Band 4 – Page 69

GRAPHS, FUR1 2015 VCAA 4 MC

A straight line is graphed below.

An equation for this straight line is

A. `2x + 5y = 20`

B. `10x + 4y = 20`

C. `2x - 5y = 20`

D. `5x + 2y = 20`

E. `4x - 10y = 20`

Show Answers Only

`A`

Show Worked Solution

`text(Solution 1)`

`x/10 + y/4`	`= 1`
`:. 2x + 5y`	`=20`

`=> A`

`text(Solution 2)`

`y text(-intercept)\ = 4`

`text(Gradient)\ =(0-4)/(10-0)=- 2/5`

`:. y`	`=- 2/5 x +4`
`5y`	`=-2x + 20`
`2x+5y`	`=20`

`=> A`

Financial Maths, 2ADV M1 SM-Bank 8 MC

There are 3000 tickets available for a concert.

On the first day of ticket sales, 200 tickets are sold.

On the second day, 250 tickets are sold.

On the third day, 300 tickets are sold.

This pattern of ticket sales continues until all 3000 tickets are sold.

How many days does it take for all of the tickets to be sold?

`5`
`6`
`8`
`34`

Show Answers Only

`C`

Show Worked Solution

`200+250+300+…`

`text(AP where)\ \ \ a`	`=200`
`d`	`=250-200=50`

`text(Find)\ \ n\ \ text(when)\ \ S_n=3000:`

`S_n`	`=n/2[2a+(n-1)d]`
`3000`	`=n/2[2×200+(n-1)50]`
	`=n/2(400+50n-50)`
	`=n/2(350+50n)`
`3000`	`=175n+25n^2`

`n^2+7n-120`	`=0`
`(n-8)(n+15)`	`=0`

`:. n=8,\ \ \ (n>0)`

`=> C`

Financial Maths, 2ADV M1 SM-Bank 7 MC

A dragster is travelling at a speed of 100 km/h.

It increases its speed by

50 km/h in the 1st second
30 km/h in the 2nd second
18 km/h in the 3rd second

and so on in this pattern.

Correct to the nearest whole number, the greatest speed, in km/h, that the dragster will reach is

`125`
`200`
`220`
`225`

Show Answers Only

`D`

Show Worked Solution

`text (Sequence of speed increases is)`

`text (50, 30, 18, …)`

`text (GP where)\ \ \ a`	`= 50, and`
`r`	`= t_2/t_1 = 30/50 = 0.6`

`text (S) text (ince)\ \ |\ r\ | < |,`

`S_oo`	`= a / (1-r)`
	`= 50/ (1 – 0.6)`
	`= 125`

`:.\ text (Max speed is 100 + 125 = 225 km/h)`

`rArr D`

Financial Maths, 2ADV M1 SM-Bank 4 MC

The first four terms of a geometric sequence are

`4, – 8, 16, – 32`

The sum of the first ten terms of this sequence is

`– 2048`
`– 1364`
`684`
`1367`

Show Answers Only

`B`

Show Worked Solution

`4, –8, 16, –32,\ …`

`text(GP where)\ \ \ a`	`= 4`
`r`	`=t_(2)/t_(1)= (–8)/4=–2`

`S_n`	`=(a(r^n – 1))/(r – 1)`
`:.S_10`	`=[4[(–2)^10 – 1]]/(–2 – 1)`
	`= – 1364`

`=> B`

Financial Maths, 2ADV M1 SM-Bank 2 MC

The first four terms of a geometric sequence are 6400, `t_2` , 8100, – 9112.5

The value of `t_2` is

`– 7250`
`– 7200`
`– 1700`
`7200`

Show Answers Only

`B`

Show Worked Solution

`text(GP is)\ \ 6400, t_2, 8100, –9112.5`

`r`	`=t_2/t_1 = t_3/t_2`
`:. t_2 / 6400`	`= (–9112.5) / 8100`
`t_2`	`= (–9112.5 × 6400) / 8100`
	`= – 7200`

`=> B`

Statistics, STD2 S5 SM-Bank 1 MC

The head circumference (in cm) of a population of infant boys is normally distributed with a mean of 49.5 cm and a standard deviation of 1.5 cm.

Four hundred of these boys are selected at random and each boy’s head circumference is measured.

The number of these boys with a head circumference of less than 48.0 cm is closest to

`3`
`10`
`64`
`272`

Show Answers Only

`C`

Show Worked Solution

`mu=49.5,\ \ sigma=1.5`

`z text{-score (49.5)`	`=(x-mu)/sigma`
	`=(48.0-49.5)/1.5`
	`=– 1`

`:.\ text(Number of boys with a head under 48.0 cm)`

`=16text(%) xx 400`

`=64`

`=> C`

GRAPHS, FUR1 2010 VCAA 6 MC

The graphs of the linear relations

`x - 2y = - 4\ \ \ and\ \ \ 3x - y = 3`

are shown below.

A point that satisfies both the inequalities

`x - 2y ≥ - 4\ \ \ and\ \ \ 3x - y ≥ 3` is

A. `(1, 2)`

B. `(1, 2.5)`

C. `(2, 4)`

D. `(3, 2)`

E. `(3, 4)`

Show Answers Only

`D`

Show Worked Solution

`x – 2y`	`≥ – 4`
`2y`	`<= x+4`
`y`	`<= 1/2 x +2`

`3x-y`	`>=3`
`y`	`<=3x-3`

`text(The inequalities represent the following region,)`

`text{The only point in the shaded area is (3, 2).}`

`=> D`

GRAPHS, FUR1 2010 VCAA 5 MC

The cost in dollars, `C`, of making `n` pottery mugs is given by the equation

`C = 150 + 6n`

A loss will result from selling

A. 60 mugs at $9.00 each.

B. 70 mugs at $8.50 each.

C. 80 mugs at $7.50 each.

D. 90 mugs at $8.00 each.

E. 100 mugs at $9.50 each.

Show Answers Only

`C`

Show Worked Solution

`C = 150 + 6n`

`R = n xx text(selling price)`

`text(Test the cost and revenue of each option,)`

`text(Consider C,)`

`n = 80, and text(selling price = $7.50)`

`C`	`= 150 + 6 (80)`
	`=630`

`R`	`= 80 xx 7.50`
	`= 600`

`:.\ text(C)text(ost exceeds revenue and a loss results.)`

`=> C`

GRAPHS, FUR1 2010 VCAA 4 MC

The manager of an office is ordering finger food for an office party.

