SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

NETWORKS, FUR2 2020 VCAA 1

The Sunny Coast Cricket Club has five new players join its team: Alex, Bo, Cameron, Dale and Emerson.

The graph below shows the players who have played cricket together before joining the team.

For example, the edge between Alex and Bo shows that they have previously played cricket together.
 

  1. How many of these players had Emerson played cricket with before joining the team?   (1 mark)

  2. Who had played cricket with both Alex and Bo before joining the team?   (1 mark)

  3. During the season, another new player, Finn, joined the team.

      

    Finn had not played cricket with any of these players before.

      

    Represent this information on the graph above.   (1 mark)

Show Answers Only
  1. `text(one)`
  2. `text(Dale)`
  3.  
Show Worked Solution

a.  `text{One (one edge connected to Emerson.)}`
  

b.  `text{Dale (he has edges directly connected to Alex and Bo.)}`

 

c.  

MATRICES, FUR2 2020 VCAA 3

An offer to buy the Westmall shopping centre was made by a competitor.

One market research project suggested that if the Westmall shopping centre were sold, each of the three centres (Westmall, Grandmall and Eastmall) would continue to have regular shoppers but would attract and lose shoppers on a weekly basis.

Let  `S_n`  be the state matrix that shows the expected number of shoppers at each of the three centres  `n`  weeks after Westmall is sold.

A matrix recurrence relation that generates values of  `S_n`  is

`S_(n+1) = T xx S_n`

`{:(quad qquad qquad qquad qquad qquad qquad qquad text(this week)),(qquad qquad qquad qquad qquad qquad quad \ W qquad quad G qquad quad \ E),(text(where)\ T = [(quad 0.80, 0.09, 0.10),(quad 0.12, 0.79, 0.10),(quad 0.08, 0.12, 0.80)]{:(W),(G),(E):}\ text(next week,) qquad qquad  S_0 = [(250\ 000), (230\ 000), (200\ 000)]{:(W),(G),(E):}):}`
 

  1. Calculate the state matrix, `S_1`, to show the expected number of shoppers at each of the three centres one week after Westmall is sold.   (1 mark)

Using values from the recurrence relation above, the graph below shows the expected number of shoppers at Westmall, Grandmall and Eastmall for each of the 10 weeks after Westmall is sold.
 


 

  1. What is the difference in the expected weekly number of shoppers at Westmall from the time Westmall is sold to 10 weeks after Westmall is sold?
  2. Give your answer correct to the nearest thousand.   (1 mark)

  3. Grandmall is expected to achieve its maximum number of shoppers sometime between the fourth and the tenth week after Westmall is sold.
  4. Write down the week number in which this is expected to occur.   (1 mark)

  5. In the long term, what is the expected weekly number of shoppers at Westmall?
  6. Round your answer to the nearest whole number.   (1 mark)

Show Answers Only
  1. `S_1 =[(240\ 700),(231\ 700),(207\ 600)]`
  2. `30\ 000`
  3. `S_6 = T^6S_0 =  [(text(__)), (233\ 708), (text(__))]`
  4. `218\ 884`
Show Worked Solution
a.   `S_1` `= TS_0`
    `= [(0.80, 0.09, 0.10),(0.12, 0.79, 0.10),(0.08, 0.12, 0.80)][(250\ 000),(230\ 000),(200\ 000)]=[(240\ 700),(231\ 700),(207\ 600)]`

 

b.   `text(Using the graph)`
  `text(Difference)` `= 250\ 000-220\ 000`
    `= 30\ 000`

 

♦♦ Mean mark part (c) 27%.

c.  `text(Testing options:)`

`S_6 = T^6S_0 = [(0.80, 0.09, 0.10),(0.12, 0.79, 0.10),(0.08, 0.12, 0.80)]^6[(250\ 000),(230\ 000),(200\ 000)] = [(text(__)), (233\ 708), (text(__))]`
 

`:.\ text(Maximum shoppers in Grandmall expected in week 6.)`


♦ Mean mark part (d) 39%.

d.  `text(Test with high integer)\ n:`

`S_50 = T^50S_0 -> text(Westmall) = 218\ 884`

MATRICES, FUR2 2020 VCAA 2

The preferred number of cafes `(x)` and sandwich bars `(y)` in Grandmall’s food court can be determined by solving the following equations written in matrix form.
 

`[(5, -9),(4, -7)][(x),(y)]=[(7), (6)]`
 

  1. The value of the determinant of the 2 × 2 matrix is 1.
  2. Use this information to explain why this matrix has an inverse.   (1 mark)

  3. Write the three missing values of the inverse matrix that can be used to solve these equations.   (1 mark)

 
`[(text( __), 9),(text( __), text( __)\ )]`
 

  1. Determine the preferred number of sandwich bars for Grandmall’s food court.   (1 mark)

Show Answers Only
  1. `text(S) text(ince determinant) = 1 != 0,`

     

    `text(the matrix has an inverse)`

  2. `[(-7, 9),(-4, 5)]`
  3. `2`
Show Worked Solution

a.  `text(S) text(ince determinant) = 1 != 0,`

♦ Mean mark part (a) 37%.

`->\ text(the matrix has an inverse)`


b.  `[(-7, 9),(-4, 5)]`

 

Mean mark part (c) 51%.
c.   `[(x), (y)] = [(-7, 9), (-4, 5)][(7), (6)] = [(7),(2)]`

 
`:.\ text(Preferred number of sandwich bars) = 2`

MATRICES, FUR2 2020 VCAA 1

The three major shopping centres in a large city, Eastmall `(E)`, Grandmall `(G)` and Westmall `(W)`, are owned by the same company.

The total number of shoppers at each of the centres at 1.00 pm on a typical day is shown in matrix `V`.

`qquad qquad qquad {:(qquad qquad qquad \ E qquad qquad G qquad qquad \  W),(V = [(2300,2700,2200)]):}`

  1. Write down the order of matrix `V`.   (1 mark)

Each of these centres has three major shopping areas: food `(F)`, clothing `(C)` and merchandise `(M)`.

The proportion of shoppers in each of these three areas at 1.00 pm on a typical day is the same at all three centres and is given in matrix `P` below

`qquad qquad qquad P = [(0.48), (0.27), (0.25)] {:(F),(C),(M):}

  1. Grandmall’s management would like to see 700 shoppers in its merchandise area at 1.00 pm.

     

    If this were to happen, how many shoppers, in total, would be at Grandmall at this time?   (1 mark)

  2. The matrix  `Q = P xx V`  is shown below. Two of the elements of this matrix are missing.
     
    `{:(quad qquad qquad qquad \ E qquad qquad G qquad qquad W), (Q = [(1104, \ text{___}, 1056 ), (621,\ text{___}, 594), (575, 675, 550)]{:(F),(C), (M):}):}`
     
    1. Complete matrix `Q` above by filling in the missing elements.   (1 mark)

    2. The element in row `i` and column `j` of matrix `Q` is `q_(ij)`.
    3. What does the element `q_23` represent?   (1 mark)

The average daily amount spent, in dollars, by each shopper in each of the three areas at Grandmall in 2019 is shown in matrix  `A_2019`  below.

`qquad qquad A_2019 = [(21.30), (34.00), (14.70)] {:(F),(C),(M):}`

On one particular day, 135 shoppers spent the average daily amount on food, 143 shoppers spent the average daily amount on clothing and 131 shoppers spent the average daily amount on merchandise.

  1. Write a matrix calculation, using matrix  `A_2019`, showing that the total amount spent by all these shoppers is $9663.20   (1 mark)

  2. In 2020, the average daily amount spent by each shopper was expected to change by the percentage shown in the table below.
      

