Free math worksheets

- How far does the herd travel in one hour? - How many groups of \(4\) miles are in \(20\) miles? - Divide distance by rate to find time.

1. Use \(\text{time} = \text{distance} \div \text{rate}\). 2. Calculate \(20\,\text{mi} \div 4\,\text{mi/h} = 5\,\text{h}\).

The herd travels for \(5\) hours.

- A unit rate compares the distance with exactly one hour. - Use one row of the table to determine how much distance corresponds to one hour. - State both the numerical rate and its meaning in context.

1. Use either row to find the distance for one hour: \(90 \div 2 = 45\). 2. The unit rate is \(45\) miles per hour. 3. This means the van travels \(45\) miles during each hour when the rate stays constant.

\(45\,\text{mi/h}\). The van travels \(45\) miles each hour at this constant rate.

- Use the price per pound and the number of pounds to find the total cost. - Keep track of decimal place value when multiplying. - Estimate first to check whether your answer is reasonable.

1. Multiply the unit price by the number of pounds: \(\$4.80/\text{lb}\times2.25\,\text{lb}\). 2. \(4.80\times2.25=10.80\). 3. The total cost is \(\$10.80\).

Mrs. Miller pays \(\$10.80\).

- Find Rider A's total distance first. - Then compare that distance with Rider B's distance. - Use place value carefully when multiplying decimals.

1. Find Rider A's distance: \(14.5\times2.25=32.625\) miles. 2. Compare the distances: \(33>32.625\). 3. Rider B is correct.

Rider B is correct. Rider A rode \(32.625\) miles, which is less than \(33\) miles.

- Divide each total price by the number of oranges. - Use the same unit, dollars per orange, for both comparisons. - Compare the two unit prices.

1. Deal A: \(\$5.85\div15=\$0.39\) per orange. 2. Deal B: \(\$4.80\div12=\$0.40\) per orange. 3. Since \(\$0.39<\$0.40\), Deal A is less expensive per orange.

Deal A is less expensive at \(\$0.39\) per orange; Deal B costs \(\$0.40\) per orange.

- Find the amount of juice needed for one person. - Use a unit rate or equivalent ratio. - Check whether doubling the number of people doubles the amount of juice.

1. Find the amount per person: \(2.5\div 25=0.1\) gallon per person. 2. Let \(x\) be the gallons needed for \(60\) people. An equation is \(x=60\times 0.1\). 3. Solve: \(x=6\).

An equation is \(x=60\times 0.1\). The class needs \(6\) gallons of juice.

- What does “price per ounce” mean? - Divide each price by the matching number of ounces. - Compare the unit prices, then subtract to find the difference.

1. Package A costs \(1.80\div 12=\$0.15\) per ounce. 2. Package B costs \(2.80\div 20=\$0.14\) per ounce. 3. Since \(0.14<0.15\), Package B is the better buy. 4. The difference is \(0.15-0.14=\$0.01\) per ounce.

Package A costs \(\$0.15\) per ounce. Package B costs \(\$0.14\) per ounce, so Package B is cheaper by \(\$0.01\) per ounce.

- Find the amount of flour for one person. - Compare \(15\) eggs with the \(3\) eggs in the original recipe. - Is this a “more means more” relationship?

1. The flour per person is \(2\div 4=0.5\) cup. 2. For \(10\) people, the flour needed is \(10\times 0.5=5\) cups. 3. The egg amount is multiplied by \(15\div 3=5\). 4. Therefore, the number of servings is \(4\times 5=20\).

a) \(5\) cups of flour. b) \(20\) people.

- Use the same comparison amount for all three cup sizes. - Count how many groups of \(4\,\text{fl oz}\) are in each cup. - After identifying the lowest rate, apply that rate to \(36\,\text{fl oz}\).

1. Small: \(8\,\text{fl oz}\) contains two groups of \(4\,\text{fl oz}\), so \(\$2.20 \div 2 = \$1.10\) per \(4\,\text{fl oz}\). 2. Medium: \(12\,\text{fl oz}\) contains three groups of \(4\,\text{fl oz}\), so \(\$3.00 \div 3 = \$1.00\) per \(4\,\text{fl oz}\). 3. Large: \(20\,\text{fl oz}\) contains five groups of \(4\,\text{fl oz}\), so \(\$4.50 \div 5 = \$0.90\) per \(4\,\text{fl oz}\). 4. The large-cup rate is lowest. 5. There are nine groups of \(4\,\text{fl oz}\) in \(36\,\text{fl oz}\), so \(9 \times \$0.90 = \$8.10\).

a) Small: \(\$1.10\) per \(4\,\text{fl oz}\); medium: \(\$1.00\) per \(4\,\text{fl oz}\); large: \(\$0.90\) per \(4\,\text{fl oz}\). b) The large cup has the lowest rate. c) \(\$8.10\)

- How many degrees does the minute hand move in one minute? - Divide each angle by the hand’s rate. - In part d, the angle is greater than one full revolution.

1. The minute hand sweeps \(360^\circ\) in \(60\) minutes, so its rate is \(360^\circ \div 60\,\text{minutes} = 6^\circ\) per minute. 2. For part a, \(48^\circ \div 6^\circ\text{ per minute} = 8\,\text{minutes}\). 3. For part b, \(162^\circ \div 6^\circ\text{ per minute} = 27\,\text{minutes}\). 4. For part c, \(210^\circ \div 6^\circ\text{ per minute} = 35\,\text{minutes}\). 5. For part d, \(540^\circ \div 6^\circ\text{ per minute} = 90\,\text{minutes}\).

a) \(8\,\text{minutes}\) b) \(27\,\text{minutes}\) c) \(35\,\text{minutes}\) d) \(90\,\text{minutes}\), or \(1\,\text{hour}\ 30\,\text{minutes}\)

- What operation gives the time when you know the total amount and the amount delivered each minute? - Divide the pool capacity by the hose's rate. - How many minutes are in one hour?

1. Divide the pool capacity by the flow rate: \(1500 \div 12=125\) minutes. 2. Convert \(125\) minutes to hours and minutes: \(125=2 \times 60+5\). Therefore, the time is \(2\) hours \(5\) minutes.

It will take \(2\) hours \(5\) minutes.

- Use the rate of \(50\) miles for each hour. - When the time doubles, the distance doubles. - For half an hour, find half of the one-hour distance. - For \(10\) hours, multiply the one-hour distance by \(10\).

1. For \(2\) hours: \(50 \times 2 = 100\) miles. 2. For \(4\) hours: \(50 \times 4 = 200\) miles. 3. For \(10\) hours: \(50 \times 10 = 500\) miles. 4. For \(\frac{1}{2}\) hour: \(50 \div 2 = 25\) miles. 5. Complete the table. <table> <tr><td>Time</td><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(4\,\text{h}\)</td><td>\(10\,\text{h}\)</td><td>\(\frac{1}{2}\,\text{h}\)</td></tr> <tr><td>Distance</td><td>\(50\,\text{mi}\)</td><td>\(100\,\text{mi}\)</td><td>\(200\,\text{mi}\)</td><td>\(500\,\text{mi}\)</td><td>\(25\,\text{mi}\)</td></tr> </table>

<table> <tr><td>Time</td><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(4\,\text{h}\)</td><td>\(10\,\text{h}\)</td><td>\(\frac{1}{2}\,\text{h}\)</td></tr> <tr><td>Distance</td><td>\(50\,\text{mi}\)</td><td>\(100\,\text{mi}\)</td><td>\(200\,\text{mi}\)</td><td>\(500\,\text{mi}\)</td><td>\(25\,\text{mi}\)</td></tr> </table>

- Use the rate of \(4\) miles for each hour. - Half an hour gives half the one-hour distance. - For part c), determine how many groups of \(4\) miles are in \(24\) miles.

1. In \(3\) hours, the hiker walks \(4 \times 3 = 12\) miles. 2. Half an hour is half of one hour, so the hiker walks \(4 \div 2 = 2\) miles. 3. For \(24\) miles, divide distance by the unit rate: \(24 \div 4 = 6\) hours.

a) \(12\,\text{mi}\) b) \(2\,\text{mi}\) c) \(6\,\text{h}\)

- What is the cost for half of an hour? - Combine the costs for one hour and half an hour when needed. - Determine how many hourly rates are included in each parking time.

1. For \(2\) hours, \(2 \times \$4 = \$8\). 2. Thirty minutes is half an hour, so \(\$4 \div 2 = \$2\). 3. For \(1\) hour \(30\) minutes, \(\$4 + \$2 = \$6\). 4. For \(3\) hours, \(3 \times \$4 = \$12\).

<table> <tr> <td>Parking time</td> <td>\(1\,\text{hr}\)</td> <td>\(2\,\text{hr}\)</td> <td>\(30\,\text{min}\)</td> <td>\(1\,\text{hr}\ 30\,\text{min}\)</td> <td>\(3\,\text{hr}\)</td> </tr> <tr> <td>Cost</td> <td>\(\$4\)</td> <td>\(\$8\)</td> <td>\(\$2\)</td> <td>\(\$6\)</td> <td>\(\$12\)</td> </tr> </table>

- How many \(15\)-minute intervals are in one hour? - Find the price for \(15\) minutes first. - Combine the prices for \(30\) minutes and \(15\) minutes to find the price for \(45\) minutes. - Break longer rental times into full hours and additional minutes.

1. For \(30\) minutes, \(\$12 \div 2 = \$6\). 2. For \(15\) minutes, \(\$12 \div 4 = \$3\). 3. For \(45\) minutes, \(\$6 + \$3 = \$9\). 4. For \(2\) hours \(15\) minutes, \(2 \times \$12 + \$3 = \$27\). 5. For \(3\) hours \(30\) minutes, \(3 \times \$12 + \$6 = \$42\).

<table> <tr> <td>Rental time</td> <td>\(30\,\text{min}\)</td> <td>\(15\,\text{min}\)</td> <td>\(45\,\text{min}\)</td> <td>\(2\,\text{hr}\ 15\,\text{min}\)</td> <td>\(3\,\text{hr}\ 30\,\text{min}\)</td> </tr> <tr> <td>Price</td> <td>\(\$6\)</td> <td>\(\$3\)</td> <td>\(\$9\)</td> <td>\(\$27\)</td> <td>\(\$42\)</td> </tr> </table>

- How many hours pass between the start and end times? - Multiply the number of hours by the hourly rate.

1. The time from \(8{:}30\) a.m. to \(11{:}30\) a.m. is \(3\) hours. 2. Multiply the hours by the hourly rate: \(3 \times \$42 = \$126\).

The total charge is \(\$126\).

- Find the elapsed time in hours and minutes. - What is half of the hourly charge? - Add the charges for the full hours and the remaining half hour.

1. The work time is \(2\) hours \(30\) minutes, or \(2.5\) hours. 2. The charge for \(2\) full hours is \(2 \times \$64 = \$128\). 3. The charge for half an hour is \(\$64 \div 2 = \$32\). 4. The total is \(\$128 + \$32 = \$160\).

The office must pay \(\$160\).

- How much does the price increase for each additional night? - Combine the prices for \(10\) nights and \(4\) nights to find the price for \(14\) nights. - Multiply the number of nights by the nightly rate.

1. Multiply each number of nights by \(\$12\) to complete the table. <table> <tr><th>Nights</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr> <tr><th>Price</th><td>\(\$12\)</td><td>\(\$24\)</td><td>\(\$36\)</td><td>\(\$60\)</td><td>\(\$96\)</td><td>\(\$120\)</td></tr> </table> 2. For \(14\) nights, \(14 \times \$12 = \$168\).

a) <table><tr><th>Nights</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr><tr><th>Price</th><td>\(\$12\)</td><td>\(\$24\)</td><td>\(\$36\)</td><td>\(\$60\)</td><td>\(\$96\)</td><td>\(\$120\)</td></tr></table> b) \(14\) nights cost \(\$168\).

- How are \(4\), \(8\), and \(2\) related? - What happens to the price when the number of scoops doubles? - What happens to the price when the number of scoops is cut in half?

1. Eight scoops is twice \(4\) scoops, so double the price: \(\$6.00 \times 2 = \$12.00\). 2. Two scoops is half of \(4\) scoops, so halve the price: \(\$6.00 \div 2 = \$3.00\).

a) \(8\) scoops cost \(\$12.00\). b) \(2\) scoops cost \(\$3.00\).

- Divide each total capacity by the number of people. - Find the rate for each boat separately. - Compare the two rates.

1. For Wave, divide the total capacity by the number of people: \(560\,\text{kg} \div 8 = 70\,\text{kg}\) per person. 2. For Anchor, divide: \(450\,\text{kg} \div 6 = 75\,\text{kg}\) per person. 3. Since \(75\,\text{kg} > 70\,\text{kg}\), Anchor has the greater capacity per person.

Wave allows \(70\,\text{kg}\) per person, and Anchor allows \(75\,\text{kg}\) per person. Anchor has the greater capacity per person.

- Would two identical pumps take more or less time than one pump? - How does the combined pumping rate compare with the rate of one pump? - What happens to the time when the rate doubles?

1. Two identical pumps working together pump water at twice the rate of one pump. 2. Doubling the rate cuts the time in half. 3. Find half of \(40\) minutes: \(40 \div 2 = 20\) minutes.

Both pumps would empty the pond in \(20\) minutes.

- Divide the total juice volume by the average volume from one tree. - Simplify the division by removing a factor of \(10\). - Use approximation language in the answer.

1. Divide the total juice volume of \(48{,}000\,\text{L}\) by the average volume of \(60\,\text{L}\) per tree. The quotient is about \(800\). 2. The quotient is approximate because the yield per tree is an average.

About \(800\) apple trees are needed.

- Divide the monthly volume by the volume in one full truck. - Simplify by dividing both numbers by \(1000\). - Use approximation language in the answer.

1. Divide the approximate monthly volume of \(900{,}000\,\text{L}\) by the truck capacity of \(15{,}000\,\text{L}\). The quotient is about \(60\). 2. The result is approximate because the monthly volume is approximate.

About \(60\) full truckloads equal the monthly amount.

- How many \(200\)-gallon amounts fit in \(1000\) gallons? - Use that relationship to find the cost of \(200\) gallons. - Half the water has half the cost.

1. Since \(1000 \div 200 = 5\), \(200\) gallons is one fifth of \(1000\) gallons. A full tank costs \(\$5.00 \div 5 = \$1.00\). 2. Half of \(200\) gallons is \(100\) gallons, so a half-tank water change costs \(\$1.00 \div 2 = \$0.50\).

a) A complete fill costs \(\$1.00\). b) One half-tank water change costs \(\$0.50\).

- First find how many bags the machine fills in one minute. - Use the unit rate to find the amount for \(10\) minutes.

1. Find the unit rate: \(12 \div 3 = 4\) bags per minute. 2. Multiply by \(10\) minutes: \(4 \times 10 = 40\) bags.

The machine fills \(40\) bags in \(10\) minutes.

- First find how much flour is used in one day. - Which operation divides an amount equally among several days? - Once you know the daily amount, determine how many times it fits into the new supply.

1. Find the daily flour use: \(35 \div 5 = 7\,\text{lb}\). 2. Find how many days the new supply will last: \(56 \div 7 = 8\) days.

a) He uses \(7\,\text{lb}\) of flour each day. b) The \(56\,\text{lb}\) supply will last \(8\) days.

- How many times does \(5\) minutes fit into \(15\) minutes? - At a constant rate, how does the output change when the time is multiplied by that factor?

1. Compare the time intervals: \(15 \div 5 = 3\), so \(15\) minutes is \(3\) times as long as \(5\) minutes. 2. Multiply the number of bags by the same factor: \(40 \times 3 = 120\) bags.

The machine packs \(120\) bags in \(15\) minutes.

- Find how many liters one container holds. - Divide the delivery volume by the volume in one container. - A ratio table can organize the two situations.

1. Find the volume in one container: \(72\,\text{L} \div 6 = 12\,\text{L}\). 2. Divide the delivery volume by the volume per container: \(156\,\text{L} \div 12\,\text{L} = 13\).

\(13\) containers are needed.

- Determine how many times as many people will be served. - Multiply the amount of apples by the same factor. - Check that your answer is larger than the original amount.

1. Find the scale factor: \(20 \div 4 = 5\). 2. Multiply the apple amount by the scale factor: \(1.5\,\text{lb} \times 5 = 7.5\,\text{lb}\).

The class needs \(7.5\,\text{lb}\) of apples.

- First find how many tons the truck carries in one load. - Then divide the new total by the amount per load. - A ratio table may help.

1. Find the amount delivered per load: \(48\,\text{tons} \div 6 = 8\,\text{tons}\) per load. 2. Find the number of loads for \(72\) tons: \(72\,\text{tons} \div 8\,\text{tons} = 9\).

The truck needs \(9\) loads.

- Determine how many times the smaller group's weight fits into the bag's weight. - Each equal-weight group contains \(4\) apples. - Multiply the number of groups by \(4\).

1. Find how many groups of \(1.25\,\text{lb}\) are in \(5\,\text{lb}\): \(5\,\text{lb} \div 1.25\,\text{lb} = 4\). 2. Each group contains \(4\) apples, so the bag contains \(4 \times 4 = 16\) apples.

The bag contains \(16\) apples.

- Determine how many \(100\)-meter intervals are in the total rise. - Decide whether the temperature increases or decreases. - Subtract the total change from the starting temperature. - The result may be a negative number.

1. Find the number of \(100\)-meter intervals: \(1800 \div 100 = 18\). 2. The predicted temperature decrease is \(18 \times 1\,^{\circ}\text{C} = 18\,^{\circ}\text{C}\). 3. Subtract the decrease from the starting temperature: \(7 - 18 = -11\). The model predicts \(-11\,^{\circ}\text{C}\).

The model predicts a temperature of \(-11\,^{\circ}\text{C}\).

- Find the price of one marker in each package. - Divide each total price by its number of markers. - Subtract the two unit prices. - Express the final difference in cents.

1. Find the unit price in the small package: \(\$4.50 \div 5 = \$0.90\) per marker. 2. Find the unit price in the value pack: \(\$9.60 \div 12 = \$0.80\) per marker. 3. Find the difference: \(\$0.90 - \$0.80 = \$0.10\), which is \(10\) cents.

Each marker costs \(10\) cents less in the value pack.

- Find how many groups of \(12\,\text{kg}\) are in \(480\,\text{kg}\). - Scale the amount of new paper by the same factor. - A ratio table can organize the quantities.

1. Find the scale factor: \(480\,\text{kg} \div 12\,\text{kg} = 40\). 2. Multiply the output by the same factor: \(40 \times 9\,\text{kg} = 360\,\text{kg}\).

The collected paper can produce \(360\,\text{kg}\) of new paper.

- How many minutes are in one hour? - How many minutes are in half an hour? - Multiply the number of bags sold per minute by the number of minutes.

1. Convert the time: half an hour is \(30\) minutes. 2. Multiply the rate by the time: \(5 \times 30 = 150\) bags.

The concession stand sells \(150\) bags of popcorn in half an hour.

- Find the total distance between the starting and ending temperatures. - Include the change on both sides of zero. - Divide the total temperature change by the hourly change.

1. Find the total temperature decrease. From \(18\) to \(0\) is \(18\,^{\circ}\text{C}\), and from \(0\) to \(-12\) is another \(12\,^{\circ}\text{C}\). The total decrease is \(18 + 12 = 30\,^{\circ}\text{C}\). 2. Divide the total decrease by the hourly rate: \(30 \div 5 = 6\).

The cooling process takes \(6\) hours.

- Use doubling to find the price for \(8\) balls. - Divide by \(4\) to find the unit price. - Multiply the unit price by any requested number of balls.

1. Eight balls are twice as many as four balls, so the cost is \(\$6.40 \times 2 = \$12.80\). 2. The unit price is \(\$6.40 \div 4 = \$1.60\) per ball. 3. The cost of \(15\) balls is \(15 \times \$1.60 = \$24.00\).

a) \(\$12.80\) b) \(\$1.60\) c) \(\$24.00\)

- Determine how many times as many servings are needed. - More servings require proportionally more milk. - Multiply the original milk amount by the scale factor.

1. Find the scale factor: \(20 \div 4 = 5\). 2. Multiply the milk amount by the same factor: \(1\,\text{qt} \times 5 = 5\,\text{qt}\).

The recipe needs \(5\,\text{qt}\) of milk.

- Compare the new number of sheets with the original number. - Determine the factor that changes \(500\) to \(100\). - Apply the same operation to the weight.

1. One hundred sheets is one-fifth of \(500\) sheets because \(500 \div 5 = 100\). 2. Divide the weight by the same factor: \(5\,\text{lb} \div 5 = 1\,\text{lb}\).

\(100\) sheets weigh \(1\,\text{lb}\).

- Find the number of pages printed in one minute. - A table of time and pages may help. - Use the unit rate to scale up or work backward.

1. Find the unit rate: \(30 \div 2 = 15\) pages per minute. 2. In \(7\) minutes, the copier prints \(7 \times 15 = 105\) pages. 3. To print \(225\) pages, it takes \(225 \div 15 = 15\) minutes.

a) \(105\) pages b) \(15\) minutes

- Find the price of one pound first. - Use the unit price to calculate either a cost or a weight.