Hot items cost $2.15 each and cold items cost $1.50 each.

Let `x` be the number of hot items ordered.
Let `y` be the number of cold items ordered.

The manager can spend no more than $5 for each of the 200 employees.

An inequality that can be used to represent this constraint is

A. `1.5x + 2.15y ≤ 5`

B. `1.5x + 2.15y ≤ 200`

C. `1.5x + 2.15y ≤ 1000`

D. `2.15x + 1.5y ≤ 200`

E. `2.15x + 1.5y ≤ 1000`

Show Answers Only

`E`

Show Worked Solution

`x = text(number of hot items)`

`y = text(number of cold items)`

`text(Total spend)`	`= 5 xx 200`
	`= 1000`

`:.\ text(Inequality is)`

`x (2.15) + y (1.50)`	`<= 1000`
`2.15x + 1.5y`	`<= 1000`

`=> E`

GRAPHS, FUR1 2010 VCAA 3 MC

An equation for the straight line that passes through the points `(10, 1)` and `(4, –2)` is

A. `x + 2y = 12`

B. `2x + y = 6`

C. `4x + y = 14`

D. `x - 4y = 14`

E. `x - 2y = 8`

Show Answers Only

`E`

Show Worked Solution

`text(Line passes through)\ (10, 1),\ (4, -2)`

`text(Gradient)`	`=(y_2-y_1)/(x_2-x_1)`
	`=(-2-1)/(4-10)`
	`=1/2`

`:.\ text(Equation has)\ \ m=1/2,\ text{and passes through (10, 1)}`

`(y – y_1)`	`= m(x – x_1)`
`y – 1`	`= 1/2(x – 10)`
`2 y – 2`	`= x – 10`
`x – 2y`	`= 8`

`=> E`

GRAPHS, FUR1 2010 VCAA 1-2 MC

The volume of water that is stored in a tank over a 24-hour period is shown in the graph below.

Part 1

What is the difference in the volume of water (in litres) in the tank between 8 am and 6 pm?

A. `50`

B. `100`

C. `120`

D. `200`

E. `400`

Part 2

The rate of increase in the volume of water in the tank (in litres/hour) between 8 am and 10 am is

A. `37.5`

B. `50`

C. `75`

D. `125`

E. `150`

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(From the graph)`

`text(Volume difference)`	`= 300 – 200`
	`= 100\ text(L)`

`=> B`

`text(Part 2)`

`text(Rate of increase)`	`= (450 – 300)/ (10 – 8)`
	`= 75\ text(L/hr)`

`=> C`

GRAPHS, FUR1 2011 VCAA 7-9 MC

Craig plays sport and computer games every Saturday.

Let `x` be the number of hours that he spends playing sport.
Let `y` be the number of hours that he spends playing computer games.

Craig has placed some constraints on the amount of time that he spends playing sport and computer games.

These constraints define the feasible region shown shaded in the graph below. The equations of the lines that define the boundaries of the feasible regions are also shown.

Part 1

One of the constraints that defines the feasible region is

A. `y ≤ 1`

B. `x ≤ 2`

C. `x + y ≥ 9`

D. `2x + y ≤ 6`

E. `4x - y ≤ 11`

Part 2

By spending Saturday playing sport and computer games, Craig believes he can improve his health.

Let `W` be the health rating Craig achieves by spending a day playing sport and computer games.

The value of `W` is determined by using the rule `W = 5x - 2y`.

For the feasible region shown in the graph above, the maximum value of `W` occurs at

A. point `A`

B. point `B`

C. point `C`

D. point `D`

E. point `E`

Part 3

By spending Saturday playing sport and computer games, Craig believes he can improve his mental alertness.

Let `M` be the mental alertness rating Craig achieves by spending a day playing sport and computer games.

For the feasible region shown in the graph above, the maximum value of `M` occurs at any point that lies on the line that joins points `A` and `B` is the feasible region.

The rule for `M` could be

A. `M = 2x - 5y`

B. `M = 5x - 2y`

C. `M = 5x - 5y`

D. `M = 5x + 2y`

E. `M = 5x + 5y`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ C`

`text(Part 3:)\ E`

Show Worked Solution

`text(Part 1)`

`text(The constraints on the feasible region are:)`

`y`	`>=1`
`x`	`>=2`
`x+y`	`<=9`
`2x+y`	`>=6`
`4x-y`	`<=11`

`=>E`

`text(Part 2)`

`text(Testing each of the boundary points in)`

`W = 5x – 2y`

`text(At)\ A(2,7),\ \ W`	`= 5 (2) – 2 (7)`
	`= – 4`
`text(At)\ B(4,5),\ \ W`	`= 5 (4) – 2 (5)`
	`=10`
`text(At)\ C(3,1),\ \ W`	`= 5 (3) – 2 (1)`
	`=13`
`text(At)\ D(2.5,1),\ \ W`	`= 5 (2.5) – 2 (1)`
	`= 10.5`
`text(At)\ E(2,2),\ \ W`	`= 5 (2) – 2 (2)`
	`=6`

`:. W_(max) = 13`

`=> C`

`text(Part 3)`

`text(Find gradient of line between)\ \ A\ (2, 7), and B\ (4, 5).`

`m_(AB)`	`= (y_2 – y_1) / (x_2 – x_1)`
	`= (7 – 5) / (2 – 4)`
	`= – 1`

`:.\ text(The rule for)\ M\ text(has a gradient)\ = – 1.`

`text(Testing each option,)`

`text(Consider)\ E,`

`M`	`= 5x + 5y`
`5y`	`= -5x + M`
`y`	`= -x + M/5`

`:.\ text(Gradient) = – 1`

`=> E`

GRAPHS, FUR1 2011 VCAA 5 MC

The cost, `$C`, of making `x` kilograms of chocolate fudge is given by `C = 60 + 5x`.

The revenue, `$R`, from selling `x` kilograms of chocolate fudge is given by `R = 15x`.

A particular quantity of chocolate fudge was made and sold. It resulted in a loss of $20.

