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NETWORKS, FUR1 2017 VCAA 8 MC

The flow of oil through a series of pipelines, in litres per minute, is shown in the network below.
 

 
The weightings of three of the edges are labelled `x`.

Five cuts labelled A–E are shown on the network.

The maximum flow of oil from the source to the sink, in litres per minute, is given by the capacity of

  1. `text(Cut A if)\ x = 1`
  2. `text(Cut B if)\ x = 2`
  3. `text(Cut C if)\ x = 2`
  4. `text(Cut D if)\ x = 3`
  5. `text(Cut E if)\ x = 3`
Show Answers Only

`B`

Show Worked Solution

`=> B`

GEOMETRY, FUR1 2017 VCAA 8 MC

Three circles of radius 50 mm are placed so that they just touch each other.
The region enclosed by the circles is shaded in the diagram below.

The area of the shaded region, in square millimetres, is closest to

  1.   `403`
  2.   `436`
  3. `1309`
  4. `2844`
  5. `4330`
Show Answers Only

`A`

Show Worked Solution

`text(Connect the centres of each circle to form)`

`text(an equilateral triangle.)`

`text(Shaded area)` `=\ text(Area of triangle − Area of sectors)`
  `=1/2 ab sinC – 3 xx (60/360 xx pi r^2)`
  `= 1/2 xx 100^2 xx sin60° – 3 xx (60/360 xx pi xx 50^2)`
  `= 403.13…`

`=> A`

GEOMETRY, FUR1 2017 VCAA 7 MC

A triangle `ABC` has:

• one side, `bar(AB)`, of length 4 cm
• one side, `bar(BC)`, of length 7 cm
• one angle, `∠ACB`, of 26°.

Which one of the following angles, correct to the nearest degree, could not be another angle in triangle `ABC`?

  1.   `24°`
  2.   `50°`
  3. `104°`
  4. `130°`
  5. `144°`
Show Answers Only

`E`

Show Worked Solution

`text(Using the sine rule,)`

`(sin x)/7` `= (sin 26)/4`
`sin x` `= (7 xx sin 26)/4`
  `= 0.767…`
   
`:. x` `= 50° text{or 130°  (nearest degree)}`

`text(If)\ \ x = 50°,\ text(other angle = 104°)`

`text(If)\ \ x = 130°,\ text(other angle = 24°)`

`=> E`

CORE, FUR1 2017 VCAA 24 MC

Xavier borrowed $245 000 to pay for a house.

For the first 10 years of the loan, the interest rate was 4.35% per annum, compounding monthly.

Xavier made monthly repayments of $1800.

After 10 years, the interest rate changed.

If Xavier now makes monthly repayments of $2000, he could repay the loan in a further five years.

The new annual interest rate for Xavier’s loan is closest to

  1.  0.35%
  2.  4.1%
  3.  4.5%
  4.  4.8%
  5.  18.7%
Show Answers Only

`B`

Show Worked Solution

`text(Find principal left after 10 years:)`

`text(By TVM Solver,)`

`N` `= 10 xx 12 = 120`
`I(%)` `= 4.35`
`PV` `= 245\ 000`
`PMT` `= −1800`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> FV = −108\ 219.1611`
 

`text(Loan can be repaid in 5 years at $2000/month.)`

`text(Find interest rate by TVM Solver;)`

`N` `= 5 xx 12 = 60`
`I(%)` `= ?`
`PV` `= 108\ 219.16`
`PMT` `= −2000`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> I(%)` `= 4.1427…`
  `= 4.14text{%  (2 d.p.)}`

`=> B`

Statistics, NAP-I3-CA09

The graph shows the origin and type of all vehicles in a city.
 

 
Which statement is most accurate based on the graph?

 
 There are more utility vehicles than trucks and vans.
 
 Utility vehicles are the most common type of vehicles
 
 There are more Australian vehicles than European vehicles.
 
