Band 3 – Page 4

Functions, 2ADV EQ-Bank 12

The braking distance of a car, in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation

`text{braking distance}\ = k xx text{(speed)}^2`

where `k` is the constant of variation.

The braking distance for a car travelling at 50 km/h is 20 m.

Find the value of `k`. (2 marks)

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What is the braking distance when the speed of the car is 90 km/h? (1 mark)

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a. `k=0.008`

b. `64.8\ text{m}`

Show Worked Solution

a. `text{braking distance}\ = k xx text{(speed)}^2`

`20`	`=k xx 50^2`
`k`	`=20/50^2=0.008`

b. `text{Find}\ d\ text{when speed = 90 km/h:}`

`d=0.008 xx 90^2=64.8\ text{m}`

Functions, 2ADV EQ-Bank 13

Energy $(E)$ stored in a spring, measured in joules, varies directly with the square of its compression distance $d$, measured in centimetres.

When a spring is compressed by 4 cm, it stores 48 joules of energy.

How much energy is stored when the spring is compressed by 7 cm? (3 marks)

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$\text{147 joules}$

Show Worked Solution

$E \propto d^2 \ \Rightarrow \ E=k d^2$

$\text{Find \(k$ given $\ E=48 \ $ when $\ d=4$:}\)

$48$	$=k \times 4^2$
$k$	$=3$

$\text{Find \(E$ when $\ d=7$:}\)

$E=3 \times 7^2=147 \ \text{joules}$

Functions, 2ADV EQ-Bank 16

Given the functions $f(x)=x-2$ and $g(x)=x^2$, determine the equation of $f(g(x))$ and sketch its graph showing all intercepts. (2 marks)

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Evaluate $g(f(-1))$. (1 mark)

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b. $g(f(-1))=9$

Show Worked Solution

a. $f(x)=x-2, \ g(x)=x^2$

$f(g(x))=x^2-2$

$\text{Intercepts:} \ \ x^2-2=0 \ \Rightarrow \ x= \pm \sqrt{2}$

b. $f(-1)=-1-2=-3$

$g(f(-1))=g(-3)=9$

Functions, 2ADV EQ-Bank 13

Without calculus, sketch the graph of $g(x)$, where $g(x)=-2-\dfrac{3}{x}$
In your answer label all asymptotes and any intercepts. (3 marks)

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Express the range of $g(x)$ in set notation. (1 mark)

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b. $\text{Range} \ \ y \in(-\infty,-2) \cup(-2, \infty)$

Show Worked Solution

a. $\text{As} \ \ x \rightarrow \infty, g(x) \rightarrow-2^{-}; \text{As} \ \ x \rightarrow-\infty, g(x) \rightarrow-2^{+}$

$\text{As} \ \ x \rightarrow 0^{+}, g(x) \rightarrow-\infty; \text{As} \ \ x \rightarrow 0^{-}, g(x) \rightarrow \infty$

$\text{Intercept} \ (x \text {-axis}):-2-\dfrac{3}{x}=0 \ \Rightarrow \ 2 x=-3 \ \Rightarrow \ x=-\dfrac{3}{2}$

b. $\text{Range} \ \ y \in(-\infty,-2) \cup(-2, \infty)$

Financial Maths, STD2 EQ-Bank 22

Mei is purchasing a new car and has a choice between two finance packages.

Package A: Deposit of $5000, monthly repayments of $1150 for 4 years.

Package B: No deposit, monthly repayments of $1280 for 5 years.

Determine the total cost of Package A. (2 marks)
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Determine the total cost of Package B. (2 marks)
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How much will Mei save by selecting the cheaper package? (1 mark)
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a. $$60\,200$

b. $$76\,800$

c. $$16\,600$

Show Worked Solution

a. $\text{Total repayments} (A)=1150\times 12\times 4=$55\,200$

$\text{Total cost of Package A}=5000+55\,200=$60\,200$

b. $\text{Total repayments (B)}=1280\times 12\times 5=$76\,800$

$\text{Total cost of Package B}=$76\,800\ \text{(No deposit)}$

c. $\text{Savings using Package A}=76\,800-60\,200=$16\,600$

Financial Maths, STD2 EQ-Bank 20

A motorcycle is for sale at $16 500. Finance is available with a $3200 deposit and monthly repayments of $420 for 3 years.

What is the total cost of the repayments? (1 mark)
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How much will the motorcycle cost if Daniel uses the finance package? (1 mark)
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What is the interest paid? (1 mark)
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a. $$15\,120$

b. $$18\,320$

c. $$1820$

Show Worked Solution

a. $\text{Number of months}=3\times 12=36\ \text{months}$

$\text{Total repayments}=36\times 420=$15\,120$

b.	$\text{Total cost}$	$=\text{Deposit}+\text{Total repayments}$
		$=$3200+$15\,120=$18\,320$

c.	$\text{Interest paid}$	$=\text{Total cost}-\text{Original cost}$
		$=$18\,320-$16\,500=$1820$

Financial Maths, STD2 EQ-Bank 18

Rachel is purchasing a new refrigerator priced at $3200. The store offers finance terms of 30% deposit and repayments of $65 per week for 40 weeks.

What is the amount of the deposit? (1 mark)
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Find the total cost of the repayments. (1 mark)
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What is the total cost of purchasing the refrigerator using the finance package? (1 mark)
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a. $$960$

b. $$2600$

c. $$3560$

Show Worked Solution

a. $\text{Deposit}=\dfrac{30}{100}\times 3200=$960$

b. $\text{Total repayments}=65\times 40=$2600$

c.	$\text{Total cost}$	$=\text{Deposit}+\text{Total repayments}$
		$=960+2600=$3560$

Financial Maths, STD2 EQ-Bank 19

Ben is purchasing a used van with a sale price of $32 600. He has arranged finance with weekly repayments of $280 for 3 years.

Calculate the total amount of the repayments. (2 marks)

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$$43\,680$

Show Worked Solution

$\text{Number of weeks in 3 years}=3\times 52=156\ \text{weeks}$

$\text{Total repayments}=280\times 156=$43\,680$

Financial Maths, STD2 EQ-Bank 4 MC

David is purchasing a motorcycle for sale at $15 600. The finance terms are weekly repayments of $180 for 2 years.

What is the total amount of the repayments?

