- Notice that the two given distances use different units. - Choose the unit requested for the answer before setting up the operation. - “Remaining” describes the whole amount with one known part removed.

1. The answer is needed in meters, so convert \(1\,\text{km}\) to \(1000\,\text{m}\). 2. The remaining distance is the whole route minus the distance already walked, so use \(1000-350\).

Convert \(1\,\text{km}\) to \(1000\,\text{m}\), then use \(1000-350\).

- The requested unit is milliliters, so rewrite the total volume in that unit first. - Equal amounts in several bottles call for equal sharing. - Write the operation after the units are consistent; no final quotient is needed.

1. Convert \(3\,\text{L}\) to \(3000\,\text{mL}\). 2. Equal sharing among \(6\) bottles means divide the total volume by \(6\), so the amount per bottle is represented by \((3000\div6)\,\text{mL}\).

\((3000\div6)\,\text{mL}\)

- How many ounces are in \(1\) pound? - For each package size, decide how many equal packages make the target weight. - You can use repeated addition or division.

1. Convert the target weight: \(1\,\text{lb} = 16\,\text{oz}\). 2. For chocolate bars, calculate \(16 \div 2 = 8\). She needs \(8\) bars. 3. For yogurt cups, calculate \(16 \div 4 = 4\). She needs \(4\) cups. 4. For bags of oats, calculate \(16 \div 8 = 2\). She needs \(2\) bags.

a) \(8\) chocolate bars b) \(4\) yogurt cups c) \(2\) bags of oats

- First determine how many kilometers Lisa walks in the complete day. - Then express that total in meters before comparing it with her claim. - An explanation should identify both the correct converted distance and the mismatch.

1. Lisa walks the \(2\,\text{km}\) route twice, so her total distance is \(4\,\text{km}\). 2. Convert to meters: \(4\,\text{km}=4000\,\text{m}\). 3. Since \(4000\,\text{m}\ne4400\,\text{m}\), Lisa is not correct.

No. Lisa walks \(4\,\text{km}=4000\,\text{m}\), not \(4400\,\text{m}\).

- Convert the liter to milliliters. - Divide the total volume equally among \(8\) cups. - Check by multiplying the cup volume by \(8\).

1. Convert the total volume: \(1\,\text{L} = 1000\,\text{mL}\). 2. Divide the volume equally among \(8\) cups: \(1000\,\text{mL} \div 8 = 125\,\text{mL}\).

Each cup holds \(125\,\text{mL}\).

- How many ounces are in \(1\) pound? - What fraction of a pound is each smaller weight? - The price changes in the same proportion as the weight.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(8\,\text{oz}\) is half a pound. Its price is \(\$4.80 \div 2 = \$2.40\). 2. Four ounces is half of \(8\,\text{oz}\), so its price is \(\$2.40 \div 2 = \$1.20\). 3. Two ounces is half of \(4\,\text{oz}\), so its price is \(\$1.20 \div 2 = \$0.60\).

\(8\,\text{oz}\) cost \(\$2.40\). \(4\,\text{oz}\) cost \(\$1.20\). \(2\,\text{oz}\) cost \(\$0.60\).

- Convert \(1\,\text{kg}\) to grams. - Use the stated \(125\)-gram estimate for each lemon. - Determine how many equal groups make the target mass.

1. Convert the target mass: \(1\,\text{kg}=1000\,\text{g}\). 2. Use multiplication to find how many groups of \(125\,\text{g}\) make \(1000\,\text{g}\): \(8\times125\,\text{g}=1000\,\text{g}\). 3. Using the stated estimate, about \(8\) lemons have a total mass of \(1\,\text{kg}\).

About \(8\) lemons.

- Determine how many \(250\)-milliliter glasses make \(1\) liter. - The bottle holds \(2\) liters. - Double the number of glasses in one liter.

1. Four \(250\,\text{mL}\) glasses hold \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. A \(2\)-liter bottle fills twice as many glasses: \(2 \times 4 = 8\).

