Band 4 – Page 79

Financial Maths, STD2 F1 2007 HSC 7 MC

Margaret has a weekly income of $900 and allocates her money according to the budget shown in the sector graph.

How long will it take Margaret to save $3600?

4 weeks
5 weeks
16 weeks
18 weeks

Show Answers Only

`D`

Show Worked Solution

`text(Savings)`	`= 80/360\ \ \ text(sector graph)`
	`= 80/360 xx 900`
	`= $200\ text(per week)`

`:.\ text(Time to save $3600)`

`= 3600/200`

`= 18\ text(weeks)`

`=> D`

Financial Maths, STD2 F1 2007 HSC 6 MC

The price of a CD is $22.00, which includes 10% GST.

What is the amount of GST included in this price?

$2.00
$2.20
$19.80
$20.00

Show Answers Only

`A`

Show Worked Solution

`text(CD costs $22.00 incl. GST`

`text(Let)\ C = text(original cost)`

`C + text(10%) xx C`	`= 22`
`1.1C`	`= 22`
`C`	`= 20`

`:.\ text(GST)`	`= 22.00 – 20.00`
	`= $2.00`

`=> A`

Financial Maths, STD2 F1 2007 HSC 3 MC

Joe is about to go on holidays for four weeks. His weekly salary is $280 and his holiday loading is 17.5% of four weeks pay.

What is Joe’s total pay for the four weeks holiday?

$196
$329
$1169
$1316

Show Answers Only

`D`

Show Worked Solution

`text(Salary)\ text{(4 weeks)}`	`= 4 xx 280`
	`= $1120`

`text(Holiday loading)`	`= 1120 xx 17.5%`
	`= $196`

`:.\ text(Total pay)`	`= 1120 + 196`
	`= $1316`

`=> D`

GEOMETRY, FUR1 2014 VCAA 6-7 MC

A cross-country race is run on a triangular course. The points `A, B` and `C` mark the corners of the course, as shown below.

The distance from `A` to `B` is 2050 m.

The distance from `B` to `C` is 2250 m.

The distance from `A` to `C` is 1900 m.

The bearing of `B` from `A` is 140°.

Part 1

The bearing of `C` from `A` is closest to

A. `032°`

B. `069°`

C. `192°`

D. `198°`

E. `209°`

Part 2

The area within the triangular course `ABC`, in square metres, can be calculated by evaluating

A. `sqrt (3100 xx 1200 xx 1050 xx 850)`

B. `sqrt (3100 xx 2250 xx 2050 xx 1900)`

C. `sqrt (6200 xx 4300 xx 4150 xx 3950)`

D. `1/2 xx 2050 xx 2250 xx sin\ (140^@)`

E. `1/2 xx 2050 xx 2250 xx sin\ (40^@)`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 43%.

`text(Using the cosine rule:)`

`cos ∠CAB`	`= ((AC)^2 + (AB)^2 – (CB)^2)/(2 xx AC xx AB)`
	`= (1900^2 + 2050^2 – 2250^2)/(2 xx 1900 xx 2050)`
	`= 0.3530…`
`/_ CAB`	`= 69.32…°`

`∴\ text(Bearing of C from A)`

`= 140 + 69.32…`

`= 209.32…°`

`=>E`

`text(Part 2)`

`text(Using Heron’s rule,)`

`text{Semi-perimeter (s)}`

`= (1900 + 2050 + 2250)/2`

`= 3100`

`∴ A`	`= sqrt{s (s-a)(s-b)(s-c)}`
	`= sqrt{3100 xx 1200 xx 1050 xx 850}`

`=> A`

Mary plans to read a book in seven days.
Each day, Mary plans to read 15 pages more than she read on the previous day.
The book contains 1155 pages.
The number of pages that Mary will need to read on the first day, if she is to finish reading the book in seven days, is

A. `112`

B. `120`

C. `150`

D. `165`

E. `180`

Show Answers Only

`B`

Show Worked Solution

`text(Sequence is)\ \ a, a+15, a + 2 xx 15, …`

`=>\ text(AP where)\ \ \d=15`

`text(Find)\ \ a\ \ text(when)\ S_7 = 1155`

`S_n`	`=n/2[2a + (n-1)d]`
`1155`	`=7/2[2a + (7-1)15]`
`1155`	`=7/2[2a + 90]`
	`=7a + 315`
`7a`	`=840`
`a`	`=120`

`=> B`

CORE, FUR1 2014 VCAA 9 MC

The equation of a least squares regression line is used to predict the fuel consumption, in kilometres per litre of fuel, from a car’s weight, in kilograms.

This equation predicts that a car weighing 900 kg will travel 10.7 km per litre of fuel, while a car weighing 1700 kg will travel 6.7 km per litre of fuel.

The slope of this least squares regression line is closest to

A. `–250`

B. `–0.005`

C. `–0.004`

D. `0.005`

E. `200`

Show Answers Only

`B`

Show Worked Solution

`text(Gradient)`	`=(y_2-y_1)/(x_2 – x_1)`
	`=(6.7 – 10.7)/(1700 – 900)`
	`=- 4/800`
	`=-0.005`

`=> B`

CORE, FUR1 2011 VCAA 13 MC

The table below shows the number of broadband users in Australia for each of the years from 2004 to 2008.

A two-point moving mean, with centring, is used to smooth the time series.

The smoothed value for the number of broadband users in Australia in 2006 is

A. `2 \ 958 \ 000`

B. `3 \ 379 \ 600`

C. `3 \ 455 \ 500`

D. `3 \ 661 \ 500`

E. `3 \ 900 \ 000`

Show Answers Only

`D`

Show Worked Solution

`text(Two point mean for)`

`text(2005/06)`	`= (2 \ 016 \ 000 + 3 \ 900 \ 000) / 2`
	`= 2 \ 958 \ 000`

`text(2006/07)`	`= (3 \ 900 \ 000 + 4 \ 830 \ 000) / 2`
	`= 4 \ 365 \ 000`

`:.\ text(Centered mean)`	`= (2 \ 958 \ 000 + 4 \ 365 \ 000) / 2`
	`= 3 \ 661 \ 500`

`=> D`

CORE, FUR1 2011 VCAA 11 MC

For a group of 15-year-old students who regularly played computer games, the correlation between the time spent playing computer games and fitness level was found to be `r = -0.56.`

On the basis of this information it can be concluded that

56% of these students were not very fit.
these students would become fitter if they if they spent less time playing computer games.
these students would become fitter if they if they spent more time playing computer games.
the students in the group who spent a short amount of time playing computer games tended to be fitter.
the students in the group who spent a large amount of time playing computer games tended to be fitter.

Show Answers Only

`D`

Show Worked Solution

`text(Negative correlation means that as time spent playing)`

`text(computer games decreases, fitness levels tend to increase.)`

`text(Note that)\ B\ text(is incorrect because it assumes one causes the)`

`text(other.)`

`=> D`

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 2011 VCAA 6-8 MC

When blood pressure is measured, both the systolic (or maximum) pressure and the diastolic (or minimum) pressure are recorded.

Table 1 displays the blood pressure readings, in mmHg, that result from fifteen successive measurements of the same person's blood pressure.

Part 1

Correct to one decimal place, the mean and standard deviation of this person's systolic blood pressure measurements are respectively

A. `124.9 and 4.4`

B. `125.0 and 5.8`

C. `125.0 and 6.0`

D. `125.9 and 5.8`

E. `125.9 and 6.0`

Part 2

Using systolic blood pressure (systolic) as the response variable, and diastolic blood pressure (diastolic) as the explanatory variable, a least squares regression line is fitted to the data in Table 1.

