Proportional relationships in tables

Learning goals

Identifies equivalent ratios in a table.
Completes missing table values using multiplicative relationships.
Determines whether the table represents a proportional relationship.
Identifies the constant of proportionality when appropriate.

Difficulty: 1 – 5

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Proportional relationships in tables 1

The table shows a proportional relationship. <table><tr><td>$x$</td><td>$1$</td><td>$3$</td><td>$5$</td></tr><tr><td>$y$</td><td>$4$</td><td>$12$</td><td>$20$</td></tr></table> What is the constant of proportionality $k$ in $y=kx$?

Hints

- Look for the column where the input is exactly $1$. - In $y=kx$, think about what happens when $x=1$. - Use the matching output from that column.

Solution

1. In a proportional relationship, the output paired with $x=1$ is the constant of proportionality. 2. The table pairs $x=1$ with $y=4$. 3. Therefore, $k=4$.

Answer

$k=4$

Proportional relationships in tables 2

The table shows a proportional relationship. Find the missing value. <table><tr><td>$x$</td><td>$2$</td><td>$6$</td><td>$10$</td></tr><tr><td>$y$</td><td>$3$</td><td>$9$</td><td>$\square$</td></tr></table>

Hints

- Use one complete column as a starting pair. - Compare the known input with the input in the missing-value column. - Apply the same scale factor to the matching output.

Solution

1. From $x=2$ to $x=10$, the input is multiplied by $5$. 2. In a proportional relationship, the corresponding output is multiplied by the same factor. 3. So the missing value is $3\cdot5=15$.

Answer

$15$

Proportional relationships in tables 3

The table shows a proportional relationship. <table><tr><td>$x$</td><td>$2$</td><td>$4$</td><td>$6$</td></tr><tr><td>$y$</td><td>$6$</td><td>$12$</td><td>$18$</td></tr></table> Write the relationship as an equation of the form $y=kx$.

Hints

- Compare one $y$-value with its matching $x$-value. - The constant $k$ tells how many units of $y$ correspond to each unit of $x$. - Put that constant in the form $y=kx$.

Solution

1. The ratio $\frac{y}{x}$ is $\frac{6}{2}=3$, so $k=3$. 2. Therefore, the equation is $y=3x$.

Answer

$y=3x$

Proportional relationships in tables 4

A table gives these pairs: <table><tr><td>$x$</td><td>$0$</td><td>$2$</td><td>$4$</td></tr><tr><td>$y$</td><td>$5$</td><td>$6$</td><td>$12$</td></tr></table> Ava says the relationship is proportional because $6\div2=12\div4$. Is Ava correct? Explain using the first column.

Hints

- Which column cannot be checked by dividing $y$ by $x$? - What does $y=kx$ predict when $x=0$? - Does that prediction match the table?

Solution

1. The two nonzero inputs have the same output-to-input ratio, $3$. 2. But the table also pairs $x=0$ with $y=5$. 3. A proportional relationship follows $y=kx$, which gives $y=0$ when $x=0$. Therefore the table is not proportional.

Answer

No. Even though the nonzero ratios are both $3$, a proportional table must include $(0,0)$ when $x=0$. Here the table has $(0,5)$.

Proportional relationships in tables 5

The table shows a proportional relationship. What input belongs in the empty cell? <table><tr><td>$x$</td><td>$3$</td><td>$9$</td><td>$\square$</td></tr><tr><td>$y$</td><td>$4$</td><td>$12$</td><td>$20$</td></tr></table> Explain how you scaled a complete column.

Hints

- Find a complete column with an output related to the last output. - What factor changes that output into the target output? - Apply the factor to the matching input.

Solution

1. Compare the output $4$ in the first column with the output $20$ in the last: the factor is $20\div4=5$. 2. Scale the matching input by $5$: $3\cdot5=15$. 3. The complete pairs then share the same input-output relationship.

Answer

$x=15$, because $y$ changes from $4$ to $20$ by a factor of $5$, so the matching $x$ changes from $3$ to $15$.