\(8\) glasses can be filled completely.

- Determine how many \(200\)-milliliter cups make \(1\) liter. - Find how many groups of that size are in \(15\) cups. - Each group represents \(1\) liter.

1. Five \(200\,\text{mL}\) cups hold \(5 \times 200\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. Fifteen cups make \(15 \div 5 = 3\) groups of \(5\) cups, so the students serve \(3\,\text{L}\).

The students serve \(3\,\text{L}\) of tea altogether.

- How many meters are in \(1\,\text{km}\)? - Convert the trail length to the smaller unit before dividing. - How many times does each divisor fit into \(1000\)?

1. Convert the trail length: \(1\,\text{km}=1000\,\text{m}\). 2. Divide \(1000\,\text{m}\) by the number of equal sections: a) \(1000 \div 2=500\) b) \(1000 \div 4=250\) c) \(1000 \div 5=200\) d) \(1000 \div 8=125\) e) \(1000 \div 10=100\)

a) \(500\,\text{m}\) b) \(250\,\text{m}\) c) \(200\,\text{m}\) d) \(125\,\text{m}\) e) \(100\,\text{m}\)

- How many minutes are in \(2\) hours? - To divide by \(4\), you can divide by \(2\) twice. - After converting the total time, divide by the number of equal sessions.

1. Convert the total time: \(2\,\text{hr}=120\,\text{min}\), because \(2 \times 60=120\). 2. Divide the total minutes by the number of equal sessions: a) \(120 \div 3=40\) b) \(120 \div 4=30\) c) \(120 \div 5=24\) d) \(120 \div 6=20\) e) \(120 \div 8=15\)

a) \(40\,\text{min}\) b) \(30\,\text{min}\) c) \(24\,\text{min}\) d) \(20\,\text{min}\) e) \(15\,\text{min}\)

- Convert the pitcher's volume to milliliters. - Use multiplication to determine how many \(125\)-milliliter cups make the total. - Use approximation language in your answer.

1. Convert the pitcher's volume: \(1\,\text{L} = 1000\,\text{mL}\). 2. Compare the approximate cup volume with the total: \(8 \times 125\,\text{mL} = 1000\,\text{mL}\). 3. Because the cup volume is approximate, the result is also approximate.

About \(8\) cups can be poured.

- Put both lengths in the same unit before calculating. - How many centimeters are in \(1\) meter? - Determine how many groups of \(3\,\text{cm}\) fit into the total length.

1. Convert the total length to centimeters: \(1.50\,\text{m} = 150\,\text{cm}\). 2. Divide the total length by the length of one paper clip: \(150 \div 3 = 50\).

The line contains \(50\) paper clips.

- How many ounces are in \(1\) pound? - What fraction of a pound is \(8\,\text{oz}\)? - A half-pound costs half as much as a full pound at the same rate.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(8\,\text{oz}\) is half a pound. Divide each price by \(2\). 2. Zucchini: \(\$1.80 \div 2 = \$0.90\). 3. Bell peppers: \(\$3.40 \div 2 = \$1.70\). 4. Carrots: \(\$1.20 \div 2 = \$0.60\). 5. Eggplant: \(\$2.60 \div 2 = \$1.30\).

Zucchini: \(\$0.90\) Bell peppers: \(\$1.70\) Carrots: \(\$0.60\) Eggplant: \(\$1.30\)

- How many \(4\)-ounce portions are in \(1\) pound? - Once you know the price of \(4\,\text{oz}\), scale it to other amounts. - Half a pound is \(8\,\text{oz}\).

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(4\,\text{oz}\) is one fourth of a pound. The price is \(\$9.00 \div 4 = \$2.25\). 2. Twelve ounces is three \(4\)-ounce portions, so \(3 \times \$2.25 = \$6.75\). 3. Half a pound is \(8\,\text{oz}\), so the price is \(\$9.00 \div 2 = \$4.50\).

a) \(\$2.25\) b) \(\$6.75\) c) \(\$4.50\)

- Convert the ribbon to centimeters first. - Divide the total number of centimeters by \(4\). - Check whether the division has a remainder.

