Band 3 – Page 40

Statistics, STD2 S1 2004 HSC 8 MC

This sector graph shows the distribution of 116 prizes won by three schools: X, Y and Z.

How many prizes were won by School X?

Show Answers Only

`B`

Show Worked Solution

`text(Centre angle of School X sector)`

`= 100^@\ text{(by measurement)}`

`:.\ text(Prizes won by school X)`

`= 100/360 xx 116`

`= 32.22\ …`

`=> B`

Statistics, STD2 S1 2004 HSC 6-7 MC

Use the set of scores 1, 3, 3, 3, 4, 5, 7, 7, 12 to answer Questions 6 and 7.

Question 6

What is the range of the set of scores?

Question 7

What are the median and the mode of the set of scores?

Median 3, mode 5
Median 3, mode 3
Median 4, mode 5
Median 4, mode 3

Show Answers Only

`text(Question 6:)\ C`

`text(Question 7:)\ D`

Show Worked Solution

`text(Question 6)`

`text(Range)`	`= text(High) – text(Low)`
	`= 12 – 1`
	`= 11`

`=> C`

`text(Question 7)`

`text(9 scores)`

`:.\ text(Median)`	`= (9 + 1) / 2`
	`=5 text(th score)`
	`= 4`

`text(Mode) = 3`

`=> D`

Measurement, STD2 M6 2004 HSC 5 MC

What is the correct expression for tan 20° in this triangle?

`a/b`
`a/c`
`c/b`
`c/a`

Show Answers Only

`D`

Show Worked Solution

`tan 20^@`	`= text(opposite)/text(adjacent)`
	`= c/a`

`=> D`

Algebra, STD2 A1 2004 HSC 3 MC

If `K = Ft^3`, `F = 5` and `t = 0.715`, what is the value of `K` correct to three significant figures?

`1.82`
`1.827`
`1.828`
`1.83`

Show Answers Only

`D`

Show Worked Solution

`K`	`= Ft^3`
	`= 5 xx (0.715)^3`
	`= 1.8276…`
	`= 1.83\ \ text{(3 sig figures)}`

`=> D`

Algebra, STD2 A2 2004 HSC 2 MC

Susan drew a graph of the height of a plant.

What is the gradient of the line?

`1`
`5`
`7.5`
`10`

Show Answers Only

`B`

Show Worked Solution

`text(2 points on graph)\ \ (0, 10),\ (1, 15)`

`text(Gradient)`	`= (y_2-y_1) / (x_2-x_1)`
	`= (15-10) / (1-0)`
	`= 5`

`=> B`

CORE*, FUR1 2007 VCAA 5 MC

A new kitchen in a restaurant cost $50 000. Its value is depreciated over time using the reducing balance method.

The value of the kitchen in dollars at the end of each year for ten years is shown in the graph below.

Which one of the following statements is true?

A. The kitchen depreciates by $4000 annually.

B. At the end of five years, the kitchen's value is less than $20 000.

C. The reducing balance depreciation rate is less than 5% per annum.

D. The annual depreciation rate increases over time.

E. The amount of depreciation each year decreases over time.

Show Answers Only

`E`

Show Worked Solution

`text(Consider) \ E,`

`text(The amount of depreciation each year is given by)`

`text(the decrease in the)\ y text(-value compared to the)`

`text(previous year. This clearly decreases over time, making)`

`E\ text(true.)`

`text(All other answers can be shown to be incorrect.)`

`=> E`

CORE*, FUR1 2012 VCAA 4 MC

Mei’s starting salary is $65 000 per annum.

After the first year her salary will increase by 2.8%.

After the second year her salary will increase by a further 3.5%.

After this second increase, her salary will be closest to

A. $66 820

B. $68 690

C. $69 030

D. $69 160

E. $69 630

Show Answers Only

`D`

Show Worked Solution

`text(Salary after 1st year)`

`= 65\ 000 + 2.8text(%) xx 65\ 000`

`= $66\ 820`

`text(Salary after 2nd year)`

`= $66\ 820 + 3.5text(%) xx 66\ 820`

`= $69\ 158.70`

`=> D`

GRAPHS, FUR1 2014 VCAA 6 MC

The Domestics Cleaning Company provides household cleaning services.

For two hours of cleaning, the cost is $55.

For four hours of cleaning, the cost is $94.

The rule for the cost of cleaning services is

`text(cost) = a + b xx text(hours)`

where `a` is a fixed charge, in dollars, and `b` is the charge per hour of cleaning, in dollars per hour.