Proportional relationships in tables 6

A currency exchange desk uses the table below to convert US dollars to Canadian dollars. <table> <tr><th>US dollars (USD)</th><td>$1$</td><td>$5$</td><td>$10$</td><td>$50$</td></tr> <tr><th>Canadian dollars (CAD)</th><td>$1.35$</td><td>$6.75$</td><td>$13.50$</td><td>$67.50$</td></tr> </table> a) Is the relationship proportional? Check the rate in every column to explain. b) Write an expression for the amount in Canadian dollars received for $x$ US dollars. c) Use your expression to find the amount received for $\$22$.

Hints

- For each column, what should you divide by what to find Canadian dollars per US dollar? - How many columns must you check before deciding whether the full table is proportional? - Use your result as the multiplier in the expression.

Solution

1. Check every column: $1.35\div1=1.35$, $6.75\div5=1.35$, $13.50\div10=1.35$, and $67.50\div50=1.35$. All four ratios match, so the table is proportional. 2. The constant of proportionality is $1.35$ Canadian dollars per US dollar. 3. For $x$ US dollars, the amount in Canadian dollars is $1.35x$. 4. Substitute $x=22$: $1.35\cdot 22=29.70$.

Answer

a) Yes: $1.35\div1=6.75\div5=13.50\div10=67.50\div50=1.35$. b) $1.35x$ CAD c) $29.70$ CAD

Proportional relationships in tables 7

Two currency exchanges convert US dollars to Canadian dollars. **Exchange A** uses the formula $y=1.34x$, where $x$ is the amount in US dollars and $y$ is the amount in Canadian dollars. **Exchange B** uses this table: <table> <tr><th>US dollars</th><td>$\$10$</td><td>$\$20$</td><td>$\$100$</td></tr> <tr><th>Canadian dollars</th><td>$13.50\,\text{CAD}$</td><td>$27.00\,\text{CAD}$</td><td>$135.00\,\text{CAD}$</td></tr> </table> a) What exchange rate, in Canadian dollars per US dollar, does Exchange B use? b) Which exchange gives you more Canadian dollars for $\$50$? Justify your answer with calculations.

Hints

- Use one column of Exchange B's table to find the amount for $\$1$. - Interpret the coefficient $1.34$ in Exchange A's formula. - Calculate both outputs for the same input amount.

Solution

1. Divide a Canadian-dollar amount by the matching US-dollar amount: $13.50\div 10=1.35$. Exchange B uses $1.35\,\text{CAD}$ per US dollar. 2. Exchange A gives $1.34\cdot 50=67.00\,\text{CAD}$. 3. Exchange B gives $1.35\cdot 50=67.50\,\text{CAD}$. 4. Since $67.50>67.00$, Exchange B gives more.

Answer

a) Exchange B uses a rate of $1.35\,\text{CAD}$ per US dollar. b) Exchange B gives more: Exchange A gives $67.00\,\text{CAD}$, and Exchange B gives $67.50\,\text{CAD}$.

Proportional relationships in tables 8

The table represents a proportional relationship. Find the constant of proportionality $k$, where $y=kx$, and find the missing values $a$ and $b$. <table> <tr><td>$x$</td><td>$0.8$</td><td>$2$</td><td>$a$</td><td>$6$</td></tr> <tr><td>$y$</td><td>$2.4$</td><td>$6$</td><td>$12$</td><td>$b$</td></tr> </table>

Hints

- Use a complete column to find the factor that maps $x$ to $y$. - Use that factor in reverse when $y$ is known. - Multiply by the factor when $x$ is known.

Solution

1. Use a complete pair: $k=2.4\div 0.8=3$. 2. Since $12=3a$, $a=12\div 3=4$. 3. Since $b=3\cdot 6$, $b=18$.

Answer

The constant of proportionality is $k=3$. The missing values are $a=4$ and $b=18$.