$4320
$8640
$18 720
$19 920

Show Answers Only

$C$

Show Worked Solution

$\text{Number of weeks in 2 years:}=2\times 52=104$

$\text{Total repayments}=104\times $180=$18\,720$

$\Rightarrow C$

Functions, 2ADV EQ-Bank 15

Show the equation of $AC$ is $3 x+4 y-16=0$. ( 2 marks)

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Show that $BC$ is not perpendicular to $AC$. (2 marks)

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a. $\text{Proof (See Worked Solutions)}$

b. $\text{Proof (See Worked Solutions)}$

Show Worked Solution

a. $A(0,4), C(4,1)$

$m_{AC}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-4}{4-0}=-\dfrac{3}{4}$

$\text{Equation of line with} \ \ m=-\dfrac{3}{4} \ \ \text{through}\ (0,4):$

$y-y_1$	$=m\left(x-x_1\right)$
$y-4$	$=-\dfrac{3}{4}(x-0)$
$y-4$	$=-\dfrac{3}{4} x$
$4 y-16$	$=-3 x$
$3 x+4 y-16$	$=0$

b. $B(3,0), C(4,1)$

$m_{BC}=\dfrac{1-0}{4-3}=1$

$m_{AC} \times m_{BC}=-\dfrac{4}{3} \times 1=-\dfrac{4}{3} \neq-1$

$\therefore AC \ \text{is not} \ \perp \text{to} \ BC.$

Functions, 2ADV EQ-Bank 15

Express $\dfrac{m^4\, n^{-3}}{m^{-2}\, n^2}$ with positive indices only. (1 mark)

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$\dfrac{m^6}{n^5}$

Show Worked Solution

$\dfrac{m^4\, n^{-3}}{m^{-2}\, n^2}=m^{(4+2)}\, n^{(-3-2)}=m^6\, n^{-5}=\dfrac{m^6}{n^5}$

Functions, 2ADV EQ-Bank 14

Simplify $8ab^2 \times 2a^{-2}$. (2 marks)

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$\dfrac{16b^2}{a}$

Show Worked Solution

$8ab^2 \times 2a^{-2}=8ab^2 \times \dfrac{2}{a^2}=\dfrac{16b^2}{a}$

Functions, 2ADV EQ-Bank 21

Simplify $x^5 \ ÷ \ x^{-3}$. (1 mark)

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$x^{8}$

Show Worked Solution

$x^5 \ ÷ \ x^{-3}=x^{(5-(-3))}=x^8$

Financial Maths, STD2 EQ-Bank 14

George has constructed a yearly budget shown below.

Income		Expenses
Salary	$68 450.00	Groceries	$7280.00
Interest	$825.60	Clothing	$3960.00
		Council Rates	$1740.00
		Electricity	$1890.00
		Entertainment	$4320.00
		Insurance	$2650.00
		Loan repayments	$15 840.00
		Motor vehicle costs	$3180.00
		Telephone	$960.00
		Work related costs	$1250.00
		Balance
Total		Total

Calculate George's total income and total expenses. (2 marks)
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Using the information from part (a), balance the budget. (1 mark)
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What is the amount of George's weekly savings? Give your answer to the nearest dollar. (1 mark)
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a. $\text{Income}=$69\,275.60,\ \text{Expenses}=$43\,070.00$

b. $\text{Balance}=$26\,205.60$

c. $\text{Savings}=$504\ \text{(nearest dollar)}$

Show Worked Solution

a. $\text{Income}=68\,450.00+825.60=$69\,275.60$

$\text{Expenses}$

$=7280.00+3960.00+1740.00+1890.00+4320.00$

$+2650.00+15840.00+3180.00+960.00+1250.00$

$=$43\,070.00$

b.	$\text{Balance}$	$=\text{Income}-\text{Expenses}$
		$=$69\,275.60-$43\,070.00=$26\,205.60$

c. $\text{Weekly savings}=\dfrac{$26\,205.60}{52}=$503.953…\ =$504\ \text{(nearest dollar)}$

Financial Maths, STD2 EQ-Bank 18

Michael has budgeted $185 per week for groceries, $95 per week for leisure, $38 per fortnight for medical expenses and $115 per week to run a car.

Calculate Michael's monthly expenses. Assume 4 weeks in a month. (2 marks)

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$$1656$

Show Worked Solution

$\text{Calculate weekly expenses:}$

$\text{Groceries} =$185 \text{per week}$

$\text{Leisure} =$95\ \text{per week}$

$\text{Medical} =\dfrac{38}{2}= \$19\ \text{per week}$

$\text{Car} =$115\ \text{per week}$

$\text{Total weekly expenses}=185+95+19+115=$414\ \text{per week}$

$\therefore\ \text{Monthly expenses (assuming 4 weeks in a month)}=414\times 4=$1656$

Financial Maths, STD2 EQ-Bank 3 MC

Sophie has the following bills: internet $90 per month, electricity $320 per quarter, insurance $840 per year and rent $280 per week.

What is the total amount Sophie should budget for the year?

$1530
$16 640
$17 760
$18 200

Show Answers Only

$C$

Show Worked Solution

$\text{Internet} =90 \times 12 = \$1080\ \text{per year}$

$\text{Electricity} =320 \times 4 = \$1280\ \text{per year}$

$\text{Insurance} =$840\ \text{per year}$

$\text{Rent} =280 \times 52 = \$14\,560\ \text{per year}$

$\therefore\ \text{Total yearly expenses}=1080+1280+840+14\,560=$17\,760$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 20

Maria drove 420 km in 6 hours. Her average speed for the first 280 km was 80 km per hour.

How long did she take to travel the last 140 km? (2 marks)

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$\text{2.5 hours}$

Show Worked Solution

$\text{First 280 km:}$

$t=\dfrac{\text{distance}}{\text{speed}}=\dfrac{280}{80}= 3.5\ \text{hours}$

$\text{Total time}=6\ \text{hours}$

$\text{Time for first 280 km}=3.5\ \text{hours}$

$\text{Time taken for last 140 km}=6-3.5=\text{2.5 hour}$

Algebra, STD2 EQ-Bank 18

The distance between Newcastle and Sydney is 165 km. A person travels from Newcastle to Sydney at an average speed of 110 km/h.

How long does it take the person to complete the journey in hours and minutes? (2 marks)

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$\text{1 hour 30 minutes}$

Show Worked Solution

$\text{Using}\ \ d=s \times t\ \ \Rightarrow\ \ t=\dfrac{d}{s}:$

$t=\dfrac{165}{110}=1.5\ \text{hours}\ =\ \text{1 hour 30 minutes}$

Measurement, STD2 EQ-Bank 19

A bus departs Sydney at 8:00 am on Monday for Perth. The bus makes five stops on the journey whose durations are given below:

Dubbo: 45 minutes
Broken Hill: 1 hour 15 minutes
Port Augusta: 50 minutes
Ceduna: 40 minutes
Norseman: 1 hour 10 minutes

The total driving time (excluding stops) is 52 hours and 30 minutes. The bus arrives in Perth at 3:25 pm on Thursday.