The equation of the least squares regression line is closest to

A. `text(systolic) = 70.3 + 0.790 xx text(diastolic)`

B. `text(diastolic) = 70.3 + 0.790 xx text(systolic)`

C. `text(systolic) = 29.3 + 0.330 xx text(diastolic)`

D. `text(diastolic) = 0.330 + 29.3 xx text(systolic)`

E. `text(systolic) = 0.790 + 70.3 xx text(diastolic)`

Part 3

From the fifteen blood pressure measurements for this person, it can be concluded that the percentage of the variation in systolic blood pressure that is explained by the variation in diastolic blood pressure is closest to

A. `25.8text(%)`

B. `50.8text(%)`

C. `55.4text(%)`

D. `71.9text(%)`

E. `79.0text(%)`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ A`

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`text{By calculator (using sample standard deviation)}`

`text{the results are: }`

`text(Mean = 125.9, std dev = 6.0)`

`=>E`

`text(Part 2)`

`text{By calculator (making sure diastolic values are}`

`text{the explanatory or}\ x text{-variable), the regression line}`

`text(can be expressed as follows,)`

`text(systolic) = 70.3 + 0.790 xx text(diastolic)`

`=>A`

`text(Part 3)`

`text{By calculator, the regression line (above) should}`

`text(have found) \ \ r^2 = 0.258,\ text(which means that)`

`text(25.8% of the variation in systolic pressure can be)`

`text(explained by variation in diastolic pressure.)`

`=>A`

CORE, FUR1 2011 VCAA 5 MC

The boxplots below display the distribution of average pay rates, in dollars per hour, earned by workers in 35 countries for the years 1980, 1990 and 2000.

Based on the information contained in the boxplots, which one of the following statements is not true?

In 1980, over 50% of the countries had an average pay rate less than $8.00 per hour.
In 1990, over 75% of the countries had an average pay rate greater than $5.00 per hour.
In 1990, the average pay rate in the top 50% of the countries was higher than the average pay rate for any of the countries in 1980.
In 1990, over 50% of the countries had an average pay rate less than the median average pay rate in 2000.
In 2000, over 75% of the countries had an average pay rate greater than the median average pay rate in 1980.

Show Answers Only

`E`

Show Worked Solution

`text(By elimination,)`

`text(In A, 1980 median is below $8.00. True)`

`text(In B, 1990 Q1 is above $5.00. True)`

`text(In C, 1990 median is above 1980 high. True)`

`text(In D, 1990 median is below 2000 median. True)`

`text(In E, 2000 Q1 is below 1980 median. NOT true)`

`=> E`

CORE, FUR1 2009 VCAA 12 MC

The mathematics achievement level (TIMSS score) for grade 8 students and the general rate of Internet use (%) for 10 countries are displayed in the scatterplot below.

To linearise the data, it would be best to plot

A. mathematics achievement against Internet use.

B. log (mathematics achievement) against Internet use.

C. mathematics achievement against log (Internet use).

D. mathematics achievement against (Internet use)².

E. ` 1/text(mathematics achievement)` against Internet use.

Show Answers Only

`C`

Show Worked Solution

`text(The shape of the data is logarithmic.)`

`:.\ text(To linearise the data, it would be best to)`

`text{plot mathematics achievement against}`

`text{log (Internet use). }`

`=> C`

CORE, FUR1 2009 VCAA 7 MC

The level of oil use in certain countries is approximately normally distributed with a mean of 42.2 units and a standard deviation of 10.2 units.

The percentage of these countries in which the level of oil use is greater than 32 units is closest to

A. 5%

B. 16%

C. 34%

D. 84%

E. 97.5%

Show Answers Only

`D`

Show Worked Solution

`bar x`

`=42.2`

`s`

`=10.2`

`z text(-score (32))`	`=(x – barx)/s`
	`=(32-42.2)/10.2`
	`=–1`

`text(68% lie between)\ z text(-score of –1 and 1)`

`=>\ text(34% lie between z-score –1 and 0)`

`text(50% lie above z-score of 0)`

`∴\ text(% above 32 units)`	`=34 + 50`
	`= 82text(%)`

`rArr D`

CORE, FUR1 2008 VCAA 10 MC

A large study of Year 12 students shows that there is a negative association between the time spent doing homework each week and the time spent watching television. The correlation coefﬁcient is `r = – 0.6`.

From this information it can be concluded that

the time spent doing homework is 60% lower than the time spent watching television.
36% of students spend more time watching television than doing homework.
the slope of the least squares regression line is 0.6.
if a student spends less time watching television, they will do more homework.
an increased time spent watching television is associated with a decreased time doing homework.

Show Answers Only

`E`

Show Worked Solution

`text(A negative correlation means that the greater the time)`

`text(spent watching television is associated with the less)`

`text(time spent doing homework.)`

`D\ text(assumes one causes the other and is therefore incorrect.)`

`=>E`

CORE, FUR1 2008 VCAA 8-9 MC

The weights (in g) and lengths (in cm) of 12 ﬁsh were recorded and plotted in the scatterplot below. The least squares regression line that enables the weight of these ﬁsh to be predicted from their length has also been plotted.

Part 1

The least squares regression line predicts that the weight (in g) of a ﬁsh of length 30 cm would be closest to

A. `240`

B. `252`

C. `262`

D. `274`

E. `310`

Part 2

The median weight (in g) of the 12 ﬁsh is closest to

A. `346`

B. `375`

C. `440`

D. `450`

E. `475`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text{The regression line crosses the 30cm length (on the}`

`xtext{-axis) at approx 262.}`

`=>C`

`text(Part 2)`

`text(12 weight data points – the median will be the average of)`

`text(the 6th and 7th.)`

`text(From the graph,)`

`text(6th highest weight = 430 g)`

`text(7th highest weight = 450 g)`

`:.\ text(Median)`	`=(430+450)/2`
	`=440\ text(g)`

`=> C`

CORE, FUR1 2008 VCAA 6-7 MC

The pulse rates of a large group of 18-year-old students are approximately normally distributed with a mean of 75 beats/minute and a standard deviation of 11 beats/minute.

Part 1

The percentage of 18-year-old students with pulse rates less than 75 beats/minute is closest to

A. 32%

B. 50%

C. 68%

D. 84%

E. 97.5%

Part 2

The percentage of 18-year-old students with pulse rates less than 53 beats/minute or greater than 86 beats/minute is closest to

A. 2.5%

B. 5%

C. 16%

D. 18.5%

E. 21%

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`barx=75,\ \ \ s=11`

`text(In a normal distribution, mean = median.)`

`:.\ text(50% of group are below mean of 75)`

`=>B`

`text(Part 2)`

♦ Mean mark 44%.
MARKERS’ COMMENT: Two applications of the 68-95-99.7% rule are required. A good strategy is to first draw a normal curve and shade the required areas.

`barx=75,\ \ \ s=11`

`z text{-score (53)}`	`=(x-barx) /s`
	`=(53-75)/11`
	`= – 2`

`z text{-score (86)}`	`= (86-75)/11`
	`=1`

`text(From the diagram, the % of students that have a)`

`z text(-score below –2 or above 1)`

`=2.5+16`

`=18.5 text(%)`

`=>D`

CORE, FUR1 2008 VCAA 5 MC

A sample of 14 people were asked to indicate the time (in hours) they had spent watching television on the previous night. The results are displayed in the dot plot below.

Correct to one decimal place, the mean and standard deviation of these times are respectively

A. `bar x=2.0\ \ \ \ \ s=1.5`

B. `bar x=2.1\ \ \ \ \ s=1.5`

C. `bar x=2.1\ \ \ \ \ s=1.6`

D. `bar x=2.6\ \ \ \ \ s=1.2`

E. `bar x=2.6\ \ \ \ \ s=1.3`

Show Answers Only

`C`

Show Worked Solution

`text(Data points are:)`

`0,0,0,1,1,2,2,2,2,3,3,4,4,5`

`text(By calculator (using sample standard deviation))`

`bar x=2.1,\ \ s=1.6`

`=> C`

CORE, FUR1 2008 VCAA 1-4 MC

The box plot below shows the distribution of the time, in seconds, that 79 customers spent moving along a particular aisle in a large supermarket.