Proportional relationships in tables 9

A farm stand sells apples at a price proportional to weight. A $5\,\text{lb}$ bag costs $\$9.00$. a) Find the price per pound. b) Complete the table. <table> <tr><td>Weight, in pounds</td><td>$1$</td><td>$3$</td><td>$5$</td><td>$10$</td></tr> <tr><td>Price</td><td></td><td></td><td></td><td></td></tr> </table> c) Write an equation for the price $y$ of $x$ pounds of apples.

Hints

- First find the price for one pound. - Use the unit price to find the cost of each weight. - In a proportional relationship, price equals unit price times weight.

Solution

1. The price per pound is $9.00\div 5=\$1.80$. 2. Multiply each weight by $1.80$: the prices are $\$1.80$, $\$5.40$, $\$9.00$, and $\$18.00$. 3. The equation is $y=1.80x$.

Answer

a) $\$1.80$ per pound. b) <table> <tr><td>Weight, in pounds</td><td>$1$</td><td>$3$</td><td>$5$</td><td>$10$</td></tr> <tr><td>Price</td><td>$\$1.80$</td><td>$\$5.40$</td><td>$\$9.00$</td><td>$\$18.00$</td></tr> </table> c) $y=1.80x$.

Proportional relationships in tables 10

A cyclist rides at a steady speed. One distance in the table was copied incorrectly. The times are correct. <table><tr><td>Hours</td><td>$2$</td><td>$5$</td><td>$8$</td></tr><tr><td>Miles</td><td>$7$</td><td>$17.5$</td><td>$29$</td></tr></table> Which distance must be corrected? What should it be? Explain using miles per hour.

Hints

- Compare the distances per hour in the table's columns. - Can one speed explain all three distances? - Use a consistent rate to check the distance that does not fit.

Solution

1. The first two columns give $7\div2=3.5$ and $17.5\div5=3.5$ miles per hour. 2. At that constant speed, $8$ hours corresponds to $3.5\cdot8=28$ miles. 3. Since only a distance can be wrong, the entry $29$ miles must be corrected to $28$ miles.

Answer

Change $29$ miles to $28$ miles. The rate is $3.5\,\text{mi/h}$ in the first two columns, so the last distance is $3.5\cdot8=28$ miles.

Proportional relationships in tables 11

At a bulk-food store, almond butter costs $\$1.80$ for every $4\,\text{oz}$. a) Explain why weight and price form a proportional relationship. b) Make a value table for $2\,\text{oz}$, $4\,\text{oz}$, $10\,\text{oz}$, and $20\,\text{oz}$. c) An empty jar weighs $6\,\text{oz}$. After it is filled with almond butter, the jar weighs $20\,\text{oz}$. Find the price of the almond butter in the jar.

Hints

- What must stay constant in a proportional weight-to-price relationship? - Find the cost of $1\,\text{oz}$. - Subtract the empty jar's weight before finding the price. - Use the unit price for the weight of the almond butter.

Solution

1. The unit price is constant: $\$1.80\div 4=\$0.45$ per ounce. Therefore, the relationship is proportional. 2. Multiply each weight by $\$0.45$ per ounce to complete the requested value table. <table><tr><td>Weight</td><td>$2\,\text{oz}$</td><td>$4\,\text{oz}$</td><td>$10\,\text{oz}$</td><td>$20\,\text{oz}$</td></tr><tr><td>Price</td><td>$\$0.90$</td><td>$\$1.80$</td><td>$\$4.50$</td><td>$\$9.00$</td></tr></table> 3. The almond butter weighs $20-6=14\,\text{oz}$. 4. Its price is $14\cdot 0.45=\$6.30$.

Answer

a) The relationship is proportional because the price per ounce is constant. b) <table><tr><td>Weight</td><td>$2\,\text{oz}$</td><td>$4\,\text{oz}$</td><td>$10\,\text{oz}$</td><td>$20\,\text{oz}$</td></tr><tr><td>Price</td><td>$\$0.90$</td><td>$\$1.80$</td><td>$\$4.50$</td><td>$\$9.00$</td></tr></table> c) The almond butter weighs $14\,\text{oz}$ and costs $\$6.30$.