Calculate the total time for the journey from Sydney to Perth, including all stops. Give your answer in hours and minutes. (2 marks)
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Convert your answer from part (a) to days, hours and minutes, and hence state the day and time the bus arrives in Perth. Ignore time zones in your calculations. (2 marks)
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a. $\text{57 hours 10 minutes}$

b. $\text{Bus arrives on Wednesday at 5:10 pm with the journey}$

$\text{taking 2 days 9 hours 10 minutes.}$

Show Worked Solution

a. $\text{Total stopping time}$

$=45+75+50+40+70$

$= 280 \text{ minutes}= 4\ \text{hours}\ 40 \text{ minutes} $

$\text{Total journey}$

$=52\ \text{hours}\ 30 \text{ minutes}+4\ \text{hours}\ 40 \text{ minutes} $

$=57\ \text{hours}\ 10 \text{ minutes} $

b. $\text{Convert 57 hours 10 minutes to days, hours and minutes:}$

$57\ \text{hours}=(2 \times 24) + 9=2\ \text{days}\ 9\ \text{hours}$

$57\ \text{ hours 10 minutes}=2\ \text{days}\ 9\ \text{hours}\ 10\ \text{minutes}$

$\text{Bus arrival in Perth:}$

$\rightarrow\ \text{Bus departs 8 am Monday}$

$\rightarrow\ \text{Monday 8:00 am + 2 days = Wednesday 8:00 am}$

$\rightarrow\ \text{Wednesday 8:00 am + 9 hours 10 minutes = Wednesday 5:10 pm}$

$\therefore\ \text{Bus arrives Wednesday at 5:10 pm.}$

Measurement, STD2 EQ-Bank 19

Bangkok is located at $ (14^{\circ}\text{N}, 100^{\circ}\text{E}) $ and Montreal is located at $ (46^{\circ}\text{N}, 74^{\circ}\text{W}) $.

Identify which city is closer to the Equator, giving reasons. (1 mark)
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What is the difference in latitude between Bangkok and Montreal? (1 mark)
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What is the difference in longitude between Bangkok and Montreal? (1 mark)
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a. $\text{Bangkok}$

b. $32^{\circ}$

c. $174^{\circ}$

Show Worked Solution

a. $\text{Bangkok is at latitude }14^{\circ}\text{N} $

$\text{Montreal is at latitude }46^{\circ}\text{N} $

$\text{Since 14° < 46°, Bangkok is closer to the Equator.}$

b. $\text{Latitude of Bangkok}=14^{\circ}\text{N}$

$\text{Latitude of Montreal}=46^{\circ}\text{N}$

$\text{Since cities are both in the northern hemisphere, subtract the latitudes.}$

$\text{Difference in latitude}=46^{\circ}-14^{\circ}=32^{\circ} $

c. $\text{Longitude of Bangkok}=100^{\circ}\text{E}$

$\text{Longitude of Montreal}=74^{\circ}\text{W}$

$\text{Since cities on opposite sides of the Prime Meridian, add the longitudes.}$

$\text{Difference in longitude}=100^{\circ}+74^{\circ}=174^{\circ} $

Measurement, STD2 EQ-Bank 18

Tokyo is located at $ (36^{\circ}\text{N}, 140^{\circ}\text{E}) $ and Lima is located at $ (12^{\circ}\text{S}, 77^{\circ}\text{W}) $.

Explain which city is closer to the Prime Meridian? (1 mark)
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What is the difference in latitude between Tokyo and Lima? (1 mark)
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What is the difference in longitude between Tokyo and Lima? (1 mark)
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a. $\text{Lima}$

b. $48^{\circ}$

c. $217^{\circ}$

Show Worked Solution

a. $\text{Tokyo is at longitude }140^{\circ}\text{E} $

$\text{Lima is at longitude }77^{\circ}\text{W} $

$\text{Since 77° < 140°, Lima is closer to the Prime Meridian.}$

b. $\text{Latitude of Tokyo}=36^{\circ}\text{N}$

$\text{Latitude of Lima}=12^{\circ}\text{S}$

$\text{Since cities on opposite sides of the equator, add the latitudes.}$

$\text{Difference in latitude}=36^{\circ}+12^{\circ}=48^{\circ} $

c. $\text{Longitude of Tokyo}=140^{\circ}\text{E}$

$\text{Longitude of Lima}=77^{\circ}\text{W}$

$\text{Since cities on opposite sides of the Prime Meridian, add the longitudes.}$

$\text{Difference in longitude}=140^{\circ}+77^{\circ}=217^{\circ} $

Measurement, STD2 EQ-Bank 3 MC

The coordinates of Cairo are $ (30^{\circ}\text{N}, 31^{\circ}\text{E}) $.

What are the coordinates of Athens if it is 23° west of Cairo and on the same line of longitude?

$ (30^{\circ}\text{N}, 54^{\circ}\text{E}) $
$ (7^{\circ}\text{N}, 31^{\circ}\text{E}) $
$ (53^{\circ}\text{N}, 31^{\circ}\text{E}) $
$ (30^{\circ}\text{N}, 8^{\circ}\text{E}) $

Show Answers Only

$D$

Show Worked Solution

$\text{Athens is 23° west of Cairo.}$

$\text{Latitude of Athens}=30^{\circ}\text{N}$

$\text{Longitude of Athens}=31^{\circ}-23^{\circ}=8^{\circ}\text{E}$

$\therefore\ \text{Coordinates of Athens are:}\ (30^{\circ}\text{N}, 8^{\circ}\text{E})$

$\Rightarrow D$

Measurement, STD2 EQ-Bank 4 MC

The coordinates of Jakarta are $ (6^{\circ}\text{S}, 107^{\circ}\text{E}) $.

Darwin and Jakarta share the same longitude but Darwin is $ 18^{\circ} $ south of Jakarta. What are the coordinates of Darwin?

$ (6^{\circ}\text{S}, 125^{\circ}\text{E}) $
$ (6^{\circ}\text{S}, 89^{\circ}\text{E}) $
$ (24^{\circ}\text{S}, 107^{\circ}\text{E}) $
$ (12^{\circ}\text{N}, 107^{\circ}\text{E}) $

Show Answers Only

$C$

Show Worked Solution

$\text{Darwin is }18^{\circ}\text{ south of Jakarta.}$

$\text{Latitude of Darwin}=6^{\circ}+18^{\circ}=24^{\circ}\text{S}$

$\text{Longitude of Darwin}=107^{\circ}\text{E}$

$\therefore\ \text{Coordinates of Darwin are:}\ (24^{\circ}\text{S}, 107^{\circ}\text{E})$

$\Rightarrow C$

Measurement, STD2 EQ-Bank 2 MC

Tokyo is 45$^{\circ}$ west of Sydney. Using longitudinal difference, what is the time in Tokyo when it is 3:00 pm in Sydney?

