Aimathic – Proportional relationships in tables

Proportional relationships in tables

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Difficulty: 1 – 5

5519267

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

$k=4$

5519277

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>$?$</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

$15$

5545467

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

$y=3x$

5118877

A currency exchange desk uses the table below to convert US dollars to Canadian dollars at a fixed rate. <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) Describe the relationship shown in the table. 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

- Compare how the second row changes when the first row doubles. - Find the amount received for $1$ US dollar. - Multiply the input amount by the constant rate.

Solution

1. Each US-dollar amount is paired with its equivalent amount in Canadian dollars. 2. The ratio of Canadian dollars to US dollars is constant: $1.35\div 1=1.35$. The relationship is proportional with constant of proportionality $1.35$. 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) The table assigns each US-dollar amount its equivalent amount in Canadian dollars at a constant rate. b) $1.35x$ CAD c) $29.70$ CAD

5118887

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 $67.50\,\text{CAD}$.

5119417

Determine whether the table represents a proportional relationship by calculating $\frac{y}{x}$ for every pair. If it is proportional, find the value of $y$ when $x=20$. <table> <tr><td>$x$</td><td>$4$</td><td>$7$</td><td>$12$</td><td>$15$</td></tr> <tr><td>$y$</td><td>$14$</td><td>$24.5$</td><td>$42$</td><td>$52.5$</td></tr> </table>

Hints

- What must be true about $\frac{y}{x}$ in a proportional relationship? - Calculate the ratio for every column. - Apply the constant ratio to the new input.

Solution

1. Calculate the ratios: $14\div 4=3.5$, $24.5\div 7=3.5$, $42\div 12=3.5$, and $52.5\div 15=3.5$. 2. Since all ratios are equal, the relationship is proportional with constant of proportionality $k=3.5$. 3. For $x=20$, $y=3.5\cdot 20=70$.

Answer

The relationship is proportional because $\frac{y}{x}=3.5$ for every pair. When $x=20$, $y=70$.

5119437

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 $b=18$.

5119657

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) $y=1.80x$.

5119837

Analyze the complete pairs in the table. Are they consistent with a proportional relationship? If so, state the constant of proportionality $k$, where $y=kx$. Then assume the missing entries are intended to continue that proportional relationship and complete them. <table> <tr><td>$x$</td><td>$2.5$</td><td>$4$</td><td>$6$</td><td>$\ldots$</td><td>$15$</td></tr> <tr><td>$y$</td><td>$6.25$</td><td>$10$</td><td>$\ldots$</td><td>$30$</td><td>$\ldots$</td></tr> </table>

Hints

- Compare $y\div x$ for each pair in which both values are shown. - Distinguish what the shown data establish from the assumption about the missing entries. - Once the constant is identified, decide whether to multiply or divide for each missing value.

Solution

1. The known complete-pair ratios are $6.25\div 2.5=2.5$ and $10\div 4=2.5$, so those pairs are consistent with $k=2.5$. 2. Under the stated assumption that the missing entries continue that proportional relationship, use $y=2.5x$. 3. For $x=6$, $y=2.5\cdot 6=15$. 4. For $y=30$, $x=30\div 2.5=12$. 5. For $x=15$, $y=2.5\cdot 15=37.5$.

Answer

The complete pairs are consistent with a proportional relationship with $k=2.5$. Under the stated continuation assumption, $y=2.5x$, and the missing values are $15$, $12$, and $37.5$, in table order.

5131507

An automatic irrigation system starts with an empty garden tank at $t=0$ and fills it with water. The table shows the water volume $V$, in gallons, at several times $t$, in minutes. <table> <tr><td>Time $t$, in minutes</td><td>$4$</td><td>$10$</td><td>$15$</td></tr> <tr><td>Volume $V$, in gallons</td><td>$30$</td><td>$75$</td><td>$112.5$</td></tr> </table> a) Use calculations to determine whether time and volume are proportional. b) Find the constant of proportionality $k$ and explain its meaning. c) If the relationship is graphed, through which special point must the line pass? Explain in context.

Hints

- Check whether volume divided by time is constant for the shown pairs. - Find the amount added in one minute. - Use the stated starting condition when deciding what point the graph must contain.

Solution

1. The ratios are $30\div 4=7.5$, $75\div 10=7.5$, and $112.5\div 15=7.5$. Together with the stated empty start, the relationship is proportional. 2. The constant is $k=7.5$ gallons per minute, the system's filling rate. 3. The graph must pass through $(0, 0)$ because the tank contains $0$ gallons at $0$ minutes.

Answer

a) Yes. Every shown ratio $V\div t$ equals $7.5$, and the tank starts at $(0,0)$. b) $k=7.5$ gallons per minute. c) The line passes through $(0, 0)$.

5119097

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 $\$6.30$.

5119537

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>

5125117

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.

5128427

While a backyard pool is being filled, two measurements are recorded. <table> <tr><td>Time $t$, in minutes</td><td>$5$</td><td>$12$</td></tr> <tr><td>Water volume $V$, in gallons</td><td>$120$</td><td>$288$</td></tr> </table> a) Use calculations to determine whether the two measurements are consistent with a proportional relationship. b) Find the filling rate in gallons per minute. c) How much water is in the pool after $25$ minutes if the process continues at the same rate?

