Aimathic – Constant of proportionality from equation

Constant of proportionality from equation

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Difficulty: 1 – 5

5139577

A hiker walks $6$ miles in $2$ hours at a constant pace. a) Explain why distance is proportional to time, and write an equation in the form $d(t)=rt$. b) How far will the hiker 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) The hiker 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) $y=11.25$

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) $40\,\text{s}$

5119157

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

A homeowner 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) A customer wants to spend exactly $\$11.25$. How many ounces of walnuts can the customer buy?

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

A runner maintains a constant pace. After $12$ minutes, the runner has traveled $2$ miles. a) Find the runner's speed in miles per hour. Also write an equation for distance $d$, in miles, after $t$ minutes. b) How far does the runner travel in $45$ minutes at the same pace? c) If the runner 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 $k=21\div 25=0.84$. 2. The green-bean weight needed is $150\div 0.84\approx 178.57$ pounds. 3. From $12.5$ pounds, the roasted weight is $12.5\cdot 0.84=10.5$ pounds.

Answer

a) $k=0.84$. 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) A customer 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

- The slope of a line through the origin is the constant of proportionality. - Density is mass divided by volume. - Use $m=\rho V$ and solve for the unknown quantity.

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 slope 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. - What happens to the volume when the time doubles? - A proportional relationship has the form $y=mx$. Which quantities correspond to $x$ and $y$ here?

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 slope $0.5$ is the number of fluid ounces poured 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 slope means that the dispenser pours $0.5\,\text{fl oz}$ each second. d) $100$ seconds.

5141327

A student works at an ice cream shop during school break. Earnings are proportional to hours worked. After $6$ hours, the student has earned $\$81.00$. a) How much will the student earn for $14$ hours of work? b) How many hours must the student 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 the student earns $\$13.50$ per hour. 2. For $14$ hours, the earnings are $14\cdot 13.50=\$189.00$. 3. To earn $\$472.50$, the student 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. 1) Write an equation for the number of silver coins in terms of the number of gold coins. 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 received for one gold coin. - What equation represents a relationship with a constant ratio? - Rearrange your equation to find gold coins from a given number of silver coins.

Solution

1. The constant of proportionality is $14 \div 4=3.5$, so $S=3.5G$. 2. For $G=16$, $S=3.5 \cdot 16=56$. 3. Solve $105=3.5G$: $G=105 \div 3.5=30$.

Answer

1) $S=3.5G$ 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 process produces about $18\,\text{kg}$ of new paper from $25\,\text{kg}$ of used paper. Assume the amount of new paper is proportional to the amount of used paper. a) Write an equation for the amount of new paper $y$, in kilograms, from $x$ kilograms of used paper. b) Find the amount of new paper produced from $100\,\text{kg}$ and from $35\,\text{kg}$ of used paper. c) A school wants at least $500\,\text{kg}$ of new paper. How much used paper must be collected? Round up to the nearest tenth of a kilogram. d) What does the slope mean in this situation?

Hints

- Find the output per kilogram of input. - Decide whether the unknown in each part is $x$ or $y$. - What output is produced from exactly $1\,\text{kg}$ of used paper?

Solution

1. The constant of proportionality is $18 \div 25=0.72$, so $y=0.72x$. 2. For $100\,\text{kg}$, $y=0.72 \cdot 100=72\,\text{kg}$. For $35\,\text{kg}$, $y=0.72 \cdot 35=25.2\,\text{kg}$. 3. Solve $500=0.72x$: $x=500 \div 0.72=694.444\ldots$. To ensure at least $500\,\text{kg}$ of new paper, round up to $694.5\,\text{kg}$. 4. The slope $0.72$ means each kilogram of used paper produces $0.72\,\text{kg}$ of new paper.

Answer

a) $y=0.72x$ b) $72\,\text{kg}$ and $25.2\,\text{kg}$ c) At least $694.5\,\text{kg}$ d) The slope means that $1\,\text{kg}$ of used paper produces $0.72\,\text{kg}$ of new 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 barrel 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. - Substitute a volume into the equation and solve for time. - What does the coefficient of $t$ represent on the graph?

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. $V(45)=12 \cdot 45=540$ gallons. 4. Solve $600=12t$: $t=600 \div 12=50$ minutes. 5. A rate of $15$ gallons per minute gives a greater slope, so the new graph would be steeper while still passing through the origin.

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 be steeper, with slope $15$.

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) A student 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.

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