Part 1

The longest time, in seconds, spent moving along this aisle is closest to

A. `40`

B. `60`

C. `190`

D. `450`

E. `500`

Part 2

The shape of the distribution is best described as

A. symmetric.

B. negatively skewed.

C. negatively skewed with outliers.

D. positively skewed.

E. positively skewed with outliers.

Part 3

The number of customers who spent more than 90 seconds moving along this aisle is closest to

A. `7`

B. `20`

C. `26`

D. `75`

E. `79`

Part 4

From the box plot, it can be concluded that the median time spent moving along the supermarket aisle is

A. less than the mean time.

B. equal to the mean time.

C. greater than the mean time

D. half of the interquartile range.

E. one quarter of the range.

Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ E`

`text(Part 3:)\ B`

`text(Part 4:)\ A`

Show Worked Solution

`text(Part 1)`

`text(Longest time is represented by the farthest right)`

`text(data point.)`

`=>D`

`text(Part 2)`

`text(Positively skewed as the tail of the distribution can)`

`text(clearly be seen to extend to the right.)`

`text(The data also clearly shows outliers.)`

`=>E`

`text(Part 3)`

♦ Mean mark 43%.
MARKERS’ COMMENT: Note that the outliers are already accounted for in the boxplot.

`text(From the box plot,)`

`text(Q)_3=90\ text{s}\ \ text{(i.e. 25% spend over 90 s)}`

`:.\ text(Customers that spend over 90 s)`

`= 25text(%) xx 79`

`=19.75`

`=>B`

`text(Part 4)`

`text(The mean is greater than the median for positively)`

`text(skewed data.)`

`=>A`

CORE*, FUR1 2010 VCAA 7 MC

Each trading day, a share trader buys and sells shares according to the rule

`T_(n+1)=0.6 T_n + 50\ 000`

where `T_n` is the number of shares the trader owns at the start of the `n`th trading day.

From this rule, it can be concluded that each day

the trader sells 60% of the shares that she owned at the start of the day and then buys another 50 000 shares.
the trader sells 40% of the shares that she owned at the start of the day and then buys another 50 000 shares.
the trader sells 50 000 of the shares that she owned at the start of the day.
the trader sells 60% of the 50 000 shares that she owned at the start of the day.
the trader sells 40% of the 50 000 shares that she owned at the start of the day.

Show Answers Only

`B`

Show Worked Solution

`T_(n+1)=0.6\ \T_n + 50\ 000`

`text(The difference equation describes a rule)`

`text(where a trader sells 40% of shares owned on)`

`text{the day before (left with 60% or 0.6}T_n text{)} `

`text(and then buys another 50 000 each day.)`

`=> B`

PATTERNS, FUR1 2012 VCAA 6 MC

The second and third terms of a geometric sequence are 100 and 160 respectively.

The sum of the first ten terms of this sequence is closest to

A. `4300`

B. `6870`

C. `11\ 000`

D. `11\ 290`

E. `11\ 350`

Show Answers Only

`E`

Show Worked Solution

`text (GP where)\ \ \ T_2 = 100, and T_3 = 160`

`:. r= T_3/T_2 = 160/100 = 1.6`

`T_2`	`= ar`
`100`	`= a xx 1.6`
`:. a`	`= 62.5`

`text (Find)\ \ S_10`

`S_n`	`= (a (r^n – 1))/ (r-1)`
`S_10`	`= (62.5 (1.6^10 – 1))/(1.6 – 1)`
	`=11\ 349.07…`

`rArr E`

PATTERNS, FUR1 2012 VCAA 3-4 MC

Use the following information to answer Parts 1 and 2.

As part of a savings plan, Stacey saved $500 the first month and successively increased the amount that she saved each month by $50. That is, in the second month she saved $550, in the third month she saved $600, and so on.

Part 1

The amount Stacey will save in the 20th month is

A. `$1450`

B. `$1500`

C. `$1650`

D. `$1950`

E. `$3050`

Part 2

The total amount Stacey will save in four years is

A. `$13\ 400`

B. `$37\ 200`

C. `$58\ 800`

D. `$80\ 400`

E. `$81\ 600`

Show Answers Only

`text (Part 1:)\ A`

`text (Part 2:)\ D`

Show Worked Solution

`text (Part 1)`

`text (Sequence is 500, 550, 600,…)`

`text (AP where)\ \ a`	`= 500, and`
`d`	`= text (550 – 500 = 50)`

`T_n`	`= a + (n – 1) d`
`T_20`	`= 500 + (20 – 1)50`
	`= 1450`

`rArr A`

`text (Part 2)`

`n`	`= 4 xx 12 = 48`
`S_n`	`= n/2 [2a + (n-1)d]`
`S_4`	`= 48/2 [2 xx 500 + (48-1)50]`
	`= 24 [1000 + 2350]`
	`= 80\ 400`

`rArr D`

GEOMETRY, FUR1 2010 VCAA 5-6 MC

A soccer goal is 7.4 metres wide.

A rectangular region `ABCD` is marked out directly in front of the goal.

In this rectangular region, `AB = DC = 11.0\ text(metres)` and `AD = BC = 5.5\ text(metres.)`

The goal line `XY` lies on `DC` and `M` is the midpoint of both `DC` and `XY`.

Part 1

Ben kicks the ball from point `B`. It travels in a straight line to the base of the goal post at point `Y` on the goal line.

Angle `CBY`, the angle that the path of the ball makes with the line `BC`, is closest to

A. `18°`

B. `33°`

C. `45°`

D. `67°`

E. `72°`

Part 2

David kicks the ball from point `D` in a straight line to Tara. Tara is standing near point `T` on the line `AB`, a distance of 4.5 metres from point `A`. Tara then kicks the ball from point `T` in a straight line to the midpoint of the goal line at `M`.

The total distance that the ball will travel in moving from point `D` to `T` to `M` is closest to

A. `5.5\ text(m)`

B. `12.1\ text(m)`

C. `12.5\ text(m)`

D. `12.7\ text(m)`

E. `12.9\ text(m)`

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`YC + 7.4 + DX`	`=11.0`
`2YC`	`= 3.6\ \ \ (DX=YC)`
`YC`	`= 1.8`

`tan\ /_ CBY`	`= 1.8/5.5 = 0.3272…`
`/_ CBY`	`= 18.12…°`

`=> A`

`text(Part 2)`

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

`DT^2`	`= 5.5^2 + 4.5^2`
	`= 50.5`
`:. DT`	`= 7.106…`

`text(In)\ Delta MFT`

`FT = 5.5 – 4.5 = 1`

`text(Using Pythagoras)`

`MT^2`	`= 1^2 + 5.5^2`
	`= 31.25`
`MT`	`= 5.59…`

`:.\ text(The distance the ball travels)`

`= 7.106… + 5.59…`

`= 12.69…\ text(m)`

`=> D`

GEOMETRY, FUR1 2010 VCAA 2 MC

A circle has a circumference of 10 cm.

The radius of this circle is closest to

A. `1.3\ text(cm)`

B. `1.6\ text(cm)`

C. `1.8\ text(cm)`

D. `3.2\ text(cm)`

E. `5.0\ text(cm)`

Show Answers Only

`B`

Show Worked Solution

`C`	`= 2πr`
`10`	`= 2 xx π xx r`
`:. r`	`= 10/(2 xx π)`
	`= 1.59…\ text(cm)`

`=> B`

GEOMETRY, FUR1 2010 VCAA 3 MC

An equilateral triangle of side length 6 cm is cut from a sheet of cardboard.

A circle is then cut out of the triangle, leaving a hole of diameter 2 cm as shown below.