     

      Area food clothing merchandise
      Expected change     increase by 5%       decrease by 15%       decrease by 1%   

     

     

    The average daily amount, in dollars, expected to be spent in each area in 2020 can be determined by forming the matrix product

  3. `qquad qquad A_2020 = K xx A_2019`
  4. Write down matrix `K`.    (1 mark)

Show Answers Only
  1. `1 xx 3`
  2. `2800`
  3.  i.   `{:(quad qquad qquad qquad \ E qquad qquad G qquad qquad W), (Q = [(1104, 1296, 1056 ), (621, 729, 594), (575, 675, 550)]{:(F),(C), (M):}):}`
     
  4. ii. `q_23\ text(represents the number of people)`
  5.    `text(in the clothing area of Westmall.)`
  6. `text(Total spent) = [(135, 143, 131)][(21.30), (34.00), (14.70)] = [9663.20]`
  7. `K = [(1.05, 0, 0),(0, 0.85, 0),(0, 0, 0.99)]`
Show Worked Solution

a.  `1 xx 3`

♦ Mean mark part (b) 45%.
b.   `0.25 xx G\ text(shoppers in)\ M` `= 700`
  `:. G\ text(shoppers in)\ M` `= 700/0.25`
    `= 2800`

 

c.i.   `{:(quad qquad qquad qquad \ E qquad qquad G qquad qquad W), (Q = [(1104, 1296, 1056 ), (621, 729, 594), (575, 675, 550)]{:(F),(C), (M):}):}`

 

c.ii.   `q_23\ text(represents the number of people)`
  `text(in the clothing area of Westmall.)`

 

d.  `text(Total spent) = [(135, 143, 131)] [(21.30), (34.00), (14.70)] = [9663.20]`

♦♦ Mean mark part (e) 24%.
 

e.   `A_2020` `= K xx [(21.30), (34.00), (14.70)]`
  `:. K` `= [(1.05, 0, 0),(0, 0.85, 0),(0, 0, 0.99)]`

NETWORKS, FUR2 2020 VCAA 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. How many different routes from `S` to `O` are possible?   (1 mark)

When considering the possible flow of people through this network, many different cuts can be made.

  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. `10`
  2. `52`
  3. `50`
Show Worked Solution
a.   `text(Routes: )` `SMO, STUNO, STUVO, STUVPO, SRQUNO,SRQUVO,`
    `SRQUVPO, SRQVO, SRQVPO, SRQPO`

♦♦ Mean mark part (a) 30%.

 
`:. 10\ text(routes)`

 

b.   `text{Capacity (Cut 1)}` `= 20 + 12 + 20`
    `= 52`

 

c.   `text(Max flow/minimum cut)`

♦♦ Mean mark part (c) 32%.

`= 20 + 10 + 20`

`= 50`
 

CORE, FUR2 2020 VCAA 10

Samuel now invests $500 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after  `n`  months,  `A_n` , can be modelled by a recurrence relation of the form

`A_0 = 500\ 000, qquad A_(n+1) = kA_n - 2000`

  1. Calculate the balance of this annuity after two months if  `k = 1.0024`.   (1 mark)

  2. Calculate the annual compound interest rate percentage for this annuity if  `k = 1.0024`.   (1 mark)

  3. For what value of  `k`  would this investment act as a simple perpetuity?   (1 mark)

Show Answers Only
  1. `$498\ 398.08`
  2. `2.88 text(%)`
  3. `1.004`
Show Worked Solution
a.   `A_1 = 1.0024 xx 500\ 000-2000 = $499\ 200`
  `A_2 = 1.0024 xx 499\ 200-2000 = $498\ 398.08`

 

♦ Mean mark 48%.
b.   `text(Monthly interest rate)` `= (1.0024-1) xx 100 = 0.24text(%)`
  `text(Annual interest rate)` `= 12 xx 0.24 = 2.88text(%)`

 

♦ Mean mark 36%.
c.   `text(Perpetuity would occur when)`
  `k xx 500\ 000-2000` `= 500\ 000`
  `k` `= (502\ 000)/(500\ 000)`
    `= 1.004`

CORE, FUR2 2020 VCAA 9

Samuel opens a savings account.

Let `B_n` be the balance of this savings account, in dollars, `n` months after it was opened.

The month-to-month value of `B_n` can be determined using the recurrence relation shown below.

`B_0 = 5000, qquad B_(n+1) = 1.003B_n`

  1. Write down the value of `B_4`, the balance of the savings account after four months.
  2. Round your answer to the nearest cent.   (1 mark)

  3. Calculate the monthly interest rate percentage for Samuel’s savings account.   (1 mark)

  4. After one year, the balance of Samuel’s savings account, to the nearest dollar, is $5183.

     

    If Samuel had deposited an additional $50 at the end of each month immediately after the interest was added, how much extra money would be in the savings account after one year?

     

    Round your answer to the nearest dollar.   (1 mark)

Show Answers Only
  1. `$5060.27`
  2. `0.3 text(%)`
  3. `$610`
Show Worked Solution
a.   `B_1` `= 1.003 (5000)`
  `B_2` `= 1.003^2 (5000)`

`vdots`

`:. B_4` `= 1.003^4 (5000)`
  `= $5060.27`

 

b.  `text(Monthly interest rate)`

`= (1.003-1) xx 100`

`= 0.3%`

 

c.   `text(Extra)\ =\ text(value of annuity after 12 months)`

`text(By TVM solver:)`

`N` `= 12`
`I(%)` `= 3.6`
`PV` `= 0`
`PMT` `= 50`
`FV` `= ?`
`text(PY)` `= text(CY) = 12`

 
`FV = 609.84`

`:.\ text(Extra money) = $610`

CORE, FUR2 2020 VCAA 7

Samuel owns a printing machine.

The printing machine is depreciated in value by Samuel using flat rate depreciation.

The value of the machine, in dollars, after `n` years, `Vn` , can be modelled by the recurrence relation

`V_0 = 120\ 000, qquad V_(n+1) = V_n-15\ 000`

  1. By what amount, in dollars, does the value of the machine decrease each year?   (1 mark)

  2. Showing recursive calculations, determine the value of the machine, in dollars, after two years.   (1 mark)

  3. What annual flat rate percentage of depreciation is used by Samuel?   (1 mark)

  4. The value of the machine, in dollars, after `n` years, `V_n`, could also be determined using a rule of the form `V_n = a + bn`.

     

    Write down this rule for `V_n`.   (1 mark)

Show Answers Only
  1. `$15\ 000`
  2. `$90\ 000`
  3. `12.5%`
  4. `V_n = 120\ 000-15\ 000n, n = 0, 1, 2, …`
Show Worked Solution

a. `$15\ 000`
  

b.   `V_1` `= 120\ 000-15\ 000 = $105\ 000`
  `V_2` `= 105\ 000-15\ 000 = $90\ 000`

 

c.   `text(Flat rate percentage` `= (15\ 000)/ (120\ 000) xx 100`
    `= 12.5 text(%)`

 

♦ Mean mark part d. 44%.

d.  `V_n = 120\ 000-15\ 000n, \ n = 0, 1, 2, …`

CORE, FUR2 2020 VCAA 6

The table below shows the mean age, in years, and the mean height, in centimetres, of 648 women from seven different age groups.
 


 

  1. What was the difference, in centimetres, between the mean height of the women in their twenties and the mean height of the women in their eighties?  (1 mark)

A scatterplot displaying this data shows an association between the mean height and the mean age of these women. In an initial analysis of the data, a line is fitted to the data by eye, as shown.
 

 

  1. Describe this association in terms of strength and direction.  (1 mark)

  2. The line on the scatterplot passes through the points (20,168) and (85,157).

     

    Using these two points, determine the equation of this line. Write the values of the intercept and the slope in the appropriate boxes below.

     

    Round your answers to three significant figures.  (1 mark)

mean height = 
 
   +  
 
   × mean age

 

  1. In a further analysis of the data, a least squares line was fitted.

     

    The associated residual plot that was generated is shown below.

     
     

          

     

    The residual plot indicates that the association between the mean height and the mean age of women is non-linear.

     

    The data presented in the table in part a is repeated below. It can be linearised by applying an appropriate transformation to the variable mean age.

     

      

     

    Apply an appropriate transformation to the variable mean age to linearise the data. Fit a least squares line to the transformed data and write its equation below.