The number of kilograms of chocolate fudge made and sold was

A. `2`

B. `4`

C. `8`

D. `12`

E. `16`

Show Answers Only

`B`

Show Worked Solution

`C = 60 + 5x`

`R = 15x`

`text(For loss of $20,)`

`C`	`= R + 20`
`60 + 5x`	`= 15x + 20`
`40`	`= 10x`
`x`	`= 4`

`:.\ text(4 kg are made and sold.)`

`=> B`

GRAPHS, FUR1 2011 VCAA 4 MC

The fare, `$F`, to travel a distance on `n` kilometres in a taxi is given by the rule

`F = a + bn`

To travel a distance of 20 kilometres, the taxi fare is $18.20

To travel a distance of 30 kilometres, the taxi fare is $25.70

The charge per kilometre, `b`, is

A. `$0.75`

B. `$0.88`

C. `$0.91`

D. `$1.33`

E. `$3.20`

Show Answers Only

`A`

Show Worked Solution

`18.20`	`= a + b (20)`
`18.20`	`= a + 20b\ \ …\ (1)`
`25.70`	`= a + b (30)`
`25.70`	`= a + 30b\ \ …\ (2)`

`(2) – (1)`

`7.50`	`= 10b`
`:. b`	`= $0.75`

`=> A`

GRAPHS, FUR1 2011 VCAA 3 MC

Two lines intersect at point `A` on a graph.

The equation of one of the lines is

`3x + 4y = 26`

The coordinates of point `A` could be

A. `(2, 5)`

B. `(3, 4)`

C. `(4, 3)`

D. `(5, 2)`

E. `(7, 22)`

Show Answers Only

`A`

Show Worked Solution

`3x + 4y = 26\ \ …\ (1)`

`text(Test the coordinates of each option in the equation.)`

`text(Consider A,)`

`text(Substitute)\ (2, 5 )\ text(into)\ (1)`

`text(LHS)`	`= 3(2) + 4(5)`
	`= 6 + 20`
	`= 26\ \ \ text{(satisfies)}`

`text(Using the same method,)\ \ B, C, D, E\ \ text(can be eliminated.)`

`=> A`

GRAPHS, FUR1 2011 VCAA 1-2 MC

The charges for posting letters that weigh 100 g or less are shown in the graph below.

Part 1

The charge for posting a 35 g letter is

A. `$0.40`

B. `$0.60`

C. `$0.90`

D. `$1.50`

E. `$2.00`

Part 2

Two letters are posted.

The total postage charge cannot be

A. `$0.80`

B. `$1.20`

C. `$1.40`

D. `$2.10`

E. `$3.00`

Show Answers Only

`text (Part 1:)\ B`

`text (Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(From the graph)`

`text(C) text(ost for 35g letter) = $0.60`

`=> B`

`text(Part 2)`

`text(Charges are $0.40, $0.60, $0.90, and $1.50)`

`text(Consider A,)`

`$0.40 xx 2 = $0.80\ \ \ text{(Not A)}`

`text(Similarly, B, D and E can be shown to be)`

`text(combinations of 2 other charges.)`

`text(Only C isn’t a possible combination of 2 charges.)`

`=> C`

GRAPHS, FUR1 2012 VCAA 8 MC

Daisey’s bread shop makes white and brown bread subject to the following constraints.

• No more than 240 loaves of bread can be made each day.
• At least five loaves of white bread will be made for every loaf of brown bread that is made.

Let `w` be the number of loaves of white bread that are made each day.

Let `b` be the number of loaves of brown bread that are made each day.

A pair of inequalities that could be written to represent these constraints is

A. `w + b ≤ 240\ \ \ \ \ \ text(and)\ \ \ \ \ \ w ≥ 5b`

B. `w + b ≤ 240\ \ \ \ \ \ text(and)\ \ \ \ \ \ w ≤ 5b`

C. `w + b < 240\ \ \ \ \ \ text(and)\ \ \ \ \ \ w > b/5`

D. `w + b < 240\ \ \ \ \ \ text(and)\ \ \ \ \ \ w < 5b`

E. `w + b ≤ 240\ \ \ \ \ \ text(and)\ \ \ \ \ \ w ≤ b/5`

Show Answers Only

`A`

Show Worked Solution

`w = text(number of white loaves.)`

`b = text(number of brown loaves.)`

`text(No more than 240 loaves/day)`

`:. w + b <= 240`

`text(At least 5 white bread for every brown bread.)`

`w >= 5b`

`=> A`

GRAPHS, FUR1 2012 VCAA 4-5 MC

Part 1

The graph above shows the volume of water, `V` litres, in a tank at time `t` minutes.

The equation of this line between `t = 50` and `t = 85` minutes is

A. `V = 1700 - 20t`

B. `V = 700 - 20t`

C. `V = 20t + 1700`

D. `V = 20t + 700`

E. `V = 35t - 700`

Part 2

During the 85 minutes that it took to empty the tank, the volume of water in the tank first decreased at the rate of 15 litres per minute and then did not change for a period of time.

The period of time, in minutes, for which the volume of water in the tank did not change is

A. `15`

B. `20`

C. `30`

D. `50`

E. `85`

Show Answers Only

`text (Part 1:)\ A`

`text (Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 49%.

`text(Find the equation of the line that passes through)`

`(50, 700) and (85, 0).`

`text(Gradient)`	`=(y_2 – y_1)/(x_2 – x_1)`
	`=(0-700)/(85-50)`
	`=-20`

`:.\ text(Equation has)\ \ m=-20,\ \ text(and passes through)\ (50,700)`

`y-y_1`	`= m(x – x_1)`
`y-700`	`= -20(x-50)`
`y-700`	`= -20x + 1000`
`y`	`= -20x + 1700, or`
`V`	`= -20t + 1700`

`=> A`

`text(Part 2)`

`text{Volume initially decreases by 300 L (from graph) at}`

`text{a rate of 15 litres per minute (given).}`

`:.\ text(Time taken for first decrease)`

`= 300 / 15 = 20\ text(minutes)`

`text(Time of second decrease)`

`= 85 – 50 = 35\ text(minutes)`

`:.\ text(Time of no volume change)`

`= 85 – 35 – 20`

`= 30\ text(minutes)`

`=> C`

GRAPHS, FUR1 2012 VCAA 1 MC

The line `y = 2x` is drawn on the graph above.

Which one of the following points satisfies the inequality: `y > 2x`

A. `(1, 1)`

B. `(2, 5)`

C. `(3, 1)`

D. `(4, 7)`

E. `(5, 10)`

Show Answers Only

`B`

Show Worked Solution

`y > 2x`

`text(From the graph, this indicates the area)`

`text(above the line)\ y = 2x.`

`text(From the multiple choices given, only)`

`(2, 5)\ \ text(is above the line.)`

`=> B`

GRAPHS, FUR1 2013 VCAA 6 MC

In one month, an energy company charges a $30 service fee plus a supply charge of two cents per megajoule (MJ) of energy used.

