Aimathic – Constant of proportionality from equation

Constant of proportionality from equation

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Difficulty: 1 – 5

5502407

The equation $y=\frac{7}{12}x$ represents a proportional relationship. What is the constant of proportionality $k$?

Hints

- Compare the given equation with the standard form of a proportional relationship. - Identify the number multiplying the input variable. - No decimal conversion is needed; the multiplier can be read directly in fractional form.

Solution

1. A proportional equation has the form $y=kx$. 2. Comparing $y=\frac{7}{12}x$ with that form gives $k=\frac{7}{12}$.

Answer

$k=\frac{7}{12}$

5519297

A hose fills a tank according to $V=3.5t$, where $V$ is the volume in gallons and $t$ is time in minutes. What is the constant of proportionality, including its units, and what does it mean?

Hints

- Identify the number multiplying the time variable. - Use the variable descriptions to determine the units of that number. - Interpret the constant as an amount for one minute.

Solution

1. A proportional equation has the form $y=kx$. 2. In $V=3.5t$, the number multiplying the input $t$ is $3.5$. 3. Therefore, the constant is $3.5$ gallons per minute, meaning the hose adds $3.5$ gallons each minute.

Answer

The constant of proportionality is $3.5$ gallons per minute. The hose adds $3.5$ gallons of water each minute.

5545477

The equation $h=\frac{x}{5}$ represents a proportional relationship. Write it in the form $h=kx$, and state the constant of proportionality $k$.

Hints

- Compare the given equation with the pattern $h=kx$. - Ask what numerical factor is effectively multiplying $x$. - The factor attached to $x$ is the constant of proportionality.

Solution

1. Rewrite $\frac{x}{5}$ as $\frac{1}{5}x$. 2. Therefore, $h=\frac{1}{5}x$ and $k=\frac{1}{5}$.

Answer

$h=\frac{1}{5}x$, so $k=\frac{1}{5}$.

5139577

Elena walks $6$ miles in $2$ hours at a constant hiking pace. a) Explain why distance is proportional to time, and write an equation in the form $d(t)=rt$. b) How far will Elena travel in $3.5$ hours at the same pace?

Hints

- What does a constant pace mean about the ratio of distance to time? - Find the distance traveled in one hour. - A proportional relationship has an equation of the form $y=kx$.

Solution

1. At a constant pace, equal time intervals correspond to equal distances, and the relationship includes $(0, 0)$. Therefore, distance is proportional to time. 2. The constant of proportionality is $r=6 \div 2=3\,\text{mi/h}$. 3. The equation is $d(t)=3t$. 4. Evaluate the equation at $t=3.5$: $d(3.5)=3 \cdot 3.5=10.5$ miles.

Answer

a) The relationship is proportional because the rate $\frac{d}{t}$ is constant. The equation is $d(t)=3t$. b) Elena will travel $10.5$ miles.

5262177

A proportional relationship has the form $y=kx$. When $x=8$, $y=6$. a) Write the equation. b) Find $y$ when $x=15$.

Hints

- What is the equation form for a proportional relationship? - How can one input-output pair be used to find $k$? - Substitute the new x-value after finding the equation.

Solution

1. Substitute $(8, 6)$: $6=8k$, so $k=6 \div 8=0.75$. 2. The equation is $y=0.75x$. 3. For $x=15$, $y=0.75 \cdot 15=11.25$.

Answer

a) $y=0.75x$ b) $y=11.25$

5501967

The equation $C=0.18m$ gives the cost $C$, in dollars, to print $m$ photos. a) What is the constant of proportionality? Include its units and explain what it means. b) Use the equation to find the cost of printing $35$ photos.

Hints

- Look for the number that multiplies the input variable in the equation. - Use the variable descriptions to determine the units of that number. - For the second part, connect the requested number of photos to the given rule.

Solution

1. In $C=0.18m$, the constant of proportionality is $0.18$ dollars per photo, so each photo costs $\$0.18$. 2. For $35$ photos, $C=0.18\cdot35=6.30$.

Answer

a) $0.18$ dollars per photo; each photo costs $\$0.18$. b) $\$6.30$

5502137

The thickness $h$, in inches, of a stack of identical foam sheets is given by $h=\frac{x}{8}$, where $x$ is the number of sheets. a) Write the equation in the form $h=kx$ and identify $k$. b) What does $k$ mean in this situation? c) Find the thickness of a stack of $20$ sheets.

Hints

- Rewrite the division in a way that makes the multiplier of $x$ visible. - Interpret the multiplier as an amount for one sheet. - Use the same relationship for the larger stack.

Solution

1. Rewrite $h=\frac{x}{8}$ as $h=\frac{1}{8}x$, so $k=\frac{1}{8}$. 2. The constant means each sheet is $\frac{1}{8}\,\text{inch}$ thick. 3. For $20$ sheets, $h=\frac{20}{8}=2.5\,\text{inches}$.

