What is the reciprocal function?

Reciprocal function y equals k over x with two branches A rectangular hyperbola with one branch in the first quadrant and a mirror branch in the third quadrant. Both axes are asymptotes; the curve approaches them but never touches. x y 0 y = k ⁄ x, k > 0 third quadrant branch first quadrant branch

What is wrong sign of k?

In a physical problem k is almost always positive. A negative k usually means a slip in the (x, y) pair.

§-Syllabus dot point

NSWMaths Standard 2Syllabus dot point

How are reciprocal functions used to model inverse variation, and what does the graph look like?

Model practical problems involving reciprocal functions and inverse variation of the form $y = \frac{k}{x}$

A focused answer to the HSC Maths Standard 2 dot point on reciprocal functions and inverse variation. Finding the constant of proportionality, graphing y = k/x stage by stage, identifying inverse variation from a constant product, the two asymptotes, and applying to speed-time, pressure-volume and shared-cost problems with worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to spot inverse variation in worded problems. The usual settings are speed and time over a fixed distance, pressure and volume of a gas at a fixed temperature, and workers and hours for a fixed job. Once you spot it, set up the reciprocal model $y = \frac{k}{x}$ , find the constant of proportionality $k$ (the fixed number the two values always make), and use the model to predict. The key idea is the constant product: in inverse variation $x$ and $y$ always multiply to the same number, so when one doubles the other halves.

The answer

Direct vs inverse variation

Direct. $y$ doubles when $x$ doubles. Equation: $y = k x$ . The graph is a straight line through the origin.
Inverse. $y$ halves when $x$ doubles. Equation: $y = \frac{k}{x}$ . The graph is a curve that never touches either axis.

In inverse variation, the product $x y = k$ is constant. That single fact is both the test (is the product the same for every pair?) and the tool (find $k$ from one pair, then predict any other).

The reciprocal function

y = \frac{k}{x}

For $k > 0$ : branches in the first and third quadrants.
For $k < 0$ : branches in the second and fourth quadrants.
Two asymptotes: the $x$ -axis ( $y = 0$ ) and the $y$ -axis ( $x = 0$ ).
The graph is a rectangular hyperbola.

Finding $k$

Given one $(x, y)$ pair from the problem, compute

k = x y.

Then write the full model $y = \frac{k}{x}$ and use it for other values. Because $k$ stays the same, one data pair is enough to fix the whole relationship. That is what makes these questions quick once you spot the inverse variation.

Why the curve never reaches the axes

An asymptote is a line the curve gets closer and closer to but never touches. The two asymptotes here are not random; they come straight from the algebra. As $x$ grows very large, $\frac{k}{x}$ shrinks towards $0$ but stays positive, so the curve hugs the $x$ -axis without touching it. As $x$ shrinks towards $0$ , $\frac{k}{x}$ grows huge, so the curve shoots up beside the $y$ -axis. And $x = 0$ is never allowed, because you cannot divide by zero. That is the algebraic reason the hyperbola has a gap at the $y$ -axis instead of crossing it.

Domain considerations

In a real-world problem the variables are usually positive: speed cannot be negative, and pressure cannot be zero. So you only use the first-quadrant branch. The domain is the set of $x$ values that make sense for the problem. Always state this practical domain when the question asks for a graph, for example $s > 0$ for a speed, or $n \geq 1$ for a whole number of workers.

Spotting inverse variation from a table: the constant product

When a table is given and you must decide whether it is inverse variation, multiply each pair.

Take the table $T = 12, 6, 4, 3, 2$ at $n = 1, 2, 3, 4, 6$ . The products are $1 \times 12 = 2 \times 6 = 3 \times 4 = 4 \times 3 = 6 \times 2 = 12$ : all equal, so this is inverse variation with $k = 12$ , and the model is $T = \frac{12}{n}$ . Contrast direct variation, where instead the ratio $\frac{y}{x}$ would be constant, or a linear model, where the difference would be constant.

Sketch the hyperbola, stage by stage

A reciprocal model is graphed the same way as any curve, but only the first-quadrant branch matters in most real problems. The build below is a job of 12 worker-hours: with $n$ workers it takes $T = \frac{12}{n}$ hours, giving table values $T = 12, 6, 4, 3, 2$ at $n = 1, 2, 3, 4, 6$ .