The graph that best models this situation is

Show Answers Only

`E`

Show Worked Solution

`text(Every month, a $30 service fee applies, so)`

`text(graph passes through)\ \ (0, 30).`

`:.\ text(Eliminate A.)`

`text(Consider the graph in)\ E,`

`text(Gradient)`	`=($130 – $30)/(5000MJ)`
	`= $0.02\ text(per)\ MJ`

`=> E`

GRAPHS, FUR1 2013 VCAA 5 MC

The step graph below shows the speeding fines that are given for exceeding the speed limit by different amounts.

A driver was fined for driving at a speed of 65 km/h in a zone with a speed limit of 40 km/h.

The fine given was

A. `$65`

B. `$150`

C. `$250`

D. `$400`

E. `$600`

Show Answers Only

`D`

Show Worked Solution

`text(km/h over speed limit)`	`= 65 – 40`
	`= 25`

`:.\ text(Fine) = $400`

`text{(Note that open circles on the graph are not}`

`text{inclusive while closed circles are inclusive.)}`

`=> D`

GRAPHS, FUR1 2013 VCAA 2 MC

A point that lies on the graph of `3x -2y = - 5` is

A. `(3, – 2)`

B. `(1, 1)`

C. `(1, – 1)`

D. `(2, – 3)`

E. `(– 1, 1)`

Show Answers Only

`E`

Show Worked Solution

`text(Test the co-ordinates of each option to see which)`

`text(satisfies the equation.)`

`text(Consider)\ E,`

`3(-1) – 2(1)= – 5\ \ \ text{(true)}`

`=>E`

GRAPHS, FUR1 2013 VCAA 1 MC

The equation of the line shown on the graph is

A. `y = x - 6`

B. `y = x + 6`

C. `y = 6 - x`

D. `y = - 6`

E. `y = 6`

Show Answers Only

`B`

Show Worked Solution

`text(From the graph, line passes through)\ \ (0,6) and (– 6,0).`

`text(Gradient) = (6 – 0) / {0 – (-6)} = 1`

`ytext(-intercept) = 6`

`:.\ text(Equation is)\ \ \ y = x + 6`

`=> B`

GEOMETRY, FUR1 2011 VCAA 7 MC

The structure of a roof frame is shown in the diagram below.

In this diagram, `AB = BC = CD = 2400\ text(mm)` and `/_ PAB = 16°.`

The length of `QD`, in mm, is closest to

A. `2741`

B. `2767`

C. `2830`

D. `3394`

E. `5201`

Show Answers Only

`B`

Show Worked Solution

`text(In)\ Delta AQC,`

`tan 16°`	`= (QC)/(AC)`
`:. QC`	`= 4800 xx tan 16°`
	`= 1376.37…\ \ text(mm)`

`text(Using Pythagoras in)\ Delta QCD,`

`QD^2`	`= (1376.37…)^2 + 2400^2`
	`= 7\ 654\ 415.99`
`QD`	`= 2766.66…`

`=> B`

GEOMETRY, FUR1 2012 VCAA 8 MC

A triangular course for a yacht race has three stages.

Stage 1 is from the Start to Marker 1; a distance of 3.5 km on a bearing of 055°.

Stage 2 is from Marker 1 to Marker 2; a distance of 4.6 km on a bearing of 145°.

Stage 3 is from Marker 2 back to the Start.

The distance travelled on Stage 3, in km, is closest to

A. `4.9`

B. `5.3`

C. `5.8`

D. `6.0`

E. `7.7`

Show Answers Only

`C`

Show Worked Solution

`text(Let point C be due south of A)`

`/_ SAC`	`= 55°\ text{(Alternate angle)}`
`/_ BAC`	`= 180 – 145\ text{(Straight angle at A)}`
	`= 35°`

`:. /_ SAB = 55 + 35 = 90°`

`text (Using Pythagoras,)`

`(BS)^2`	`= (SA)^2 + (AB)^2`
	`= 3.5^2 + 4.6^2`
	`= 33.41`
`:. BS`	`= 5.78…\ text(km)`

`=> C`

GEOMETRY, FUR1 2012 VCAA 2 MC

`PQRS` is a square of side length 4 m as shown in the diagram below.

The distance `ST` is 1 m.

The shaded area `PQTS` shown in the diagram, in m², is closest to

A. `6`

B. `8`

C. `9`

D. `10`

E. `12`

Show Answers Only

`D`

Show Worked Solution

`text(Area of)\ Delta QRT`	`=1/2 xx RT xx QR`
	`=1/2 xx 3 xx 4`
	`=6\ text(m²)`

`:.\ text(Shaded Area) = 4 xx4 -6 = 10\ text(m²)`

`=> D`

GEOMETRY, FUR1 2012 VCAA 1 MC

The size of the angle `x°` is

A. `68°`

B. `88°`

C. `92°`

D. `112°`

E. `116°`

Show Answers Only

`C`

Show Worked Solution

`y`	`=24°\ \ \ text{(vertically opposite)}`
`x`	`=24+68\ \ \ text{(external angle of triangle)`
	`=92°`

`=> C`

GEOMETRY, FUR1 2015 VCAA 7 MC

A 50 cent coin has 12 sides of equal length.

Two 50 cent coins are balanced next to each other on a table so that they meet along one edge, as shown below.

The angle, `theta`, is

A. `12^@`

B. `30^@`

C. `36^@`

D. `60^@`

E. `72^@`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ theta=2x`

`x`	`= (360^@)/12\ \ text{(sum of external angles = 360°)}`
	`= 30^@`
`:. theta`	`= 60^@`

`=> D`

GEOMETRY, FUR1 2015 VCAA 6 MC

A cylindrical block of wood has a diameter of 12 cm and a height of 8 cm.

A hemisphere is removed from the top of the cylinder, 1 cm from the edge, as shown below.

The volume of the block of wood, in cubic centimetres, after the hemisphere has been removed is closest to

A. `452`

B. `606`

C. `643`

D. `1167`

E. `1357`

Show Answers Only

`C`

Show Worked Solution

`text{Volume (cylinder)}`	`=pi r^2 h`
	`=pi xx 6^2 xx 8`
	`=904.77…`
`text{Volume (hemisphere)}`	`=1/2 xx 4/3 xx pi xx r^3`
	`=1/2 xx 4/3 xx pi xx 5^3`
	`=261.79…`

`:.\ text(Remaining Volume)`	`= 904.77… -261.79…`
	`=642.97…\ text(cm³)`

`=> C`

GEOMETRY, FUR1 2015 VCAA 5 MC

A company logo is in the shape of a regular hexagon with side length 2 cm, as shown below.

