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Graphing Proportional Relationships: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Honey by the jar

A market sells jars of honey. Two jars cost $9$ dollars, three jars cost $13.5$ dollars, and seven jars cost $31.5$ dollars.

Is the cost proportional to the number of jars? If it is, give the constant of proportionality with its units, and write an equation for the cost $y$ of $x$ jars.
1. Hint 1
  A table is proportional when one particular ratio comes out the same for every pair.
2. Hint 2
  Divide each cost by its number of jars.
3. Hint 3
  If the three results agree, that shared number is the cost of a single jar.
Answer

Yes. $k = 4.5$ dollars per jar, and $y = 4.5x$ .

Full solution

Divide each cost by its number of jars.
Two jars give
$\frac{9}{2} = 4.5.$
Three jars give
$\frac{13.5}{3} = 4.5.$
Seven jars give
$\frac{31.5}{7} = 4.5.$

All three ratios are equal, so the cost is proportional to the number of jars.
The shared ratio is the constant of proportionality, $k = 4.5$ .

Its units are the units of $y$ per unit of $x$ , so $k = 4.5$ dollars per jar.
The equation is $y = 4.5x$ .

Answer

Yes. $k = 4.5$ dollars per jar, and $y = 4.5x$ .

Key idea

A table is proportional when the ratio of output to input is the same nonzero number wherever the input is not $0$ , and the output is $0$ wherever the input is $0$ ; that ratio is the unit rate.
Problem 2 The garden sprinkler

A sprinkler uses water at a steady rate. In $4$ minutes it uses $26$ liters.

Write an equation for the liters $y$ used in $x$ minutes. How many liters does it use in $10$ minutes? How long does it take to use $91$ liters?
1. Hint 1
  Start with the amount of water used in one minute.
2. Hint 2
  Divide the liters by the minutes to get the unit rate, then write $y$ as that rate times $x$ .
3. Hint 3
  For the time, the liters are known and the minutes are not, so solve an equation for $x$ .
Answer

$y = 6.5x$ . In $10$ minutes it uses $65$ liters, and $91$ liters take $14$ minutes.

Full solution

The rate is steady, so liters are proportional to minutes.
The unit rate is $26 \div 4 = 6.5$ liters per minute.

So the equation is $y = 6.5x$ .

For $10$ minutes, $y = 6.5 \times 10 = 65$ liters.

For $91$ liters, solve $6.5x = 91$ .
Divide both sides by $6.5$ to get $x = 14$ .
Check that $6.5 \times 14 = 91$ , so the time is $14$ minutes.

Answer

$y = 6.5x$ . In $10$ minutes it uses $65$ liters, and $91$ liters take $14$ minutes.

Key idea

Once you know $y = kx$ , multiply by $k$ to go from input to output, and divide by $k$ to go back.
Problem 3 The conveyor belt

A box rides a conveyor belt at a steady speed. After $0$ , $5$ , $10$ and $15$ seconds it has moved $0$ , $4$ , $8$ and $12$ meters.

Write the four ordered pairs, with seconds as $x$ and meters as $y$ , and plot them on a coordinate plane. Describe where the points lie. Then find how far the box has moved after $7.5$ seconds, and when it has moved $10$ meters.
1. Hint 1
  Each column of the table becomes one point, with the time first.
2. Hint 2
  Check whether the ratio of meters to seconds is the same for every point after the first.
3. Hint 3
  That ratio gives an equation for the line, and the equation gives the values between the points exactly.
Answer

$(0, 0)$ , $(5, 4)$ , $(10, 8)$ , $(15, 12)$ . They lie on a straight line through the origin, $y = 0.8x$ . After $7.5$ seconds: $6$ meters. It has moved $10$ meters at $12.5$ seconds.

Full solution

The ordered pairs are $(0, 0)$ , $(5, 4)$ , $(10, 8)$ and $(15, 12)$ .

For the points after the origin, the ratios are $\frac{4}{5}$ , $\frac{8}{10}$ and $\frac{12}{15}$ .
Each equals $0.8$ , and an input of $0$ gives $0$ .
So the relationship is proportional with $k = 0.8$ meters per second, and $y = 0.8x$ .

The points of a proportional relationship lie on a straight line through the origin.
On the plot, each step from one point to the next is $5$ across and $4$ up.