1. Find the unit price: \(\$6.00 \div 4 = \$1.50\) per pound. 2. Seven pounds cost \(7 \times \$1.50 = \$10.50\). 3. For \(\$12.00\), the customer can buy \(\$12.00 \div \$1.50 = 8\,\text{lb}\).

a) \(\$10.50\) b) \(8\,\text{lb}\)

- Find the cost for one student. - Use that unit cost for the four additional students. - The question asks for the additional amount, not the new total.

1. Find the cost per student: \(\$108 \div 18 = \$6\). 2. Find the amount paid by the four additional students: \(4 \times \$6 = \$24\).

The four additional students must pay \(\$24\) altogether.

- Find how many times \(16\,\text{kg}\) fits into \(80\,\text{kg}\). - Multiply both quantities in the ratio by the same factor. - A ratio table may help organize the amounts.

1. Find the scale factor from \(16\,\text{kg}\) to \(80\,\text{kg}\): \(80\,\text{kg} \div 16\,\text{kg} = 5\). 2. Multiply the used-paper amount by the same factor: \(20\,\text{kg} \times 5 = 100\,\text{kg}\).

\(100\,\text{kg}\) of used paper are needed.

- Convert kilometers to meters. - Express one-half hour in minutes. - Divide the total distance by the total time.

1. Convert the distance: \(9\,\text{km} = 9000\,\text{m}\). 2. One-half hour is \(30\) minutes. 3. Find the unit rate: \(9000\,\text{m} \div 30\,\text{min} = 300\,\text{m/min}\).

The cyclist travels an average of \(300\,\text{m}\) per minute.

- Start with the cost of \(8\) notebooks. - Divide to find the cost of one notebook. - Multiply the unit price by \(15\).

1. Find the price of one notebook: \(\$11.20 \div 8 = \$1.40\). 2. Find the price of \(15\) notebooks: \(15 \times \$1.40 = \$21.00\). 3. Complete the table. <table><tr><th>Number of notebooks</th><th>Price</th></tr><tr><td>\(8\)</td><td>\(\$11.20\)</td></tr><tr><td>\(1\)</td><td>\(\$1.40\)</td></tr><tr><td>\(15\)</td><td>\(\$21.00\)</td></tr></table>

<table><tr><th>Number of notebooks</th><th>Price</th></tr><tr><td>\(8\)</td><td>\(\$11.20\)</td></tr><tr><td>\(1\)</td><td>\(\$1.40\)</td></tr><tr><td>\(15\)</td><td>\(\$21.00\)</td></tr></table>

- Compare \(6\) apples with \(3\) apples. - When the number of apples doubles, determine what happens to the cost. - You can also find the cost of one apple first. - A ratio table can help match numbers of apples with costs.

1. Six apples are twice as many as \(3\) apples because \(6 \div 3 = 2\). 2. Double the cost: \(2 \times 90 = 180\) cents. 3. Convert: \(180\) cents is \(\$1.80\).

Six apples cost \(\$1.80\).

- Express all times in hours to match the rate unit. - Use distance equals rate times time. - Half the rate \(x\) is \(\frac{x}{2}\).

1. Distance equals rate times time. The expressions are \(4x\), \(0.5x\), and \(tx\), because \(30\) minutes is \(0.5\) hour. 2. At \(4.5\) miles per hour for \(3\) hours, the distance is \(4.5\times 3=13.5\) miles. 3. The first part is \(2x\), and the second part is \(\frac{x}{2}\). The total is \(2x+\frac{x}{2}=2.5x\).

a) \(4x\); \(0.5x\); \(tx\) b) \(13.5\) miles c) \(2.5x\) miles

- Average speed compares distance with time. - Divide the distance by the number of hours. - Check that your answer has units of miles per hour.

1. Average speed is distance divided by time, so \(v=\frac{s}{t}\). 2. For the first trip, \(v=\frac{52.5}{2.5}=21\) miles per hour. 3. For the second trip, \(v=\frac{18.9}{1.5}=12.6\) miles per hour.

a) \(v=\frac{s}{t}\) b) 1) \(21\) miles per hour 2) \(12.6\) miles per hour

- Divide the total cost by the number of apples to find the unit price. - Multiply the number of items by the price per item. - Substitute each pair of values into the formula.

1. The unit price is \(\$4.50\div 6=\$0.75\) per apple. 2. The total cost is \(G=np\). 3. For part a), \(G=15\times 0.60=9.00\), so the cost is \(\$9.00\). 4. For part b), \(G=8\times 1.10=8.80\), so the cost is \(\$8.80\).

1) \(\$0.75\) 2) \(G=np\) 3) a) \(\$9.00\) b) \(\$8.80\)

- Find each section's distance by multiplying speed by time. - Add the two section distances. - For part 3, multiply one speed by the total time.

1. The section distances are \(v_1t_1\) and \(v_2t_2\), so \(s=v_1t_1+v_2t_2\). 2. Substituting the values gives \(s=3\times 2+2.5\times 3=6+7.5=13.5\) miles. 3. At one constant speed for the full time, \(s=v(t_1+t_2)\).

1) \(s=v_1t_1+v_2t_2\) 2) \(13.5\) miles 3) \(s=v(t_1+t_2)\)

- Multiply the number of megabytes by the unit price. - Convert gigabytes to megabytes before calculating. - Use the same multiplication rule with the variable \(m\).

1. Multiply the data amount by \(\$0.05\) per megabyte. 2. For \(10\,\text{MB}\), the cost is \(10\times 0.05=0.50\), or \(\$0.50\). 3. For \(100\,\text{MB}\), the cost is \(100\times 0.05=5.00\), or \(\$5.00\). 4. For \(250\,\text{MB}\), the cost is \(250\times 0.05=12.50\), or \(\$12.50\). 5. Since \(1.5\,\text{GB}=1500\,\text{MB}\), the cost is \(1500\times 0.05=75.00\), or \(\$75.00\). 6. For \(m\,\text{MB}\), the cost is \(0.05m\) dollars.

a) \(\$0.50\) b) \(\$5.00\) c) \(\$12.50\) d) \(\$75.00\) e) \(0.05m\) dollars

- Read the two number lines at matching horizontal positions. - Find the tick for \(4\) minutes on the time line. - Move to the corresponding position on the distance line and read its value.

1. The aligned lines pair \(2\) minutes with \(6\) miles, so each additional \(2\) minutes corresponds to another \(6\) miles. 2. The point aligned with \(4\) minutes on panel a) is \(12\) miles on panel b).

\(12\) miles

- The route length and the distance number line must use the same unit before you compare them. - Locate the converted distance on panel b). - Read the aligned time from panel a), working backward from distance to time.

1. Convert the route length: \(4500\,\text{m}=4.5\,\text{km}\). 2. On panel b), \(4.5\) kilometers aligns with \(6\) minutes on panel a). 3. Therefore the cyclist takes \(6\) minutes to travel the route at the shown constant rate.

\(6\) minutes

- Express all volumes in the same unit. - Find the scale factor from the pitcher to the dispenser. - Determine the fraction of the pitcher that is apple juice. - Apply the same ratio to the larger container.

1. Convert the total volumes to milliliters: \(1.5\,\text{L}=1500\,\text{mL}\) and \(6\,\text{L}=6000\,\text{mL}\). 2. The apple-juice fraction in the pitcher is \(\frac{600}{1500}=\frac{2}{5}\). 3. The dispenser has the same ratio, so its apple-juice amount is \(6000\times\frac{2}{5}=2400\,\text{mL}\). Equivalently, the dispenser is \(4\) times as large, so \(600\times4=2400\).

The dispenser contains \(2400\,\text{mL}\), or \(2.4\,\text{L}\), of apple juice.

- Find the value of the beans that were removed. - Use the removed value and weight to find the unit rate. - In b), the full bag's weight stays the same while the price per pound changes.

1. The removed beans were worth \(\$144-\$108=\$36\). 2. The unit price was \(\$36\div 3\,\text{lb}=\$12\) per pound. 3. The full bag weighed \(\$144\div(\$12\text{ per pound})=12\,\text{lb}\). 4. The reduced price is \(\$12-\$2=\$10\) per pound. 5. At the reduced price, the full bag would cost \(12\,\text{lb}\times\$10\text{ per pound}=\$120\).

a) \(12\,\text{lb}\) b) \(\$120\)

- Use the meaning of miles per gallon to connect distance and fuel. - For part a), ask how many groups of \(25\) miles fit into \(15\) miles. - For part b), use the distance traveled for each gallon.

1. For a), divide distance by miles per gallon: \(15\div25=0.6\) gallon. 2. For b), multiply the amount of gas by the distance per gallon: \(21.25\times25=531.25\) miles.

a) \(0.6\) gallon b) \(531.25\) miles

- Find each rider's distance per minute first. - Use \(60\) minutes per hour to convert to miles per hour. - For part b), use \(5280\) feet per mile and \(3600\) seconds per hour.

1. Luke's speed is \(5.4\div15=0.36\) mile per minute, so \(0.36\times60=21.6\) mph. 2. Sarah's speed is \(4.5\div12=0.375\) mile per minute, so \(0.375\times60=22.5\) mph. 3. Since \(22.5>21.6\), Sarah is faster. 4. Sarah travels \(22.5\times5280=118{,}800\) feet in \(3600\) seconds. Thus \(118{,}800\div3600=33\) ft per second.

a) Sarah is faster: \(22.5\) mph compared with Luke's \(21.6\) mph. b) \(33\) ft per second

- Divide distance by time to find each speed. - Use the meaning of meters per second to find a distance over a shorter time. - Convert meters to centimeters at the end.

1. Drone Alpha: \(45.5\div3.5=13\,\text{m/s}\). 2. Drone Beta: \(61.2\div4.8=12.75\,\text{m/s}\). 3. Since \(13>12.75\), Drone Alpha is faster. 4. In \(0.01\) second, Alpha travels \(13\times0.01=0.13\,\text{m}=13\,\text{cm}\).

a) Drone Alpha is faster at \(13\,\text{m/s}\); Drone Beta travels \(12.75\,\text{m/s}\). b) \(13\,\text{cm}\)

- Find how much water each hose delivers in one minute. - Compare the two unit rates. - Divide the faster rate by the number of seconds in one minute for part b).

1. Hose A: \(25.5\div1.5=17\) gal/min. 2. Hose B: \(32.4\div1.8=18\) gal/min. 3. Hose B has the greater flow rate. 4. Since one minute has \(60\) seconds, \(18\div60=0.3\) gallon per second.

a) Hose B, at \(18\) gal/min; Hose A has \(17\) gal/min. b) \(0.3\) gallon per second

- Convert the basin volume to liters first. - Divide the total amount by the amount pumped per minute. - Convert the resulting minutes to hours.

1. Since \(1\,\text{m}^3=1000\,\text{L}\), the basin holds \(450\times1000=450{,}000\,\text{L}\). 2. The pumping time is \(450{,}000\,\text{L}\div(1500\,\text{L/min})=300\,\text{min}\). 3. Convert minutes to hours: \(300\div60=5\,\text{h}\).

a) \(450{,}000\,\text{L}\) b) \(5\,\text{h}\)

- Find the water volume at the target depth. - Convert cubic meters to liters. - Divide the total liters by the rate, then convert minutes to hours and minutes.

1. The water volume is \(10\times5\times1.6=80\,\text{m}^3=80{,}000\,\text{L}\). 2. At \(40\,\text{L}\) per minute, the filling time is \(80{,}000\div40=2000\) minutes. 3. Since \(2000=33\times60+20\), the time is \(33\) hours \(20\) minutes.

a) The pool will contain \(80{,}000\,\text{L}\). b) Filling it will take \(33\) hours \(20\) minutes.

- A larger decrease per \(1000\,\text{ft}\) leads to a lower final temperature. - How many \(1000\,\text{ft}\) intervals are in \(10{,}000\,\text{ft}\)? - Find each predicted temperature separately, then compare them.

1. Model B predicts the lower temperature because it has the greater temperature decrease for each \(1000\,\text{ft}\). 2. An altitude of \(10{,}000\,\text{ft}\) contains \(10\) intervals of \(1000\,\text{ft}\). Model A predicts \(59-10\times3.5=59-35=24^\circ\text{F}\). 3. Model B predicts \(59-10\times5.5=59-55=4^\circ\text{F}\). 4. The difference is \(24-4=20^\circ\text{F}\).

Model B predicts the lower temperature. The predictions differ by \(20^\circ\text{F}\): Model A gives \(24^\circ\text{F}\) and Model B gives \(4^\circ\text{F}\).

- Follow the order of operations. - Which number could represent the starting temperature? - Which numbers could represent elevations? - Since the hiker climbs to a higher elevation, should the temperature change be added to or subtracted from the starting temperature?

1. Find the change in altitude: \(9000-3000=6000\,\text{ft}\). 2. Find the number of \(1000\,\text{ft}\) intervals: \(6000\div1000=6\). 3. Find the total temperature decrease: \(6\times3.5=21^\circ\text{F}\). 4. Subtract from the starting temperature: \(50-21=29\). The expression equals \(29\). 5. One possible situation is a hike that starts at \(3000\,\text{ft}\) with a temperature of \(50^\circ\text{F}\) and ends at \(9000\,\text{ft}\), with temperature decreasing \(3.5^\circ\text{F}\) for every \(1000\,\text{ft}\) gained.

a) \(29\) b) Example: A hike starts at \(3000\,\text{ft}\) when the temperature is \(50^\circ\text{F}\) and ends at \(9000\,\text{ft}\). The temperature decreases \(3.5^\circ\text{F}\) for every \(1000\,\text{ft}\) of elevation gain.

- Convert the trail distance to feet first. - Find Leon's number of steps by using distance per step. - For part b), use the sister's total number of steps and the same trail distance.

1. Convert the trail length: \(2.5\times5280=13{,}200\) ft. 2. Leon's number of steps is \(13{,}200\div2.2=6000\). 3. His sister takes \(6000+2000=8000\) steps. 4. Her average step length is \(13{,}200\div8000=1.65\) ft.

a) \(6000\) steps b) \(1.65\) ft

- Find the total cost of each type of candy. - Add the costs and the weights separately. - Divide the total cost by the total weight.

1. The strawberry candy costs \(3\times\$14=\$42\). 2. The lemon candy costs \(2\times\$9=\$18\). 3. The total cost is \(\$42+\$18=\$60\), and the total weight is \(3+2=5\) pounds. 4. The average cost is \(\$60\div5=\$12\) per pound.

The mixture costs an average of \(\$12\) per pound.

- Use the rate and time to find distance. - For part b), you can scale the first distance by the speed factor. - Keep track of decimal place value in each product.

1. For a), use distance equals rate times time: \(0.14\,\text{m/min}\times7.5\,\text{min}=1.05\,\text{m}\). 2. The second snail travels \(1.3\) times as far in the same time, so \(1.05\times1.3=1.365\,\text{m}\).

a) \(1.05\,\text{m}\) b) \(1.365\,\text{m}\)

- Make sure the time and rate use compatible units. - Recall how many minutes are in one hour. - Multiply the rate per minute by the total number of minutes.

1. Convert the time to minutes: \(4.5\times60=270\) minutes. 2. Multiply the time by the leak rate: \(270\,\text{min}\times0.045\,\text{gal/min}=12.15\,\text{gal}\). 3. The faucet wastes \(12.15\) gallons.

\(12.15\) gallons of water are wasted.

- Convert the entire time to seconds first. - Determine how many pool lengths make up \(400\,\text{m}\). - Divide the total time by the number of equal lengths.

1. Convert the total time to seconds: \(4\times60+24.8=264.8\) seconds. 2. The swimmer completes \(400\div50=8\) pool lengths. 3. Divide the total time by the number of lengths: \(264.8\div8=33.1\) seconds per length.

\(33.1\) seconds per pool length

- For part a), find the amount filled in one minute. - For part b), convert the total time to seconds first. - Divide total time by total gallons to find time per gallon.

1. For a), \(112.5\div15=7.5\) gal/min. 2. Convert \(15\) minutes to \(900\) seconds. 3. For b), \(900\div112.5=8\) seconds per gallon.

a) \(7.5\) gal/min b) \(8\) seconds per gallon

- Use distance divided by time to find speed. - Convert the swift's speed so both birds are measured in the same unit. - Compare only after the units match.

1. For a), \(336 \div 12 = 28\,\text{m/s}\). 2. Convert to kilometers per hour: \(28 \times 3.6 = 100.8\,\text{km/h}\). 3. Since \(324 > 100.8\), the peregrine falcon is faster in this comparison.

a) \(28\,\text{m/s}\) b) The peregrine falcon is faster in this comparison; the swift's speed is \(100.8\,\text{km/h}\).

- Divide distance by time to find speed. - Convert the speeds to the same unit before comparing. - Check which decimal value is greater after conversion.

1. For a), \(188\div40=4.7\,\text{m/s}\). 2. Convert to kilometers per hour: \(4.7\times3.6=16.92\,\text{km/h}\). 3. Since \(16.92>16.2\), the first car is faster.

a) \(4.7\,\text{m/s}\) b) The first car is faster at \(16.92\,\text{km/h}\), compared with \(16.2\,\text{km/h}\).

- Divide total fuel by total distance to find fuel used per mile. - For part b), determine how many \(0.04\)-gallon portions fit into \(6\) gallons. - A rate table can help keep the quantities organized.

1. For a), \(12\div300=0.04\) gallon per mile. 2. For b), \(6\div0.04=150\) miles.

a) \(0.04\) gal/mi b) \(150\) miles

- Find the price of one notebook in each package. - Compare the two unit prices. - Use the lower unit price for the larger quantity.

1. Package A costs \(6.40\div 8=\$0.80\) per notebook. 2. Package B costs \(9.00\div 12=\$0.75\) per notebook. 3. Since \(0.75<0.80\), Package B is the better deal. 4. At \(\$0.75\) each, \(20\) notebooks cost \(20\times 0.75=\$15.00\).

a) Package B is cheaper at \(\$0.75\) per notebook; Package A costs \(\$0.80\) per notebook. b) \(20\) notebooks would cost \(\$15.00\).

- Find the amount of gasoline used per mile. - Find the current amount of gasoline as a percent of tank capacity. - Find the gasoline needed for the planned distance. - Compare the amount available with the amount needed.

1. The car uses \(6 \div 150 = 0.04\) gallon per mile. 2. The tank currently contains \(15 \times 0.40 = 6\) gallons. 3. The planned trip requires \(140 \times 0.04 = 5.6\) gallons. 4. Since \(5.6 \le 6\), there is enough gasoline.

Yes. The tank contains \(6\) gallons, and the trip requires \(5.6\) gallons.

- First find the price of one serving. - Does buying more servings make the total price increase or decrease? - What happens to the price when the number of servings doubles?

1. One serving costs \(13.50\div 3=\$4.50\). 2. Five servings cost \(5\times 4.50=\$22.50\). 3. The number of servings for \(\$31.50\) is \(31.50\div 4.50=7\). 4. The relationship is proportional because the unit price is constant and doubling the number of servings doubles the total price.

a) \(\$22.50\). b) \(7\) servings. c) Proportional, because the price per serving is constant.

- First find how many flyers one copier prints in one hour. - Then find the combined hourly rate of five copiers. - Divide the total number of flyers by the combined rate.

1. One copier's hourly rate is \(\frac{1200}{3\times4}=100\) flyers per hour. 2. Five copiers print \(5\times100=500\) flyers per hour. 3. The time needed is \(\frac{3000}{500}=6\) hours.

The \(5\) copiers will take \(6\) hours.

- Find the gallons per minute for one hose. - Find the combined rate of three hoses. - Use the total volume and combined rate to find the time. - Compare the calculated time with \(45\) minutes.

1. One hose has a rate of \(120\div 10=12\) gallons per minute. 2. Three hoses have a combined rate of \(3\times 12=36\) gallons per minute. 3. Filling \(1800\) gallons takes \(1800\div 36=50\) minutes. 4. Since \(50>45\), the claim is false.

No. Three hoses take \(50\) minutes, which is longer than \(45\) minutes.

- How many hours does the hour hand take to complete one revolution? - Find its angle per hour. - Convert \(15\) minutes to a fraction of an hour. - Find the total elapsed time in part c.

1. The hour hand completes a \(360^\circ\) revolution in \(12\) hours, so it moves \(360^\circ \div 12 = 30^\circ\) per hour. 2. Fifteen minutes is one-fourth of an hour, so the hand moves \(30^\circ \div 4 = 7.5^\circ\). 3. The elapsed time from \(8{:}00\) a.m. to \(11{:}30\) a.m. is \(3.5\) hours. 4. The hour hand turns \(3.5 \times 30^\circ = 105^\circ\).

a) \(30^\circ\) b) \(7.5^\circ\) c) \(105^\circ\)

- Use the relationship among distance, speed, and time. - Convert between hours and minutes carefully. - Subtract the break from the total allowed time. - Divide the distance by the available riding time.