The hexagon is divided into six equilateral triangles. Every second triangle is shaded.

The shaded area of the logo, in square centimetres, is closest to

A. `1.7`

B. `2.0`

C. `5.2`

D. `6.0`

E. `10.4`

Show Answers Only

`C`

Show Worked Solution

`text(Each triangle has 3 sides of 2 cm and 3)`

`text(internal angles of 60°.)`

`text(Shaded Area)`	`= 3 xx (1/2 ab sinC )`
	`= 3 xx (1/2 xx 2 xx 2 xx sin60^@)`
	`=5.196…\ text(cm²)`

`=> C`

GEOMETRY, FUR1 2015 VCAA 4 MC

Town A is due west of town B.

Town C is due south of town B.

The bearing of town A from town C is

A. between `000^@` and `090^@`

B. between `090^@` and `180^@`

C. exactly `135^@`

D. between `180^@` and `270^@`

E. between `270^@` and `360^@`

Show Answers Only

`E`

Show Worked Solution

`=> E`

GEOMETRY, FUR1 2015 VCAA 2 MC

A one-on-one basketball court is a composite shape made up of a rectangle and a semicircle, as shown below.

A boundary line is painted around the perimeter of the shape.

The total length of the boundary line, in metres, is closest to

A. `38.8`

B. `57.7`

C. `66.8`

D. `76.5`

E. `85.7`

Show Answers Only

`A`

Show Worked Solution

`text(Perimeter)`	`= (1/2 xx pi xx 12) + (4 + 12 + 4)`
	`=18.849… + 20`
	`= 38.849…\ text(m²)`

`=> A`

CORE*, FUR1 2015 VCAA 7 MC

The following graph shows the depreciating value of a van.

The graph could represent the van being depreciated using

flat rate depreciation with an initial value of $35 000 and a depreciation rate of $25 per year.
flat rate depreciation with an initial value of $35 000 and a depreciation rate of 25 cents per year.
reducing balance depreciation with an initial value of $35 000 and a depreciation rate of 2.5% per annum.
unit cost depreciation with an initial value of $35 000 and a depreciation rate of 25 cents per kilometre travelled.
unit cost depreciation with an initial value of $35 000 and a depreciation rate of $25 per kilometre travelled.

Show Answers Only

`D`

Show Worked Solution

`text{Initial value = $35 000 (from graph)}`

`text(Depreciation looks constant, and using the van)`

`text(valuation of $25 000 at 40 000 kms travelled.)`

`:.\ text(Depreciation Rate)`	`= (35\ 000 – 25\ 000)/(40\ 000 – 0)`
	`= $0.25\ text(per km)`

`=> D`

CORE*, FUR1 2015 VCAA 5 MC

The purchase price of a car is $20 000.

A deposit of $5000 is paid.

The balance will be repaid with 60 monthly repayments of $400.

The total amount of interest charged is

A. $1000

B. $4000

C. $9000

D. $19 000

E. $24 000

Show Answers Only

`C`

Show Worked Solution

`text(Amount borrowed)\ =20\ 000 – 5000 = $15\ 000`

`text(Total paid in 60 repayments)`

`=400 xx 60 = $24\ 000`

`:.\ text(Interest charged)`	`= 24\ 000 – 15\ 000`
	`= $9000`

`=> C`

CORE*, FUR1 2015 VCAA 4 MC

Mary invests $1200 for two years.

Interest is calculated at the rate of 3.35% per annum, compounding monthly.

The amount of interest she earns in two years is closest to

A. `$6.71`

B. `$40.82`

C. `$80.40`

D. `$81.75`

E. `$83.03`

Show Answers Only

`E`

Show Worked Solution

`text(Monthly interest rate)\ = 3.35/12 = 0.27916…`

`text(Total after 2 years)`	`= 1200(1 + (0.27916…)/100)^24`
	`=$1283.03…`

`:.\ text(Interest earned)`	`= 1283.03 – 1200`
	`= $83.03`

`=> E`

CORE*, FUR1 2015 VCAA 2 MC

An investment property was purchased for $600 000.

Over a 10-year period, its value increased to $850 000.

The increase in value, as a percentage of the purchase price, is closest to

A. `text(4.2%)`

B. `text(25.0%)`

C. `text(29.4%)`

D. `text(41.7%)`

E. `text(70.6%)`

Show Answers Only

`D`

Show Worked Solution

`text{Increase (%)}`	`=(850\ 000 – 600\ 000)/(600\ 000)`
	`=0.4166…`
	`=41.66…text(%)`

`=> D`

CORE*, FUR1 2015 VCAA 6 MC

Miki is competing as a runner in a half-marathon.

After 30 minutes, his progress in the race is modelled by the difference equation

`K_(n + 1) = 0.99K_n + 250,\ \ \ \ \ \ K_30 = 7550`

where `n ≥ 30` and `K_n` is the total distance Miki has run, in metres, after `n` minutes.

Using this difference equation, the total distance, in metres, that Miki is expected to have run 32 minutes after the start of the race is closest to

A. 7650

B. 7725

C. 7800

D. 7900

E. 8050

Show Answers Only

`D`

Show Worked Solution

`K_(n + 1) = 0.99K_n + 250`

`text(S)text(ince)\ \ K_30 = 7550,`

`K_31`	`= 0.99 xx 7550 + 250`
	`= 7724.5`
`K_32`	`= 0.99 xx 7724.5 + 250`
	`= 7897.255`

`=> D`

Calculus, SPEC2 2014 VCAA 13 MC

Using the substitution `u = sqrt(x + 1)` then `int_0^2(dx)/((x + 2) sqrt (x + 1))` can be expressed as

A. `int_1^sqrt 3 1/(sqrt u (u^2 + 1))\ du`

B. `int_0^2 2/(u^2 + 1)\ du`

C. `int_1^3 1/(sqrt u (u + 1))\ du`

D. `1/4 int_0^2 1/(u^2(u^2 + 1))\ du`

E. `2 int_1^sqrt 3 1/(u^2 + 1)\ du`

Show Answers Only

`E`

Show Worked Solution

`u`	`=sqrt(x + 1)`
`(du)/(dx)`	`= 1/(2 sqrt (x + 1))\ \ =>\ \ dx=2sqrt(x + 1)\ du`

