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Test for proportional relationships

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5519307
Which equation represents a proportional relationship? a) \(y=5x\) b) \(y=x+5\) c) \(y=5\)

Hints

- Recall the standard equation form for a proportional relationship. - Look for an equation where the output is a constant multiple of the input. - Check whether any added constant remains when the input is \(0\).

Solution

1. A proportional relationship can be written in the form \(y=kx\). 2. Choice a), \(y=5x\), has exactly that form. 3. Choices b) and c) do not have the form \(y=kx\).

Answer

a) \(y=5x\)
5519317
The graph shows four plotted pairs in a relationship. Is the relationship proportional? Explain using the graph.
Figure for problem 551931

Hints

- Look at the overall pattern made by the plotted points. - Check whether the origin is one of the plotted pairs. - Recall the graph shape associated with a proportional relationship.

Solution

1. The plotted pairs are \((0,0)\), \((1,3)\), \((2,6)\), and \((3,9)\). 2. The points line up on a straight line that passes through the origin. 3. Therefore, the relationship is proportional.

Answer

Yes. The plotted points lie on one straight line through the origin, so the relationship is proportional.
5545487
The table shows two pairs. <table><tr><td>\(x\)</td><td>\(2\)</td><td>\(4\)</td></tr><tr><td>\(y\)</td><td>\(3\)</td><td>\(6\)</td></tr></table> Does the table represent a proportional relationship? Give one numerical check that supports your answer.

Hints

- A proportional table must use the same value of \(\frac{y}{x}\) for every pair. - Compare how the second pair scales from the first pair. - One consistent multiplicative scale factor should work for both coordinates.

Solution

1. When \(x\) doubles from \(2\) to \(4\), \(y\) also doubles from \(3\) to \(6\). Equivalently, \(\frac{3}{2}=\frac{6}{4}=1.5\). 2. Therefore, the two-pair table represents a proportional relationship with constant of proportionality \(1.5\).

Answer

Yes. For example, \(\frac{3}{2}=\frac{6}{4}=1.5\), so both pairs use the same constant of proportionality.
5119307
Consider each geometric relationship. Decide whether it is proportional. a) Side length of an equilateral triangle \(\to\) perimeter of the triangle. b) Radius of a circle \(\to\) area of the circle. c) Edge length of a cube \(\to\) total length of all its edges. d) Edge length of a cube \(\to\) surface area of the cube.

Hints

- Write a formula for each relationship. - Check whether multiplying the input by a factor multiplies the output by the same factor. - In particular, compare what happens when the input is doubled.

Solution

1. a) Proportional: \(P=3s\), so the perimeter is always three times the side length. 2. b) Not proportional: \(A=\pi r^2\). Doubling the radius multiplies the area by \(4\), not by \(2\). 3. c) Proportional: a cube has \(12\) edges, so \(L=12s\). 4. d) Not proportional: \(S=6s^2\). Doubling the edge length multiplies the surface area by \(4\), not by \(2\).

Answer

a) Proportional. b) Not proportional. c) Proportional. d) Not proportional.
5119357
A farm stand sells apples in different bags: - A \(3\,\text{lb}\) bag costs \(\$4.20\). - A \(6\,\text{lb}\) bag costs \(\$7.80\). - A \(10\,\text{lb}\) bag costs \(\$12.00\). Use calculations to determine whether the relationship weight \(\to\) price is proportional. Justify your answer.

Hints

- What must be true about the price per pound in a proportional relationship? - Find the price for exactly \(1\,\text{lb}\) for each bag. - Compare the unit prices.

Solution

1. Find the price per pound for each bag: \(4.20\div 3=\$1.40\) per pound, \(7.80\div 6=\$1.30\) per pound, and \(12.00\div 10=\$1.20\) per pound. 2. Since the unit prices are not equal, the relationship is not proportional.

Answer

The relationship is not proportional because the unit prices are different: \(\$1.40\), \(\$1.30\), and \(\$1.20\) per pound.
5119387
A car-sharing service charges a fixed \(\$5.00\) booking fee plus \(\$0.50\) for each mile driven. a) Find the total cost for a \(10\)-mile trip and for a \(20\)-mile trip. b) Is the relationship between miles driven and total cost proportional? Justify your answer using your results from part a).

Hints

- In a proportional relationship, what happens to the output when the input is doubled? - Calculate both total costs before comparing them. - Separate the fixed part of the cost from the part that changes with distance.