6:00 am
9:00 am
6:00 pm
12:00 pm

Show Answers Only

$D$

Show Worked Solution

$15^{\circ}\ =\text{1 hour time difference}$

$\text{Longitudinal distance}=45^{\circ}$

$\text{Time Difference}=\dfrac{45}{15}=3\ \text{hours}$

$\text{Time in Sydney}\ =\ 3:00\ \text{pm}$

$\text{Since Sydney is East of Tokyo:}$

$\text{Time in Tokyo}=\ 3:00\ \text{pm}-3\ \text{hours}=\ 12:00\ \text{pm}$

$\Rightarrow D$

Measurement, STD2 EQ-Bank 26

Use the train timetable below to answer this question.

Emma lives in Berowra and travels to Wyong for an appointment. After her appointment, she needs to attend a meeting in Gosford before returning home to Berowra.

Emma catches the train that departs Wyong at 09:16. How long does this train take to reach Gosford? (1 mark)
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Emma's meeting in Gosford starts at 10:15 am at a venue that is 7 minutes walk from Gosford station. What is the latest train she can catch from Wyong to arrive at her meeting on time? (2 marks)
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After her meeting finishes at 11:20 am, Emma walks back to Gosford station (taking 7 minutes). What is the earliest train she can catch from Gosford to return home to Berowra, and what time will she arrive home? (2 marks)
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a. $17\ \text{minutes}$

b. $09:37\ \text{train}$

c. $\text{11:34 am Gosford train, arrives at Berowra at 12:37 pm}$

Show Worked Solution

a. $\text{Elapsed time:}\ =09:33-09:16 = 17 \text{ minutes} $

b. $\text{Step 1: Meeting starts at 10:15 am}$

$\rightarrow\ 7\text{ minutes to walk from Gosford station to the venue.}$

$\rightarrow\ \text{At Gosford Station by: }\ 10:15-7 \text{ minutes} = 10:08 \text{ am}$

$\text{Step 2: Find latest train arriving at Gosford by 10:08 am}$

$\rightarrow\ \text{Arrival times: } 08:12, 08:33, 08:58, 09:33, 09:58, 10:33$

$\rightarrow\ \text{She must catch the }09:58.$

$\text{Step 3: When does 09:58 depart Wyong?}$

$\rightarrow\ 09:37\ \text{train}$

c. $\text{Step 1: Determine when Emma arrives back at Gosford station}$

$\rightarrow\ \text{Meeting finishes at }11:20 \text{ am}$

$\text{Walking time back to station: 7 minutes}$

$\text{Emma arrives at Gosford station at:}\ 11:20 + 7 \text{ minutes} = 11:27 \text{ am} $

$\text{Step 1: Find the earliest train departing Gosford after 11:27 am}$

$\rightarrow\ \text{Meeting finishes at }11:20 \text{ am}$

$\rightarrow\ \text{Relevant departure times:} …10:34, 10:59, 11:34, 11:59$

$\rightarrow\ \text{Earliest train after 11:27 am is the }11:34 \text{ am}$

$\text{Step 3: Find when this train arrives in Berowra}$

$\rightarrow\ \text{11:34 am Gosford train, arrives at Berowra at }12:37 \text{ pm}$

Measurement, STD2 EQ-Bank 4 MC

On Tuesday Sarah started work at 6:45 am and finished at 3:12 pm.

For how many hours and minutes did Sarah work on Tuesday?

8 hours 27 minutes
8 hours 33 minutes
9 hours 27 minutes
9 hours 33 minutes

Show Answers Only

$A$

Show Worked Solution

$\text{Break into steps}$

$\text{From 6:45 am to 7:00 am: }\ 60-45 = 15 \text{ minutes} $

$\text{From 7:00 am to 3:12 pm: }\ 8\ \text{hours}\ 12 \text{ minutes} $

$\text{Total: }\ 8\ \text{hours}+12 \text{ minutes} +15 \text{ minutes}=8\ \text{hours}\ 27\ \text{minutes}$

$\Rightarrow A$

Measurement, STD2 EQ-Bank 3 MC

A bus departs at 8:23 am and arrives at its destination at 11:47 am.

How long was the bus journey?

3 hours 24 minutes
3 hours 34 minutes
4 hours 24 minutes
4 hours 34 minutes

Show Answers Only

$A$

Show Worked Solution

$\text{Method 1: Break into steps}$

$\rightarrow\text{From 8:23 am to 9:00 am: }\ 60-23 = 37 \text{ minutes} $

$\rightarrow\text{From 9:00 am to 11:47 am: }\ 2\ \text{hours}\ 47 \text{ minutes} $

$\rightarrow\text{Total: }\ 37 \text{ minutes} + 2\ \text{hours}+\ 47 \text{ minutes}=2\ \text{hours}\ 84\ \text{minutes}$

$=3\ \text{hours}\ 24\ \text{minutes}$

$\text{Method 2: Calculate hours and minutes separately}$

$\rightarrow\text{Hours: } \ 11-8 = 3 \text{ hours} $

$\rightarrow\text{Minutes: }\ 47-23 = 24 \text{ minutes} $

$\rightarrow\text{Total: }\ 3 \text{ hours } 24 \text{ minutes} $

$\Rightarrow B$

Measurement, STD2 EQ-Bank 18

A concert runs for 5400 seconds.

Convert this time to hours and minutes. (2 marks)
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If the concert finishes at 10:45 pm, what time did it start? (2 marks)
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a. $\text{1 hours 30 minutes}$

b. $\text{9:15 pm}$

Show Worked Solution

a. $\text{Step 1: Convert seconds to minutes}$

$5400 \text{ seconds} = \dfrac{5400}{60} \text{ minutes} = 90 \text{ minutes} $

$\text{Step 2: Convert minutes to hours and minutes}$

$90 \text{ minutes} = \dfrac{90}{60} \text{ hours} = 1.5 \text{ hours} $

$\therefore\ 90 \text{ minutes} = 1 \text{ hour } 30 \text{ minutes} $

b. $\text{Finish time: 10:45 pm}$

$\text{Concert duration: 1 hours 30 minutes}$

$10:45\ \text{pm}-1\ \text{hour}=9:45\ \text{pm}$

$9:45\ \text{pm}-30\ \text{minutes}=9:15\ \text{pm}$

$\text{Start time: 9:15 pm}$

Measurement, STD2 EQ-Bank 16

A movie runs for 8400 seconds.

