between 0 and `z` for different values of `z`.

 

The probability values given in the table for different values of `z` are represented by the shaded area in the following diagram.
 

  1. Using the table, show that the probability that a value from a random variable that is normally distributed with mean 0 and standard deviation 1 is greater than 0.3 is equal to 0.3821.  (1 mark)

  2. Birth weights are normally distributed with a mean of 3300 grams and a standard deviation of 570 grams. By first calculating a `z`-score, find how many babies, out of 1000 born, are expected to have a birth weight greater than 3471 grams.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `382`
Show Worked Solution

a.   `P(z>0)=0.5`

♦♦♦ Mean mark part (a) 1%.
COMMENT: Note the Std2 and Advanced questions varied slightly but used the same table and graph.

`P(0<z<0.3)=0.1179`

`P(z>0.3) = 0.5-0.1179=0.3821`
 

b.   `z text{-score (3471)}` `=(x-mu)/sigma`
    `=(3471-3300)/570`
    `=0.3`

 
`P(z>0.3) = 0.3821\ \ text{(see part (a))}`
  

`:.\ text(Number of babies > 3471 grams)`  
`=1000 xx 0.3821`  
`=382\ \ text{(nearest whole)}`  

Calculus, 2ADV C3 2021 HSC 26

A particle is shot vertically upwards from a point 100 metres above ground level.

The position of the particle, `y` metres above the ground after `t` seconds, is given by

`y(t) = −5t^2 + 70t + 100`.

  1. Find the maximum height above ground level reached by the particle.  (2 marks)

  2. Find the velocity of the particle, in metres per second, immediately before it hits the ground, leaving your answer in the form  `asqrtb`,  where  `a`  and  `b`  are integers.  (3 marks)

Show Answers Only
  1. `345\ text(m)`
  2. `-20sqrt69\ text(m/s)`
Show Worked Solution

a.    `y(t) = -5t^2 + 70t + 100`

`y^(′)(t) = -10t + 70`

`y^(″)(t) = -10`

`text(Max height occurs when)\ \ y^(′)(t) = 0:`

`-10t + 70` `= 0`
`10t` `= 70`
`t` `= 7`

 

`:.\ text(Max height)\ ` `= -5(7)^2 + 70 xx 7 + 100`
  `= 345\ text(m)`

 

b.   `text(Particle hits ground when)\ \ y = 0:`

♦ Mean mark 38%.
`0` `= -5t^2 + 70t + 100`
`0` `= t^2 – 14t – 20`

 
`text(Using quadratic formula:)`

`t` `= (14 ± sqrt(14^2 + 4*20))/2`
  `= (14 + sqrt(276))/2\ \ \ (t > 0)`
  `= 7 + sqrt69`

 

`:. text(Velocity)\ (y^(′)(t))` `= -10(7 + sqrt69) + 70`
  `= -10sqrt69\ text(m/s)`

 
`:.a=-10 and b=69`

Calculus, 2ADV C4 2021 HSC 24

The curve  `y = 3/(x - 1)`  intersects the line  `y = 3/2 x`  at the point (2, 3).

The region bounded by the curve  `y = 3/(x - 1)`, the line  `y = 3/2 x`, the  `x`-axis and the line  `x = 4`  is shaded in the diagram.
 

Find the exact area of the shaded region.  (3 marks)

Show Answers Only

`3 + 3 log_e 3\ text(u)²`

Show Worked Solution
`text(Area)` `= int_0^2 3/2 x\ dx + int_2^4 3/(x – 1)\ dx`
  `= [3/4 x^2]_0^2 + 3[log_e(x – 1)]_2^4`
  `= 12/4 + 3[log_e 3 – log_e 1]`
  `= 3 + 3 log_e 3\ \ text(u²)`

Trigonometry, 2ADV T3 2021 HSC 20

For what values of `x`, in the interval  `0 <= x <= pi/4`,  does the line  `y = 1`  intersect the graph of  `y = 2sin4x`?  (2 marks)

Show Answers Only

`pi/24, (5pi)/24`

Show Worked Solution

`text(Find)\ x\ text(such that:)`

`2sin4x` `= 1`
`sin4x` `= 1/2`
`4x` `= sin^(-1)\ 1/2`
`4x` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, …`
`:. x` `= pi/24, (5pi)/24\ \ \ (0 <= x <= pi/4)`

Algebra, STD2 A4 2021 HSC 24

A population,  `P`,  is to be modelled using the function  `P = 2000 (1.2)^t`,  where `t` is the time in years.