Using this rule, the cost for five hours of cleaning is

A. `$19.50`

B. `$97.50`

C. `$99.50`

D. `$113.50`

E. `$121.50`

Show Answers Only

`D`

Show Worked Solution

`text(C)text(ost)`	`= a + b xx text(hours)`
`55`	`= a + 2b\ \ \ …\ (1)`
`94`	`= a + 4b\ \ \ …\ (2)`

`(2) – (1)`

`39`	`= 2b`
`b`	`= 19.5`

`text(Substitute)\ b = 19.5\ text(into)\ (1)`

`a + (2 xx 19.5)`	`= 55`
`a`	`= 16`

`:.\ text(C)text(ost of 5 Hours)`

`= 16 + (5 xx 19.5)`

`= text($) 113.50`

`=> D`

GRAPHS, FUR1 2014 VCAA 1 MC

The graph below shows the altitude, in metres, of a balloon over a six-hour flight.

Over the six-hour period, the length of time, in hours, where the altitude of the balloon was at least 1500 m is

A. `3`

B. `4`

C. `5`

D. `6`

E. `7`

Show Answers Only

`B`

Show Worked Solution

`text{The graph is higher than 1500 metres (on the}\ y text(-axis)`

`text{between the 1 hour mark and the 5 hour mark).}`

`:.\ text(It spends 4 hours at or above 1500 m.)`

`=> B`

CORE*, FUR1 2014 VCAA 1 MC

This month, a business charges $1500 to install a water tank.

Next month, the charge will increase by 3.5%.

The charge next month will be

A. `$45.00`

B. `$52.50`

C. `$1545.00`

D. `$1552.50`

E. `$1950.00`

Show Answers Only

`D`

Show Worked Solution

`text(Next month’s charge)`

`= 1500 + 3.5text(%) xx 1500`

`= 1500 + 52.5`

`= $1552.5`

`=> D`

Measurement, 2UG MM6 SM-Bank 02 MC

This is a sketch of a sector of a circle.

Find the value of `theta` to the nearest degree.

(A) `47°`

(B) `48°`

(C) `68°`

(D) `69°`

Show Answers Only

`D`

Show Worked Solution

`text(Arc Length)`	`= theta/360 xx 2 xx pi xx r`
`12`	`= theta/360 xx 2 xx pi xx 10`
`theta/360`	`= 12/(20 pi)`
`theta`	`= (12 xx 360)/(20 pi)`
	`= 68.75…°`

`=> D`

GRAPHS, FUR1 2007 VCAA 2 MC

A builder's fee, `C` dollars, can be determined from the rule `C = 60 + 55n`, where `n` represents the number of hours worked.

According to this rule, the builder's fee will be

A. $60 for 1 hour of work.

B. $110 for 2 hours of work.

C. $500 for 8 hours of work.

D. $550 for 10 hours of work.

E. $1150 for 10 hours of work.

Show Answers Only

`C`

Show Worked Solution

`C = 60 + 55n`

`text(Consider)\ A\ text(when)\ n = 1`

`C`	`= 60 + 55(1)`
	`= 115`

`:.\ text(Eliminate)\ A`

`text(Consider)\ B\ text(when)\ n = 2`

`C`	`= 60 + 55(2)`
	`= 170`

`:. text(Eliminate)\ B`

`text(Consider)\ C\ text(when)\ n = 8`

`C`	`= 60 + 55(8)`
	`= 500`

`:.\ text(True)`

`D\ text(and)\ E\ text(can be similarly ruled out.)`

`=> C`

Calculate the cost of a call of duration `6` minutes and `20` seconds, given that there is a connection fee of `35` cents and a call rate of `37` cents per `30` second block or part thereof. (2 marks)

Show Answers Only

`$5.16`

Show Worked Solution

`text(Connection fee) = 35\ text(cents)`

`text(30 second blocks in 6m 20s)=13`

`=> text(Call rate total)`	`= 13 xx 37`
	`= 481\ text(cents)`

`:.\ text(Total cost of call)`	`= 35 + 481`
	`= 516\ text(cents)`
	`= $5.16`

FS Comm, 2UG SM-Bank 04

Murray is a photographer and has recently purchased an external hard drive with a `500 text(GB)` capacity.