 There are more Asian vehicles than European vehicles.
Show Answers Only

`text(There are more Asian vehicles)`

`text(than European vehicles.)`

Show Worked Solution

`text(Consider the 4th option:)`

`text(Total Asian vehicles)`

`=5+7+7=19`

`text(Total European vehicles)`

`=4+3+3=10`

`:.\ text(There are more Asian vehicles than European vehicles.)`

Statistics, NAP-A3-NC09

This graph shows the UV index for the town of Coonamble during one day.
 

 
When was the UV index always in the high range?

 
 between 8 am and 6 pm
 
 between 11 am and 12 pm
 
 after 2 pm
 
 between 2 pm and 4 pm

​

Show Answers Only

`text(between 11 am and 12 pm)`

Show Worked Solution

`text(UV index is on high)`

`text(– between 11 am – 12 pm and)`

`text(– between 2 pm – 3 pm)`

`:.\ text(between 11 am and 12 pm)`

Probability, NAP-A3-NC08

Ryan has white and black marbles in his bag.

If he chooses one without looking he is likely, but not certain, to get a white marble.

Which is Ryan's bag?

 
 
 
 

​

Show Answers Only

Show Worked Solution

`text(The 1st option gives a)\ \ 3/5\ \ text(chance for a)`

`text{white marble (likely but not certain).}`

Algebra, NAP-A3-NC05

Flynn is making a pattern with soccer balls.
 

 
This table shows the number of balls he needs for each shape in the pattern.
 

  Pattern # 1 2 3 4 5
  Number of balls 1 4 9 16 ?

 
How many balls will Flynn need for Pattern #5?

`17` `23` `25` `30`
 
 
 
 
Show Answers Only

`25`

Show Worked Solution

`text(Solution 1)`

`text(By drawing the next shape, pattern #5 is)`

`:. 25\ text(balls needed for Pattern #5.)`

 

`text(Solution 2)`

`text(The table shows a pattern where the balls in each)`

`text(shape are the square of the pattern number:)`

`:.\ text(Balls in pattern #5) = 5^2 = 25`

Statistics, NAP-A3-CA04

This table summarises the time Tutty spent training her parrot over five days.
 

 
What was the average (mean) time for training the parrot each day?

`text(30 minutes)` `text(56 minutes)` `text(66 minutes)` `text(280 minutes)`
 
 
 
 
Show Answers Only

`text(56 minutes)`

Show Worked Solution
`text(Average)` `= (25 + 55 + 60 + 94 + 46)/5`
  `= 280/5`
  `= 56\ text(minutes)`

Algebra, NAP-A4-NC01

Flynn is making a pattern with soccer balls.
 

 
This table shows the number of balls he needs for each shape in the pattern.
 

  Pattern # 1 2 3 4 5
  Number of balls 1 4 9 16 ?

 
How many balls will Flynn need for Pattern #5?

`17` `23` `25` `30`
 
 
 
 
Show Answers Only

`25`

Show Worked Solution

`text(Pattern #5 is)`

`:. 25\ text(balls needed for Pattern #5.)`

Statistics, NAP-A4-CA04

This table summarises the time Tutty spent training her parrot over five days.
 

 
What was the average (mean) time for training the parrot each day?

`text(30 minutes)` `text(56 minutes)` `text(66 minutes)` `text(280 minutes)`
 
 
 
 
Show Answers Only

`text(56 minutes)`

Show Worked Solution
`text(Average)` `= (25 + 55 + 60 + 94 + 46)/5`
  `= 280/5`
  `= 56\ text(minutes)`

Statistics, NAP-A4-CA03

A dealership sells new and used cars.

The graph shows the price of 2 similar cars and their age in years
 

 
Which one of these statements is true?

 
 Car Q is older and less expensive than Car P.
 
 Car P is newer and less expensive than Car Q.
 
 Car P is older and more expensive than Car Q.
 
