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Statistics, STD1 S3 2019 HSC 27

A set of bivariate data is collected by measuring the height and arm span of eight children. The graph shows a scatterplot of these measurements.
 

  1. On the graph, draw a line of best fit by eye.  (1 mark)

  2. Robert is a child from the class who was absent when the measurements were taken. He has an arm span of 147 cm. Using your line of best fit from part (a), estimate Robert’s height.  (1 mark)

Show Answers Only
  1.   
  2. `text(Robert’s height ≈ 151.1 cm)`
Show Worked Solution

a.     
       

♦ Mean mark (a) 38%.

b.   `text(Robert’s height ≈ 151.1 cm)`

`text{(Answers can vary slightly depending on line of best fit drawn).}`

Financial Maths, STD1 F1 2019 HSC 26

Scott decided to have a shed built at his home. He agreed to the following costs:

  • Materials
– Timber $5400
– Roof $1800
– Nails $160
– Paint $375
  • $70 per hour for the builder when working Monday to Friday
  • $30 per hour for the labourer when working Monday to Friday
  • Builder and labourer paid time-and-a-half when working on Saturday.

It took six days to build the shed. The builder and labourer both worked from 8 am until 4 pm each day from Monday to Friday. On each of these days, they both took a 1 hour unpaid lunch break at 12 noon.

The builder and labourer also both worked 4 hours on Saturday, without a break, to finish the job.

What was the total cost to build the shed?  (4 marks)

Show Answers Only

`$11\ 835`

Show Worked Solution
`text(Material cost)` `= 5400 + 1800 + 160 + 375`
  `= $7735`

 
`text(Labour cost:)`

`text{Hours per weekday = 8 am to 4 pm (less 1 hour) = 7}`

`text(Builder cost)` `= 5 xx 7 xx 70 + 4 xx 1.5 xx 70`
  `= $2870`
`text(Labourer cost)` `= 5 xx 7 xx 30 + 4 xx 1.5 xx 30`
  `= $1230`

 

`:.\ text(Total cost)` `= 7735 + 2870 + 1230`
  `= $11\ 835`

Probability, STD1 S2 2019 HSC 24

The faces on a biased six-sided die are labelled 1, 2, 3, 4, 5 and 6. The die was rolled twenty times. The relative frequency of rolling a 6 was 30% and the relative frequency of rolling a 2 was 15%. The number 3 was the only other number rolled in the twenty rolls.

How many times was the number 3 rolled in the twenty rolls of the die?  (3 marks)

 
Show Answers Only

`11`

Show Worked Solution

`text(Number of 6’s) = 30/100 xx 20 = 6`

`text(Number of 2’s) = 15/100 xx 20 = 3`

`:.\ text(Number of 3’s)` `= 20 – (6 + 3)`
  `= 11`

Measurement, STD1 M5 2019 HSC 20

The plan of the lower level of a small house is shown.


 

  1. How many windows are shown on the plan?  (1 mark)

  2. What is the actual perimeter, in metres, of the shaded part of the kitchen floor?  (2 marks)

Show Answers Only
  1. `text(6 windows)`
  2. `6.5\ text(metres)`
Show Worked Solution

a.   `text(6 windows)`

 

b.    `text(Plan Perimeter)` `= 2 xx 3\ text(units) + 2 xx 3.5\ text(units)`
    `= 13\ text(units)`

 

`:.\ text(Actual perimeter)` `= 13 xx 0.5`
  `= 6.5\ text(metres)`

Functions, EXT1 F2 2019 HSC 14b

The diagram shows the graph of  `y = 1/(x - k)`, where  `k`  is a positive real number.
 


 

By considering the graphs of  `y = x^2`  and  `y = 1/(x - k)`, explain why the function  `f(x) = x^3 - kx^2 - 1`  has exactly one real zero.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Draw)\ \ y = x^2\ \ text(on the diagram:)`
 

`
 

`text(One point of intersection occurs when)`

`x^2` `= 1/(x – k)`
`x^3 – kx^2` `= 1`
`x^3 – kx^2 – 1` `= 0`

 
`text(S)text(ince only 1 point of intersection)`

`=> x^3 – kx^2 – 1 = 0\ \ text(has exactly 1 zero)`

Financial Maths, STD1 F3 2019 HSC 21

A new car is bought for $24 950. Each year the value of the car depreciates by 14%.

Using the declining-balance method, calculate the salvage value of the car at the end of 10 years.  (2 marks)

Show Answers Only

`$5521.47\ \ (text(nearest cent))`

Show Worked Solution

`V_0 = 24\ 950, \ r = 0.14, \ n = 10`

`S` `= V_0(1 – r)^n`
  `= 24\ 950(1 – 0.14)^10`
  `= $5521.47\ \ (text(nearest cent))`

Measurement, STD1 M4 2019 HSC 18

The travel graph displays Nikau's car trip along a straight road from home and back again. The trip has been broken into four separate sections: `A`, `B`, `C`  and  `D`.
 


 

  1. How far did Nikau travel in total?  (1 mark)

  2. In which section of the trip, `A`, `B`, `C` and `D`, did Nikau travel the fastest?  (1 mark)

Show Answers Only
  1. `400\ text(km)`
  2. `text(S)text(ection)\ D`
Show Worked Solution
a.    `text(Distance travelled)` `= 2 xx 200`
    `= 400\ text(km)`

 

♦ Mean mark part (b) 42%.

b.   `text{Fastest section has the steepest slope (in either direction).}`

`:. text(S)text(ection)\ D\ text(was the fastest.)`

Mechanics, EXT2* M1 2019 HSC 13d

The point  `O`  is on a sloping plane that forms an angle of 45° to the horizontal. A particle is projected from the point  `O`. The particle hits a point  `A`  on the sloping plane as shown in the diagram.
 


 

The equation of the line  `OA`  is  `y = -x`. The equations of motion of the particle are

`x = 18t`

`y = 18 sqrt(3t) - 5t^2,`

where  `t`  is the time in seconds after projection. Do NOT prove these equations.