     

    Round the values of the intercept and the slope to four significant figures.  (2 marks)


Show Answers Only
  1. `10.4\ text(cm)`
  2. `text(Strong and negative)`
  3. `text(mean height) = 171 – 0.169 xx text(mean age)`
  4. `text(mean height) = 167.9 – 0.001621 xx text{(mean age)}^2`
Show Worked Solution
a.   `text(Difference)` `= 167.1 – 156.7`
    `= 10.4\ text(cm)`

 

Mean mark part b. 51%.

b.  `text(Strong and negative.)`

 

♦♦ Mean mark part c. 23%.

c.   `text(Slope) = (157 – 168)/(85 – 20) = -0.169`

`text(Equation of line)`

`y – 168` `= -0.1692 (x – 20)`
`y` `= -0.169x + 171`

 
`:.\ text(mean height) = 171 – 0.169 xx text(mean age)`

 

D.    `text(By CAS)`

`text(mean height) = 167.9 – 0.001621 xx text{(mean age)}^2`

CORE, FUR2 2020 VCAA 5

The scatterplot below shows body density, in kilograms per litre, plotted against waist measurement, in centimetres, for 250 men.

When a least squares line is fitted to the scatterplot, the equation of this line is

body density = 1.195 – 0.001512 × waist measurement

  1. Draw the graph of this least squares line on the scatterplot above.   (1 mark)

  2. Use the equation of this least squares line to predict the body density of a man whose waist measurement is 65 cm.
  3. Round your answer to two decimal places.   (1 mark)

  4. When using the equation of this least squares line to make the prediction in part b., are you extrapolating or interpolating?   (1 mark)

  5. Interpret the slope of this least squares line in terms of a man’s body density and waist measurement.   (1 mark)

  6. In this study, the body density of the man with a waist measurement of 122 cm was 0.995 kg/litre.
  7. Show that, when this least squares line is fitted to the scatterplot, the residual, rounded to two decimal places, is –0.02   (1 mark)

  8. The coefficient of determination for this data is 0.6783
  9. Write down the value of the correlation coefficient `r`.
  10. Round your answer to three decimal places.   (1 mark)

  11. The residual plot associated with fitting a least squares line to this data is shown below.
     
       

     

    Does this residual plot support the assumption of linearity that was made when fitting this line to this data? Briefly explain your answer.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `1.10\ text(kg/litre)`
  3. `text(extrapolating)`
  4. `text(See Worked Solutions)`
  5. `-0.02`
  6. `-0.824`
  7. `text(See Worked Solutions)`
Show Worked Solution

a.   `text(LSRL passes through)\ (60, 1.1043) and (130, 0.998)`

♦ Mean mark part a. 41%.

b.   `text(body density)` `= 1.195-0.001512 xx 65`
    `= 1.09672`
    `= 1.10\ text{kg/litre (to 2 d.p.)}`
♦ Mean mark part c. 38%.
c.   `text(A waist of 65 cm is outside the)`
  `text(range of the existing data set.)`

 
`:.\ text(Extrapolating)`

 

♦ Mean mark part d. 44%.
d.   `text(Body density decreases by 0.001512 kg/litre)`
  `text(for each increase in waist size of 1 cm.)`

 

e.   `text{Body density (predicted)}`

`= 1.195-0.001512 xx 122`

`~~ 1.0105\ text(kg/litre)`
 

`text(Residual)` `= text(Actual-predicted)`
  `~~ 0.995-1.0105`
  `~~ -0.0155`
  `~~ -0.02\ text{(to 2 d.p.)}`

 

♦♦ Mean mark part f. 25%.

f.   `r` `= -sqrt(0.6783)`
    `=-0.8235…`
    `= -0.824\ text{(to 3 d.p.)}`

 

g.  `text(The residual plot has no pattern and is centred around zero.)`

`:.\ text(It supports the assumption of linearity of the LSRL.)`

CORE, FUR2 2020 VCAA 4

The age, in years, body density, in kilograms per litre, and weight, in kilograms, of a sample of 12 men aged 23 to 25 years are shown in the table below.
 

          Age       
(years)

        Body density        
(kg/litre)

        Weight        
(kg)
  23 1.07 70.1
  23 1.07 90.4
  23 1.08 73.2
  23 1.08 85.0
  24 1.03 84.3
  24 1.05 95.6
  24 1.07 71.7
  24 1.06 95.0
  25 1.07 80.2
  25 1.09 87.4
  25 1.02 94.9
  25 1.09 65.3
     
  1. For these 12 men, determine
  2.  i. their median age, in years.   (1 mark)

  3. ii. the mean of their body density, in kilograms per litre.   (1 mark)

  4. A least squares line is to be fitted to the data with the aim of predicting body density from weight.
  5.  i. Name the explanatory variable for this least squares line.   (1 mark)

  6. ii. Determine the slope of this least squares line.
  7.     Round your answer to three significant figures.   (1 mark)

  8. What percentage of the variation in body density can be explained by the variation in weight?
  9. Round your answer to the nearest percentage.   (1 mark)

Show Answers Only
  1.  i. `24`
  2. ii. `1.065\ text(kg/litre)`
  3. i. `text(Weight)`
  4. ii. `text(Slope) = -0.00112\ text{(by CAS)}`
  5. `29 text(%)`
Show Worked Solution
a.i.   `n = 12`  
  `text(Median)` `= (text{6th + 7th})/2`
    `= (24 + 24)/2`
    `= 24`

 

a.ii.   `text(Mean)` `= (∑\ text{body density})/12`
    `= 1.065\ text(kg/litre)`

 

b.i.   `text(Weight)`

♦ Mean mark b.ii. 29%.
MARKER’S COMMENT: Most students did not round correctly.

b.ii.   `text(Slope) = -0.00112\ text{(by CAS)}`

 

c.   `r` `= -0.53847\ text{(by CAS)}`
  `r^2` `= 0.289…`

 
`:. 29 text(%)`

CORE, FUR2 2020 VCAA 2

The neck size, in centimetres, of 250 men was recorded and displayed in the dot plot below.
 

  1. Write down the modal neck size, in centimetres, for these 250 men.   (1 mark)

  2. Assume that this sample of 250 men has been drawn at random from a population of men whose neck size is normally distributed with a mean of 38 cm and a standard deviation of 2.3 cm.
  3.  i. How many of these 250 men are expected to have a neck size that is more than three standard deviations above or below the mean? Round your answer to the nearest whole number.   (1 mark)

  4. ii. How many of these 250 men actually have a neck size that is more than three standard deviations above or below the mean?   (1 mark)

  5. The five-number summary for this sample of neck sizes, in centimetres, is given below.
     

    Use the five-number summary to construct a boxplot, showing any outliers if appropriate, on the grid below.   (2 marks)

Show Answers Only

a.    `text(Mode) = 38\ text(cm)`

b.i.   `1`

b.ii. `1`

c.    `text(See Worked Solutions)`

Show Worked Solution

a.   `text(Mode) = 38\ text(cm)`

 

♦ Mean mark part b.i. 40%.
b.i.   `text(Expected number of men)` `= (1-0.997) xx 250`
    `= 0.75`
    `= 1\ text{(nearest whole)}`

 

♦ Mean mark part b.ii. 42%.
b.ii.   `text(When)\ \ z = +- 3`
  `text(Neck size limits)` `= 38 +- (2.3 xx 3)`
    `= 44.9 or 31.1`

 
`:.\ text(1 man has neck size outside 3 s.d.)`
 

c.   `IQR = 39-36=3`

`text(Upper fence)\ =Q_3 + 1.5 xx 3=39 + 4.5=43.5`

`text(Lower fence)\ =Q_1-1.5 xx 3=36-4.5=31.5`

 

CORE, FUR2 2020 VCAA 1

Body mass index (BMI), in kilograms per square metre, was recorded for a sample of 32 men and displayed in the ordered stem plot below.
  