Solution

1. For \(10\) miles, the cost is \(5.00+10\cdot 0.50=\$10.00\). 2. For \(20\) miles, the cost is \(5.00+20\cdot 0.50=\$15.00\). 3. If the relationship were proportional, doubling the miles would double the cost. Doubling \(\$10.00\) gives \(\$20.00\), not \(\$15.00\). 4. Therefore, the relationship is not proportional.

Answer

a) The costs are \(\$10.00\) for \(10\) miles and \(\$15.00\) for \(20\) miles. b) The relationship is not proportional because doubling the distance does not double the total cost.
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 \(x=20\), \(y=70\).
5119427
Two tables are shown. Decide which table represents a proportional relationship and which does not. Justify your decision by comparing \(\frac{y}{x}\) values. Table A: <table> <tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr> <tr><td>\(y\)</td><td>\(1.5\)</td><td>\(3\)</td><td>\(4.5\)</td><td>\(6\)</td></tr> </table> Table B: <table> <tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr> <tr><td>\(y\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> </table>

Hints

- Calculate the ratio of \(y\) to \(x\) for each pair in both tables. - What condition defines a proportional relationship? - Is checking only the first pair enough?

Solution

1. For Table A, the ratios are \(1.5\div 1=1.5\), \(3\div 2=1.5\), \(4.5\div 3=1.5\), and \(6\div 4=1.5\). Table A is proportional. 2. For Table B, \(2\div 1=2\), while \(3\div 2=1.5\). Since the ratios are not equal, Table B is not proportional.

Answer

Table A is proportional because every ratio \(\frac{y}{x}\) equals \(1.5\). Table B is not proportional because its ratios are not constant.
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.
5120547
A square has side length \(a\). Consider these two relationships: Relationship 1: side length \(a\) \(\rightarrow\) perimeter \(P\) Relationship 2: side length \(a\) \(\rightarrow\) area \(A\) Use a doubling test to explain why Relationship 1 is proportional but Relationship 2 is not. Use \(a=3\,\text{cm}\) and \(a=6\,\text{cm}\) in your explanation.

Hints

- Recall the formulas for the perimeter and area of a square. - In a proportional relationship, doubling the input doubles the output. - Compare each value at \(a=3\) with the corresponding value at \(a=6\).

Solution

1. When \(a=3\,\text{cm}\), the perimeter is \(P=4\cdot3=12\,\text{cm}\). When the side length doubles to \(6\,\text{cm}\), the perimeter is \(P=4\cdot6=24\,\text{cm}\). The perimeter also doubles, so the relationship is proportional. 2. When \(a=3\,\text{cm}\), the area is \(A=3^2=9\,\text{cm}^2\). When the side length doubles to \(6\,\text{cm}\), the area is \(A=6^2=36\,\text{cm}^2\). The area is multiplied by \(4\), not \(2\), so the relationship is not proportional.

Answer

Relationship 1 is proportional because doubling \(a\) from \(3\,\text{cm}\) to \(6\,\text{cm}\) doubles \(P\) from \(12\,\text{cm}\) to \(24\,\text{cm}\). Relationship 2 is not proportional because doubling \(a\) from \(3\,\text{cm}\) to \(6\,\text{cm}\) multiplies \(A\) by \(4\), from \(9\,\text{cm}^2\) to \(36\,\text{cm}^2\).
5131147
Consider two everyday situations. a) Number of equally priced admission tickets with no quantity discount \(\to\) total price. b) Miles traveled in a taxi \(\to\) fare when the taxi charges a fixed starting fee plus a fixed price per mile. For each situation, decide whether the relationship is proportional. Explain how the taxi's starting fee affects proportionality.

Hints

- Check whether doubling the input exactly doubles the output. - Does a taxi charge anything before the first mile is traveled? - How does adding a fixed starting value affect a graph?

Solution

1. The ticket relationship is proportional because the price per ticket is constant. Doubling the number of tickets doubles the total price. 2. The taxi relationship is not proportional because the fixed starting fee is charged even when \(0\) miles are traveled. 3. The starting fee makes the fare-to-distance ratio vary and makes the graph start above the origin.

Answer

a) Proportional, because the price per ticket is constant. b) Not proportional, because the fixed starting fee prevents the graph from passing through the origin.
5131357
Decide whether each situation represents a proportional relationship. Briefly justify each decision. a) Number of movie tickets sold and total revenue when every ticket has the same price. b) Age of a tree and its current height. c) Side length of a square and its perimeter. d) Travel time and distance for a train moving at a constant speed of \(60\) miles per hour.