Answer

a) $h=\frac{1}{8}x$, so $k=\frac{1}{8}$. b) Each sheet is $\frac{1}{8}\,\text{inch}$ thick. c) $2.5\,\text{inches}$

5502167

Two 3D printers use filament according to the equations $F_A=1.25t$ and $F_B=\frac{3}{2}t$, where $F$ is filament used in grams and $t$ is time in minutes. a) Which printer has the greater constant of proportionality? b) How much more filament does that printer use each minute?

Hints

- Identify the multiplier of time in each equation. - Put the two constants in comparable numerical forms. - Compare the rates before finding their difference.

Solution

1. Printer A has constant $1.25$ grams per minute. Printer B has constant $\frac{3}{2}=1.5$ grams per minute. 2. Since $1.5>1.25$, Printer B has the greater constant. 3. The difference is $1.5-1.25=0.25$ gram per minute.

Answer

a) Printer B b) Printer B uses $0.25$ gram more filament per minute.

5119147

An empty cylindrical glass is placed under a faucet that runs at a constant rate. After $12\,\text{s}$, the water is $4.2\,\text{cm}$ deep. a) How deep will the water be after a total of $30\,\text{s}$? b) Write an equation for the water depth $h$, in centimeters, after $t$ seconds. c) The glass is $14\,\text{cm}$ tall. After how many seconds will the water reach the rim?

Hints

- First find how many centimeters the water level rises each second. - A cylinder has a constant cross-sectional area, so a constant flow produces a constant rise in height. - Use the unit rate as the coefficient of $t$. - To find the time from a known height, undo the multiplication.

Solution

1. Find the unit rate: $4.2\div 12=0.35$ centimeter per second. 2. After $30$ seconds, $h=0.35\cdot 30=10.5$, so the water is $10.5\,\text{cm}$ deep. 3. The proportional relationship is $h=0.35t$. 4. Set $h=14$: $14=0.35t$, so $t=14\div 0.35=40$.

Answer

a) $10.5\,\text{cm}$ b) $h=0.35t$ c) $40\,\text{s}$

5119158

Two empty cylindrical containers, A and B, are filled by identical faucets with the same constant flow rate. The base area of container A is exactly three times the base area of container B. a) In which container does the water level rise faster? Explain. b) In container B, the water level rises $15\,\text{cm}$ in one minute. How many centimeters does the water level rise in container A during the same time? c) Write an equation for the water depth $h$, in centimeters, after $t$ minutes in each container.

Hints

- Imagine pouring the same amount of water into a narrow container and a wide container. - Use the relationship $V=Bh$ between volume, base area, and height. - For the same volume, a larger base area produces a proportionally smaller height increase.

Solution

1. The same volume of water spreads across a larger base in container A, so the water level rises faster in container B. 2. For equal volumes, $V=Bh$. Since the base area of A is three times the base area of B, its height increase is one-third as great: $15\div 3=5$. Container A rises $5\,\text{cm}$ per minute. 3. The equations are $h_A=5t$ and $h_B=15t$.

Answer

a) Container B, because it has the smaller base area. b) $5\,\text{cm}$ c) $h_A=5t$ and $h_B=15t$

5119237

A common rule estimates how far away lightning is: “Count the seconds between the flash and the thunder, then divide by $5$. The result is the distance in miles.” a) Write an equation for the distance $d$, in miles, after a delay of $t$ seconds. b) Find the estimated distances for delays of $5\,\text{s}$, $15\,\text{s}$, and $25\,\text{s}$. c) How far away is the lightning when the delay is $7.5\,\text{s}$? d) Explain why this rule gives only an estimate. Use a speed of sound of about $1125\,\text{ft/s}$.

Hints

- Translate “divide by $5$” directly into an algebraic expression. - Substitute each time value for $t$. - Compare the distance sound travels in $5$ seconds with $5280$ feet in one mile.

Solution

1. Dividing the time by $5$ gives $d=\frac{t}{5}$. 2. For $t=5,15,25$, the distances are $1$, $3$, and $5$ miles. 3. For $t=7.5$, $d=\frac{7.5}{5}=1.5$ miles. 4. In $5$ seconds, sound travels about $5\cdot 1125=5625$ feet. One mile is $5280$ feet, so the rule rounds this distance to about $1$ mile. The speed of sound also varies with air conditions.

Answer

a) $d=\frac{t}{5}$ b) $1$ mile, $3$ miles, and $5$ miles c) $1.5$ miles d) In $5$ seconds, sound travels about $5625$ feet, which is close to but not exactly $1$ mile, and the speed varies with conditions.

5119457

Nadia fills an empty rain barrel with a hose. After $6$ minutes, the barrel contains $15$ gallons of water. The flow rate is constant. a) How many gallons enter the barrel each minute? b) Write a formula for the volume $V$, in gallons, after $t$ minutes. c) How long will it take for the barrel to contain $50$ gallons? d) Explain how you know the relationship is proportional.