1. At \(12\,\text{mph}\), the riding time is \(18 \div 12 = 1.5\) hours, or \(90\) minutes. 2. Including the break, the total time is \(90 + 15 = 105\) minutes, which is \(1\) hour \(45\) minutes. 3. To arrive in \(1\) hour with a \(15\)-minute break, Leah has \(45\) minutes of riding time. 4. Convert \(45\) minutes to \(0.75\) hour. 5. The required speed is \(18 \div 0.75 = 24\,\text{mph}\).

a) The trip will take \(1\) hour \(45\) minutes. b) Leah would need to ride at \(24\,\text{mph}\).

- Find the total round-trip distance. - Find the gasoline used per mile. - Multiply the number of gallons by the price per gallon.

1. The round-trip distance is \(2\times 27=54\) miles. 2. The gasoline rate is \(1.5\div 60=0.025\) gallon per mile. 3. The trip uses \(54\times 0.025=1.35\) gallons. 4. The cost is \(1.35\times 3.60=\$4.86\).

a) \(1.35\) gallons. b) \(\$4.86\).

- Divide the full-tank price by the tank capacity. - Find how many miles the scooter travels per gallon. - First find the gasoline needed for \(90\) miles, then its cost. - Compare the calculated cost with the budget.

1. The price per gallon is \(7.20\div 2=\$3.60\). 2. The scooter travels \(60\div 1.2=50\) miles per gallon. With \(2\) gallons, it can travel \(2\times 50=100\) miles. 3. The trip uses \(90\times \frac{1.2}{60}=1.8\) gallons. 4. The cost is \(1.8\times 3.60=\$6.48\). Since \(6.48<6.50\), the money is enough.

a) \(\$3.60\) per gallon. b) \(100\) miles. c) Yes. The trip costs \(\$6.48\).

- How many \(15\)-second intervals are in one minute? - Use the hose's gallons-per-minute rate. - Find the cost per gallon from the cost per \(1000\) gallons. - Convert the final time from minutes to hours.

1. One minute contains four \(15\)-second intervals, so the rate is \(4\times 0.75=3\) gallons per minute. 2. Filling \(180\) gallons takes \(180\div 3=60\) minutes, or \(1\) hour. 3. The pool costs \(\frac{180}{1000}\times 6.00=\$1.08\) to fill. 4. A cost of \(\$2.70\) represents \(\frac{2.70}{6.00}\times 1000=450\) gallons. At \(3\) gallons per minute, that takes \(450\div 3=150\) minutes, or \(2.5\) hours.

a) \(3\) gallons per minute. b) \(60\) minutes, or \(1\) hour. c) \(\$1.08\). d) \(2.5\) hours, or \(2\) hours \(30\) minutes.

- First find the difference in gallons used per minute. - Use the cost per \(1000\) gallons to find the cost of the saved water. - For part c, find the total water used during the \(7\)-minute shower first.

1. The savings per minute is \(3 - 1.75 = 1.25\) gallons. 2. In \(8\) minutes, the water saved is \(1.25 \times 8 = 10\) gallons. 3. The cost savings is \(\frac{10}{1000} \times 6.00 = \$0.06\). 4. A \(7\)-minute shower at \(3\) gallons per minute uses \(7 \times 3 = 21\) gallons. With the water-saving showerhead, the time is \(21 \div 1.75 = 12\) minutes.

a) \(10\) gallons b) \(\$0.06\) c) \(12\) minutes

- Find the volume added during the first \(10\) minutes. - Find how much space remains in the tank. - Divide the remaining volume by the filling rate and compare with \(30\) minutes.

1. In \(10\) minutes, the volume increases by \(35-10=25\) gallons, so the filling rate is \(25\div10=2.5\) gallons per minute. 2. The tank still needs \(140-35=105\) gallons. 3. At \(2.5\) gallons per minute, the remaining time is \(105\div2.5=42\) minutes. 4. The owner's estimate of \(30\) more minutes is too short. The tank needs \(42\) more minutes.

The claim is not realistic. The filling rate is \(2.5\) gallons per minute, and the tank needs another \(42\) minutes to fill.

- Express how much water each hose supplies during the same unknown number of minutes. - The aquarium is full when the amounts from the two hoses together equal its capacity. - After finding the common running time, use the first hose's rate for part c).

1. In \(x\) minutes, the first hose supplies \(12x\) liters and the second supplies \(8x\) liters. 2. Their combined amount must equal \(500\) liters, so \(12x + 8x = 500\). 3. Combine like terms: \(20x = 500\), so \(x = 25\). 4. The first hose supplies \(12 \times 25 = 300\) liters.

a) \(12x + 8x = 500\) b) \(25\) minutes c) \(300\,\text{L}\)

- Find the total percentage-point increase during the \(30\)-minute interval. - Divide the increase by the number of minutes to find the unit rate. - Use the constant rate to find the time needed for \(100\) percentage points.

1. The battery level increases by \(75\% - 15\% = 60\) percentage points. 2. The elapsed time is \(30\) minutes. 3. The charging rate is \(60 \div 30 = 2\) percentage points per minute. 4. A full charge takes \(100 \div 2 = 50\) minutes.

a) \(2\) percentage points per minute b) \(50\) minutes

- First find what one machine produces in one hour. - Find the hourly production required for the new goal. - Break the calculation into smaller rate steps. - A per-minute rate can be a useful starting point.

1. The rate per minute is \(180\div 12=15\) jars per minute. 2. The hourly rate is \(15\times 60=900\) jars per hour. 3. In \(7.5\) hours, one machine fills \(900\times 7.5=6750\) jars. 4. The target rate is \(13{,}500\div 3=4500\) jars per hour. 5. The number of machines is \(4500\div 900=5\).

a) \(900\) jars per hour. b) \(6750\) jars. c) \(5\) machines.

- Find the cost of one copy at each shop. - Use the unit price for a larger number of copies. - When money is given, divide by the price per copy.

1. Shop A costs \(6.40\div 80=\$0.08\) per copy. Shop B costs \(9.00\div 150=\$0.06\) per copy. 2. For \(400\) copies, Shop A costs \(400\times 0.08=\$32.00\), and Shop B costs \(400\times 0.06=\$24.00\). 3. For \(\$12.00\), Shop A provides \(12.00\div 0.08=150\) copies, and Shop B provides \(12.00\div 0.06=200\) copies.

a) Shop A: \(\$0.08\) per copy. Shop B: \(\$0.06\) per copy. b) Shop A: \(\$32.00\). Shop B: \(\$24.00\). c) Shop A: \(150\) copies. Shop B: \(200\) copies.

- Convert each rate to gasoline used for the same distance. - Use a per-\(100\)-mile rate for comparison. - Determine how many groups of \(100\) miles are in the trip.

1. Car A uses \((6\div 120)\times 100=5\) gallons per \(100\) miles. 2. Car B uses \((8\div 128)\times 100=6.25\) gallons per \(100\) miles. 3. Car A uses less gasoline. 4. For \(1400\) miles, Car A uses \(14\times 5=70\) gallons, and Car B uses \(14\times 6.25=87.5\) gallons.

a) Car A uses \(5\) gallons per \(100\) miles; Car B uses \(6.25\) gallons per \(100\) miles. Car A uses less. b) Car A uses \(70\) gallons; Car B uses \(87.5\) gallons.

- Interpret “faster” as more parts produced per hour. - Divide the target number of parts by each hourly rate. - Compare each time with the deadline.

1. Machine A makes \(480\div 4=120\) parts per hour. Machine B makes \(525\div 5=105\) parts per hour. Machine A is faster. 2. Machine A takes \(1050\div 120=8.75\) hours, or \(8\) hours \(45\) minutes. 3. Machine B takes \(1050\div 105=10\) hours. 4. Only Machine A meets the \(9\)-hour deadline.

a) Machine A, at \(120\) parts per hour; Machine B makes \(105\) parts per hour. b) Machine A: \(8.75\) hours. Machine B: \(10\) hours. c) Only Machine A.

- First find the number of pages printed in one minute. - Use the unit rate to scale up or work backward. - Think about technical or human interruptions during a very large job.

1. The copier's rate is \(150\div 4=37.5\) pages per minute. 2. In \(10\) minutes, it prints \(37.5\times 10=375\) pages. 3. Printing \(525\) pages takes \(525\div 37.5=14\) minutes. 4. A very large job may require paper refills, toner replacement, clearing jams, or cooling pauses, so the rate may not remain constant.

a) \(375\) pages. b) \(14\) minutes. c) Possible reasons include paper refills, toner replacement, paper jams, or overheating.

- Find the change in volume. - Divide by the elapsed number of minutes. - Convert a per-minute rate to a per-hour rate.

1. The volume increases by \(27-12=15\) gallons. 2. The elapsed time is \(12\) minutes, so the rate is \(15\div12=1.25\) gallons per minute. 3. Multiply by \(60\) minutes per hour: \(1.25\times60=75\) gallons per hour.

\(75\) gallons per hour

- Express both pumping rates in the same unit before comparing them. - How can you convert gallons per minute to gallons per hour? - What happens to the total rate when both pumps work at the same time?

1. Pump A removes \(180 \div 2=90\,\text{gal/min}\). Its hourly rate is \(90 \times 60=5400\,\text{gal/h}\). 2. Pump B is faster because \(6600\,\text{gal/h}>5400\,\text{gal/h}\). 3. Together, the pumps remove \(5400+6600=12{,}000\,\text{gal/h}\). 4. The time is \(12{,}000 \div 12{,}000=1\) hour.

a) Pump B is faster. Its rate is \(6600\,\text{gal/h}\), compared with Pump A's rate of \(5400\,\text{gal/h}\). b) Together, the pumps will remove the water in \(1\) hour.

- Use distance divided by time for each average speed. - For the whole hike, use total distance divided by total time rather than averaging the two speeds. - In part c, express the total time as the fixed outbound time plus the variable return time.

1. The speed to the shelter is \(v_1=\frac{6}{1.5}=4\,\text{mph}\). 2. The return time is \(t_2=\frac{9}{4.5}=2\) hours. 3. The total distance is \(15\) miles, and the total time is \(1.5+2=3.5\) hours. 4. The average speed for the entire hike is \(v_g=\frac{15}{3.5}\approx4.29\,\text{mph}\). 5. If the return time is \(t_2\), the total time is \(1.5+t_2\), so \(v_g=\frac{15}{1.5+t_2}\).

a) \(v_1=4\,\text{mph}\) b) \(t_2=2\) hours, and \(v_g\approx4.29\,\text{mph}\) c) \(v_g=\frac{15}{1.5+t_2}\)

- First determine how much water the pump delivers in one minute. - How many minutes are in one hour? - Divide the pool capacity by the hourly rate.

1. In one minute, the pump delivers \(125 \times 2=250\,\text{gal}\). 2. In one hour, it delivers \(250 \times 60=15{,}000\,\text{gal}\). 3. The filling time is \(60{,}000 \div 15{,}000=4\) hours. 4. Four hours after \(7{:}45\) a.m. is \(11{:}45\) a.m.

a) \(15{,}000\,\text{gal/h}\) b) \(4\) hours c) \(11{:}45\) a.m.

- Subtract the unwanted fraction from the total amount. - Convert kilograms to grams before finding the number of bottles. - Divide the total clean-glass mass by the mass of one bottle.

1. The unwanted material weighs \(\frac{2}{25}\times2500=200\,\text{kg}\). 2. The clean glass weighs \(2500-200=2300\,\text{kg}\). 3. Convert to grams: \(2300\,\text{kg}=2{,}300{,}000\,\text{g}\). 4. Divide by the mass of one bottle: \(2{,}300{,}000\div400\approx5750\).

a) \(2300\,\text{kg}\) of clean glass b) About \(5750\) bottles

- First find the area used for vegetables. - Then find the stated fraction of the vegetable area. - Divide the tomato area by the area needed for one plant.

1. The vegetable area is \(\frac{3}{8}\times480=180\,\text{m}^2\). 2. The tomato area is \(\frac{1}{5}\times180=36\,\text{m}^2\). 3. The number of plants is \(36\div0.25=144\).

a) \(180\,\text{m}^2\) b) \(144\) tomato plants

- A decrease in the distance remaining tells how far the group traveled between signs. - Compare each interval's traveled distance with its own elapsed time. - Put both interval rates in miles per hour before comparing them.

1. From Sign 1 to Sign 2, the remaining distance decreases by \(21 - 17 = 4\) miles, so the group travels \(4\) miles in \(1\) hour. The average speed is \(4\,\text{mi/h}\). 2. From Sign 2 to Sign 3, the remaining distance decreases by \(17 - 12 = 5\) miles. The average speed is \(5 \div 1.25 = 4\,\text{mi/h}\). 3. The two average speeds are equal.

a) Sign 1 to Sign 2: \(4\,\text{mi/h}\); Sign 2 to Sign 3: \(4\,\text{mi/h}\). b) Yes. The average speed is the same on both intervals.

- Decide which days belong in the denominator for each requested rate. - Both parts use the same total distance but not the same number of days. - Compare the meanings of “per riding day” and “per calendar day.”

1. The riding-day rate is \(520 \div 32 = 16.25\) miles per riding day. 2. The calendar-day rate is \(520 \div 40 = 13\) miles per calendar day. 3. The rates use different wholes for the time quantity: one counts only \(32\) riding days, while the other counts all \(40\) calendar days. The same total distance is spread across different numbers of days.

a) \(16.25\) miles per riding day b) \(13\) miles per calendar day c) The denominators count different kinds of days: \(32\) riding days versus all \(40\) calendar days.

- How many minutes are in one hour? - How many hours are in one day? - First find the hourly amount, then the daily amount.

1. Find the approximate amount leaked in one hour: \(15\,\text{L} \times 60 \approx 900\,\text{L}\). 2. Find the approximate amount leaked in \(24\) hours: \(900\,\text{L} \times 24 \approx 21{,}600\,\text{L}\).

The pipe leaks about \(21{,}600\,\text{L}\) in one day.

- Divide the total capacity by the maximum number of people. - For part b, check the number-of-people limit and the mass limit separately. - Both conditions must be satisfied.

1. Find the average capacity per person: \(1200\,\text{kg} \div 15 = 80\,\text{kg}\) per person. 2. For part b, the group has \(12\) people, and \(12 < 15\). 3. Their combined mass is \(980\,\text{kg}\), and \(980\,\text{kg} < 1200\,\text{kg}\). 4. The group meets both limits, so all \(12\) workers may ride together.

a) The average capacity is \(80\,\text{kg}\) per person. b) Yes. Both the number of people and their combined mass are below the elevator's limits.

- How many groups of \(8\) gallons fit in \(400\) gallons? - Multiply the number of groups by the distance traveled for each group.

1. Find how many \(8\)-gallon portions are in the tank: \(400 \div 8 = 50\). 2. Each portion covers \(100\) miles, so the total distance is \(50 \times 100\,\text{mi} = 5000\,\text{mi}\).

The truck can travel \(5000\,\text{mi}\) on a full tank.

- Find the total number of nuts first. - Determine how many groups of \(100\) nuts are in the total. - Use the sample rate, then convert grams to kilograms.

1. Estimate the total number of nuts: \(20 \times 1500 \approx 30{,}000\). 2. Estimate the number of groups of \(100\) nuts: \(30{,}000 \div 100 \approx 300\). 3. Estimate the total mass: \(300 \times 240\,\text{g} \approx 72{,}000\,\text{g}\). 4. Convert to kilograms: \(72{,}000\,\text{g} = 72\,\text{kg}\), so the estimated mass is about \(72\,\text{kg}\).

The bushes produce about \(72\,\text{kg}\) of hazelnuts.

- Divide the container's volume by the daily amount per child. - Then divide the total number of children by the number supplied by one container. - Use approximation language because the daily amount is an average.

1. Estimate the number of children supplied by one container: \(24\,\text{L} \div 3\,\text{L} \approx 8\) children. 2. Estimate the number of containers for \(120\) children: \(120 \div 8 \approx 15\) containers.

a) One container supplies about \(8\) children for one day. b) The camp needs about \(15\) containers per day.

- Determine how many \(15\)-kilogram groups are in \(150\,\text{kg}\). - Use the crates-per-tree rate for all \(200\) trees. - Check the multiplication by working backward.

1. Estimate the number of crates per tree: \(150\,\text{kg} \div 15\,\text{kg} \approx 10\). 2. Estimate the total for \(200\) trees: \(10 \times 200 \approx 2000\) crates.

a) One tree fills about \(10\) crates. b) The orchard fills about \(2000\) crates.

- Divide the modeled daily water amount by the amount in one modeled trunkful. - Multiply one adult's daily amount by \(5\). - Multiply the newborn's daily amount by \(7\). - Use approximation language throughout.

1. Estimate the number of modeled trunkfuls: \(80\,\text{L} \div 16\,\text{L} \approx 5\). 2. Estimate the herd's daily water: \(80\,\text{L} \times 5 \approx 400\,\text{L}\). 3. Estimate the newborn's weekly milk: \(10\,\text{L} \times 7 \approx 70\,\text{L}\).

a) About \(5\) modeled trunkfuls b) About \(400\,\text{L}\) c) About \(70\,\text{L}\)

- Use the elephant’s modeled hourly rate to find the time for \(20\) miles. - For part c), first find each distance traveled in \(4\) hours. - Subtract the two distances to find how much farther one traveled.

1. For a), \(20\,\text{mi} \div 5\,\text{mi/h} = 4\,\text{h}\). 2. For b), \(4\,\text{h} \times 4\,\text{mi/h} = 16\,\text{mi}\). 3. In \(4\) hours, the elephant travels \(4\,\text{h} \times 5\,\text{mi/h} = 20\,\text{mi}\). 4. The difference is \(20\,\text{mi} - 16\,\text{mi} = 4\,\text{mi}\).

a) \(4\) hours b) \(16\) miles c) \(4\) miles farther

- Divide the modeled daily distance by the walking rate. - Add the time spent walking and eating. - Subtract that total from the \(24\) hours in a day.

1. Find the walking time: \(18\,\text{mi} \div 6\,\text{mi/h} = 3\,\text{h}\). 2. Walking and eating take \(3\,\text{h} + 18\,\text{h} = 21\,\text{h}\). 3. The remaining time is \(24\,\text{h} - 21\,\text{h} = 3\,\text{h}\).

a) The elephant walks for \(3\) hours. b) It has \(3\) hours left for other activities.

- Multiply the savings per minute by the daily shower time. - Use \(7\) days for one week. - Compare the total gallons saved with groups of \(1000\) gallons.

1. The daily savings are \(1\,\text{gal/min} \times 30\,\text{min} = 30\,\text{gal}\). 2. The weekly savings are \(30\,\text{gal} \times 7 = 210\,\text{gal}\). 3. In \(10\) weeks, the family saves \(210\,\text{gal} \times 10 = 2100\,\text{gal}\). 4. At \(\$4.00\) per \(1000\) gallons, \(2100\) gallons cost \(2.1 \times \$4.00 = \$8.40\).

a) \(30\) gallons b) \(210\) gallons c) \(\$8.40\)

- Find the water saved with one flush. - Scale the per-flush savings to one person per day. - Then scale by the number of people and days. - Use approximation language because the daily number of flushes is an average.

1. Find the savings per flush: \(12\,\text{L} - 5\,\text{L} = 7\,\text{L}\). 2. Estimate one person's daily savings: \(6 \times 7\,\text{L} \approx 42\,\text{L}\). 3. Estimate the family's daily savings: \(4 \times 42\,\text{L} \approx 168\,\text{L}\). 4. Estimate the weekly savings: \(7 \times 168\,\text{L} \approx 1176\,\text{L}\). 5. Estimate the yearly savings: \(365 \times 168\,\text{L} \approx 61{,}320\,\text{L}\).

a) About \(42\,\text{L}\) b) About \(1176\,\text{L}\) c) About \(61{,}320\,\text{L}\)

- Convert \(\$2.00\) to cents first. - Determine the cost of one tenth of \(1000\) gallons. - Multiply the daily use by \(30\). - Break the monthly amount into convenient groups of gallons.

1. Since \(\$2.00 = 200\) cents, \(100\) gallons cost \(200 \div 10 = 20\) cents. 2. The monthly use is \(80\,\text{gal} \times 30 = 2400\,\text{gal}\). 3. Two thousand gallons cost \(\$4.00\), and \(400\) gallons cost \(4 \times \$0.20 = \$0.80\). 4. The total cost is \(\$4.00 + \$0.80 = \$4.80\).

a) \(20\) cents b) \(2400\) gallons c) \(\$4.80\)

- Find the difference between the two water-use amounts. - Remember that Tim brushes in the morning and evening. - Scale the daily savings to \(100\) days. - Use the price per \(1000\) gallons to find the monetary savings.

1. The savings per brushing are \(15\,\text{gal} - 1\,\text{gal} = 14\,\text{gal}\). 2. The daily savings are \(14\,\text{gal} \times 2 = 28\,\text{gal}\). 3. In \(100\) days, Tim saves \(28\,\text{gal} \times 100 = 2800\,\text{gal}\). 4. At about \(\$3.00\) per \(1000\) gallons, the savings are \(2.8 \times \$3.00 \approx \$8.40\).

a) \(28\) gallons b) \(2800\) gallons c) About \(\$8.40\)

- Find the cost of \(20\) gallons from the cost of \(40\) gallons. - Compare the costs of a bath and a shower. - How many baths are replaced during \(52\) weeks?