Hints

- If the first quantity doubles, must the second also double? - Check whether the ratio of the two quantities stays constant. - Consider what the second quantity should be when the first is \(0\).

Solution

1. a) Proportional. Revenue equals ticket price times number of tickets. 2. b) Not proportional. Trees do not grow the same amount each year, and height is not a constant multiple of age. 3. c) Proportional. The perimeter is \(P=4s\). 4. d) Proportional. At constant speed, \(d=60t\).

Answer

a) Proportional. b) Not proportional. c) Proportional. d) Proportional.
5131417
During a long trip, a delivery van's cumulative gasoline use is recorded. <table> <tr><td>Distance \(s\), in miles</td><td>\(150\)</td><td>\(280\)</td><td>\(420\)</td><td>\(550\)</td></tr> <tr><td>Gasoline \(G\), in gallons</td><td>\(5.1\)</td><td>\(9.5\)</td><td>\(14.3\)</td><td>\(18.7\)</td></tr> </table> a) Use calculations to determine whether \(s\to G\) is exactly proportional. b) Give one reason real measurements may differ slightly from exact proportionality. What does \(\frac{G}{s}\) mean in this context?

Hints

- Calculate gallons per mile for every pair. - What must be true of all ratios for exact proportionality? - Think of conditions besides distance that affect gasoline use. - Identify the units of gallons divided by miles.

Solution

1. The ratios are \(5.1\div 150=0.034\), \(9.5\div 280\approx 0.03393\), \(14.3\div 420\approx 0.03405\), and \(18.7\div 550=0.034\) gallon per mile. 2. Because the ratios are not exactly equal, the measured relationship is not exactly proportional. 3. Small differences may result from traffic, hills, wind, driving style, or measurement error. 4. The ratio \(G\div s\) is average gasoline use per mile.

Answer

a) No. The ratios are very close but not identical. b) Factors such as traffic, terrain, wind, or driving style can change gasoline use. The ratio \(G\div s\) is gallons used per mile.
5131477
Which equations represent proportional relationships? Briefly justify your choices using the general form of a proportional function. a) \(y=3.5x\) b) \(y=\frac{x}{2}\) c) \(y=2x+1\) d) \(y=x^2\) e) \(y=\frac{4}{x}\)

Hints

- What equation form gives a line through the origin? - Does doubling \(x\) always double \(y\)? - Check whether each equation can be written as a constant times \(x\).

Solution

1. A proportional relationship has the form \(y=kx\). 2. a) \(y=3.5x\) has this form, so it is proportional. 3. b) \(y=\frac{1}{2}x\) has this form, so it is proportional. 4. c) The added \(1\) prevents the graph from passing through the origin, so it is not proportional. 5. d) The ratio \(y\div x=x\) is not constant, so it is not proportional. 6. e) The variable is in the denominator, so the equation is not of the form \(y=kx\).

Answer

a) Proportional: \(y=3.5x\) has the form \(y=kx\). b) Proportional: \(y=\frac{1}{2}x\) has the form \(y=kx\). c) Not proportional: the added \(1\) gives a nonzero initial value. d) Not proportional: \(\frac{y}{x}=x\) is not constant. e) Not proportional: the equation is not of the form \(y=kx\).
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 \((0, 0)\).
5138437
Yuna claims, “When a circle's diameter doubles, its circumference also doubles. Therefore, circumference is proportional to diameter.” Use \(C(d)=\pi d\) to test Yuna's claim. Show algebraically how the circumference changes when the diameter changes from \(d\) to \(2d\), and explain whether the relationship is proportional.

Hints

- What happens to a product when one factor doubles? - For a proportional relationship, what should happen to the output when the input doubles? - Is the ratio \(\frac{C}{d}\) constant?

Solution

1. The circumference at diameter \(d\) is \(C(d)=\pi d\). 2. At diameter \(2d\), \(C(2d)=\pi(2d)=2\pi d=2C(d)\). 3. Also, \(\frac{C}{d}=\pi\) for every positive diameter, so the ratio is constant. 4. Therefore, circumference is proportional to diameter, and Yuna's claim is correct.

Answer

Yuna's claim is correct. Since \(C(2d)=2C(d)\) and \(\frac{C}{d}=\pi\) is constant, circumference is proportional to diameter.
5502087
Graphs a) and b) show two relationships between \(x\) and \(y\). Which graph represents a proportional relationship? Explain using a feature of the graph.
Figure for problem 550208

Hints

- Recall the graph feature that every proportional relationship must have. - Check whether each straight line passes through the origin. - A straight-line pattern alone is not enough to make a relationship proportional.

