smc-6249-10-Find k

Algebra, STD2 EQ-Bank 11

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. $W=kt$

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

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

b. $162.5\ \text{minutes}$

Show Worked Solution

a. $W=kt$

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

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

b. $W = 4t$

$\text{When } W = 650:$

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

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

Algebra, STD2 EQ-Bank 10

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}$
	$=400\ \text{m}$

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

Algebra, STD2 EQ-Bank 09 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

Show Answers Only

$C$

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 08 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

$E \propto m$

$E=km$

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

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

$\therefore\ E=4.2m$

$\Rightarrow A$

Algebra, STD2 EQ-Bank 07 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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$D$

Show Worked Solution

$d \propto F$

$d=kF$

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

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

$\Rightarrow D$

Algebra, STD2 EQ-Bank 06 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

$C \propto n$

$C=kn$

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

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

$\Rightarrow B$

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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`text{Show Worked Solutions}`
`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, STD2 A2 2019 HSC 34

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)`

Show Worked Solution

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

♦ Mean mark 42%.

`p = 4/7 d`

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

`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 SM-Bank 2

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)`

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

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

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

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

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

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

\(3.5\)	\(=k \times 28\)
\(k\)	\(=\dfrac{3.5}{28}\)
\(k\)	\(=0.125\)

\(P\)	\(=0.125\times 42\)
	\(=5.25\)

\(2100\)	\(=k \times 500\)
\(k\)	\(=\dfrac{2100}{500}\)
\(k\)	\(=4.2\)