 Car Q is newer and more expensive than Car P.
Show Answers Only

`text(Car Q is older and less expensive than Car P.)`

Show Worked Solution

`text(Car)\ Q\ text(is further right on)\ xtext(-axis)`

`=> Q\ text(is older)`

`text(Car)\ Q\ text(is lower on)\ ytext(-axis)`

`=> Q\ text(is less expensive)`

 

`:. text(Car)\ Q\ text(is older and less expensive than Car)\ P.`

Number, NAP-A4-CA01

At 7 am the temperature in Merewether was 15.9°C.

At midday it was 12.8°C warmer.

At 7 pm it was 13.9°C cooler than at midday.

What was the temperature at 7 pm?

`14.8^@text(C)` `15^@text(C)` `17^@text(C)` `42.6^@text(C)`
 
 
 
 
Show Answers Only

`14.8^@text(C)`

Show Worked Solution

`text(Temperature at 6 pm)`

`= 15.9 + 12.8 – 13.9`

`= 14.8^@text(C)`

Number, NAP-B4-CA04

Shelly and Carly collect dolls.

The ratio of the number of dolls Shelly owns compared to Carly is  3 : 2.

Shelley owns 12 dolls.

How many dolls does Carly own?

`2` `6` `8` `18`
 
 
 
 
Show Answers Only

`8`

Show Worked Solution

`text(Ratio 3 : 2)`

`text(Let)\ \ x=\ text(Number of Carly’s dolls)`

`x/12` `= 2/3`
`x` `= (2 xx 12)/3`
  `= 8\ text(dolls)`

Quadratic, EXT1 2017 HSC 14b

Let  `P(2p, p^2)`  be a point on the parabola  `x^2 = 4y`.

The tangent to the parabola at `P` meets the parabola  `x^2 = −4ay`, `a > 0`, at `Q` and `R`. Let `M` be the midpoint of `QR`.

  1. Show that the `x` coordinates of `R` and `Q` satisfy
  2. `qquadx^2 + 4apx - 4ap^2 = 0`.  (2 marks)

  3. Show that the coordination of `M` are  `(−2ap, −p^2(2a + 1))`.  (2 marks)
  4. Find the value of `a` so that the point `M` always lies on the parabola  `x^2 = −4y`.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `1 + sqrt2, a > 0`
Show Worked Solution

(i)   `x^2 = 4y,\ \ =>y = (x^2)/4`

`(dy)/(dx) = x/2`

`text(At)\ P(2p, p^2),`

`(dy)/(dx) = p`

`text(Equation of tangent:)`

`y = px – p^2`

 

`R and Q\ text(at intersection)`

`y` `= px – p^2\ …\ (1)`
`y` `= −(x^2)/(4a)\ …\ (2)`

 

`text(Subtract)\ (1) – (2)`

`px – p^2 + (x^2)/(4a)` `= 0`
`x^2 + 4apx – 4ap^2` `= 0\ …\ text(as required)`

 

(ii)   `x^2 + 4apx – 4ap^2 = 0`

♦♦ Mean mark 30%.
`x=` `\ (−4ap ± sqrt((4ap)^2 – 4 · 1 · (−4ap^2)))/2`
`=`  `\ (−4ap ± sqrt(16ap^2(a + 1)))/2`
`=`  `\ −2ap ± 2psqrt(a(a + 1))`

 

`=> xtext(-coordinate of)\ M\ text(is)\ −2ap.`

 

`M\ text(lies on tangent)\ \ y = px – p^2`

`:.y` `= p(−2ap) – p^2`
  `= −2ap^2 – p^2`
  `= −p^2(2a + 1)`

`:. M\ text(has coordinates)\ \ (−2ap, −p^2(2a + 1))`

 

(iii)   `text(If)\ M\ text(always lies on)\ \ x^2 = −4y`

♦♦ Mean mark 23%.
`(−2ap)^2` `= −4(−p^2(2a + 1))`
`4a^2p^2` `= 4p^2(2a + 1)`
`a^2` `= 2a + 1`
`a^2 – 2a – 1` `= 0`
`:. a` `= (+2 ± sqrt(4 + 4 · 1· 1))/2`
  `= 1 ± sqrt2`
  `= 1 + sqrt2, \ a > 0`

Mechanics, EXT2* M1 2017 HSC 13c

A golfer hits a golf ball with initial speed `V\ text(ms)^(−1)` at an angle `theta` to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake.
 