Stage 1, draw and scale the axes. Number of workers $n$ runs along the horizontal axis and time $T$ up the vertical axis, both from 0 to 12. Only positive values are drawn, since you cannot have a negative number of workers or a negative time.

Stage 2, plot the table points. Plot $(1, 12)$ , $(2, 6)$ , $(3, 4)$ , $(4, 3)$ and $(6, 2)$ . Multiply the coordinates of any point and you always get $12$ : that constant product $n \times T = 12$ is the signature of inverse variation.

Stage 3, join the points with one smooth curve. Join the dots with a single smooth curve that drops steeply for small $n$ and flattens for large $n$ . This shape is a rectangular hyperbola; it never becomes a straight line and never crosses either axis.

Stage 4, mark the asymptotes and the constant product. Both axes are asymptotes. As $n$ grows the curve hugs the horizontal axis $T = 0$ (more workers, ever less time, but never zero); as $n$ shrinks towards $0$ the curve shoots up alongside the vertical axis $n = 0$ . The dashed rectangle at $(3, 4)$ shows the constant product: its area $3 \times 4 = 12$ is the same for every point on the curve.

For $k < 0$ the same construction reflects into the second and fourth quadrants, but Standard 2 contexts are almost always positive, so the first-quadrant branch above is the one you will draw.

Practical applications

Speed and time at fixed distance. $t = \frac{d}{s}$ . The constant is the trip distance.
Pressure and volume of a gas at fixed temperature (Boyle's Law). $P V = k$ .
Workers and time for a fixed job. $T = \frac{W}{n}$ where $W$ is the total worker-hours and $n$ is the number of workers.
Per-unit cost and number of units when total cost is fixed. Cost per person when splitting a $240 booking $n$ ways is $\frac{240}{n}$ .

How exam questions ask about reciprocal models

The wording shifts but each version is the same underlying task:

" $y$ varies inversely as $x$ " or " $y$ is inversely proportional to $x$ ." Write $y = \frac{k}{x}$ straight away; the constant $k$ is what you solve for next.
"Find the constant of variation / the constant $k$ ." Multiply a given pair: $k = xy$ .
"Find $y$ when $x = \ldots$ " (after a pair is given). Find $k$ first, then substitute the new $x$ into $y = \frac{k}{x}$ .
"Find $x$ for a given $y$ ." Rearrange $y = \frac{k}{x}$ to $x = \frac{k}{y}$ and substitute.
"Which model fits this table?" Test the product $xy$ : a constant product means inverse variation (a constant ratio would be direct variation instead).
"Sketch the graph" or "describe the graph." Draw the first-quadrant branch of a hyperbola, name both axes as asymptotes, and state the practical domain.
"What happens as $x$ gets very large / very small?" This is an asymptote question: as $x \to$ large, $y \to 0$ ; as $x \to 0$ , $y$ grows without bound.

Reciprocal modelling: speed, shared cost and a solve-back

Three contexts illustrate how a fixed quantity divided by a variable produces a reciprocal (hyperbolic) relationship, finishing with a solve-back for an unknown.

Travel time as a reciprocal of speed

Because distance equals speed multiplied by time, rearranging gives time as distance divided by speed. The drive from Sydney to Newcastle is about $160$ km, so driving time $t$ (in hours) at average speed $s$ (in km/h) is modelled by a reciprocal function.

t = \frac{160}{s}

Evaluate the model at three speeds: Substituting each speed into the formula shows how quickly the time savings shrink as speed increases, which is the hallmark of diminishing returns on a hyperbola.

Speed $s$ (km/h)	Time $t$ (hours)
$80$	$t = 2$
$100$	$t = 1.6$
$110$	$t \approx 1.45$

Even doubling your speed from $80$ to $160$ km/h only halves the time, illustrating the diminishing returns of speeding.

Cost per person of a shared holiday house

When the total amount is fixed, the cost per person falls as the number of people rises. A group hires a holiday house for $1200, so the cost per person $c$ when $p$ people share it follows the same reciprocal structure.

c = \frac{1200}{p}

Evaluate the shared-cost model at three group sizes: Substituting values confirms the per-person cost drops as the group grows, and that the product (people times cost) stays at $1200, the constant $k$ .