Hints

- Find the amount added in one minute. - Use the constant rate to connect volume and time. - Divide the target volume by the amount added each minute. - What happens to volume when time is doubled?

Solution

1. The flow rate is $15\div 6=2.5$ gallons per minute. 2. Since the barrel starts empty, the formula is $V=2.5t$. 3. Solve $50=2.5t$: $t=50\div 2.5=20$ minutes. 4. The relationship is proportional because $V\div t=2.5$ is constant and the model has no initial amount.

Answer

a) $2.5$ gallons per minute. b) $V=2.5t$. c) $20$ minutes. d) The ratio of volume to time is constant, and the barrel starts with $0$ gallons.

5119467

A copier in a school office works at a constant rate. It prints $120$ pages in $8$ minutes. a) Find the constant of proportionality for time $\to$ number of pages. What does this value mean in context? b) How many pages can the copier print during a $45$-minute class period? c) A project requires $1200$ pages. How many hours and minutes must the copier run?

Hints

- Find how many pages the copier prints in one minute. - Use the one-minute rate for $45$ minutes. - Divide the total number of pages by the pages per minute. - Remember that $60$ minutes equals $1$ hour.

Solution

1. The constant of proportionality is $120\div 8=15$, meaning the copier prints $15$ pages per minute. 2. In $45$ minutes, it prints $45\cdot 15=675$ pages. 3. Printing $1200$ pages takes $1200\div 15=80$ minutes. 4. Since $80$ minutes is $1$ hour $20$ minutes, that is the required running time.

Answer

a) $15$ pages per minute. b) $675$ pages. c) $1$ hour $20$ minutes.

5119677

An empty swimming pool is filled at a constant rate. A hose adds $225$ gallons of water in $15$ minutes. a) Find the flow rate in gallons per minute. b) Write an equation for the amount of water $V$, in gallons, after $t$ minutes. c) The pool holds $4500$ gallons. How long will it take to fill? Give the answer in hours. d) If a second identical hose is used at the same time, how does the constant of proportionality for the total flow change? Explain.

Hints

- Find how much water enters in one minute. - Connect the amount of water to time using the constant rate. - Divide total capacity by the rate, then convert minutes to hours. - Determine what two equal flow rates produce together.

Solution

1. The flow rate is $225\div 15=15$ gallons per minute. 2. Since the pool starts empty, $V=15t$. 3. The fill time is $4500\div 15=300$ minutes, which is $300\div 60=5$ hours. 4. A second identical hose doubles the total rate to $30$ gallons per minute.

Answer

a) $15$ gallons per minute. b) $V=15t$. c) $5$ hours. d) The constant doubles to $30$ gallons per minute.

5120587

A community garden charges annual rent proportional to plot area. A $150\,\text{ft}^2$ plot costs $\$60$ per year. a) Write an equation for the annual rent $C$, in dollars, for a plot with area $A$, in square feet. Then find the annual rent for a $250\,\text{ft}^2$ plot. b) The price per square foot increases by $\$0.10$. Write the new equation and find what plot area could be rented for $\$60$.

Hints

- Find the original cost per square foot and use it as the coefficient of $A$. - Substitute $A=250$ into the original equation. - Increase the coefficient by $\$0.10$, then solve the new equation when $C=60$.

Solution

1. The original constant of proportionality is $60\div 150=\$0.40$ per square foot, so $C=0.40A$. 2. A $250\,\text{ft}^2$ plot costs $C=0.40\cdot 250=\$100$. 3. The new rate is $0.40+0.10=\$0.50$ per square foot, so the new equation is $C=0.50A$. 4. For $C=60$, solve $60=0.50A$: $A=60\div 0.50=120\,\text{ft}^2$.

Answer

a) $C=0.40A$, and a $250\,\text{ft}^2$ plot costs $\$100$. b) The new equation is $C=0.50A$, and $\$60$ rents $120\,\text{ft}^2$.

5125127

A market sells walnuts by weight. Price $P$ is proportional to weight $w$. An $8\,\text{oz}$ portion costs $\$4.50$. a) Write an equation for price $P$, in dollars, as a function of weight $w$, in ounces. b) Find the price of $12\,\text{oz}$ and of $2\,\text{lb}$. c) For a purchase totaling exactly $\$11.25$, how many ounces of walnuts are bought?

Hints

- Find the price of one ounce. - Convert all weights to ounces before using the equation. - Divide the total amount of money by the price per ounce.

Solution

1. The price per ounce is $4.50\div 8=\$0.5625$, so $P=0.5625w$. 2. For $12\,\text{oz}$, $P=0.5625\cdot 12=\$6.75$. 3. Convert $2\,\text{lb}$ to $32\,\text{oz}$. Then $P=0.5625\cdot 32=\$18.00$. 4. For $\$11.25$, $w=11.25\div 0.5625=20\,\text{oz}$.