  1. Find the distance  `OA`  between the point of projection and the point where the particle hits the sloping plane.  (2 marks)

  2. What is the size of the acute angle that the path of the particle makes with the sloping plane as the particle hits the point  `A`?  (3 marks)

Show Answers Only
  1. `(324(sqrt 2 + sqrt 6))/5\ text(units)`
  2. `30^@`
Show Worked Solution
i.    `x` `= 18t`
  `y` `= 18 sqrt 3 t – 5t^2`

 
`text(Particle hits slope when)\ \ y = -x`

`18 sqrt 3 t – 5t^2` `= -18t`
`5t^2 – 18t – 18 sqrt3 t` `= 0`
`t(5t – 18 – 18 sqrt 3)` `= 0`
`5t – 18 – 18 sqrt 3` `= 0`
`5t` `= 18 + 18 sqrt 3`
`t` `= (18 + 18 sqrt 3)/5`

 
`text(When)\ t = (18 + 18 sqrt 3)/5,`

`x = 18 xx ((18 + 18 sqrt 3)/5)`

`text{Using Pythagoras (isosceles Δ):}`

`OA` `= sqrt(2 xx 18^2 xx((18 + 18 sqrt 3)/5)^2)`
  `= sqrt 2 xx 18 xx ((18 + 18 sqrt 3)/5)`
  `= (324(sqrt 2 + sqrt 6))/5\ text(units)`

 

ii.    `x` `= 18t => dot x = 18`
  `y` `= 18 sqrt 3 t – 5t^2 => dot y = 18 sqrt 3 – 10t`

 
`text(When)\ \ t = (18 + 18 sqrt 3)/5,`

`dot y` `= 18 sqrt 3 – 10 ((18 + 18 sqrt 3)/5)`
  `= 18 sqrt 3 – 36 – 36 sqrt 3`
  `= -18 sqrt 3 – 36`
  `= -18(sqrt 3 + 2)`

 

`text(Find angle with the horizontal at impact:)`

♦ Mean mark part (ii) 39%.
 

 

`tan theta` `= (18(sqrt 3 + 2))/18`
  `= sqrt 3 + 2`
`theta` `= tan^(-1)(sqrt 3 + 2)`
  `= 75^@`

 
`:.\ text(Angle made with the slope)`

`= 75 – 45`

`= 30^@`

Mechanics, EXT2* M1 2019 HSC 13c

A particle moves in a straight line. At time  `t`  seconds the particle has a displacement of `x` m, a velocity of  `v\ text(m s)^(-1)`  and acceleration  `a\ text(m s)^(-2)`.

Initially the particle has displacement  0 m  and velocity  `2\ text(m s)^(-1)`. The acceleration is given by  `a = -2e^(-x)`. The velocity of the particle is always positive.

  1. Show that  `v = 2e^((-x)/2)`.  (2 marks)

  2. Find an expression for `x` as a function of  `t`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `x = 2 ln(t + 1)`
Show Worked Solution
i.    `a` `= -2e^(-x)`
  `d/(dx)(1/2v^2)` `= -2e^(-x)`
  `1/2 v^2` `= int -2e^(-x) dx`
    `= 2e^(-x) + C`

 
`text(When)\ \ x = 0,\ \ v = 2:`

`1/2 ⋅ 2^2` `= 2e^0 + C`
`C` `= 0`
`1/2 v^2` `= 2e^(-x)`
`v^2` `= 4e^(-x)`
`v` `= +-(4e^(-x))^(1/2)`
  `= +-2e^(-x/2)`

 
`text(S)text(ince)\ \ v = 2\ \ text(when)\ \ x = 0,`

`v = 2e^((-x)/2)`

 

ii.    `(dx)/(dt)` `= 2e^((-x)/2)`
  `(dt)/(dx)` `= (e^(x/2))/2`
  `t` `= 1/2 int e^(x/2) dx`
    `= 1/2 xx 2 xx e^(x/2) + C`
    `= e^(x/2) + C`

 
`text(When)\ \ t = 0,\ \ x = 0:`

♦ Mean mark part (ii) 46%.

`0` `= e^0 + C`
`C` `= -1`
`t` `= e^(x/2) – 1`
`e^(x/2)` `= t + 1`
`x/2` `= ln (t + 1)`
`:. x` `= 2 ln (t + 1)`

Calculus, EXT1 C2 2019 HSC 13a

Use the substitution  `u = cos^2 x`  to evaluate  `int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx`.  (3 marks)

Show Answers Only

`ln {:10/9`

Show Worked Solution
`u` `= cos^2 x`
`(du)/(dx)` `= -2 sin x cos x`
  `= -sin 2x`
`du` `= -sin 2x\ dx`

 
`text(When)\ \ x = pi/4,\ \ u = 1/2`

`text(When)\ \ x = 0,\ \ u = 1`

`:. int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx` `= -int_1^(1/2) (du)/(4 + u)`
  `= -[ln (4 + u)]_1^(1/2)`
  `= -(ln 4.5 – ln 5)`
  `= -ln {:9/10`
  `= ln {:10/9`

Calculus, EXT1 C1 2019 HSC 12d

A refrigerator has a constant temperature of 3°C. A can of drink with temperature 30°C is placed in the refrigerator.

After being in the refrigerator for 15 minutes, the temperature of the can of drink is 28°C.

The change in the temperature of the can of drink can be modelled by  `(dT)/(dt) = k(T - 3)`,  where `T` is the temperature of the can of drink, `t` is the time in minutes after the can is placed in the refrigerator and `k` is a constant.

  1. Show that  `T = 3 + Ae^(kt)`, where `A` is a constant, satisfies

     

    `qquad(dT)/(dt) = k(T - 3)`.  (1 mark)

  2. After 60 minutes, at what rate is the temperature of the can of drink changing?   (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `-0.10^@\ text{C per minute (decreasing)}`
Show Worked Solution

a.    `T = 3 + Ae^(kt)`

`(dT)/(dt)` `= k ⋅ Ae^(kt)`
  `= k (3 + Ae^(kt) – 3)`
  `= k(T – 3)`

 

b.   `text(When)\ \ t = 0,\ \ T = 30,`

`30` `= 3 + Ae^0`
`A` `= 27`

 
`text(When)\ \ t = 15,\ \ T = 28,`

`28` `= 3 + 27e^(15k)`
`25` `= 27e^(15k)`
`e^(15k)` `= 25/27`
`15k` `= ln (25/27)`
`k` `= 1/15 xx ln (25/27)`

 
`text(Find)\ \ (dT)/(dt)\ \ text(when)\ \ t = 60:`

`(dT)/(dt)` `= 1/15 xx ln (25/27) xx 27e^(60 xx 1/15 ln (25/27))`
  `= 27/15 xx ln (25/27) xx e^(4 ln (25/27))`
  `=27/15 xx ln(25/27) xx (25/27)^4`
  `= -0.1018…`
  `= -0.10^@ text{C per minute (decreasing)}`

Mechanics, EXT2* M1 2019 HSC 12b

A particle is moving along the `x`-axis in simple harmonic motion. The position of the particle is given by

`x = sqrt 2 cos 3t + sqrt 6 sin 3t,` for  `t >= 0` 

  1. Write  `x`  in the form  `R cos(3t - alpha)`, where  `R > 0`  and  `0 < alpha < pi/2`.  (2 marks)

  2. Find the two values for  `x`  where the particle comes to rest.   (1 mark)

  3. When is the first time that the speed of the particle is equal to half of its maximum speed?  (2 marks)

Show Answers Only
  1. `x = 2 sqrt 2 cos (3t – pi/3)`
  2. `2 sqrt 2 or -2 sqrt 2`
  3. `t = pi/18`
Show Worked Solution

i.    `x = sqrt 2 cos 3t + sqrt 6 sin 3t`

`R cos (3t – alpha) = R cos alpha cos 3t + R sin alpha sin 3t`

`=> R cos alpha = sqrt 2`

`=> R sin alpha = sqrt 6`

`R^2 cos^2 alpha + R^2 sin^2 alpha` `= 2 + 6`
`R^2 (cos^2 alpha + sin^2 alpha)` `= 8`
`R` `=2sqrt2`

  

`2 sqrt 2 cos alpha` `= sqrt 2`
`cos alpha` `= 1/2`
`alpha` `= pi/3`

 
`:. x = 2 sqrt 2 cos (3t – pi/3)`

♦ Mean mark part (ii) 46%.