1. Twenty gallons costs half of \(\$0.20\), or \(\$0.10\). Therefore, about \(60\) gallons costs \(\$0.20 + \$0.10 \approx \$0.30\). 2. The savings per shower are approximately \(\$0.30 - \$0.10 \approx \$0.20\). 3. The weekly savings are approximately \(2 \times \$0.20 \approx \$0.40\). 4. The yearly savings are approximately \(52 \times \$0.40 \approx \$20.80\).

a) About \(\$0.30\) b) About \(\$0.20\) c) About \(\$20.80\)

- Find the total water used by each method. - Use the cost per \(1000\) gallons to find the sprinkler’s cost. - Find the daily difference, then scale it to \(30\) days.

1. Watering cans use \(12 \times 2\,\text{gal} = 24\,\text{gal}\). The sprinkler uses \(3\,\text{gal/min} \times 10\,\text{min} = 30\,\text{gal}\). 2. At \(\$2.00\) per \(1000\) gallons, \(30\) gallons costs \(\frac{30}{1000} \times \$2.00 = \$0.06\). 3. The watering cans save \(30\,\text{gal} - 24\,\text{gal} = 6\,\text{gal}\) each day. 4. In \(30\) days, the savings are \(6\,\text{gal} \times 30 = 180\,\text{gal}\). 5. The monetary savings are \(\frac{180}{1000} \times \$2.00 = \$0.36\).

a) Watering cans: \(24\) gallons; sprinkler: \(30\) gallons b) \(\$0.06\) c) \(\$0.36\)

- Convert \(\$4.00\) to cents. - Determine how many \(10\)-gallon groups fit in \(1000\) gallons and in \(200\) gallons. - Find the total volume and cost of \(25\) carts. - Compare either the water amounts or their costs.

1. Since \(\$4.00 = 400\) cents, \(10\) gallons costs \(400 \div 100 = 4\) cents. 2. The pond uses \(200 \div 10 = 20\) groups of \(10\) gallons, so it costs \(20 \times 4 = 80\) cents. 3. Twenty-five carts hold \(25 \times 10\,\text{gal} = 250\,\text{gal}\) and cost \(25 \times 4 = 100\) cents. This is more than the \(80\)-cent pond filling.

a) \(4\) cents b) \(80\) cents c) Twenty-five carts would cost more because they use \(250\) gallons and cost \(100\) cents.

- Find the total water used in \(100\) flushes. - Compare that amount with groups of \(100\) gallons. - You can first find the water saved by one flush. - Convert cents to dollars at the end.

1. The older toilet uses \(100 \times 3\,\text{gal} = 300\,\text{gal}\). Since \(300\) gallons is three groups of \(100\) gallons, the cost is \(3 \times 30 = 90\) cents, or \(\$0.90\). 2. The newer toilet saves \(3\,\text{gal} - 1.5\,\text{gal} = 1.5\,\text{gal}\) per flush. 3. Over \(100\) flushes, it saves \(150\) gallons. At \(30\) cents per \(100\) gallons, the savings are \(1.5 \times 30 = 45\) cents, or \(\$0.45\).

a) \(\$0.90\) b) \(\$0.45\)

- Find the cost of a convenient amount such as \(10\) or \(100\) gallons. - Use that smaller unit rate for \(60\) and \(150\) gallons. - For part c), first find the total gallons used.

1. Since \(\$2.00 = 200\) cents, \(10\) gallons costs \(2\) cents. 2. A \(60\)-gallon cleaning costs \(6 \times 2 = 12\) cents. 3. A \(150\)-gallon wash costs \(15 \times 2 = 30\) cents. 4. Ten automatic washes use \(10 \times 150\,\text{gal} = 1500\,\text{gal}\), which costs \(\$3.00\).

a) \(12\) cents b) \(30\) cents c) \(\$3.00\)

- How many minutes are in one hour? - Use the minutes per mile to find how many miles fit in one hour. - Multiply the time for one mile by \(10\). - Convert the final minutes to hours.

1. Group A walks \(60 \div 12 = 5\) miles in one hour. 2. Group B walks \(60 \div 10 = 6\) miles in one hour. 3. Group A needs \(10 \times 12\,\text{min} = 120\,\text{min}\). 4. Since \(120\,\text{min} = 2\,\text{hr}\ 0\,\text{min}\), the time is \(2\) hours \(0\) minutes.

a) \(5\) miles b) \(6\) miles c) \(2\) hours \(0\) minutes

- How many seconds are in one minute? - If the express train travels \(1\) mile in \(30\) seconds, how far does it go in \(60\) seconds? - Use the result from part b) for \(60\) minutes.

1. The regional train travels \(60 \div 2 = 30\) miles in one hour. 2. Since one minute is \(60\) seconds, the express train travels \(60 \div 30 = 2\) miles in one minute. 3. In one hour, the express train travels \(60 \times 2 = 120\) miles.

a) \(30\) miles b) \(2\) miles c) \(120\) miles

- What fraction of \(100\) cards is \(25\) cards? - Use that fraction to find the weight of \(25\) cards. - Find the difference between the two modeled water-use rates per pound.

1. Since \(25\) cards is one fourth of \(100\) cards, \(25\) cards weigh \(4\,\text{lb} \div 4 = 1\,\text{lb}\). 2. The modeled water savings per pound are \(40\,\text{gal} - 10\,\text{gal} = 30\,\text{gal}\). 3. Therefore, making \(25\) cards with the recycled-paper process saves \(30\) gallons in this model.

\(30\) gallons of water

- First find the total number of minutes from the word total and the words-per-minute rate. - How many minutes are in one hour? - How many hours are in one day? - Use remainders to keep track of leftover minutes and hours.

1. Find the total number of minutes: \(1{,}000{,}000 \div 125 = 8000\,\text{min}\). 2. Convert minutes to hours: \(8000 \div 60 = 133\) remainder \(20\), so the time is \(133\,\text{hr}\) and \(20\,\text{min}\). 3. Convert hours to days: \(133 \div 24 = 5\) remainder \(13\), so the time is \(5\) days, \(13\) hours, and \(20\) minutes.

The narrator needs \(5\) days, \(13\) hours, and \(20\) minutes of reading time.

- Add the two running rates to find how quickly the gap closes. - Use the closing rate for \(10\) seconds. - Subtract the closed distance from the original distance. - Divide the original distance by the closing rate to find the meeting time.

1. Their combined closing rate is \(2\,\text{ft/s} + 4\,\text{ft/s} = 6\,\text{ft/s}\). 2. In \(10\) seconds, the distance decreases by \(10\,\text{s} \times 6\,\text{ft/s} = 60\,\text{ft}\). 3. Their remaining distance is \(120\,\text{ft} - 60\,\text{ft} = 60\,\text{ft}\). 4. They meet after \(120\,\text{ft} \div 6\,\text{ft/s} = 20\,\text{s}\).

a) \(6\) feet per second b) \(60\) feet c) \(20\) seconds

- Find Lucas’s speed on the trip to the lake. - Subtract \(4\) miles per hour to find his return speed. - Divide the distance by the return speed. - Convert the difference in hours to minutes.

1. His speed to the lake is \(24\,\text{mi} \div 2\,\text{h} = 12\,\text{mi/h}\). 2. His return speed is \(12\,\text{mi/h} - 4\,\text{mi/h} = 8\,\text{mi/h}\). 3. The return time is \(24\,\text{mi} \div 8\,\text{mi/h} = 3\,\text{h}\). 4. The return trip is \(3\,\text{h} - 2\,\text{h} = 1\,\text{h}\), or \(60\) minutes, longer.

The return trip takes \(3\) hours and is \(60\) minutes longer.

- Find how many feet the painter completes in one hour. - Use that rate to find Tuesday's distance. - Add the amounts from Monday and Tuesday.

1. Find the unit rate: \(35 \div 5 = 7\) feet per hour. 2. On Tuesday, the painter completes \(3 \times 7 = 21\) feet. 3. Altogether, the painter completes \(35 + 21 = 56\) feet.

The painter completes \(56\,\text{ft}\) of fence during both days altogether.

- First find how much all four children pick in one hour. - Divide that hourly amount equally among the four children. - Use two division steps.

1. The four children pick \(24\,\text{lb} \div 3\,\text{h} = 8\,\text{lb/h}\) together. 2. Divide that rate among four children: \(8\,\text{lb/h} \div 4 = 2\,\text{lb/h}\) per child.

One child picks an average of \(2\,\text{lb}\) per hour.

- First find the rate for one machine in one hour. - Use that unit rate to scale to a different number of machines. - Then scale the hourly amount to the required number of hours.

1. Find the number of posters all \(5\) machines make in one hour: \(600 \div 4 = 150\) posters per hour. 2. Find the rate for one machine: \(150 \div 5 = 30\) posters per hour. 3. Find the hourly rate for \(3\) machines: \(30 \times 3 = 90\) posters per hour. 4. Find the total for \(7\) hours: \(90 \times 7 = 630\) posters.

a) One machine makes \(30\) posters per hour. b) Three machines make \(630\) posters in \(7\) hours.

- Find the total water in the four cans. - Use the \(5\)-minute amount to find the hose’s flow rate. - Apply the same rate for \(15\) minutes.

1. In \(5\) minutes, the hose delivers \(4 \times 2\,\text{gal} = 8\,\text{gal}\). 2. The flow rate is \(8\,\text{gal} \div 5\,\text{min} = 1.6\,\text{gal/min}\). 3. In \(15\) minutes, the hose delivers \(1.6\,\text{gal/min} \times 15\,\text{min} = 24\,\text{gal}\).

The barrel holds \(24\) gallons.

- Find the price per pound first. - Determine how many pounds the second family receives. - Multiply each family’s weight by the unit price.

1. Find the price per pound: \(\$28 \div 14\,\text{lb} = \$2\) per pound. 2. The Miller family pays \(6\,\text{lb} \times \$2/\text{lb} = \$12\). 3. The Schmidt family receives \(14\,\text{lb} - 6\,\text{lb} = 8\,\text{lb}\). 4. The Schmidt family pays \(8\,\text{lb} \times \$2/\text{lb} = \$16\).

The Miller family pays \(\$12\), and the Schmidt family pays \(\$16\).

- First find the price of \(1\,\text{lb}\) of oranges. - How much of Leon’s total was for bananas? - Use that amount to find the price of \(1\,\text{lb}\) of bananas. - Then find and add Sarah’s two costs.

1. Find the price per pound of oranges: \(\$6.00 \div 3 = \$2.00\) per pound. 2. Find what Leon paid for the bananas: \(\$11.00 - \$6.00 = \$5.00\). 3. Find the price per pound of bananas: \(\$5.00 \div 2 = \$2.50\) per pound. 4. Find Sarah’s cost for oranges: \(1 \times \$2.00 = \$2.00\). 5. Find Sarah’s cost for bananas: \(3 \times \$2.50 = \$7.50\). 6. Add Sarah’s costs: \(\$2.00 + \$7.50 = \$9.50\).

Sarah pays \(\$9.50\).

- How many pounds are sorted before the machine begins helping? - How many pounds remain after that first period? - At the new hourly rate, how long will the remaining work take? - Add the time before and after the machine begins helping.

1. Find how many pounds are sorted during the first \(4\) hours: \(4 \times 45 = 180\,\text{lb}\). 2. Find the amount still to be sorted: \(450 - 180 = 270\,\text{lb}\). 3. Find the time needed after the machine begins helping: \(270 \div 90 = 3\) hours. 4. Add the two time periods: \(4 + 3 = 7\) hours.

The entire job takes \(7\) hours.

- One method can begin by finding the total number of hours worked. - Another method can begin by finding the number of pounds harvested each day. - For each method, use the information step by step to find an hourly rate.

1. Method 1: Find the total work time: \(5 \times 6 = 30\) hours. Then find the hourly rate: \(450 \div 30 = 15\,\text{lb}\) per hour. 2. Method 2: Find the amount harvested each day: \(450 \div 5 = 90\,\text{lb}\) per day. Then find the hourly rate: \(90 \div 6 = 15\,\text{lb}\) per hour.

The grower harvests an average of \(15\,\text{lb}\) of apples per hour.

- Find the total operating hours for each machine. - Divide the total number of posters by the total number of hours. - Subtract the smaller hourly rate from the larger hourly rate.

1. The older machine runs for \(5 \times 8 = 40\) hours. Its rate is \(4000 \div 40 = 100\) posters per hour. 2. The newer machine runs for \(4 \times 8 = 32\) hours. Its rate is \(4000 \div 32 = 125\) posters per hour. 3. Compare the rates: \(125 - 100 = 25\) posters per hour. The newer machine is faster.

The older machine prints \(100\) posters per hour, and the newer machine prints \(125\) posters per hour. The newer machine is faster by \(25\) posters per hour.

- First find the rate for one minute. - Increase that rate by \(20\) bottles per minute. - Multiply the new rate by \(8\) minutes.

1. Find the first machine’s unit rate: \(240 \div 4 = 60\) bottles per minute. 2. Find the newer machine’s rate: \(60 + 20 = 80\) bottles per minute. 3. Find the number filled in \(8\) minutes: \(80 \times 8 = 640\) bottles.

a) The first machine fills \(60\) bottles per minute. b) The newer machine fills \(640\) bottles in \(8\) minutes.

- Find how far each runner travels in \(1\) minute. - Compare the two unit rates. - Use the faster runner’s rate to find the distance in \(10\) minutes.

1. Find Jonas’s rate: \(800 \div 4 = 200\,\text{m}\) per minute. 2. Find Sarah’s rate: \(500 \div 2 = 250\,\text{m}\) per minute. 3. Since \(250 > 200\), Sarah runs faster. 4. Find Sarah’s distance in \(10\) minutes: \(250 \times 10 = 2500\,\text{m}\).

Sarah runs faster. At the same rate, she would run \(2500\,\text{m}\) in \(10\) minutes.

- Compare each new amount or time with the original pair. - Use the same scale factor for time and number of bottles. - Half an hour is \(30\) minutes.

1. For \(60\) minutes, the scale factor is \(60 \div 20 = 3\). Therefore, \(12 \times 3 = 36\) bottles. 2. Six bottles is half of \(12\), so the time is half of \(20\) minutes: \(20 \div 2 = 10\) minutes. 3. Thirty minutes is \(1.5\) times \(20\) minutes, so \(12 \times 1.5 = 18\) bottles.

a) \(36\) bottles b) \(10\) minutes c) \(18\) bottles

- Compare the robots over the same amount of time. - Find how many cars each robot makes in \(60\) minutes. - Add their hourly amounts for part b).

1. Find Robot A’s hourly rate: \(60 \div 30 = 2\), so \(15 \times 2 = 30\) cars per hour. 2. Find Robot B’s hourly rate: \(60 \div 15 = 4\), so \(10 \times 4 = 40\) cars per hour. 3. Since \(40 > 30\), Robot B works faster. 4. Add the hourly rates: \(30 + 40 = 70\) cars per hour.

a) Robot B works faster. Robot A makes \(30\) cars per hour, and Robot B makes \(40\) cars per hour. b) Together, the robots make \(70\) cars in one hour.

- Compare each new apple amount with \(5\,\text{lb}\). - If the amount of apples is multiplied by a number, the amount of cider is multiplied by the same number. - A ratio table with apples in one column and cider in the other may help. - For a target cider amount, determine the scale factor from \(3\,\text{qt}\).

1. For \(20\,\text{lb}\) of apples, the scale factor is \(20 \div 5 = 4\). The cider amount is \(4 \times 3 = 12\,\text{qt}\). 2. For \(15\,\text{qt}\) of cider, the scale factor is \(15 \div 3 = 5\). The apple amount is \(5 \times 5 = 25\,\text{lb}\). 3. For \(9\,\text{qt}\) of cider, the scale factor is \(9 \div 3 = 3\). The apple amount is \(3 \times 5 = 15\,\text{lb}\).

a) \(12\,\text{qt}\) of cider b) \(25\,\text{lb}\) of apples c) \(15\,\text{lb}\) of apples

- Find the amount of ink used for one poster. - Multiply the unit rate to find the ink for \(200\) posters. - Divide each available ink amount by the unit rate.

1. Find the ink used per poster: \(120\,\text{mL} \div 40 = 3\,\text{mL}\) per poster. 2. For part a, multiply: \(200 \times 3\,\text{mL} = 600\,\text{mL}\). 3. For part b, divide: \(300\,\text{mL} \div 3\,\text{mL} = 100\) posters. 4. For part c, divide: \(30\,\text{mL} \div 3\,\text{mL} = 10\) posters.

a) \(600\,\text{mL}\) b) \(100\) posters c) \(10\) posters

- First find how many miles Paul rides in one hour. - How many times does that hourly distance fit into the morning distance? - How many miles remain for the afternoon? - Check that the two riding times add to the total time.

1. Find Paul’s speed: \(72 \div 6 = 12\,\text{mi/h}\). 2. Find the morning riding time: \(24 \div 12 = 2\) hours. 3. Find the afternoon distance: \(72 - 24 = 48\,\text{mi}\). 4. Find the afternoon riding time: \(48 \div 12 = 4\) hours.

Paul rides for \(2\) hours in the morning and \(4\) hours in the afternoon.

- Find the price of \(1\,\text{lb}\) of each fruit. - Determine how many bags of each fruit make \(12\,\text{lb}\). - Find the two total costs, and then compare them.

1. Find the price per pound of apples: \(\$4 \div 2 = \$2\) per pound. 2. Find the cost of \(12\,\text{lb}\) of apples: \(12 \times \$2 = \$24\). 3. Find the price per pound of pears: \(\$9 \div 3 = \$3\) per pound. 4. Find the cost of \(12\,\text{lb}\) of pears: \(12 \times \$3 = \$36\). 5. Find the difference: \(\$36 - \$24 = \$12\).

The pears will cost \(\$12\) more than the apples.

- Find how many chairs Group A finishes during the first phase. - Subtract to find how many chairs remain. - Divide the remaining chairs by the last \(3\) hours. - Compare that rate with Group B’s rate.

1. Find the chairs Group A sets up in the first \(2\) hours: \(2 \times 45 = 90\) chairs. 2. Find the remaining chairs: \(240 - 90 = 150\) chairs. 3. Find Group A’s rate during the last \(3\) hours: \(150 \div 3 = 50\) chairs per hour. 4. Compare the rates: \(50 - 48 = 2\) chairs per hour. Group A works faster during the second phase.

a) Group A must set up \(50\) chairs per hour during the last \(3\) hours. b) Group A works faster by \(2\) chairs per hour.

- Find the difference between the two elevations. - Determine how many \(200\)-meter intervals are in that difference. - Subtract the total temperature decrease from the starting temperature.

1. Find the elevation gain: \(2250\,\text{m} - 650\,\text{m} = 1600\,\text{m}\). 2. Find the number of \(200\)-meter intervals: \(1600 \div 200 = 8\). 3. The temperature decreases by \(8\,^{\circ}\text{C}\). 4. Subtract the decrease from the starting temperature: \(2 - 8 = -6\). The model predicts \(-6\,^{\circ}\text{C}\).

The model predicts a temperature of \(-6\,^{\circ}\text{C}\).

- Find each rider’s mileage charge separately. - Add the base fare to each mileage charge. - Compare the two total fares.

1. Find Ms. Maier’s mileage charge: \(12 \times \$2 = \$24\). 2. Add the base fare: \(\$24 + \$4.50 = \$28.50\). 3. Find Mr. Schmidt’s mileage charge: \(10 \times \$3 = \$30\). 4. Add the base fare: \(\$30 + \$4.50 = \$34.50\). 5. Since \(\$34.50 > \$28.50\), Mr. Schmidt pays more.

Ms. Maier pays \(\$28.50\), and Mr. Schmidt pays \(\$34.50\). Mr. Schmidt pays more.

- Find the number of miles the hiker walks in one hour. - Use the unit rate to find the total time for \(20\) miles. - Subtract the \(2\) hours already completed.

1. Find the unit rate: \(8 \div 2 = 4\) miles per hour. 2. Find the total time for \(20\) miles: \(20 \div 4 = 5\) hours. 3. Subtract the time already walked: \(5 - 2 = 3\) hours.

The hiker must walk for \(3\) more hours.

- First find the total number of hours the boat travels. - Use the total distance and total time to find the distance traveled in one hour. - Break the problem into two calculations.

1. Find the total travel time: \(6 \times 8 = 48\) hours. 2. Find the average distance per hour: \(864 \div 48 = 18\,\text{mi/h}\).

The boat travels an average of \(18\,\text{mi}\) per hour.

- Find the driver’s total number of hours for each week. - Find the average miles per hour for each week. - Compare the two unit rates.

1. For the first week, find the total driving time: \(4 \times 8 = 32\) hours. Then find the average speed: \(960 \div 32 = 30\,\text{mi/h}\). 2. For the second week, find the total driving time: \(5 \times 6 = 30\) hours. Then find the average speed: \(990 \div 30 = 33\,\text{mi/h}\). 3. Since \(33 > 30\), his average speed is greater during the second week.

His average speed is greater during the second week: \(33\,\text{mi/h}\) compared with \(30\,\text{mi/h}\).

- Find the unit weight of one tomato crate. - Use the total-weight comparison to find the pepper crates' combined weight. - Find one pepper crate's weight and compare the two unit weights.