     

Neglecting the effects of air resistance, the equations describing the motion of the ball are

`x = Vt costheta`

`y = Vt sintheta - 1/2 g t^2`,

where `t` is the time in seconds after the ball is hit and `g` is the acceleration due to gravity in `text(ms)^(−2)`. Do NOT prove these equations.

  1. Show that the horizontal range of the golf ball is
     
         `(V^2sin 2theta)/g` metres.  (2 marks)

  2. Show that if  `V^2 < 100 g`  then the horizontal range of the ball is less than 100 m.  (1 mark)

It is now given that  `V^2 = 200 g`  and that the horizontal range of the ball is 100 m or more.

  1. Show that  `pi/12 <= theta <= (5pi)/12`.  (2 marks)

  2. Find the greatest height the ball can achieve.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `text(See Worked Solution)`
  4. `25(2 – sqrt 3)\ text(metres)`
Show Worked Solution

i.   `text(Find)\ \ t\ \ text(when)\ \ y = 0:`

`1/2 g t^2` `= Vtsintheta`
`1/2 g t` `= Vsintheta`
`t` `= (2Vsintheta)/g`

 

`text(Horizontal range)\ (x)\ text(when)\ \ t = (2Vsintheta)/g :`

`x` `= V · (2Vsintheta)/g costheta`
  `= (V^2 2sintheta costheta)/g`
  `= (V^2sin2theta)/g\ … text(as required)`

 

ii.   `text(If)\ \ V^2 < 100 g`

♦ Mean mark 44%.
`x` `< (100 g sin2theta)/g`
`x` `< 100 sin2theta`

 

`text(S)text(ince)\ −1 <= 2theta <= 1,`

`x < 100\ text(metres)`

 

iii.   `V^2 = 200g,\ \ x >= 100`

`(200 g · sin2theta)/g` `>= 100`
`sin2theta` `>= 1/2`

`:. pi/6 <= 2theta <= (5pi)/6`

`:. pi/12 <= theta <= (5pi)/12\ …\ text(as required)`

 

iv.   `text(Max height occurs when)`

♦♦ Mean mark 35%.
`t` `= 1/2 xx text(time of flight)`
  `= (Vsintheta)/g`

 
`text(Find)\ \ y\ \ text(when)\ \ t = (Vsintheta)/g`

`y` `= V · (Vsintheta)/g · sintheta – 1/2 g ((Vsintheta)/g)^2`
  `= (V^2 sin^2theta)/g – 1/2 · (V^2 sin^2 theta)/g`
  `= (V^2 sin^2theta)/(2g)`

 

`text(Max height when)\ theta = (5pi)/12\ (text(steepest angle)), V^2 = 200 g\ (text(given))`

`y_text(max)` `= (200 g · sin^2 ((5pi)/12))/(2g)`
  `= 100 sin^2 ((5pi)/12)`
  `= 50(1 – cos((5pi)/6))`
  `= 50(1 + sqrt3/2)`
  `= 25(2 + sqrt 3)\ text(metres)`

Plane Geometry, 2UA 2017 HSC 16c

In the triangle `ABC`, the point `M` is the mid-point of `BC`. The point `D` lies on `AB` and

`BD = DA + AC`.

The line that passes through the point `C` and is parallel to `MD` meets `BA` produced at `E`.

Copy or trace this diagram into your writing booklet.

  1. Prove that `Delta ACE` is isosceles.  (3 marks)
  2. The point `F` is chosen on `BC` so that `AF` is parallel to `DM`.
    `qquad` 
    Show that `AF` bisects `/_ BAC`.  (2 marks)

 

 

 

 

 

 

 

 

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

`text(Prove)\ Delta ACE\ text(is isosceles)`

♦♦♦ Mean mark 21%.