Answer

a) $P=0.5625w$. b) $12\,\text{oz}$ costs $\$6.75$, and $2\,\text{lb}$ costs $\$18.00$. c) $20\,\text{oz}$.

5128137

Mateo maintains a constant running pace. After $12$ minutes, he has traveled $2$ miles. a) Find Mateo's speed in miles per hour. Also write an equation for distance $d$, in miles, after $t$ minutes. b) How far does Mateo travel in $45$ minutes at the same pace? c) If Mateo wants to travel twice the distance, how does the required time change? Justify your answer using the equation or a property of proportional relationships.

Hints

- Find the distance traveled in one minute and use it as the coefficient of $t$. - Scale the per-minute rate to $60$ minutes to find miles per hour. - Substitute $t=45$ into the equation. - In a proportional relationship, what happens to one quantity when the other doubles?

Solution

1. The rate per minute is $2\div 12=\frac{1}{6}$ mile per minute, so the equation is $d=\frac{1}{6}t$. 2. Multiply the per-minute rate by $60$: $\frac{1}{6}\cdot 60=10$ miles per hour. 3. In $45$ minutes, the distance is $d=\frac{1}{6}\cdot 45=7.5$ miles. 4. In $d=\frac{1}{6}t$, doubling $d$ requires doubling $t$. Therefore, the required time doubles.

Answer

a) $10$ miles per hour, and $d=\frac{1}{6}t$. b) $7.5$ miles. c) The required time doubles.

5128457

Coffee beans lose weight during roasting. From $25\,\text{lb}$ of green coffee beans, a roaster produces $21\,\text{lb}$ of roasted beans. Assume the roasted weight is proportional to the green-bean weight. a) Find the constant of proportionality $k$, in pounds of roasted coffee per pound of green coffee. b) How many pounds of green coffee are needed to produce $150\,\text{lb}$ of roasted coffee? Round to the nearest hundredth. c) How much roasted coffee is produced from $12.5\,\text{lb}$ of green coffee?

Hints

- Since roasting reduces weight, should the green-bean amount be greater or less than the roasted amount? - Write a rule connecting green-bean weight and roasted weight. - In part b, identify whether you are solving for the input or the output.

Solution

1. The constant is $k=21 \div 25=0.84$ pounds of roasted coffee per pound of green coffee. 2. The green-bean weight needed is $150 \div 0.84\approx178.57$ pounds. 3. From $12.5$ pounds, the roasted weight is $12.5\cdot0.84=10.5$ pounds.

Answer

a) $k=0.84$ pounds of roasted coffee per pound of green coffee. b) Approximately $178.57\,\text{lb}$. c) $10.5\,\text{lb}$.

5131177

A market sells cherries at a fixed price per ounce. A $12\,\text{oz}$ bag costs $\$4.50$, and a $20\,\text{oz}$ bag costs $\$7.50$. a) Explain why the relationship is proportional. b) Write a function for price $y$, in dollars, as a function of weight $x$, in ounces. c) What is the price of $32\,\text{oz}$ of cherries? How many ounces can be purchased for $\$18.00$?

Hints

- Check whether doubling weight would double price. - Find the price of one ounce. - Interpret the constant of proportionality in context. - Rearrange the equation when weight is unknown.

Solution

1. The unit prices are $4.50\div 12=\$0.375$ per ounce and $7.50\div 20=\$0.375$ per ounce, so the ratio of price to weight is constant. 2. The function is $y=0.375x$. 3. For $32$ ounces, $y=0.375\cdot 32=\$12.00$. 4. For $\$18.00$, $x=18\div 0.375=48$ ounces.

Answer

a) The unit price is constant at $\$0.375$ per ounce. b) $y=0.375x$. c) $32\,\text{oz}$ costs $\$12.00$, and $\$18.00$ buys $48\,\text{oz}$.

5131347

For one type of wood trim, weight is proportional to length. A $10$-foot piece weighs $5.6$ pounds. a) Write a function for weight $y$, in pounds, as a function of length $x$, in feet. b) Priya claims, “A piece that is twice as long weighs twice as much.” Use a property of proportional functions to determine whether the claim is correct. c) How long is a piece that weighs exactly $14$ pounds?

Hints

- Find the weight of one foot of trim. - What happens to a product when one factor doubles? - Divide the total weight by the weight per foot. - Recall the doubling property of a line through the origin.

Solution

1. The constant of proportionality is $5.6\div 10=0.56$ pound per foot. 2. The function is $y=0.56x$. 3. A piece of length $2x$ has weight $0.56(2x)=2(0.56x)$, so the claim is correct. 4. Solve $14=0.56x$: $x=14\div 0.56=25$ feet.

Answer

a) $y=0.56x$. b) The claim is correct; doubling length doubles weight. c) $25$ feet.

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.