 

ii.    `text(At the extremities of the amplitude,)`
 

`text(the particle stops and reverses.)`

`:. v = 0\ \ text(when)\ \ x = 2 sqrt 2 or -2 sqrt 2`

 

iii.   `x = 2 sqrt 2 cos (3t – pi/3)`

♦ Mean mark part (iii) 49%.

`(dx)/(dt) = -6 sqrt 2 sin(3t – pi/3)`

 
`text(Max speed) = 6 sqrt 2`

`text(Find)\ \ t\ \ text(when)\ \ (dx)/(dt) = +-3 sqrt 2`

`-6 sqrt 2 sin (3t – pi/3)` `= 3 sqrt 2`
`sin(3t – pi/3)` `= -1/2`
`3t – pi/3` `= -pi/6`
`3t` `= pi/6`
`t` `= pi/18`

 

`-6 sqrt 2 sin (3t – pi/3)` `= -3 sqrt 2`
`sin(3t – pi/3)` `= 1/2`
`3t – pi/3` `= pi/6`
`3t` `= pi/2`
`t` `= pi/6`

 
`:. t = pi/18\ text{(1st time)}`

Statistics, EXT1 S1 2019 HSC 11f

Prize-winning symbols are printed on 5% of ice-cream sticks. The ice-creams are randomly packed into boxes of 8.

  1. What is the probability that a box contains no prize-winning symbols?  (1 mark)

  2. What is the probability that a box contains at least 2 prize-winning symbols?  (2 marks)

Show Answers Only
  1. `0.95^8`
  2. `5.72 text(%)`
Show Worked Solution

i.   `text(Chances of any stick winning:)`

`P(W) = 0.05`

`P(barW) = 0.95`

`P (text{In box of 8, all}\ barW)`

`= 0.95^8`

 

ii.   `P\ text{(at least two winners in a box)}`

`= 1 – P text{(1 winner)} – P text{(0 winners)}`

`= 1 – \ ^8 C_1 xx 0.95^7 xx 0.05^1 – \ ^8 C_0 xx 0.95^8`

`= 0.05724…`

`= 5.72 text{%   (to 2 d.p.)}`

Functions, EXT1 F2 2019 HSC 7 MC

Let  `P(x) = qx^3 + rx^2 + rx + q`  where `q` and `r` are constants, `q != 0`. One of the zeros of  `P(x)`  is  `-1`.

Given that  ` alpha`  is a zero of  `P(x),\ alpha != -1`, which of the following is also a zero?

A.     `-1/alpha`

B.     `-q/alpha`

C.     `1/alpha`

D.     `q/alpha`

Show Answers Only

`C`

Show Worked Solution

`text(Roots:)\ \ alpha,\ beta,\ -1`

Mean mark 51%.

`alpha beta(-1) = -d/a = -q/q = -1`

`- alpha beta` `= -1`
`:. beta` `= 1/alpha`

 
`=>  C`

Financial Maths, 2ADV M1 2019 HSC 16a

A person wins $1 000 000 in a competition and decides to invest this money in an account that earns interest at 6% per annum compounded quarterly. The person decides to withdraw $80 000 from this account at the end of every fourth quarter. Let  `A_n`  be the amount remaining in the account after the `n`th withdrawal.

  1.  Show that the amount remaining in the account after the withdrawal at the end of the eighth quarter is

     

    `qquad A_2 = 1\ 000\ 000 xx 1.015^8 - 80\ 000(1 + 1.015^4)`.  (2 marks)

  2.  For how many years can the full amount of $80 000 be withdrawn?  (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `23\ text(years)`
Show Worked Solution

i.  `6 text(% p.a.)= 1.5 text(%)\ text(per quarter)`

  `A_1` `= 1\ 000\ 000 xx 1.015^4 – 80\ 000`
  `A_2` `= A_1 xx 1.015^4 – 80\ 000`
    `= (1\ 000\ 000 xx 1.015^4 – 80\ 000) xx 1.015^4 – 80\ 000`
    `= 1\ 000\ 000 xx 1.015^8 – 80\ 000 xx 1.015^4 – 80\ 000`
    `= 1\ 000\ 000 xx 1.015^8 – 80\ 000 (1 + 1.015^4)`

 

ii.    `A_3` `= 1\ 000\ 000 xx 1.015^12 – 80\ 000 (1 + 1.015^4 + 1.015^8)`
  `vdots`  
  `A_n` `= 1\ 000\ 000 xx 1.015^(4n) – 80\ 000\ underbrace{(1 + 1.015^4 + … + 1.015^(4n – 4))}_{text(GP where)\ a = 1,\ r = 1.015^4}`
    `= 1\ 000\ 000 xx 1.015^(4n) – 80\ 000 [(1(1.015^(4n) – 1))/(1.015^4 – 1)]`
    `= 1\ 000\ 000 xx 1.015^(4n) – 1\ 303\ 706 (1.015^(4n) – 1)`

♦♦♦ Mean mark part (ii) 19%.

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

`0` `= 1\ 000\ 000 xx 1.015^(4n) – 1\ 303\ 706 xx 1.015^(4n) + 1\ 303\ 706`
`-1\ 303\ 706` `> -303\ 706 xx 1.015^(4n)`
`1.015^(4n)` `> (1\ 303\ 706)/(303\ 706)`
`4n` `> (ln((1\ 303\ 706)/(303\ 706)))/(ln 1.015)`
  `> 97.853…`
`n` `> 24.46…`

 
`:.\ text(Full amount can be drawn for 24 years.)`

Probability, 2ADV S1 2019 HSC 15d

The probability that a person chosen at random has red hair is 0.02

  1. Two people are chosen at random.

     

    What is the probability that at least ONE has red hair?  (2 marks)

  2. What is the smallest number of people that can be chosen at random so that the probability that at least ONE has red hair is greater than 0.4?  (2 marks)

Show Answers Only
  1. `0.0396`
  2. `26\ text(people)`
Show Worked Solution
a.   `P(R)` `= 0.02`
  `P(barR)` `= 0.98`

 
`P\ text{(At least 1 has red hair)}`

`= 1 – P(barR, barR)`

`= 1 – 0.98 xx 0.98`

`= 0.0396`

 

b.  `text(Find)\ \ n\ \ text(such that)`

♦♦ Mean mark 24%.