If the average size of his photographic files is `4.0 text(MB)`, how many should he expect to fit on the hard drive? (2 marks)

Show Answers Only

`text(128 000 files)`

Show Worked Solution

`text(Hard drive capacity) = 500 xx 2^30\ text(bytes)`

`text(Average file size) = 4.0 xx 2^20\ text(bytes)`

`:.\ text(# Files)`	`= (500 xx 2^30)/(4.0 xx 2^20)`
	`= 128\ 000`

`:.\ text(Murray should expect to fit 128 000 files.)`

Algebra, 2UG AM3 SM-Bank 04

Solve these simultaneous equations to find the values of `x` and `y`.

`y = 3x - 2`

`x + 2y + 18 = 0` (3 marks)

Show Answers Only

`x = -2 ,\ y = -8`

Show Worked Solution

`text(Solution 1 – Substitution)`

`y = 3x – 2\ \ \ \ \ …\ text{(i)}`

`x + 2y + 18 = 0\ \ \ \ \ …\ text{(ii)}`

`text(Substitute)\ \ y = 3x – 2\ \ text{into (ii)}`

`x + 2(3x – 2) + 18`	`= 0`
`x + 6x – 4 + 18`	`= 0`
`7x`	`= -14`
`x`	`= -2`

`text(Substitute)\ \ x = -2\ \ text{into (i)}`

`y`	`= 3(-2) – 2`
	`= -8`

`:.\ text(Solution is)\ x = -2,\ y = -8`

`text(Alternative Solution – Elimination)`

`y = 3x – 2\ \ \ \ \ …\ text{(i)}`

`x + 2y + 18 = 0\ \ \ \ \ …\ text{(ii)}`

`text{Multiply (i)} xx 2`

`2y`	`= 6x – 4`
`6x – 2y – 4`	`= 0\ …\ text{(iii)}`

`text{(ii) + (iii)}`

`7x + 14`	`= 0`
`x`	`= -2`
`y`	`= -8\ \ text{(see Solution 1)}`

Algebra, 2UG AM3 SM-Bank 03

Simplify `(a(b^2)^3)/(a^2b)` (2 marks)

Show Answers Only

`b^5/a`

Show Worked Solution

`(a(b^2)^3)/(a^2b)`	`= (ab^6)/(a^2b)`
	`= b^5/a`

Algebra, 2UG AM3 SM-Bank 02

Solve for `w` given

`w/3 - w/7 = -1` (2 marks)

Show Answers Only

`w = -5.25`

Show Worked Solution

`21 (w/3) – 21 (w/7)`	`= 21 xx -1`
`7w – 3w`	`= -21`
`4w`	`= -21`
`w`	`= -5.25`

Algebra, 2UG AM3 SM-Bank 01

Solve for `x` given

`x/6 + x/4 = 5` (2 marks)

Show Answers Only

`x = 12`

Show Worked Solution

`x/6 + x/4`	`= 5`
`24 (x/6) + 24 (x/4)`	`= 24 xx 5`
`4x + 6x`	`= 120`
`10x`	`= 120`
`x`	`= 12`

Algebra, STD2 A4 2007 HSC 27a

A rectangular playing surface is to be constructed so that the length is 6 metres more than the width.

Give an example of a length and width that would be possible for this playing surface. (1 mark)

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Write an equation for the area (`A`) of the playing surface in terms of its length (`l`). (1 mark)

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A graph comparing the area of the playing surface to its length is shown.
Why are lengths of 0 metres to 6 metres impossible? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What would be the dimensions of the playing surface if it had an area of 135 m²? (2 marks)

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Company `A` constructs playing surfaces.
Draw a graph to represent the cost of using Company `A` to construct all playing surface sizes up to and including 200 m².

Use the horizontal axis to represent the area and the vertical axis to represent the cost. (2 marks)

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Company `B` charges a rate of $360 per square metre regardless of size.
Which company would charge less to construct a playing surface with an area of 135 m²

Justify your answer with suitable calculations. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(One possibility is a length of 10 m, and a width)`

`text{of 4 m (among many possibilities).}`
`A = l (l – 6)\ \ text(m²)`
`text(Given the length must be 6 m more than)`

`text(the width, it follows that the length)`

`text(must be greater than 6 m.)`
`text(15 m × 9 m)`
`text(Proof)\ \ text{(See worked solutions)}`