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

  2. Find the population after 5 years.  (1 mark)

  3. On the axes below, draw the graph of the population against time, showing the points at  `t = 0`  and at  `t = 5`.  (2 marks)
      

Show Answers Only
  1. `2000`
  2. `4977`
  3. `text{See Worked Solutions}`
Show Worked Solution
a.    `text{Initial population occurs when}\ \  t = 0.`
`P` `= 2000 (1.2)^0`  
  `= 2000`  

 

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

`P` `= 2000 (1.2)^5`  
  `= 4976.64`  
  `= 4977 \ text{(nearest whole)}`  

 

♦ Mean mark (c) 48%.

c. 

Networks, STD1 N1 2021 HSC 17

The network diagram shows the travel times in minutes along roads connecting a number of different towns.
 


 

  1. Draw a minimum spanning tree for this network and determine its length.  (3 marks)

  2. How long does it take to travel from Queentown to Underwood using the fastest route?  (1 mark)

Show Answers Only
  1. `\text{See Worked Solutions}`
  2. `65 \ \text{minutes}`
Show Worked Solution

a.   `text{Using Prim’s algorithm (starting at}\ W):`

♦ Mean mark part (a) 40%.

`text(1st edge:)\ \ WM`

`text(2nd edge:)\ \ WP`

`text(3rd/4th edges:)\ \ MU\ text(and)\ \ WF`

`text(5th edge:)\ \ MK`

`text(6th edge:)\ \ KQ`

`text(7th edge:)\ \ PC`

 

`text{Length of minimum spanning tree} \ = 160` 
 
b.   `text{Fastest route}\ (Q\ text(to)\ U)` `= 45 + 20`
    `= 65 \ text{minutes}`

Networks, STD2 N2 2021 HSC 23

The network diagram shows the travel times in minutes along roads connecting a number of different towns.
 


 

  1. Draw a minimum spanning tree for this network and determine its length.  (3 marks)

  2. How long does it take to travel from Queentown to Underwood using the fastest route?  (1 mark)

Show Answers Only
  1. `\text{See Worked Solutions}`
  2. `65 \ \text{minutes}`
Show Worked Solution

a.   `text{Using Prim’s algorithm (starting at}\ W):`

`text(1st edge:)\ \ WM`

`text(2nd edge:)\ \ WP`

`text(3rd/4th edges:)\ \ MU\ text(and)\ \ WF`

`text(5th edge:)\ \ MK`

`text(6th edge:)\ \ KQ`

`text(7th edge:)\ \ PC`

 

`text{Length of minimum spanning tree} \ = 160` 
 
b.   `text{Fastest route}\ (Q\ text(to)\ U)` `= 45 + 20`
    `= 65 \ text{minutes}`

Calculus, 2ADV C3 2021 HSC 13

Find the exact gradient of the tangent to the curve  `y = x tan x`  at the point where  `x = pi/3`.  (3 marks)

Show Answers Only

`sqrt3 + (4pi)/3`

Show Worked Solution

`y = x tan x`

`(dy)/(dx) = tan x + x sec^2 x`

`text(Find)\ \ m\ \ text(when)\ \ x = pi/3:`

`(dy)/(dx)` `= tan\ pi/3 + pi/3 · 1/(cos^2\ pi/3)`
  `= sqrt3 + pi/3 · 1/(1/4)`
  `= sqrt3 + (4pi)/3`

Financial Maths, STD2 F1 2021 HSC 22

The table shows the income tax rates for the 2020-2021 financial year.
 

   

William has a gross annual salary of $84 000. He has allowable tax deductions of $900 for home-office equipment and $474 for union fees. William must also pay a Medicare Levy of 2% of his taxable income.

Calculate the total tax payable by William including the Medicare Levy.  (3 marks)

Show Answers Only

`$18\ 972.97`

Show Worked Solution
`text(Taxable income)` `=84\ 000-900-474`  
  `=$82\ 626`  
`text(Income tax)` `=5092+0.325(82\ 626 – 45\ 000)`  
  `=$17\ 320.45`  
`text(Medicare Levy)` `=0.02 xx 82\ 626`  
  `=$1652.52`  

 

`:.\ text(Total tax payable)` `=17\ 320.45 + 1652.52`  
  `=$18\ 972.97`  

Measurement, STD2 M1 2021 HSC 16

The volume, `V`, of a sphere is given by the formula

`V = frac{4}{3} pi r^3,`

where `r` is the radius of the sphere.

