Direct Variation

Algebra, STD2 EQ-Bank 6 MC

Which graph shows a relationship that is a direct variation?

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$B$

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$\text{Direct variation requires a linear relationship of the form }\ y = kx.$

$\text{Hence the line must be straight and pass through the origin.}$

$\text{By elimination:}$

$\text{Eliminate C and D – not linear}$

$\text{Eliminate A – linear but not passing through origin}$

$\Rightarrow B$

Algebra, STD2 EQ-Bank 17

It is known that a quantity $N$ varies directly with another quantity $Q$.

The relationship can be modelled by the equation $N= k \times Q$, where $k$ is a constant.

If $N = 18$ when $Q=4:$

Show the value of $k=4.5$. (1 mark)

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Hence find the value of $Q$ when $N=63$. (1 mark)

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a. $\text{Using}\ \ N=k \times Q:$

$18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5$

b. $Q=14$

Show Worked Solution

a. $\text{Using}\ \ N=k \times Q:$

$18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5$

b. $\text{Find}\ Q\ \text{when}\ \ N=63:$

$63$	$=4.5 \times Q$
$Q$	$=\dfrac{63}{4.5}=14$

Algebra, STD2 EQ-Bank 25

Two quantities, $M$ and $t$, have a relationship such that $M$ varies directly with $t$ and $M=2.4$ when $t=8$.

Find the value of $t$ when $M=5.1$. (2 marks)

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$t=17$

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$M \propto t \ \ \Rightarrow \ \ M=kt$

$\text{Given} \ \ M=2.4 \ \ \text{when} \ \ t=8:$

$2.4=8k \ \ \Rightarrow \ \ k=\dfrac{2.4}{8}=0.3$

$\text{When} \ \ M=5.1:$

$5.1$	$=0.3 \times t$
$t$	$=\dfrac{5.1}{0.3}=17$

Algebra, STD2 EQ-Bank 26

$F$ varies directly with $m$ and $F=14$ when $m=3.5$.

Find the value of $m$ when $F=50$. (2 marks)

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$m=12.5$

Show Worked Solution

$F \propto m \ \ \Rightarrow \ \ F=km$

$\text{Given} \ \ F=14 \ \ \text{when} \ \ m=3.5:$

$14=3.5k \ \ \Rightarrow \ \ k=4$

$\text{When} \ \ F=50:$

$50$	$=4 \times m$
$m$	$=\dfrac{50}{4} = 12.5$

Algebra, STD2 EQ-Bank 7 MC

$d$ varies directly with $h$ and it is known that $d=1.2$ when $h=5$.

The value of $h$ when $d=48$ is

$11.52$
$57.6$
$200$
$240$

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$C$

Show Worked Solution

$d \propto h \ \ \Rightarrow \ \ d=k \times h$

$\text{Find \(k$ given $\ d=1.2$ when $\ h=5$:}\)

$1.2$	$=5 \times k$
$k$	$=\dfrac{1.2}{5}=0.24$

$\text{Find} \ h \ \text{when} \ \ d=48:$

$48$	$=0.24 \times h$
$h$	$=\dfrac{48}{0.24}=200$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 8 MC

$Q$ varies directly with $r$ and $Q=2$ when $r=16$.

The value of $r$ when $Q=13$ is

$1.625$
$26$
$76$
$104$

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$D$

Show Worked Solution

$Q \propto r \ \Rightarrow \ Q=kr$

$\text{Find \(k$ given $\ Q=2$ given $\ r=16$:}\)

$2=16 \times k \ \Rightarrow \ k=\dfrac{1}{8}$

$\text{Find \(r$ when $\ Q=13:$}\)

$13$	$=\dfrac{1}{8} \times r$
$r$	$=13 \times 8=104$

$\Rightarrow D$

Algebra, STD2 EQ-Bank 28

The amount of water ($W$) in litres used by a garden irrigation system varies directly with the time ($t$) in minutes that the system operates.

This relationship is modelled by the formula $W=kt$, where $k$ is a constant.

The irrigation system uses 96 litres of water when it operates for 24 minutes.

