- 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.

- 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.

- 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) \(300\,\text{L}\)

- 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.

- 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.

- 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

- 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.

- 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.

- 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

- Both vehicles travel for the same length of time. - Think about how far each vehicle goes in four hours. - How many extra miles does the faster vehicle gain each hour?

1. The car travels \(55\times4=220\,\text{mi}\). 2. The motorcycle travels \(62\times4=248\,\text{mi}\). 3. Their gap is \(248-220=28\,\text{mi}\). 4. Check using the difference in speeds: \((62-55)\times4=7\times4=28\,\text{mi}\).

\(28\,\text{mi}\)

- 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.

- Convert the speeds to the same time unit. - Use each speed to calculate the distance traveled in four minutes. - Compare the distances, not the numerical speed labels alone.

1. Sarah runs \(3\times60=180\,\text{m}\) per minute. 2. Sarah is faster since \(180>150\). 3. In four minutes Sarah covers \(180\times4=720\,\text{m}\), and Liam covers \(150\times4=600\,\text{m}\). 4. Sarah is \(720-600=120\,\text{m}\) ahead.

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

- 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.

- 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.

- 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 \(\$11{,}250.50\).

- Does a floating leaf move at the same speed as the kayak when seen from shore? - Compare how far both travel during the same amount of time. - Is there any part of their motion that both share?

1. The kayak travels downstream at \(5+2=7\,\text{mi/h}\) relative to the shore. The leaf travels at \(2\,\text{mi/h}\). 2. Thirty minutes is \(0.5\) hour. The kayak covers \(7\times0.5=3.5\) miles, and the leaf covers \(2\times0.5=1\) mile. 3. The kayak is \(3.5-1=2.5\) miles ahead. The shared current adds the same motion to both.

\(2.5\) miles. In half an hour, the kayak covers \(3.5\) miles along the shore while the leaf covers \(1\) mile.

- 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.

- 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 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 \(15\,\text{mph}\).

- 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 \(3{:}00\,\text{p.m.}\).

- 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) \(175\,\text{m}\)

- 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.

- What information would let you calculate the ring's downstream speed? - Which facts about the boat's motion stay the same when it turns around? - How could you test whether a possible total drifting time fits the boat catching the ring?

1. Imagine watching from water that moves with the current. The ring stays in place in this view. 2. The boat travels away from the ring for \(15\) minutes and returns at the same speed through the water, so its return takes \(15\) minutes too. 3. The ring drifts for \(30\) minutes, or \(0.5\) hour. 4. The current carries it \(0.75\) mile, so its speed is \(0.75\div0.5=1.5\,\text{mi/h}\).

\(1.5\,\text{mi/h}\)