5131467

Two metal wires, Wire A and Wire B, are compared. Each wire's mass is proportional to its length. Wire A has a mass of $125$ grams for $5$ meters. Wire B has a mass of $192$ grams for $8$ meters. a) Find each wire's constant of proportionality and write an equation for mass $m$, in grams, as a function of length $l$, in meters. Which wire has the greater mass per meter? b) What is the mass of a $20$-meter piece of Wire A? c) How long would a piece of Wire B need to be to have a mass of $125$ grams? Round to the nearest tenth of a meter.

Hints

- Find the mass of one meter of each wire and use it as the coefficient of $l$. - Substitute the given length into Wire A's equation. - Set Wire B's equation equal to the given mass and solve for length.

Solution

1. Wire A's constant is $125\div 5=25$ grams per meter, so $m_A=25l$. Wire B's constant is $192\div 8=24$ grams per meter, so $m_B=24l$. Wire A has the greater mass per meter. 2. A $20$-meter piece of Wire A has mass $m_A=25\cdot 20=500$ grams. 3. For Wire B, solve $125=24l$: $l=125\div 24\approx 5.2$ meters.

Answer

a) Wire A: $m_A=25l$. Wire B: $m_B=24l$. Wire A has the greater mass per meter. b) $500$ grams. c) Approximately $5.2$ meters.

5131527

Students test how far a spring stretches when different masses are attached. The spring is rated for loads up to $500$ grams. - With $20$ grams, the spring stretches $3.0$ centimeters. - With $50$ grams, the spring stretches $7.5$ centimeters. a) Find the constant of proportionality for mass to stretch, including units, and write an equation for stretch $s$, in centimeters, as a function of mass $m$, in grams. b) Predict the stretch for a mass of $120$ grams. c) Would the proportional model remain reasonable for a mass of $5$ kilograms? Explain using the spring's material limitations.

Hints

- Find stretch per gram and use it as the coefficient of $m$. - Substitute the new mass into the equation. - Compare $5$ kilograms with the spring's rated maximum. - What can happen to an elastic object under too much force?

Solution

1. The ratios are $3.0\div 20=0.15$ and $7.5\div 50=0.15$, so $k=0.15\,\text{cm/g}$ and $s=0.15m$. 2. For $120$ grams, the predicted stretch is $s=0.15\cdot 120=18$ centimeters. 3. A $5$-kilogram mass is $5000$ grams, far above the $500$-gram rating. The spring would likely deform permanently or break, so the proportional model would no longer apply.

Answer

a) $k=0.15\,\text{cm/g}$, and $s=0.15m$. b) $18$ centimeters. c) No. A $5$-kilogram load is far beyond the spring's rated range and could permanently deform or break it.

5139587

In a chemistry lab, the mass $m$ of different volumes $V$ of a liquid is measured. The data lie on a line through the origin. A volume of $60\,\text{cm}^3$ has a mass of $48\,\text{g}$. a) Find the liquid's density $\rho$, the constant of proportionality, in $\text{g/cm}^3$. b) What volume of this liquid has a mass of $100\,\text{g}$?

Hints

- For a proportional relationship, the constant can be found by comparing the output with the corresponding input. - Density is mass divided by volume. - Use $m=\rho V$ and identify which quantity is unknown in each part.

Solution

1. Find the density by dividing mass by volume: $\rho=\frac{m}{V}=\frac{48}{60}=0.8\,\text{g/cm}^3$. 2. The proportional relationship is $m=0.8V$. 3. Solve for volume: $V=\frac{m}{0.8}$. 4. For a mass of $100\,\text{g}$, $V=100 \div 0.8=125\,\text{cm}^3$.

Answer

a) The density is $0.8\,\text{g/cm}^3$. b) The volume is $125\,\text{cm}^3$.

5139847

Lemonade flows from a beverage dispenser at a constant rate. In $12$ seconds, the dispenser pours $6\,\text{fl oz}$. a) Create a table that gives the volume $V$, in fluid ounces, after $10$, $20$, $30$, and $60$ seconds. b) Write an equation for the proportional relationship. c) Explain what the constant of proportionality means in this situation. d) How long will it take to empty a dispenser containing $50\,\text{fl oz}$ of lemonade?

Hints

- First find how many fluid ounces flow in one second. - Use that unit rate to connect each time in the table to its volume. - A proportional relationship has the form $y=kx$; identify what $k$ means for this dispenser. - For the final part, decide whether the unknown is the input or the output of the proportional relationship.

Solution

1. The flow rate is $6 \div 12=0.5\,\text{fl oz/s}$. 2. Use $V=0.5t$: $V(10)=5$, $V(20)=10$, $V(30)=15$, and $V(60)=30$, in fluid ounces. 3. The equation is $V(t)=0.5t$. 4. The constant of proportionality $0.5$ means the dispenser pours $0.5\,\text{fl oz}$ each second. 5. Solve $50=0.5t$: $t=50 \div 0.5=100$ seconds.