`1 – 0.98^n` `> 0.4`
`0.98^n` `< 0.6`
`ln 0.98^n` `< ln 0.6`
`n ln 0.98` `< ln 0.6`
`n` `> (ln 0.6)/(ln 0.98),\ \ \ (ln 0.98 <0)`
  `> 25.28…`

 
`:. 26\ text(people must be chosen.)`

Financial Maths, STD2 F4 2019 HSC 37

A new car is bought for $24 950. Each year the value of the car is depreciated by the same percentage.

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

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{End of year}\rule[-1ex]{0pt}{0pt} & \textit{Value}\\
\hline
\rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & \$21\ 457.00 \\
\hline
\rule{0pt}{2.5ex}2\rule[-1ex]{0pt}{0pt} & \$18\ 453.02 \\
\hline
\rule{0pt}{2.5ex}3\rule[-1ex]{0pt}{0pt} & \$15\ 869.60 \\
\hline
\end{array}

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

Show Answers Only

`$5521.47`

Show Worked Solution

`text(Find the depreciation rate:)`

`S` `= V_0(1-r)^n`
`21\ 457` `= 24\ 950(1-r)^1`
`1-r` `= (21\ 457)/(24\ 950)`
`1-r` `= 0.86`
`r` `= 0.14`

 
`:.\ text(Value after 10 years)`

`= 24\ 950(1-0.14)^10`

`= 5521.474…`

`= $5521.47\ \ (text(nearest cent))`

Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.
 

  1. Write the direct variation equation relating British pounds to Australian dollars in the form  `p = md`. Leave `m` as a fraction.  (1 mark)

  2. The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation  `y = 76d`.

     

    Convert 93 100 Japanese yen to British pounds.  (2 marks)

Show Answers Only
  1. `p = 4/7 d`
  2. `93\ 100\ text(Yen = 700 pounds)`
Show Worked Solution

a.   `m = text(rise)/text(run) = 4/7`

♦ Mean mark 42%.

`p = 4/7 d`

 

b.   `text(Yen to Australian dollars:)`

`y` `=76d`
`93\ 100` `= 76d`
`d` `= (93\ 100)/76`
  `= 1225`

 
`text(Aust dollars to pounds:)`

`p` `= 4/7 xx 1225`
  `= 700\ text(pounds)`

 
`:. 93\ 100\ text(Yen = 700 pounds)`

Algebra, STD2 A4 2019 HSC 33

The time taken for a car to travel between two towns at a constant speed varies inversely with its speed.

It takes 1.5 hours for the car to travel between the two towns at a constant speed of 80 km/h.

  1. Calculate the distance between the two towns.  (1 mark)

  2. By first plotting four points, draw the curve that shows the time taken to travel between the two towns at different constant speeds.  (3 marks)

Show Answers Only
  1. `120\ text(km)`
  2.  
Show Worked Solution
a.    `D` `= S xx T`
    `= 80 xx 1.5`
    `= 120\ text(km)`

 
b.   

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ s\ \  \rule[-1ex]{0pt}{0pt} & 20 & 40 & 60 & 80 \\
\hline
\rule{0pt}{2.5ex} t \rule[-1ex]{0pt}{0pt} & 6 & 3 & 2 & 1.5 \\
\hline
\end{array}

Financial Maths, STD2 F1 2019 HSC 32

The table shows the income tax rates for the 2018-2019 financial year.

The Medicare levy is calculated as 2% of taxable income.

For the 2018-2019 financial year, Charlie pays a Medicare levy of $1934.80.

Calculate the tax payable on Charlie's taxable income.  (3 marks)

Show Answers Only

`$23\ 290.80`

Show Worked Solution
`text(2% taxable income)` `= 1934.80`
`text(taxable income)` `= 1934.80/0.02`
  `= $96\ 740`

 

`:.\ text(Tax payable)` `= 20\ 797 + 0.37 xx (96\ 740 – 90\ 000)`
  `= 20\ 797 + 2493.80`
  `= $23\ 290.80`

Calculus, 2ADV C4 2019 HSC 14a

A particle is moving along a straight line. The particle is initially at rest. The acceleration of the particle at time  `t`  seconds is given by  `a = e^(2t)-4`, where  `t >= 0`.

Find an expression, in terms of  `t`, for the velocity of the particle.  (2 marks)

Show Answers Only

`v = 1/2e^(2t)-4t-1/2`

Show Worked Solution

`a = (dv)/(dt) = e^(2t)-4`

`v` `= int e^(2t)-4\ dt`
  `= 1/2 e^(2t)-4t + c`

 
`text(When)\ t = 0,\ v = 0`

`0 = 1/2 e^0-0 + c`

`c = -1/2`

`:. v = 1/2e^(2t)-4t-1/2`

Functions, 2ADV F1 2019 HSC 13e

  1. Sketch the graph of  `y = |\ x - 1\ |`  for  `-4 <= x <= 4`.  (1 mark)

  2. Using the sketch from part i, or otherwise, solve  `|\ x - 1\ | = 2x + 4`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(-1, 2)`
Show Worked Solution
(i)   

 

ii.    `text(By inspection, intersection when)\ x = -1`

`text(Test:)`

`|-1-1|` `= -2 + 4`
`2` `= 2`

 
`:.\ text(Intersection at)\ (-1, 2)`

Networks, STD2 N2 2019 HSC 30

The network diagram shows the tracks connecting 8 picnic sites in a nature park. The vertices `A` to `H` represents the picnic sites. The weights on the edges represent the distance along the tracks between the picnic sites, in kilometres.
 


 

  1. Each picnic site needs to provide drinking water. The main water source is at site `A`.

     

    Draw a minimum spanning tree and calculate the minimum length of water pipes required to supply water to all the sites if the water pipes can only be laid along the tracks.  (2 marks)

  2. One day, the track between `C` and `H` is closed. State the vertices that identify the shortest path from `C` to `E` that avoids the closed track.  (1 mark)

Show Answers Only
  1. `text(25 km)`
  2. `CGHE`
Show Worked Solution

a.   `text(One strategy – using Prim’s algorithm:)`

`text(Starting at)\ A`

`text(1st edge -)\ AB,\ \ text(2nd edge -)\ BC`

`text(3rd edge -)\ CH,\ \ text(4th edge -)\ HG`

`text(5th edge -)\ GF,\ \ text(6th edge -)\ HD`

`text(7th edge -)\ DE\ text(or)\ HE`
 

`text(Maximum length = 4 + 5 + 3 + 2 + 1 + 5 + 5 = 25 km)`

 

b.   `text(Shortest Path is)\ CGHE`

Calculus, EXT1* C3 2019 HSC 13d

The diagram shows the region bounded by the curve  `y = x - x^3`, and the `x`-axis between  `x = 0`  and  `x = 1`. The region is rotated about the `x`-axis to form a solid.
 