Convert this time to hours and minutes. (2 marks)
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If the movie starts at 7:15 pm, what time will it finish? (1 mark)
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a. $\text{2 hours 20 minutes}$

b. $\text{9:35 pm}$

Show Worked Solution

a. $\text{Step 1: Convert seconds to minutes}$

$8400 \text{ seconds} = \dfrac{8400}{60} \text{ minutes} = 140 \text{ minutes} $

$\text{Step 2: Convert minutes to hours and minutes}$

$140 \text{ minutes} = \dfrac{140}{60} \text{ hours} = 2.333… \text{ hours} $

$\text{To find the minutes part:}\ 140 = 2 \times 60 + 20 $

$\therefore\ 140 \text{ minutes} = 2 \text{ hours } 20 \text{ minutes} $

b. $\text{Starting time: 7:15 pm}$

$\text{Movie duration: 2 hours 20 minutes}$

$\text{Finish time: 9:35 pm}$

Measurement, STD2 EQ-Bank 5 MC

Jeremy watched a movie at the cinema that began at 11:35 am. If it finished at 2:05 pm, what was the duration of the movie, in minutes?

25 minutes
125 minutes
150 minutes
250 minutes

Show Answers Only

$C$

Show Worked Solution

$\text{11:35 to 2:05 = 2.5 hours}$

$\text{To convert hours to minutes multiply by 60.}$

$ 2.5 \text{ hours} = 2.5 \times 60 \text{ minutes} = 150 \text{ minutes} $

$\Rightarrow C$

Statistics, STD2 EQ-Bank 3 MC

A dataset has a lower quartile (Q₁) of 25 and an upper quartile (Q₃₎ of 35.

Which of the following values would be classified as an outlier?

Show Answers Only

$A$

Show Worked Solution

$\text{IQR}=Q_3-Q_1$

$=35-25=10$

$\text{Lower boundary}=25-1.5\times 10=10$

$\text{Upper boundary}=35+1.5\times 10=50$

$\text{Check each value:} $

$2 < 10 \text{ (outlier)} $

$10 = 10 \text{ (on boundary, not an outlier)}$

$48 < 50 \text{ (not an outlier)} $

$50 = 50 \text{ (on boundary, not an outlier)}$

$\therefore\ 2\ \text{is the only outlier}$

$\Rightarrow A$

Statistics, STD2 EQ-Bank 1 MC

A dataset has the following values arranged in order:

12, 15, 18, 21, 24, 27, 30, 33, 36

What is the interquartile range (IQR) of this dataset?

Show Answers Only

$B$

Show Worked Solution

$\text{Number of data values: }n=9$

$\text{Median }(Q_2): =24$

$\text{Lower half of data: }12, 15, 18, 21$

$Q_1=\dfrac{15+18}{2}\ =16.5$

$\text{Upper half of data: } 27, 30, 33, 36$

$Q_3=\dfrac{30+33}{2}\ =31.5$

$\text{IQR}=Q_3-Q_1$

$=31.5-16.5=15$

$\Rightarrow C$

Statistics, STD2 EQ-Bank 2 MC

In a small business, the seven employees earn the following wages per week:

$\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970$

What is the population standard deviation of the wages, correct to 2 decimal places?

$187.25
$191.04
$201.58
$215.73

Show Answers Only

$B$

Show Worked Solution

$\text{Using a calculator to find the population standard deviation:}$

$\sigma = 191.044…= \$191.04\ \text{(to 2 d.p.)}$

$\Rightarrow B$

Functions, 2ADV 2011 HSC 3c

The diagram shows a line `l_1`, with equation `3x + 4y-12 = 0`, which intersects the `y`-axis at `B`.

A second line `l_2`, with equation `4x-3y = 0`, passes through the origin `O` and intersects `l_1` at `E`.

Show that the coordinates of `B` are `(0, 3)`. (1 mark)
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Show that `l_1` is perpendicular to `l_2`. (2 marks)
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a. `\text{See Worked Solution}`

b. `\text{See Worked Solution}`

Show Worked Solution

a. `B\ text(is)\ y text(-intercept of)\ l_1`

`text(When)\ x = 0:`

`(3 xx 0) + 4y-12=0\ \ =>\ \ y=3`

`:.\ B\ text(is)\ (0,3)`

b.	`l_1:\ \ 3x + 4y -12`	`= 0`
	`4y`	`= -3x + 12`
	`y`	`= -3/4x + 3`

`m(l_1)=-3/4`

`l_2:\ \ 4x-3y`	`= 0`
`3y`	`= 4x`
`y`	`= 4/3 x`

`m(l_2)=4/3`

`m (l_1) xx m (l_2)= -3/4 xx 4/3= -1`

`:.\ l_1\ text(and)\ l_2\ text(are perpendicular)`

Measurement, STD2 EQ-Bank 3 MC

A square pyramid has a base with side length 9 cm and a perpendicular height of 8 cm.

Calculate the volume of the pyramid.

72 cm³
144 cm³
192 cm³
216 cm³

Show Answers Only

$D$

Show Worked Solution

$A=9^2=81$

$V=\frac{1}{3}Ah=\frac{1}{3}\times 81\times 8=216\ \text{cm}^3$

$\Rightarrow D$

Measurement, STD2 EQ-Bank 5 MC

A conical water tank has a radius of 3 metres and a perpendicular height of 5 metres.

Calculate the volume of water the tank can hold in kilolitres. (Note: 1 m³ = 1 kL)

15.71 kL
23.56 kL
47.12 kL
141.37 kL

Show Answers Only

$C$

Show Worked Solution

$V$	$=\frac{1}{3}\pi r^2h$
	$=\frac{1}{3}\times \pi \times (3)^2\times 5$
	$=47.123\dots$

$\text{Using: 1 m}^3 = 1\ \text{kL:} $

$\text{47.12 m}^3= 47.12\ \text{kL}$

$\Rightarrow C$

Measurement, STD2 EQ-Bank 4 MC

A cone has a radius of 6 cm and a height of 8 cm.

Calculate the volume of the cone, giving your answer in cubic centimetres correct to 1 decimal place.

100.5 cm³
150.8 cm³
301.6 cm³
904.8 cm³

Show Answers Only

$C$

Show Worked Solution

$V$	$=\frac{1}{3}\pi r^2h$
	$=\frac{1}{3}\times \pi \times (6)^2\times 8$
	$=301.592\dots=301.6\ \text{cm}^3 \text{(1 d.p.)}$

$\Rightarrow C$

Measurement, STD2 EQ-Bank 17

The cone below has a diameter of 13.5 centimetres and a height of 0.4 metres.