Solution

1. Graph a) is a straight line that passes through \((0,0)\), so it represents a proportional relationship. 2. Graph b) does not pass through the origin, so it is not proportional.

Answer

Graph a) is proportional because its line passes through \((0,0)\). Graph b) is not proportional because its line does not pass through the origin.
5545317
Rosa sees that a relationship contains the point \((3,12)\). She says, “The relationship must be proportional with constant of proportionality \(4\), because \(12\div3=4\).” a) Is Rosa's conclusion justified from that one point alone? Explain. b) A second point from the same relationship is \((6,30)\). Use both points to determine whether the relationship is proportional.

Hints

- Ask what must be true about \(\frac{y}{x}\) for every nonzero point in a proportional relationship. - One data pair can suggest a constant, but consider what information it does not provide. - Compare the ratio from the second point with the ratio from the first.

Solution

1. The point \((3,12)\) gives the candidate ratio \(12\div3=4\), but one point alone does not establish that the same ratio holds for every point in the relationship. 2. For the second point, \(30\div6=5\). 3. Since the ratios \(4\) and \(5\) are different, the two points cannot belong to one proportional relationship of the form \(y=kx\).

Answer

a) No. One point gives a possible value of \(k\), but it does not prove that the ratio is constant for the whole relationship. b) The relationship is not proportional because \(12\div3=4\) while \(30\div6=5\).
5118897
A bank charges a fixed \(\$5\) service fee when converting US dollars to Canadian dollars. Only the money left after the fee is converted at a rate of \(1\) US dollar to \(1.35\) Canadian dollars. The bank accepts deposits of at least \(\$5\). a) Make a table showing the Canadian-dollar amounts received for deposits of \(\$20\), \(\$50\), and \(\$100\). b) Write an expression for the Canadian-dollar amount \(D\) received from a deposit of \(x\) US dollars. c) Is the relationship between the deposit and the amount received proportional? Justify your answer mathematically.

Hints

- Decide whether the fee is subtracted before or after the currency conversion. - Compare the output-to-input ratios for at least two deposits. - In a proportional relationship, that ratio must stay constant.

Solution

1. For \(\$20\): \((20-5)\cdot1.35=20.25\,\text{CAD}\). For \(\$50\): \((50-5)\cdot1.35=60.75\,\text{CAD}\). For \(\$100\): \((100-5)\cdot1.35=128.25\,\text{CAD}\). 2. For \(x\ge5\), the rule is \(D(x)=1.35(x-5)\). 3. The relationship is not proportional because the output-to-input ratio is not constant. For example, \(20.25 \div 20=1.0125\), while \(60.75 \div 50=1.215\).

Answer

a) <table><tr><th>Deposit (US dollars)</th><td>\(\$20\)</td><td>\(\$50\)</td><td>\(\$100\)</td></tr><tr><th>Amount received (CAD)</th><td>\(20.25\,\text{CAD}\)</td><td>\(60.75\,\text{CAD}\)</td><td>\(128.25\,\text{CAD}\)</td></tr></table> b) \(D(x)=1.35(x-5)\) for \(x\ge5\). c) No. The output-to-input ratio is not constant, so the relationship is not proportional.
5119317
A bike-rental shop offers two pricing plans: Plan A: no starting fee and \(\$2.50\) per hour. Plan B: a \(\$5.00\) starting fee and \(\$1.00\) per hour. For each plan, determine whether the relationship between rental time, in hours, and total price is proportional. Use specific numerical examples to justify your conclusions.

Hints

- Make a small table for \(1\) hour and \(2\) hours under each plan. - Check whether price divided by hours stays constant. - Consider how a fixed starting fee affects the ratio.

Solution

1. For Plan A, the rule is \(C=2.50h\). For example, \(1\) hour costs \(\$2.50\), and \(2\) hours cost \(\$5.00\). The price per hour is \(\$2.50\) in both cases, so Plan A is proportional. 2. For Plan B, \(1\) hour costs \(5+1=\$6.00\), and \(2\) hours cost \(5+2=\$7.00\). 3. The unit rates are \(\$6.00\) per hour and \(\$3.50\) per hour, which are different, so Plan B is not proportional.