A tank consists of the bottom half of a sphere of radius 2 metres, as shown.
 

Find the volume of the tank in cubic metres, correct to one decimal place.   (2 marks)

Show Answers Only

`16.8\ text{m}^3`

Show Worked Solution
 `V` `= frac{1}{2} times frac{4}{3} pi r^3`
  `= frac{1}{2} times frac{4}{3} times pi times 2^3`
  `= 16.755…`
  `= 16.8\ text{m}^3\ \ text{(1 d.p.)}`

Functions, 2ADV F1 2021 HSC 8 MC

The graph of  `y = f(x)`  is shown.

Which of the following could be the equation of this graph?

  1. `y = (1 - x)(2 + x)^3`
  2. `y = (x + 1)(x - 2)^3`
  3. `y = (x + 1)(2 - x)^3`
  4. `y = (x - 1)(2 + x)^3`
Show Answers Only

`C`

Show Worked Solution

`text(By elimination:)`

`text(A single negative root occurs when)\ \ x =–1`

`->\ text(Eliminate A and D)`

`text(When)\ \ x = 0, \ y > 0`

`->\ text(Eliminate B)`

`=> C`

Calculus, 2ADV C3 2021 HSC 7 MC

The diagram shows part of  `y = f(x)`  which has a local minimum at  `x = –2`  and a local maximum at  `x = 3`.
 

Which of the following shows the correct relationship between  `f^(″)(–2), \ f(0)`  and  `f^(′)(3)`?

  1. `f(0) < f^(′)(3) < f^(″)(–2)`
  2. `f(0) < f^(″)(–2) < f^(′)(3)`
  3. `f^(″)(–2) < f^(′)(3) < f(0)`
  4. `f^(″)(–2) < f(0) < f^(′)(3)`
Show Answers Only

`A`

Show Worked Solution

Mean mark 52%.
`f^(″)(–2)` `> 0\ \ \ (text(concave up at)\ \  x = –2)`
`f(0)` `< 0\ \ \ (text(see graph))`
`f^(′)(3)` `= 0\ \ \ (text(S.P.))`

 
`:. f(0) < f^(′)(3) < f^(″)(–2)`

`=> A`

Measurement, STD2 M1 2021 HSC 12 MC

A block of land is represented by the shaded region on the number plane. All measurements are in kilometres. 
 

Which of the following is the approximation for the area of this block of land in square kilometres, using two applications of the trapezoidal rule?

  1. 99
  2. 19.8
  3. 39.6
  4. 72
Show Answers Only

`B`

Show Worked Solution

`\text{Area}` `≈ \frac{6}{2} (1.2 + 2 \times 2 + 1.4)`
  `≈ 3 (6.6)`
  `≈ 19.8 \ \text{km}^2`

 
`=> B`

Probability, 2ADV S1 2021 HSC 6 MC

There are 8 chocolates in a box. Three have peppermint centres (P) and five have caramel centres (C).

Kim randomly chooses a chocolate from the box and eats it. Sam then randomly chooses and eats one of the remaining chocolates.

A partially completed probability tree is shown.
 

What is the probability that Kim and Sam choose chocolates with different centres?

  1. `\frac{15}{64}`
  2. `\frac{15}{56}`
  3. `\frac{15}{32}`
  4. `\frac{15}{28}`
Show Answers Only

`D`

Show Worked Solution

 

`Ptext{(different centres)}` `= P text{(PC)} + P text{(CP)}`
  `=\frac{3}{8} · \frac{5}{7} + \frac{5}{8} · \frac{3}{7}`
  `= \frac{15}{56} + \frac{15}{56}`
  `= \frac{15}{28}`

 
`=> D`

Functions, 2ADV F2 2021 HSC 5 MC

Which of the following best represents the graph of  `y = 10 (0.8)^x`?
 

Show Answers Only

`A`

Show Worked Solution

`\text{By elimination:}`

`\text{When} \ x = 0 \ , \ y = 10(0.8) ^0 = 10`

`-> \ text{Eliminate B and D}`

`text(As)\ \ x→oo, \ y→0`

`-> \ text{Eliminate C}`

`=> A`

Algebra, STD1 A2 2021 HSC 8 MC

A student is thinking of a number. Let the number be `x`.