1. Find the weight of one tomato crate: \(60\,\text{lb} \div 5 = 12\,\text{lb}\). 2. Find the combined weight of the pepper crates: \(60\,\text{lb} + 12\,\text{lb} = 72\,\text{lb}\). 3. Find the weight of one pepper crate: \(72\,\text{lb} \div 8 = 9\,\text{lb}\). 4. Find the difference: \(12\,\text{lb} - 9\,\text{lb} = 3\,\text{lb}\).

One tomato crate is \(3\,\text{lb}\) heavier than one pepper crate.

- Multiply the rate by the number of hours. - For part b), find the distance already traveled. - Add the distance traveled and the distance remaining.

1. In \(4\) hours, the ferry travels \(12 \times 4 = 48\) miles. 2. In \(3\) hours, it travels \(12 \times 3 = 36\) miles. 3. The entire route is \(36 + 24 = 60\) miles.

a) \(48\,\text{mi}\) b) \(60\,\text{mi}\)

- Add the two rates for part a). - Multiply the red machine's rate by \(10\) for part b). - For part c), compare the rates and use the \(10\)-minute time.

1. Together in one minute: \(5 + 9 = 14\) bottles. 2. The red machine in \(10\) minutes: \(9 \times 10 = 90\) bottles. 3. The rate difference is \(9 - 5 = 4\) bottles per minute, so after \(10\) minutes the difference is \(4 \times 10 = 40\) bottles.

a) \(14\) bottles b) \(90\) bottles c) \(40\) more bottles

- Find the distance each rider travels. - Use the relationship between rate, time, and distance. - Subtract the two distances to find the difference.

1. Find Mia’s distance: \(3 \times 12 = 36\,\text{mi}\). 2. Find Jonas’s distance: \(4 \times 8 = 32\,\text{mi}\). 3. Since \(36 > 32\), Mia travels farther. 4. Find the difference: \(36 - 32 = 4\,\text{mi}\).

Mia travels farther. She rides \(4\,\text{mi}\) farther than Jonas.

- How many miles does the truck travel during the first hour? - Subtract that distance from the total distance. - Use the remaining distance and time to find the average speed.

1. Find the distance traveled during the first hour: \(1 \times 70 = 70\,\text{mi}\). 2. Find the remaining distance: \(180 - 70 = 110\,\text{mi}\). 3. Find the average speed during the second part: \(110 \div 2 = 55\,\text{mi/h}\).

The truck’s average speed during the second part is \(55\,\text{mi/h}\).

- Restate what each part asks you to compare. - Multiply each step length by the number of steps. - A table can help match steps with distances. - For part b), determine how many of Mia's steps fit the father's distance.

1. Mia travels \(10 \times 24\,\text{in} = 240\,\text{in}\). 2. Her father travels \(10 \times 36\,\text{in} = 360\,\text{in}\). 3. Her father travels \(2 \times 36\,\text{in} = 72\,\text{in}\) in \(2\) steps. 4. Mia needs \(72 \div 24 = 3\) steps to travel the same distance.

a) Mia travels \(240\,\text{in}\), and her father travels \(360\,\text{in}\). b) Mia must take \(3\) steps.

- How many hours does Lukas ride in all? - How far does he ride in one hour? - Another method is to find each day’s distance separately.

1. Find the total riding time: \(4 + 5 = 9\) hours. 2. Find the total distance: \(14 \times 9 = 126\,\text{mi}\). Alternative method: 1. Find the first-day distance: \(14 \times 4 = 56\,\text{mi}\). 2. Find the second-day distance: \(14 \times 5 = 70\,\text{mi}\). 3. Add the distances: \(56 + 70 = 126\,\text{mi}\).

Lukas rides \(126\,\text{mi}\) in all.

- How far does each vehicle travel in \(4\) hours? - Which vehicle is farther ahead? - How much does the motorcycle gain each hour? - You can use the difference in their speeds to find the gap directly.

1. Find the car’s distance: \(55 \times 4 = 220\,\text{mi}\). 2. Find the motorcycle’s distance: \(62 \times 4 = 248\,\text{mi}\). 3. Find the difference: \(248 - 220 = 28\,\text{mi}\). Alternative method: 1. Find the difference in their speeds: \(62 - 55 = 7\,\text{mi/h}\). 2. Find how much the gap grows in \(4\) hours: \(7 \times 4 = 28\,\text{mi}\).

After \(4\) hours, the vehicles are \(28\,\text{mi}\) apart.

- Add the two hourly rates first. - Multiply the combined rate by \(7\) hours. - Compare the result with the order amount. - Subtract to find the shortage or extra amount.

1. Find the combined hourly rate: \(320 + 280 = 600\) bottles per hour. 2. Find the output for \(7\) hours: \(600 \times 7 = 4200\) bottles. 3. Compare with the order: \(4500 - 4200 = 300\). The shift does not produce enough bottles.

a) The lines fill \(4200\) bottles during the shift. b) One shift is not enough. The factory is \(300\) bottles short.

- First find how much water each pump removes in one minute. - When both pumps run at the same time, which operation combines their rates? - Check each division carefully.

1. Find Pump 1’s rate: \(2538 \div 9 = 282\,\text{gal}\) per minute. 2. Find Pump 2’s rate: \(1734 \div 6 = 289\,\text{gal}\) per minute. 3. Add the rates: \(282 + 289 = 571\,\text{gal}\) per minute.

Together, the pumps remove \(571\,\text{gal}\) per minute.

- Find the amount picked in one day for each field. - Once you have both daily averages, subtract to find how much greater one is. - Check each division by \(6\) carefully.

1. Find the daily average for Field A: \(1350 \div 6 = 225\,\text{lb}\) per day. 2. Find the daily average for Field B: \(2088 \div 6 = 348\,\text{lb}\) per day. 3. Find the difference: \(348 - 225 = 123\,\text{lb}\) per day.

Field A averages \(225\,\text{lb}\) per day, and Field B averages \(348\,\text{lb}\) per day. The difference is \(123\,\text{lb}\) per day.

- Use the constant drops-per-minute rate to scale to an hour and then a day. - For part c, determine how many groups of \(20\) drops are in the daily total. - Keep track of the units at each step.

1. a) In one hour, the faucet loses about \(12\times60=720\) drops. 2. b) In one day, it loses about \(720\times24=17{,}280\) drops. 3. c) Convert drops to milliliters: \(17{,}280\div20=864\). The faucet loses about \(864\,\text{mL}\) per day.

a) About \(720\) drops b) About \(17{,}280\) drops c) About \(864\,\text{mL}\)

- First find the number of pages read each day. - Scale the daily amount to one week and then to \(52\) weeks. - Divide the \(52\)-week page total by the number of pages per book.

1. Lucas reads \(20\times2=40\) pages per day. 2. a) In one week, he reads \(40\times7=280\) pages. 3. In \(52\) weeks, he reads \(280\times52=14{,}560\) pages. 4. b) The number of complete books is \(14{,}560\div160=91\).

a) \(280\) pages b) \(91\) complete books

- Find the total distance of the first trip. - If the same distance is covered in fewer days, should the daily distance be greater or less? - Divide the total distance by the number of return-trip days.

1. Find the total distance: \(6 \times 15 = 90\,\text{mi}\). 2. Divide the same distance among \(5\) days: \(90 \div 5 = 18\,\text{mi}\) per day.

The group must hike an average of \(18\,\text{mi}\) per day on the return trip.

- First find the tank’s total capacity. - Use the stronger pump’s rate to find its filling time. - Subtract the two times to find how much shorter the wait is.

1. Find the tank’s capacity: \(6 \times 120 = 720\,\text{gal}\). 2. Find the time needed by the stronger pump: \(720 \div 180 = 4\) hours. 3. Find the time saved: \(6 - 4 = 2\) hours.

The stronger pump takes \(4\) hours and reduces the wait by \(2\) hours.

- How far does the truck carry the package? - How many miles remain for the plane? - Use the plane’s distance and time to find its average speed. - Subtract the two speeds.

1. Find the distance traveled by truck: \(6 \times 60 = 360\,\text{mi}\). 2. Find the remaining distance: \(600 - 360 = 240\,\text{mi}\). 3. Find the plane’s average speed: \(240 \div 2 = 120\,\text{mi/h}\). 4. Find the difference in speeds: \(120 - 60 = 60\,\text{mi/h}\).

The plane’s average speed is \(60\,\text{mi/h}\) faster than the truck’s speed.

- Find how many gallons each pump moves in one minute. - Compare the two per-minute rates. - Which operation tells how many times one amount is another?

1. Find the large pump’s rate: \(2240 \div 7 = 320\,\text{gal}\) per minute. 2. Find the small pump’s rate: \(48 \div 6 = 8\,\text{gal}\) per minute. 3. Compare the rates: \(320 \div 8 = 40\).

The large pump moves \(40\) times as much water per minute as the small pump.

- Find how many rolls each machine packs in one hour. - Divide the same total amount by each machine’s time. - Subtract the smaller hourly rate from the larger one.

1. Find Machine A’s rate: \(960 \div 4 = 240\) rolls per hour. 2. Find Machine B’s rate: \(960 \div 6 = 160\) rolls per hour. 3. Find the difference: \(240 - 160 = 80\) rolls per hour.

Machine A packs \(80\) more rolls per hour than Machine B.

- Divide each total amount by the number of trips. - Compare the two rates. - Subtract the smaller rate from the larger rate.

1. Find Strong's amount per trip: \(435\,\text{tons} \div 5 = 87\,\text{tons}\) per trip. 2. Find Swift's amount per trip: \(656\,\text{tons} \div 8 = 82\,\text{tons}\) per trip. 3. Strong carries more per trip because \(87\,\text{tons} > 82\,\text{tons}\). 4. Find the difference: \(87\,\text{tons} - 82\,\text{tons} = 5\,\text{tons}\) per trip.

Strong carries \(5\) tons more per trip than Swift.

- Determine how many groups of \(3\) crates are in \(345\) crates. - Each group corresponds to \(5\) baskets. - Multiply the number of groups by \(5\).

1. Find the number of groups of \(3\) apple crates: \(345 \div 3 = 115\). 2. Each group is equivalent to \(5\) strawberry baskets, so \(115 \times 5 = 575\) baskets.

\(575\) strawberry baskets would have the same total weight.

- Determine the total number of slices. - Add all expenses. - Add the desired profit to the expenses. - Divide the required revenue by the number of slices.

1. Find the number of slices: \(8 \times 8 = 64\). 2. Find the total expenses: \(\$14.50 + \$26.80 + \$3.50 = \$44.80\). 3. Add the desired profit to find the required revenue: \(\$44.80 + \$51.20 = \$96.00\). 4. Divide the required revenue by the number of slices: \(\$96.00 \div 64 = \$1.50\).

Each pizza slice should cost \(\$1.50\).

- Add the two operating times. - Use the total flyers and combined time to find the shared unit rate. - Multiply that rate by each machine’s operating time.

1. Add the operating times: \(14 + 12 = 26\) minutes. 2. Find the shared printing rate: \(1560 \div 26 = 60\) flyers per minute. 3. Find the first machine’s output: \(14 \times 60 = 840\) flyers. 4. Find the second machine’s output: \(12 \times 60 = 720\) flyers.

The first machine prints \(840\) flyers, and the second machine prints \(720\) flyers.

- Think about the units that belong with each number. - What does dividing a total cost by the number of pounds find? - Once you know the price per pound, how can you find the cost of another amount?

1. The first calculation divides the total cost by the total number of pounds, so it finds the unit price: \(\$2\) per pound. 2. The second calculation multiplies the unit price by \(7\), so it shows that \(7\,\text{lb}\) of apples cost \(\$14\). 3. Find the cost of \(10\,\text{lb}\): \(10 \times \$2 = \$20\).

a) It finds the price per pound, which is \(\$2\). b) It means that \(7\,\text{lb}\) of apples cost \(\$14\). c) \(10\,\text{lb}\) of apples cost \(\$20\).

- Find each machine’s unit rate first. - Use the faster unit rate for part b). - For part c), divide the target number of pages by Machine A’s rate. - Fifteen minutes is one quarter of an hour.

1. Find Machine A’s rate: \(120 \div 4 = 30\) pages per minute. 2. Find Machine B’s rate: \(150 \div 6 = 25\) pages per minute. 3. Since \(30 > 25\), Machine A is faster. 4. Find Machine A’s output in \(15\) minutes: \(30 \times 15 = 450\) pages. 5. Find the time for \(300\) pages: \(300 \div 30 = 10\) minutes.

a) Machine A prints more pages per minute. b) Machine A prints \(450\) pages in \(15\) minutes. c) Machine A takes \(10\) minutes to print \(300\) pages.

- First find each difference for one square yard. - How does the difference change for an area \(100\) times as large? - Be sure to compare the correct two yields in each part.

1. Find the difference per square yard between this year and last year: \(5 - 3 = 2\,\text{lb}\). 2. Find the difference for \(100\,\text{yd}^2\): \(2 \times 100 = 200\,\text{lb}\). 3. Find the difference per square yard between this year and before the switch: \(5 - 1 = 4\,\text{lb}\). 4. Find the difference for \(100\,\text{yd}^2\): \(4 \times 100 = 400\,\text{lb}\).

a) The \(100\,\text{yd}^2\) area produces \(200\,\text{lb}\) more than last year. b) It produces \(400\,\text{lb}\) more than before the switch.

- How many seconds are in one minute? - How many \(10\)-second intervals are in one minute? - Find both distances for the same amount of time. - Subtract the shorter distance from the longer distance.

1. Convert the time: \(1\) minute is \(60\) seconds. 2. There are \(60 \div 10 = 6\) ten-second intervals, so the car travels \(6 \times 25 = 150\,\text{m}\). 3. The robot travels \(60 \times 2 = 120\,\text{m}\). 4. Since \(150 > 120\), the car wins. 5. Find the winning margin: \(150 - 120 = 30\,\text{m}\).

a) The car travels \(150\,\text{m}\). b) The robot travels \(120\,\text{m}\). c) The car wins by \(30\,\text{m}\).

- How many seconds are in \(2\) minutes? - Find how far the package travels in that time. - Compare the result with \(300\,\text{m}\).

1. Convert the time to seconds: \(2\) minutes is \(120\) seconds. 2. Find the distance traveled in that time: \(120 \times 2 = 240\,\text{m}\). 3. Since \(240 < 300\), the package does not reach the end. 4. It is \(300 - 240 = 60\,\text{m}\) short.

No. In \(2\) minutes, the package travels \(240\,\text{m}\), so it is \(60\,\text{m}\) short of the end.

- Find the elapsed time from \(5{:}00\) p.m. to \(10{:}00\) p.m. - Multiply the number of hours by the hourly temperature decrease. - For b), find the temperature difference from \(-4\) to \(-10\). - Divide that difference by the hourly rate.

1. Five hours pass from \(5{:}00\) p.m. to \(10{:}00\) p.m. The total decrease is \(5 \times 3\,^{\circ}\text{C} = 15\,^{\circ}\text{C}\). Therefore, the temperature at \(10{:}00\) p.m. is \(-4 - 15 = -19\,^{\circ}\text{C}\). 2. The change from \(-4\,^{\circ}\text{C}\) to \(-10\,^{\circ}\text{C}\) is a decrease of \(6\,^{\circ}\text{C}\). At \(3\,^{\circ}\text{C}\) per hour, this takes \(6 \div 3 = 2\) hours. Two hours after \(5{:}00\) p.m. is \(7{:}00\) p.m.

a) \(-19\,^{\circ}\text{C}\) b) \(7{:}00\) p.m.

- Find the elapsed time. - Find the total distance between the starting and ending temperatures. - Divide the total temperature change by the number of hours.

1. From \(8{:}00\) a.m. to \(2{:}00\) p.m. is \(6\) hours. 2. The total temperature decrease is \(5 - (-19) = 24\,^{\circ}\text{C}\). 3. Divide the total decrease by the elapsed time: \(24 \div 6 = 4\). The temperature decreased by \(4\,^{\circ}\text{C}\) per hour.

The temperature decreased by \(4\,^{\circ}\text{C}\) per hour.

- How much more water is warmed in one hour on a sunny day? - Use the one-hour difference to find the difference after several hours. - The same hourly difference applies to both time periods.

1. Find the difference per hour: \(28 - 12 = 16\,\text{gal}\). 2. Find the difference after \(6\) hours: \(16 \times 6 = 96\,\text{gal}\). 3. Find the difference after \(9\) hours: \(16 \times 9 = 144\,\text{gal}\).

After \(6\) hours, the difference is \(96\,\text{gal}\). After \(9\) hours, the difference is \(144\,\text{gal}\).

- Find the total number of flowers in the project. - Determine how many days the gardeners actually worked. - Divide the total flowers by the actual days. - Compare the actual daily rate with the planned rate.

1. Find the total number of flowers: \(180 \times 12 = 2160\) flowers. 2. Find the actual number of workdays: \(12 - 4 = 8\) days. 3. Find the actual daily rate: \(2160 \div 8 = 270\) flowers per day. 4. Find the increase: \(270 - 180 = 90\) flowers per day.

The gardeners planted an average of \(90\) more flowers per day than planned.

- Compare \(24\) with the given group of \(8\). - Find the cost of half a package. - Decompose \(12\) into quantities whose costs you know.

1. Twenty-four hot dogs are \(3\) groups of \(8\), so they cost \(3 \times \$6 = \$18\). 2. Four hot dogs are half of \(8\), so they cost half of \(\$6\), which is \(\$3\). 3. Twelve hot dogs are \(8 + 4\), so they cost \(\$6 + \$3 = \$9\). 4. Finding the cost of \(4\) helps because \(12\) can be composed as \(8 + 4\), or as \(3\) groups of \(4\).

a) \(24\) hot dogs cost \(\$18\). b) \(12\) hot dogs cost \(\$9\). c) Four hot dogs cost \(\$3\), and \(12 = 8 + 4\), so \(\$6 + \$3 = \$9\).

- Compare \(\$12\) with the price of one group of \(20\) marbles. - Determine how many groups of \(20\) make \(100\). - Compare the cost of \(100\) marbles with Tim’s budget.

1. Since \(\$12\) is \(3\) times \(\$4\), the number of marbles is \(3 \times 20 = 60\). 2. One hundred marbles are \(5\) groups of \(20\), so they cost \(5 \times \$4 = \$20\). 3. Since \(\$20 < \$25\), Tim has enough money.

a) You can buy \(60\) marbles for \(\$12\). b) Yes. The \(100\) marbles cost \(\$20\), so Tim has enough money.

- Compare \(900\,\text{mi}\) with the given \(300\,\text{mi}\). - For part b, compare \(30\,\text{gal}\) with \(6\,\text{gal}\). - How is \(150\,\text{mi}\) related to \(300\,\text{mi}\)?

1. The distance \(900\,\text{mi}\) is \(3\) times \(300\,\text{mi}\), so the gasoline use is \(3 \times 6 = 18\,\text{gal}\). 2. The amount \(30\,\text{gal}\) is \(5\) times \(6\,\text{gal}\), so the distance is \(5 \times 300 = 1500\,\text{mi}\). 3. The distance \(150\,\text{mi}\) is half of \(300\,\text{mi}\), so the gasoline use is \(6 \div 2 = 3\,\text{gal}\).

a) \(18\,\text{gal}\) b) \(1500\,\text{mi}\) c) \(3\,\text{gal}\)

- Find how many strips fit across the wall's width. - Find how many wall-height strips can be cut from one roll. - The purchase must be a whole number of rolls.

1. Find the number of vertical strips needed across the wall: \(12\div2=6\) strips. 2. Each strip must be \(8\,\text{ft}\) long. One roll provides \(32\div8=4\) strips. 3. One roll gives 4 of the 6 needed strips, leaving 2 strips still to cover. 4. A second roll provides those remaining strips, so \(2\) rolls are needed.

\(2\) rolls are needed.

- Think of covering the floor with long strips placed side by side. - Find how many strips are needed across the room. - Find the total strip length and compare it with one roll's length.

1. Find the number of strips needed across the room: \(12\div6=2\) strips. 2. Each strip must be \(15\,\text{ft}\) long, so the total length needed is \(2\times15\,\text{ft}=30\,\text{ft}\). 3. One roll is \(36\,\text{ft}\) long. Since \(30\,\text{ft}<36\,\text{ft}\), one roll is enough.

\(1\) roll is needed.

- Convert the two squirrel measurements to the same unit. - Divide the jump distance by the body length. - Multiply the child's height by the same factor.

1. Convert \(5\,\text{m}\) to centimeters: \(5\,\text{m}=500\,\text{cm}\). 2. Find the ratio: \(500\,\text{cm}\div 25\,\text{cm}=20\). The character jumps \(20\) times its body length. 3. Apply the same factor to the child's height: \(1.45\,\text{m}\times 20=29\,\text{m}\).

a) \(20\) times its body length b) \(29\,\text{m}\)

- Find how many equal burned sections make up the full candle. - Multiply that number by \(15\) minutes. - Convert the final time from minutes to hours.

1. Find how many \(12\)-millimeter sections are in the candle: \(144\,\text{mm} \div 12\,\text{mm} = 12\). 2. Each section takes \(15\) minutes to burn, so the total time is \(12 \times 15 = 180\) minutes. 3. Convert to hours: \(180 \div 60 = 3\) hours.