Answer

Plan A is proportional. Plan B is not proportional.
5119367
A candy shop sells loose candy. A \(5\,\text{oz}\) bag costs \(\$1.50\). a) Find what a \(12\,\text{oz}\) bag would cost if price were proportional to weight. b) The \(12\,\text{oz}\) bag actually costs only \(\$3.30\). Show that the two bag prices are not proportional to weight, and explain one reason a store might charge a lower price per ounce for a larger bag.

Hints

- First find the price for \(1\,\text{oz}\) in each bag. - Compare the two unit prices. - Think about why a store might encourage customers to buy a larger bag.

Solution

1. The unit price for the \(5\,\text{oz}\) bag is \(1.50\div 5=\$0.30\) per ounce. 2. A proportional price for \(12\,\text{oz}\) is \(12\cdot 0.30=\$3.60\). 3. The actual unit price for the \(12\,\text{oz}\) bag is \(3.30\div 12=\$0.275\) per ounce. Since \(\$0.275\) per ounce is not equal to \(\$0.30\) per ounce, the prices are not proportional to weight. 4. A store may offer a quantity discount to encourage larger purchases, or packaging and handling may cost less per ounce for larger bags.

Answer

a) The proportional price would be \(\$3.60\). b) The actual unit prices are \(\$0.30\) per ounce and \(\$0.275\) per ounce, so the prices are not proportional to weight. A larger bag may have a quantity discount or lower packaging and handling costs per ounce.
5119407
Orchard manager Marisol claims, “The relationship between the number of apple trees and the total harvest, in pounds, is proportional.” a) Under what idealized conditions would Marisol be correct? b) Give two real-world reasons why this relationship usually is not exactly proportional.

Hints

- When would the number of pounds per tree always be the same? - Are all living plants identical in their production? - What resources might not be equally available to every tree?

Solution

1. The relationship would be proportional if every tree produced exactly the same number of pounds of apples. 2. In reality, trees differ in age, size, health, and exposure to pests. 3. Soil quality, sunlight, water, and available nutrients may also vary from tree to tree. 4. If trees are planted too closely, competition for light or nutrients can reduce the yield per tree.

Answer

a) It would be proportional if every tree had the same yield. b) Possible reasons include differences in tree age or health, pests, soil quality, sunlight, water, and competition between trees.
5119487
A copy shop offers two pricing plans: Plan A: each copy costs \(\$0.10\). Plan B: a monthly fee of \(\$2.50\), plus \(\$0.05\) for each copy. a) Write an expression for the total cost \(C\) under each plan as a function of the number of copies \(x\). b) Determine which plan represents a proportional relationship. Justify your answer using the expressions or a property of proportional relationships.

Hints

- What should the cost be for \(0\) copies in a proportional relationship? - Which expression has the form \(y=kx\)? - Check whether doubling the number of copies doubles the cost under each plan.

Solution

1. Plan A is \(C_A=0.10x\). Plan B is \(C_B=0.05x+2.50\). 2. A proportional relationship has the form \(y=kx\) and has output \(0\) when the input is \(0\). 3. Plan A is proportional because its cost per copy is constant and \(C_A(0)=0\). 4. Plan B is not proportional because \(C_B(0)=2.50\).

Answer

a) \(C_A=0.10x\) and \(C_B=0.05x+2.50\). b) Only Plan A is proportional because only its cost is \(\$0\) for \(0\) copies and its output-to-input ratio is constant.
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.
5131277
Kojo wants to know whether he walked at a constant speed during three parts of a trip. - Part 1: \(3\) miles in \(1.5\) hours - Part 2: \(5\) miles in \(2.5\) hours - Part 3: \(7.5\) miles in \(4\) hours Use the constant of proportionality to determine whether distance and time are proportional over all three parts and whether Kojo's speed was constant.

Hints

- Find speed by dividing distance by time. - What must be true about distance divided by time for constant speed? - Check whether Kojo covers the same distance each hour in every part.

Solution

1. Part 1 speed: \(3\div 1.5=2\) miles per hour. 2. Part 2 speed: \(5\div 2.5=2\) miles per hour. 3. Part 3 speed: \(7.5\div 4=1.875\) miles per hour. 4. Since the third rate differs, the speed was not constant and the three pairs do not form one proportional relationship.