  1. Describe the shape of the distribution.   (1 mark)

  2. Determine the median BMI for this group of men.   (1 mark)

  3. People with a BMI of 25 or over are considered to be overweight.
  4. What percentage of these men would be considered to be overweight?   (1 mark)

Show Answers Only
  1. `text(Positively skewed)`
  2. `24.55`
  3. `37.5 text(%)`
Show Worked Solution

a.   `text(Positively skewed)`
 

b.   `32\ text(data points)`

`text(Median)` `= text(16th + 17th)/2`
  `= (24.5 + 24.6)/2`
  `= 24.55`

 

c.    `text(Percentage)` `= 12/32 xx 100`
    `= 37.5%`

Statistics, NAPX-p167881v02

A menswear store recorded the number of items it sold in four different categories.

What category makes up 60% of the total number of items sold?
 


  

 
   Hoodies
 
   Shorts
 
   T-shirts
 
   Pants
Show Answers Only

`text{Pants}`

Show Worked Solution

`text{Pants represent 60% of the graph.}`

Number, NAPX-p167756v02

`8 xx` 
 
  `- 13 = 75`

What value would make the number sentence right?

 
   11
 
     9
 
   13
 
     7
Show Answers Only

`11`

Show Worked Solution

`text{Check each option (multiply then add):}`

`8 xx 11 -13 = 75 \ text{(Correct)}`

`8 3 xx 9 -13 = 59 \ text{(Incorrect)}`

`8 3 xx 9 -13 = 91 \ text{(Incorrect)}`

`8  xx 7 -13 = 43 \ text{(Incorrect)}`

Number, NAPX-p167756v01

`12 + 3 xx ` 
 
  `= 39`

 
For this number sentence to be true, what is the missing value?

 
   15
 
   11
 
     9
 
     7
Show Answers Only

`9`

Show Worked Solution

`text{Check each option (multiply then add):}`

`12 + 3 xx 15= 57 \ text{(Incorrect)}`

`12 + 3 xx 11= 57 \ text{(Incorrect)}`

`12 + 3 xx 9 = 57 \ text{(Correct)}`

`12 + 3 xx 7= 57 \ text{(Incorrect)}`

Number, NAPX-p167749v02

The return trip from a school to the museum is 15.97 kilometres.

How far does a student need to travel if they visit the museum 6 times?

 
   89.53 km
 
   95.82 km
 
   57.39 km
 
   64.72 km
Show Answers Only

`95.82 \ text{km}`

Show Worked Solution
`:.\ text(Distance)` `= 6 xx 15.97 \ text{km}`  
  `= 95.82 \ text{km}`  

Number, NAPX-p167749v01

A return trip from your house to the grocery store is 9.82 kilometres.

If you go to the store 23 times, how far do you need to travel?

 
    426.95 km
 
   234.21 km
 
   225.86 km
 
   382.23 km
Show Answers Only

`225.86 \ text{km}`

Show Worked Solution
`:.\ text(Distance)` `= 23 xx 9.82 \ text{km}`  
  `= 225.86 \ text{km}`  

Functions, MET1 2010 VCAA 4b

Solve the equation  `sqrt 3 sin (x) = cos (x)`  for  `x in [-pi, pi]`.   (2 marks)

Show Answers Only

`x = pi/6,\ \ \ -(5 pi)/6`

Show Worked Solution

`text(Divide both sides by)\ \ cos(x) :`

MARKER’S COMMENT: Many students who found the base angle correctly could not solve within the restrictions.
`sqrt 3 sin x` `=cos x`
`sqrt 3 tan x` `= 1`
`tan x` `= 1/sqrt 3`
`=>\ text(Base angle)\ = pi/6`

 

`:. x = pi/6\ \ text(or)\ \ -(5 pi)/6,\ \ \ x in [-pi, pi].`

Mechanics, SPEC2 2020 VCAA 5

Two objects, each of mass `m` kilograms, are connected by a light inextensible strings that passes over a smooth pulley, as shown below. The object on the platform is initially at point A and, when it is released, it moves towards point C. The distance from point A to point C is 10 m. The platform has a rough surface and, when it moves along the platform, the object experiences a horizontal force opposing the motion of magnitude `F_1` newtons in the section AB and a horizontal force opposing the motion of magnitude `F_2` newtons when it moves in the section BC.
 

  1. On the diagram above, mark all forces that act on each object once the object on the platform has been released and the system is in motion.  (2 marks)

The force `F_1` is given by  `F_1 = kmg, \ k ∈ R^+`.

  1.  i. Show that an expression for the acceleration, in `text(ms)^(−2)`, of the object on the platform, in terms of `k`, as it moves from point A to point B is given by  `(g(1 - k))/2`.  (2 marks)
  2. ii. The system will only be in motion for certain values of `k`.
  3.     Find these values of `k`.  (1 mark)

Point B is midway between points A and C.

  1. Find, in terms of `k`, the time taken, is seconds, for the object on the platform to reach point B.  (2 marks)
  2. Express, in terms of `k`, the speed `v_B`, in `text(ms)^(−1)`, of the object on the platform when it reaches point B.  (2 marks)
  3. When the object on the platform is at point B, the string breaks. The velocity of the object at point B is  `v_B = 2.5\ text(ms)^(−1)`. The force that opposes motion from point B to point C is  `F_2 = 0.075 mg + 0.4 mv^2`, where `v` is the velocity of the object when it is a distance of `x` metres from point B. The object on the platform comes to rest before point C.
  4. Find the object's distance from point C when it comes to rest. Give your answer in metres, correct to two decimal places.  (4 marks)
Show Answers Only
  1.  
  2.  i. `text(See Worked Solutions)`
  3. ii. `k ∈ (0, 1), k ∈ R^+`
  4. `2sqrt(5/(g(1 – k)))`
  5. `v_text(B) = sqrt(5g(1 – k))`
  6. `3.15\ text(m)`
Show Worked Solution
a.   

 

b. i.   `text(Horizontally:)`

`ma = T – F_1 = T – kmg\ …\ (1)`

`text(Vertically:)`

`ma = mg – T\ …\ (2)`
 

`text(Add)\ \ (1) + (2) :`

`2ma = mg – kmg`

`:. a` `= (g – kg)/2`
  `= (g(1 – k))/2`
♦ Mean mark (b)(ii) 45%.

 

b. ii.   `text(System in motion when)\ a > 0`

`(g(1 – k))/2 > 0`

`:. k ∈ (0, 1), \ k ∈ R^+`

 

c.   `text(AB) = 5\ (text(given)), u = 0\ (text(given))`

`text(Find)\ t\ text(when)\ s = 5:`

`s = ut + 1/2at^2`

`5 = 0 + 1/2 · (g(1 – k))/2 · t^2`

`t^2` `= 20/(g(1 – k))`
`t` `= sqrt(20/(g(1 – k)))`
  `= 2sqrt(5/(g(1 – k)))`

 

d.   `text(At B,)\ s = 5`

`v_text(B)^2` `= u^2 + 2as`
  `= 0 + 2 · (g(1 – k))/2 · 5`
  `= 5g(1 – k)`

 
`:. v_text(B) = sqrt(5g(1 – k))`

 

e.   `text(Acceleration is against the direction of motion.)`

♦♦ Mean mark (e) 35%.
`a` `= −F/m`
  `= −0.075g – 0.4v^2`
  `= −0.4(0.1875g + v^2)`

 

`d/(dx)(1/2 v^2)` `= −0.4(0.1875g + v^2)`
`d/(dx)(v^2)` `= −0.8(0.1875g + v^2)`
`(dx)/(d(v^2))` `= −1.25(1/(0.1875g + v^2))`
`:. x` `= −1.25 int_(2.5^2)^0 1/(0.1875 + v^2)\ dv^2`
  `= 1.85\ text(m)`

 

`:.\ text(Distance from C)` `= 5 – 1.85`
  `= 3.15\ text(m)`

Vectors, SPEC2 2020 VCAA 4

A pilot is performing at an air show. The position of her aeroplane at time `t` relative to a fixed origin `O` is given by 

   `underset~r_text(A) (t) = (450-150sin((pit)/6))underset~i + (400-200cos((pit)/6))underset~j`,

where  `underset~i`  is a unit vector in a horizontal direction, `underset~j`  is a unit vector vertically up, displacement components are measured in metres and time `t` is measured in seconds where  `t >= 0`.