 

Find the exact value of the volume of the solid formed.  (3 marks)

Show Answers Only

`(8 pi)/105\ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 (x – x^3)^2 dx`
  `= pi int_0^1 x^2 – 2x^4 + x^6\ dx`
  `= pi [x^3/3 – 2/5 x^5 + 1/7 x^7]_0^1`
  `= pi(1/3 – 2/5 + 1/7)`
  `= (8 pi)/105\ text(u³)`

Trigonometry, 2ADV T1 2019 HSC 13b

The diagram shows a circle with centre `O` and radius 20 cm.

The points `A` and `B` lie on the circle such that  `∠AOB = 70^@`.
 


 

Find the perimeter of the shaded segment, giving your answer correct to one decimal place.  (3 marks)

Show Answers Only

`47.4\ text{cm}`

Show Worked Solution
`text(Arc)\ AB` `= 70/360 xx 2pi xx 20`
  `= 24.43…`

 
`text(Using cosine rule,)`

`AB^2` `= 20^2 xx 20^2 – 2 ⋅ 20 ⋅ 20 xx cos 70`
  `= 526.383…`
`AB` `= 22.94…`

 

`:.\ text(Perimeter)` `= 24.43 + 22.94…`
  `= 47.37`
  `= 47.4\ text{cm (1 d.p.)}`

Calculus, 2ADV C4 2019 HSC 12d

The diagram shows the graph of  `y = (3x)/(x^2 + 1)`.
 

 
The region enclosed by the graph, the `x`-axis and the line  `x = 3`  is shaded.

Calculate the exact value of the area of the shaded region.  (3 marks)

Show Answers Only

`3/2 ln 10\ text(u²)`

Show Worked Solution
`text(Area)` `= int_0^3 (3x)/(x^2 + 1)\ dx`
  `= 3/2 int_0^3 (2x)/(x^2 + 1)\ dx`
  `= 3/2[ln(x^2 + 1)]_0^3`
  `= 3/2(ln 10 – ln 1)`
  `= 3/2 ln10\ \ text(u²)`

Calculus, EXT1* C1 2019 HSC 12c

The number of leaves, `L(t)`, on a tree `t` days after the start of autumn can be modelled by

`L(t) = 200\ 000e^(-0.14t)`

  1. What is the number of leaves on the tree when  `t = 31`?  (1 mark)

  2. What is the rate of change of the number of leaves on the tree when  `t = 31`?  (2 marks)

  3. For what value of `t` are there 100 leaves on the tree?  (2 marks)

Show Answers Only
  1. `2607\ text(leaves)`
  2. `-365.02…`
  3. `54.3\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(When)\ \ t = 31`

`L(t)` `= 200\ 000 xx e^(-0.14(31))`
  `=2607.305…`
  `= 2607\ text(leaves)`

 

ii.    `L` `= 2000\ 000^(-0.14t)`
  `(dL)/(dt)` `= -0.14 xx 200\ 000 e^(0.14t`
    `= -28\ 000e^(-0.14t)`

 
`text(When)\ \ t = 31,`

`(dL)/(dt)` `= -28\ 000 xx e^(-0.14(31))`
  `= -365.02…`

 
`:.\ text(365 leaves fall per day.)`

 

iii.   `text(Find)\ t\ text(when)\ \ L = 100:`

`100` `= 200\ 000 e^(-0.14t)`
`e^(-0.14t)` `= 0.0005`
`e^(-0.14t)` `= ln 0.0005`
`t` `= (ln 0.0005)/(-0.14)`
  `= 54.292…`
  `= 54.3\ text{(1 d.p.)}`

Functions, EXT1* F1 2019 HSC 11g

The parabola  `y = x^2`  meets the line  `y = x + 2`  at the points  `(-1, 1)`  and  `(2, 4)`. Do NOT prove this.

By first sketching the graphs of  `y = x^2`  and  `y = x + 2`, shade the region which simultaneously satisfies the two inequalities  `y >= x^2`  and  `y >= x + 2`.  (2 marks)

Show Answers Only

Show Worked Solution

Algebra, STD2 A1 2019 HSC 28

The formula below is used to calculate an estimate for blood alcohol content `(BAC)` for females.

`BAC_text(female) = (10N - 7.5H)/(5.5M)`

The number of hours required for a person to reach zero `BAC` after they stop consuming alcohol is given by the following formula.

`text(Time) = (BAC)/0.015`

The number of standard drinks in a glass of wine and a glass of spirits is shown.
 

Hannah weighs 60 kg. She consumed 3 glasses of wine and 4 glasses of spirits between 6:15 pm and 12:30 am the following day. She then stopped drinking alcohol.

Using the given formulae, calculate the time in the morning when Hannah's `BAC` should reach zero.  (4 marks)

Show Answers Only

`text(6:23 am)`

Show Worked Solution

`text(Standard drinks consumed)\ (N) = 3 xx 1.2 + 4 = 7.6`

`text(Hours drinking)\ (H) = text(6 h 15 min = 6.25 hours)`

`BAC(text(Hannah))` `= (10 xx 7.6 – 7.5 xx 6.25)/(5.5 xx 60)`
  `= 0.08825…`

COMMENT: Convert a decimal answer into hours and minutes using the calculator degree/minute function.

`text(Time(to zero))` `= (0.08825…)/0.015`
  `= 5.883…\ text(hours)`
  `= 5\ text(hours 53 minutes)`

 
`:.\ text(Hannah should reach zero)\ BAC\ text(at 6:23 am)`

Networks, STD2 N3 2019 HSC 26

A project requires activities `A` to `F` to be completed. The activity chart shows the immediate prerequisite(s) and duration for each activity.

  1. By drawing a network diagram, determine the minimum time for the project to be completed.  (3 marks)

  2. Determine the float time of the non-critical activity.  (1 mark)

Show Answers Only
  1. `text(15 hours)`
  2. `text(3 hours)`
Show Worked Solution
a.   

 

`text(Scanning forwards:)`

`text(Minimum time = 2 + 6 + 2 + 4 + 1 = 15 hours)`

`text{(Scanning forwards and backwards is highly recommended but not}`

 `text{required in the network diagram.)}`

 

b.   `text(Critical Path is)\ ABDEF.`

♦♦ Mean mark 30%.

`text(Non-critical activity is)\ C.`

`text(Float time)` `= 5 – 2`
  `= 3\ text(hours)`

Algebra, STD2 A2 2019 HSC 14 MC

Last Saturday, Luke had 165 followers on social media. Rhys had 537 followers. On average, Luke gains another 3 followers per day and Rhys loses 2 followers per day.