Show Worked Solution

i. `text(One possibility is a length of 10 m, and a width)`

`text{of 4 m (among many possibilities).}`

ii. `text(Length) = l\ text(m)`

`text(Width) = (l – 6)\ text(m)`

`:.\ A`

`= l (l – 6)`

iii. `text(Given the length must be 6m more than the width,)`

`text(it follows that the length must be greater than 6 m)`

`text(so that the width is positive.)`

iv. `text(From the graph, an area of 135 m² corresponds to)`

`text(a length of 15 m.)`

`:.\ text(The dimensions would be 15 m × 9 m.)`

vi. `text(Company)\ A\ text(cost) = $50\ 000`

`text(Company)\ B\ text(cost)`	`= 135 xx 360`
	`= $48\ 600`

`:.\ text(Company)\ B\ text(would charge $1400 less)`

`text(than Company)\ A.`

GEOMETRY, FUR1 2011 VCAA 4 MC

In triangle `ACB`, `/_ CAB = 60^@` and `/_ ABC = 80^@`

The length of side `AB = 50\text(m.)`

The length of side `AC` is closest to

A. `57\ text(m)`

B. `67\ text(m)`

C. `77\ text(m)`

D. `81\ text(m)`

E. `100\ text(m)`

Show Answers Only

`C`

Show Worked Solution

`/_ ACB`	` = 180 – (60 + 80) \ \ text{(angle sum of}\ Delta ABC text{)}`
	` = 40 ^@`

`text(Using sine rule:)`

`(AC) / (sin 80^@)`	`= 50/(sin 40^@)`
`:. AC`	`= (50 xx sin 80^@) / (sin 40^@)`
	`= 76.60…\ \ text(m)`

`=> C`

GEOMETRY, FUR1 2013 VCAA 2 MC

The distances from a kiosk to points `A` and `B` on opposite sides of a pond are found to be 12.6 m and 19.2 m respectively.

The angle between the lines joining these points to the kiosk is 63°.

The distance, in m, across the pond between points `A` and `B` can be found by evaluating

A. `1/2 xx 12.6 xx 19.2 xx sin(63°)`

B. `{19.2 xx sin(63°)}/12.6`

C. `sqrt(12.6^2 + 19.2^2)`

D. `sqrt(12.6^2 + 19.2^2 - 2 xx 12.6 xx 19.2 xx cos(63°)`

E. `sqrt{s(s - 12.6)(s - 19.2)(s - 63)} , text(where)\ s = 1/2 (12.6 + 10.2 + 63)`

Show Answers Only

`D`

Show Worked Solution

`text(Using the cosine rule:)`

`(AB)^2`	`= 12.6^2 xx 19.2^2 – 2 xx 12.6 xx 19.2 xx cos(63°)`
`:. AB`	`= sqrt{12.6^2 xx 19.2^2 – 2 xx 12.6 xx 19.2 xx cos(63°)}`

`=>D`

GEOMETRY, FUR1 2013 VCAA 1 MC

The perimeter of a regular pentagon is 100 cm.

The side length of this pentagon, in cm, is

A. `5`

B. `10`

C. `20`

D. `25`

E. `50`

Show Answers Only

`C`

Show Worked Solution

`text(A pentagon has 5 sides)`

`∴ text(Side length) = 100/5 = 20\ \ text(cm)`

`=>C`

Probability, STD2 S2 2007 HSC 25c

In a stack of 10 DVDs, there are 5 rated PG, 3 rated G and 2 rated M.

A DVD is selected at random. What is the probability that it is rated M? (1 mark)

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Grant chooses two DVDs at random from the stack. Copy or trace the tree diagram into your writing booklet.

Complete the tree diagram by writing the correct probability on each branch. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---
Calculate the probability that Grant chooses two DVDs with the same rating. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`1/5`
`14/45`

Show Worked Solution

i. `text(5 PG, 3 G, 2 M)`

`P text{(M)} = 2/10 = 1/5`

ii.

iii. `P text{(same rating)}`

`= P text{(PG, PG)} + P text{(G, G)} + P text{(M, M)}`

`= (1/2 xx 4/9) + (3/10 xx 2/9) + (1/5 xx 1/9)`

`= 2/9 + 1/15 + 1/45`

`= 14/45`

Statistics, STD2 S1 2007 HSC 24d

Barry constructed a back-to-back stem-and-leaf plot to compare the ages of his students.

Write a brief statement that compares the distribution of the ages of males and females from this set of data. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What is the mode of this set of data? (1 mark)

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Liam decided to use a grouped frequency distribution table to calculate the mean age of the students at Barry’s Ballroom Dancing Studio.

For the age group 30 - 39 years, what is the value of the product of the class centre and the frequency? (2 marks)

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Liam correctly calculated the mean from the grouped frequency distribution table to be 39.5.