Answer

Kojo's speed was not constant. It was \(2\) miles per hour for the first two parts and \(1.875\) miles per hour for the third, so the relationship is not proportional across all three parts.
5131287
A laboratory tests three metal cylinders that are claimed to be made of the same material. For this problem, treat the listed measurements as exact and assume that cylinders made of the same material must have exactly the same mass-to-volume ratio. - Cylinder A: volume \(20\,\text{cm}^3\), mass \(178\,\text{g}\) - Cylinder B: volume \(50\,\text{cm}^3\), mass \(445\,\text{g}\) - Cylinder C: volume \(12\,\text{cm}^3\), mass \(105\,\text{g}\) a) Use calculations to determine whether the three measurements are consistent with that assumption. b) What mass would Cylinder C need to have the same constant of proportionality as Cylinders A and B?

Hints

- Compare mass divided by volume for each cylinder. - Under the stated assumption, what must be true about those ratios? - For part b, use the common ratio from Cylinders A and B with Cylinder C's volume.

Solution

1. Cylinder A's density is \(178 \div 20=8.9\,\text{g/cm}^3\). 2. Cylinder B's density is \(445 \div 50=8.9\,\text{g/cm}^3\). 3. Cylinder C's density is \(105 \div 12=8.75\,\text{g/cm}^3\). 4. Since Cylinder C's ratio differs, the three measurements are not consistent with one exact mass-to-volume constant. 5. At \(8.9\,\text{g/cm}^3\), Cylinder C would need mass \(12\cdot8.9=106.8\,\text{g}\).

Answer

a) No. Cylinders A and B have density \(8.9\,\text{g/cm}^3\), while Cylinder C has density \(8.75\,\text{g/cm}^3\), so the three exact measurements do not share one constant ratio. b) \(106.8\,\text{g}\).
5131377
Two companies rent electric scooters using different pricing plans. - Company A: no starting fee; each minute costs \(\$0.20\). - Company B: a \(\$1.00\) activation fee for each ride, then \(\$0.10\) per minute. Determine which plan represents a proportional relationship between ride time and total cost. Justify your answer using properties of a line through the origin.

Hints

- Write each cost as an equation. - What is the cost at \(0\) minutes? - Where must the graph begin for a proportional relationship?

Solution

1. Company A has equation \(y=0.20x\). Company B has equation \(y=0.10x+1.00\). 2. A proportional relationship has a graph that is a line through \((0, 0)\). 3. For Company A, \(0\) minutes costs \(\$0\), so the graph passes through the origin. 4. For Company B, \(0\) minutes still costs \(\$1.00\), so its graph does not pass through the origin.

Answer

Only Company A is proportional. Company B's activation fee gives a nonzero cost at \(0\) minutes.
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.
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.
5131457
A laboratory measures the mass \(m\) of a liquid for several volumes \(V\). <table> <tr><td>Volume \(V\), in \(\text{cm}^3\)</td><td>\(50\)</td><td>\(150\)</td><td>\(300\)</td></tr> <tr><td>Mass \(m\), in grams</td><td>\(40\)</td><td>\(120\)</td><td>\(240\)</td></tr> </table> a) Determine whether mass is proportional to volume. b) Write the function \(m(V)=kV\). What does \(k\) represent physically? c) Find the mass of \(450\,\text{cm}^3\) of the liquid.

Hints

- What table test shows that two quantities are proportional? - What physical quantity is mass divided by volume? - Use the constant to calculate a new output.

Solution

1. The ratios are \(40\div 50=0.8\), \(120\div 150=0.8\), and \(240\div 300=0.8\), so mass is proportional to volume. 2. The function is \(m(V)=0.8V\). The constant \(0.8\,\text{g/cm}^3\) is the liquid's density. 3. \(m(450)=0.8\cdot 450=360\) grams.

Answer

a) Yes, because \(m\div V=0.8\) for every pair. b) \(m(V)=0.8V\); \(k\) is the density. c) \(360\) grams.
5131487
A relationship \(x\mapsto y\) is proportional when \(\frac{y}{x}\) has the same value for every \(x\ne0\). This value is the constant of proportionality. Use the ratio test to decide whether each relationship is proportional. a) \(y=12x\) b) \(y=3x+6\) c) \(y=kx\), where \(k\) is a fixed number

Hints

- For a proportional relationship, the output-to-input ratio must stay the same. - For a rule that is not already in the form \(y=kx\), test two convenient nonzero inputs. - One pair of different output-to-input ratios is enough to show that a relationship is not proportional.

Solution

1. a) \(\frac{y}{x}=\frac{12x}{x}=12\), which is constant, so the relationship is proportional. 2. b) When \(x=1\), \(y=9\), so \(\frac{y}{x}=9\). When \(x=2\), \(y=12\), so \(\frac{y}{x}=6\). The ratios differ, so the relationship is not proportional. 3. c) \(\frac{y}{x}=\frac{kx}{x}=k\), which is constant, so the relationship is proportional.