When the student subtracts 8 from this number and multiplies the result by 3, the answer is 2 more than `x`.

Which equation can be used to find `x`?

  1. `3(x-8)=2x`
  2. `3x-8=2x`
  3. `3(x-8)=x+2`
  4. `3x-8=x+2`
Show Answers Only

`C`

Show Worked Solution

`text(The description defines the following equation:)`

`(x-8) xx 3` `= x + 2`
`3(x-8)` `=x+2`

 
`=>C`

Financial Maths, STD1 F3 2021 HSC 4 MC

Three years ago an appliance was valued at $2467. Its value has depreciated by 15% each year, based on the declining-balance method.

What is the salvage value today, to the nearest dollar?

  1. $952
  2. $1110
  3. $1357
  4. $1515
Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 50%.
`S` `= V_0 (1 – r)^n`
  `= 2467 (1 – 0.15)^3`
  `= 2467 (0.85)^3`
  `= $1515`

 
`=>  D`

Networks, STD2 N2 2021 HSC 2 MC

Consider the network diagram.
 

What is the sum of the degrees of all the vertices in this network?

  1.  5
  2.  8
  3.  14
  4.  16
Show Answers Only

`D`

Show Worked Solution

`text(Working from)\ A\ text(to)\ E:`

COMMENT: This simple question caused problems for many with mean mark just 58%.
`text{Sum of degrees}` `= 4 + 3 + 4 + 2 + 3`
  `= 16`

 
`=> D`

Measurement, STD2 M7 2021 HSC 27

The price and the power consumption of two different brands of television are shown.

The average cost for electricity is 25c/kWh. A particular family watches an average of 3 hours of television per day.

  1. The annual cost of electricity for Television A for this family is $48.18.
  2. For this family, what is the difference in the annual cost of electricity between Television A and Television B?  (2 marks)

  3. For this family, how many years will it take for the total cost of buying and using Television A to be equal to the cost of buying and using Television B?  (2 marks)

Show Answers Only
  1. `$4.38`
  2. `5\ text(years)`
Show Worked Solution
a.   `text{Annual power usage (B)}` `= 160 xx 3 xx 365`
    `=175\ 200`
    `=175.2\ text(kWh)`

♦ Mean mark part (a) 49%.
  `text{Annual cost (B)}` `= 175.2 xx 0.25`
    `=$43.80`

 

  `text{Difference in cost}` `= 48.18 – 43.80`
    `=$4.38`

♦♦ Mean mark part (b) 23%.
b.   `text{Difference in price}` `= 921.90 – 900`
    `=$21.90`

 

  `text(Years to even out cost)` `=21.90/4.38`
    `=5\ text{years}`

Financial Maths, STD2 F4 2021 HSC 26

Nina plans to invest $35 000 for 1 year. She is offered two different investment options.

Option A:  Interest is paid at 6% per annum compounded monthly.

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

  1. Calculate the future value of Nina's investment after 1 year if she chooses Option A.  (2 marks)

  2. Find the value of `r` in Option B that would give Nina the same future value after 1 year as for Option A. Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `$37\ 158.72`
  2. `6.17text(%)`
Show Worked Solution
a.   `r` `= text(6%)/12= text(0.5%) = 0.005\ text(per month)`
  `n` `=12`

 

`FV` `= PV(1 + r)^n`
  `= 35\ 000(1 + 0.005)^(12)`
  `= $37\ 158.72`

♦♦ Mean mark part (b) 36%.
b.   `I` `=Prn`
  `2158.72` `=35\ 000 xx r xx 1`
  `r` `=2158.72/(35\ 000)`
    `=0.06167…`
    `=6.17 text{% (to 2 d.p.)}`

Measurement, STD2 M2 2021 HSC 20

City A is in Sweden and is located at (58°N, 16°E). Sydney, in Australia, is located at (33°S, 151°E).

Robert lives in Sydney and needs to give an online presentation to his colleagues in City A starting at 5:00 pm Thursday, local time in Sweden.

What time and day, in Sydney, should Robert start his presentation?

It is given that 15° = 1 hour time difference. Ignore daylight saving.  (3 marks)

Show Answers Only

`text(2 am Friday)`

Show Worked Solution

`text{Angular difference}\ = 151 – 16 = 135°`

Mean mark 52%.