Answer

a) <table><tr><th>Time</th><td>$10\,\text{s}$</td><td>$20\,\text{s}$</td><td>$30\,\text{s}$</td><td>$60\,\text{s}$</td></tr><tr><th>Volume</th><td>$5\,\text{fl oz}$</td><td>$10\,\text{fl oz}$</td><td>$15\,\text{fl oz}$</td><td>$30\,\text{fl oz}$</td></tr></table> b) $V(t)=0.5t$. c) The constant of proportionality means that the dispenser pours $0.5\,\text{fl oz}$ each second. d) $100$ seconds.

5141327

Malik works at an ice cream shop during school break. His earnings are proportional to the hours he works. After $6$ hours, Malik has earned $\$81.00$. a) How much will Malik earn for $14$ hours of work? b) How many hours must Malik work to earn $\$472.50$ for a tablet? c) What happens to the earnings if the number of hours worked is tripled? Explain using a property of proportional relationships.

Hints

- Find the amount earned per hour. - Once you know the hourly wage, how can you find earnings for any number of hours? - What does proportional mean about multiplying one quantity by a scale factor?

Solution

1. The hourly wage is $81.00 \div 6=13.50$, so Malik earns $\$13.50$ per hour. 2. For $14$ hours, the earnings are $14\cdot 13.50=\$189.00$. 3. To earn $\$472.50$, Malik must work $472.50 \div 13.50=35$ hours. 4. In a proportional relationship $y=kx$, replacing $x$ with $3x$ gives $k(3x)=3kx=3y$. Therefore, tripling the hours triples the earnings.

Answer

a) $\$189.00$ b) $35$ hours c) The earnings triple because both quantities change by the same scale factor in a proportional relationship.

5141337

In a physics experiment, a spring's stretch $s$, in centimeters, is proportional to the attached mass $m$, in grams. A mass of $100\,\text{g}$ stretches the spring $2.5\,\text{cm}$. a) Write an equation in the form $s(m)=km$. b) Find the stretch when the mass is $250\,\text{g}$. c) Find the attached mass when the stretch is exactly $10\,\text{cm}$.

Hints

- A proportional relationship has the form $y=kx$. Which variables represent $x$ and $y$ here? - Use the known mass and stretch to find $k$. - Substitute the given value into your equation, then solve for the unknown.

Solution

1. Find the constant of proportionality: $k=2.5 \div 100=0.025\,\text{cm/g}$. The equation is $s(m)=0.025m$. 2. For $m=250$, $s(250)=0.025 \cdot 250=6.25\,\text{cm}$. 3. For a stretch of $10\,\text{cm}$, solve $10=0.025m$: $m=10 \div 0.025=400\,\text{g}$.

Answer

a) $s(m)=0.025m$ b) $6.25\,\text{cm}$ c) $400\,\text{g}$

5226997

An online game uses two currencies: gold coins $G$ and silver coins $S$. The exchange rate is fixed: $4$ gold coins can be exchanged for $14$ silver coins. Exchanges use whole coins, and the game allows only even numbers of gold coins to be exchanged. 1) Write an equation for the number of silver coins in terms of the number of gold coins. State the allowed values of $G$. 2) How many silver coins does a player receive for $16$ gold coins? 3) How many gold coins must be exchanged to receive exactly $105$ silver coins?

Hints

- Find the number of silver coins corresponding to one gold coin in the proportional equation. - Check which whole-number values of $G$ make sense under the exchange rule. - Use the equation in reverse when the number of silver coins is given.

Solution

1. The constant of proportionality is $14 \div 4=3.5$, so $S=3.5G$. Because exchanges use whole coins and only even numbers of gold coins are allowed, $G$ is a nonnegative even integer. 2. For $G=16$, $S=3.5 \cdot 16=56$. 3. Solve $105=3.5G$: $G=105 \div 3.5=30$, which is an allowed even whole number.

Answer

1) $S=3.5G$, where $G$ is a nonnegative even integer. 2) $56$ silver coins 3) $30$ gold coins

5233397

A compact car uses an average of $3$ gallons of gasoline for every $100$ miles driven. Assume fuel use is proportional to distance. Let $x$ be distance in miles and $y$ be gasoline used in gallons. a) Find the constant of proportionality and write an equation for fuel use. b) Create a value table for $50$, $150$, $200$, $250$, and $500$ miles. c) What does the point $(100, 3)$ mean in this situation? d) How far can the car travel using $15$ gallons?

Hints

- First find the amount of gasoline used for one mile. - In a proportional relationship, how does fuel use change when distance is multiplied by a factor? - Interpret an ordered pair by matching each coordinate to its variable. - Divide the available gasoline by the amount used per mile.

Solution

1. The constant of proportionality is $k=3 \div 100=0.03\,\text{gal/mi}$, so $y=0.03x$. 2. Substitution gives $1.5$, $4.5$, $6$, $7.5$, and $15$ gallons for the listed distances. 3. The point $(100, 3)$ means the car uses $3$ gallons to travel $100$ miles. 4. Solve $15=0.03x$: $x=15 \div 0.03=500$ miles.