1. Convert the ribbon's length: \(3\,\text{m} = 300\,\text{cm}\). 2. Divide the total length equally among four children: \(300 \div 4 = 75\). 3. Each child receives \(75\,\text{cm}\), with no ribbon left over.

Each child receives \(75\,\text{cm}\) of ribbon, and there is no remainder.

- How many days are in one week? - Multiply the daily amount by the number of days. - Recall that \(1000\,\text{kg}\) equals \(1\) metric ton.

1. Estimate the weekly amount: \(150\,\text{kg} \times 7 \approx 1050\,\text{kg}\). 2. Since \(1000\,\text{kg}\) is \(1\) metric ton, the estimate is \(1\) metric ton \(50\,\text{kg}\).

The elephant eats about \(1050\,\text{kg}\), or about \(1\) metric ton \(50\,\text{kg}\), in one week.

- Convert \(4\) metric tons to kilograms. - Think of the number of newborn elephants as a missing factor in a multiplication equation. - Check the number you find by multiplying it by \(100\,\text{kg}\).

1. Convert the adult elephant's mass: \(4\) metric tons \(= 4000\,\text{kg}\). 2. Find how many \(100\)-kilogram groups make \(4000\,\text{kg}\). Since \(40 \times 100 = 4000\), about \(40\) newborn elephants have the same total mass.

\(40\) newborn elephants have about the same total mass as one adult elephant.

- How much length does one unit interval use? - How many unit intervals are needed to reach \(2500\)? - Use the relationship between \(1000\,\text{mm}\) and \(1\,\text{m}\) to express the final length in meters.

1. The number line needs \(2500\) unit intervals, so its length is \(2500 \times 4\,\text{mm} = 10{,}000\,\text{mm}\). 2. Since \(10\times1000\,\text{mm}=10{,}000\,\text{mm}\), the length is \(10\,\text{m}\).

The number line must be at least \(10\,\text{m}\) long.

- How many centimeters are in one meter? - Use the same unit for the length and the difference. - Does 'shorter than' mean adding or subtracting here?

1. Convert Lucas's rope length: \(2\,\text{m}=200\,\text{cm}\). 2. Subtract: \(200-45=155\,\text{cm}\).

Tim's rope is \(155\,\text{cm}\) long.

- How many fluid ounces are in \(1\) quart? - How many times does the smaller capacity fit into the larger capacity? - Use division to find the number of equal cupfuls.

1. Convert the bowl's capacity: \(1\,\text{qt} = 32\,\text{fl oz}\). 2. Divide by the capacity of one measuring cup: \(32\,\text{fl oz} \div 2\,\text{fl oz} = 16\). 3. Therefore, \(16\) full measuring cups are needed.

\(16\) full measuring cups are needed.

- How far does the scooter move during one wheel rotation? - Multiply that distance by the number of rotations. - Use \(100\,\text{cm} = 1\,\text{m}\) to convert the result.

1. Find the total distance in centimeters: \(9 \times 80\,\text{cm} = 720\,\text{cm}\). 2. Since \(100\,\text{cm} = 1\,\text{m}\), \(720\,\text{cm} = 7\,\text{m}\ 20\,\text{cm}\).

The scooter travels \(720\,\text{cm}\), or \(7\,\text{m}\ 20\,\text{cm}\).

- Express the rolling distance and both circumferences in centimeters. - Think of each rotation as one circumference of travel. - Use a missing-factor multiplication equation for each wheel.

1. Convert the total distance: \(4\,\text{m}=400\,\text{cm}\). 2. For the small wheel, find the number of \(20\)-centimeter groups in \(400\,\text{cm}\). Since \(20 \times 20\,\text{cm}=400\,\text{cm}\), it makes \(20\) rotations. 3. For the large wheel, since \(10 \times 40\,\text{cm}=400\,\text{cm}\), it makes \(10\) rotations.