Caitlyn correctly used the original data in the back-to-back stem-and-leaf plot and calculated the mean to be 38.2.

What is the reason for the difference in the two answers? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(More males attend than females and a higher proportion)`
`text(of those are younger males, with the distribution being)`
`text(positively skewed. Female attendees are generally older)`
`text(and have a negatively skewed distribution.)`
`text(Mode) = 64\ \ \ text{(4 times)}`
`172.5`
`text(The difference in the answers is due to the class)`
`text(centres used in group frequency tables distorting)`
`text(the mean value from the exact data.)`

Show Worked Solution

i.	`text(More males attend than females and a higher proportion)`
	`text(of those are younger males, with the distribution being)`
	`text(positively skewed. Female attendees are generally older)`
	`text(and have a negatively skewed distribution.)`

ii.

`text(Mode) = 64\ \ \ text{(4 times)}`

iii.	`text(Class centre)`	`= (30 + 39)/2`
		`= 34.5`
	`text(Frequency) = 5`

`:.\ text(Class centre) xx text(frequency)`

`= 34.5 xx 5`

`= 172.5`

iv.	`text(The difference in the answers is due to the class)`
	`text(centres used in group frequency tables distorting)`
	`text(the mean value from the exact data.)`

Algebra, STD2 A1 2007 HSC 24b

The distance in kilometres (`D`) of an observer from the centre of a thunderstorm can be estimated by counting the number of seconds (`t`) between seeing the lightning and first hearing the thunder.

Use the formula `D = t/3` to estimate the number of seconds between seeing the lightning and hearing the thunder if the storm is 1.2 km away. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`3.6\ text(seconds)`

Show Worked Solution

`D = t/3`

`text(When)\ \ D = 1.2,`

`t/3`	`= 1.2`
`t`	`= 3.6\ text(seconds)`

Statistics, STD2 S1 2007 HSC 24a

Consider the following set of scores:

`3, \ 5, \ 5, \ 6, \ 8, \ 8, \ 9, \ 10, \ 10, \ 50.`

Calculate the mean of the set of scores. (1 mark)

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What is the effect on the mean and on the median of removing the outlier? (2 marks)

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Show Answers Only

`11.4`
`text{If the outlier (50) is removed, the mean}`

`text(would become lower.)`
`text(Median will NOT change.)`

Show Worked Solution

i. `text(Total of scores)`

`= 3 + 5 + 5 + 6 + 8 + 8 + 9 + 10 + 10 +50`

`= 114`

`:.\ text(Mean) = 114/10 = 11.4`

ii. `text(Mean)`

`text{If the outlier (50) is removed, the mean}`

`text(would become lower.)`

`text(Median)`

`text(The current median (10 data points))`

`= text(5th + 6th)/2 = (8 + 8)/2 = 8`

`text(The new median (9 data points))`

`=\ text(5th value)`

`= 8`

`:.\ text(Median will NOT change.)`

Measurement, STD2 M1 2007 HSC 23b

A cylindrical water tank, of height 2 m, is placed in the ground at a school.

The radius of the tank is 3.78 metres. The hole is 2 metres deep. When the tank is placed in the hole there is a gap of 1 metre all the way around the side of the tank.

When digging the hole for the water tank, what volume of soil was removed? Give your answer to the nearest cubic metre. (3 marks)

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Sprinklers are used to water the school oval at a rate of 7500 litres per hour.

The water tank holds 90 000 litres when full.

For how many hours can the sprinklers be used before a full tank is emptied? (1 mark)

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Water is to be collected in the tank from the roof of the school hall, which has an area of 400 m².

During a storm, 20 mm of rain falls on the roof and is collected in the tank.

How many litres of water were collected? (2 marks)

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Show Answers Only

`144\ text(m³)\ \ text{(nearest m³)}`
`text(12 hours)`
`8000\ text(litres)`

Show Worked Solution

i. `V = pi r^2 h\ \ \ \ text(where)`

`h = 2\ text(and)\ r = 4.78\ text(m)`

`:.\ V`	`= pi xx 4.78^2 xx 2`
	`= 143.56…`
	`= 144\ text(m³)\ \ text{(nearest m³)}`

ii. `text(Total water) = 90\ 000\ text(litres)`

`text(Usage) = 7500\ \ text(litres/hr)`

`:.\ text(Hours before it is empty)`

`= (90\ 000)/7500`

`= 12\ text(hours)`

iii. `text(Water collected)`

`= 400 xx 0.020`

`= 8\ text(m²)`

`= 8000\ text(litres)`

GEOMETRY, FUR1 2009 VCAA 2 MC

The area (in m²) of triangle `XYZ` can be found using Heron’s formula `A = sqrt(s(s−a)(s−b)(s−c))`, with `a = 1.92`, `b = 8.24`, `c = 9.20` and `s =`