People $p$	Cost each $c$
$2$	$c = 600$ , i.e. $600
$4$	$c = 300$ , i.e. $300
$6$	$c = 200$ , i.e. $200

Solve a reciprocal model for an unknown

To find an unknown in a reciprocal model, multiply both sides by the unknown and then divide by the known value. A pump empties a tank in $T$ hours when $n$ pumps run together, modelled by $T = \frac{12}{n}$ (so $1$ pump takes $12$ hours). To find how many pumps empty the tank in $4$ hours, substitute $T = 4$ and solve.

4 = \frac{12}{n} \implies 4n = 12 \implies n = \frac{12}{4} = 3 \text{ pumps}

The constant $k = 12$ pump-hours is the total work needed.

Final answer: the Newcastle drive takes $2$ hours at $80$ km/h and about $1.45$ hours at $110$ km/h; sharing the $1200 house among $6$ people costs $200 each; and $3$ pumps empty the tank in $4$ hours.

Common traps

Confusing direct and inverse variation: Read carefully. "Doubles when the other doubles" is direct ( $y = kx$ ). "Halves when the other doubles" is inverse ( $y = k/x$ ).
Wrong sign of $k$: In a physical problem $k$ is almost always positive. A negative $k$ usually means a slip in the $(x, y)$ pair.
Plotting through the origin: The hyperbola $y = k/x$ never passes through the origin; it has both axes as asymptotes.
Ignoring the practical domain: Speed cannot be zero or negative; the number of workers is a positive whole number. Only the first-quadrant branch matters.
Treating $k$ as something other than the product: In inverse variation $k = xy$ is constant. Use it to check every answer: the product should come out the same each time.
Forgetting to invert when solving for $x$: From $y = \frac{k}{x}$ , solving for $x$ gives $x = \frac{k}{y}$ , not $x = ky$ .

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC-style3 marksThe time

t

(in hours) for a car to travel

480

km varies inversely as its speed

s

(in km/h). Write the relationship and find the speed needed to complete the trip in

6

hours.

Show worked answer →

Inverse variation: $t = \frac{k}{s}$ .

The constant $k$ is total distance: $k = t \cdot s = 480$ .

So $t = \frac{480}{s}$ .

For $t = 6$ : $6 = \frac{480}{s}$ , giving $s = \frac{480}{6} = 80$ km/h.

Markers reward the inverse variation form, the constant identified as $480$ km, and the speed.

2021 HSC-style3 marksThe pressure

P

of a gas varies inversely as its volume

V

. When

V = 5

litres,

P = 200

kPa. Find

P

when

V = 8

litres.

Show worked answer →

Inverse variation: $P = \frac{k}{V}$ .

Find $k$ from the given pair: $200 = \frac{k}{5}$ , so $k = 1000$ .

Then $P = \frac{1000}{V}$ .

At $V = 8$ : $P = \frac{1000}{8} = 125$ kPa.

Markers reward identification of the inverse relationship, calculation of the constant $k$ , and the final pressure with units.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksThe quantity

y

varies inversely with

x

, and

y = 15

when

x = 2

. Find the value of

y

when

x = 5

Show worked solution →

Find the constant $k$ . Inverse variation means $y = \dfrac{k}{x}$ , where the constant is the product $k = xy$ . Substitute the given pair $x = 2$ , $y = 15$ :

k = 2 \times 15 = 30.

So the model is $y = \dfrac{30}{x}$ .

Find $y$ when $x = 5$ . Substitute $x = 5$ :

y = \frac{30}{5} = 6.

State the answer. Since $k = 30$ , the model is $y = \dfrac{30}{x}$ , so $y = 6$ when $x = 5$ . Answer: $y = 6$ (check: $5 \times 6 = 30$ ).

foundation2 marksA cyclist rides a fixed route of

60

km, and the riding time

t

(in hours) varies inversely with the average speed

s

(in km/h), so

t = \dfrac{60}{s}

. Find the time taken at an average speed of

30

km/h.