The candle will burn for \(3\) hours.

- Determine the scale factor from \(250\) flyers to \(1000\) flyers. - Multiply the minutes and seconds by that factor. - Convert any complete group of \(60\) seconds to minutes.

1. The number of flyers is multiplied by \(1000 \div 250 = 4\). 2. Multiply the time by \(4\): \(4 \times 6\) minutes is \(24\) minutes, and \(4 \times 15\) seconds is \(60\) seconds. 3. Since \(60\) seconds is \(1\) minute, the total time is \(24 + 1 = 25\) minutes.

The copier will take \(25\) minutes.

- Find the cost of one ounce. - Divide the money available by the unit price. - Keep the relationship between price and weight proportional.

1. Find the unit price: \(\$3.20 \div 16 = \$0.20\) per ounce. 2. Divide Leo's money by the price per ounce: \(\$1.20 \div \$0.20 = 6\).

Leo can buy \(6\,\text{oz}\) of strawberries.

- Find the total weight of all the beads. - Divide the total cost by the total weight to find the unit price. - Multiply the unit price by each person's weight.

1. Find the total weight: \(12 + 20 + 18 = 50\,\text{oz}\). 2. Find the unit price: \(\$12.50 \div 50 = \$0.25\) per ounce. 3. Mia's share is \(20 \times \$0.25 = \$5.00\). 4. Tom's share is \(18 \times \$0.25 = \$4.50\).

Mia should repay \(\$5.00\), and Tom should repay \(\$4.50\).

- Determine how many of each offered package are needed. - Find the pasta and sauce costs separately for each store. - Add each store's two costs and compare the totals.

1. Fresh Market: \(12 \div 3 = 4\) pasta offers, costing \(4 \times \$2.40 = \$9.60\). The sauce requires \(4 \div 2 = 2\) offers, costing \(2 \times \$2.50 = \$5.00\). The total is \(\$14.60\). 2. City Shop: \(12 \div 4 = 3\) pasta offers, costing \(3 \times \$3.12 = \$9.36\). The four jars of sauce cost \(\$4.80\). The total is \(\$14.16\). 3. Super Saver: \(12 \div 6 = 2\) pasta offers, costing \(2 \times \$4.68 = \$9.36\). Four jars of sauce cost \(4 \times \$1.29 = \$5.16\). The total is \(\$14.52\). 4. Since \(\$14.16 < \$14.52 < \$14.60\), City Shop is least expensive.

Fresh Market: \(\$14.60\) City Shop: \(\$14.16\) Super Saver: \(\$14.52\) City Shop is least expensive.

- Calculate the full cost under each offer separately. - Determine how many apple groups are needed to make \(15\) apples. - Compare the two total costs.

1. Offer 1: Fifteen apples require \(15 \div 3 = 5\) groups, so they cost \(5 \times \$1.20 = \$6.00\). The oranges cost \(5 \times \$2.45 = \$12.25\). The total is \(\$18.25\). 2. Offer 2: Fifteen apples require \(15 \div 5 = 3\) groups, so they cost \(3 \times \$1.90 = \$5.70\). The oranges cost \(\$12.00\). The total is \(\$17.70\). 3. The difference is \(\$18.25 - \$17.70 = \$0.55\).

Offer 2 is less expensive. It costs \(\$17.70\), which is \(\$0.55\) less than Offer 1.

- Divide the four-week total by \(4\). - Add the two weekly work times. - Divide weekly earnings by weekly hours to find the hourly wage. - Calculate Mr. Webb's four-week earnings separately.

1. Their combined weekly earnings are \(\$2240 \div 4 = \$560\). 2. Together, they work \(20 + 15 = 35\) hours per week. 3. Their hourly wage is \(\$560 \div 35 = \$16\) per hour. 4. Mr. Webb earns \(15 \times 4 \times \$16 = \$960\) over four weeks. Since \(\$960 < \$1000\), his claim is incorrect.

a) \(\$560\) b) \(\$16\) per hour c) No. Mr. Webb earns \(\$960\) in \(4\) weeks.

- For part a, compare \(45\,\text{kg}\) with \(15\,\text{kg}\). - For part b, find the amount needed for a convenient number of jars, such as \(10\). - Keep the honey-to-jars ratio equivalent.

1. For part a, find the scale factor: \(45\,\text{kg} \div 15\,\text{kg} = 3\). 2. Scale the number of jars: \(30 \times 3 = 90\) jars. 3. For part b, find the honey needed for \(10\) jars: \(15\,\text{kg} \div 3 = 5\,\text{kg}\). 4. Scale from \(10\) jars to \(100\) jars: \(5\,\text{kg} \times 10 = 50\,\text{kg}\).

a) The beekeeper can fill \(90\) jars. b) The beekeeper needs \(50\,\text{kg}\) of honey.

- Find the fuel estimate for one passenger over \(120\,\text{km}\). - Compare \(120\,\text{km}\) with the reference distance of \(100\,\text{km}\). - Multiply the per-passenger result by the number of passengers.

1. A \(120\)-kilometer trip is \(120 \div 100 = 1.2\) times the reference distance. 2. The fuel estimate per passenger is \(6\,\text{L} \times 1.2 = 7.2\,\text{L}\). 3. For \(850\) passengers, the total is \(850 \times 7.2\,\text{L} = 6120\,\text{L}\).

The estimated fuel use is \(6120\,\text{L}\).

- Use the same distance unit for both rates. - Convert kilometers to meters. - Divide each distance by its time.

1. Convert Ms. Miller's distance: \(1.2\,\text{km} = 1200\,\text{m}\). Her unit rate is \(1200\,\text{m} \div 15\,\text{min} = 80\,\text{m/min}\). 2. The dog's unit rate is \(450\,\text{m} \div 5\,\text{min} = 90\,\text{m/min}\). 3. Since \(90 > 80\), the dog travels more meters per minute.

The dog travels farther per minute: \(90\,\text{m/min}\), compared with Ms. Miller's \(80\,\text{m/min}\).

- Find the mosquito's wingbeats in one second. - Convert \(2\) minutes to seconds. - Multiply each per-second rate by the total number of seconds.

1. The mosquito's unit rate is \(15{,}000 \div 30 = 500\) wingbeats per second. Since \(500 > 80\), the mosquito has the greater rate. 2. Convert the flight time: \(2\) minutes is \(120\) seconds. 3. The hummingbird makes \(80 \times 120 = 9600\) wingbeats. 4. The mosquito makes \(500 \times 120 = 60{,}000\) wingbeats.

a) The mosquito, at \(500\) wingbeats per second compared with \(80\) for the hummingbird. b) Hummingbird: \(9600\) wingbeats; mosquito: \(60{,}000\) wingbeats.

- Express both rates using minutes or both using seconds. - One minute has \(60\) seconds. - One-half hour has \(30\) minutes. - Use each per-minute rate to find the half-hour total.

1. Model A makes \(1200 \div 2 = 600\) revolutions per minute. Model B makes \(15 \times 60 = 900\) revolutions per minute. Therefore, Model B rotates faster. 2. One-half hour is \(30\) minutes. Model A makes \(600 \times 30 = 18{,}000\) revolutions. 3. Model B makes \(900 \times 30 = 27{,}000\) revolutions.

a) Model B rotates faster. b) Model A: \(18{,}000\) revolutions; Model B: \(27{,}000\) revolutions.

- Subtract the shorter travel time from the longer travel time. - Express \(1\) hour \(30\) minutes as hours or use proportional reasoning. - Divide distance by time to find average speed.

1. The time saved is \(1\) hour \(30\) minutes minus \(1\) hour \(15\) minutes, which is \(15\) minutes. 2. The driving time is \(1.5\) hours. The average speed is \(120\,\text{km} \div 1.5\,\text{h} = 80\,\text{km/h}\).

a) \(15\) minutes b) \(80\,\text{km/h}\)

- Convert both speeds to the same time unit. - Find each runner's distance after \(4\) minutes. - Subtract the two distances.

1. Convert Sarah's rate to meters per minute: \(3 \times 60 = 180\,\text{m/min}\). 2. Since \(180 > 150\), Sarah is faster. 3. In \(4\) minutes, Sarah travels \(180 \times 4 = 720\,\text{m}\), while Liam travels \(150 \times 4 = 600\,\text{m}\). 4. The lead is \(720 - 600 = 120\,\text{m}\).

Sarah is faster, and after \(4\) minutes she is \(120\,\text{m}\) ahead.

- Divide the volume by the time to find the per-minute rate. - Divide the tank capacity by that rate. - Regroup every \(60\) minutes as one hour.

1. The flow rate is \(120\,\text{L} \div 15\,\text{min} = 8\,\text{L/min}\). 2. The filling time is \(800\,\text{L} \div 8\,\text{L/min} = 100\) minutes. 3. Since \(100\) minutes is \(1\) hour \(40\) minutes, that is the total filling time.

a) \(8\,\text{L/min}\) b) \(1\) hour \(40\) minutes

- Use time equals distance divided by speed. - Convert \(0.5\) hour to minutes. - Calculate Ben's two travel times separately. - Add each total travel time to the departure time.

1. Alex's travel time is \(48\,\text{mi} \div 12\,\text{mph} = 4\) hours. He arrives at \(6{:}00\) p.m. 2. Ben's time for the first section is \(24\,\text{mi} \div 16\,\text{mph} = 1.5\) hours, or \(1\) hour \(30\) minutes. 3. His time for the second section is \(24\,\text{mi} \div 8\,\text{mph} = 3\) hours. 4. Ben's total time is \(4\) hours \(30\) minutes, so he arrives at \(6{:}30\) p.m.

Alex arrives at \(6{:}00\) p.m. Ben arrives at \(6{:}30\) p.m.

- Count both the trip to work and the trip home. - Find the total weekly mileage. - Divide miles by miles per gallon to find gallons used. - Multiply gallons by the price per gallon.

1. The daily round trip is \(50 \times 2 = 100\,\text{mi}\). 2. The weekly distance is \(100 \times 4 = 400\,\text{mi}\). 3. The car uses \(400 \div 25 = 16\) gallons. 4. The weekly cost is \(16 \times \$4.10 = \$65.60\).

Mr. Webb spends \(\$65.60\) on gasoline each week.

- For Route B, calculate each section separately. - Convert decimal hours to minutes. - Add each travel time to the departure time. - A faster second half does not necessarily cancel a slower first half.

1. Route A takes \(120\,\text{mi} \div 60\,\text{mph} = 2\) hours, so the van arrives at \(10{:}00\) a.m. 2. The first part of Route B takes \(60 \div 40 = 1.5\) hours, or \(1\) hour \(30\) minutes. 3. The second part takes \(60 \div 80 = 0.75\) hour, or \(45\) minutes. 4. Route B takes \(2\) hours \(15\) minutes total, so the van arrives at \(10{:}15\) a.m.

Route A: \(10{:}00\) a.m. Route B: \(10{:}15\) a.m.

- How much time passes between the two clock times? - How long does one full revolution of a second hand take? - What happens between \(12\) and \(1\) during each revolution? - How many seconds are in one minute?

1. The elapsed time from \(2{:}45\) p.m. to \(3{:}15\) p.m. is \(30\) minutes. 2. A second hand makes one full revolution each minute, so it makes \(30\) revolutions. 3. During each revolution, the hand spends \(5\) seconds between \(12\) and \(1\). The total is \(30 \times 5 = 150\) seconds. 4. Convert \(150\) seconds: \(150\,\text{s} = 2\,\text{min}\,30\,\text{s}\).

a) \(30\) revolutions b) \(2\,\text{min}\,30\,\text{s}\)

- Find the duration of the second part. - Use its rate to find the distance covered during that part. - Subtract that distance from the total ascent.

1. The second part of the ascent lasts \(40 - 10 = 30\) minutes. 2. During the second part, the submarine rises \(30 \times 25 = 750\,\text{m}\). 3. The first part accounts for the remaining distance: \(1200 - 750 = 450\,\text{m}\).

The submarine rises \(450\,\text{m}\) during the first \(10\) minutes.

- Divide distance by rate to find travel time. - Count backward from the required arrival time. - For b), find the elapsed time and divide distance by time.

1. The trip to her grandmother's home takes \(1500 \div 60 = 25\) minutes. 2. Twenty-five minutes before \(2{:}00\) p.m. is \(1{:}35\) p.m. 3. The return trip lasts \(30\) minutes. 4. Her average return rate is \(1500\,\text{m} \div 30\,\text{min} = 50\,\text{m/min}\).

a) \(1{:}35\) p.m. b) \(50\,\text{m/min}\)

- Find the total distance first. - Divide the total distance by the total time. - Compare the result with the guide's statement.

1. Add the three distances: \(14.6 + 12.9 + 17.5 = 45.0\,\text{km}\). 2. Divide the total distance by the total hiking time: \(45.0\,\text{km} \div 9\,\text{h} = 5\,\text{km/h}\). 3. The calculated average speed matches the guide's statement.

Yes. The group traveled \(45.0\,\text{km}\) in \(9\) hours, for an average speed of \(5\,\text{km/h}\).

- First find the distance Lukas travels in the full \(4\) minutes. - Subtract Lukas’s distance from the total distance between the homes. - Use Mia’s distance and time to find her per-minute rate.

1. Find Lukas’s distance: \(4 \times 140 = 560\,\text{m}\). 2. Find Mia’s distance: \(1200 - 560 = 640\,\text{m}\). 3. Find Mia’s rate: \(640 \div 4 = 160\,\text{m}\) per minute. 4. Since \(640 > 560\), Mia rides farther.

a) Lukas rides \(560\,\text{m}\). b) Mia rides \(160\,\text{m}\) per minute. c) Mia rides farther.

- Are the groups moving toward each other or away from each other? - Add the distances each group travels in one hour. - Look for a constant hourly increase in their separation.

1. Their separation increases by \(4 + 5 = 9\,\text{mi}\) each hour. 2. After \(3\) hours, they are \(9 \times 3 = 27\,\text{mi}\) apart. 3. After \(5\) hours, they are \(9 \times 5 = 45\,\text{mi}\) apart.

a) After \(3\) hours, they are \(27\,\text{mi}\) apart. b) After \(5\) hours, they are \(45\,\text{mi}\) apart. c) Their distance increases by \(9\,\text{mi}\) each hour.

- Determine the factor that changes \(2\) pounds to \(8\) pounds. - Apply the same factor to both quantities. - Recheck the multiplication on the price side.

1. The first scale-down step is correct: dividing both quantities by \(10\) gives \(2\,\text{lb}\) for \(\$3\). 2. To change \(2\,\text{lb}\) to \(8\,\text{lb}\), multiply by \(4\). The price must also be multiplied by \(4\). 3. Since \(\$3 \times 4 = \$12\), not \(\$9\), the correct price is \(\$12\).

The error is in the final step. Leo should calculate \(\$3 \times 4 = \$12\). Eight pounds of potatoes cost \(\$12\).

- Check whether the unit rate itself is correct. - Compare \(30\) seconds with one full minute. - Ask whether traveling for less than one minute should produce more or less than the one-minute distance.

1. The unit rate \(12 \div 4 = 3\,\text{m/min}\) is correct. 2. The conversion \(30\,\text{s} = 0.5\,\text{min}\) is also correct. 3. The first incorrect step is \(3 \div 0.5\). A distance equals rate multiplied by time, so the distance is \(3 \times 0.5 = 1.5\,\text{m}\). 4. Dividing by \(0.5\) would make the distance larger even though the car travels for less than one minute, which is inconsistent with the rate.

The first incorrect step is \(3 \div 0.5 = 6\,\text{m}\). The correct distance is \(1.5\,\text{m}\).

- Recalculate \(18 \div 12\). - Multiply the proposed unit price by \(12\) to check it. - Use the correct unit price to find the cost of \(5\) quarts.

1. The unit price must be found by dividing \(\$18\) by \(12\), not by \(10\). 2. The correct unit price is \(\$18 \div 12 = \$1.50\) per quart. 3. Five quarts cost \(5 \times \$1.50 = \$7.50\).

Mateo divided incorrectly. The unit price is \(\$1.50\) per quart, so \(5\,\text{qt}\) cost \(\$7.50\).

- Find the price of one pound first. - Use the unit price to find the cost of \(8\) pounds. - Compare your result with the customer's claim.

1. Find the unit price: \(\$10 \div 5 = \$2\) per pound. 2. Find the price of \(8\) pounds: \(8 \times \$2 = \$16\). 3. Since \(\$16 \ne \$18\), the customer is not correct.

No. The customer is not correct because \(8\) pounds of potatoes cost \(\$16\).

- The requested unit rate is for one baker and one hour. - First reduce one of the two “per” quantities, then reduce the other. - Keep track of the units after each division.

1. The whole team’s rate is \(150\div5=30\) loaves per hour. 2. Dividing that rate equally among \(6\) bakers gives \(30\div6=5\) loaves per baker per hour.

One baker makes an average of \(5\) loaves per hour.

- Find the distance traveled during the first \(3\) hours. - Find the afternoon distance in the same way. - Add the two distances.

1. Find the morning distance: \(3 \times 58 = 174\,\text{mi}\). 2. Find the afternoon distance: \(4 \times 65 = 260\,\text{mi}\). 3. Add the distances: \(174 + 260 = 434\,\text{mi}\).

The van travels \(434\,\text{mi}\) that day.

- Find the distance traveled by bus. - Find the distance traveled by train. - Compare the distances and subtract to find the difference.

1. Find the bus distance: \(4 \times 55 = 220\,\text{mi}\). 2. Find the train distance: \(3 \times 80 = 240\,\text{mi}\). 3. Since \(240 > 220\), the train portion is longer. 4. Find the difference: \(240 - 220 = 20\,\text{mi}\).

The train portion is \(20\,\text{mi}\) longer.

- How many hours does the class hike each day? - Use the hourly rate to find one day’s distance. - Then find the distance for all \(3\) days.

1. Find the hiking time each day: \(4 + 2 = 6\) hours. 2. Find the distance each day: \(6 \times 3 = 18\,\text{mi}\). 3. Find the total distance for \(3\) days: \(3 \times 18 = 54\,\text{mi}\).

The class hikes \(54\,\text{mi}\) in all.

- Find Lukas’s total weekly distance. - Find Sarah’s total weekly distance. - Compare the two distances. - Be careful to use the correct number of workdays for each person.

1. Find Lukas’s total riding time: \(5 \times 5 = 25\) hours. 2. Find Lukas’s distance: \(25 \times 12 = 300\,\text{mi}\). 3. Find Sarah’s total riding time: \(4 \times 7 = 28\) hours. 4. Find Sarah’s distance: \(28 \times 10 = 280\,\text{mi}\). 5. Since \(300 > 280\), Lukas travels farther.

Lukas travels farther during the week.

- Find the total number of yards purchased. - Use the total cost and total length to find the price per yard. - Use the unit price to find the cost of each piece.

1. Find the total length: \(2 + 6 = 8\,\text{yd}\). 2. Find the price per yard: \(\$72 \div 8 = \$9\) per yard. 3. Find the cost of the \(2\,\text{yd}\) piece: \(2 \times \$9 = \$18\). 4. Find the cost of the \(6\,\text{yd}\) piece: \(6 \times \$9 = \$54\).

The \(2\,\text{yd}\) piece costs \(\$18\), and the \(6\,\text{yd}\) piece costs \(\$54\).

- Find the distance traveled during each time interval. - Add the two distances already traveled. - Find one-fourth of that distance. - Add the traveled and remaining distances.

1. During the first part, the train travels \(6\,\text{h} \times 60\,\text{mph} = 360\,\text{mi}\). 2. During the second part, it travels \(4\,\text{h} \times 40\,\text{mph} = 160\,\text{mi}\). 3. The distance already traveled is \(360 + 160 = 520\,\text{mi}\). 4. The remaining distance is \(520 \div 4 = 130\,\text{mi}\). 5. The entire route is \(520 + 130 = 650\,\text{mi}\).

The entire route is \(650\,\text{mi}\).

- Determine how many groups of \(10\) rotations fit into \(150\) rotations. - Each group of rotations corresponds to the same distance. - Multiply the number of groups by \(4\,\text{m}\).

1. Find how many groups of \(10\) rotations are in \(150\) rotations: \(150 \div 10 = 15\). 2. Each group corresponds to \(4\,\text{m}\), so the distance is \(15 \times 4\,\text{m} = 60\,\text{m}\).

The car can travel \(60\,\text{m}\).

- Find how many \(10\)-meter sections are in the full path. - Use each student's constant number of steps per section. - Compare the two totals.

1. Find the number of \(10\)-meter sections: \(120\,\text{m} \div 10\,\text{m} = 12\). 2. Lucas takes \(12 \times 15 = 180\) steps. 3. Emilia takes \(12 \times 18 = 216\) steps. 4. Find the difference: \(216 - 180 = 36\) steps.

Emilia takes \(36\) more steps than Lucas.

- For part a, determine how many \(3\)-minute intervals fit into \(15\) minutes. - For part b, determine how many groups of \(50\) worksheets make \(300\). - Use the constant rate for each equal interval.