Answer

a) Proportional; \(\frac{y}{x}=12\). b) Not proportional; for example, the ratios are \(9\) at \(x=1\) and \(6\) at \(x=2\). c) Proportional; \(\frac{y}{x}=k\).
5131517
A copy shop lists these prices for black-and-white copies: - \(10\) copies cost \(\$0.80\). - \(50\) copies cost \(\$4.00\). - \(100\) copies cost \(\$7.50\). a) Determine whether price is proportional to number of copies. Show your calculations. b) Give one real-world reason a store might use pricing that is not proportional.

Hints

- Find the price of one copy for each offer. - What must stay constant in a proportional relationship? - Think about discounts for buying larger quantities.

Solution

1. The unit prices are \(0.80\div 10=\$0.08\), \(4.00\div 50=\$0.08\), and \(7.50\div 100=\$0.075\) per copy. 2. Since the unit prices are not all equal, the relationship is not proportional. 3. The store may offer a quantity discount for large orders.

Answer

a) No. The first two prices are \(\$0.08\) per copy, but \(100\) copies cost \(\$0.075\) per copy. b) A possible reason is a quantity discount.
5131697
Test the pairs in the table for proportionality. <table> <tr><td>\(x\)</td><td>\(3\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr> <tr><td>\(y\)</td><td>\(10\)</td><td>\(16\)</td><td>\(25\)</td><td>\(31\)</td></tr> </table> Justify your conclusion mathematically. If the relationship is not proportional, describe how \(\frac{y}{x}\) changes as \(x\) increases.

Hints

- Calculate the ratio for every column. - Is matching in only two columns enough? - Read the ratio values from left to right and describe the pattern.

Solution

1. The ratios are \(10\div 3\approx 3.33\), \(16\div 5=3.2\), \(25\div 8=3.125\), and \(31\div 10=3.1\). 2. Since the ratios are not equal, the relationship is not proportional. 3. As \(x\) increases, the ratio \(y\div x\) decreases.

Answer

The relationship is not proportional because the ratios \(y\div x\) are not equal. The ratio decreases as \(x\) increases.
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.
5138447
A cylinder has a fixed height \(h\), so its volume as a function of radius is \(V(r)=\pi r^2h\). Determine how the volume changes when the radius is tripled. Use your result to explain why the relationship between \(r\) and \(V\) is not proportional.

Hints

- Substitute \(3r\) into the formula and square the entire expression. - Compare the new volume with the original volume. - What output change would a proportional relationship require when the input is tripled?

Solution

1. Substitute \(3r\) for \(r\): \(V(3r)=\pi(3r)^2h\). 2. Simplify: \(V(3r)=9\pi r^2h=9V(r)\). 3. Tripling the radius multiplies the volume by \(9\). 4. In a proportional relationship, tripling the input must triple the output. Since the volume is multiplied by \(9\) instead, the relationship is not proportional.

Answer

The volume is multiplied by \(9\): \(V(3r)=9V(r)\). The relationship is not proportional because tripling \(r\) does not triple \(V\).
5138457
A circular flower bed is enlarged from radius \(r_1=2\,\text{ft}\) to radius \(r_2=5\,\text{ft}\). 1) For each radius, calculate the ratio of area to radius, \(\frac{A}{r}\). 2) Use the ratios to decide whether the relationship “radius \(\to\) area” is proportional. 3) By what factor does the area increase when the radius is multiplied by \(2.5\)?

Hints

- First calculate each circle's area. - Divide each area by its matching radius and compare the ratios. - A proportional relationship must have a constant output-to-input ratio. - How does a length scale factor affect area?

Solution

1. For \(r_1=2\,\text{ft}\), \(A_1=\pi(2\,\text{ft})^2=4\pi\,\text{ft}^2\), so \(\frac{A_1}{r_1}=2\pi\,\text{ft}\approx6.28\,\text{ft}\). 2. For \(r_2=5\,\text{ft}\), \(A_2=\pi(5\,\text{ft})^2=25\pi\,\text{ft}^2\), so \(\frac{A_2}{r_2}=5\pi\,\text{ft}\approx15.71\,\text{ft}\). 3. The ratios are not equal, so area is not proportional to radius. 4. Since area depends on the square of the radius, a radius factor of \(2.5\) gives an area factor of \(2.5^2=6.25\).