The small wheel makes \(20\) rotations, and the large wheel makes \(10\) rotations.

- Convert the \(2\)-meter target length to centimeters. - Think of the number of stones as the missing factor in a multiplication equation. - For part b), multiply the length of one stone by the number of stones.

1. Convert the target length: \(2\,\text{m}=200\,\text{cm}\). 2. For part a), \(10 \times 20\,\text{cm}=200\,\text{cm}\), so \(10\) stones are needed. 3. For part b), \(5 \times 20\,\text{cm}=100\,\text{cm}\). 4. Since \(100\,\text{cm}=1\,\text{m}\), the shorter row is \(1\,\text{m}\) long.

a) \(10\) stones b) \(100\,\text{cm}\), or \(1\,\text{m}\)

- How many grams are in \(1\) kilogram? - Convert the total weight to grams first. - What must you subtract to find the weight of only the apples?

1. Convert the total weight: \(1\,\text{kg} = 1000\,\text{g}\). 2. Subtract the empty crate's weight: \(1000\,\text{g} - 245\,\text{g} = 755\,\text{g}\).

The apples weigh \(755\,\text{g}\).

- Convert both distances to meters. - Compare only after using the same unit. - Subtract to find how much longer one hike is.

1. Convert Ms. Lee's class's distance: \(1\,\text{km}\ 200\,\text{m}=1200\,\text{m}\). 2. Since \(1500>1200\), Mr. Ortiz's class hikes farther. 3. The difference is \(1500-1200=300\,\text{m}\).

Mr. Ortiz's class hikes farther, by \(300\,\text{m}\).

- Convert both movie lengths to the same unit. - How many minutes are in \(1\) hour? - Subtract the shorter time from the longer time.

1. Convert Movie A's length to minutes: \(1\,\text{hour}\,35\,\text{minutes} = 60 + 35 = 95\) minutes. 2. Compare the lengths: \(110 > 95\), so Movie B is longer. 3. Find the difference: \(110 - 95 = 15\) minutes.

Movie B is longer by \(15\) minutes.

- The starting length and the cut piece need to be expressed in the same unit. - After the units match, decide which operation gives the amount left. - The remaining piece should be shorter than the original rope.

1. Convert the starting length: \(2\,\text{m} = 200\,\text{cm}\). 2. Subtract: \(200 - 65 = 135\).

\(135\,\text{cm}\)

- Put the entire workshop duration into one unit before dividing it. - The three sessions have equal lengths. - Check that three copies of your session length reproduce the full duration.

1. Convert the full workshop time to minutes: \(60 + 30 = 90\) minutes. 2. Divide equally: \(90 \div 3 = 30\).

\(30\) minutes

- Look at the unit attached to each number before deciding what can be added directly. - Express both lengths in one common unit. - After adding, decide whether the result can be regrouped into meters and centimeters.

1. Kai treated \(175\,\text{cm}\) as if it were \(175\,\text{m}\), so he added quantities with different units as though the units matched. 2. Convert \(2\,\text{m}\,40\,\text{cm}\) to centimeters: \(240\,\text{cm}\). 3. Add: \(240 + 175 = 415\,\text{cm}\). 4. Regroup: \(415\,\text{cm} = 4\,\text{m}\,15\,\text{cm}\).

Kai added unlike units as though they were the same. The correct total is \(4\,\text{m}\,15\,\text{cm}\).

- Find the total amount in all the bottles first. - Compare the bottle total and the cooler capacity using one common unit. - After deciding whether it fits, find the difference between the two amounts.

1. The eight bottles contain \(8 \times 600 = 4800\,\text{mL}\). 2. Convert the cooler capacity: \(5\,\text{L} = 5000\,\text{mL}\). 3. Since \(4800 < 5000\), all the liquid fits. 4. The unused capacity is \(5000 - 4800 = 200\,\text{mL}\).