`text(S) text(ince)\ DM text(||) EC,`

`∠BDM = ∠BEC\ \ text{(corresponding angles)}`

`∠DBM\ \ text(is common)`

`:. Delta BDM\ text(|||)\ Delta BEC\ \ text{(AAA)}`

`text(Using ratios of similar)\ Delta text(s):`

`(BM)/(BD)` `= (BC)/(BE)`
`(BM)/(BC)` `= (BD)/(BE) = 1/2\ \ text{(}M\ text(is midpoint of)\ BC text{)}`

 

`=> D\ text(is the midpoint of)\ BE.`

`=> BD = DE`

`AE` `= DE – DA`
`AC` `= BD – DA\ text{(given)}`
  `= DE – DA`
`:.  AE` `= AC`

`:. Delta ACE\ text(is isosceles)`

 

(ii)  

`text(Show)\ AF\ text(bisects)\ /_ BAC`

♦♦ Mean mark 31%.

`/_ BAF = /_ AEC\ \ \ text{(corresponding angles)}`

`text{From part (i)}`

`/_ AEC` `= /_ ECA\ \ \ text{(}Delta AEC\ text{is isosceles)}`
`/_ FAC` `= /_ ECA\ \ \ text{(alternate angles)}`
`:. /_ BAF` `= /_ FAC`

`:. AF\ text(bisects)\  /_ BAC\ \ \ text(… as required)`

Calculus, 2ADV C3 2017 HSC 16a

John’s home is at point `A` and his school is at point `B`. A straight river runs nearby.

The point on the river closest to `A` is point `C`, which is 5 km from `A`.

The point on the river closest to `B` is point `D`, which is 7 km from `B`.

The distance from `C` to `D` is 9 km.

To get some exercise, John cycles from home directly to point `E` on the river, `x` km from `C`, before cycling directly to school at `B`, as shown in the diagram.
 


  

The total distance John cycles from home to school is `L` km.

  1. Show that  `L = sqrt (x^2 + 25) + sqrt (49 + (9 - x)^2)`.   (1 mark)

  2. Show that if  `(dL)/(dx) = 0`, then  `sin alpha = sin beta`.   (3 marks)

  3. Find the value of `x` that makes  `sin alpha = sin beta`.   (2 marks)

  4. Explain why this value of `x` gives a minimum for `L`.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `45/12`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

`text(Using Pythagoras:)`

`L` `= AE + EB`
  `= sqrt (5^2 + x^2) + sqrt (7^2 + (9 – x)^2)`
  `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)\ text(… as required)`

 

ii.  `text(From diagram):`

♦ Mean mark 41%.

`sin alpha = x/sqrt(25 + x^2) and sin beta = (9 – x)/sqrt(49 + (9 – x)^2)`

`L` `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)`
`(dL)/(dx)` `= (2x)/sqrt(25 + x^2) – (2(9 – x))/sqrt(49 + (9 – x)^2)`

 

`text(If)\ \ (dL)/(dx) = 0,`

`=> (2x)/sqrt(25 + x^2)` `= (2(9 – x))/sqrt(49 + (9 – x)^2)`
`x/sqrt(25 + x^2)` `= (9 – x)/sqrt(49 + (9 – x)^2)`
`sin alpha` `= sin beta\ text(… as required)`

 

iii.  `text(If)\ sin alpha = sin beta,\ text(then)\ alpha = beta and`

♦♦ Mean mark 29%.

`Delta ACE\ text(|||)\ Delta BDE`

`text(Using corresponding sides of similar triangles:)`

`x/5` `= (9 – x)/7`
`7x` `= 45 – 5x`
`12x` `= 45`
`:. x` `= 45/12\ text(km)`
♦♦♦ Mean mark 9%.