1. For part a, find how many \(3\)-minute intervals are in \(15\) minutes: \(15 \div 3 = 5\). Then \(5 \times 50 = 250\), so the copier prints \(250\) worksheets. 2. For part b, find how many groups of \(50\) worksheets are in \(300\): \(300 \div 50 = 6\). Then \(6 \times 3 = 18\), so printing \(300\) worksheets takes \(18\) minutes.

a) The copier prints \(250\) worksheets. b) It takes \(18\) minutes.

- Use the distance traveled in one hour. - Multiply the hourly distance by the number of hours. - Subtract the distance already traveled from the total distance. - Determine how many hourly distances fit into the full trip.

1. Find the distance after \(2\) hours: \(12 \times 2 = 24\,\text{mi}\). 2. Find the distance after \(3\) hours: \(12 \times 3 = 36\,\text{mi}\). 3. Find the remaining distance: \(60 - 36 = 24\,\text{mi}\). 4. Find the total riding time: \(60 \div 12 = 5\) hours.

a) The group has traveled \(24\,\text{mi}\). b) The group is \(24\,\text{mi}\) from the lake. c) The entire trip requires \(5\) hours of riding.

- Find half of the total distance. - Work with each vehicle separately. - Use multiplication to find a distance after a given time. - Use division to find how many hourly distances fit into half the trip.

1. Find the truck’s distance after \(3\) hours: \(65 \times 3 = 195\,\text{mi}\). 2. Find half the total distance: \(260 \div 2 = 130\,\text{mi}\). 3. Find the time to travel half the distance: \(130 \div 65 = 2\) hours. 4. Find the van’s distance after \(3\) hours: \(80 \times 3 = 240\,\text{mi}\). 5. Find the van’s remaining distance: \(260 - 240 = 20\,\text{mi}\).

a) The truck has traveled \(195\,\text{mi}\). b) It reaches halfway after \(2\) hours. c) The van is \(20\,\text{mi}\) from the city.

- Combine all costs charged per square foot. - Multiply the total unit cost by the lot area. - Subtract the purchase cost from the starting balance. - Align decimal points when subtracting money.

1. Combine the costs per square foot: \(\$16 + \$4 = \$20\) per square foot. 2. Find the total cost: \(7200 \times \$20 = \$144{,}000\). 3. Subtract from the account balance: \(\$155{,}250.50 - \$144{,}000 = \$11{,}250.50\).

The remaining account balance is \(\$11{,}250.50\).

- Express both yields using the same area unit. - Determine how many groups of \(1000\,\text{ft}^2\) each farm uses. - Divide each harvest by its number of area groups.

1. Sunny Acres produces \(480 \div 12 = 40\,\text{lb}\) per \(1000\,\text{ft}^2\). 2. Fresh Fields produces \(900 \div 20 = 45\,\text{lb}\) per \(1000\,\text{ft}^2\). 3. Since \(45 > 40\), Fresh Fields has the greater yield.

Fresh Fields has the greater yield: \(45\,\text{lb}\) per \(1000\,\text{ft}^2\), compared with \(40\,\text{lb}\) per \(1000\,\text{ft}^2\) at Sunny Acres.

- Find each rectangular area first. - Divide each total price by its area. - Compare the three unit prices.

1. Plot 1 has area \(40 \times 50 = 2000\,\text{ft}^2\), so its unit price is \(\$12{,}000 \div 2000 = \$6.00\) per square foot. 2. Plot 2 has area \(60 \times 30 = 1800\,\text{ft}^2\), so its unit price is \(\$9000 \div 1800 = \$5.00\) per square foot. 3. Plot 3 has area \(25 \times 80 = 2000\,\text{ft}^2\), so its unit price is \(\$11{,}000 \div 2000 = \$5.50\) per square foot. 4. Since \(\$5.00 < \$5.50 < \$6.00\), the order is Plot 2, Plot 3, Plot 1.

Plot 1: \(\$6.00/\text{ft}^2\) Plot 2: \(\$5.00/\text{ft}^2\) Plot 3: \(\$5.50/\text{ft}^2\) Order: Plot 2, Plot 3, Plot 1

- First find the amount of water remaining. - Convert cubic meters to liters before finding the daily amount. - Divide the total remaining amount equally among the \(d\) days.

1. The remaining volume is \(V-v\) cubic meters. 2. Convert the remaining amount to liters: \(1000(V-v)\). 3. Divide by the number of remaining days: \(\frac{1000(V-v)}{d}\) liters per day.

\(\frac{1000(V-v)}{d}\) liters per day

- Consider what happens to a quotient when its positive divisor increases. - Imagine sharing the same amount among more groups. - Use parentheses so the remaining amount is found before conversion and division.

1. The remaining amount is \(k-m\) kilograms. 2. Convert to grams: \(1000(k-m)\). 3. Divide equally among \(n\) days: \(\frac{1000(k-m)}{n}\) grams per day. 4. With a fixed numerator, increasing the positive denominator makes the quotient smaller. The same amount of pasta must last more days, so less is available each day.

a) \(\frac{1000(k-m)}{n}\) grams per day b) The value decreases because the same remaining amount is divided among more days.

- Find their combined closing rate. - Determine how far each rider travels in \(t\) hours. - Subtract the total distance traveled from the original separation. - Keep the time and rate units consistent.

1. In \(t\) hours, Lucas travels \(v_Lt\) miles, and Maya travels \(v_Mt\) miles. 2. Together, they reduce the distance by \((v_L+v_M)t\). 3. Therefore, \(e=s-(v_L+v_M)t\). 4. Substitute the values: \(e=38-(16+14)\times 0.5=38-15=23\) miles.

a) \(e=s-(v_L+v_M)t\) b) \(23\) miles

- Find how many meters Lucas gains each minute. - Divide the entire head start by the rate at which the gap closes. - Write the head start and the difference in speeds as a quotient.

1. Lucas closes the gap at \(v_1-v_2\) meters per minute. 2. Time equals distance divided by rate, so \(t=\frac{d}{v_1-v_2}\). 3. Substitute the values: \(t=\frac{450}{80-50}=\frac{450}{30}=15\) minutes.

a) \(t=\frac{d}{v_1-v_2}\) b) \(15\) minutes

- Convert feet to inches before dividing by a step length. - Divide the total distance by the length of one step. - Decide which person takes more steps before writing the difference.

1. Convert the walking distance to inches: \(12s\) inches. 2. Ava takes \(\frac{12s}{L}\) steps. 3. Lucas's step length is \(L-2\) inches, so he takes \(\frac{12s}{L-2}\) steps. 4. Lucas takes \(\frac{12s}{L-2}-\frac{12s}{L}\) more steps than Ava. 5. For \(s=4200\) and \(L=30\), Ava takes \(\frac{12\times 4200}{30}=1680\) steps. Lucas takes \(\frac{12\times 4200}{28}=1800\) steps. The difference is \(1800-1680=120\) steps.

a) Ava: \(\frac{12s}{L}\) steps; Lucas: \(\frac{12s}{L-2}\) steps b) \(\frac{12s}{L-2}-\frac{12s}{L}\) steps c) Ava takes \(1680\) steps, Lucas takes \(1800\) steps, and Lucas takes \(120\) more steps.

- First determine how many pages the copier prints in one minute. - Divide the number of pages by the printing rate to find the time. - Substitute the expression from part a) into the formula for time.

1. The printing rate is pages divided by minutes: \(v=\frac{n}{m}\). 2. The time for \(x\) pages is \(T=\frac{x}{v}=\frac{xm}{n}\). 3. For \(n=180\) and \(m=4.5\), \(v=180\div 4.5=40\) pages per minute. 4. For \(x=540\), \(T=540\div 40=13.5\) minutes.

a) \(v=\frac{n}{m}\) pages per minute b) \(T=\frac{x}{v}=\frac{xm}{n}\) minutes c) \(40\) pages per minute d) \(13.5\) minutes

- Average speed describes the entire distance compared with the entire travel time. - Find the distance traveled during each part of the ride. - Check whether the cyclist spends equal amounts of time at the two speeds before treating them equally.

1. The cyclist travels \(14 \times 1.5 = 21\) miles during the first part. 2. The cyclist travels \(10 \times 0.5 = 5\) miles during the second part. 3. The total distance is \(21 + 5 = 26\) miles, and the total time is \(1.5 + 0.5 = 2\) hours. 4. The average speed is \(26 \div 2 = 13\) miles per hour. 5. Jordan's method is not valid because the cyclist spends different amounts of time at the two speeds, so the two speeds do not contribute equally to the whole trip.

Jordan is not correct. The average speed is \(13\,\text{mi/h}\). The two speeds cannot simply be averaged because the cyclist rides at them for different lengths of time.

- Break the trip into riding intervals and the rest interval before finding accumulated distance. - During the break, elapsed time increases but distance traveled does not. - For a target distance, first decide which riding interval contains that target. - Whole-trip average speed uses total distance and total elapsed time, including the break.

1. At \(9{:}00\) a.m., the distance is \(0\). At \(10{:}00\) a.m., it is \(12\) miles. During the next \(1.5\) hours, she travels \(1.5\times8=12\) miles, so at \(11{:}30\) a.m. she has traveled \(24\) miles. The distance remains \(24\) miles through the break until \(12{:}00\) p.m. In the final \(0.5\) hour, she travels \(0.5\times16=8\) miles, so the final distance is \(32\) miles. 2. From \(10{:}00\) a.m. to \(11{:}00\) a.m., she travels \(8\) more miles, so the total is \(12+8=20\) miles. 3. After \(10{:}00\) a.m., she needs \(16-12=4\) more miles at \(8\,\text{mph}\). This takes \(4\div8=0.5\) hour, so she reaches \(16\) miles at \(10{:}30\) a.m. 4. The total elapsed time is \(3.5\) hours. The average speed is \(32\div3.5=\frac{64}{7}\approx9.1\,\text{mph}\).

1) <table><tr><td>Time</td><td>\(9{:}00\) a.m.</td><td>\(10{:}00\) a.m.</td><td>\(11{:}30\) a.m.</td><td>\(12{:}00\) p.m.</td><td>\(12{:}30\) p.m.</td></tr><tr><td>Distance (mi)</td><td>\(0\)</td><td>\(12\)</td><td>\(24\)</td><td>\(24\)</td><td>\(32\)</td></tr></table> 2) \(20\) miles 3) \(10{:}30\) a.m. 4) About \(9.1\,\text{mph}\)

- Use the image to compare the number of intervals from \(0\) to \(A\) with the total number of intervals. - Determine how many times the distance from \(0\) to \(A\) fits in the whole scale. - Multiply the value at \(A\) by that scale factor.

1. From the image, point \(A\) is \(4\) of the \(20\) equal intervals from \(0\) to the endpoint. Thus, the whole scale is \(20 \div 4 = 5\) times the value at \(A\). 2. For a), the endpoint is \(5 \times 0.8 = 4\). 3. For b), the endpoint is \(5 \times 0.08 = 0.4\). 4. For c), the endpoint is \(5 \times \frac{2}{5} = 2\).

a) \(4\) b) \(0.4\) c) \(2\)

- Read the wall dimensions from the diagram. - Find the wall’s area first. - Divide the wall area by each roll’s coverage. - Multiply the number of rolls by the price per roll.

1. Read the wall dimensions from the diagram: \(15\,\text{ft}\) by \(10\,\text{ft}\). 2. The wall area is \(15\,\text{ft} \times 10\,\text{ft} = 150\,\text{ft}^2\). 3. For Type A, the number of rolls is \(150 \div 50 = 3\). The cost is \(3 \times \$12.50 = \$37.50\). 4. For Type B, the number of rolls is \(150 \div 75 = 2\). The cost is \(2 \times \$18.00 = \$36.00\). 5. Since \(\$36.00 < \$37.50\), Type B costs less.

Type A costs \(\$37.50\). Type B costs \(\$36.00\). Type B costs less.

- Read one convenient time-and-pages point from the graph. - Use that point to determine how many pages correspond to \(1\) minute. - Use the same constant rate both forward and backward.

1. From the graph, the printer produces \(120\) pages in \(4\) minutes, so its unit rate is \(120 \div 4 = 30\) pages per minute. 2. In \(6\) minutes, it produces \(30 \times 6 = 180\) pages. 3. To produce \(150\) pages, the time is \(150 \div 30 = 5\) minutes.

a) \(30\) pages per minute b) \(180\) pages c) \(5\) minutes

- Choose a clear point on each line whose coordinates you can read exactly. - For each line, compare distance with time in miles per hour. - After finding both unit rates, compare them and then find their difference.

1. Line a passes through \((4,20)\), so its unit rate is \(20 \div 4=5\,\text{mi/h}\). 2. Line b passes through \((4,24)\), so its unit rate is \(24 \div 4=6\,\text{mi/h}\). 3. Since \(6>5\), hiker b is faster. 4. The speed difference is \(6-5=1\,\text{mi/h}\).

a) Hiker a: \(5\,\text{mi/h}\); hiker b: \(6\,\text{mi/h}\) b) Hiker b is faster by \(1\,\text{mi/h}\).

- Find gallons per one minute from a nonzero table row. - Read point P using the graph's x-axis for time and y-axis for water. - Compare the two unit rates, not just the sizes of the displayed numbers.

1. The table gives \(50 \div 2 = 25\,\text{gal/min}\), which agrees with \(125 \div 5 = 25\,\text{gal/min}\). 2. The graph's marked point is \(P=(4,120)\), so its rate is \(120 \div 4 = 30\,\text{gal/min}\). 3. The rates are different, \(25\,\text{gal/min}\) and \(30\,\text{gal/min}\), so the table and graph do not represent the same constant-rate relationship.

a) \(25\,\text{gal/min}\) b) \(P=(4,120)\), so \(30\,\text{gal/min}\) c) No. The table shows \(25\,\text{gal/min}\), while the graph shows \(30\,\text{gal/min}\).

- Look for what stays the same across the nonzero rows of the table. - Compare bottles per minute rather than adding the same number to quantities with different units. - Use the constant multiplicative relationship to extend the table to \(8\) minutes.

1. The table shows a multiplicative constant rate: \(10 \div 2=5\) bottles per minute. 2. Talia used the same additive change for quantities with different units. The table does not say that adding \(2\) minutes means adding \(2\) bottles. 3. At the constant rate, \(8\) minutes gives \(5 \times 8=40\) bottles.

Talia’s additive reasoning is incorrect. The constant rate is \(5\) bottles per minute, so the machine fills \(40\) bottles in \(8\) minutes.

- Find the numerical distance between points \(A\) and \(B\). - Use the relationship between that numerical distance and \(5\,\text{cm}\) to find the length of one whole unit. - Convert \(2\,\text{cm}\) back to a numerical amount before adding it to \(B\).

1. Find the numerical distance: \(1\frac{1}{4}-\frac{5}{6}=\frac{5}{4}-\frac{5}{6}=\frac{15}{12}-\frac{10}{12}=\frac{5}{12}\). 2. A numerical distance of \(\frac{5}{12}\) corresponds to \(5\,\text{cm}\), so \(\frac{1}{12}\) corresponds to \(1\,\text{cm}\). Therefore, one whole unit corresponds to \(12\,\text{cm}\). 3. A distance of \(2\,\text{cm}\) corresponds to \(\frac{2}{12}=\frac{1}{6}\). 4. Add this value to point \(B\): \(\frac{5}{4}+\frac{1}{6}=\frac{15}{12}+\frac{2}{12}=\frac{17}{12}=1\frac{5}{12}\).

a) \(12\,\text{cm}\) b) \(\frac{17}{12}=1\frac{5}{12}\)

- Write the large numbers in full or use place-value reasoning. - Recall how many kilograms are in one metric ton. - Simplify the final division by canceling common powers of \(10\).

1. Household waste is \(\frac{1}{2}\) of \(12{,}000{,}000\) metric tons, which is \(6{,}000{,}000\) metric tons. 2. Preventable household waste is \(\frac{2}{5}\) of \(6{,}000{,}000\), which is \(2{,}400{,}000\) metric tons. 3. Convert to kilograms: \(2{,}400{,}000\) metric tons is \(2{,}400{,}000{,}000\,\text{kg}\). 4. Divide by the population: \(2{,}400{,}000{,}000\div80{,}000{,}000=30\).

The average is \(30\,\text{kg}\) of preventable household food waste per person per year.

- Convert all time values to seconds when useful. - How many seconds are in \(10\) minutes? - Use the cyclist's distance per second to find a longer-distance total. - Can you find the runner's distance by comparing the time intervals without first finding a decimal speed?

1. For a), convert the cyclist's speed: \(23.4\div3.6=6.5\,\text{m/s}\). 2. Ten minutes is \(600\) seconds, so the cyclist travels \(6.5\times600=3900\,\text{m}\). 3. The runner needs \(75\) seconds for \(400\,\text{m}\). In \(600\) seconds there are \(600\div75=8\) such intervals, so the runner travels \(8\times400=3200\,\text{m}\). 4. Since \(3900>3200\), the cyclist travels farther.

a) \(6.5\,\text{m/s}\) b) The cyclist travels farther: \(3900\,\text{m}\) versus \(3200\,\text{m}\).

- First determine how many person-hours are needed for one page. - Use the unit rate to find the total number of person-hours needed for 18 pages. - How many students will share the work on the next issue?

1. Find the total work used to create the first issue: \(3 \times 4 = 12\) person-hours. 2. Find the work needed per page: \(12 \div 12 = 1\) person-hour per page. 3. Find the total work needed for 18 pages: \(18 \times 1 = 18\) person-hours. 4. Divide the work between the two remaining students: \(18 \div 2 = 9\) hours per student.

Each of the two remaining students must work \(9\,\text{hours}\).

- Find the amount of soil one excavator moves in one hour. - Use that unit rate to find how much four excavators move in eight hours. - For part b, think about what happens when several large machines share a limited workspace.

1. One excavator moves \(\frac{150}{2\times5}=15\,\text{yd}^3\) per hour. 2. Four excavators working for \(8\) hours would move \(4\times8\times15=480\,\text{yd}^3\). 3. Because \(480<600\), four excavators are not enough. The number required by the ideal model is \(\frac{600}{15\times8}=5\). 4. In reality, limited space, interference between machines, or delays in hauling soil away could reduce the combined rate.

a) No. Four excavators would move only \(480\,\text{yd}^3\); the ideal model requires \(5\) excavators. b) Possible reasons include limited space, interference between excavators, or delays in hauling away the soil.

- Find the total amount of work in painter-days. - Subtract the work completed during the first four days. - Find how long the remaining painters need, then compare that time with the original remaining time.

1. The entire job requires \(12\times10=120\) painter-days. 2. In the first \(4\) days, the team completes \(12\times4=48\) painter-days of work. 3. The remaining work is \(120-48=72\) painter-days. 4. Eight painters remain, so they need \(72\div8=9\) more days. 5. The original schedule allowed \(10-4=6\) more days. The delay is \(9-6=3\) days.

The renovation will be delayed by \(3\) days.

- Express the entire project as a number of person-days. - How many person-days of work are completed during the first 12 days? - Divide the remaining person-days by the 9 days that are left.

1. Find the total amount of work in person-days: \(4 \times 30 = 120\) person-days. 2. Find the work completed during the first 12 days: \(4 \times 12 = 48\) person-days. 3. Find the remaining work: \(120 - 48 = 72\) person-days. 4. Let \(x\) be the number of programmers needed for the remaining 9 days. Then \(9x = 72\). 5. Divide both sides by \(9\): \(x = 8\).

A total of 8 programmers must work on the project starting on day 13.

- How many minutes does the break last? - Use the number of degrees each hand moves per minute. - For part c, find each hand’s position measured clockwise from 12.

1. The break lasts \(55 - 35 = 20\) minutes. 2. The minute hand moves \(6^\circ\) per minute, so it sweeps through \(20 \times 6^\circ = 120^\circ\). 3. The hour hand moves \(0.5^\circ\) per minute, so it sweeps through \(20 \times 0.5^\circ = 10^\circ\). 4. At \(9{:}55\), the minute hand is \(55 \times 6^\circ = 330^\circ\) clockwise from 12. 5. The hour hand is \(9 \times 30^\circ + 55 \times 0.5^\circ = 297.5^\circ\) clockwise from 12. 6. The smaller angle between the hands is \(330^\circ - 297.5^\circ = 32.5^\circ\).

a) \(120^\circ\) b) \(10^\circ\) c) \(32.5^\circ\)

- Add the speeds when two riders move toward each other. - Use distance divided by speed to find time. - Convert hours to minutes. - Divide Mia's distance by the total distance to find the percentage.

1. Their combined closing speed is \(20 + 12 = 32\,\text{mph}\). 2. The meeting time is \(\frac{24}{32} = 0.75\) hour. 3. Convert to minutes: \(0.75 \times 60 = 45\) minutes. 4. Mia travels \(20 \times 0.75 = 15\) miles. 5. Her share of the total distance is \(\frac{15}{24} \times 100\% = 62.5\%\).

a) They meet after \(45\) minutes. b) Mia travels \(62.5\%\) of the total distance.

- Find the walking speed when the walkway is off. - Find the combined speed when the walkway is on. - Subtract the walking speed from the combined speed.