A. `4.40`

B. `6.45`

C. `9.20`

D. `9.68`

E. `19.36`

Show Answers Only

`D`

Show Worked Solution

`s`	`= (a + b + c) / 2`
	`= (1.92 + 8.24 + 9.20) /2`
	`= 9.68`

`=> D`

Financial Maths, STD2 F1 2007 HSC 23a

Lilly and Rose each have money to invest and choose different investment accounts.

The graph shows the values of their investments over time.

How much was Rose’s original investment? (1 mark)

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At the end of 6 years, which investment will be worth the most and by how much? (2 marks)

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Lilly’s investment will reach a value of $20 000 first.
How much longer will it take Rose’s investment to reach a value of $20 000? (1 mark)

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Show Answers Only

`$5000`
`text(Rose’s is worth $2000 more.)`
`text(It takes Lilly 14 years to reach $20 000 and it takes)`

`text{Rose 1 year longer (15 years) to reach the same value}`

Show Worked Solution

i. `$5000\ text{(} y text(-intercept) text{)}`

ii. `text(After 6 years,)`

`text(Lilly’s investment)`	`= $9000`
`text(Rose’s investment)`	`= $11\ 000`
`:.\ text(Rose’s is worth $2000 more.)`

iii. `text(It takes Lilly 14 years to reach $20 000 and it)`

`text{takes Rose 1 year longer (15 years) to reach the}`

`text(same value.)`

PATTERNS, FUR1 2009 VCAA 4 MC

The sum of the infinite geometric sequence 96, – 48, 24, –12, 6 . . . is equal to

A. `64`

B. `66`

C. `68`

D. `144`

E. `192`

Show Answers Only

`A`

Show Worked Solution

`text(Sequence is)\ \ \ 96, –48, 24, –12, 6, …`

`text(GP where)\ \ \ a`	`=96, and`
`r`	`=t_2/t_1 = (–48)/96 = –1 /2`

`text(S)text(ince)\ \ |\ r\ | < 1`

`S_∞`	`=a/(1-r)`
	`=96/(1-(-1 /2))`
	`=64`

`=> A`

PATTERNS, FUR1 2009 VCAA 3 MC

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

The value of `t_2` is

A. `– 7250`

B. `– 7200`

C. `–1700`

D. `7200`

E. `7250`

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`

Measurement, STD2 M6 2007 HSC 8 MC

What is the length of the side `MN` in the following triangle, correct to two decimal places?

9.19 cm
10.07 cm
15.66 cm
18.67 cm

Show Answers Only

`C`

Show Worked Solution

`sin 50^@`	`= 12/(MN)`
`MN`	`= 12/sin50^@`
	`= 15.66…`

`=> C`

Probability, STD2 S2 2007 HSC 2 MC

Each student in a class is given a packet of lollies. The teacher records the number of red lollies in each packet using a frequency table.

What is the relative frequency of a packet of lollies containing more than three red lollies?

`4/19`
`4/15`
`11/19`
`11/15`

Show Answers Only

`A`

Show Worked Solution

`text(# Packets with more than 3)`

`= 3 + 1 = 4`

`text(Total packets) = 19`

`:.\ text(Relative Frequency) = 4/19`

`=> A`

Measurement, STD2 M1 2007 HSC 1 MC

What is `0.000\ 000\ 326` mm expressed in scientific notation?

`0.326 xx 10^-6\ \ text(mm)`
`3.26 xx 10^-7\ \ text(mm)`
`0.326 xx 10^6\ \ text(mm)`
`3.26 xx 10^7\ \ text(mm)`

Show Answers Only

`B`

Show Worked Solution

`0.000\ 000\ 326\ text(mm)`

`= 3.26 xx 10^-7\ \ text(mm)`

`=> B`

GEOMETRY, FUR1 2014 VCAA 3 MC

The diagram below shows the location of three boats, `A`, `B` and `C`.

Boat `B` is on a bearing of 110° from boat `A`.

Boat `B` is also on a bearing of 035° from boat `C`.

Boat `A` is due north of boat `C`.