 

iv.  

`text(If point)\ B\ text(is reflected across the)`

`text(river),\ AEB\ text(will be a straight line.)`

`text(If any other point is chosen,)\ AEB`

`text(would not be straight and the distance)`

`text(would be longer.)`

Measurement, 2UG 2017 HSC 30e

A solid is made up of a sphere sitting partially inside a cone.

The sphere, centre `O`, has a radius of 4 cm and sits 2 cm inside the cone. The solid has a total height of 15 cm. The solid and its cross-section are shown.
 


 

What is the volume of the cone, correct to the nearest cm³?  (3 marks)

Show Answers Only

`113\ text{cm³  (nearest cm³)}`

Show Worked Solution

`V = 1/3 xx text(base of cone × height)`

`text(Consider the base area of the cone,)`

♦♦ Mean mark 34%.

`text(Need to find)\ x:`

`x^2` `= 4^2 – 2^2 = 16 – 4 = 12`
`x` `= sqrt12\ text(cm)`

 

`:. V` `= 1/3 xx pi xx (sqrt12)^2 xx (15 – 6)`
  `= 1/3 xx pi xx 12 xx 9`
  `= 113.097…`
  `= 113\ text{cm³  (nearest cm³)}`

Statistics, STD2 S5 2017 HSC 29d

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 


 

  1. Find the median test mark.  (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.

     

    Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)

  3. A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context.  (1 mark)

Show Answers Only
  1. `6`
  2. `text(43%)`
  3. `text(The statement assumes the data is normally)`
    `text(distributed which is not the case here.)`
Show Worked Solution
♦ Mean mark 50%.
i.    `text(Median)` `= text(15th + 16th score)/2`
    `= (4 + 8)/2`
    `= 6`

 

ii.   `text(Lower limit) = 5.4 – 4.22 = 1.18`

♦♦ Mean mark 34%.

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{%  (nearest %)}`

 

iii.   `text(The statement assumes the data is normally)`

♦♦♦ Mean mark 13%.

`text(distributed which is not the case here.)`

Measurement, 2UG 2017 HSC 29a

A new 200-metre long dam is to be built.
The plan for the new dam shows evenly spaced cross-sectional areas.
 


 

  1. Using TWO applications of Simpson’s rule, show that the volume of the dam is approximately  44 333 m³.  (2 marks)
  2. It is known that the catchment area for this dam is 2 km².
    Calculate how much rainfall is needed, to the nearest mm, to fill the dam.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution
(i)    `text(Volume)` `~~ h/3 (y_0 + 4y_1 + y_2)\ \ …\ text(applied twice)`
    `~~ 50/3(0 + 4 xx 140 + 270) + 50/3(270 + 4 xx 300 + 360)`
    `~~ 50/3(830) + 50/3(1830)`
    `~~ 44\ 333\ text(m³)\ …\ \ text(as required)`

 

(ii)   `text(Convert 2km² to m²:)`

♦♦♦ Mean mark 11%.
`text(2 km²)` `= 2 xx 1000 xx 1000`
  `= 2\ 000\ 000\ text(m²)`
`:.\ text(Rainfall needed)` `= (44\ 333)/(2\ 000\ 000)`
  `= 0.0221…\ text(m)`
  `= 22.1…\ text(mm)`
  `= 22\ text{mm  (nearest mm)}`

Algebra, 2UG 2017 HSC 28a

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation  `F = (9C)/5 + 32`.

  1. Calculate the temperature in degrees Fahrenheit when it is  −20 degrees Celsius.  (1 mark)
  2. Solve the following equations simultaneously, using either the substitution method or the elimination method.  (2 marks)
    `qquadF = (9C)/5 + 32`

    `qquadF = C`

     

  3. The graphs of  `F = (9C)/5 + 32`  and  `F = C`  are shown below.
  4.  


  5. What does the result from part (ii) mean in the context of the graph?  (1 mark)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Show Answers Only
  1. `−4^@F`
  2. `C = −40, F = −40`
  3. `text(It means the two graphs intersect)`
    `text(at)\ (−40,−40).`

 

 

Show Worked Solution
(i)   `F` `= (9(−20))/5 + 32`
    `= −4^@F`

 

(ii)   `F` `= (9C)/5 + 32` `…\ (1)`
  `F` `= C` `…\ (2)`

 

♦♦ Mean mark 31%.
MARKER’S COMMENT: An area that requires attention.