After $7.5$ seconds, $y = 0.8 \times 7.5 = 6$ meters.
For $10$ meters, solve $0.8x = 10$ , so $x = 10 \div 0.8 = 12.5$ seconds.

Answer

$(0, 0)$ , $(5, 4)$ , $(10, 8)$ , $(15, 12)$ . They lie on a straight line through the origin, $y = 0.8x$ . After $7.5$ seconds: $6$ meters. It has moved $10$ meters at $12.5$ seconds.

Key idea

Plotted pairs of a proportional table line up through the origin, and the equation reads off any point between them exactly.
Problem 4 Two straight lines

Line $P$ is straight and passes through $(0, 0)$ and $(8, 10)$ . Line $Q$ is straight and passes through $(0, 3)$ and $(4, 9)$ .

For each line, decide whether it is the graph of a proportional relationship. For any line that is, write its equation.
1. Hint 1
  A proportional graph has two features, and both must be present.
2. Hint 2
  Look at where each line meets the vertical axis.
3. Hint 3
  For a line through the origin, the point that is not the origin gives $k$ as a ratio.
Answer

Line $P$ is proportional, with $y = 1.25x$ . Line $Q$ is not proportional, because it meets the y-axis at $(0, 3)$ instead of the origin.

Full solution

A graph shows a proportional relationship when its points lie on a straight line through the origin, other than the axes.

Line $P$ is straight and passes through the origin, so it is the graph of $y = kx$ .
Its other point gives $k = \frac{10}{8} = 1.25$ , so its equation is $y = 1.25x$ .

Line $Q$ is straight, but at $x = 0$ its height is $3$ .
A proportional relationship gives $y = k \times 0 = 0$ at $x = 0$ .
So line $Q$ is not proportional.

Answer

Line $P$ is proportional, with $y = 1.25x$ . Line $Q$ is not proportional, because it meets the y-axis at $(0, 3)$ instead of the origin.

Key idea

A straight line is proportional only if it also passes through the origin.
Problem 5 Cooling a drink

A drink is put in a freezer for $50$ minutes. Its temperature falls steadily by $4$ degrees Celsius every $10$ minutes. Let $y$ be the change in temperature, in degrees, after $x$ minutes.

Write an equation for $y$ , and give the constant with its units. Find the change after $35$ minutes, and the drink's temperature then if it started at $22$ degrees. Describe the graph for these $50$ minutes.
1. Hint 1
  A fall in temperature is a negative change.
2. Hint 2
  Divide the change by the minutes to find the change in one minute, keeping its sign.
3. Hint 3
  The situation only runs from $0$ to $50$ minutes, so only that part of the line belongs to the graph.
Answer

$y = -0.4x$ , with $k = -0.4$ degrees per minute. After $35$ minutes the change is $-14$ degrees, so the drink is at $8$ degrees. The graph is the piece of the line from $(0, 0)$ to $(50, -20)$ .

Full solution

The temperature changes by $-4$ degrees in $10$ minutes.
So $k = \frac{-4}{10} = -0.4$ degrees per minute, and $y = -0.4x$ .

After $35$ minutes the change is $y = -0.4 \times 35 = -14$ degrees.
The temperature is then $22 + (-14) = 8$ degrees.

The minutes run from $0$ to $50$ , every time in between included.
At $x = 50$ the change is
$-0.4 \times 50 = -20.$
So the graph is a solid piece of the line $y = -0.4x$ , from $(0, 0)$ to $(50, -20)$ .

It falls from left to right, because $k$ is negative.
Apart from its starting point at the origin, it lies in Quadrant IV.

Answer

$y = -0.4x$ , with $k = -0.4$ degrees per minute. After $35$ minutes the change is $-14$ degrees, so the drink is at $8$ degrees. The graph is the piece of the line from $(0, 0)$ to $(50, -20)$ .

Key idea

A negative $k$ makes the line fall, and the situation decides which part of that line is the graph.
Problem 6 Ravi's parking table

A parking lot charges $5$ dollars for $1$ hour, $7$ dollars for $2$ hours, $9$ dollars for $3$ hours and $11$ dollars for $4$ hours.

Ravi says the cost is proportional to the time, because each extra hour adds the same $2$ dollars. Is he right? Explain, say what the lot would charge for $0$ hours if the pattern continued, and describe the graph of the four points.
1. Hint 1
  Equal steps and equal ratios are two different things.
2. Hint 2
  Divide each cost by its hours and compare the results.
3. Hint 3
  Run the pattern back one hour, from $1$ hour to $0$ hours.
Answer

No. The ratios are $5$ , $3.5$ , $3$ and $2.75$ , so the cost is not proportional. The pattern gives $3$ dollars for $0$ hours. The points lie on a straight line that meets the y-axis at $(0, 3)$ , not the origin.