The angle `ABC` is

A. `35°`

B. `65°`

C. `70°`

D. `75°`

E. `110°`

Show Answers Only

`D`

Show Worked Solution

`/_CAB`	`= 180 – 110\ \ \ text{(straight angle at A)}`
	`= 70°`

`:./_ABC`	`= 180 – (70 + 35)\ \ text{(Angle sum of}\ Delta ABC text{)}`
	`= 75°`

`=> D`

GEOMETRY, FUR1 2014 VCAA 1 MC

The top of a ladder that is 4.50 m long rests 3.25 m up a wall, as shown in the diagram below.

The angle, `theta`, that the ladder makes with the wall is closest to

A. `36°`

B. `44°`

C. `46°`

D. `50°`

E. `54°`

Show Answers Only

`B`

Show Worked Solution

`cos theta`	`=text(adj)/text(hyp)`
	`=3.25/4.50`
	`=0.722…`
`∴ theta`	`= 43.76°`

`=>B`

CORE*, FUR1 2014 VCAA 6 MC

Consider the following sequence.

`2,\ 1,\ 0.5\ …`

Which of the following difference equations could generate this sequence?

A.	`t_(n + 1) = t_n - 1`	`t_1 = 2`
B.	`t_(n + 1) = 3 - t_n`	`t_1 = 2`
C.	`t_(n + 1) = 2 × 0.5^(n – 1)`	`t_1 = 2`
D.	`t_(n + 1) = - 0.5t_n + 2`	`t_1 = 2`
E.	`t_(n + 1) = 0.5t_n`	`t_1 = 2`

Show Answers Only

`E`

Show Worked Solution

`text(Sequence is)\ \ 2, 1, 0.5, …`

NOTE: “GP” is used as an abbreviation of “geometric sequence”.

`=>\ text(Geometric sequence where common ratio = 0.5)`

`∴\ text(Difference equation is)`

`t_(n + 1) = 0.5t_n`

`=> E`

CORE*, FUR1 2014 VCAA 4 MC

On day 1, Vikki spends 90 minutes on a training program.

On each following day, she spends 10 minutes less on the training program than she did the day before.

Let `t_n` be the number of minutes that Vikki spends on the training program on day `n`.

A difference equation that can be used to model this situation for `1 ≤ n ≤ 10` is

A. `t_(n + 1) = 0.90t_n`	`t_1 = 90`
B. `t_(n + 1) = 1.10 t_n`	`t_1 = 90`
C. `t_(n + 1) = t_n - 0.10`	`t_1 = 90`
D. `t_(n + 1) = 1 - 10 t_n`	`t_1 = 90`
E. `t_(n + 1) = t_n - 10`	`t_1 = 90`

Show Answers Only

`E`

Show Worked Solution

`text(Difference equation where each term is 10 minutes)`

`text(less than the preceding term.)`

`∴\ text(Equation)\ \ \t_(n+1) = t_n-10, t_1 = 90`

`=> E`

PATTERNS, FUR1 2014 VCAA 3 MC

A city has a population of 100 000 people in 2014.
Each year, the population of the city is expected to increase by 4%.
In 2018, the population is expected to be closest to

A. `108\ 000`

B. `112\ 000`

C. `115\ 000`

D. `117\ 000`

E. `122\ 000`

Show Answers Only

`D`

Show Worked Solution

`text(Population is a geometric sequence with)`

`a = 100\ 000,\ \ \ \ r = 1.04`

`P_1 (2014)`	`= a = text(100 000)`
`P_2 (2015)`	`= ar^1 = text(100 000) (1.04) = text(104 000)`
`vdots`
`P_5 (2018)`	`= ar^4 = text(100 000) (1.04)^4 = text(116 985.8…)`

`∴\ text(In 2018, the population will be closest)`

`text(to)\ 117\ 000.`

`=> D`

CORE, FUR1 2014 VCAA 10-11 MC

The seasonal indices for the first 11 months of the year, for sales in a sporting equipment store, are shown in the table below.

Part 1

The seasonal index for December is

A. `0.89`

B. `0.97`

C. `1.02`

D. `1.23`

E. `1.29`

Part 2

In May, the store sold $213 956 worth of sporting equipment.