Calculate the volume of the cone, giving your answer in cubic centimetres correct to 1 decimal place. (2 marks)

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$1908.5\ \text{cm}^3 \text{(1 d.p.)}$

Show Worked Solution

$\text{Radius}\ =\dfrac{13.5}{2}=6.75\ \text{cm}$

$0.4\ \text{m}\ =\ 40\ \text{cm}$

$V$	$=\frac{1}{3}\pi r^2h$
	$=\frac{1}{3}\times \pi \times (6.75)^2\times 40$
	$=1908.517\dots$
	$=1908.5\ \text{cm}^3 \text{(1 d.p.)}$

Measurement, STD2 EQ-Bank 18

The cone below has a radius of 4.2 metres and a height of 6.1 metres.

Calculate the volume of the cone, giving your answer in cubic metres correct to 1 decimal point. (2 marks)

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$112.7\ \text{m}^3 \text{(1 d.p.)}$

Show Worked Solution

$V$	$=\frac{1}{3}\pi r^2h$
	$=\frac{1}{3}\times \pi \times (4.2)^2\times 6.1$
	$=112.682\dots$
	$=112.7\ \text{m}^3 \text{(1 d.p.)}$

Measurement, STD2 EQ-Bank 16

The square pyramid below, has a side measurement of 120 metres and a perpendicular height $(h)$ of 65 metres.

Find the volume of the pyramid in cubic metres. (2 marks)

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$312\, 000\ \text{m}^3$

Show Worked Solution

$V$	$=\frac{1}{3}Ah$
	$=\frac{1}{3}\times 120\times 120\times 65$
	$=312\, 000\ \text{m}^3$

Financial Maths, STD2 EQ-Bank 34

David and Mary are a couple living together. They each receive the maximum Age Pension payment of $888.50 per fortnight (which includes basic rate, Pension Supplement and Energy Supplement).

Their combined income from part-time work is $520 per fortnight. Their Age Pension is reduced by 25 cents each for every dollar of combined income over $380 per fortnight..

Calculate the reduction in each person's Age Pension payment. (2 marks)
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Calculate their total household fortnightly income including their reduced Age Pension payments and earnings. (2 marks)
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What percentage of their total household income comes from their Age Pension payments? Give your answer to 1 decimal place. (1 mark)
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a. $$35$

b. $$2227$

c. $76.6\%\ \text{(to 1 d.p.)}$

Show Worked Solution

a. $\text{Combined income over free area} = 520-380=$140$

$\text{Reduction for each person} = 0.25 \times 140= $35$

$\therefore\ \text{Each person’s pension is reduced by \$35}$

b. $\text{Reduced pension for each person}=888.50-35= $853.50$

$\text{Combined Age Pension} = 2 \times 853.50=$1707$

$\text{Total household income} = 1707+520=$2227$

c. $\text{Percentage from Age Pension} =\dfrac{1707}{2227}\times 100=76.6\%\ \text{(to 1 d.p.)}$

$\therefore\ 76.6\%\text{ of their household income comes from Age Pension}$

Financial Maths, STD2 EQ-Bank 30

Baron is 19 years old, single with no children, and lives away from his parents' home to study. He receives the maximum Youth Allowance payment of $663.30 per fortnight.

Baron supplements his payments by working part-time and earns $15.50 per hour in a retail position. His Youth Allowance is reduced by 50 cents for each dollar earned over $236 per fortnight.

Baron works 22 hours per fortnight. Calculate his total fortnightly income including his reduced Youth Allowance payment and earnings. (2 marks)
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What is the maximum number of hours Baron can work per fortnight before his Youth Allowance payment is reduced to zero? Give your answer to the nearest whole hour. (2 marks)
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a. $\text{\$951.80 per fortnight}$

b. $\text{101 hours per fortnight}$

Show Worked Solution

a. $\text{Baron’s earnings} = 15.50\times 22=$341$

$\text{Income over free area} = 341-236 = $105$

$\text{Reduction in payment} = 0.50\times 105= $52.50$

$\text{Reduced Youth Allowance} = 663.30-52.50=$610.80$

$\text{Total fortnightly income} = 610.80+341=$951.80$

b. $\text{For payment to reduce to zero, reduction needed} = $663.30$

$\text{Income over free area} = \dfrac{663.30}{0.50}=$1326.60$

$\text{Total earnings needed} = 1326.60+236=$1562.60$

$\text{Hours needed}=\dfrac{1562.60}{15.50}=100.8$

$\therefore\ \text{Baron can work a maximum of 101 hours per fortnight}$

Financial Maths, STD2 EQ-Bank 29

Yuki is single with no children and receives the maximum JobSeeker Payment of $793.60 per fortnight.

Her JobSeeker Payment is reduced by 50 cents for each dollar earned over $150 per fortnight.

Yuki earns $380 per fortnight from casual work. Calculate her reduced JobSeeker Payment. (2 marks)
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How much would Yuki need to earn per fortnight for her JobSeeker Payment to be reduced to $600? (2 marks)
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a. $$678.60\ \text{per fortnight}$

b. $$537.20\ \text{per fortnight}$

Show Worked Solution

a. $\text{Income over free area} = 380-150=$230$

$\text{Reduction in payment} = 0.50\times 230 = $115$

$\text{Reduced JobSeeker Payment} = 793.60-115 = $678.60$

$\therefore\ \text{Yuki receives \$678.60 per fortnight}$

b. $\text{Required reduction} = 793.60-600=$193.60$

$\text{Income over free area} = \dfrac{193.60}{0.50}=$387.20$

$\text{Total income needed} = 387.20+150=$537.20$

$\therefore\ \text{Yuki needs to earn \$537.20 per fortnight}$

Financial Maths, STD1 EQ-Bank 25

The table shows Youth Allowance payment rates per fortnight.