If  `x`  represents the number of days since last Saturday and  `y`  represents the number of followers, which pair of equations model this situation?

A.  `text(Luke:)\ \ y = 165x + 3`

 

`text(Rhys:)\ \ y = 537x - 2`
B. `text(Luke:)\ \ y = 165 + 3x`

 

`text(Rhys:)\ \ y = 537 - 2x`
C. `text(Luke:)\ \ y = 3x + 165`

 

`text(Rhys:)\ \ y = 2x - 537`
D. `text(Luke:)\ \ y = 3 + 165x`

 

`text(Rhys:)\ \ y = 2 - 537x`
Show Answers Only

`B`

Show Worked Solution

`text(Luke starts with 165 and adds 3 per day:)`

`y = 165 + 3x`

`text(Rhys starts with 537 and loses 2 per day:)`

`y = 537 – 2x`

`=> B`

Financial Maths, STD2 F4 2019 HSC 13 MC

The graph show the future values over time of  `$P`, invested at three different rates of compound interest.
 


 

Which of the following correctly identifies each graph?

A. B.
C. D.
Show Answers Only

`C`

Show Worked Solution

`text(Values increase quicker)`

`text(- higher compounding interest rate)`

`text(- same rate but more frequent compounding period)`

`:. W = 10text(% quarterly)`

`X = 10text(% annually)`

`Y = 5text(% annually)`

 
`=> C`

Measurement, STD2 M2 2019 HSC 5 MC

The Coordinated Universal Time (UTC) of Auckland is +12 hours and the UTC of Chicago is −5 hours.

When the time in Chicago is 2 pm, Thursday, what is the time in Auckland?

  1. 9 pm, Wednesday
  2. 7 am, Thursday
  3. 9 pm, Thursday
  4. 7 am, Friday
Show Answers Only

`D`

Show Worked Solution

`text(Auckland is 17 hours ahead of Chicago.)`

`text(2 pm Thursday + 17 hours)`

`text(= 2 am Friday + 5 hours)`

`text(= 7 am Friday)`

`=> D`

Financial Maths, STD2 F4 2019 HSC 3 MC

Chris opens a bank account and deposits $1000 into it. Interest is paid at 3.5% per annum, compounding annually.

Assuming no further deposits or withdrawals are made, what will be the balance in the account at the end of two years?

  1. $1070.00
  2. $1071.23
  3. $1822.50
  4. $2070.00
Show Answers Only

`=> B`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 1000(1 + 0.035)^2`
  `= $1071.23`

 
`=> B`

Statistics, STD2 S4 2019 HSC 23

A set of bivariate data is collected by measuring the height and arm span of seven children. The graph shows a scatterplot of these measurements.
 


 

  1. Calculate Pearson's correlation coefficient for the data, correct to two decimal places.  (1 mark)

  2. Identify the direction and the strength of the linear association between height and arm span.  (1 mark)

  3. The equation of the least-squares regression line is shown.
     
               Height = 0.866 × (arm span) + 23.7
     
    A child has an arm span of 143 cm.

     

    Calculate the predicted height for this child using the equation of the least-squares regression line.  (1 mark)

Show Answers Only
  1. `0.98\ \ (text(2 d.p.))`
  2. `text(Direction: positive)`
    `text(Strength: strong)`
  3. `147.538\ text(cm)`
Show Worked Solution

a.   `text{Use  “A + Bx”  function (fx-82 calc):}`

♦ Mean mark 40%.
COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 correlation” .

`r` `= 0.9811…`
  `= 0.98\ \ (text(2 d.p.))`

 

b.   `text(Direction: positive)`

`text(Strength: strong)`

 

c.    `text(Height)` `= 0.866 xx 143 + 23.7`
    `= 147.538\ text(cm)`

Measurement, STD2 M6 2019 HSC 22

Two right-angled triangles, `ABC` and `ADC`, are shown.
 

Calculate the size of angle `theta`, correct to the nearest minute.  (3 marks)

Show Answers Only

`41°4^{′}\ \ text{(nearest minute)}`

Show Worked Solution

`text(Using Pythagoras in)\ DeltaACD:`

Mean mark 51%.

`AC^2` `= 2.5^2 + 6^2`
  `= 42.25`
`:.AC` `= 6.5\ text(cm)`

 
`text(In)\ DeltaABC:`

`costheta` `= 4.9/6.5`
`theta` `= cos^(−1)\ 4.9/6.5`
  `= 41.075…`
  `= 41°4^{′}31^{″}`
  `= 41°5^{′}\ \ text{(nearest minute)}`

Financial Maths, STD2 F4 2019 HSC 21

A person owns 1526 shares with a market value of $8.75 per share. The total dividend received for these shares is $1068.20.

Calculate the percentage dividend yield.  (2 marks)

Show Answers Only

`8text(%)`

Show Worked Solution
`text(Value of shares)` `= 1526 xx 8.75`
  `= $13\ 352.50`

 

`:.\ text(Dividend yield)` `= 1068.20/(13\ 352.50)`
  `= 0.08`
  `= 8text(%)`

Probability, STD2 S2 2019 HSC 20

A roulette wheel has the numbers 0, 1, 2, …, 36 where each of the 37 numbers is equally likely to be spun.
 

 
If the wheel is spun 18 500 times, calculate the expected frequency of spinning the number 8.  (2 marks)

Show Answers Only

`500`

Show Worked Solution

`P(8) = 1/37`

`:.\ text(Expected Frequency (8))`

`= 1/37 xx 18\ 500`

`= 500`

Statistics, STD2 S1 2019 HSC 19

The heights, in centimetres, of 10 players on a basketball team are shown.

170, 180, 185, 188, 192, 193, 193, 194, 196, 202

Is the height of the shortest player on the team considered an outlier? Justify your answer with calculations. (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`Q_1 = 185, \ Q_3 = 194`

Mean mark 51%.
COMMENT: The last statement must be made to achieve full marks here!

`IQR = 194 – 185 = 9`

`text(Shortest player = 170)`

`text(Outlier height:)`

`Q_1 – 1.5 xx IQR ` `= 185 – 1.5 xx 9`
  `= 171.5`

 
`:.\ text(S)text(ince 170 < 171.5, 170 is an outlier.)`

Measurement, STD2 M7 2019 HSC 18

Andrew, Brandon and Cosmo are the first three batters in the school cricket team. In a recent match, Andrew scored 30 runs, Brandon scored 25 runs and Cosmo scored 40 runs.