`u^2`	`= x + 1`
`u^2 + 1`	`= x + 2`

`u(0)`	`= sqrt (0 + 1) = 1`
`u(2)`	`= sqrt (2 + 1) = sqrt 3`

`:. int_0^2 (dx)/((x + 2) sqrt(x + 1))`

`=int_1^sqrt3 (2sqrt(x + 1))/((u^2 + 1) xx sqrt(x + 1))\ du`

`= int_1^sqrt 3 2/(u^2 + 1) du`

`=> E`

Calculus, SPEC2 2014 VCAA 12 MC

If `(dy)/(dx) = sqrt((2x^6 + 1))` and `y = 5` when `x = 1`, then the value of `y` when `x = 4` is given by

`int_1^4(sqrt((2x^6 + 1)) + 5)\ dx`
`int_1^4sqrt((2x^6 + 1))\ dx`
`int_1^4 sqrt((2x^6 + 1))\ dx + 5`
`int_1^4sqrt((2x^6 + 1))\ dx - 5`
`int_1^4(sqrt((2x^6 + 1)) - 5)\ dx`

Show Answers Only

`C`

Show Worked Solution

`F(b)-F(a) `	`= int_a^b sqrt((2x^6 + 1))\ dx`
`F(b)`	`= int_a^b sqrt((2x^6 + 1))\ dx + F(a)`
`:. F(4)`	`= int_1^4 sqrt((2x^6 + 1))\ dx + F(1)`
	`= int_1^4 sqrt((2x^6 + 1))\ dx + 5`

`=> C`

CORE*, FUR1 2008 VCAA 8 MC

A loan of $300 000 is taken out to finance a new business venture.

The loan is to be repaid fully over twenty years with quarterly payments of $6727.80.

Interest is calculated quarterly on the reducing balance.

The annual interest rate for this loan is closest to

A. 4.1%

B. 6.5%

C. 7.3%

D. 19.5%

E. 26.7%

Show Answers Only

`B`

Show Worked Solution

`text(Using a TVM Solver,)`

`N`	`=4 xx 20=80`
`I (%)`	`= ?`
`PV`	`= – 300\ 000`
`PMT`	`= 6727.80`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.I = 6.50text(%)`

`=> B`

CORE*, FUR1 2009 VCAA 7 MC

A loan of $17 500 is to be paid back over four years at an interest rate of 6.25% per annum on a reducing monthly balance.

The monthly repayment, correct to the nearest cent, will be

A. $364.58

B. $413.00

C. $802.08

D. $1156.77

E. $5079.29

Show Answers Only

`B`

Show Worked Solution

`text(By TVM solver:)`

`N`	`= 4 × 12 = 48`
`I(%)`	`= 6.25text(%)`
`PV`	`= – 17\ 500`
`PMT`	`= ?`
`FV`	`=0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. PMT = $412.99…`

`=> B`

CORE*, FUR1 2010 VCAA 7 MC

A loan of $300 000 is to be repaid over a period of 20 years. Interest is charged at the rate of 7.25% per annum compounding quarterly.

The quarterly repayment to the nearest cent is

A. $2371.13

B. $5511.46

C. $7113.39

D. $7132.42

E. $7156.45

Show Answers Only

`D`

Show Worked Solution

`text(By TVM Solver:)`

`N`	`= 4 xx 20=80`
`I (%)`	`= 7.25text(%)`
`PV`	`= – 300\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 4`

`:.PMT = $ 7132.41…`

`=> D`

CORE*, FUR1 2010 VCAA 5 MC

A file server costs $30 000.

The file server depreciates by 20% of its value each year.

After three years its value is

A. `$6000`

B. `$12\ 000`

C. `$15\ 360`

D. `$19\ 200`

E. `$24\ 000`

Show Answers Only

`C`

Show Worked Solution

`P = 30\ 000,\ \ \ r = 0.2,\ \ \ n = 3`

`A`	`= P (1-r)^n`
`A`	`= 30\ 000 (1-0.2)^3`
	`= $15\ 360`

`=> C`

CORE*, FUR1 2012 VCAA 5 MC

A second-hand car is purchased for $9000.

A deposit of $2500 is paid.

Interest is calculated at the rate of 14.95% per annum on the reducing monthly balance.

The balance and interest will be repaid over two years with equal monthly payments.

The monthly payment is closest to

A. $315

B. $415

C. $436

D. $575

E. $587

Show Answers Only

`A`

Show Worked Solution

`text(Balance owing after deposit)`

`= $ 9000 – 2500`

`= $ 6500`

`text(By TVM Solver:)`

`N`	`= 2 xx 12=24`
`I (%)`	`= 14.95`
`PV`	`= – 6500`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:.PMT = $315.01`

`=> A`

CORE, FUR1 2010 VCAA 13 MC

A garden supplies outlet sells water tanks. The monthly seasonal indices for the revenue from the sale of water tanks are given below.

The seasonal index for September is missing.

The revenue from the sale of water tanks in September 2009 was $104 500.

The deseasonalised revenue for September 2009 is closest to

A. `$42\ 800`

B. `$74\ 100`

C. `$104\ 500`

D. `$141\ 000`

E. `$147\ 300`

Show Answers Only

`B`

Show Worked Solution

`text(Seasonal index for September)`	`=12 -\ text(sum of others)`
	`=12-10.59`
	`=1.41`

`:.\ text(Deseasonalised Revenue for September)`

`= (text(actual))/(text(index))`

`= (104\ 500)/(1.41)`

`= $74\ 113.48`

`=> B`

CORE, FUR1 2010 VCAA 10 MC

For a set of bivariate data that involves the variables `x` and `y`, with `y` as the response variable

`r = – 0.644, \ \ barx = 5.30, \ \ bary = 5.60, \ \ s_x = 3.06, \ \ s_y = 3.20`

The equation of the least squares regression line is closest to

A. `y = 9.2 - 0.7x`

B. `y = 9.2 + 0.7x`

C. `y = 2.0 - 0.6x`

D. `y = 2.0 - 0.7x`

E. `y = 2.0 + 0.7x`

Show Answers Only

`A`

Show Worked Solution

`y = a + bx\ \ text(where)`

`b`	`= r(s_y)/(s_x)`
	`=– 0.644 xx 3.20/3.06`
	`= – 0.6734…`

`a`	`= bary – b barx`
	`=5.30 – (– 0.6734… xx 5.60)`
	`= 9.07…`

`:. y ~~ 9.2 – 0.7x`

`=> A`

PATTERNS, FUR1 2015 VCAA 4 MC

The amount added to a new savings account each month follows a geometric sequence.