Show worked solution →

Read off the model. The route distance is fixed at $60$ km, so $t = \dfrac{60}{s}$ with constant $k = 60$ km (distance equals speed multiplied by time).

Substitute the speed. Put $s = 30$ into the model:

t = \frac{60}{30} = 2.

State the answer. At an average speed of $30$ km/h the ride takes $2$ hours. Answer: $t = 2$ hours (check: $30 \times 2 = 60$ km).

foundation3 marksIt is known that

y

varies inversely with

x

, and

y = 8

when

x = 3

. (a) Write an equation connecting

y

and

x

, using

k

as the constant of variation. (b) Find the value of

y

when

x = 6

. (c) Find the value of

y

when

x = 12

Show worked solution →

Part (a), write the equation and find $k$ . Inverse variation means $y = \dfrac{k}{x}$ , where the constant is the product $k = xy$ . Substitute the data pair $x = 3$ , $y = 8$ :

k = 3 \times 8 = 24.

So the equation is $y = \dfrac{24}{x}$ .

Part (b), find $y$ when $x = 6$ . Substitute $x = 6$ :

y = \frac{24}{6} = 4.

Part (c), find $y$ when $x = 12$ . Substitute $x = 12$ :

y = \frac{24}{12} = 2.

State the answer. The equation is $y = \dfrac{24}{x}$ , with $y = 4$ when $x = 6$ and $y = 2$ when $x = 12$ . Check the constant product: $6 \times 4 = 24$ and $12 \times 2 = 24$ , and notice that doubling $x$ from $3$ to $6$ halves $y$ from $8$ to $4$ , the signature of inverse variation.

foundation3 marksA group hires a minibus for a day trip for a total of $360, and the cost is shared equally among the people who go. Let

c

be the cost per person (in dollars) when

p

people share the hire. (a) Explain why

c

varies inversely with

p

and write the equation. (b) Find the cost per person when

6

people go. (c) Find the cost per person when

8

people go.

Show worked solution →

Part (a), write the equation. The total hire is fixed at $360, so the cost per person is the total divided by the number of people: $c = \dfrac{360}{p}$ . This is inverse variation because the product $c \times p = 360$ is constant (the fixed total), so as the number of people rises the cost each falls.

Part (b), cost per person for $6$ people. Substitute $p = 6$ :

c = \frac{360}{6} = 60.

So each person pays $60.

Part (c), cost per person for $8$ people. Substitute $p = 8$ :

c = \frac{360}{8} = 45.

So each person pays $45.

State the answer. The model is $c = \dfrac{360}{p}$ ; $6$ people pay $60 each and $8$ people pay $45 each. Check the constant product: $6 \times 60 = 360$ and $8 \times 45 = 360$ , the fixed hire cost.

core4 marksA water tank is filled by opening taps, and the table shows the time

T

(in minutes) to fill the tank using

n

taps. |

n

| 1 | 2 | 3 | 6 | | --- | --- | --- | --- | --- | |

T

| 48 | 24 | 16 | 8 | (a) Show that

T

varies inversely with

n

and state the constant of variation. (b) Write the equation connecting

T

and

n

. (c) Find the time to fill the tank with

4

taps.

Show worked solution →

Part (a), test the product. For inverse variation the product $n \times T$ must be the same for every column. Compute it for each pair:

1 \times 48 = 48, \qquad 2 \times 24 = 48, \qquad 3 \times 16 = 48, \qquad 6 \times 8 = 48.

The product is the constant $48$ every time, so $T$ varies inversely with $n$ with constant of variation $k = 48$ (the total tap-minutes of work to fill the tank).

Part (b), write the equation. Put $k = 48$ into $T = \dfrac{k}{n}$ :

T = \frac{48}{n}.

Part (c), time with $4$ taps. Substitute $n = 4$ :

T = \frac{48}{4} = 12.

State the answer. The constant product $48$ confirms inverse variation ( $T = \dfrac{48}{n}$ ), and $4$ taps fill the tank in $12$ minutes. Check: $4 \times 12 = 48$ .

core4 marksThe pressure

P

(in kPa) of a fixed mass of gas varies inversely with its volume

V

(in litres) at constant temperature (Boyle's Law). When

V = 2.5

litres,

P = 120

kPa. (a) Find the constant of variation

k

. (b) Write the equation connecting

P

and

V

. (c) Find the pressure when

V = 6

litres. (d) Find the volume when the pressure is

400

kPa.