`=>\ text{Time difference}\ = 135/15 = 9\ text(hours)`

`text(Sydney is east of Sweden → ahead)`
 

`text{Presentation time (Sydney)}` `=\ text(5 pm Thurs + 9 hours)`  
  `=\ text(2 am Friday)`  

Statistics, STD2 S1 2021 HSC 17

The five-number summary of a dataset is given.

Lowest score = 1

Lowest quartile (`Q_1`) = 4

Median (`Q_2`) = 7

Upper quartile (`Q_3`) = 10

Highest score = 20

Is 20 an outlier? Justify your answer with calculations.  (2 marks)

Show Answers Only

`text(20 is an outlier – See Worked Solution)`

Show Worked Solution

`IQR = Q_3 – Q_1 = 10-4=6`

COMMENT: Stating 20 > 19 is necessary to justify this answer.
`text(Upper Fence)` `= Q_3 + 1.5 xx IQR`
  `=10 + 1.5 xx 6`
  `=19`

  
`text{S}text{ince 20 > 19, 20 is an outlier.}`

Algebra, STD2 A2 2021 HSC 9 MC

A student is thinking of a number. Let the number be `x`.

When the student subtracts 8 from this number and multiplies the result by 3, the answer is 2 more than `x`.

Which equation can be used to find `x`?

  1. `3(x-8)=2x`
  2. `3x-8=2x`
  3. `3(x-8)=x+2`
  4. `3x-8=x+2`
Show Answers Only

`C`

Show Worked Solution

`text(The description defines the following equation:)`

`(x-8) xx 3` `= x + 2`
`3(x-8)` `=x+2`

 
`=>C`

Statistics, STD2 S5 2021 HSC 8 MC

On a test, Zac's mark corresponded to a `z`-score of 2. The test scores had a mean of 63 and a standard deviation of 8.

What was Zac's actual mark on the test?

  1. 65
  2. 67
  3. 73
  4. 79
Show Answers Only

`D`

Show Worked Solution

`text{Method 1 (quicker)}`

`text(Actual mark)` `= mu + 2 xx sigma`
  `= 63 + 2 xx 8`
  `= 79`

 

`text(Method 2)`

`ztext(-score)` `= (x – mu)/sigma`
`2` `= (x-63)/8`
`16` `=x-63`
`x` `=79`

 
`=>D`

Financial Maths, STD2 F2 2021 HSC 5 MC

Peter currently earns $21.50 per hour. His hourly wage will increase by 2.1% compounded each year for the next four years.

What will his hourly wage be after four years?

  1. `21.50(1.21)^4`
  2. `21.50(1.021)^4`
  3. `21.50 + 21.50 xx 0.21 xx 4`
  4. `21.50 + 21.50 xx 0.021 xx 4`
Show Answers Only

`B`

Show Worked Solution

`text(Wage after 1 year) = 21.50 xx 1.021`

`text(Wage after 2 years) = 21.50 xx 1.021 xx 1.021 = 21.50(1.021)^2`

`vdots`

`text(Wage after 4 years) = 21.50(1.021)^4`

`=>  B`

Measurement, NAP-E3-NC03v1

At the start of the day, a farmer drove around his property inspecting the fences and livestock.

He started the drive at 6:15 and finished at 7:04.

How long did the farmer drive for?

`text(49 minutes)` `text(51 minutes)` `text(71 minutes)` `text(89 minutes)`
 
 
 
 
Show Answers Only

`text(49 minutes)`

Show Worked Solution

`text(49 minutes)`

Measurement, NAP-K3-CA01v1

Fiona is nurse who is administering vaccines to patients using a needle

Which unit would be the most appropriate to measure the volume of vaccine she needs to inject?

milligrams joules millimetres millilitres
 
 
 
 
Show Answers Only

`text(millilitres)`

Show Worked Solution

`text(A needle injects liquid vaccine into people.)`

`text(The most appropriate measure: millilitres)`

Measurement, NAP-E2-05v1

Helene started her walk to work at 2:15 pm.

She arrived at her work at 3:03 pm.

How long did Helene walk for?

`text(12 minutes)` `text(48 minutes)` `text(72 minutes)` `text(88 minutes)`
 
 
 
 
Show Answers Only

`text(48 minutes)`

Show Worked Solution

`text(One Strategy:)`

`text(2:15 pm to 3:00 pm = 45 minutes)`

`text(3:00 pm to 3:03 pm = 3 minutes)`

`text(Walking time = 45 + 3 = 48 minutes)`

Number and Algebra, NAP-I2-01v1

Kranskie delivers brochures by hand and is paid 10 cents for every brochure he delivers.