In the first month, $64 was added to the account.

In the second month, $80 was added to the account.

In the third month, $100 was added to the account.

Assuming this sequence continues, the total amount that will have been added to this savings account after five months is closest to

A. `$155`

B. `$195`

C. `$370`

D. `$400`

E. `$525`

Show Answers Only

`E`

Show Worked Solution

`text(GP where)\ \ \ a`	`= 64, and`
`r`	`= t_2/t_1 = 80/64=1.25`

`S_n`	`= (a(r^n – 1))/(r – 1)`
`:. S_5`	`= (64(1.25^5 – 1))/(0.25)`
	`= 525.25`

`=> E`

CORE*, FUR1 2015 VCAA 3 MC

A town has a population of 200 people when a company opens a large mine.

Due to the opening of the mine, the town’s population is expected to increase by 50% each year.

Let `P_n` be the population of the town `n` years after the mine opened.

The expected growth in the town’s population can be modelled by

A. `P_(n + 1) = P_n + 100`	`\ \ \ \ \ P_0 = 200`
B. `P_(n + 1) = P_n + 100`	`\ \ \ \ \ P_1= 300`
C. `P_(n + 1) = 0.5P_n`	`\ \ \ \ \ P_0 = 200`
D. `P_(n + 1) = 1.5P_n`	`\ \ \ \ \ P_0 = 300`
E. `P_(n + 1) = 1.5P_n`	`\ \ \ \ \ P_1 = 300`

Show Answers Only

`E`

Show Worked Solution

`text(After 1 year,)`

`P_1`	`= 1.5 xx P_0`
	`= 1.5 xx 200`
	`=300`

`=> E`

CORE, FUR1 2015 VCAA 12 MC

The time series plot below charts the number of calls per year to a computer help centre over a 10-year period.

Using five-median smoothing, the smoothed number of calls in year 6 was closest to

A. `3500`

B. `3700`

C. `3800`

D. `4000`

E. `4200`

Show Answers Only

`D`

Show Worked Solution

`text(The median of the 5 data points from year 4 – year 8)`

`text(is 4000.)`

`=> D`

CORE, FUR1 2015 VCAA 10 MC

For a set of bivariate data that involves the variables `x` and `y`:

`r = –0.47`, `barx = 1.8`, `s_x = 1.2`, `bary = 7.2`, `s_y = 0.85`

Given the information above, the least squares regression line predicting `y` from `x` is closest to

A. `y = 8.4 - 0.66x`

B. `y = 8.4 + 0.66x`

C. `y = 7.8 - 0.33x`

D. `y = 7.8 + 0.33x`

E. `y = 1.8 + 5.4x`

Show Answers Only

`C`

Show Worked Solution

`text(Regression line has the form,)`

`y = a + bx\ \ \ text(where)`

`b`	`=r s_y/s_x`
	`= – 0.47 xx (0.85)/1.2`
	`= -0.332…`

`a`	`=bar y – b bar x`
	`= 7.2 – (-0.332… xx 1.8)`
	`= 7.799…`

`=> C`

CORE, FUR1 2015 VCAA 9 MC

A least squares regression line has been fitted to the scatterplot above to enable distance, in kilometres, to be predicted from time, in minutes.

The equation of this line is closest to

A. distance `= 3.5 + 1.6 ×`time

B. time `= 3.5 + 1.6 ×`distance

C. distance `= 1.6 + 3.5 ×`time

D. time `= 1.8 + 3.5 ×`distance

E. distance `= 3.5 + 1.8 ×`time

Show Answers Only

`A`

Show Worked Solution

`text{Line passes through (0, 3.5) and (50, 82)`

`:.\ text(Gradient) `	`=(y_2-y_1)/(x_2-x_1)`
	`~~(82 – 3.5)/(50 – 0)`
	`~~ 1.57`

`text(Distance is the dependent variable and the)`

`y text(-intercept is approximately 3.5.)`

`=> A`

CORE, FUR1 2015 VCAA 6-7 MC

The following information relates to Parts 1 and 2.

In New Zealand, rivers flow into either the Pacific Ocean (the Pacific rivers) or the Tasman Sea (the Tasman rivers).

The boxplots below can be used to compare the distribution of the lengths of the Pacific rivers and the Tasman rivers.

Part 1

The five-number summary for the lengths of the Tasman rivers is closest to

`32, 48, 64, 76, 108`
`32, 48, 64, 76, 180`
`32, 48, 64, 76, 322`
`48, 64, 97, 169, 180`
`48, 64, 97, 169, 322`

Part 2

Which one of the following statements is not true?

The lengths of two of the Tasman rivers are outliers.
The median length of the Pacific rivers is greater than the length of more than 75% of the Tasman rivers.
The Pacific rivers are more variable in length than the Tasman rivers.
More than half of the Pacific rivers are less than 100 km in length.
More than half of the Tasman rivers are greater than 60 km in length.

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 46%.

`text(Outliers are inputs into a five-number summary,)`

`text(including the maximum and minimum values.)`

`:.\ text(A maximum length of 180 km is part of the Tasman)`

`text(river summary.)`

`=> B`

`text(Part 2)`

`text(Consider)\ D,`

`D\ text(would be true if its median value was less than)`

`text(100 km, which is not the case.)`

`=> D`

CORE, FUR1 2015 VCAA 4-5 MC

The foot lengths of a sample of 2400 women were approximately normally distributed with a mean of 23.8 cm and a standard deviation of 1.2 cm.

Part 1

The expected number of these women with foot lengths less than 21.4 cm is closest to

A. `60`

B. `120`

C. ` 384`

D. `2280`

E. `2340`

Part 2

The standardised foot length of one of these women is `z` = – 1.3

Her actual foot length, in centimetres, is closest to

A. `22.2`

B. `22.7`

C. `25.3`

D. `25.6`

E. `31.2`

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`z text{-score (21.4)}`	`=(x – barx)/s`
	`=(21.4-23.8)/1.2`
	`= – 2`

`:.\ text(Number of women with foot length under 21.4 cm)`

`=2.5text(%) xx 2400 = 60`

`=> A`

`text(Part 2)`

`text(If)\ \ z = – 1.3`

`– 1.3`	`=(x-23.8)/1.2`
`x-23.8`	`= – 1.3 xx 1.2`
`:. x`	`=23.8 – 1.56`
	`=22.24`

`=> A`

CORE, FUR1 2015 VCAA 3 MC

The dot plot below displays the difference between female and male life expectancy, in years, for a sample of 20 countries.