Yes. All eight bottles' contents fit, with \(200\,\text{mL}\) of capacity remaining.

- Draw a path with the part already walked marked on it. - Could the whole and the finished part be shown on the same meter scale?

1. Two kilometers equal \(2000\,\text{m}\). 2. Subtract the distance already walked: \(2000-1500=500\,\text{m}\).

\(500\,\text{m}\)

- Change the first part of the walk into meters. - Add the two parts after the units match.

1. One kilometer is \(1000\,\text{m}\). 2. Add the last part: \(1000+200=1200\,\text{m}\).

\(1200\,\text{m}\)

- Express the amount in the can in milliliters. - Does using water make the amount increase or decrease?

1. Two liters is \(2000\,\text{mL}\). 2. Subtract the water used: \(2000-500=1500\,\text{mL}\).

\(1500\,\text{mL}\)

- Change the original string length to inches. - Equal pieces must all have the same length.

1. Two feet equal \(24\,\text{in}\). 2. Each piece is \(24\div3=8\,\text{in}\) long.

\(8\,\text{in}\)

- Add the lengths of both pieces first. - Compare with the number of inches in two feet.

1. Two pieces measure \(12+12=24\,\text{in}\). 2. Since one foot has twelve inches, two feet have twenty-four inches.

Yes. Both lengths are \(24\,\text{in}\).

- Change the apples' weight into ounces. - The grapes' weight is already in the requested unit.

1. The apples weigh \(2\times16=32\,\text{oz}\). 2. Add the grapes: \(32+8=40\,\text{oz}\).

\(40\,\text{oz}\)

- You know the end time, not the start time. - Work backward by the whole hour and then the extra minutes.

1. Go back one hour from \(5{:}00\) p.m. to \(4{:}00\) p.m. 2. Go back another twenty minutes to \(3{:}40\) p.m.

\(3{:}40\) p.m.

- Could two numbers with different units be compared as if they counted the same thing? - What would the lengths look like side by side on an inch ruler?

1. Three feet equal \(36\,\text{in}\); the other plank is \(30\,\text{in}\). 2. Their difference is \(36-30=6\,\text{in}\), so Mira is wrong: the first plank is longer.

No. The first plank is \(6\,\text{in}\) longer.

- Compare the backpack before and after the books were removed. - Express both weights in the same unit first. - Think about which part of the original weight is no longer included. - Check that the books weigh less than the full backpack.

1. Convert both weights to ounces: \(7\,\text{lb}\ 4\,\text{oz} = 116\,\text{oz}\) and \(4\,\text{lb}\ 12\,\text{oz} = 76\,\text{oz}\). 2. Find the difference: \(116\,\text{oz} - 76\,\text{oz} = 40\,\text{oz}\). 3. Convert the difference to pounds and ounces: \(40\,\text{oz} = 2\,\text{lb}\ 8\,\text{oz}\).

\(2\,\text{lb}\ 8\,\text{oz}\)

- Picture the scarf as one long segment. Which part is finished, and which part is still missing? - First express both lengths in the same unit. - Decide whether centimeters or meters will give the clearest answer. - Check that the finished part and the missing part add to the full scarf length.

1. Convert the total length to centimeters: \(2\,\text{m} = 200\,\text{cm}\). 2. Subtract the finished length from the total length: \(200\,\text{cm} - 78\,\text{cm} = 122\,\text{cm}\).

\(122\,\text{cm}\)

- Recall how many milliliters equal one liter. - Determine how many \(200\)-milliliter cups make \(1\) liter. - Multiply the cups per liter by \(40\).

1. Convert the aquarium's capacity: \(40\,\text{L} = 40{,}000\,\text{mL}\). 2. Five \(200\,\text{mL}\) cups hold \(5 \times 200\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 3. Multiply the \(5\) cups needed per liter by \(40\) liters: \(5 \times 40 = 200\) cups.

Maya needs \(200\) full cups of water.