Show that the value of $k$ is 4. (1 mark)
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The water tank for the irrigation system contains 650 litres of water. Calculate how many minutes the irrigation system can operate before the tank is empty. (2 marks)
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a. $W=kt$

$\text{When } W = 96 \text{ and } t = 24:$

$96$	$=k \times 24$
$k$	$=\dfrac{96}{24}=4\ \text{(as required)}$

b. $162.5\ \text{minutes}$

Show Worked Solution

a. $W=kt$

$\text{When } W = 96 \text{ and } t = 24:$

$96$	$=k \times 24$
$k$	$=\dfrac{96}{24}=4\ \text{(as required)}$

b. $W = 4t$

$\text{When } W = 650:$

$650$	$=4\times t$
$t$	$ =\dfrac{650}{4}=162.5\ \text{minutes}$

$\therefore\ \text{The irrigation system can operate for 162.5 minutes.}$

Algebra, STD2 EQ-Bank 30

The cost ($C$) of copper wire varies directly with the length ($L$) in metres of the wire.

This relationship is modelled by the formula $C = kL$, where $k$ is a constant.

A 250 metre roll of copper wire costs $87.50.

Show that the value of $k$ is 0.35. (1 mark)
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A builder has a budget of $140 for copper wire. Calculate the maximum length of wire that can be purchased. (2 marks)
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a. $C=kL$

$\text{When } C = 87.50 \text{ and } L = 250:$

$87.50$	$=k \times 250$
$k$	$=\dfrac{87.50}{250}$
$k$	$=0.35\ \text{(as required)}$

b. $400\ \text{m}$

Show Worked Solution

a. $C=kL$

$\text{When } C = 87.50 \text{ and } L = 250:$

$87.50$	$=k \times 250$
$k$	$=\dfrac{87.50}{250}$
$k$	$=0.35\ \text{(as required)}$

b. $C = 0.35L$

$\text{When } C = 140:$

$140$	$=0.35\times L$
$L$	$ =\dfrac{140}{0.35}=400\ \text{m}$

$\therefore\ \text{The builder can purchase 400 metres of wire.}$

Algebra, STD2 EQ-Bank 13 MC

The amount of paint ($P$) needed to cover a wall varies directly with the area ($A$) of the wall.

A painter uses 3.5 litres of paint to cover a wall with an area of 28 square metres.

How much paint is needed to cover a wall with an area of 42 square metres?

4.2 L
4.75 L
5.25 L
5.75 L

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$C$

Show Worked Solution

$P \propto A$

$P=kA$

$\text{When } P = 3.5 \text{ and } A = 28:$

$3.5$	$=k \times 28$
$k$	$=\dfrac{3.5}{28}$
$k$	$=0.125$

$\text{When } A = 42:$

$P=0.125\times 42=5.25$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 10 MC

The energy ($E$) required to heat water varies directly with the mass ($m$) of the water.

It takes 2100 joules of energy to heat 500 grams of water by 1°C.

Which equation represents the relationship between $E$ and $m$?

$E = 4.2m$
$E = 0.24m$
$m = 4.2E$
$m = 0.24E$

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$A$

Show Worked Solution

$E \propto m$

$E=km$

$\text{When } E = 2100 \text{ and } m = 500:$

$2100$	$=k \times 500$
$k$	$=\dfrac{2100}{500}=4.2$

$\therefore\ E=4.2m$

$\Rightarrow A$

Algebra, STD2 EQ-Bank 9 MC

The distance ($d$) a spring stretches varies directly with the force ($F$) applied to it.

When a force of 18 newtons is applied, the spring stretches 27 mm.

What is the value of the constant of variation ($k$)?

$\dfrac{2}{5}$
$\dfrac{5}{2}$
$\dfrac{2}{3}$
$\dfrac{3}{2}$

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$D$

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$d \propto F$

$d=kF$

$\text{When } d = 27 \text{ and } F = 18:$

$27$	$=k \times 18$
$k$	$=\dfrac{27}{18}=\dfrac{3}{2}$

$\Rightarrow D$

Algebra, STD2 EQ-Bank 4 MC

The cost ($C$) of printing flyers varies directly with the number of flyers ($n$) printed.

A printing company charges $45 to print 300 flyers.

What is the value of the constant of variation ($k$)?