Proportional relationships in tables 12

A coffee roaster sells specialty beans at a fixed price per ounce. A $16\,\text{oz}$ bag costs $\$12.00$. a) What type of relationship connects weight and price? Justify your answer. Give the constant of proportionality in dollars per ounce and the price per pound. b) Find the prices for $4\,\text{oz}$, $8\,\text{oz}$, $24\,\text{oz}$, and $2.5\,\text{lb}$, and display them in a table.

Hints

- Check whether doubling the amount doubles the price. - Find the price of $1\,\text{oz}$. - Convert the final weight to ounces before using the unit price.

Solution

1. Because every ounce has the same price, the relationship is proportional. 2. The unit price is $12.00\div 16=\$0.75$ per ounce. Since $16\,\text{oz}=1\,\text{lb}$, the price per pound is $\$12.00$. 3. The prices are $4\cdot 0.75=\$3.00$, $8\cdot 0.75=\$6.00$, and $24\cdot 0.75=\$18.00$. 4. Convert $2.5\,\text{lb}$ to $40\,\text{oz}$. Its price is $40\cdot 0.75=\$30.00$.

Answer

a) The relationship is proportional. The constant of proportionality is $\$0.75$ per ounce, and the price is $\$12.00$ per pound. b) <table> <tr><td>Weight</td><td>$4\,\text{oz}$</td><td>$8\,\text{oz}$</td><td>$24\,\text{oz}$</td><td>$2.5\,\text{lb}$</td></tr> <tr><td>Price</td><td>$\$3.00$</td><td>$\$6.00$</td><td>$\$18.00$</td><td>$\$30.00$</td></tr> </table>

Proportional relationships in tables 13

A leaking faucet drips at a constant rate. The table shows time $t$, in minutes, and the volume $V$, in milliliters, collected in a measuring cup. <table> <tr><td>Time $t$, in minutes</td><td>$4$</td><td>$8$</td><td>$12$</td><td>$20$</td></tr> <tr><td>Volume $V$, in milliliters</td><td>$60$</td><td>$120$</td><td>$180$</td><td>$300$</td></tr> </table> a) Write an equation for $V$ in terms of $t$. b) How much water is collected after $45$ minutes and after $60$ minutes? c) How long does it take to collect exactly $1.5$ liters?

Hints

- Find the number that multiplies time to produce volume. - Keep time and volume units consistent. - Rearrange the equation when volume is known and time is unknown.

Solution

1. The ratios are constant: $60\div 4=120\div 8=180\div 12=300\div 20=15$. 2. The equation is $V=15t$. 3. At $45$ minutes, $V=15\cdot 45=675\,\text{mL}$. At $60$ minutes, $V=15\cdot 60=900\,\text{mL}$. 4. Convert $1.5$ liters to $1500\,\text{mL}$. Then $t=1500\div 15=100$ minutes.

Answer

a) $V=15t$. b) $675\,\text{mL}$ after $45$ minutes and $900\,\text{mL}$ after $60$ minutes. c) $100$ minutes, or $1$ hour $40$ minutes.

Proportional relationships in tables 14

The table represents a proportional relationship. <table> <tr><td>$x$</td><td>$1.5$</td><td>$4$</td><td>$7.5$</td><td>$10$</td></tr> <tr><td>$y$</td><td>$4.5$</td><td>$12$</td><td>$22.5$</td><td>$30$</td></tr> </table> a) Find the constant of proportionality $k$ and write the equation $y=kx$. b) Find $y$ when $x=0.5$ and when $x=100$. c) Explain how the graph would change if the constant of proportionality were doubled.

Hints

- Divide each $y$-value by its matching $x$-value. - Once $k$ is known, substitute any input into $y=kx$. - If $k$ doubles, what happens to the corresponding $y$-value for the same $x$?