  1. Find the maximum speed of the aeroplane. Give your answer in `text(ms)^(−1)`.  (3 marks)

  2.  i. Use  `underset~r_text(A)(t)`  to show that the cartesian equation of the path of the aeroplane is given by
  3.     `((x-450)^2)/(22\ 500) + ((y-400)^2)/(40\ 000) = 1`.  (2 marks)

  4. ii. Sketch the path of the aeroplane on the axes provided below. Label the position of the aeroplane when  `t = 0`, using coordinates, and use an arrow to show the direction of motion of the aeroplane.  (3 marks)

  5.   

         
     

A friend of the pilot launches an experimental jet-powered drone to take photographs of the air show. The position of the drone at time `t` relative to the fixed origin is given by  `underset~r_text(D)(t) = (30t)underset~i + (−t^2 + 40t)underset~j`, where `t` is in seconds and  `0 <= t <= 40, underset~i`  is a unit vector in the same horizontal direction, `underset~j`  is a unit vector vertically up, and displacement components are measured in metres.

  1. Sketch the path of the drone on the axes provided in part b.ii. Using coordinates, label the points where the path of the drone crosses the path of the aeroplane, correct to the nearest metre.  (3 marks)

  2. Determine whether the drone will make contact with the aeroplane. Give reasons for your answer.  (3 marks)

Show Answers Only

  1. `(100pi)/3 text(ms)^(−1)`
  2.  i. `text(See Worked Solutions)`
  3. ii.
  4.  

     
  5. `text(See Worked Solutions)`

Show Worked Solution

a.    `underset~r_text(A)′(t)` `= −25picos((pit)/60)underset~i + 100/3pisin((pit)/6)underset~j`
    `= (25pi)/3(−3cos((pit)/6) + 4sin((pit)/6))`
`|underset~r_text(A)′(t)|` `= (25pi)/3 sqrt(9cos^2((pit)/6) + 16sin^2((pit)/6))`
  `= (25pi)/3 sqrt(9 + 7sin^2((pit)/6))`
`:. |underset~r_text(A)′(t)|_text(max)` `= (25pi)/3 sqrt(9 + 7)`
  `= (100pi)/3\ text(ms)^(−1)`

 

b. i.   `x = 450-150 sin((pit)/6) \ => \ sin((pit)/6) = (450-x)/150`

`sin^2((pit)/6) = ((x-450)^2)/(22\ 500)`
 

`y = 400-200cos((pit)/6) \ => \ cos((pit)/6) = (400-y)/200`

`cos^2((pit)/6) = ((y-400)^2)/(40\ 000)`
 

`text(Using)\ \ sin^2theta + cos^2theta = 1:`

`((x-450)^2)/(22\ 500) + ((y-400)^2)/(40\ 000) = 1`

 

b. ii.   `text(When)\ x = 450, \ y = 200, 600`

`text(When)\ y = 400, \ x = 300, 600`

`text(As)\ \ t -> 1, \ x↓, \ y↑`

`:.\ text(Motion is clockwise.)`
 

 

c.   `x = 30t \ => \ t = x/30`

`y` `= −t^2 + 40t`
  `= −(x^2)/900 + 4/3 x`

 

 

d.   `text(The graph shows 2 points where the paths cross.)`

`text(Consider)\ (316, 310):`

`text(Drone passes through when)`

`t = 316/30 ~~ 10.53\ text(seconds)`

`text(Plane passes through when)`

`450-150sin((pit)/6) = 316 \ => \ t ~~ 12.85\ text{seconds (no contact)}`
 

`text(Similarly, consider)\ (600, 400):`

`text(Drone passes through when)`

`t = 600/30 = 20\ text(seconds)`

`text(Plane passes through when)`

`400-200cos((pit)/6) = 400 \ => \ t = 10\ text(or)\ ~~ 14.7\ text{seconds (no contact)}`
  

`:.\ text(Drone will not make contact with the plane.)`

Calculus, SPEC2 2020 VCAA 3

Let  `f(x) = x^2e^(−x)`.

  1. Find an expression for  `f′(x)`  and state the coordinates of the stationary points of  `f(x)`.  (2 marks)
  2. State the equation(s) of any asymptotes of  `f(x)`.  (1 mark)
  3. Sketch the graph of  `y = f(x)`  on the axes provided below, labelling the local maximum stationary point and all points of inflection with their coordinates, correct to two decimal places.  (3 marks)
     
         

Let  `g(x) = x^n e^(−x)`, where  `n ∈ Z`.

  1. Write down an expression for  `g″(x)`.  (1 mark)
  2.  i. Find the non-zero values of `x` for which  `g″(x) = 0`.  (1 mark)
  3. ii. Complete the following table by stating the value(s) of `n` for which the graph of  `g(x)`  has the given number of points of inflection.  (2 marks)
     
       
         
Show Answers Only
  1. `f(0) = 0; f(2) = 4e^(−2) text(and SP’s at)\ (0, 0), (2, 4e^(−2))`
  2. `y = 0`
  3.   
  4. `g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`
  5.  i. `x = x ± sqrtn`
  6. ii.
       
Show Worked Solution

a.   `f′(x) = 2xe^(−x) – x^2e^(−x)`

`text(SP’s when)\ \ f′(x) = 0:`

`x^2e^(−x)` `= 2xe^(−x)`
`x` `= 2\ \ text(or)\ \ 0`

 
`f(0) = 0; \ f(2) = 4e^(−2)`

`:. text(SP’s at)\ \ (0, 0) and  (2, 4e^(−2))`

 

b.   `text(As)\ \ x -> ∞, \ f(x) -> 0^+`

`:. text(Horizontal asymptote at)\ \ y = 0`

 

c.   

`text(POI when)\ \ f″(x) = 0`

`:. text(POI’s:)\ (0.59, 0.19), \ (3.41, 0.38)`

 

d.   `g′(x) = x^(n – 1) e^(−x)(n – x)`

`g″(x) = x^(n – 2) e^(−x)(x^2 – 2xn + n^2 – n)`

 

e.i.   `text(Solve:)\ \ x^2 – 2xn + n^2 – n = 0`

`x = n ± sqrtn`

♦♦♦ Mean mark (e)(ii) 9%.

 

e.ii.   

Complex Numbers, EXT2 N2 2020 SPEC2 2

Two complex numbers, `u` and `v`, are defined as  `u = −2 - i`  and  `v = −4 - 3i`.

  1. Express the relation  `|z - u| = |z - v|`  in the cartesian form  `y = mx + c`, where  `m, c ∈ R`.  (3 marks)

  2. Plot the points that represent `u` and `v` and the relation `|z - u| = |z - v|` on the Argand diagram below.  (2 marks)
     
         
     
  3. State a geometrical interpretation of the graph of  `|z - u| = |z - v|`  in relation to the points that represent `u` and `v`.  (1 mark)

  4.  i. Sketch the ray given by  `text(Arg)(z - u) = pi/4`  on the Argand diagram in part b.  (1 mark)

  5. ii. In Cartesian form, write down the function that describes the ray  `text(Arg)(z - u) = pi/4`.  (1 mark)

Show Answers Only
  1. `y = −x – 5`
  2.  
  3. `|z – u| = |z – v|\ text(is the perpendicular bisector of the line joining)\ u and v.`
  4.  i.
  5. ii. `f: (−2, ∞) ->, f(x) = x + 1`
Show Worked Solution

a.   `text(Let)\ \ z = x + iy`

`z – u = x + 2 + iy + i`

`z – v = x + 4 + iy + 3i`

`|z – u| = |z – v|`

`(x + 2)^2 + (y + 1)^2` `= (x + 4)^2 + (y + 3)^2`
`x^2 + 4x + 4 + y^2 + 2y + 1` `= x^2 + 8x + 16 + y^2 + 6y + 9`
`-4y` `= 4x + 20`
`y` `= −x – 5`

 

b.   

 

c.   `|z – u| = |z – v|\ text(is the graph of the perpendicular bisector of the)`

`text(line joining)\ u and v.`

 

d.i.   