$\dfrac{1}{15}$
$\dfrac{3}{20}$
$\dfrac{20}{3}$
$15$

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$B$

Show Worked Solution

$C \propto n$

$C=kn$

$\text{When } C = 45 \text{ and } n = 300:$

$45$	$=k \times 300$
$k$	$=\dfrac{45}{300}=\dfrac{3}{20}$

$\Rightarrow B$

Algebra, STD2 EQ-Bank 21

Jordan visits Italy on his holidays. He pays €180 (180 euros) for a pair of Italian leather boots.

How much is €180 in Australian dollars if AUD1 is worth €0.58? Give your answer correct to the nearest cent. (2 marks)

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$$310.34$

Show Worked Solution

$0.58\ $€ $\ =\ \text{AUD 1}$

$\Rightarrow\ 1\ $€$\ =$$\ \text{AUD }$$\dfrac{1}{0.58}$

$\Rightarrow\ 180\ $€$=180\times\dfrac{1}{0.58}=310.344…$

$\therefore\ \text{Jordan’s boots cost }$310.34\ \text{in Australian dollars.}$

Algebra, STD1 A2 2025 HSC 16

The mass $(M )$ of a box with a square base, in grams, is directly proportional to the area of its base, in cm².

A box with a square base of side length 5 cm has a mass of 500 g.

What is the mass of a similar box with a square base of side length 3 cm? (3 marks)

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$M=180\ \text{grams}$

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$\text{Area of square base }= s^2$

$M \propto s^2\ \ \Rightarrow \ \ M=k\times s^2$

$\text{Find \(k$ given $\ s=5\ $ when $\ M=500$:}\)

$500$	$=k\times 5^2$
$25k$	$=500$
$k$	$=\dfrac{500}{25}=20$

$\text{Find \(M$ when $s=3$:}\)

$M=20\times 3^2=180\ \text{grams}$

♦♦♦ Mean mark 21%.

Algebra, STD1 A2 2020 HSC 20

The weight of a bundle of A4 paper (`W` kg) varies directly with the number of sheets (`N`) of A4 paper that the bundle contains.

This relationship is modelled by the formula `W = kN`, where `k` is a constant.

The weight of a bundle containing 500 sheets of A4 paper is 2.5 kilograms.

Show that the value of `k` is 0.005. (1 mark)

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A bundle of A4 paper has a weight of 1.2 kilograms. Calculate the number of sheets of A4 paper in the bundle. (2 marks)

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a. `text{Show Worked Solutions}`

b. `240 \ text{sheets}`

Show Worked Solution

a. `W = 2.5\ text{kg when} \ N = 500:`

`2.5`	`= k xx 500`
`therefore \ k`	`= frac{2.5}{500}= 0.005`

b. `text{Find}\ \ N \ \ text{when} \ \ W = 1.2\ text{kg:}`

♦ Mean mark 50%.

`1.2`	`= 0.005 xx N`
`therefore N`	`= frac{1.2}{0.005}= 240 \ text{sheets}`

Algebra, STD1 A2 2019 HSC 33

smc-6514

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.

Write the direct variation equation relating British pounds to Australian dollars in the form `p = md`. Leave `m` as a fraction. (1 mark)

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The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation `y = 76d`.

Convert 93 100 Japanese yen to British pounds. (2 marks)

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`p = 4/7 d`
`93\ 100\ text(Yen = 700 pounds)`

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a. `m = text(rise)/text(run) = 4/7`

♦♦♦ Mean mark part (a) 8%.

`p = 4/7 d`

b. `text(Yen to Australian dollars:)`

♦♦ Mean mark part (b) 16%.

`y`	`=76d`
`93\ 100`	`= 76d`
`d`	`= (93\ 100)/76`
	`= 1225`

`text(Aust dollars to pounds:)`

`p`	`= 4/7 xx 1225`
	`= 700\ text(pounds)`

`:. 93\ 100\ text(Yen = 700 pounds)`

Algebra, STD2 A2 EQ-Bank 37

The weight of a steel beam, `w`, varies directly with its length, `L`.

A 1200 mm steel beam weighs 144 kg.

Calculate the weight of a 750 mm steel beam. (2 marks)

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`90\ text(kg)`

Show Worked Solution

`w propto L\ \ =>\ \ w=kL`

`text(When)\ \ w = 144\ text(kg),\ \ L = 1200\ text(mm)`

`144`	`= k xx 1200`
`k`	`= 144/1200= 3/25`

`text(When)\ \ L = 750\ text(mm:)`

`w= 3/25 xx 750= 90\ text(kg)`

Algebra, STD2 A2 2004 HSC 22 MC

John knows that

• one Australian dollar is worth 0.62 euros
• one Vistabella dollar `text{($V)}` is worth 1.44 euros.