Show worked solution →

Part (a), find $k$ . Inverse variation gives $P = \dfrac{k}{V}$ , which rearranges to $k = P \times V$ . Substitute the data pair $V = 2.5$ , $P = 120$ :

k = 120 \times 2.5 = 300.

So $k = 300$ (in kPa litres).

Part (b), write the equation. Put $k = 300$ into $P = \dfrac{k}{V}$ :

P = \frac{300}{V}.

Part (c), pressure when $V = 6$ . Substitute $V = 6$ :

P = \frac{300}{6} = 50.

So the pressure is $50$ kPa.

Part (d), volume when $P = 400$ . Now the pressure is known and the volume is wanted, so rearrange to $V = \dfrac{k}{P}$ and substitute $P = 400$ :

V = \frac{300}{400} = 0.75.

So the volume is $0.75$ litres.

State the answer. Here $k = 300$ and $P = \dfrac{300}{V}$ ; the pressure is $50$ kPa when $V = 6$ litres, and the volume is $0.75$ litres when $P = 400$ kPa. Check the constant product: $6 \times 50 = 300$ and $400 \times 0.75 = 300$ .

core4 marksFor a fixed road trip, the time

t

(in hours) varies inversely with the average speed

s

(in km/h). At an average speed of

90

km/h the trip takes

6

hours. (a) Find the constant of variation

k

and state what it represents. (b) Write the equation connecting

t

and

s

. (c) Find the average speed needed to complete the trip in

4.5

hours. (d) Describe what happens to the travel time as the average speed becomes very large.

Show worked solution →

Part (a), find $k$ and interpret it. Inverse variation gives $t = \dfrac{k}{s}$ , so $k = t \times s$ . Substitute $s = 90$ , $t = 6$ :

k = 6 \times 90 = 540.

So $k = 540$ . Because $k = t \times s$ is time multiplied by speed, it is the fixed trip distance of $540$ km.

Part (b), write the equation. Put $k = 540$ into $t = \dfrac{k}{s}$ :

t = \frac{540}{s}.

Part (c), speed for $4.5$ hours. The time is known and the speed is wanted, so rearrange to $s = \dfrac{k}{t}$ and substitute $t = 4.5$ :

s = \frac{540}{4.5} = 120.

So an average speed of $120$ km/h is needed.

Part (d), as the speed gets very large. As $s$ grows, $\dfrac{540}{s}$ shrinks towards $0$ but stays positive, so the travel time keeps falling and gets close to $0$ without ever reaching it. The horizontal axis $t = 0$ is an asymptote: no finite speed makes the trip instant.

State the answer. Here $k = 540$ km, $t = \dfrac{540}{s}$ , and a speed of $120$ km/h completes the trip in $4.5$ hours. Check the constant product: $90 \times 6 = 540$ and $120 \times 4.5 = 540$ .

core4 marksA fencing contractor finds that the time

T

(in hours) to build a fixed length of fence varies inversely with the number of workers

n

on the job. With

4

workers the fence takes

9

hours. (a) Find the constant of variation

k

and state what it represents. (b) Write the equation connecting

T

and

n

. (c) Find the time taken by

6

workers. (d) How many workers are needed to finish in

4.5

hours?

Show worked solution →

Part (a), find $k$ and interpret it. Inverse variation gives $T = \dfrac{k}{n}$ , so $k = T \times n$ . Substitute $n = 4$ , $T = 9$ :

k = 9 \times 4 = 36.

So $k = 36$ , the total worker-hours of work needed to build the fence.

Part (b), write the equation. Put $k = 36$ into $T = \dfrac{k}{n}$ :

T = \frac{36}{n}.

Part (c), time with $6$ workers. Substitute $n = 6$ :

T = \frac{36}{6} = 6.

So $6$ workers finish in $6$ hours.

Part (d), workers for $4.5$ hours. The time is known and the number of workers is wanted, so rearrange to $n = \dfrac{k}{T}$ and substitute $T = 4.5$ :

n = \frac{36}{4.5} = 8.

