- Compare what happens in one minute with the next minute. - Does a total over three minutes tell you what happened in each minute? - Think about what “steady” means.

1. Theo packs the same number in each one-minute period, so his rate is steady. 2. Lena's one-minute counts change from \(4\) to \(2\) to \(6\), so her rate is not steady. 3. The students can have the same total and still have different patterns over time.

Theo has a steady rate because he packs \(4\) bookmarks each minute. Lena does not: her counts change between one-minute periods.

- Can you compare distance traveled over the same length of time? - Would the total distance alone show whether the speed changed?

1. Equal times would give equal distances at a steady speed. 2. She goes \(2\), \(2\), and \(3\) miles in equal periods, so the speed is not steady. 3. The total distance is \(2+2+3=7\) miles.

No; she covers different distances in equal times. She travels \(7\) miles in all.

- Read the distances at the start, after two hours, and after four hours. - How far did the cyclist travel during each of those time intervals? - Equal lengths of time do not always mean equal distances.

1. The cyclist goes from \(0\) to \(4\) miles in the first \(2\) hours, so the speed is \(4\div2=2\,\text{mi/h}\). 2. From hour \(2\) to hour \(4\), the distance grows from \(4\) to \(6\) miles. The speed is \((6-4)\div2=1\,\text{mi/h}\). 3. The rates differ, so the speed is not constant.

First \(2\) hours: \(2\,\text{mi/h}\); next \(2\) hours: \(1\,\text{mi/h}\). No, the speeds differ.

- Which distance and time belong to the entire trip? - What does an average say about a trip with changing speeds? - Compare the distance for each one-hour part.

1. Average speed is total distance divided by total time: \(60\div3=20\,\text{mi/h}\). 2. The three hourly distances are different, so they did not drive at a steady \(20\,\text{mi/h}\) throughout.

a) \(20\,\text{mi/h}\) b) No. Their hourly distances were \(15\), \(25\), and \(20\) miles.

- 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 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 rate for the first part using its distance and time. - The speed for the whole hike comes from the total distance and total time, not the average of two speed numbers. - In part c), which part of the total time can change?

1. The average speed to the shelter is \(6\div1.5=4\,\text{mi/h}\). 2. The return takes \(9\div4.5=2\) hours. 3. The total trip lasts \(1.5+2=3.5\) hours, so its average speed is \(15\div3.5\approx4.29\,\text{mi/h}\). 4. If the return takes \(t\) hours, the whole hike takes \(1.5+t\) hours. The average speed is \(\frac{15}{1.5+t}\,\text{mi/h}\).

a) \(4\,\text{mi/h}\) b) \(2\) hours; approximately \(4.29\,\text{mi/h}\) c) \(\frac{15}{1.5+t}\,\text{mi/h}\)

- 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.

- 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.

- 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.

- 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.

- 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 \(55\,\text{mi/h}\).

- 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.

- 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.

- 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.

- The submarine travels the full starting depth to reach the surface. - Find how long the second stage lasts and how far it goes. - Think about what distance must be left for the first stage.

1. The second part lasts \(30\) minutes. 2. The submarine rises \(30\times25=750\,\text{m}\) in that part. 3. The remaining distance is \(1200-750=450\,\text{m}\), covered in the first part.

\(450\,\text{m}\)

- 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 \(5\,\text{km/h}\).

- 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.

- 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 \(650\,\text{mi}\).

- 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.

- 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 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.

- 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.

- Mark the time when each ride or break starts or ends. - Distance stays the same during a break, even though time passes. - Find which ride contains the target distance in part 3). - Average speed for a trip counts its full elapsed time.

1. At \(9{:}00\) a.m. the distance is \(0\). At \(10{:}00\) a.m. it is \(12\) miles. Another \(1.5\times8=12\) miles brings the total to \(24\) miles at \(11{:}30\) a.m. The break ends at noon with no change. The last \(0.5\) hour adds \(0.5\times16=8\) miles, for \(32\) miles at \(12{:}30\) p.m. 2. At \(11{:}00\) a.m. she has traveled \(12+8=20\) miles. 3. After \(10{:}00\) a.m., she needs \(16-12=4\) miles at \(8\,\text{mph}\), taking half an hour. The time is \(10{:}30\) a.m. 4. The trip lasts \(3.5\) hours including the break. The average speed is \(32\div3.5\approx9.1\,\text{mph}\).

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 \(9.1\,\text{mph}\)