John changes 25 Australian dollars to Vistabella dollars.

How many Vistabella dollars will he get?

`text($V10.76)`
`text($V22.32)`
`text($V28.00)`
`text($V58.06)`

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`A`

Show Worked Solution

`text(John has 25 Aust dollars.)`

`text(Converting to Euros:)`

`text(25 Aust)= 25 xx 0.62= 15.5\ text(Euros)`

`text(Converting to Vistabella dollars:)`

`text(15.5 Euros)= 15.5/1.44=\ text($V10.76)`

`=> A`

Algebra, STD2 A2 2007 HSC 24c

Sandy travels to Europe via the USA. She uses this graph to calculate her currency conversions.

After leaving the USA she has US$150 to add to the A$600 that she plans to spend in Europe.

She converts all of her money to euros. How many euros does she have to spend in Europe? (3 marks)

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If the value of the euro falls in comparison to the Australian dollar, what will be the effect on the gradient of the line used to convert Australian dollars to euros? (1 mark)

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a. `480\ €`

b. `text(If the value of the euro falls against the A$, then 1 A$)`

`text(will buy more euros than before and the gradient used)`

`text{to convert the currencies will steepen (increase).}`

Show Worked Solution

a. `text(From graph:)`

`75\ text(US$)`	`=\ text(100 A$)`
`=> 150\ text(US$)`	`=\ text(200 A$)`

`:.\ text(Sandy has a total of 800 A$)`

`text(Converting A$ to €:)`

`text(100 A$)`	`= 60\ €`
`:.\ text(800 A$)`	`= 8 xx 60= 480\ €`

b. `text(If the value of the euro falls against the A$, then 1 A$)`

`text(will buy more euros than before and the gradient used)`

`text{to convert the currencies will steepen (increase).}`

Algebra, STD2 A2 2014 HSC 26f

The weight of an object on the moon varies directly with its weight on Earth. An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.

A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth. (2 marks)

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`14\ 694\ text(kg)`

Show Worked Solution

`W_text(moon) prop W_text(earth)\ \ => \ \ W_text(m) = k xx W_text(e)`

`text(Find)\ k\ text{given}\ W_text(e) = 84\ text{when}\ W_text(m) = 14:`

`14`	`= k xx 84`
`k`	`= 14/84 = 1/6`

`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`

`2449`	`= 1/6 xx W_text(e)`
`W_text(e)`	`= 14\ 694\ text(kg)`

`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`

Algebra, STD2 A2 2012 HSC 13 MC

Conversion graphs can be used to convert from one currency to another.

Sarah converted 60 Australian dollars into Euros. She then converted all of these Euros
into New Zealand dollars.

How much money, in New Zealand dollars, should Sarah have?

$26
$45
$78
$135

Show Answers Only

`C`

Show Worked Solution

`text(Using the graphs)`

`$60\ text(Australian)=46\ text(Euro)`

`46 \ text(Euro)=$78\ text(New Zealand)`

`=> C`

\(63\)	\(=4.5 \times Q\)
\(Q\)	\(=\dfrac{63}{4.5}=14\)

\(5.1\)	\(=0.3 \times t\)
\(t\)	\(=\dfrac{5.1}{0.3}=17\)

\(50\)	\(=4 \times m\)
\(m\)	\(=\dfrac{50}{4} = 12.5\)

\(1.2\)	\(=5 \times k\)
\(k\)	\(=\dfrac{1.2}{5}=0.24\)

\(48\)	\(=0.24 \times h\)
\(h\)	\(=\dfrac{48}{0.24}=200\)

\(13\)	\(=\dfrac{1}{8} \times r\)
\(r\)	\(=13 \times 8=104\)

\(96\)	\(=k \times 24\)
\(k\)	\(=\dfrac{96}{24}=4\ \text{(as required)}\)

\(96\)	\(=k \times 24\)
\(k\)	\(=\dfrac{96}{24}=4\ \text{(as required)}\)

\(650\)	\(=4\times t\)
\(t\)	\( =\dfrac{650}{4}=162.5\ \text{minutes}\)