The area of cardboard remaining, as shown by the shaded region in the diagram above, is closest to

A. `3\ text(cm²)`

B. `9\ text(cm²)`

C. `12\ text(cm²)`

D. `15\ text(cm²)`

E. `16\ text(cm²)`

Show Answers Only

`C`

Show Worked Solution

`text(Equilateral triangle)\ =>\ 3 xx 60°\ text(angles.)`

`text(Area of)\ Delta`	`= 1/2 ab sin C`
	`= 1/2 xx 6 xx 6 xx sin 60°`
	`= 15.58\ text(cm²)`

`text(Area of circle)`	`= pir^2`
	`= pi xx 1^2`
	`= 3.14\ text(cm²)`

`:.\ text(Area of cardboard remaining)`

`= 15.58- 3.14`

`= 12.44\ text(cm²)`

`=> C`

CORE, FUR1 2010 VCAA 7-9 MC

The height (in cm) and foot length (in cm) for each of eight Year 12 students were recorded and displayed in the scatterplot below.
A least squares regression line has been fitted to the data as shown.

Part 1

By inspection, the value of the product-moment correlation coefficient `(r)` for this data is closest to

`0.98`
`0.78`
`0.23`
`– 0.44`
`– 0.67`

Part 2

The explanatory variable is foot length.

The equation of the least squares regression line is closest to

height = –110 + 0.78 × foot length.
height = 141 + 1.3 × foot length.
height = 167 + 1.3 × foot length.
height = 167 + 0.67 × foot length.
foot length = 167 + 1.3 × height.

Part 3

The plot of the residuals against foot length is closest to

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ B`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`text(The correlation is positive and strong.)`

`text(Eliminate)\ C, D\ text(and)\ E.`

`r= 0.98\ text(is too strong. Eliminate)\ A.`

`=> B`

`text(Part 2)`

♦♦ Mean mark 35%.
STRATEGY: An alternate but less efficient strategy could be to find 2 points and calculate the gradient and then use the point gradient formula.

`text(The intercept with the height axis)\ (ytext{-axis)}`

`text{is below 167 because that is the height when}`

`text{foot length = 20 cm.}`

`text(Eliminate)\ C, D\ text(and)\ E.`

`text(The gradient is approximately 1.3, by observing)`

`text(the increase in height values when the foot)`

`text(length increases from 20 to 22 cm.)`

`=> B`

`text(Part 3)`

`text(First residual is positive. Eliminate)\ A, D, E.`

`text(The next 3 residuals are negative. Eliminate)\ C`

`=> B`

Trig Ratios, EXT1 2008 HSC 6a

From a point `A` due south of a tower, the angle of elevation of the top of the tower `T`, is 23°. From another point `B`, on a bearing of 120° from the tower, the angle of elevation of `T` is 32°. The distance `AB` is 200 metres.

Copy or trace the diagram into your writing booklet, adding the given information to your diagram. (1 mark)
Hence find the height of the tower. (3 marks)

Show Answers Only

`96\ text(m)`

Show Worked Solution

(i)

(ii) `text(Find)\ \ OT = h`

`text(Using the cosine rule in)\ Delta AOB :`

`200^2 = OA^2 + OB^2 – 2 * OA * OB * cos 60\ …\ text{(*)}`

`text(In)\ Delta OAT,\tan 23^@= h/(OA)`

`=> OA= h/(tan 23^@)\ …\ (1)`

`text(In)\ Delta OBT,\ tan 32^@= h/(OB)`

`=> OB= h/(tan 32^@)\ \ \ …\ (2)`

`text(Substitute)\ (1)\ text(and)\ (2)\ text(into)\ text{(*)}`

`200^2`	`= (h^2)/(tan^2 23^@) + (h^2)/(tan^2 32^@) – 2 * h/(tan 23^@) * h/(tan 32^@) * 1/2`
	`= h^2 (1/(tan^2 23^@) + 1/(tan^2 32^@) + 1/(tan23^@ * tan32^@) )`
	`= h^2 (4.340…)`
`h^2`	`= (40\ 000)/(4.340…)`
	`= 9214.55…`
`:. h`	`= 95.99…`
	`= 96\ text(m)\ \ \ text{(to nearest m)}`

Functions, EXT1 F1 2008 HSC 5a

Let `f(x) = x-1/2 x^2` for `x <= 1`. This function has an inverse, `f^(-1) (x)`.

Sketch the graphs of `y = f(x)` and `y = f^(-1) (x)` on the same set of axes. (Use the same scale on both axes.) (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Find an expression for `f^(-1) (x)`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---
Evaluate `f^(-1) (3/8)`. (1 mark)

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

Show Answers Only

`y = 1-sqrt(1-2x)`
`1/2`

Show Worked Solution

ii.

`y = x-1/2 x^2,\ \ \ x <= 1`

`text(Inverse function: swap)\ \ x↔y,`

`x`	`= y-1/2 y^2,\ \ \ y <= 1`
`2x`	`= 2y-y^2`
`y^2-2y + 2x`	`= 0`

`y`	`= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2`
	`= (2 +- sqrt(4-8x))/2`
	`= (2 +- 2 sqrt(1-2x))/2`
	`= 1 +- sqrt (1-2x)`

`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`

iii.	`f^(-1) (3/8)`	`= 1-sqrt(1 -2(3/8))`
		`= 1-sqrt(1-6/8)`
		`= 1-sqrt(1/4)`
		`= 1-1/2`
		`= 1/2`

CORE, FUR1 2012 VCAA 11-12 MC

Use the following information to answer Parts 1 and 2.

The table below shows the long-term average rainfall (in mm) for summer, autumn, winter and spring. Also shown are the seasonal indices for summer and autumn. The seasonal indices for winter and spring are missing.

Part 1

The seasonal index for spring is closest to

A. `0.90`

B. `1.03`

C. `1.13`

D. `1.15`

E. `1.17`

Part 2

In 2011, the rainfall in autumn was 48.9 mm.

The deseasonalised rainfall (in mm) for autumn is closest to

A. `48.4`

B. `48.9`

C. `49.4`

D. `50.9`

E. `54.0`

Show Answers Only

`text (Part 1:)\ C`

`text (Part 2:)\ A`

Show Worked Solution

`text (Part 1)`

`text (Average Seasonal Rainfall)`

`= (52.0 + 54.5 + 48.8 + 61.3)/4`

`=54.15`

`:.\ text {Seasonal index (Spring)}`

`= 61.3/54.15`

`= 1.132…`

`rArr C`

`text (Part 2)`

`:.\ text {Deseasonalised Rainfall (Autumn)}`

`= 48.9/1.01`

`=48.415`

`rArr A`

CORE, FUR1 2012 VCAA 10 MC

Which one of the following statistics is never negative?

A. a median

B. a residual

C. a standardised score

D. an interquartile range

E. a correlation coefficient

Show Answers Only

`D`

Show Worked Solution

`text (S) text(ince IQR)\ = Q_3 – Q_1, and`

`Q_1\ text(is always less than)\ Q_3,`

`text(IQR is always positive.)`

`rArr D`

CORE, FUR1 2012 VCAA 8 MC

The maximum wind speed and maximum temperature were recorded each day for a month. The data is displayed in the scatterplot below and a least squares regression line has been fitted. The response variable is temperature. The explanatory variable is wind speed.

The equation of the least squares regression line is closest to

A. `text(temperature) = 25.7 - 0.191 xx text(wind speed)`

B. `text(wind speed) = 25.7 - 0.191 xx text(temperature)`

C. `text(temperature) = 0.191 + 25.7 xx text(wind speed)`

D. `text(wind speed) = 25.7 + 0.191 xx text(temperature)`

E. `text(temperature) = 25.7 + 0.191 xx text(wind speed)`

Show Answers Only

`A`

Show Worked Solution

`text (Using the form)\ \ y = mx +b\ \ text(where)`

`y rArr text (temperature)`

`x rArr text (wind speed)`

`text (b = 25.7 (y intercept))`

`text (Gradient is negative because temperature decreases as)`

`text(wind speed increases.)`

`:.\ text (Equation must take the form of A.)`

`rArr A`

CORE, FUR1 2012 VCAA 7 MC

The table below shows the percentage of students in two age groups (15–19 years and 20–24 years) who regularly use the internet at one or more of three locations.

at home
at an educational institution
at work

For the students surveyed, which one of the following statements, by itself, supports the contention that the location of internet use is associated with the age group of the internet user?