`text(Substitute)\ \ F = C\ \ text{from (2) into (1)}`

`C` `= (9C)/5 + 32`
`(9C)/5 – C` `= −32`
`(4C)/5 – C` `= −32`
`C` `= −32 xx 5/4 = −40`

 

`text{From (2),}`

`F = −40`

`text{(i.e. when}\ C = −40, F = −40)`

♦♦♦ Mean mark 20%.

 

(iii)   `text(It means the two graphs intersect)`

`text{at (−40,−40).}`

Probability, STD2 S2 2017 HSC 24 MC

A deck of 52 playing cards contains 12 picture cards. Two cards from the deck are drawn at random and placed on a table.

What is the probability, correct to four decimal places, that exactly one picture card is on the table?

A.     `0.0498`

B.     `0.1810`

C.     `0.3550`

D.     `0.3620`

Show Answers Only

`text(D)`

Show Worked Solution

`P(text(exactly 1 picture card))`

`= P(text(picture)) xx P(text(no picture)) + P(text(no picture)) xx P(text(picture))`

`= 12/52 xx 40/51quad+quad40/52 xx 12/51`

♦♦♦ Mean mark 21%.

`= 2((12 xx 40)/(52 xx 51))`

`= 0.36199…`

`=>\ text(D)`

FS Comm, 2UG 2017 HSC 23 MC

How many bits are there in 2 terabytes?

A.     `2^40`

B.     `2^41`

C.     `2^43`

D.     `2^44`

Show Answers Only

`text(D)`

Show Worked Solution

`text(Bytes in 2 terabytes)`

♦♦ Mean mark 34%.

`= 2 xx 2^40\ text(bytes)`

`:.\ text(Bits)` `= 8 xx 2 xx 2^40`
  `= 16 xx 2^40`
  `= 2^4 xx 2^40`
  `= 2^44\ text(bits)`

`=>\ text(D)`

Number and Algebra, NAP-J2-19

A company has 3706 computers in a warehouse.

Another 423 computers are delivered to the warehouse.

Which of these could be used to calculate the correct number of computers in the warehouse?

 
 3000 + 4000 + 700 + 20 + 6 + 3
 
 3000 + 700 + 400 + 20 + 60 + 3
 
 3000 + 700 + 400 + 20 + 6 + 3
 
 3000 + 400 + 70 + 20 + 6 + 3
Show Answers Only

`3000 + 700 + 400 + 20 + 6 + 3`

Show Worked Solution

`3706 + 423`

`= 3000 + 700 + 6 + 400 + 20 + 3`

`=3000 + 700 + 400 + 20 + 6 + 3`

Statistics, NAP-J3-CA17

The bottles in Renee's fridge are pictured below.

Renee decides to make a graph where each bar represents one type of bottle in her fridge.
  

Renee makes an error when creating the graph.

What should Renee do to correct the error?

 
 Make each category bar a different colour.
 
 Change the title to 'Number of bottles in the fridge by volume'.
 
 Change the 'Number of bottles' label to 'Volume of bottles'.
 
 Remove the 'Juice' category since orange juice and apple juice are already shown.
Show Answers Only

`text(Remove the ‘Juice’ category since orange juice and)`

`text(apple juice are already shown.)`

Show Worked Solution

`text(Remove the ‘Juice’ category since orange juice and)`

`text(apple juice are already shown.)`

Number, NAP-J3-CA14

Madison uses the number sentence  15 × 12 = 180  to solve a problem.