Helen is single and receives the maximum Age Pension payment. She is entitled to the Pension Supplement and Energy Supplement. Calculate her total Age Pension payment per fortnight. (1 mark)
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Helen earns $290 per fortnight from part-time work. Her pension is reduced by 50 cents for each dollar earned over $218 per fortnight. Calculate Helen's reduced Age Pension payment. (2 mark)
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a. $$1178.70$

b. $$1,142.70\ \text{per fortnight}$

Show Worked Solution

a. $\text{Total Age Pension}= 1079.70+84.90+14.10=$1178.70$

b. $\text{Income over free area} =290-218=$72$

$\text{Reduction in pension} = 0.50\times 72=$36$

$\text{Reduced Age Pension} = 1178.70-36=$1,142.70$

$\therefore\ \text{Helen receives }$1,142.70\ \text{per fortnight}$

Financial Maths, STD1 EQ-Bank 24

The table shows Youth Allowance payment rates per fortnight.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Your situation} \rule[-1ex]{0pt}{0pt} & \textbf{Maximum fortnightly payment} \\
\hline
\rule{0pt}{2.5ex} \text{Single, no children, younger than 18, at parents' home} \rule[-1ex]{0pt}{0pt} & $410.30 \\
\hline
\rule{0pt}{2.5ex} \text{Single, no children, younger than 18, living away} \rule[-1ex]{0pt}{0pt} & $663.30 \\
\hline
\rule{0pt}{2.5ex} \text{Single, no children, 18 or older, at parents' home} \rule[-1ex]{0pt}{0pt} & $472.50 \\
\hline
\rule{0pt}{2.5ex} \text{Single, no children, 18 or older, living away} \rule[-1ex]{0pt}{0pt} & $663.30 \\
\hline
\rule{0pt}{2.5ex} \text{Single, with children} \rule[-1ex]{0pt}{0pt} & $836.60 \\
\hline
\end{array}

Lisa is 17 years old, single with no children, and lives away from her parents' home to study. What is the maximum Youth Allowance payment she can receive per fortnight? (1 mark)
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Lisa earns $320 per fortnight from part-time work. Her Youth Allowance is reduced by 50 cents for each dollar earned over $236 per fortnight. Calculate Lisa's reduced Youth Allowance payment. (1 mark)
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a. $$663.30$

b. $$621.30\ \text{per fortnight}$

Show Worked Solution

a. $\text{From table: Single, no children, younger than 18, living away} = $663.30$

b. $\text{Income over free area} = 320-236=$84$

$\text{Reduction in payment} = 0.50\times 84=$42$

$\text{Reduced Youth Allowance} = 663.30-42=$621.30$

$\therefore\ \text{Lisa receives }$621.30\ \text{per fortnight}$

Financial Maths, STD1 EQ-Bank 23

The table shows JobSeeker Payment rates per fortnight.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Your situation} \rule[-1ex]{0pt}{0pt} & \textbf{Maximum fortnightly payment} \\
\hline
\rule{0pt}{2.5ex} \text{Single, no children} \rule[-1ex]{0pt}{0pt} & $793.60 \\
\hline
\rule{0pt}{2.5ex} \text{Single, with a dependent child or children} \rule[-1ex]{0pt}{0pt} & $849.90 \\
\hline
\rule{0pt}{2.5ex} \text{Partnered} \rule[-1ex]{0pt}{0pt} & $726.50 \\
\hline
\end{array}

James is single with a dependent child and receives the maximum JobSeeker Payment. How much does he receive per fortnight? (1 mark)
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James earns $280 per fortnight from casual work. His JobSeeker Payment is reduced by 50 cents for each dollar earned over $150 per fortnight. Calculate James's reduced JobSeeker Payment. (1 mark)
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a. $$849.90$

b. $$784.90\ \text{per fortnight}$

Show Worked Solution

a. $\text{From table: Single, with a dependent child} = $849.90$

b. $\text{Income over free area} = 280-150=$130$

$\text{Reduction in payment} = 0.50\times 130=$65$

$\text{Reduced JobSeeker Payment} = 849.90-65=$784.90$

$\therefore\ \text{James receives }$784.90\ \text{per fortnight}$

Financial Maths, STD1 EQ-Bank 3 MC

The table shows Youth Allowance payment rates from 1 January 2025.

Marcus is 19 years old, single with no children, and lives at home with his parents.

What is the maximum Youth Allowance payment Marcus can receive per fortnight?

$410.30
$472.50
$663.30
$836.60

Show Answers Only

$B$

Show Worked Solution

$\text{Marcus is 19 years old (18 or older)}$

$\text{Marcus is single with no children}$

$\text{Marcus lives at his parents’ home}$

$\text{From table: Single, no children, 18 or older, at parents’ home} = $472.50$

$\Rightarrow B$

Financial Maths, STD1 EQ-Bank 2 MC

The table shows Age Pension payment rates per fortnight.

Robert is single and receives the maximum Age Pension payment. He is entitled to the Pension Supplement and Energy Supplement.

How much does Robert receive per fortnight?

$888.50
$1079.70
$1178.70
$1777.00

Show Answers Only

$C$

Show Worked Solution

$\therefore\ \text{Total Payment (Single)}$	$=\text{Basic rate} + \text{Pension Supplement} + \text{Energy Supplement}$
	$=1079.70+84.90+14.10=$1,178.70$

$\Rightarrow C$

Functions, EXT1′ F1 2012 HSC 11fi

Sketch the function $y = \abs{x}- 1$, showing the $x$- and $y$-intercepts. (2 marks)

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`text(See Worked Solutions.)`

Show Worked Solution

Functions, EXT1′ F1 2018 HSC 12di

The diagram shows the graph of the function $f(x) = \dfrac{x}{x-1}$.

Draw a graph of $y = \abs{f(x)}$, showing all asymptotes and intercepts. (2 marks)

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Show Worked Solution

Functions, EXT1 EQ-Bank 18

A cubic function is given by $f(x)=\left(2 x^2+3 x-5\right)(x+2)$

Find the zeros of $f(x)$. (1 mark)

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Hence, or otherwise, solve $f(x) \geqslant 0$, giving your answer in set notation. (2 marks)

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a. $\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2$

b. $x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)$

Show Worked Solution

a. $f(x)=\left(2 x^2+3 x-5\right)(x+2)=(2 x+5)(x-1)(x+2)$

$\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2$

b. $\text{Find} \ x \ \text{such that} \ \ f(x) \geqslant 0.$

$\text{At} \ \ x=0: \ (5)(-1)(2)<0$

$\therefore \text{Graph}\ (f(x)) \geqslant 0 \ \ \text{for} \ \ x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)$

Functions, 2ADV EQ-Bank 14

The function $y=f(x)$ is defined by:

\begin{align*}
f(x)= \begin{cases}|x+1|-2, & \text { for }\ x \leqslant 1 \\ x^2-4, & \text { for }\ x>1\end{cases}
\end{align*}

Sketch $y=f (x)$ (3 marks)

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For what values of $x$ is $f(x)=0$? (1 mark)

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a.

b. $f(x)=0\ \ \text{when}\ \ x=-3,2$.

Show Worked Solution

a.

b. $f(x)=0\ \ \text{when}\ \ x=-3,2$.

Financial Maths, STD2 EQ-Bank 19

Michael is a children's book author. His publisher printed 25000 copies of his new book and after 6 months there are 9200 copies left in stock. Michael receives 12% of the retail price as royalties.