Solution

1. The ratios are $4.5\div 1.5=3$, $12\div 4=3$, and $30\div 10=3$, so $k=3$. 2. The equation is $y=3x$. 3. For $x=0.5$, $y=3\cdot 0.5=1.5$. For $x=100$, $y=3\cdot 100=300$. 4. Doubling $k$ from $3$ to $6$ doubles every corresponding $y$-value. The new graph is still a line through the origin, but it rises more quickly as $x$ increases.

Answer

a) $k=3$, so $y=3x$. b) $y=1.5$ when $x=0.5$, and $y=300$ when $x=100$. c) The graph would still be a line through the origin, and each $y$-value for the same positive $x$ would be twice as large.

Proportional relationships in tables 15

A hiking group travels at a constant speed. The table shows distance $y$, in miles, after time $x$, in hours. <table> <tr><td>Time $x$ in hours</td><td>$1$</td><td>$2$</td><td>$3$</td><td>$4$</td><td>$5$</td></tr> <tr><td>Distance $y$ in miles</td><td>$3$</td><td>$6$</td><td>$9$</td><td>$12$</td><td>$15$</td></tr> </table> 1) Choose any two time values $x_1$ and $x_2$. Compare $\frac{x_1}{x_2}$ with $\frac{y_1}{y_2}$ for the matching distances. What do you notice? 2) Find the constant of proportionality and write an equation relating $x$ and $y$. 3) Find the distance traveled after $6.5$ hours. 4) Describe the graph of this relationship.

Hints

- Check whether corresponding values change by the same scale factor. - Find the distance traveled in one hour. - A proportional relationship has an equation of the form $y=kx$. - What special point does every proportional graph contain?

Solution

1. For example, using $x_1=2$ and $x_2=4$, $\frac{2}{4}=0.5$. The corresponding distances are $6$ and $12$, and $\frac{6}{12}=0.5$. The ratios are equal. 2. The constant of proportionality is $k=3\div 1=3\,\text{mi/h}$, so $y=3x$. 3. After $6.5$ hours, $y=3\cdot 6.5=19.5$ miles. 4. The graph is a straight line through $(0, 0)$. For each increase of $1$ hour, the distance increases by $3$ miles.

Answer

1) The ratios are equal; for example, $\frac{2}{4}=\frac{6}{12}=0.5$. 2) $k=3\,\text{mi/h}$; $y=3x$ 3) $19.5$ miles 4) A straight line through the origin; distance increases by $3$ miles for each additional hour.

Proportional relationships in tables 16

A copier prints at a constant speed. The table shows the number of fully completed pages $y$ at certain times $x$, in seconds. <table><tr><td>Time $x$ in seconds</td><td>$10$</td><td>$20$</td><td>$35$</td><td>$50$</td></tr><tr><td>Completed pages $y$</td><td>$4$</td><td>$8$</td><td>$14$</td><td>$20$</td></tr></table> a) Find $\frac{y}{x}$ for each pair. What does this show about the table? b) Write an equation relating $x$ and $y$ at the moments when a page is completed. c) How many pages will the copier complete in one minute? d) Suppose a graph shows only the moments when a whole page finishes. Should these be isolated points or a connected line? Explain. What would a connected line $y=0.4x$ represent instead?

Hints

- Check the ratio of completed pages to seconds in each column. - Convert one minute to seconds before using the rate. - Think about what happens between the moments when pages finish. - Does a connected straight line count only completed whole pages?

Solution

1. The ratios are $4\div10=0.4$, $8\div20=0.4$, $14\div35=0.4$, and $20\div50=0.4$; the table pairs follow a proportional pattern. 2. At page-completion moments, $y=0.4x$. 3. One minute is $60$ seconds; $0.4\cdot60=24$ completed pages. 4. When only page-completion moments are plotted, use isolated points. A connected straight line would instead model continuous printing progress, including parts of a page.

Answer

a) Each ratio is $0.4$. The table follows a proportional pattern. b) $y=0.4x$ at page-completion moments. c) $24$ pages d) Isolated points, because the graph includes only times when pages finish. A connected line models progress, including parts of a page.

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