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Proportional relationships in tables

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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 table. Does it represent a proportional relationship? If so, state the constant of proportionality \(k\), where \(y=kx\), and complete the missing values. <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

- Check whether \(y\div x\) is constant for the complete pairs. - Once you know the constant, use it for each missing value. - Decide whether to multiply or divide based on which variable is missing.

Solution

1. The known ratios are \(6.25\div 2.5=2.5\) and \(10\div 4=2.5\), so the table is proportional with \(k=2.5\). 2. For \(x=6\), \(y=2.5\cdot 6=15\). 3. For \(y=30\), \(x=30\div 2.5=12\). 4. For \(x=15\), \(y=2.5\cdot 15=37.5\).

Answer

Yes. The relationship is proportional with \(k=2.5\), so \(y=2.5x\). The missing values are \(15\), \(12\), and \(37.5\), in table order.
5131507
An automatic irrigation system fills a garden tank. 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. - Find the amount added in one minute. - What has happened at the starting time?

Solution

1. The ratios are \(30\div 4=7.5\), \(75\div 10=7.5\), and \(112.5\div 15=7.5\). 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 at \(0\) minutes, \(0\) gallons have been added.

Answer

a) Yes. Every ratio \(V\div t\) equals \(7.5\). 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. The prices are \(2\cdot 0.45=\$0.90\), \(4\cdot 0.45=\$1.80\), \(10\cdot 0.45=\$4.50\), and \(20\cdot 0.45=\$9.00\). 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) \(2\,\text{oz}\to\$0.90\), \(4\,\text{oz}\to\$1.80\), \(10\,\text{oz}\to\$4.50\), and \(20\,\text{oz}\to\$9.00\). 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) Determine whether the output-to-input ratios are approximately equal. What does “approximately” mean in this context? b) Find a suitable constant of proportionality by calculating the mean of the four gasoline-per-mile ratios. Round to the nearest ten-thousandth. c) Use the model 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. - Decide whether the ratios are identical or merely close. - Average the four ratios rather than selecting one of them. - Multiply the model's unit rate by 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. These values are close but not identical. “Approximately” allows small differences caused by measurement error and changing driving conditions. 3. The mean is \(\frac{0.0325+0.032666\ldots+0.032291\ldots+0.032580\ldots}{4}\approx 0.0325\) gallon per mile. 4. The estimate is \(600\cdot 0.0325=19.5\) gallons.

Answer

a) Yes. The ratios are close to one another; small differences can come from measurement and driving conditions. b) \(k\approx 0.0325\) 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 \(y=kx\). - How does increasing the coefficient of \(x\) affect slope?

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 the slope, so the line through the origin becomes steeper.

Answer

a) \(k=3\), so \(y=3x\). b) \(y=1.5\) when \(x=0.5\), and \(y=300\) when \(x=100\). c) The line would be steeper because its slope would be \(6\) instead of \(3\).
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
A cyclist rides at a constant speed. The table shows distance \(d\) as a function of time \(t\). a) Determine whether the relationship is proportional. b) Find the missing values in the table. <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

- What does constant speed mean about the ratio of distance to time? - Find the miles traveled per hour for each known pair. - Once you know the unit rate, use it to find each missing value.

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 relationship is proportional because \(\frac{d}{t}=18\,\text{mi/h}\) for every known pair. 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 \((0, 0)\) with slope \(3\).

Answer

1) The ratios are equal; for example, \(\frac{2}{4}=\frac{6}{12}=0.5\). 2) \(k=3\); \(y=3x\) 3) \(19.5\) miles 4) A line through the origin with slope \(3\)
5244157
A copier prints at a constant speed. The table shows the number of pages \(y\) printed after \(x\) seconds. | Time \(x\) in seconds | \(10\) | \(20\) | \(35\) | \(50\) | | :--- | :---: | :---: | :---: | :---: | | Pages \(y\) | \(4\) | \(8\) | \(14\) | \(20\) | 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. c) How many pages does the copier print in one minute? d) Describe the graph and explain why it represents a proportional relationship.

Hints

- Check whether the ratio of pages to seconds is constant. - How does the page count change when time doubles? - Convert one minute to seconds. - Where is the graph when no time has passed?

Solution

1. The ratios are \(4 \div 10=0.4\), \(8 \div 20=0.4\), \(14 \div 35=0.4\), and \(20 \div 50=0.4\). The constant ratio shows that the relationship is proportional. 2. The equation is \(y=0.4x\). 3. One minute is \(60\) seconds, so \(y=0.4 \cdot 60=24\) pages. 4. The graph is a straight line through the origin. A straight-line graph through the origin represents a proportional relationship.

Answer

a) Every ratio equals \(0.4\), so the relationship is proportional. b) \(y=0.4x\) c) \(24\) pages d) It is a straight line through the origin, so it represents a proportional relationship.

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