Hints

- How can a table show that two quantities are proportional? - Find the gallons per minute in both columns. - Use the unit rate for the new time.

Solution

1. The rates are $120\div 5=24$ and $288\div 12=24$, so the measurements are consistent with a proportional relationship. 2. The filling rate is $24$ gallons per minute. 3. After $25$ minutes, the volume is $24\cdot 25=600$ gallons.

Answer

a) Yes. Both ratios $V\div t$ equal $24$. b) $24$ gallons per minute. c) $600$ gallons.

5131127

A car travels at a constant speed on a highway. The table shows distance $s$, in miles, and gasoline used $V$, in gallons. <table> <tr><td>Distance $s$, in miles</td><td>$100$</td><td>$250$</td><td>$400$</td><td>$550$</td></tr> <tr><td>Gasoline $V$, in gallons</td><td>$4$</td><td>$10$</td><td>$16$</td><td>$22$</td></tr> </table> Determine whether gasoline used is proportional to distance. If so, find the constant of proportionality and explain its meaning in context.

Hints

- Compare gasoline used per mile for every column. - What units result when gallons are divided by miles? - Interpret that unit rate for the car.

Solution

1. Calculate $V\div s$: $4\div 100=0.04$, $10\div 250=0.04$, $16\div 400=0.04$, and $22\div 550=0.04$. 2. Since all ratios are equal, the relationship is proportional. 3. The constant of proportionality is $k=0.04$ gallon per mile, equivalent to $4$ gallons per $100$ miles.

Answer

The relationship is proportional because $V\div s=0.04$ for every pair. The constant $0.04$ means the car uses $0.04$ gallon per mile, or $4$ gallons per $100$ miles.

5131387

The table shows measured gasoline use for a delivery truck. <table> <tr><td>Distance, in miles</td><td>$80$</td><td>$150$</td><td>$240$</td><td>$310$</td></tr> <tr><td>Gasoline used, in gallons</td><td>$2.6$</td><td>$4.9$</td><td>$7.75$</td><td>$10.1$</td></tr> </table> a) Calculate the gasoline-per-mile ratio for each measurement. Do all four ratios fall between $0.032$ and $0.033$ gallon per mile? Explain what this suggests about using a proportional model for estimation. b) Find a suitable estimated unit rate by calculating the mean of the four gasoline-per-mile ratios. Round to the nearest ten-thousandth. c) Use a proportional model with that estimated unit rate to estimate the gasoline used for $600$ miles. Round to the nearest tenth of a gallon.

Hints

- Find gasoline used per mile for every pair. - Compare every calculated ratio with the interval stated in part a). - Average the four ratios rather than selecting one of them. - Use the estimated unit rate for the new distance.

Solution

1. The ratios are $2.6\div 80=0.0325$, $4.9\div 150\approx 0.0327$, $7.75\div 240\approx 0.0323$, and $10.1\div 310\approx 0.0326$ gallon per mile. 2. All four ratios lie between $0.032$ and $0.033$ gallon per mile. They are close but not identical, so a proportional model is reasonable for estimation but does not describe the measurements exactly. 3. The mean is $\frac{0.0325+0.032666\ldots+0.032291\ldots+0.032580\ldots}{4}\approx 0.0325$ gallon per mile. This is an estimated unit rate for the model, not an exact constant shared by all four measurements. 4. The estimate is $600\cdot 0.0325=19.5$ gallons.

Answer

a) Yes. All four ratios are between $0.032$ and $0.033$ gallon per mile. This supports using a proportional model for estimation, even though the measured ratios are not exactly equal. b) The estimated unit rate is approximately $0.0325$ gallon per mile. c) Approximately $19.5$ gallons.

5131397

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.

5131427

A copper-wire manufacturer measures the length $l$ and mass $m$ of several rolls. <table> <tr><td>Length $l$, in meters</td><td>$10$</td><td>$25$</td><td>$40$</td><td>$60$</td><td>$100$</td></tr> <tr><td>Mass $m$, in grams</td><td>$185$</td><td>$462.5$</td><td>$740$</td><td>$1110$</td><td>$1850$</td></tr> </table> a) Determine whether length and mass form a proportional relationship. b) Find the constant of proportionality and explain its physical meaning for the wire.

Hints

- Find mass divided by length for each column. - Check whether this ratio stays constant. - Interpret the units grams per meter.

Solution

1. Calculate $m\div l$: $185\div 10=18.5$, $462.5\div 25=18.5$, $740\div 40=18.5$, $1110\div 60=18.5$, and $1850\div 100=18.5$. 2. Since every ratio is equal, the relationship is proportional. 3. The constant $18.5\,\text{g/m}$ is the wire's mass per meter, or linear density.

Answer

a) Yes. The ratio $m\div l$ is constant at $18.5$. b) $k=18.5\,\text{g/m}$, meaning each meter of wire has a mass of $18.5$ grams.