 

d.ii.   `text(Arg)(z – u) = pi/4 =>\ text(gradient) = 1, ytext(-intercept at)\ (0, 1)`

`:. f: (−2, ∞) -> RR, \ f(x) = x + 1`

Complex Numbers, SPEC2 2020 VCAA 2

Two complex numbers, `u` and `v`, are defined as  `u = -2-i`  and  `v = −4-3i`.

  1. Express the relation  `|z-u| = |z-v|`  in the cartesian form  `y = mx + c`, where  `m, c ∈ R`.   (3 marks)

  2. Plot the points that represent `u` and `v` and the relation `|z-u| = |z-v|` on the Argand diagram below.   (2 marks)

     
         
     

  3. State a geometrical interpretation of the graph of  `|z-u| = |z-v|`  in relation to the points that represent `u` and `v`.   (1 mark)

  4.  i. Sketch the ray given by  `text(Arg)(z-u) = pi/4`  on the Argand diagram in part b.   (1 mark)

  5. ii. Write down the function that describes the ray  `text(Arg)(z-u) = pi/4`, giving the rule in cartesian form.   (1 mark)

  6. The points representing `u` and `v` and  `−5i`  lie on the circle given by  `|z-z_c| = r`, where `z_c` is the centre of the circle and `r` is the radius.
  7. Find `z_c` in the form  `a + ib`, where  `a, b ∈ R`, and find the radius `r`.   (3 marks)

Show Answers Only
  1. `y = −x-5`
  2.  
  3. `|z-u| = |z-v|\ text(is the perpendicular bisector of the line joining)\ u and v.`
  4.  i.
  5. ii. `f: (−2, ∞) ->, f(x) = x + 1`
  6. `z_c = −5/3-10/3 i and r = (5sqrt2)/3`
Show Worked Solution

a.   `text(Let)\ \ z = x + iy`

`z-u = x + 2 + iy + i`

`z-v = x + 4 + iy + 3i`

`|z-u| = |z-v|`

`(x + 2)^2 + (y + 1)^2` `= (x + 4)^2 + (y + 3)^2`
`x^2 + 4x + 4 + y^2 + 2y + 1` `= x^2 + 8x + 16 + y^2 + 6y + 9`
`-4y` `= 4x + 20`
`y` `= −x-5`

 

b.   

 

c.   `|z-u| = |z-v|\ text(is the graph of the perpendicular bisector of the)`

`text(line joining)\ u and v.`

 

d.i.   

♦♦ Mean mark (d.ii.) 25%.

 

d.ii.   `text(Arg)(z-u) = pi/4 =>\ text(gradient) = 1, ytext(-intercept at)\ (0, 1)`

`:. f: (−2, ∞) -> RR, \ f(x) = x + 1`

 

e.   `|z_c-u| = |z_c-v| = |z_c-(−5i)|`

♦ Mean mark (e) 40%.
`z_c-u` `= (a + 2) + (b + 1)i`
`z_c-v` `= (a + 4) + (b + 3)i`
`z_c +5i` `= a + (b + 5)i`

 
`a^2 + (b + 5)^2 = (a + 2)^2 + (b + 1)^2\ …\ (1)`

`a^2 + (b + 5)^2 = (a + 4)^2 + (b + 3)^2\ …\ (2)`

`a = −5/3, b = −10/3\ \ text{(by CAS)}`
 

`:.z_c = −5/3-10/3 i`

`:.r` `= |z_c-(−5i)|`
  `= sqrt((−5/3)^2 + (−10/3 + 5)^2)`
  `= (5sqrt2)/3`

Vectors, SPEC2 2020 VCAA 1

A particle moves in the  `x\ – y`  plane such that its position in terms of `x` and `y` metres at `t` seconds is given by the parametric equations

`x = 2sin(2t)`

`y = 3cos(t)`

where `t >= 0`

  1. Find the distance, in metres, of the particle from the origin when  `t = pi/6`.   (2 marks)

  2.   i. Express  `(dy)/(dx)`  in terms of `t` and, hence, find the equation of the tangent to the path of the particle at  `t = pi`  seconds.   (3 marks)

  3.  ii. Find the velocity, `underset ~ v`, in `text(ms)^(−1)`, of the particle when  `t = pi`.   (2 marks)

  4. iii. Find the magnitude of the acceleration, in `text(ms)^(−2)`, when  `t = pi`.   (2 marks)

  5. Find the time, in seconds, when the particle first passes through the origin.   (1 mark)

  6. Express the distance, `d` metres, travelled by the particle from  `t = 0`  to  `t = pi/6`  as a definite integral and find this distance correct to three decimal places.   (2 marks)

Show Answers Only
  1. `sqrt39/2\ text(metres)`
  2.   i. `y = −3`
  3.  ii. `4underset~i`
  4. iii. `3\ text(ms)^(−2)`
  5. `pi/2\ \ (text(1st time))`
  6. `1.804\ \ (text(to 3 d.p.))`
Show Worked Solution

a.   `text(At)\ \ t = pi/6,`

`x = 2sin\ pi/3 = sqrt3`

`y = 3cos\ pi/6 = (3sqrt3)/2`

`:.\ text(Distance)` `= sqrt((sqrt3)^2 + ((3sqrt3)/2)^2)`
  `= sqrt39/2\ text(metres)`

 

b.i.   `(dx)/(dt) = 4cos(2t),\ \ (dy)/(dt) = −3sin(t)`

`(dy)/(dx)` `= (dy)/(dt) · (dt)/(dx)`
  `= (−3sin(t))/(4cos(2t))`

 
`text(When)\ t = pi :`

`(dy)/(dx)` `= (−3sin(pi))/(4cos(2pi))=0`

 
`text(Equation of tangent where)\ \ m = 0,\ text(through)\ (0, −3):`

`y = −3`
  

b.ii.   `underset~r(t)` `= 2sin(2t)underset~i + 3cos(t)underset~j`
  `underset~v(t)` `= 4cos(2t)underset~i-3sin(t)underset~j`
  `underset~v(pi)` `= 4cos(2pi)underset~i-3sin(pi)underset~i`
    `= 4underset~i`

 

b.iii.  `underset~a(t) = −8sin(2t)underset~i-3cos(t)underset~j`

♦ Mean mark (b)(iii) 48%.
`underset~a(pi)` `= −8sin(2pi)underset~i-3cos(pi)underset~j`
  `= 3underset~j`
`|underset~a(pi)|` `= sqrt(0^2 + 3^2)`
  `= 3\ text(ms)^(−2)`

 

c.   `text(Find)\ t\ text(when)\ \ x = 2sin(2t) = 0\ \ text(and)\ \ y = 3cos(t) = 0:`

`t = pi/2\ \ (text(1st time))`

 

d.    `text(Distance)` `= int_0^(pi/6) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
    `= int_0^(pi/6) sqrt((4cos(2t))^2 + (−3sin(t))^2)\ dt`
    `~~ 1.804\ \ (text(to 3 d.p.))`

Number, NAPX-p110532v02

Ana recycles 3 kilograms of plastic bottles in her first week of recycling.

She then recycles twice as much as she recycled the previous week for the next 5 weeks.

The total weight, in kilograms, of the plastic bottles she recycles in any given week will be

 
  always even
 
  always odd
 
  sometimes even or sometimes odd
 
  none of the above
Show Answers Only

`text{Always odd}`

Show Worked Solution

`text{1st week: 3 kg (odd)}`

`text{2nd week: 3 + 6 = 9 kg (odd)}`

`text{3rd week: 9 + 18 = 27 kg (odd)}`

   `vdots`

`text{Any week: odd + even = odd}`

Number, NAPX-p110532v01

Leo puts $15 into his piggy bank to start saving money.

He then puts the same amount into the piggy bank at the end of each month.