Answer

a) $k=0.03\,\text{gal/mi}$; $y=0.03x$ b) <table> <tr><td>$x$ in miles</td><td>$50$</td><td>$150$</td><td>$200$</td><td>$250$</td><td>$500$</td></tr> <tr><td>$y$ in gallons</td><td>$1.5$</td><td>$4.5$</td><td>$6$</td><td>$7.5$</td><td>$15$</td></tr> </table> c) The car uses $3$ gallons to travel $100$ miles. d) $500$ miles

5233407

A factory uses two machines to produce screws. Machine A produces $15$ screws per minute. Machine B produces $12$ screws in $45$ seconds. a) Find each machine's production rate in screws per minute. Which machine is faster? b) Write an equation for the faster machine, where $y$ is the number of screws produced in $x$ minutes. c) How long does the faster machine take to produce exactly $1000$ screws? Give the answer in minutes and seconds. d) Maintenance increases the faster machine's production rate by $20\%$. Write the new equation.

Hints

- Convert both time measurements to the same unit before comparing rates. - How many seconds are in one minute? - A faster production rate gives a larger constant of proportionality. - Increasing a value by $20\%$ means multiplying it by $1.20$.

Solution

1. Machine A produces $15$ screws per minute. Machine B produces $12$ screws in $45$ seconds, or $0.75$ minute, so its rate is $12 \div 0.75=16$ screws per minute. Machine B is faster. 2. The equation for Machine B is $y=16x$. 3. The time for $1000$ screws is $1000 \div 16=62.5$ minutes, or $62$ minutes $30$ seconds. 4. The increased rate is $16 \cdot 1.20=19.2$ screws per minute, so the new equation is $y=19.2x$.

Answer

a) Machine A: $15$ screws per minute; Machine B: $16$ screws per minute. Machine B is faster. b) $y=16x$ c) $62$ minutes $30$ seconds d) $y=19.2x$

5237997

A recycling model uses the estimate that $25\,\text{kg}$ of used paper produces $18\,\text{kg}$ of new paper and assumes the amounts are proportional. a) Write an equation for the amount of new paper $y$, in kilograms, from $x$ kilograms of used paper. b) According to the model, find the amount of new paper produced from $100\,\text{kg}$ and from $35\,\text{kg}$ of used paper. c) According to the model, how much used paper is needed for a predicted output of at least $500\,\text{kg}$ of new paper? Round up to the nearest tenth of a kilogram. d) What does the constant of proportionality mean in this model?

Hints

- Find the model's output per kilogram of used paper. - Decide whether each part gives the input or the output of the proportional model. - For the threshold, think about why rounding down would not meet the modeled target.

Solution

1. The constant of proportionality is $18 \div 25=0.72$, so $y=0.72x$. 2. For $100\,\text{kg}$, $y=0.72\cdot100=72\,\text{kg}$. For $35\,\text{kg}$, $y=0.72\cdot35=25.2\,\text{kg}$. 3. Solve $500=0.72x$: $x=500 \div 0.72=694.444\ldots$. Under the model, rounding up gives $694.5\,\text{kg}$ to reach a predicted output of at least $500\,\text{kg}$. 4. The constant of proportionality $0.72$ means the model predicts $0.72\,\text{kg}$ of new paper for each kilogram of used paper.

Answer

a) $y=0.72x$ b) $72\,\text{kg}$ and $25.2\,\text{kg}$ c) $694.5\,\text{kg}$ according to the model d) The model predicts $0.72\,\text{kg}$ of new paper per $1\,\text{kg}$ of used paper.

5241667

A pump fills an empty water-storage tank at a constant rate. After $6$ minutes, the tank contains $72$ gallons. 1) Make a table for volume $V$, in gallons, at $t=0$, $5$, $10$, $15$, and $20$ minutes. 2) Explain why the relationship is proportional and write an equation $V(t)$. 3) How much water is in the tank after $45$ minutes? 4) The tank holds $600$ gallons. After how many minutes must the pump be turned off? 5) How would the graph change if a stronger pump delivered $15$ gallons per minute?

Hints

- Find the number of gallons pumped in one minute. - A proportional relationship has a constant rate and begins at zero. - For the capacity question, identify which variable is unknown in the equation. - Compare the two gallons-per-minute rates to decide how the graphs differ.

Solution

1. The rate is $72 \div 6=12\,\text{gal/min}$. The volumes are $0$, $60$, $120$, $180$, and $240$ gallons. 2. The rate is constant and $V(0)=0$, so the relationship is proportional. The equation is $V(t)=12t$. 3. $V(45)=12\cdot45=540$ gallons. 4. Solve $600=12t$: $t=600 \div 12=50$ minutes. 5. With a rate of $15$ gallons per minute, the graph would still pass through the origin but would rise faster; for each positive time, it would show a greater volume than the original graph.