85% of students aged 15–19 years used the internet at an educational institution.
95% of students aged 15–19 years used the internet at home, but only 38% of 15–19 year olds used it at work.
95% of students aged 15–19 years used the internet at home and 18% of 20–24 year olds used the internet at an educational institution.
The percentage of students who used the internet at an educational institution decreased from 85% for those aged 15–19 years to 18% for those aged 20–24 years.
The percentage of students who used the internet at home was 95% for those aged 15–19 years and 95% for those aged 20–24 years.

Show Answers Only

`D`

Show Worked Solution

`text (The contention requires that the two age groups are)`

`text(compared at the same location.)`

`:.\ text(Eliminates A, B, and C.)`

`text (Considering E,)`

`text(the usage at home is THE SAME for both groups)`

`text(and therefore age group doesn’t matter.)`

`text (Considering D,)`

`text(the same location is used and the usage differs)`

`text(between groups.)`

`rArr D`

CORE, FUR1 2012 VCAA 5 MC

The temperature of a room is measured at hourly intervals throughout the day.

The most appropriate graph to show how the temperature changes from one hour to the next is a

A. boxplot.

B. stem plot.

C. histogram.

D. time series plot.

E. two-way frequency table.

Show Answers Only

`D`

Show Worked Solution

` text (A time series plot is best because the temperature)`

`text(is measured at regular time intervals.)`

`rArr D`

CORE, FUR1 2010 VCAA 5-6 MC

The lengths of the left feet of a large sample of Year 12 students were measured and recorded. These foot lengths are approximately normally distributed with a mean of 24.2 cm and a standard deviation of 4.2 cm.

Part 1

A Year 12 student has a foot length of 23 cm.
The student’s standardised foot length (standard `z` score) is closest to

A. –1.2

B. –0.9

C. –0.3

D. 0.3

E. 1.2

Part 2

The percentage of students with foot lengths between 20.0 and 24.2 cm is closest to

A. 16%

B. 32%

C. 34%

D. 52%

E. 68%

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`bar(x) = 24.2,`	`s=4.2`
`z text{-score (23)}`	`=(x – bar(x))/s`
	`= (23 – 24.2)/4.2`
	`= -0.285…`

`=> C`

`text(Part 2)`

`z text{-score (20)}`	`=(20- 24.2)/4.2`
	`= -1`
`z text{-score (24.2)}`	`= 0`

`text(68% have a)\ z text(-score between –1 and 1)`

`:.\ text(34% have a)\ z text(-score between –1 and 0)`

`=> C`

CORE, FUR1 2010 VCAA 1-3 MC

To test the temperature control on an oven, the control is set to 180°C and the oven is heated for 15 minutes.
The temperature of the oven is then measured. Three hundred ovens were tested in this way. Their temperatures were recorded and are displayed below using both a histogram and a boxplot.

Part 1

A total of 300 ovens were tested and their temperatures were recorded.

The number of these temperatures that lie between 179°C and 181°C is closest to

A. `40`

B. `50`

C. `70`

D. `110`

E. `150`

Part 2

The interquartile range for temperature is closest to

A. `1.3°text(C)`

B. `1.5°text(C)`

C. `2.0°text(C)`

D. `2.7°text(C)`

E. `4.0°text(C)`

Part 3

Using the 68–95–99.7% rule, the standard deviation for temperature is closest to

A. `1°text(C)`

B. `2°text(C)`

C. `3°text(C)`

D. `4°text(C)`

E. `6°text(C)`

Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ D`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`text(22% of ovens had temperatures between 179 – 180°)`

`text{and 16% between 180 – 181° (from bar chart).}`

`:.\ text(Number of ovens between 179° and 181°)`
	`=\ text{(22% + 16%)} xx 300`
	`= 38text(%) xx 300`
	`= 114`

`=> D`

`text(Part 2)`

`text(IQR)`	`=\ text(Q3 – Q1)`
	`= 181.5- 179`
	`= 2.5text(%)`

`=> D`

`text(Part 3)`

♦ Mean mark 43%.

`text(The percentage of ovens between 179 – 181°)`

`=21 + 16 = 38text(%)`

`text(Taking another bar column either side, we have)`

`text{178 – 179° (13%) and 181–182° (15%).}`

`:.\ text(178 – 182° accounts for approximately 66% of all values.)`

`:.\ text(1 standard deviation is approximately 2°.)`

`=> B`

CORE, FUR1 2009 VCAA 9-10 MC

The table below lists the average life span (in years) and average sleeping time (in hours/day) of 12 animal species.

Part 1

Using sleeping time as the independent variable, a least squares regression line is fitted to the data.

The equation of the least squares regression line is closest to

A. life span = 38.9 – 2.36 × sleeping time.

B. life span = 11.7 – 0.185 × sleeping time.

C. life span = – 0.185 – 11.7 × sleeping time.

D. sleeping time = 11.7 – 0.185 × life span.

E. sleeping time = 38.9 – 2.36 × life span.

Part 2

The value of Pearson’s product-moment correlation coefficient for life span and sleeping time is closest to

A. `–0.6603`

B. `–0.4360`

C. `–0.1901`

D. `0.4360`

E. `0.6603`

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 49%.
MARKERS’ COMMENT: Almost a quarter of students incorrectly assumed the independent variable was in the first column!

`text{By calculator (with “life span” as the}`

`text{dependent variable), the equation is:}`

`text(life span = 38.9 – 2.36 × sleeping time.)`

`=>A`

`text(Part 2)`

`text (By calculator)`

`=>A`

CORE, FUR1 2012 VCAA 4 MC

A class of students sat for a Biology test and a Legal Studies test. Each test had a possible maximum score of 100 marks. The table below shows the mean and standard deviation of the marks obtained in these tests.

The class marks in each subject are approximately normally distributed.

Sashi obtained a mark of 81 in the Biology test.

The mark that Sashi would need to obtain on the Legal Studies test to achieve the same standard score for both Legal Studies and Biology is

A. 81

B. 82

C. 83

D. 87

E. 95

Show Answers Only

`D`

Show Worked Solution

`z text {-score (Biology)}`	`= ( x – bar x)/ s`
	`= (81-54)/15`
	`= 1.8`

`text(Legal Studies mark must have a) \ z text(-score of 1.8:)`

`1.8`	`= (x-78)/5`
`9`	`= x – 78`
`x`	`= 87`

`rArr D`

CORE, FUR1 2008 VCAA 11-13 MC

The time series plot below shows the number of users each month of an online help service over a twelve-month period.

Part 1

The time series plot has

A. no trend.

B. no variability.

C. seasonality only.

D. an increasing trend with seasonality.

E. an increasing trend only.

Part 2

The data values used to construct the time series plot are given below.

2008 12

A four-point moving mean with centring is used to smooth timeline series.
The smoothed value of the number of users in month number 5 is closest to

A. `357`

B. `359`

C. `360`

D. `365`

E. `373`

Part 3

A least squares regression line is ﬁtted to the time series plot.
The equation of this least squares regression line is

number of users = 346 + 2.77 × month number

Let month number 1 = January 2007, month number 2 = February 2007, and so on.

Using the above information, the regression line predicts that the number of users in December 2009 will be closest to

A. `379`

B. `412`

C. `443`

D. `446`

E. `448`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ C`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 39%.
MARKERS’ COMMENT: 50% of students incorrectly read the three large random fluctuations in monthly sales as seasonality, which can’t be determined over only 12 months.