The total amount of money in his piggy bank will be which of the following:

 
  Always an even number
 
  Always an odd number
 
  Sometimes odd sometimes even
 
  None of the above
Show Answers Only

`text{Sometimes odd sometimes even}`

Show Worked Solution

`text{Starting amount = $15 (odd)}`

`text{End of 1st month = 15 + 15 = $30 (even)}`

`text{End of 2nd month = 30 + 15 = $45 (odd)}`

`:.\ text{Total amount will be either an odd or even number.}`

Geometry, NAPX-p110494v02

Michael’s company is shown in the map.
 

   
 

Michael rides home 5 kilometres west, walks 3 kilometres south and walks another 1 kilometre west.

In which cell on the map is Michael’s home?

D1 E2 C1 D2
 
 
 
 
Show Answers Only

`text{D1}`

Show Worked Solution

`text{Each grid width (or height) on the map represents 1 kilometre.}`

`therefore \ text{Michaels’s house lies at D1 on the map.}`

Geometry, NAPX-p110494v01

Phil’s home is shown in the map below.

Everyday he drives 3 kilometres east and 2 kilometres south from home to work

In which cell on the map is Phil’s office?

A5 B5 E5 D5
 
 
 
 
Show Answers Only

`text{E5}`

Show Worked Solution

`text{Each box  in the map represents 1 kilometre.}`

`therefore \ text{Phil’s office lies on E5 on the map.}`

Algebra, NAPX-p116707v03

Louise donated money to two charities

She donated $125 to the first charity and donated $45 to the second charity.

Which number sentence could be used to find total amount of money Louise donated?

 
  120 + 40 = 160
 
  130 + 50 = 180
 
  120 + 40 + 5 = 165
 
  130 + 50 – 10 = 170

Show Answers Only

`130 + 50 – 10 = 170`

Show Worked Solution

`text(Total amount donated to charity)`

`= 125 + 45`

`= 130 – 5 + 50 – 5`

`= 130 + 50 – 10`

`= 170`

Algebra, NAPX-p116707v02

Shane played two basketball games.

She scored 35 points in the first game.

In the second game she scored 25 points.

Which number sentence could be used to find the total number of points Shane scored in the first two games?

30 + 20 + 5 = 55 40 + 30 = 70 30 + 20 + 10 = 60 30 + 20 = 50
 
 
 
 
Show Answers Only

`30 + 20 + 10 = 60`

Show Worked Solution

`text(Total points in first 2 games)`

`= 35 + 25`

`= 30 + 5 + 20 + 5`

`= 30 + 20 + 10`

`= 60`

Number, NAPX-p116663v04

Danica decided to travel a total of two thousand, nine hundred and nine kilometres from Brisbane to Adelaide.
 


 

The distance travelled can be written as:

2090 km 2099 km 2909 km 2990 km
 
 
 
 
Show Answers Only

`2909 \ text{km}`

Show Worked Solution

`text{Two thousand, nine hundred and nine = 2909.}`

Number, NAPX-p116663v03

A man travelled a total of three thousand four hundred and six kilometres from Perth to Melbourne.

The distance travelled can be written as:

3046 km 3460 km 3406 km 4306 km
 
 
 
 
Show Answers Only

`3406 \ text{km}`

Show Worked Solution

`text{Three thousand four hundred and six = 3406}`

Geometry, NAPX-p116636v03

Gary folded this piece of paper along the dashed lines to make a solid figure.
 

 
 

Which of these models did Gary make?

 
 
 
 
Show Answers Only

Show Worked Solution

`text{By folding the paper along the dashed lines we could obtain a figure of:}`

Geometry, NAPX-p116636v02

Lance folded this piece of card board along the dotted line to make a solid figure.
 


 

Which of these models did Lance made?

 
 
 
 
Show Answers Only

Show Worked Solution

`text{By folding the cardboard along the dotted lines we could obtain a figure of:}`

Statistics, NAPX-p116609v03

In a school, a group of students were given a list of 4 subjects and asked to pick their favourite.

A table of their responses is shown below.
 


 

How many students were asked this question altogether?
Show Answers Only

`44`

Show Worked Solution

`text{Mathematics} = 5 + 5 + 5 + 2 = 17`

`text{Science} = 5 + 5 = 10`

`text{English} = 5 + 1 = 6`

`text{History} = 5 + 5 +  1 = 11`

`:.\ text{Total}  = 17 + 10 + 6 + 11 = 44`

Statistics, NAPX-p116609v02

Some people were asked which type of music they like to listen to most.

The table below shows their responses.

In total, how many people were asked which type of music they like?
Show Answers Only

`51`

Show Worked Solution

`text{Pop} = 5 + 5 + 5 + 2 = 17`

`text{Rock} = 5 + 5 + 5 + 5 + 1 = 21`

`text{Jazz} = 5 + 5 + 3 = 13`

`:.\ text{Total}  = 17 + 21 + 13 = 51`

Mechanics, EXT2 M1 2013 SPEC2 18*

A particle moves in a straight line such that its acceleration is given by  `a = sqrt(v^2 - 1)` , where `v` is its velocity and `x` is its displacement from a fixed point.

Given that  `v = sqrt2`  when  `x = 0`, find the velocity `v` in terms of `x`.   (4 marks)

Show Answers Only

`v = sqrt(1 + (1 + x)^2)`

Show Worked Solution
`v(dv)/(dx)` `= sqrt(v^2 – 1)`
`(dv)/(dx)` `= sqrt(v^2 – 1)/v`
`(dx)/(dv)` `= v/sqrt(v^2 – 1)`
`x` `=int v/sqrt(v^2 – 1)\ dv`

 
`text(Integration by substitution:)`

`text(Let)\ \ u=v^2-1`

`(du)/(dv) = 2v \ => \ 1/2 du = v\ dv`

`x` `= 1/2 int u^(- 1/2) \ du`
  `= 1/2 * 2 u^(1/2) + c`
  `= sqrt(v^2 – 1) + c`

 

 
`text(When)\ \ x=0, \ v = sqrt2\ \ =>\ \ c = −1`

`x` `= sqrt(v^2 – 1) -1`
`v^2` `= 1 + (1 + x)^2`

 
`:. v = sqrt(1 + (1 + x)^2)`

Mechanics, EXT2 M1 2015 SPEC1 6

The acceleration `a` ms¯² of a body moving in a straight line in terms of the velocity `v` ms¯¹ is given by  `a = 4v^2.`

Given that  `v = e`  when  `x = 1`, where `x` is the displacement of the body in metres, find the velocity of the body when  `x = 2.`  (4 marks)

Show Answers Only

`e^5`

Show Worked Solution
`v* (dv)/(dx)` `= 4v^2`
`(dv)/(dx)` `= 4v`
`(dx)/(dv)` `= 1/(4v)`
`x` `=int 1/(4v) \ dv`
  `=1/4 ln (4v)+c`

 
`text(When)\ \ x=1, v=e`

`1` `=1/4 ln(4e)+c`  
`:.c` `=1-1/4 ln(4e)`  

 
`text(Find)\ \ v\ \ text(when)\ \ x=2:`

COMMENT: Strong log and exponential calculation ability required here.

`1/4 ln (4v)+1-1/4 ln(4e)` `= 2`
`1/4 ln(4v)` `= 1/4 ln (4e)+1`
`ln(4v)` `= ln(4e)+4`
`e^(ln4v)` `= e^(ln(4e) +4)`
`4v` `= e^(ln(4e)) * e^4`
`4v` `=4e xx e^4`
`:.v` `=e^5`

Vectors, SPEC2 2020 VCAA 19 MC

A cricket ball of mass 0.02 kg, moving with velocity  `2underset~i - 10 underset~j\ \ text(ms)^(−1)`,  is hit and after impact travels with velocity  `2underset~i - 7underset~j\ \ text(ms) ^(−1)`.

The magnitude of the change in momentum of the cricket ball, in  `text(kg ms)^(−1)`, is closest to

  1. 0.04
  2. 0.06
  3. 0.10
  4. 0.24
  5. 0.34
Show Answers Only

`B`

Show Worked Solution
`Delta underset~p` `= m Delta underset~v`
  `= 0.02((2underset~i – 7underset~j) – (2underset~i – 10underset~j))`
  `= 0.02(3underset~j)`
  `= 0.06underset~j`

 
`|Delta underset~p| = 0.06\ text(kg ms)^(−1)`

`=>B`

Mechanics, SPEC2 2020 VCAA 18 MC

A particle of mass `m` kilogram hangs from a string that is attached to a fixed point. The particle is acted on by a horizontal force of magnitude `F` newtons. The system is in equilibrium when the string makes an angle `alpha` to the horizontal, as shown in the diagram below. The tension in the string has magnitude `T` newtons.
 