So $8$ workers are needed.

State the answer. Here $k = 36$ worker-hours and $T = \dfrac{36}{n}$ ; $6$ workers take $6$ hours and $8$ workers finish in $4.5$ hours. Answer: $k = 36$ , $T = \dfrac{36}{n}$ , $6$ hours for $6$ workers, $8$ workers for $4.5$ hours (check: $6 \times 6 = 36$ and $8 \times 4.5 = 36$ ).

exam5 marksA diving instructor uses Boyle's Law: for a fixed mass of air at constant temperature the pressure

P

(in kPa) varies inversely with the volume

V

(in litres). A sample of air has a volume of

4

litres at a pressure of

150

kPa. (a) Find the constant of variation

k

. (b) Write the equation connecting

P

and

V

. (c) Find the pressure when the volume is compressed to

10

litres. (d) Find the volume when the pressure is

240

kPa. (e) Describe what happens to the pressure as the volume becomes very large.

Show worked solution →

Part (a), find $k$ . Inverse variation gives $P = \dfrac{k}{V}$ , so $k = P \times V$ . Substitute the data pair $V = 4$ , $P = 150$ :

k = 150 \times 4 = 600.

So $k = 600$ (in kPa litres).

Part (b), write the equation. Put $k = 600$ into $P = \dfrac{k}{V}$ :

P = \frac{600}{V}.

Part (c), pressure when $V = 10$ . Substitute $V = 10$ :

P = \frac{600}{10} = 60.

So the pressure is $60$ kPa.

Part (d), volume when $P = 240$ . The pressure is known and the volume is wanted, so rearrange to $V = \dfrac{k}{P}$ and substitute $P = 240$ :

V = \frac{600}{240} = 2.5.

So the volume is $2.5$ litres.

Part (e), as the volume gets very large. As $V$ grows, $\dfrac{600}{V}$ shrinks towards $0$ but stays positive, so the pressure keeps falling and approaches $0$ without ever reaching it. The horizontal axis $P = 0$ is an asymptote.

State the answer. Here $k = 600$ kPa litres and $P = \dfrac{600}{V}$ ; the pressure is $60$ kPa at $V = 10$ litres, and the volume is $2.5$ litres at $P = 240$ kPa. Answer: $k = 600$ , $P = \dfrac{600}{V}$ , $60$ kPa, $2.5$ litres (check: $10 \times 60 = 600$ and $240 \times 2.5 = 600$ ).

exam5 marksA house-painting job needs a fixed total of work. The time

T

(in hours) to finish varies inversely with the number of painters

n

. With

5

painters the job takes

18

hours. (a) Find the constant of variation

k

. (b) Write the equation connecting

T

and

n

. (c) Find the time taken by

6

painters. (d) How many painters are needed to finish in

7.5

hours? (e) State a suitable domain for

n

if a sketch of

T

against

n

were drawn, and explain why.

Show worked solution →

Part (a), find $k$ . Inverse variation gives $T = \dfrac{k}{n}$ , so $k = T \times n$ . Substitute $n = 5$ , $T = 18$ :

k = 18 \times 5 = 90.

So $k = 90$ (the total painter-hours of work).

Part (b), write the equation. Put $k = 90$ into $T = \dfrac{k}{n}$ :

T = \frac{90}{n}.

Part (c), time with $6$ painters. Substitute $n = 6$ :

T = \frac{90}{6} = 15.

So $6$ painters finish in $15$ hours.

Part (d), painters for $7.5$ hours. The time is known and the number of painters is wanted, so rearrange to $n = \dfrac{k}{T}$ and substitute $T = 7.5$ :

n = \frac{90}{7.5} = 12.

So $12$ painters are needed.

Part (e), suitable domain. A suitable domain is $n \geq 1$ where $n$ is a whole number. You cannot have a negative or zero number of painters, and you cannot have a fraction of a painter, so only positive whole numbers make sense and only the first-quadrant branch of the hyperbola is drawn.