  1. What is the ratio of Andrew's to Brandon's to Cosmo's runs scored, in simplest form?  (2 marks)

  2. In this match, the ratio of the total number of runs scored by Andrew, Brandon and Cosmo to the total number of runs scored by the whole team is `19:36`.
  3. How many runs were scored by the whole team?  (2 marks)

Show Answers Only
  1. `6:5:8`
  2. `180\ text(runs)`
Show Worked Solution
a.    `A:B:C` `= 30:25:40`
    `= 6:5:8`

 

b.    `text(Total runs by)\ A,B,C` `= 30 + 25 + 40`
    `= 95`

 

`text(Let)\ R` `=\ text(team runs)`
`R/95` `= 36/19`
`:. R` `= (36 xx 95)/19`
  `= 180\ text(runs)`

Measurement, STD2 M6 2019 HSC 17

The diagram shows a triangle with sides of length `x` cm, 11 cm and 13 cm and an angle of 80°.
 


 

Use the cosine rule to calculate the value of `x`, correct to two significant figures.  (3 marks)

Show Answers Only

`16\ text(cm  (2 sig. fig.))`

Show Worked Solution
`x^2` `= 11^2 + 13^2 – 2 xx 11 xx 13 xx cos80°`
  `= 240.336…`

 

`:.x` `= 15.502…`
  `= 16\ text(cm  (2 sig. fig.))`

Measurement, STD2 M1 2019 HSC 16

A bowl is in the shape of a hemisphere with a diameter of 16 cm.
 

What is the volume of the bowl, correct to the nearest cubic centimetre?  (2 marks)

Show Answers Only

`1072\ text(cm)^3`

Show Worked Solution
`V` `= 1/2 xx 4/3pir^3`
  `= 1/2 xx 4/3 xx pi xx 8^3`
  `= 1072.3…`
  `= 1072\ text{cm}^3\ text{(nearest cm}^3 text{)}`

Calculus, 2ADV C1 2019 HSC 8 MC

A particle is moving along a straight line. The graph shows the acceleration of the particle.
 


 

For what value of `t` is the velocity `v` a maximum?

  1. `1`
  2. `2`
  3. `3`
  4. `5`
Show Answers Only

`C`

Show Worked Solution

`text(Velocity increases when)\ \ a > 0.`

`:. v_text(max)\ \ text(occurs when)\ \ t = 3.`

`=>  C`

Trigonometry, 2ADV T3 2019 HSC 7 MC

The diagram shows part of the graph of  `y = a sin(bx) + 4`.
 

What are the values of `a` and `b`?

  1. `a = 3 qquad qquad b = 1/2`
  2. `a = 3 qquad qquad b = 2`
  3. `a = 1.5 qquad \ b = 1/2`
  4. `a = 1.5 qquad \ b = 2`
Show Answers Only

`D`

Show Worked Solution

`a = 1/2 (5.5 – 2.5) = 1.5`

`text(S)text(ince graph passes through)\ \ (pi/4, 5.5):`

`5.5 = 1.5 sin(b xx pi/4) + 4`

`sin (b xx pi/4)` `= 1`
`b xx pi/4` `= pi/2`
`:. b` `= 2`

 
`=>  D`

Probability, 2ADV S1 2019 HSC 6 MC

A game is played by tossing an ordinary 6-sided die and an ordinary coin at the same time. The game is won if the uppermost face of the die shows an even number or the uppermost face of the coin shows a tail (or both).

What is the probability of winning this game?

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

`C`

Show Worked Solution

`text(Game lost only if an odd and a head show.)`

`:. P(W)` `= 1 – P text{(odd)} ⋅ P text{(head)}`
  `= 1 – 3/6 ⋅ 1/2`
  `= 3/4`

 
`=>  C`

Vectors, EXT1 V1 SM-Bank 8

A projectile is fired horizontally off a cliff at an initial speed of  `V`  metres per second.
 

 

The projectile strikes the water, `l`  metres from the base of the cliff.

Let `g` be the acceleration due to gravity and assume air resistance is negligible.

  1.  Show the projectile hits the water when
     
    `qquadt = sqrt((2d)/g)`  (2 marks)

  2.  If  `l`  equals twice the height of the cliff, at what angle does the projectile hit the water?  (2 marks)

  3.  Show that the speed at which the projectile hits the water is
     
    `qquad2sqrt(dg)`  metres per second.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `45°`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `underset~v` `= Vcosthetaunderset~i + (Vsintheta – g t)underset~j`
    `= Vcos0 underset~i + (Vsin0 – g t)underset~j`
    `= Vunderset~i – g tunderset~j`
`underset~s` `= intunderset~v\ dt`
  `= Vtunderset~i – 1/2g t^2 underset~j + c`

 
`text(When)\ t = 0, underset~s = 0, c = 0`
 

`text(Time of flight:)`

`underset~j\ text(component of)\ underset~s = −d`

`−1/2 g t^2` `= −d`
`t^2` `= (2d)/g`
`t` `= sqrt((2d)/g)`

 

ii.   `l = 2d\ \ (text(given))`

`text(Projectile hits water at)\ theta:`
 

`overset.y` `= underset~j\ text(component of)\ underset~v`
  `= −g t`
  `= −g · sqrt((2d)/g)`
  `= −sqrt(2dg)`
   
`overset.x` `= underset~i\ text(component of)\ underset~v`
  `= V`

 
`text(When)\ \ t = sqrt((2d)/g),`

`underset~i\ text(component of)\ underset~s = 2d`

`2d` `= V · sqrt((2d)/g)`
`V` `= (2dsqrtg)/sqrt(2d) = sqrt(2dg)`

 

`tantheta` `= (|overset.y|)/(|overset.x|)= sqrt(2dg)/sqrt(2dg)=1`
`:. theta` `= 45°`

 

iii.   `text(Speed = magnitude of velocity)`

`|underset~v|` `= sqrt((sqrt(2dg))^2 + (sqrt(2dg))^2)`
  `= sqrt(4dg)`
  `= 2sqrt(dg)`

Vectors, EXT1 V1 SM-Bank 7

A basketball player aims to throw a basketball through a ring, the centre of which is at a horizontal distance of 4.5 m from the point of release of the ball and 3 m above floor level. The ball is released at a height of 1.75 m above floor level, at an angle of projection `alpha` to the horizontal and at a speed of  `V\ text(ms)^(-1)`. Air resistance is assumed to be negligible.
 


 

The position vector of the centre of the ball at any time, `t` seconds, for  `t >= 0`, relative to the point of release is given by 
 
`qquad underset ~s(t) = Vt cos (alpha) underset ~i + (Vt sin(alpha) - 4.9t^2) underset ~j`,
 
where  `underset ~i`  is a unit vector in the horizontal direction of motion of the ball and  `underset ~j`  is a unit vector vertically up. Displacement components are measured in metres.