`text(The time series is clearly trending upwards with)`

`text(higher lows and higher highs occurring.)`

`text(The large fluctuations are random and should)`

`text(not be confused with seasonality.)`

`=>E`

`text(Part 2)`

`text(Mean for months 3-6)`

`=(354+356+373+353)/4`

`=359`

`text(Mean for months 4-7)`

`=(356+373+353+364)/4`

`=361.5`

`:.\ text(Four point moving mean with centring)`

`=(359+361.5)/2`

`=360.25`

`=> C`

`text(Part 3)`

`text(December 2009 will be month number 36.)`

`:.\ text(Number of users)`	`= 346+2.77xx36`
	`= 445.72`

`=> D`

CORE, FUR1 2012 VCAA 1-2 MC

The following bar chart shows the distribution of wind directions recorded at a weather station at 9.00 am on each of 214 days in 2011.

Part 1

According to the bar chart, the most frequently observed wind direction was

A. south-east.

B. south.

C. south-west.

D. west.

E. north-west.

Part 2

According to the bar chart, the percentage of the 214 days on which the wind direction was observed to be east or south-east is closest to

A. `10text(%)`

B. `16text(%)`

C. `25text(%)`

D. `33text(%)`

E. `35text(%)`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ B`

Show Worked Solution

`text(Part 1)`

`text{North-west (highest bar)}`

`=>E`

`text(Part 2)`

`text (# Days with East or South East wind)`

`= 10 + 25`

`= 35`

`:.\ text(% Days)`	`= 35/text (Total Days) xx 100`
	`= 35/214 xx 100`
	`= 16.355…text(%)`

`=> B`

PATTERNS, FUR1 2013 VCAA 4 MC

The vertical distance, in m, that a hot air balloon rises in each successive minute of its flight is given by the geometric sequence

`64.0,\ \ 60.8,\ \ 57.76\ …`

The total vertical distance, in m, that the balloon rises in the first 10 minutes of its flight is closest to

A. `38`

B. `40`

C. `473`

D. `514`

E. `1280`

Show Answers Only

`D`

Show Worked Solution

`text(Series 64.0, 60.8, 57.76 …)`

`text(GP where)\ \ a`	`=64.0`
`r`	`=60.8/64.0=0.95`

`S_n`	`=(a(1-r^n))/(1−r)`
`S_10`	`=(64 (1−0.95^10))/(1−0.95)`
	`=513.616…`

`=> D`

PATTERNS, FUR1 2013 VCAA 2 MC

The graph above shows the first six terms of a sequence.

This sequence could be

A. an arithmetic sequence that sums to one.

B. an arithmetic sequence with a common difference of one.

C. a Fibonacci-related sequence whose first term is one.

D. a geometric sequence with an infinite sum of one.

E. a geometric sequence with a common ratio of one.

Show Answers Only

`E`

Show Worked Solution

`text(Series is 1, 1, 1, …)`

♦ Mean mark 41%.

`text(The only possibility within the choices)`

`text(is a geometric sequence where)\ \ r=1.`

`=> E`

CORE, FUR1 2013 VCAA 12-13 MC

The time series plot below displays the number of guests staying at a holiday resort during summer, autumn, winter and spring for the years 2007 to 2012 inclusive.

Part 1

Which one of the following best describes the pattern in the time series?

A. random variation only

B. decreasing trend with seasonality

C. seasonality only

D. increasing trend only

E. increasing trend with seasonality

Part 2

The table below shows the data from the times series plot for the years 2007 and 2008.

Using four-mean smoothing with centring, the smoothed number of guests for winter 2007 is closest to

A. `85`

B. `107`

C. `183`

D. `192`

E. `200`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The pattern in the time series is seasonal only,)`

`text(with peaks appearing in Summer. There is no)`

`text(apparent year-to-year trend.)`

`=> C`

`text(Part 2)`

`text{Mean of guests (Season 1-4)}`

`=(390+126+85+130)/4`

`=182.75`

`text{Mean of guests (Season 2-5)}`

`=(126+85+85+130+460)/4`

`=200.25`

`:.\ text(Four-mean smoothing with centring)`

`=(182.75+200.25)/2`

`=191.5`

`=> D`

CORE, FUR1 2013 VCAA 11 MC

A least squares regression line is fitted to data in a scatterplot, as shown below.

The corresponding residual plot is closest to

Show Answers Only

`A`

Show Worked Solution

`text(By Elimination)`

`text(The first 2 and last 3 data points are below regression line,)`

`text(and are therefore negative residuals.)`

`:.\ text(Cannot be)\ B, D\ text(or)\ E.`

`text(The first two data points will have similar negative)`

`text(residuals.)`

`:.\ text(Cannot be)\ C.`

`=> A`

CORE, FUR1 2013 VCAA 10 MC

The data in the scatterplot below shows the width, in cm, and the surface area, in cm², of leaves sampled from 10 different trees. The scatterplot is non-linear.

To linearise the scatterplot, (width)² is plotted against area and a least squares regression line is then fitted to the linearised plot.

The equation of this least squares regression line is

(width)² = 1.8 + 0.8 × area

Using this equation, a leaf with a surface area of 120 cm² is predicted to have a width, in cm, closest to

A. 9.2

B. 9.9

C. 10.6

D. 84.6

E. 97.8

Show Answers Only

`B`

Show Worked Solution

`text(Substituting)`

`text(Width)^2`	`=1.8+0.8 xx area`
	`=1.8+(0.8×120)`
	`=97.8`
`:.\ text(Width)`	`=\sqrt97.8`
	`=9.889…\ text(cm)`

`=> B`

CORE, FUR1 2013 VCAA 7 MC

For a city, the correlation coefficient between

population density and distance from the centre of the city is `r` = – 0.563
house size and distance from the centre of the city is `r` = 0.357.

Given this information, which one of the following statements is true?

Around 31.7% of the variation observed in house size in the city can be explained by the variation in distance from the centre of the city.
Population density tends to increase as the distance from the centre of the city increases.
House sizes tend to be larger as the distance from the centre of the city decreases.
The slope of a least squares regression line relating population density to distance from the centre of the city is positive.
Population density is more strongly associated with distance from the centre of the city than is house size.

Show Answers Only

`E`

Show Worked Solution

`text(Population density)`

♦ Mean mark 49%.

`r^2=(-0.563)^2=0.3169…~~31.7%`

`text(House size)`

`r^2=(0.357)^2 = 0.1274…~~12.7%`

`text(In A, 12.7% of the variability is explained. Not true.)`

`text(In B,)\ r = –0.563. text(The negative correlation means population)`

`text(density increases as distance decreases. Not true.)`

`text(In C, statement assumes a negative correlation which is not the case)`

`text(given)\ r = 0.357.\ \ text(Not true.)`

`text(In D, negative correlation would have a negatively slope. Not true.)`

`text{In E, true since}\ \ | \ r^2\ |\ \ text{for house size is greater.}`

`=>\ E`

Quadratic, EXT1 2008 HSC 4c

The points `P(2ap, ap^2)`, `Q(2aq, aq^2)` lie on the parabola `x^2 = 4ay`. The tangents to the parabola at `P` and `Q` intersect at `T`. The chord `QO` produced meets `PT` at `K`, and `/_PKQ` is a right angle.