The deseasonalised value of these sales was closest to

A. `$165\ 857`

B. `$190\ 420`

C. `$209\ 677`

D. `$218\ 322`

E. `$240\ 400`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Sum of seasonal indices) = 12`

`:.\ text(December’s seasonal index)`

`=12 – text{(1.23 + 0.96 + 1.12 + 1.08 + 0.89}`

`text{+ 0.98 + 0.86 + 0.76 + 0.76 + 0.95 + 1.12)}`

`=1.29`

`=>E`

`text(Part 2)`

`text(May Index)`	`=\ text(Actual Sales)/text{Deseasonalised Sales (D)}`
`0.89`	`= (213\ 956)/(text{D})`
`:.\ text(D)`	`= (213\ 956)/0.89`
	`= $240\ 400`

`=> E`

CORE, FUR1 2011 VCAA 9-10 MC

The length of a type of ant is approximately normally distributed with a mean of 4.8 mm and a standard deviation of 1.2 mm.

Part 1

From this information it can be concluded that around 95% of the lengths of these ants should lie between

A. `text(2.4 mm and 6.0 mm)`

B. `text(2.4 mm and 7.2 mm)`

C. `text(3.6 mm and 6.0 mm)`

D. `text(3.6 mm and 7.2 mm)`

E. `text(4.8 mm and 7.2 mm)`

Part 2

A standardised ant length of `z = text(−0.5)` corresponds to an actual ant length of

A. `text(2.4 mm)`

B. `text(3.6 mm)`

C. `text(4.2 mm)`

D. `text(5.4 mm)`

E. `text(7.0 mm)`

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(95% of scores lie between ±2 std dev)`

`bar x = 4.8, \ \ \ s = 1.2`

`text(Lower limit)`	`= bar x – 2text(s)`
	`= 4.8 – 2(1.2)`
	`= 2.4\ text(mm)`
`text(Upper limit)`	`= bar x + 2text(s)`
	`= 4.8 + 2(1.2)`
	`= 7.2\ text(mm)`

`=>B`

`text(Part 2)`

`z`	`= \ \ (x – bar x)/s`
`-0.5`	`= \ \ (x – 4.8)/1.2`
`-0.6`	`= \ \ x – 4.8`
`x`	`= \ \ 4.2\ text(mm)`

`=>C`

CORE, FUR1 2009 VCAA 4-6 MC

The percentage histogram below shows the distribution of the fertility rates (in average births per woman) for 173 countries in 1975.

Part 1

In 1975, the percentage of these 173 countries with fertility rates of 4.5 or greater was closest to

A. `12text(%)`

B. `35text(%)`

C. `47text(%)`

D. `53text(%)`

E. `65text(%)`

Part 2

In 1975, for these 173 countries, fertility rates were most frequently

A. less than 2.5

B. between 1.5 and 2.5

C. between 2.5 and 4.5

D. between 6.5 and 7.5

E. greater than 7.5

Part 3

Which one of the boxplots below could best be used to represent the same fertility rate data as displayed in the percentage histogram?

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ D`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`text(Adding up the histogram bars from 4.5.)`

`%`	`= 12 + 19 + 28 + 5 + 1`
	`= 65text(%)`

`=> E`

`text(Part 2)`

`text(Fertility rates between 6.5 and 7.5 were 28%)`

`text(which is greater than any other range given.)`

`=> D`

`text(Part 3)`

♦ Mean mark 43%.
MARKERS’ COMMENT: A systemic approach where students calculated the median, `Q_1` and `Q_3` was most successful.

`text(The boxplots have the same range, therefore)`

`text(consider the values of)\ Q_1,\ Q_3\ text(and median.)`

`text(By elimination,)`

`Q_1\ text{estimate is slightly below 3.5 (the first 2}`

`text{bars add up to 29%), therefore not A, D or E.}`

`Q_3\ text(estimate is around 7. Eliminate C.)`

`=> B`

`I`	`= (PrT)/100`
`910`	`= ((6500)(3.5)(T))/100`
`:. T`	`= (910 xx 100)/(6500xx3.5)`
	`= 4\ \ text(years)`

`A`	`=PR^n`
	`= 22\ 000 xx 1.035^3`
	`= 24\ 391.79…`

`text(Price after discount)`	`= 215 – 25text(%) xx 215`
	`= $215 -53.75`
	`= $161.25`

`N`	`=20 xx 12=240`
`I(%)`	`= 6.95`
`PV`	`= -90000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`60 + 0.2n`	`= n`
`0.8n`	`= 60`
`n`	`= 75`

`text(Value of truck)`	`= $65\ 000 − 132\ 600 xx 0.22`
	`= $35\ 828`

`text(GST)`	`= text(10%) xx 300`
	`= 30`

`:.\ text(Total price)`	`= 300+ 30`
	`= 330`

`:.\ text(Salary)`	`= 1971.75 − 1529.55`
	`= $442.2`