5131707

Two runners, Lucas and Simon, are training for a marathon. Their distances were measured at different times: Lucas: <table> <tr><td>Time $t$ (in $\text{min}$)</td><td>$10$</td><td>$25$</td><td>$40$</td></tr> <tr><td>Distance $d$ (in $\text{m}$)</td><td>$2200$</td><td>$5500$</td><td>$8800$</td></tr> </table> Simon: <table> <tr><td>Time $t$ (in $\text{min}$)</td><td>$12$</td><td>$20$</td><td>$45$</td></tr> <tr><td>Distance $d$ (in $\text{m}$)</td><td>$2700$</td><td>$4500$</td><td>$10{,}125$</td></tr> </table> a) Show that distance is proportional to time for each runner. b) Who runs faster? Justify your answer by comparing the constants of proportionality. c) How far would the faster runner travel in one hour at the same speed?

Hints

- What does the value of distance divided by time represent? - How is speed related to the constant of proportionality? - Pay attention to the units when finding the distance traveled in one hour.

Solution

1. For Lucas, divide each distance by its time: $2200 \div 10=220$, $5500 \div 25=220$, and $8800 \div 40=220$. The constant ratio is $220$, so the relationship is proportional. 2. For Simon, $2700 \div 12=225$, $4500 \div 20=225$, and $10{,}125 \div 45=225$. The constant ratio is $225$, so this relationship is also proportional. 3. The constants of proportionality are $220\,\text{m/min}$ for Lucas and $225\,\text{m/min}$ for Simon. Simon runs faster because $225>220$. 4. In $60$ minutes, Simon travels $225 \cdot 60=13{,}500\,\text{m}$, or $13.5\,\text{km}$.

Answer

a) Both relationships are proportional because the ratio $\frac{d}{t}$ is constant for each runner: $220$ for Lucas and $225$ for Simon. b) Simon runs faster because $225\,\text{m/min}>220\,\text{m/min}$. c) Simon would run $13{,}500\,\text{m}$, or $13.5\,\text{km}$, in one hour.

5141987

Darius records several time-distance pairs from a bicycle ride. The table shows distance $d$ as a function of time $t$. a) Do the three complete pairs support the claim that distance is proportional to time? Justify your answer using the ratios $\frac{d}{t}$. b) Complete the missing entries so the table continues that proportional relationship. <table> <tr><td>Time $t$ (in $\text{h}$)</td><td>$1.5$</td><td>$2$</td><td>$3.5$</td><td>$5$</td><td></td></tr> <tr><td>Distance $d$ (in $\text{mi}$)</td><td>$27$</td><td>$36$</td><td>$63$</td><td></td><td>$108$</td></tr> </table>

Hints

- Test the claim from the numerical pairs rather than from the story context. - Find $\frac{d}{t}$ for each complete column and compare the results. - If one constant ratio works for all complete pairs, use that same ratio to fill the missing entries.

Solution

1. Divide distance by time for the known pairs: $27 \div 1.5=18$, $36 \div 2=18$, and $63 \div 3.5=18$. The constant rate is $18\,\text{mi/h}$, so the relationship is proportional. 2. At $5$ hours, the distance is $18 \cdot 5=90$ miles. 3. To travel $108$ miles, the time is $108 \div 18=6$ hours.

Answer

a) Yes. The three complete pairs support the claim because $\frac{d}{t}=18\,\text{mi/h}$ for each one. b) The missing distance is $90$ miles, and the missing time is $6$ hours.

5241617

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.

5244157

A copier prints at a constant speed. The table shows the number of fully completed pages $y$ after $x$ seconds. <table><tr><td>Time $x$ in seconds</td><td>$10$</td><td>$20$</td><td>$35$</td><td>$50$</td></tr><tr><td>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 relationship? b) Write an equation for pages as a function of time at the listed page-completion times. c) How many pages does the copier complete in one minute? d) If a graph records only fully completed pages, should the plotted values be shown as isolated points or as one connected line? Explain. How does the line $y=0.4x$ relate to those points if printing progress is modeled continuously between page completions?

Hints

- Check whether the ratio of completed pages to seconds is constant. - Convert one minute to seconds before using the rate. - For the graph question, decide whether a completed-page count can take values such as $4.5$. - Separate a graph of completed whole pages from a continuous model of printing progress.

Solution

1. The ratios are $4\div10=0.4$, $8\div20=0.4$, $14\div35=0.4$, and $20\div50=0.4$. The constant ratio shows that the listed values follow a proportional relationship. 2. The equation is $y=0.4x$ at page-completion times. 3. One minute is $60$ seconds, so $y=0.4\cdot60=24$ pages. 4. If $y$ counts only fully completed pages, the graph should show discrete points because page counts are whole numbers. Those points lie on the line $y=0.4x$. A connected line can instead represent an idealized continuous model of printing progress, including partial-page progress between completions.

Answer

a) Every ratio equals $0.4$, so the listed values are proportional. b) $y=0.4x$ at page-completion times c) $24$ pages d) Use isolated points for fully completed page counts. The points lie on $y=0.4x$; the connected line represents an idealized continuous printing-progress model that includes partial pages.

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