Find the gradient of `QO`, and hence show that `pq = –2`. (2 marks)
The chord `PO` produced meets `QT` at `L`. Show that `/_PLQ` is a right angle. (1 mark)
Let `M` be the midpoint of the chord `PQ`. By considering the quadrilateral `PQLK`, or otherwise, show that `MK = ML`. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `Q(2aq, aq^2),\ \ \ \ O(0,0)`

`m_(QO)`	`= (y_2 – y_1)/(x_2 – x_1)`
	`= (aq^2 – 0)/(2aq – 0)`
	`= q/2`

`text(Gradient of tangent at)\ P = p`

`text(We know the extension of chord)\ \ QO _|_ PT\ \ \ text{(given)}`

`m_(QO) xx m_(PT)`	`= -1`
`q/2 xx p`	`= -1`
`:. pq`	`= -2\ \ \ text(… as required.)`

(ii) `m_(PO) = (ap^2-0)/(2ap-0) = p/2`

`m\ text(of tangent)\ QT = q`

`m_(PO) xx m_(QT)`	`= p/2 xx q`
	`= (pq)/2`
	`= (-2)/2\ \ \ text{(from (i))}`
	`= -1`

`:.\ /_PLQ\ text(is a right angle.)`

(iii) `PQ\ text(subtends 2 right-angles at)\ L\ text(and)\ K`

`=>\ PQLK\ text(is a cyclic quad, with)\ \ PQ`

`text(a diameter.)`

`=> text(Midpoint of)\ PQ=M\ \ \ text{(centre of the circle)}`

`:.\ MK = ML\ \ text{(radii)}`

Combinatorics, EXT1 A1 2008 HSC 4b

Barbara and John and six other people go through a doorway one at a time.

In how many ways can the eight people go through the doorway if John goes through the doorway after Barbara with no-one in between? (1 mark)

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Find the number of ways in which the eight people can go through the doorway if John goes through the doorway after Barbara. (1 mark)

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Show Answers Only

`5040`
`20\ 160`

Show Worked Solution

i. `text(Barbara and John can be treated as one person)`

`:.\ text(# Combinations)`	`= 7!`
	`= 5040`

ii. `text(Solution 1)`

`text(Total possible combinations)=8!`

`text(Barbara has an equal chance of being behind as)`

`text(being in front of John.)`

`:.\ text{# Combinations (John before Barbara)}`

`=(8!)/2`

`=20\ 160`

`text(Solution 2)`

`text(If)\ B\ text(goes first,)\ J\ text(has 7 options and the other)`

`text(6 people can go in any order.)`

`=> text(# Combinations) = 7 xx 6!`

`text(If)\ B\ text(goes second,)\ J\ text(has 6 options)`

`=> text(# Combinations) = 6 xx 6!`

`text(If)\ B\ text(goes third,)\ J\ text(has 5 options)`

`=> text(# Combinations) = 5 xx 6!`

`text(And so on…)`

`:.\ text(Total Combinations)`

`= 7 xx 6! + 6 xx 6! + … + 1 xx 6!`

`= 6! ( 7 + 6 + 5 + 4 + 3 + 2 + 1)`

`= 6! ( 28)`

`= 20\ 160`

Calculus, EXT1 C1 2008 HSC 3c

A race car is travelling on the `x`-axis from `P` to `Q` at a constant velocity, `v`.

A spectator is at `A` which is directly opposite `O`, and `OA = l` metres. When the car is at `C`, its displacement from `O` is `x` metres and `/_OAC = theta`, with `- pi/2 < theta < pi/2`.

Show that `(d theta)/(dt) = (vl)/(l^2 + x^2)`. (2 marks)

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Let `m` be the maximum value of `(d theta)/dt`.

Find the value of `m` in terms of `v` and `l`. (1 mark)

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There are two values of `theta` for which `(d theta)/(dt) = m/4`.

Find these two values of `theta`. (2 marks)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`v/l`
`+- pi/3`

Show Worked Solution

i. `text(Show)\ \ (d theta)/(dt) = (vl)/(l^2 + x^2)`

`(d theta)/(dt)`	`= (d theta)/(dx) * (dx)/(dt)\ \ \ \ \ \ \ …\ (1)`
`v`	`= (dx)/(dt)\ \ \ text{(given)}`

`text(Find)\ \ (d theta)/(dx),`

MARKER’S COMMENT: The most successful students used `theta=tan^-1(x/l)` and the chain rule to find `(d theta)/(dt)`.

`tan theta`	`= x/l`
`theta`	`= tan^-1 (x/l)`
`(d theta)/(dx)`	`= l/(l^2+x^2)`

`text{Substituting into (1),}`
`(d theta)/(dt)`	`= l/(l^2+x^2) * v`
	`= (vl)/(l^2 + x^2)\ \ \ text(… as required)`

ii. `(d theta)/(dt)\ \ text(is a MAX when)\ \ x = 0`

`:. m`	`= (vl)/(l^2 + 0^2)`
	`= v/l`

iii. `text(Find)\ \ theta\ \ text(when)\ \ (d theta)/(dt) = m/4`

MARKER’S COMMENT: Many students lost marks by not giving their answer in the specified range. Be careful!

`=> (d theta)/(dt)`	`= v/(4l)`
`(vl)/(l^2 +x^2)`	`= v/(4l)`
`l^2 + x^2`	`= 4l^2`
`x^2`	`= 3l^2`
`x`	`= +- sqrt3 l`

`tan theta`	`= x/l`
	`= +- sqrt 3`
`:. theta`	`= +- pi/3,\ \ \ \ \ (- pi/2 < theta < pi/2)`

Proof, EXT1 P1 2008 HSC 3b

Use mathematical induction to prove that, for integers `n >= 1`,

`1 xx 3 + 2 xx 4 + 3 xx 5 + ... + n(n+2) = n/6 (n + 1)(2n + 7)`. (3 marks)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ 1 xx 3 + 2 xx 4 + 3 xx 5 + … + n(n+2)`

`= n/6 (n+1)(2n + 7)\ \ \ text(for)\ n >= 1`

`text(If)\ \ n = 1`

`text(LHS) = 1 xx 3 = 3`

`text(RHS) = 1/6 (2)(9) = 3 = text(LHS)`

`:.\ text(True for)\ \ n = 1`

`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ 1 xx 3 + 2 xx 4 + … + k (k + 2)`

`= k/6 (k + 1)(2k + 7)`

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ 1 xx 3 + 2 xx 4 + … + k(k+2) + (k + 1)(k + 3)`

`= ((k+1))/6 (k + 2)(2k + 9)`

`text(LHS)`	`= k/6 (k + 1)(2k + 7) + (k + 1)(k + 3)`
	`= 1/6 (k + 1)[k(2k + 7) + 6(k + 3)]`
	`= 1/6 (k + 1)[2k^2 + 7k + 6k + 18]`
	`= 1/6 (k + 1)[2k^2 + 13k + 18]`
	`= 1/6 (k +1)(k + 2)(2k + 9)\ \ \ \ text(… as required)`

`=>\ text(True for)\ n = k + 1`

`:.\ text(S) text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Functions, EXT1 F1 2008 HSC 3a

Sketch the graph of `y = |\ 2x - 1\ |`. (1 mark)

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Hence, or otherwise, solve `|\ 2x - 1\ | <= |\ x - 3\ |`. (3 marks)

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Show Answers Only

`-2 <= x <= 4/3`

Show Worked Solution

ii. `text(Solving for)\ \ |\ 2x – 1\ | <= |\ x – 3\ |`

`text(Graph shows the statement is TRUE)`

`text(between the points of intersection.)`

`=>\ text(Intersection occurs when)`

`(2x – 1)`	`= (x – 3)\ \ \ text(or)\ \ \ `	`-(2x – 1)`	`= x – 3`
`x`	`= -2`	`-2x + 1`	`= x – 3`
		`-3x`	`= -4`
		`x`	`= 4/3`

`:.\ text(Solution is)\ \ {x: -2 <= x <= 4/3}`

Measurement, STD2 M7 SM-Bank 3

Bronwyn needs to have 3.0 litres of intravenous liquid given to her over a period of 4 hours.

What is the required flow rate in mL per minute? (2 marks)

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Show Answers Only

`12.5\ text(mL/minute)`

Show Worked Solution

`text(Total liquid)`	`= 3.0 xx 1000`
	`= 3000\ text(mL)`

`text(Total minutes)`	`= 4 xx 60`
	`= 240`

`:.\ text(Flow Rate)`	`= 3000/240`
	`= 12.5\ text(mL/minute)`

Measurement, STD2 M7 SM-Bank 2

A medication is available in both tablet and liquid form. A tablet contains 50 mg of the active ingredient while the liquid form contains 60 mg per 10 mL.