Full solution

The ratios of cost to hours are $\frac{5}{1} = 5$ , $\frac{7}{2} = 3.5$ , $\frac{9}{3} = 3$ and
$\frac{11}{4} = 2.75.$
They are not all equal, so the cost is not proportional to the time.

Ravi is right that each hour adds $2$ dollars, but equal steps are not the test.
Running the pattern back from $1$ hour to $0$ hours subtracts $2$ dollars, giving $5 - 2 = 3$ dollars.

That $3$ dollars is a starting charge.
The four points rise by $2$ for each $1$ across, so they lie on a straight line.
That line meets the y-axis at $(0, 3)$ instead of passing through the origin.

Answer

No. The ratios are $5$ , $3.5$ , $3$ and $2.75$ , so the cost is not proportional. The pattern gives $3$ dollars for $0$ hours. The points lie on a straight line that meets the y-axis at $(0, 3)$ , not the origin.

Key idea

A table can rise by equal steps and still fail to be proportional, because a starting amount spoils the ratios.
Problem 7 Three points on one line

Consider the relationship $y = 2.5x$ . Find the points of its graph with $x = 0$ , $x = 2$ and $x = 6$ .

Explain why the three points must lie on one straight line. Then explain why the graph of every equation of the form $y = kx$ contains the first of them.
1. Hint 1
  Think of reaching the point for $x = 2$ from the origin as one trip across and up.
2. Hint 2
  Compare the trip to the point for $x = 6$ with three of those trips in a row.
3. Hint 3
  For the last part, substitute $x = 0$ into $y = kx$ without choosing a value for $k$ .
Answer

The points are $(0, 0)$ , $(2, 5)$ and $(6, 15)$ . They lie on one straight line through the origin. Every graph of $y = kx$ contains $(0, 0)$ , because $k \times 0 = 0$ .

Full solution

Substituting gives $y = 2.5 \times 0 = 0$ , $y = 2.5 \times 2 = 5$ and
$y = 2.5 \times 6 = 15.$
So the points are $(0, 0)$ , $(2, 5)$ and $(6, 15)$ .

Going from the origin to $(2, 5)$ is a trip of $2$ across and $5$ up.
The point $(6, 15)$ is $6$ across and $15$ up, which is three of those trips laid end to end.

Each new trip starts where the last one ended and heads in exactly the same direction, so the path never bends.
So the origin, $(2, 5)$ and $(6, 15)$ all lie on one straight line.

For any $k$ , putting $x = 0$ into $y = kx$ gives
$y = k \times 0 = 0.$
So every such graph contains the origin, $(0, 0)$ .

Answer

The points are $(0, 0)$ , $(2, 5)$ and $(6, 15)$ . They lie on one straight line through the origin. Every graph of $y = kx$ contains $(0, 0)$ , because $k \times 0 = 0$ .

Key idea

Every point of $y = kx$ is copies of one trip from the origin, so the points share one straight line through the origin.
Problem 8 A line through a point on the left

A straight line passes through the origin and the point $(-5, 8)$ .

Write its equation, and find $y$ when $x = 10$ . Then decide whether a single straight line through the origin can pass through both $(3, 7)$ and $(6, 13)$ , and explain.
1. Hint 1
  The line is straight and goes through the origin, so it has the form $y = kx$ .
2. Hint 2
  Find $k$ from the point that is not the origin, and keep track of the sign.
3. Hint 3
  For the second part, compare the ratio $\frac{y}{x}$ at the two points.
Answer

$y = -1.6x$ , and $y = -16$ when $x = 10$ . No line through the origin passes through both points, because their ratios $\frac{7}{3}$ and $\frac{13}{6}$ differ.

Full solution

A straight line through the origin, other than an axis, is the graph of $y = kx$ .
Its point $(-5, 8)$ gives
$k = \frac{8}{-5} = -1.6.$
So the equation is $y = -1.6x$ .

When $x = 10$ ,
$y = -1.6 \times 10 = -16.$

Every point other than the origin on the line $y = kx$ has the same ratio $\frac{y}{x}$ , namely $k$ .
For $(3, 7)$ that ratio is $\frac{7}{3}$ , and for $(6, 13)$ it is $\frac{13}{6}$ .