Kranskie delivers 79 brochures in his first hour of work.

How much money will he be paid for this?

`79¢` `$7.90` `$79` `$790`
 
 
 
 
Show Answers Only

`$7.90`

Show Worked Solution

`79 xx 10\ text(cents)`

`= 790\ text(cents)`

`= $7.90`

Number and Algebra, NAP-C1-13v1

Ernie has collected 93 bottles for recycling.

Grover has collected 88 bottles for recycling.

In total, how many bottles can Ernie and Grover deliver to the recycling depot?

`165` `171` `177` `181`
 
 
 
 
Show Answers Only

`181`

Show Worked Solution

`text(One strategy:)`

`text(Total bottles)` `=93+88`
  `=90 + 80 + 3 + 8`
  `=170 + 11`
  `=181`

Networks, STD2 N3 FUR2 4

Training program 1 has the cricket team starting from exercise station `S` and running to exercise station `O`.

For safety reasons, the cricket coach has placed a restriction on the maximum number of people who can use the tracks in the fitness park.

The directed graph below shows the capacity of the tracks, in number of people per minute.
 


 

  1. Determine the capacity of Cut 1, shown above.  (1 mark)

  2. What is the maximum flow from `S` to `O`, in number of people per minute?  (1 mark)

Show Answers Only
  1. `52`
  2. `50`
Show Worked Solution
a.   `text{Capacity (Cut 1)}` `= 20 + 12 + 20`
    `= 52`

 

b.   `text(Max flow/minimum cut)`

♦♦ Mean mark part (c) 32%.

`= 20 + 10 + 20`

`= 50`
 

NETWORKS, FUR2 2020 VCAA 3

A local fitness park has 10 exercise stations: `M` to `V`.

The edges on the graph below represent the tracks between the exercise stations.

The number on each edge represents the length, in kilometres, of each track.
 


 

The Sunny Coast cricket coach designs three different training programs, all starting at exercise station `S`.

  Training program 
number		 Training details
  1 The team must run to exercise station `O`.
  2 The team must run along all tracks just once.
  3 The team must visit each exercise station and return to  exercise station `S`.

 

  1. What is the shortest distance, in kilometres, covered in training program 1?  (1 mark)

  2.  i. What mathematical term is used to describe training program 2?   (1 mark)

  3. ii. At which exercise station would training program 2 finish?  (1 mark)

  4. To complete training program 3 in the minimum distance, one track will need to be repeated.

     

    Complete the following sentence by filling in the boxes provided.  (1 mark)


    This track is between exercise station   
     
    		 and exercise station   
     
Show Answers Only
  1. `3.2\ text(km)`
  2.  i. `text(Eulerian trail)`
  3. ii. `text(Station)\ P`
  4. `S and T`
Show Worked Solution
a.   `text(Shortest distance)` `= STUVO`
    `= 0.6 + 1.2 + 0.6 + 0.8`
    `= 3.2\ text(km)`

 

b.i.  `text(Eulerian trail)`
 

b.ii.  `text(Station)\ P\ text{(only other vertex with}\ S\ text{to have odd degree)} `

♦♦ Mean mark part (c) 25%.

 

c.   `S and T`

NETWORKS, FUR2 2020 VCAA 2

A cricket team has 11 players who are each assigned to a batting position.

Three of the new players, Alex, Bo and Cameron, can bat in position 1, 2 or 3.

The table below shows the average scores, in runs, for each player for the batting positions 1, 2 and 3.

 

      Batting position
      1 2 3
     Player          Alex              22                     24                     24          
    Bo 25 25 21
    Cameron    24 25 19

 

Each player will be assigned to one batting position.

To which position should each player be assigned to maximise the team’s score? Write your answer in the table below.   (1 mark)

 

       Player           Batting position     
     Alex  
     Bo  
     Cameron  

 

Show Answers Only

`text{Bo (1), Cameron (2), Alex (3)}`

Show Worked Solution

`text(Test different combinations:)`

`text(CBA) = 24 + 25 + 24 = 73`

`text(BCA) = 25 + 25 + 24 = 74`

`text(BAC) = 25 + 24 + 19 = 68`

`:.\ text{Combination for max score: Bo (1), Cameron (2), Alex (3)}`
 

       Player           Batting position     
     Alex 3
     Bo 1
     Cameron 2