Which of the following could be the problem?

 
 Madison buys 15 showbags. How much does each showbag cost?
 
 Madison spends $15 on 180 showbags. How much does she spend?
 
 Madison buys 15 showbags that cost $12 each. How many showbags does she buy?
 
 Madison buys 12 showbags that cost $15 each. How much does she spend?
Show Answers Only

`text(Madison buys 12 showbags that cost $15 each.)`

`text(How much does she spend?)`

Show Worked Solution

`text($15 per showbag × 12 showbags = $180)`

`:.\ text(Madison buys 12 showbags that cost $15 each.)`

`text(How much does she spend?)`

Number, NAP-J3-CA09

Dinesh is a teacher and buys pencils for his class.

He purchases the pencils in two different sized packets.

Dinesh buys 10 packet A's and 4 packet B's.

He then divides all the pencils equally among his 9 students.

How many pencils does each student receive?

`6` `8` `12` `54` `72`
 
 
 
 
 
Show Answers Only

`8`

Show Worked Solution
`text(Total pencils)` `=(10 xx 6)+(4 xx3)`
  `=72`

 

`:.\ text(Pencils per student)` `=72 -: 9`
  `=8`

Number, NAP-J3-CA06 SA

There are 48 Year 7 students at a high school.

Each student is asked if they own a bike or not and the results are recorded.

`3/4` of the students said they owned a bike.

How many Year 7 students at the school own a bike?
Show Answers Only

`36`

Show Worked Solution

`text(Students who own a bike)`

`= 3/4 xx 48`

`= 36`

Algebra, NAP-J3-CA04 SA

A school teacher allocates pieces of cardboard to class groups depending on the number of students in each group.

The table below is used.
  

   
Using the pattern in the table, how many pieces of cardboard should a group of 3 students receive?
Show Answers Only

`9`

Show Worked Solution

`text(The pattern shows that each student receives)`

`text(2 pieces of cardboard.)`

`:.\ text(A group of 3 will be given 9 pieces.)`

Algebra, MET2 2007 VCAA 21 MC

`{x: cos^2(x) + 2cos (x) = 0} =`

  1. `{x : cos (x) = 0}`
  2. `{x : cos(x) = -1/2}`
  3. `{x : cos(x) = 1/2}`
  4. `{x: cos (x) = 0} uu { x : cos (x) = -1/2}`
  5. `{x: cos (x) = 1/2} uu { x : cos (x) = -1/2}`
Show Answers Only

`A`

Show Worked Solution

`text(Factorise:)`

♦♦♦ Mean mark 27%.

`cos (x) (cos x + 2) = 0`
 

 `:. cos x = 0,\ \ text(or)`

`cos x = -2 -> text(no solution)`

`=>   A`

Graphs, MET2 2008 VCAA 18 MC

Let  `f: [0, pi/2] -> R,\ f(x) = sin(4x) + 1.` The graph of  `f`  is transformed by a reflection in the `x`-axis followed by a dilation of factor 4 from the `y`-axis.

The resulting graph is defined by

  1. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = -1 - 4 sin (4x)`
  2. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (16x)`
  3. `g: [0, pi/2] -> R\ \ \ \ \ \ g(x) = 1 - sin (x)`
  4. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = 1 - sin (4x)`
  5. `g: [0, 2 pi] -> R\ \ \ \ \ \ \ g(x) = -1 - sin (x)`
Show Answers Only

`E`

Show Worked Solution

`f(x) = sin(4x)+1`

♦♦ Mean mark 36%.

`text(Reflecting in the)\ x text(-axis,)`

`=> h(x) = – sin (4x) – 1`

 

`text(Dilation by a factor of 4 from the)\ \ y text(-axis,)`

`=> g(x) = -sin(x)-1`

`text(Dilation factor: adjust the domain from)\ \ [0, pi/2]`

`text(to)\ \ [0xx4, pi/2 xx 4]=[0,2pi].`

`=>   E`