How many copies of Michael's book were sold? (1 mark)
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Calculate Michael's royalty payment if the retail price is $19.95. (2 marks)
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The publisher decides to sell the remaining stock at a reduced price of $15.00 per book. Calculate Michael's total royalty payment if all the remaining books are sold at this price. (2 marks)
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a. $15\,800\ \text{copies}$

b. $$37\,825.20$

c. $$54\,385.20$

Show Worked Solution

a. $\text{Copies sold}\ =25\,000-9200\ =15\,800\ \text{copies}$

b. $\text{Total sales}\ =15\,800\times 19.95=$315\,210$

$\text{Royalty payment}$	$=12\% \times 315\,210$
	$= 0.12\times 315\,210$
	$=$37\,825.20$

c. $\text{Sales from remaining books}\ =9200\times 15.00\ =$138\,000$

$\text{Total royalty on remaining books}\ =0.12\times 138\,000=$16\,560$

$\text{Total royalty when all books sold}\ = 37\,825.20 + 16\,560\ =$54\,385.20$

$\therefore\ \text{Michael’s total royalty payment is }$54\,385.20$

Financial Maths, STD2 EQ-Bank 2 MC

A musician receives a royalty of 8% on album sales.

If 3500 albums were sold at $18.00 each, what is the musician's royalty payment?

$4320
$5040
$5600
$6300

Show Answers Only

$B$

Show Worked Solution

$\text{Total sales}\ =3500 \times 18.00\ =$63\,000$

$\text{Royalty payment}$	$=8\% \times 63\,000$
	$= 0.08\times 63\,000=$5040$

$\Rightarrow B$

Financial Maths, STD2 EQ-Bank 28

Ian works in a packaging factory and is paid $0.85 for each box he packs. Last month he worked 160 hours and packed 8960 boxes.

Calculate Ian's total earnings for the month. (1 mark)
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The following month the factory decides to pay Ian a flat hourly rate of $44.50. What percentage increase/decrease is this from his equivalent hourly wage of the previous month. Give your answer correct to 1 decimal place. (2 marks)
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a. $$7616$

b. $\text{6.5% decrease}$

Show Worked Solution

a. $\text{Total Earnings}=0.85\times 8960=$7616$

b. $\text{Hourly rate (last month)}=\dfrac{7616}{160}=$47.60$

$\text{New hourly wage} = $44.50\ \text{(given)}$

$\text{% decrease}=\dfrac{47.60-44.50}{47.60}=0.0651… = 6.5\%\ \text{decrease}$

$\therefore\ \text{Ian’s hourly rate has decreased 6.5%.}$

Financial Maths, STD2 EQ-Bank 17

Roberto works as a furniture assembler and is paid $18.50 for each bookshelf he completes. In one week he worked 40 hours and earned $1110.

How many bookshelves did Roberto complete that week? (1 mark)
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Calculate his hourly rate of pay. (2 marks)
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a. $60\ \text{bookshelves}$

b. $$27.75\text{/hour}$

Show Worked Solution

a.	$\text{Number of Bookshelves}$	$=\dfrac{1110}{18.50}$
		$=60\ \text{bookshelves}$

b.	$\text{Hourly rate}$	$=\dfrac{\text{Total earnings}}{\text{Hours worked}}$
		$=\dfrac{1110}{40}=$27.75$

$\therefore\ \text{Roberto’s hourly rate is }$27.75$

Financial Maths, STD2 EQ-Bank 3 MC

Maria works in a textile factory and is paid $1.20 for each scarf she completes. Last week Maria earned $456.

How many scarves did she complete?

Show Answers Only

$B$

Show Worked Solution

$\text{Number of scarves}$	$=\dfrac{\text{Total earnings}}{\text{Payment per scarf}}$
	$= \dfrac{456}{1.20} =380$

$\Rightarrow B$

Functions, EXT1 EQ-Bank 12

Find all values of $x$ for which $(2 x+1)(x-3)(x-1) \leqslant 0$. (2 marks)

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$x \leqslant-\dfrac{1}{2}\ \cup \ 1 \leqslant x \leqslant 3$

Show Worked Solution

$(2 x+1)(x-3)(x-1) \leqslant 0$

$\text{Zeros at} \ \ x=-\dfrac{1}{2}, x=3 \ \ \text{and} \ \ x=1$

$\text{At} \ \ x=0: \ (1)(-3)(-1)>0$

$\therefore \text{Graph} \ \leqslant 0 \ \ \text{for} \ \ x \leqslant-\dfrac{1}{2}\ \cup \ 1 \leqslant x \leqslant 3$

Functions, EXT1 EQ-Bank 11

Solve the inequality $(x-4)(x+2)(x-1) \geqslant 0$. (2 marks)

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$-2 \leqslant x \leqslant 1\ \cup\ x \geqslant 4$

Show Worked Solution

$(x-4)(x+2)(x-1) \geqslant 0$

$\text{Zeros at} \ \ x=4, x=-2 \ \ \text{and}\ \ x=1$

$\text{At} \ \ x=0: \ (-4)(2)(-1)>0$

$\therefore \text{Graph}\ \geqslant 0 \ \text {for} \ -2 \leqslant x \leqslant 1\ \cup\ x \geqslant 4$

\(48\)	\(=k \times 4^2\)
\(k\)	\(=3\)

b.	\(\text{Total cost}\)	\(=\text{Deposit}+\text{Total repayments}\)
		\(=$3200+$15\,120=$18\,320\)

c.	\(\text{Interest paid}\)	\(=\text{Total cost}-\text{Original cost}\)
		\(=$18\,320-$16\,500=$1820\)

c.	\(\text{Total cost}\)	\(=\text{Deposit}+\text{Total repayments}\)
		\(=960+2600=$3560\)

`b^2-4ac`	`> 0`
`4-4 ⋅ 1 ⋅ (-k)`	`> 0`
`4k`	`> -4`
`k`	`> -1`

\(y-y_1\)	\(=m\left(x-x_1\right)\)
\(y-4\)	\(=-\dfrac{3}{4}(x-0)\)
\(y-4\)	\(=-\dfrac{3}{4} x\)
\(4 y-16\)	\(=-3 x\)
\(3 x+4 y-16\)	\(=0\)

b.	\(\text{Balance}\)	\(=\text{Income}-\text{Expenses}\)
		\(=$69\,275.60-$43\,070.00=$26\,205.60\)

\(V\)	\(=\frac{1}{3}\pi r^2h\)
	\(=\frac{1}{3}\times \pi \times (3)^2\times 5\)
	\(=47.123\dots\)

\(V\)	\(=\frac{1}{3}\pi r^2h\)
	\(=\frac{1}{3}\times \pi \times (6)^2\times 8\)
	\(=301.592\dots=301.6\ \text{cm}^3 \text{(1 d.p.)}\)