 

The value of  `tan(alpha)`  is

  1. `(mg)/T`
  2. `T/(mg)`
  3. `T/F`
  4. `F/(mg)`
  5. `(mg)/F`
Show Answers Only

`E`

Show Worked Solution

`text(Resolving forces vertically:)`

`mg = Tsin(alpha)`
 

`text(Resolving forces horizontally:)`

`F = Tcos(alpha)`

`:. tan(alpha) = (mg)/F`

`=>E`

Mechanics, EXT2 M1 2020 SPEC2 17 MC

The velocity, `v` ms`\ ^(−1)`, of a particle at time  `t >= 0`  seconds and at position  `x >= 1`  metre from the origin is  `v = 1/x`.

The acceleration of the particle, in `text(ms)^(−2)`, when  `x = 2`  is

  1. `−1/4`
  2. `−1/8`
  3. `1/8`
  4. `1/4`
Show Answers Only

`B`

Show Worked Solution

`v = 1/x`

`a` `= v · (dv)/(dx)`
  `= 1/x · −1/(x^2)`
  `= −1/(x^3)`

 

`text(When)\ \ x = 2:`

`a = −1/(2^3) = −1/8`

`=>B`

Calculus, SPEC2 2020 VCAA 17 MC

The velocity, `v` ms`\ ^(−1)`, of a particle at time  `t >= 0`  seconds and at position  `x >= 1`  metre from the origin is  `v = 1/x`.

The acceleration of the particle, in `text(ms)^(−2)`, when  `x = 2`  is

  1. `−1/4`
  2. `−1/8`
  3. `1/8`
  4. `1/2`
  5. `1/4`
Show Answers Only

`B`

Show Worked Solution

`v = 1/x`

`a` `= v · (dv)/(dx)`
  `= 1/x · −1/(x^2)`
  `= −1/(x^3)`

 

`text(When)\ x = 2:`

`a = −1/(2^3) = −1/8`

`=>B`

Vectors, EXT2 V1 2020 SPEC2 16 MC

Let  `underset~a = underset~i + 2underset~j + 2underset~k`  and  `underset~b = 2underset~i - 4underset~j + 4underset~k`, where the acute angle between these vectors is  `theta`.

The value of  `sin(2theta)`  is

  1. `1/9`
  2. `(4sqrt5)/9`
  3. `(4sqrt5)/81`
  4. `(8sqrt5)/81`
Show Answers Only

`D`

Show Worked Solution

`underset~a = ((1),(2),(2)), |underset~a| = sqrt(1 + 4 + 4) = 3`

`underset~b = ((2),(−4),(4)) , |underset~b| = sqrt(4 + 16 + 16) = 6`

`underset~a · underset~b = ((1),(2),(2))((1),(−4),(4)) = 2 – 8 + 8 = 2`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 2/(3 xx 6) = 1/9`
 

`sintheta = sqrt(81 – 1)/9 = (4sqrt5)/9`

`:. sin(2theta)` `= 2 xx 1/9 xx (4sqrt5)/9`
  `= (8sqrt5)/81`

 
`=>D`

Vectors, SPEC2 2020 VCAA 16 MC

Let  `underset~a = underset~i + 2underset~j + 2underset~k`  and  `underset~b = 2underset~i - 4underset~j + 4underset~k`, where the acute angle between these vectors is  `theta`.

The value of  `sin(2theta)`  is

  1. `1/9`
  2. `(4sqrt5)/9`
  3. `(4sqrt5)/81`
  4. `(8sqrt5)/81`
  5. `(2sqrt46)/25`
Show Answers Only

`D`

Show Worked Solution

`underset~a = ((1),(2),(2)), |underset~a| = sqrt(1 + 4 + 4) = 3`

`underset~b = ((2),(−4),(4)) , |underset~b| = sqrt(4 + 16 + 16) = 6`

`underset~a · underset~b = ((1),(2),(2))((1),(−4),(4)) = 2 – 8 + 8 = 2`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 2/(3 xx 6) = 1/9`
 

`sintheta = sqrt(81 – 1)/9 = (4sqrt5)/9`

`:. sin(2theta)` `= 2 xx 1/9 xx (4sqrt5)/9`
  `= (8sqrt5)/81`

 
`=>D`

Vectors, SPEC2 SM-Bank 22

`ABCDEFGH` are the vertices of a rectangular prism.
  


 

  1. Show that the internal diagonals of the prism, `AG` and `DF`, intersect.  (2 marks)

  2. Calculate the acute angle, `theta`, between the diagonals, to the nearest minute.  (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `83^@37′`

Show Worked Solution

i.    `A(2, text{−2}, 0),`   `G(text{−2}, 2, 2)`
  `D(2, 2, 0),`   `F (text{−2}, text{−2}, 2)`

 

`text(Midpoint)\ AG = ((1/2 (2-2)),(1/2 (text{−2} + 2)),(1/2 (0 + 2))) = ((0), (0), (1))`

`text(Midpoint)\ DF = ((1/2 (2-2)),(1/2 (2-2)),(1/2 (0 + 2))) = ((0), (0), (1))`
 

`text(S) text(ince midpoints are the same), AG and DF\ text(intersect.)`

 

ii.    `vec(AG) = ((text{−2}), (2), (2))-((2), (text{−2}), (0)) = ((text{−4}), (4), (2))`
  `vec(DF) = ((text{−2}), (text{−2}), (2))-((2), (2), (0)) = ((text{−4}), (text{−4}), (2))`

 

`vec (AG) ⋅ vec (DF) = |\ vec (AG)\ | ⋅ |\ vec(DF)\ |\  cos theta`

`((text{−4}), (4), (2)) ⋅ ((text{−4}), (text{−4}), (2)) = sqrt 36 sqrt 36 cos theta`

`16-16 + 4` `= 36 cos theta`
`cos theta` `= 1/9`
`theta` `= 83.62…`
  `= 83^@37′`

Vectors, SPEC2 SM-Bank 23

A cube with side length 3 units is pictured below.
 

     
 

  1. Calculate the magnitude of vector `vec(AG)`.  (1 mark)

  2. Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`.  (3 marks)

Show Answers Only

  1. `3 sqrt 3\ text(units)`
  2. `70^@32′`

Show Worked Solution

i.   `A(3, 0 , 0), \ \ G(0, 3, 3)`

  `vec(AG)` `= ((0), (3), (3))-((3), (0), (0)) = ((text{−3}), (3), (3))`
  `|\ vec(AG)\ |` `= sqrt (9 + 9 + 9)`
    `= 3 sqrt 3\ text(units)`

 

ii.    `H (3, 3, 3)`
  `vec(BH) = ((3), (3), (3))`
`vec(AG) ⋅ vec(BH)` `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta`
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta`
`-9 + 9 + 9` `= 27 cos theta`
`cos theta` `= 1/3`
`theta` `= 70.52…`
  `= 70^@32′`

Vectors, SPEC2 2020 VCAA 14 MC

The magnitude of the component of the force  `underset~F = underset~i + 6underset~j - 18underset~k`  that acts in the direction  `underset~d = 2underset~i - 3underset~j - 6underset~k`  is

  1. `92/19`
  2. `92/7`
  3. `124/7`
  4. `92/11`
  5. `18/7`
Show Answers Only

`B`

Show Worked Solution
`|underset~d|` `= sqrt(2^2 + (−3)^2 + (−6)^2) = 7`
`underset~overset^d` `= 1/7(2underset~i – 3underset~j – 6underset~k)`
`underset~F · underset~overset^d` `= 1/7((1),(6),(−18))((2),(−3),(−6))`
  `= 92/7`

 
`=>B`