1. Mr. Miller's walking speed is \(300 \div 200 = 1.5\,\text{ft/s}\). 2. His combined speed while walking on the moving walkway is \(300 \div 100 = 3\,\text{ft/s}\). 3. The combined speed equals his walking speed plus the walkway speed. 4. The walkway speed is \(3 - 1.5 = 1.5\,\text{ft/s}\).

The moving walkway travels at \(1.5\,\text{ft/s}\).

- Find the escalator's rate in steps per second. - Add Julia's climbing rate to the escalator's rate. - Multiply Julia's own rate by the time she spends on the escalator.

1. The escalator moves at \(60 \div 30 = 2\) steps per second. 2. Julia's combined upward rate is \(2 + 1 = 3\) steps per second. 3. Her time to the top is \(60 \div 3 = 20\) seconds. 4. Julia climbs \(20 \times 1 = 20\) steps herself. 5. During those \(20\) seconds, the escalator carries her the equivalent of \(40\) steps, so she does not need to climb all \(60\) steps.

a) Julia takes \(20\) seconds. b) She climbs \(20\) steps herself. c) The moving escalator carries her through the remaining \(40\) steps.

- Find the number of minutes in one day. - Use the leak rate to find the daily volume. - Compare multiples of the daily cost with the replacement-part price.

1. One day has \(24\times 60=1440\) minutes. 2. The daily waste is \(0.5\times 1440=720\) gallons. 3. The daily cost is \(\frac{720}{1000}\times 6.00=\$4.32\). 4. After \(4\) days, the cost is \(4\times 4.32=\$17.28\), which is below \(\$18.50\). After \(5\) days, it is \(5\times 4.32=\$21.60\), which exceeds \(\$18.50\).

a) \(720\) gallons per day. b) \(\$4.32\) per day. c) After \(5\) complete days.

- Find Luke's total walking time before the proposed catch. - Calculate how far Luke would travel by then. - Jake must cover that same distance in only \(2\) minutes. - Compare the required speed with Jake's stated maximum speed.

1. When Jake catches him, Luke would have walked for \(12 + 2 = 14\) minutes. 2. Convert the time to hours: \(14\) minutes is \(\frac{14}{60}\) hour. 3. Luke travels \(3 \times \frac{14}{60} = 0.7\,\text{mi}\). 4. Jake has \(2\) minutes, or \(\frac{1}{30}\) hour, to travel the same distance. 5. Jake's required speed is \(0.7 \div \frac{1}{30} = 21\,\text{mph}\). 6. Because \(21\,\text{mph} > 15\,\text{mph}\), Jake cannot catch Luke in exactly \(2\) minutes under the stated limit.

No. Jake would need to run at \(21\,\text{mph}\), which is above his stated maximum of \(15\,\text{mph}\).

- Use \(1\,\text{mi}=5280\,\text{ft}\) and \(1\,\text{h}=3600\,\text{s}\). - How far must the front of the train travel before the back of the train leaves the bridge? - What fraction of an hour is \(12\) minutes?

1. Convert the speed to feet per second: \(60 \times \frac{5280}{3600}=88\,\text{ft/s}\). 2. To clear the bridge, the front of the train travels \(1320+440=1760\,\text{ft}\). 3. The crossing time is \(1760 \div 88=20\) seconds. 4. Convert \(12\) minutes to hours: \(12 \div 60=0.2\) hour. The distance is \(60 \times 0.2=12\) miles.

a) The train travels \(88\,\text{ft/s}\). b) The train takes \(20\) seconds to completely cross the bridge. c) The passenger travels \(12\) miles.

- Use total round-trip distance and the desired average speed to find the total time allowed. - Subtract the fixed outbound time to determine the time left for the return trip. - For part c, check whether the proposed average leaves any positive amount of time for the return trip.

1. The trip to work takes \(0.5\) hour, so \(v_1=\frac{20}{0.5}=40\,\text{mph}\). 2. A \(50\,\text{mph}\) average over the \(40\)-mile round trip requires a total time of \(\frac{40}{50}=0.8\) hour. 3. The return trip must therefore take \(0.8-0.5=0.3\) hour. 4. The required return speed is \(v_2=\frac{20}{0.3}=\frac{200}{3}\,\text{mph}\approx66.67\,\text{mph}\). 5. An \(80\,\text{mph}\) average over \(40\) miles would require a total time of \(\frac{40}{80}=0.5\) hour. 6. The outbound trip already takes all \(0.5\) hour, leaving \(0\) hour for the return trip. That would require an unattainable infinite speed, so the proposed round-trip average is not possible.

a) \(v_1=40\,\text{mph}\) b) \(v_2=\frac{200}{3}\,\text{mph}\approx66.67\,\text{mph}\) c) No. It would require a return time of \(0\) hours.

- First find the number of gallons lost each minute. - How long does it take the leak to fill a \(3\)-gallon bucket? - Remember that the bucket must also be empty at the end of the day.

1. The leak rate is \(0.2 \div 5=0.04\,\text{gal/min}\). 2. One week has \(7 \times 24 \times 60=10{,}080\) minutes. The weekly loss is \(10{,}080 \times 0.04=403.2\) gallons. 3. In one day, the leak produces \(24 \times 60 \times 0.04=57.6\) gallons. 4. The bucket fills every \(3 \div 0.04=75\) minutes. It becomes completely full \(19\) times during the day, accounting for \(19 \times 3=57\) gallons. 5. At the end of the day, \(57.6-57=0.6\) gallon remains. Including the required final emptying, the bucket must be emptied \(20\) times.

a) \(403.2\) gallons b) The bucket must be emptied \(20\) times: \(19\) times when full and once more at the end of the day.

- First find the driving time without a break. - What would the arrival time be if the driver did not stop? - What additional event is described during the trip? - Is its duration given?

1. Find the driving time without the break: \(400 \div 80 = 5\) hours. 2. Without a break, the truck would arrive at \(12{:}30\) p.m. 3. The problem states that the driver takes a break but does not give the break’s duration. 4. Therefore, the exact arrival time cannot be determined.

The exact arrival time cannot be determined because the duration of the break is not given.

- Add the two original work times first. - To find time saved, subtract the new time from the original time. - For the same job, divide the longer time by the shorter time to compare the work rates.

1. Find the original total time: \(165 + 75 = 240\) minutes. 2. Find the time saved: \(240 - 40 = 200\) minutes. 3. For the same amount of work, compare the times: \(240 \div 40 = 6\). The machine works at \(6\) times the manual rate.

a) The work originally took \(240\) minutes. b) The machine saves \(200\) minutes. c) The machine works at \(6\) times the manual rate.

- Find how much the temperature increased. - Use the temperature change to determine the vertical distance traveled. - An elevation becomes lower during a descent. - Subtract the descent from the summit elevation.

1. Find the temperature increase: \(-6 - (-14) = 8\). The temperature rose by \(8\,^{\circ}\text{C}\). 2. Each \(1\,^{\circ}\text{C}\) increase represents \(250\,\text{m}\) of descent, so the climber descended \(8 \times 250\,\text{m} = 2000\,\text{m}\). 3. Subtract the descent from the summit elevation: \(4100\,\text{m} - 2000\,\text{m} = 2100\,\text{m}\).

The shelter is at an elevation of \(2100\,\text{m}\).

- How many bookmarks does Class A finish before Class B joins? - How many bookmarks remain at \(10{:}00\,\text{a.m.}\)? - Add the two hourly rates. - Use the combined rate to find the remaining work time.

1. Class A works alone for \(2\) hours and makes \(2 \times 40 = 80\) bookmarks. 2. Find the number still needed: \(530 - 80 = 450\) bookmarks. 3. Find the combined rate after \(10{:}00\,\text{a.m.}\): \(40 + 50 = 90\) bookmarks per hour. 4. Find the time needed for the remaining bookmarks: \(450 \div 90 = 5\) hours. 5. Five hours after \(10{:}00\,\text{a.m.}\) is \(3{:}00\,\text{p.m.}\).

All the bookmarks will be finished at \(3{:}00\,\text{p.m.}\).

- Find the total number of passengers. - Divide each bus price by its number of seats to compare unit rates. - Test combinations with enough total seats. - Compare the costs of all valid combinations.

1. There are \(88+4=92\) people. 2. a) Bus A costs \(\$130\div20=\$6.50\) per seat. Bus B costs \(\$180\div30=\$6.00\) per seat. The claim is correct. 3. b) For each possible number of Bus B buses from \(0\) through \(4\), use the fewest Bus A buses that provide at least \(92\) seats: five Bus A buses cost \(\$650\); four Bus A buses and one Bus B bus cost \(\$700\); two buses of each type cost \(\$620\); one Bus A bus and three Bus B buses cost \(\$670\); and four Bus B buses cost \(\$720\). 4. The least expensive valid combination is two buses of each type for \(\$620\).

a) Yes. Bus B costs \(\$6.00\) per seat, compared with \(\$6.50\) for Bus A. b) Use \(2\) buses of each type. The total cost is \(\$620\).

- Divide each package price by its number of apples. - A lower unit rate does not automatically mean using only that package gives the least total cost. - Test combinations that reach at least \(115\) apples and compare their costs.

1. a) Bag A costs \(\$3.00\div15=\$0.20\) per apple. Bag B costs \(\$4.00\div25=\$0.16\) per apple, so Bag B has the lower unit rate. 2. b) For \(0,1,2,3,4\), or \(5\) Bag B packages, the least-cost valid combinations have total costs of \(\$24.00, \$22.00, \$23.00, \$21.00, \$19.00\), and \(\$20.00\), respectively. 3. The \(\$19.00\) combination uses \(4\) Bag B packages and \(1\) Bag A package. It gives \(4\times25+15=115\) apples. 4. Therefore, buy \(4\) Bag B packages and \(1\) Bag A package.

a) Bag B, at \(\$0.16\) per apple b) Buy \(4\) Bag B packages and \(1\) Bag A package for \(\$19.00\).

- Compare costs for the same number of sleeping spaces. - Check whether one tent type can provide exactly \(54\) spaces. - Compare any mixed arrangement with an arrangement using the lower-rate tent.

1. a) Three small tents provide \(12\) spaces and cost \(3\times\$15=\$45\). Two large tents also provide \(12\) spaces and cost \(2\times\$20=\$40\). Therefore, large tents have the lower cost per space. 2. b) Nine large tents provide exactly \(9\times6=54\) spaces and cost \(9\times\$20=\$180\). 3. Replacing two large tents with three small tents keeps the same number of spaces but increases the cost by \(\$5\). Other mixtures either cost more or require extra unused space. Therefore, nine large tents give the least cost.

a) Large tents have the lower cost per sleeping space. b) Rent \(9\) large tents for \(\$180\).

- First find the combined price for the known combined weight. - What is the price of \(1\,\text{lb}\) of soil? - Use the unit price to find each of the first two weights. - Use the same unit price for the third bag.

1. Find the combined cost of the first two bags: \(\$10.50 + \$12.00 = \$22.50\). 2. Express the total in cents: \(\$22.50 = 2250\) cents. 3. Find the price per pound: \(2250 \div 75 = 30\) cents per pound. 4. Find the first bag’s weight: \(1050 \div 30 = 35\,\text{lb}\). 5. Find the second bag’s weight: \(1200 \div 30 = 40\,\text{lb}\). 6. Find the third bag’s cost: \(50 \times 30 = 1500\) cents, or \(\$15.00\).

a) The first bag weighs \(35\,\text{lb}\), and the second bag weighs \(40\,\text{lb}\). b) The third bag costs \(\$15.00\).

- Use the difference in revenue and the difference in the number of bags to find one bag’s price. - Find the number of bags sold in each part of the day. - Add the numbers of bags. - Convert the total number of bags to a total weight.

1. Find the difference in sales revenue: \(\$264 - \$180 = \$84\). 2. Since the difference represents \(7\) bags, find the price per bag: \(\$84 \div 7 = \$12\). 3. Find the number of bags sold in the morning: \(\$180 \div \$12 = 15\) bags. 4. Find the number sold in the afternoon: \(\$264 \div \$12 = 22\) bags. 5. Find the total number of bags: \(15 + 22 = 37\). 6. Find the total weight: \(37 \times 10 = 370\,\text{lb}\).

The farm sells \(370\,\text{lb}\) of potatoes that day.

- First determine the distance each group walks. - Subtract Group B’s extra distance from the combined distance; the remainder can be split equally. - Use distance and speed to find each group’s time.

1. Find Group A’s distance by removing the \(4\,\text{mi}\) difference and dividing the remainder equally: \((46 - 4) \div 2 = 21\,\text{mi}\). 2. Find Group B’s distance: \(21 + 4 = 25\,\text{mi}\). 3. Find Group A’s time: \(21 \div 3 = 7\) hours. 4. Find Group B’s time: \(25 \div 5 = 5\) hours.

Group A walks for \(7\) hours, and Group B walks for \(5\) hours.

- First split the combined amount using the \(60\)-ton difference. - Then divide each truck's total by its tons-per-trip rate. - Check that the two totals add to \(1260\) tons.

1. Find Blue's total amount: \((1260\,\text{tons} - 60\,\text{tons}) \div 2 = 600\,\text{tons}\). 2. Find Yellow's total amount: \(600\,\text{tons} + 60\,\text{tons} = 660\,\text{tons}\). 3. Find Blue's trips: \(600\,\text{tons} \div 15\,\text{tons} = 40\). 4. Find Yellow's trips: \(660\,\text{tons} \div 22\,\text{tons} = 30\).

Blue makes \(40\) trips, and Yellow makes \(30\) trips.

- First find how much the gardener paid for the \(12\,\text{ft}\) piece. - Use that amount to find the price per foot. - Determine how many one-foot prices fit into the full roll’s value.

1. Find the value of the \(12\,\text{ft}\) piece sold: \(\$630 - \$450 = \$180\). 2. Find the price per foot: \(\$180 \div 12 = \$15\) per foot. 3. Find the original length: \(\$630 \div \$15 = 42\,\text{ft}\).

The roll originally held \(42\,\text{ft}\) of cable.

- Divide the distance by the time to find meters per second. - Convert \(10\,\text{km}\) to meters. - Convert the resulting seconds to minutes and seconds. - Distinguish a mathematical constant-rate model from realistic endurance.

1. The average speed is \(200\,\text{m} \div 25\,\text{s} = 8\,\text{m/s}\). 2. Convert the distance: \(10\,\text{km} = 10{,}000\,\text{m}\). At \(8\,\text{m/s}\), the time would be \(10{,}000 \div 8 = 1250\) seconds. 3. Convert \(1250\) seconds: \(1250 = 20 \times 60 + 50\), so the time is \(20\) minutes \(50\) seconds. This is less than \(25\) minutes, so the claim is mathematically correct under the constant-pace assumption. 4. In reality, a runner generally cannot maintain a short-sprint pace over \(10\,\text{km}\), so the actual time would be longer.

a) \(8\,\text{m/s}\) b) The claim is mathematically correct under the stated assumption; the calculated time is \(20\) minutes \(50\) seconds. c) A short-distance sprint pace is not sustainable over \(10\,\text{km}\).

- Find Hannah's total time first. - How long would Tom take without breaks? - How much total time is left for Tom's breaks? - Count how many \(45\)-minute moving intervals are completed before Tom finishes.

1. Hannah's total time is \(4 \times 12 + 92 + 5 \times 14 = 48 + 92 + 70 = 210\) minutes. 2. Tom's moving time is \(15 \times 10 = 150\) minutes. 3. Tom takes breaks after \(45\), \(90\), and \(135\) minutes of moving, so he takes \(3\) breaks before finishing. 4. His total break time is \(210 - 150 = 60\) minutes. 5. Each break lasts \(60 \div 3 = 20\) minutes.

Each of Tom's breaks lasts \(20\) minutes.

- Compare rates with the same units: seedlings per child per hour. - For each group, account for both the number of hours and the number of children. - Compare the two unit rates only after their units match.

1. Group A plants \(48\div3=16\) seedlings per hour, or \(16\div4=4\) seedlings per child per hour. 2. Group B plants \(42\div2=21\) seedlings per hour, or \(21\div3=7\) seedlings per child per hour. 3. The difference is \(7-4=3\) seedlings per child per hour.

One child in Group B plants an average of \(3\) more seedlings per hour than one child in Group A.

- Sketch the gate, fountain, and meeting point. - Add the distances traveled by both runners. - Their combined distance covers the gate-to-fountain distance twice.

1. Tom runs \(3\,\text{m/s} \times 50\,\text{s} = 150\,\text{m}\). 2. Sarah runs \(4\,\text{m/s} \times 50\,\text{s} = 200\,\text{m}\). 3. Together, their traveled distances equal one complete trip from the gate to the fountain and back: \(150 + 200 = 350\,\text{m}\). 4. The one-way distance is \(350 \div 2 = 175\,\text{m}\).

a) \(150\,\text{m}\) b) \(175\,\text{m}\)

- Determine how many groups of \(1000\,\text{ft}^2\) are in the lawn. - Calculate the minimum and maximum total seed amounts separately. - Remember that only whole bags can be purchased and every bag is fully used.

1. The lawn contains \(15\) groups of \(1000\,\text{ft}^2\). 2. The minimum seed amount is \(15 \times 2 = 30\,\text{lb}\), which requires \(30 \div 2 = 15\) bags. 3. The maximum recommended amount is \(15 \times 3 = 45\,\text{lb}\). 4. Twenty-two bags contain \(44\,\text{lb}\), which is within the recommendation. Twenty-three bags contain \(46\,\text{lb}\), which exceeds it. Therefore, the maximum is \(22\) bags.

She can buy at least \(15\) bags and at most \(22\) bags.

- Find the area of the rectangular field. - Determine how many \(1000\)-square-foot groups are in the field. - Calculate the required fertilizer weight. - Round the number of bags up because only whole bags can be purchased.

1. Find the field area: \(315 \times 200 = 63{,}000\,\text{ft}^2\). 2. There are \(63{,}000 \div 1000 = 63\) groups of \(1000\,\text{ft}^2\). 3. The required fertilizer amount is \(63 \times 4 = 252\,\text{lb}\). 4. Since \(252 \div 25 = 10\) remainder \(2\), the buyer needs \(11\) whole bags. 5. The total cost is \(11 \times \$32 = \$352\).

The fertilizer costs \(\$352\).

- Convert the per-acre payment to a per-square-foot rate. - Use the total payment to find the strip's area. - Divide the rectangular area by its width to find its length.

1. Find the payment per square foot: \(\$4356 \div 43{,}560 = \$0.10\) per square foot. 2. Find the strip area: \(\$140 \div \$0.10 = 1400\,\text{ft}^2\). 3. Divide area by width to find length: \(1400\,\text{ft}^2 \div 4\,\text{ft} = 350\,\text{ft}\).

The field-edge strip is \(350\,\text{ft}\) long.

- Express each filling time as a fraction of the aquarium filled per minute. - The net rate equals the pump rate minus the drain rate. - After finding the drain rate, use its reciprocal to find the drain time.

1. The pump fills \(\frac{1}{20}\) of the aquarium per minute. 2. With the drain open, the net filling rate is \(\frac{1}{50}\) of the aquarium per minute. 3. Let \(r\) be the drain rate. Write \(\frac{1}{20} - r = \frac{1}{50}\). 4. Solve: \(r = \frac{1}{20} - \frac{1}{50} = \frac{3}{100}\) of the aquarium per minute. 5. The drain time is the reciprocal, \(\frac{100}{3}\) minutes, or \(33\frac{1}{3}\) minutes. 6. One-third of a minute is \(20\) seconds, so the time is \(33\) minutes \(20\) seconds.

The drain would empty the aquarium in \(33\) minutes \(20\) seconds.

- Count the intervals between point \(A\) and the tick labeled \(2\). - For each case, divide the value difference by that number of intervals. - Use the interval value to determine how far and in which direction to move from \(A\) to \(0\).

1. There are \(8\) intervals from tick \(12\) to tick \(20\). 2. For a), the values from \(0.4\) to \(2\) span \(1.6\), so each interval is \(0.2\). Two intervals left of \(0.4\) reaches \(0\), so \(0\) is at tick \(10\). 3. For b), the values from \(1.2\) to \(2\) span \(0.8\), so each interval is \(0.1\). Twelve intervals left of \(1.2\) reaches \(0\), so \(0\) is at tick \(0\). 4. For c), moving from \(-2\) to \(2\) spans \(4\) units across the same \(8\) intervals, so each interval is \(0.5\). Four intervals right of \(-2\) reaches \(0\), so \(0\) is at tick \(16\).

a) Tick \(10\) b) Tick \(0\) c) Tick \(16\)

- Think about the boat and fender from the point of view of the moving water. - How long does the boat take to return to the same parcel of water? - Find the fender's total drifting time. - Use rate equals distance divided by time.

1. View the motion relative to the moving water. In this frame, the fender stays in one place while the boat travels away from it for \(15\) minutes. 2. Because the boat returns at the same speed relative to the water, it takes another \(15\) minutes to reach the fender. 3. The fender therefore drifts for a total of \(30\) minutes, or \(0.5\) hour. 4. Let \(r\) be the current's speed in miles per hour. The fender travels with the current, so \(0.5r = 0.75\). 5. Divide by \(0.5\): \(r = 1.5\).

The river current flows at \(1.5\,\text{mph}\).