As fractions over $6$ , these are $\frac{14}{6}$ and $\frac{13}{6}$ , which are not equal.
So no single straight line through the origin passes through both points.

Answer

$y = -1.6x$ , and $y = -16$ when $x = 10$ . No line through the origin passes through both points, because their ratios $\frac{7}{3}$ and $\frac{13}{6}$ differ.

Key idea

Points whose x-coordinates are not zero share a line through the origin exactly when their ratios $\frac{y}{x}$ are equal.
Problem 9 Two pumps, one tank

Pump A moves $84$ liters of water in $12$ minutes. Pump B moves $y = 6.5x$ liters in $x$ minutes. Both run at steady rates.

On one graph of liters against minutes, whose line is steeper, and why? How long does each pump take to move $182$ liters?
1. Hint 1
  Steepness on these graphs comes from the amount moved in one minute.
2. Hint 2
  Find pump A's liters per minute, and compare it with the number in pump B's equation.
3. Hint 3
  For the times, solve one equation for each pump, with $182$ as the output.
Answer

Pump A's line is steeper, since $7$ liters per minute is more than $6.5$ . Pump A takes $26$ minutes and pump B takes $28$ minutes.

Full solution

Pump A moves $84 \div 12 = 7$ liters per minute, so its equation is $y = 7x$ .
Pump B moves $6.5$ liters per minute.

Both lines pass through the origin, and each climbs its unit rate for every $1$ minute across.
Since $7 > 6.5$ , pump A's line climbs more and is steeper.

For pump A, solve $7x = 182$ , so $x = 182 \div 7 = 26$ minutes.

For pump B, solve $6.5x = 182$ , so $x = 182 \div 6.5 = 28$ minutes.

Answer

Pump A's line is steeper, since $7$ liters per minute is more than $6.5$ . Pump A takes $26$ minutes and pump B takes $28$ minutes.

Key idea

For positive rates, the larger unit rate gives the steeper line through the origin.
Problem 10 Stamps and ribbon

A shop sells stamps at $0.75$ dollars each, and ribbon at $0.75$ dollars per meter, cut to any length. A customer buys up to $8$ stamps, or up to $8$ meters of ribbon. In both cases the cost is $y = 0.75x$ dollars.

Describe how the graph for stamps differs from the graph for ribbon. Then decide, for each graph, whether it contains the point $(3.5, 2.625)$ and whether it contains the point $(-2, -1.5)$ .
1. Hint 1
  The equation is the same, so the difference must come from which inputs each situation allows.
2. Hint 2
  Stamps are counted, while ribbon is measured.
3. Hint 3
  Check each point against the equation first, then against the allowed inputs.
Answer

Stamps: separate dots at $x = 0, 1, \ldots, 8$ on $y = 0.75x$ . Ribbon: the solid piece of $y = 0.75x$ from $(0, 0)$ to $(8, 6)$ . $(3.5, 2.625)$ is on the ribbon graph only. $(-2, -1.5)$ is on neither.

Full solution

Both situations follow $y = 0.75x$ , so every point of either graph lies on that line through the origin.

Stamps come in whole numbers from $0$ to $8$ .
So the stamp graph is nine separate dots, at $x = 0, 1, 2, \ldots, 8$ , ending at $(8, 6)$ .

Ribbon can be any length from $0$ to $8$ meters.
So the ribbon graph is the solid piece of the line from $(0, 0)$ to $(8, 6)$ .

The point $(3.5, 2.625)$ fits the equation, since
$0.75 \times 3.5 = 2.625.$
Three and a half meters of ribbon is allowed, so it is on the ribbon graph.
Three and a half stamps cannot be bought, so it is not on the stamp graph.

The point $(-2, -1.5)$ also fits the equation, since
$0.75 \times (-2) = -1.5.$
But nobody buys a negative amount, so it is on neither graph.

Answer

Stamps: separate dots at $x = 0, 1, \ldots, 8$ on $y = 0.75x$ . Ribbon: the solid piece of $y = 0.75x$ from $(0, 0)$ to $(8, 6)$ . $(3.5, 2.625)$ is on the ribbon graph only. $(-2, -1.5)$ is on neither.

Key idea

The equation fixes the line, and the situation decides which of its points belong to the graph.

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