State the answer. Here $k = 90$ painter-hours and $T = \dfrac{90}{n}$ ; $6$ painters take $15$ hours, $12$ painters finish in $7.5$ hours, and the domain is whole-number $n \geq 1$ . Check the constant product: $6 \times 15 = 90$ and $12 \times 7.5 = 90$ .

exam6 marksA class hires a coach for an excursion at a fixed total cost of $2400, and the cost is shared equally among the students who attend. Let

c

be the cost per student (in dollars) when

w

students attend. (a) Write the equation connecting

c

and

w

. (b) Find the cost per student if

10

students attend. (c) Find the cost per student if

16

students attend. (d) How many students would need to attend for the cost to be exactly $40 each? (e) A student claims that doubling the number who attend doubles the cost each person pays. Using your model, explain why this is wrong and state what actually happens.

Show worked solution →

Part (a), write the equation. The coach cost is fixed at $2400 and is shared equally, so the cost per student is the total divided by the number attending: $c = \dfrac{2400}{w}$ . This is inverse variation, with constant product $c \times w = 2400$ .

Part (b), cost for $10$ students. Substitute $w = 10$ :

c = \frac{2400}{10} = 240.

So each student pays $240.

Part (c), cost for $16$ students. Substitute $w = 16$ :

c = \frac{2400}{16} = 150.

So each student pays $150.

Part (d), students for a $40 cost. The cost is known and the number attending is wanted, so rearrange to $w = \dfrac{2400}{c}$ and substitute $c = 40$ :

w = \frac{2400}{40} = 60.

So $60$ students give $40 each.

Part (e), why doubling is wrong. The relationship is inverse, not direct, so doubling $w$ does not double $c$ ; it halves it. For example, from part (b) ten students pay $240 each, and doubling to twenty students gives $c = \dfrac{2400}{20} = 120$ , which is half of $240, not double. The product $c \times w$ stays fixed at $2400$ , so when more students attend the cost each falls.

State the answer. The model is $c = \dfrac{2400}{w}$ ; $10$ students pay $240 each, $16$ students pay $150 each, and $60$ students pay $40 each. Doubling the number attending halves the cost each. Answer: $c = \dfrac{2400}{w}$ , $240, $150, $60$ students, and doubling halves the cost (check: $10 \times 240 = 2400$ , $16 \times 150 = 2400$ and $60 \times 40 = 2400$ ).

exam6 marksA school sets aside a fixed prize budget of $1800 to be shared equally among the winners of a competition. Let

c

be the prize per winner (in dollars) when there are

w

winners. (a) Write the equation connecting

c

and

w

. (b) Find the prize per winner if there are

5

winners. (c) Find the prize per winner if there are

12

winners. (d) How many winners would give a prize of exactly $75 each? (e) A student claims that doubling the number of winners doubles each prize. Using your model, explain why this is wrong and state what actually happens.

Show worked solution →

Part (a), write the equation. The budget is fixed at $1800 and is shared equally, so the prize per winner is the budget divided by the number of winners: $c = \dfrac{1800}{w}$ . This is inverse variation, with constant product $c \times w = 1800$ .

Part (b), prize for $5$ winners. Substitute $w = 5$ :

c = \frac{1800}{5} = 360.

So each winner gets $360.

Part (c), prize for $12$ winners. Substitute $w = 12$ :

c = \frac{1800}{12} = 150.

So each winner gets $150.

Part (d), winners for a $75 prize. The prize is known and the number of winners is wanted, so rearrange to $w = \dfrac{1800}{c}$ and substitute $c = 75$ :

w = \frac{1800}{75} = 24.

So $24$ winners give $75 each.

Part (e), why doubling is wrong. The relationship is inverse, not direct, so doubling $w$ does not double $c$ ; it halves it. For example, from part (b) five winners receive $360 each, and doubling to ten winners gives $c = \dfrac{1800}{10} = 180$ , which is half of $360, not double. In inverse variation the product $c \times w$ stays fixed at $1800$ , so when the number of winners goes up the prize each goes down.

State the answer. The model is $c = \dfrac{1800}{w}$ ; $5$ winners get $360 each, $12$ winners get $150 each, and $24$ winners get $75 each. Doubling the winners halves each prize. Check the constant product: $5 \times 360 = 1800$ , $12 \times 150 = 1800$ and $24 \times 75 = 1800$ .

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The answer

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Domain considerations

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Practical applications

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Exam-style practice questions

Practice questions

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