For the player’s first shot at goal, `V = 7\ text(ms)^(-1)` and  `alpha = 45^@`

  1. Find the time, in seconds, taken for the ball to reach its maximum height. Give your answer in the form  `(a sqrt b)/c`, where  `a, b` and `c` are positive integers.  (2 marks) 

  2. Find the maximum height, in metres, above floor level, reached by the centre of the ball.  (2 marks)

  3. Find the distance of the centre of the ball from the centre of the ring one second after release. Give your answer in metres, correct to two decimal places.  (2 marks)

Show Answers Only
  1. `(5 sqrt 2)/14`
  2. `3\ text(m)`
  3. `128\ text(m)`
Show Worked Solution
a.i.    `underset ~s(t)` `=7t\ cos 45^@ underset ~i+(7t\ sin 45^@ – 4.9t^2) underset ~j`
    `=(7sqrt2 t)/2 underset ~i + ((7sqrt2)/2 t – 4.9t^2) underset ~j`
     

 `text(Maximum height  ⇒  find)\ \t\ text(when)\ \ underset~j\ \ text(component of)\ \ underset ~V(t) = 0:`

`underset~V = (7sqrt2)/2 underset ~i + ((7sqrt2)/2 – 9.8t) underset ~j`

`(7sqrt2)/2 – 9.8t` `= 0`
`t` `= (5 sqrt 2)/14\ \ text(seconds)`

 

a.ii.  `text(Max height)\ =>\ text(Find)\ \ underset~j\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t=(5 sqrt 2)/14`

  `underset ~s_(underset ~j)((5 sqrt 2)/14)` `= 7 xx (5 sqrt 2)/14 xx 1/sqrt 2 – 4.9 xx ((5 sqrt 2)/14)^2`
    `= 1.25\ text(m)`

 
`:.\ text(Height above floor) = 1.25 + 1.75 = 3\ text(m)`

 

a.iii.  `underset ~s_text(ring)= 4.5 underset ~i + 1.25 underset ~j`

`text(After 1 second,)`

`underset~s_text(ball) = 7/sqrt 2 underset ~i + ((7sqrt2)/2 – 4.9)underset ~j`

  `:.d` `= |underset ~s_text(ring) – underset ~s(1)|`
    `= sqrt((4.5 – 7/sqrt 2)^2 + (1.25 – 7/sqrt 2 + 4.9)^2)`
    `~~ 1.28\ text(m)`

Vectors, EXT1 V1 SM-Bank 6

A cricketer hits a ball at time  `t = 0`  seconds from an origin `O` at ground level across a level playing field.

The position vector  `underset ~s(t)`, from `O`, of the ball after `t` seconds is given by
 
  `qquad underset ~s(t) = 15t underset ~i + (15 sqrt 3 t - 4.9t^2)underset ~j`,
 
where,  `underset ~i`  is a unit vector in the forward direction, `underset ~j`  is a unit vector vertically up and displacement components are measured in metres.

  1. Find the initial velocity of the ball and the initial angle, in degrees, of its trajectory to the horizontal.  (2 marks)

  2. Find the maximum height reached by the ball, giving your answer in metres, correct to two decimal places.  (2 marks)

  3. Find the time of flight of the ball. Give your answer in seconds, correct to three decimal places.  (1 mark)

  4. Find the range of the ball in metres, correct to one decimal place.  (1 mark)

  5. A fielder, more than 40 m from `O`, catches the ball at a height of 2 m above the ground.

     

    How far horizontally from `O` is the fielder when the ball is caught? Give your answer in metres, correct to one decimal place.  (2 marks)

Show Answers Only
  1. `underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`

     

    `theta = pi/3 = 60^@`

  2. `34.44\ text(m)`
  3. `5.302\ text(s)`
  4. `79.5\ text(m)`
  5. `78.4\ text(m)`
Show Worked Solution

a.   `underset ~v (t) = underset ~ dot s (t) = 15 underset ~i + (15 sqrt 3 – 9.8t)underset ~j`

`text(Initial velocity occurs when)\ \ t=0:`

`:. underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`
 

`text(Let)\ \ theta = text(Initial trajectory,)`

`tan theta` `=(15sqrt3)/15`  
  `=sqrt3`  
`:. theta` `=pi/3\ \ text(or)\ \ 60^@`  

 

b.  `text(Max height)\ =>underset~j\ \ text(component of)\ \ underset ~v=0.`

`15 sqrt 3 – 9.8t` `=0`
`t` `=(15 sqrt 3)/9.8`
  `=2.651…`

 
`text(Find max height when)\ \ t = 2.651…`

`:.\ text(Max height)` `= 15 sqrt 3 xx 2.651 – 4.9 xx (2.651)^2`
  `~~ 34.44\ text(m)`

 

c.   `text(Ball travels in symmetrical parabolic path.)`

`:.\ text(Total time of flight)`

`= 2 xx (15 sqrt 3)/9.8`

`= (15 sqrt 3)/4.9`

`~~ 5.302\ text(s)`
 

d.  `text(Range)\ =>underset~i\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t= (15 sqrt 3)/4.9`

`:.\ text(Range)` `= 15 xx (15 sqrt 3)/4.9`
  `= (225 sqrt 3)/4.9`
  `~~ 79.5\ text(m)`


e.   `text(Find)\ t\ text(when height of ball = 2 m:)`

`15 sqrt 3 t – 4.9t^2` `=2`  
`4.9t^2 – 15 sqrt 3 t + 2` `=0`  

 

  `t=(15 sqrt 3 +- sqrt ((15 sqrt 3)^2 – 4 xx 4.9 xx2))/(2 xx 4.9)`  
     

`t ~~ 0.078131\ \ text(or)\ \ t ~~ 5.22406`

 
`text(When)\ \ t=0.0781,`

`x= 15 xx 0.0781 = 1.17\ text(m)\ \ text{(no solution →}\ x<40 text{)}`
 

 `text(When)\ \ t=5.2241,`

`x=15 xx 5.2241 = 78.4\ text(m)`
 

`:.\ text(Ball is caught 78.4 m horizontally from)\ O.`

Statistics, EXT1 S1 SM-Bank 5

A manufacturer makes torches that have a probability of 0.03 of being defective.

Let  `overset^p`  be the random variable that represents the sample proportion of torches for samples of size `n` drawn from production.

Find the smallest integer value of  `n`  such that the standard deviation of  `overset^p`  is less than  `1/50`.  (2 marks)

Show Answers Only

`73`

Show Worked Solution

`text(Let)\ \ X =\ text(number of defective torches)`

`X\ ~\ text(Bin)(n, p)\ ~\ text(Bin)(n, 0.03)`

`text(Var)(overset^p)` `= (p(1 – p))/n`
`σ(overset^p)` `= sqrt((0.03(0.97))/n)`
  `= sqrt(291/(10\ 000n))`

 

`sqrt(291/(10\ 000n))` `< 1/50`
`291/(10\ 000n)` `< 1/2500`
`10\ 000n` `> 2500 xx 291`
`n` `> (2500 xx 291)/(10\ 000)`
  `> 72.75`

 
`:.n_text(min) = 73`