Answer

1) \(\frac{A_1}{r_1}=2\pi\,\text{ft}\approx6.28\,\text{ft}\) and \(\frac{A_2}{r_2}=5\pi\,\text{ft}\approx15.71\,\text{ft}\) 2) The relationship is not proportional. 3) The area is multiplied by \(6.25\).
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.
5141997
Two cell phone plans sell mobile data. Determine which plan shows a proportional relationship between data and price. Justify your answer, then find the price of \(120\,\text{GB}\) for the proportional plan. **Plan A:** <table> <tr><td>Data (in \(\text{GB}\))</td><td>\(10\)</td><td>\(50\)</td></tr> <tr><td>Price</td><td>\(\$1.00\)</td><td>\(\$5.00\)</td></tr> </table> **Plan B:** <table> <tr><td>Data (in \(\text{GB}\))</td><td>\(10\)</td><td>\(50\)</td></tr> <tr><td>Price</td><td>\(\$5.50\)</td><td>\(\$7.50\)</td></tr> </table>

Hints

- Check whether the price is multiplied by the same factor as the data amount. - Find the price per gigabyte for each pair in each table. - What must be constant in a proportional relationship?

Solution

1. For Plan A, \(1.00 \div 10=0.10\) and \(5.00 \div 50=0.10\). The price per gigabyte is constant, so Plan A is proportional. 2. For Plan B, \(5.50 \div 10=0.55\) and \(7.50 \div 50=0.15\). The ratios differ, so Plan B is not proportional. 3. For Plan A, \(120\,\text{GB}\) costs \(0.10\cdot 120=\$12.00\).

Answer

Plan A is proportional because its unit price is always \(\$0.10\) per gigabyte. Plan B is not proportional because its unit prices are different. Under Plan A, \(120\,\text{GB}\) costs \(\$12.00\).
5241627
A copy shop charges the prices shown. <table><tr><td>Number of pages \(n\)</td><td>\(10\)</td><td>\(20\)</td><td>\(50\)</td><td>\(100\)</td></tr><tr><td>Total cost \(C\)</td><td>\(\$0.80\)</td><td>\(\$1.60\)</td><td>\(\$4.00\)</td><td>\(\$8.00\)</td></tr></table> 1) Show that cost is proportional to the number of pages. 2) Write an equation \(C(n)\). 3) How many pages were printed if the bill was \(\$14.40\)? 4) Suppose the shop also charged a one-time \(\$2.00\) fee per order. Would total cost still be proportional to the number of pages? Explain.

Hints

- What must be constant for all pairs in a proportional relationship? - Use an equation of the form \(y=kx\). - Use the inverse operation to find the input from the output. - Would a fixed fee make the graph pass through \((0,0)\)?

Solution

1. The ratios are \(0.80 \div 10=0.08\), \(1.60 \div 20=0.08\), \(4.00 \div 50=0.08\), and \(8.00 \div 100=0.08\). Since the ratios are equal, the relationship is proportional. 2. The equation is \(C(n)=0.08n\). 3. Solve \(14.40=0.08n\): \(n=14.40 \div 0.08=180\) pages. 4. With a fee, the equation would be \(C(n)=0.08n+2\). It would not be proportional because \(C(0)=2\), so the graph would not pass through the origin.

Answer

1) Every ratio \(\frac{C}{n}\) equals \(0.08\). 2) \(C(n)=0.08n\) 3) \(180\) pages 4) No. With the fixed fee, \(0\) pages would correspond to a cost of \(\$2.00\), so the graph would not pass through the origin.
5359177
A rectangular prism has a fixed width of \(2\,\text{inches}\) and a fixed height of \(5\,\text{inches}\). Its length \(x\) can vary. Determine algebraically whether each relationship is proportional. a) \(x\mapsto\) volume \(V\) b) \(x\mapsto\) surface area \(S\)

Hints

- Separate the dimensions that stay fixed from the dimension that changes. - Consider what happens to each output when the length is multiplied by the same factor. - Compare each resulting rule with the defining form of a proportional relationship.

Solution

1. The volume is \(V(x)=x \cdot 2 \cdot 5=10x\). This has the form \(V=kx\), so volume is proportional to length. 2. The surface area is \(S(x)=2(2x+5x+2 \cdot 5)=14x+20\). 3. The surface-area relationship is not proportional because \(S(0)=20\), not \(0\), and the rule is not of the form \(S=kx\).

Answer

a) Proportional, because \(V(x)=10x\). b) Not proportional, because \(S(x)=14x+20\).

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