Michael likes taking tablets and Georgia prefers liquid medicines. If they each need 0.2 g of the active ingredient, what dosages do they take? (3 marks)

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Show Answers Only

`text(Michael needs to take 4 tablets)`

`text(and Georgia needs to take 33.3mL.)`

Show Worked Solution

`text(Michael – tablets)`

`0.2\ text(g) = 200\ text(mg)`

`:.\ text(# Tablets)`	`= 200/50`
	`= 4`

`text(Georgia – liquid)`

`60\ text(mg in)\ 10\ text(mL)`

`=> 1\ text(mg) = 10/60 = 0.166…\ text(mL)`

`=> 200\ text(mg)`	`= 200 xx 0.166…`
	`= 33.33…\ text(mL)`
	`= 33.3\ text(mL)\ \ text{(to 1 d.p.)}`

`:.\ text(Michael needs to take 4 tablets and Georgia)`

`text(needs to take 33.3 mL.)`

Financial Maths, STD2 F5 SM-Bank 4

Dominique wants to save $15 000 to use as spending money when she travels overseas in 2 years' time.

If she invests $3500 at the end of every 6 months into an account earning 4% p.a., compounded half-yearly, will she have enough?

Use the table below to justify your answer. (2 marks)

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`text(Dominique will not have saved)`

`text(enough to reach)\ $15\ 000.`

Show Worked Solution

`text(Interest rate)\ text{(6 monthly)} = text(4%) -: 2 = text(2%)`

`n = 4\ \ \ text{(6 month periods in 2 years)}`

`=>\ text(FVA factor) = 4.122\ \ \ text{(from Table)}`

`text(FVA)`	`= 3500 xx 4.122`
	`= $14\ 427`

`:.\ text(Dominique will not have saved enough to)`

`text(reach)\ $15\ 000.`

Financial Maths, STD2 F5 SM-Bank 3

Camilla buys a car for $21 000 and repays it over 4 years through equal monthly instalments.

She pays a 10% deposit and interest is charged at 9% p.a. on the reducing balance loan.

Using the Table of present value interest factors below, where `r` represents the monthly interest and `N` represents the number of repayments

Calculate the monthly repayment, `$P`, that Camilla must pay to complete the loan after 4 years (to the nearest $). (3 marks)

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Calculate the total interest paid over the life of the loan. (1 mark)

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Show Answers Only

`text(Camilla must repay $470 per month)`
`$3660`

Show Worked Solution

i. `text(Deposit) = 10text(%) xx 21\ 000 = 2100`

`text(Loan Value)`	`= 21\ 000 – 2100`
	`= 18\ 900`

`text(Monthly interest rate) = text(9%)/12 = 0.0075`

`text(# Repayments) = 4 xx 12 = 48`

`=>\ text(PVA Factor) = 40.18478\ \ text{(from Table)}`

`text(Monthly repayment)\ ($P)`	`= (18\ 900)/(40.18478)`
	`= 470.32…`
	`= 470\ text{(nearest $)}`

`:.\ text(Camilla must repay $470 per month.)`

ii. `text(Total Repayments)`

`= 48 xx 470`

`= $22\ 560`

`:.\ text(Interest paid over loan)`

`= 22\ 560 – 18\ 900`

`= $3660`

Functions, EXT1 F2 2008 HSC 2c

The polynomial `p(x)` is given by `p(x) = ax^3 + 16x^2 + cx - 120`, where `a` and `c` are constants.

The three zeros of `p(x)` are `– 2`, `3` and `beta`.

Find the value of `beta`. (3 marks)

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Show Answers Only

`- 5`

Show Worked Solution

`p(x) = ax^3 + 16x^2 + cx – 120`

`text(Roots:)\ \ – 2, \ 3, \ beta`

`-2 + 3 + beta`	`= -B/A`
`beta + 1`	`= -16/a`
`beta`	`= -16/a – 1\ \ \ \ \ …\ (1)`

`-2 xx 3 xx beta`	`= -D/A`
`-6 beta`	`= 120/a`
`beta`	`= -20/a\ \ \ \ \ …\ (2)`

MARKER’S COMMENT: Many students displayed significant inefficiencies in solving simultaneous equations.

`- 16/a – 1`	`= -20/beta`
`-16 – a`	`= -20`
`a`	`= 4`

`text(Substitute)\ \ a = 4\ \ text(into)\ (1)`

`:. beta`	`= – 16/4 – 1`
	`= -5`

Mechanics, EXT2* M1 2008 HSC 2b

A particle moves on the `x`-axis with velocity `v`. The particle is initially at rest at `x = 1`. Its acceleration is given by `ddot x = x + 4`.

Using the fact that `ddot x = d/dx (1/2 v^2)`, find the speed of the particle at `x = 2`. (3 marks)

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Show Answers Only

`sqrt11`

Show Worked Solution

`ddot x`

`= d/dx (1/2 v^2)=x+4`

`1/2 v^2`	`= int ddot x\ dx`
	`= int x + 4\ dx`
	`= 1/2 x^2 + 4x + c`

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

`0`	`= 1/2 + 4 + c`
`c`	`= -4\ ½`

`:.\ 1/2 v^2`	`= 1/2 x^2 + 4x – 4 1/2`
`v^2`	`= x^2 + 8x – 9`

`text(When)\ \ x = 2`

`v^2`	`= 2^2 + 8 * 2 – 9`
	`= 11`
`v`	`= +- sqrt 11`

`:.\ text(When)\ x = 2,\ \ text(Speed) = sqrt 11`

Calculus, EXT1 C2 2008 HSC 2a

Use the substitution `u = log_e x` to evaluate `int_e^(e^2) 1/(x (log_e x)^2)\ dx`. (3 marks)

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Show Answers Only

`1/2`

Show Worked Solution

`u`	`= log_e x`
`(du)/(dx)`	`= 1/x`
`:. du`	`= 1/x\ dx`

`text(When)\ \ \ `	`x = e^2,\ \ `	`u = log_e e^2 = 2`
	`x = e,`	`u = 1`

`:. int_e^(e^2) 1/(x (log_e x)^2)\ dx`

`= int_1^2 1/(u^2)\ du`

WARNING: Most errors were made in the last stage of substitution in this question. Be careful!

`= int_1^2 u^(-2)\ du`

`= [-1/u]_1^2`

`= [(-1/2) – (-1)]`

`= 1/2`

L&E, EXT1 2008 HSC 1f

Let `f(x) = log_e [(x - 3)(5 - x)]`.

What is the domain of `f(x)`? (2 marks)

Show Answers Only

`text(Domain)\ {x:\ 3<x<5}`

Show Worked Solution

`f(x) = log_e [(x – 3)(5 – x)]`

`text(Find the domain)`

`(x – 3)(5 – x)>0`

`text(If)\ \ x = 4,`

`(4 – 3)(5 – 4) = 1 > 0`

`:.\ text(Domain)\ {x:\ 3<x<5}`

Trig Calculus, EXT1 2008 HSC 1e

Evaluate `int_0^(pi/4) cos theta sin^2 theta\ d theta`. (2 marks)

Show Answers Only

`sqrt2/12`

Show Worked Solution

`int_0^(pi/4) cos theta sin^2 theta\ d theta`

`= [1/3 sin^3 theta]_0^(pi/4)`

`= 1/3 (sin (pi/4))^3 – 1/3 (sin 0)^3`

`= 1/3 * (1/sqrt2)^3`

`= 1/3 * 1/(2 sqrt2)`

`= 1/(6 sqrt 2) xx sqrt2/sqrt2`

`= sqrt2/12`

`x^2`	`= 16^2 + 12^2`
	`= 400`
`:. x`	`= 20`

`d^2`	`= 8^2 + 20^2`
	`= 464`
`∴ d`	`= 21.54…\ text(cm)`

`∴ t_5 – t_4`	`= 2p + 3q – (p + 2q)`
	`= p + q`