The mean (`barx`) and standard deviation (`s`) for this data are

A. `text(mean)\ = 2.32`	`\ \ \ \ \ text(standard deviation)\ = 5.25`
B. `text(mean)\ = 2.38`	`\ \ \ \ \ text(standard deviation)\ = 5.25`
C. `text(mean)\ = 5.0`	`\ \ \ \ \ text(standard deviation)\ = 2.0`
D. `text(mean)\ = 5.25`	`\ \ \ \ \ text(standard deviation)\ = 2.32`
E. `text(mean)\ = 5.25`	`\ \ \ \ \ text(standard deviation)\ = 2.38`

Show Answers Only

`E`

Show Worked Solution

`text(By calculator.)`

`=> E`

CORE, FUR1 2015 VCAA 1 MC

The stem plot below displays the average number of decayed teeth in 12-year-old children from `31` countries.

Based on this stem plot, the distribution of the average number of decayed teeth for these countries is best described as

negatively skewed with a median of 15 decayed teeth and a range of 45
positively skewed with a median of 15 decayed teeth and a range of 45
approximately symmetric with a median of 1.5 decayed teeth and a range of 4.5
negatively skewed with a median of 1.5 decayed teeth and a range of 4.5
positively skewed with a median of 1.5 decayed teeth and a range of 4.5

Show Answers Only

`E`

Show Worked Solution

`text(Median = 16th value)\ = 1.5`

`text(Range)\ = 4.7-0.2=4.5`

`text(The clear tail to the upper end of values shows that the)`

`text(data is positively skewed.)`

`=> E`

CORE*, FUR1 2006 VCAA 7 MC

The values of the first five terms of a sequence are plotted on the graph shown below.

The first order difference equation that could describe the sequence is

A. `t_(n+1) = t_n + 5,`	`\ \ \ \ \ t_1 = 4`
B. `t_(n+1) = 2t_n + 1,`	`\ \ \ \ \ t_1 = 4`
C. `t_(n+1) = t_n - 3,`	`\ \ \ \ \ t_1 = 4`
D. `t_(n+1) = t_n + 3,`	`\ \ \ \ \ t_1 = 4`
E. `t_(n+1) = 3t_n,`	`\ \ \ \ \ t_1 = 4`

Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(There is no common difference between terms,)`

`:.\ text(Cannot be A, C or D.)`

`text(The equation in B has)\ \ t_2=9,\ \ text(while the equation)`

`text(in C has)\ \ t_2=12.`

`rArr B`

CORE*, FUR1 2006 VCAA 5 MC

A difference equation is defined by

`f_(n+1) - f_n = 5\ \ \ \ \ text (where)\ \ f_1 =– 1`

The sequence `f_1, \ f_2, \ f_3, ...` is

A. `5, 4, 3\ …`

B. `4, 9, 14\ …`

C. `– 1, – 6, – 11\ …`

D. `– 1, 4, 9\ …`

E. `– 1, 6, 11\ …`

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ \ f_(n+1) = f_n +5`

`text(Consider)\ \ f_1 = – 1,`

`f_2= – 1 + 5=4`

`f_3 = 4+5=9`

`rArr D`

CORE, FUR1 2006 VCAA 11-13 MC

The following information relates to Parts 1, 2 and 3.

The table shows the seasonal indices for the monthly unemployment numbers for workers in a regional town.

Part 1

The seasonal index for October is missing from the table.

The value of the missing seasonal index for October is

A. `0.93`

B. `0.95`

C. `0.96`

D. `0.98`

E. `1.03`

Part 2

The actual number of unemployed in the regional town in September is 330.

The deseasonalised number of unemployed in September is closest to

A. `310`

B. `344`

C. `351`

D. `371`

E. `640`

Part 3

A trend line that can be used to forecast the deseasonalised number of unemployed workers in the regional town for the first nine months of the year is given by

deseasonalised number of unemployed = 373.3 – 3.38 × month number

where month 1 is January, month 2 is February, and so on.

The actual number of unemployed for June is predicted to be closest to

A. `304`

B. `353`

C. `376`

D. `393`

E. `410`

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ C`

`text (Part 3:)\ A`

Show Worked Solution

`text (Part 1)`

`text(Oct index)`	`=12-text(sum of other 11)`
	`=12-11.02`
	`=0.98`

`rArr D`

`text (Part 2)`

`text(Deseasonalised number)`	`= text(Actual)/text(Index)`
	`=330/0.94`
	`=351.06…`

`rArr C`

`text (Part 3)`

♦♦ Mean mark 29%.
MARKERS’ COMMENT: 59% of students correctly found the deseasonalised number but failed to convert it to the actual.

`text(Deseasonalised number)`

`=373.3 – 3.38 xx 6`

`=353.02`

`text(Actual number)`	`= text(Deseasonalised) xx text(Index)`
	`=353.02 xx 0.86`
	`=303.59…`

`rArr A`

CORE, FUR1 2006 VCAA 10 MC

The table below displays the total monthly rainfall (in mm) in a reservoir catchment area over a one-year period.

Using three mean moving average smoothing, the smoothed value for the total rainfall in April is closest to

A. `65`

B. `66`

C. `70`

D. `75`

E. `88`

Show Answers Only

`C`

Show Worked Solution

`text(Three mean average for April)`

`=(35+99+75)/3`

`=69.66…`

`rArr C`

CORE, FUR1 2006 VCAA 8 MC

The waist measurement (cm) and weight (kg) of 12 men are displayed in the table below.

Using this data, the equation of the least squares regression line that enables weight to be predicted from waist measurement is

`text(weight = – 20 + 1.11 × waist)`

When this equation is used to predict the weight of the man with a waist measurement of 80 cm, the residual value is closest to

A. `–11\ text(kg)`

B. `11\ text(kg)`

C. `–2\ text(kg)`

D. `2\ text(kg)`

E. `69\ text(kg)`

Show Answers Only

`C`

Show Worked Solution

`text(Residual)`	`=\ text(Actual – Predicted)`
	`=67-(- 20+1.11 xx 80)`
	`= – 1.8\ text(kg)`

`rArr C`