Answer

1) <table><tr><th>Time $t$, in minutes</th><td>$0$</td><td>$5$</td><td>$10$</td><td>$15$</td><td>$20$</td></tr><tr><th>Volume $V$, in gallons</th><td>$0$</td><td>$60$</td><td>$120$</td><td>$180$</td><td>$240$</td></tr></table> 2) The constant rate and zero starting volume make the relationship proportional; $V(t)=12t$. 3) $540$ gallons. 4) $50$ minutes. 5) The new graph would still pass through the origin but would rise faster because its constant of proportionality is $15\,\text{gal/min}$ instead of $12\,\text{gal/min}$.

5241857

A recipe uses $w$ cups of water for $k>0$ pounds of flour. Write a proportion to find the amount of water $x$, in cups, needed for $m$ pounds of flour while keeping the same mixture. Then solve the proportion for $x$.

Hints

- If the amount of flour increases, what must happen to the amount of water? - Write each water-to-flour ratio as a fraction. - Use multiplication or cross multiplication to isolate the unknown.

Solution

1. The water-to-flour ratio must remain constant, so the relationship is proportional. 2. A proportion is $\frac{w}{k}=\frac{x}{m}$. 3. Multiply by $m$ to isolate $x$: $x=\frac{wm}{k}$.

Answer

The proportion is $\frac{w}{k}=\frac{x}{m}$, and $x=\frac{wm}{k}$.

5245027

The perimeter $P$ of a regular hexagon is proportional to its side length $s$. 1) Write the function $P(s)$. 2) What constant value does $\frac{P}{s}$ have for every $s>0$? 3) Luca claims that a regular hexagon with side length $4.5\,\text{cm}$ has perimeter $27.5\,\text{cm}$. Use the equation to check the claim. 4) How does the perimeter change when the side length doubles? Explain generally.

Hints

- How many equal sides does a regular hexagon have? - Substitute the function expression for perimeter into the ratio. - Evaluate the function at the given side length. - Try replacing $s$ with $2s$.

Solution

1. A regular hexagon has six equal sides, so $P(s)=6s$. 2. $\frac{P}{s}=\frac{6s}{s}=6$. 3. $P(4.5)=6 \cdot 4.5=27\,\text{cm}$. Since $27\ne27.5$, the claim is false. 4. $P(2s)=6(2s)=2(6s)=2P(s)$, so doubling the side length doubles the perimeter.

Answer

1) $P(s)=6s$ 2) $6$ 3) The claim is false; the perimeter is $27\,\text{cm}$. 4) The perimeter doubles.

5502127

A proportional relationship between travel time $t$, in hours, and distance $d$, in miles, is given by $d=\frac{9}{2}t$. a) Identify the constant of proportionality and explain its meaning. b) How far is traveled in $\frac{2}{3}\,\text{hour}$?

Hints

- In a proportional equation, look at the factor multiplying the input. - Use the variable descriptions to attach meaning and units to that factor. - For the second part, substitute the given fraction of an hour into the relationship.

Solution

1. The constant of proportionality is $\frac{9}{2}$ miles per hour, or $4.5$ miles per hour. 2. For $t=\frac{2}{3}$, $d=\frac{9}{2}\cdot\frac{2}{3}=3\,\text{miles}$.

Answer

a) $\frac{9}{2}$ miles per hour, or $4.5$ miles per hour. b) $3\,\text{miles}$

5502147

Four identical LED strips are installed on each stage panel. Each strip is $0.35\,\text{meter}$ long. The total strip length $L$, in meters, for $p$ panels is modeled by $L=4(0.35)p$. a) Rewrite the equation in the form $L=kp$ and identify the constant of proportionality. b) How many panels can be outfitted with $11.2\,\text{meters}$ of strip?

Hints

- Simplify the constant factors before identifying the amount associated with one panel. - Use the variable descriptions to interpret the units of the constant. - In the second part, work backward from the available total length.

Solution

1. Multiply the constant factors: $4\cdot0.35=1.4$, so $L=1.4p$ and $k=1.4$ meters per panel. 2. Set $11.2=1.4p$. Then $p=11.2 \div 1.4=8$ panels.

Answer

a) $L=1.4p$, so $k=1.4$ meters per panel. b) $8$ panels

5502157

Amina looks at the proportional equation $y=0.4x$ and says, “The constant of proportionality is $2.5$ because $\frac{x}{y}=2.5$.” Evaluate Amina's statement and explain the correct constant of proportionality.

Hints

- Pay attention to which variable is the input and which is the output. - Compare the equation with the standard form for a proportional relationship. - A reciprocal ratio describes the relationship in the opposite direction.

Solution

1. In the form $y=kx$, the constant of proportionality is the factor multiplying $x$. 2. Therefore, $k=0.4$. 3. The value $2.5$ is the reciprocal rate $\frac{x}{y}$, not the constant for the mapping from $x$ to $y$.

Answer

Amina's statement is incorrect. The constant of proportionality is $0.4$. The value $2.5$ is the reciprocal ratio $\frac{x}{y}$.

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