- Express both vacation lengths in days. - How many days are in \(3\) weeks? - Subtract to find how much longer one vacation is.

1. Convert Anton's vacation to days: \(3 \times 7 = 21\) days. 2. Compare the vacations: Lena has \(25\) days, and Anton has \(21\) days. 3. Find the difference: \(25 - 21 = 4\) days.

Lena has the longer vacation. It is \(4\) days longer than Anton's vacation.

- How many centimeters are in one meter? - Find the height of the stack and the estimated height of the building in the same unit. - Look for how many \(50\)-centimeter steps make \(1000\) centimeters.

1. Using the estimated book thickness, the stack height is \(1000\times1\,\text{cm}\approx1000\,\text{cm}=10\,\text{m}\). 2. Using the estimated story height, the building height is \(3\times3\,\text{m}\approx9\,\text{m}\). The book stack would be about \(1\,\text{m}\) taller. 3. For the step estimate, \(20\times50\,\text{cm}=1000\,\text{cm}\), so the distance would take about \(20\) steps. 4. The distance is exactly \(1000\,\text{cm}=10\,\text{m}\).

a) Yes. The stack would be about \(10\,\text{m}\) high, which is about \(1\,\text{m}\) taller than the building. b) About \(20\) steps. The distance is \(10\,\text{m}\).

- A common unit makes the additions and subtractions easier to manage. - After finding each height in inches, regroup it into feet and inches for the requested answer. - Check whether either calculation crosses a \(12\)-inch boundary.

1. Convert Leo's height to inches: \(4\times12+8=56\,\text{in}\). 2. Mia is \(56-11=45\,\text{in}\), which is \(3\,\text{ft}\ 9\,\text{in}\). 3. Felix is \(45+16=61\,\text{in}\), which is \(5\,\text{ft}\ 1\,\text{in}\). 4. Since \(61>56>45\), Felix is tallest.

Mia is \(3\,\text{ft}\ 9\,\text{in}\) tall. Felix is \(5\,\text{ft}\ 1\,\text{in}\) tall. Felix is tallest.

- Standing on the books makes Jonah taller. Add the height of the books first. - He is still shorter than Lucas, so account for the remaining difference. - Convert the final number of inches to feet and inches.

1. Find Jonah's height while he is standing on the books: \(52\,\text{in} + 6\,\text{in} = 58\,\text{in}\). 2. Add the remaining difference: \(58\,\text{in} + 2\,\text{in} = 60\,\text{in}\). 3. Convert inches to feet and inches: \(60\,\text{in} = 5\,\text{ft}\ 0\,\text{in}\).

Lucas is \(5\,\text{ft}\ 0\,\text{in}\) tall.

- Remember that \(1\,\text{kg}=1000\,\text{g}\). - Find each total mass in grams first. - For the missing number of peaches, look for a multiplication fact with \(200\) and \(1000\).

1. a) \(8\times125\,\text{g}=1000\,\text{g}\). Since \(1000\,\text{g}=1\,\text{kg}\), the apples weigh \(1\,\text{kg}\). 2. b) Convert \(1\,\text{kg}=1000\,\text{g}\). Since \(5\times200\,\text{g}=1000\,\text{g}\), \(5\) peaches weigh \(1\,\text{kg}\). 3. c) The apples weigh \(6\times125\,\text{g}=750\,\text{g}\), and the peaches weigh \(4\times200\,\text{g}=800\,\text{g}\). Since \(800\,\text{g}>750\,\text{g}\), the \(4\) peaches are heavier.

a) \(1000\,\text{g}=1\,\text{kg}\) b) \(5\) peaches c) The \(4\) peaches are heavier.

- Add the weights of all the packed items first. - How much more is needed to reach \(1000\,\text{g}\)? - Add the hundreds, tens, and ones carefully.

1. Add the item weights: \(450\,\text{g} + 320\,\text{g} + 180\,\text{g}\). 2. Calculate the total: \(450 + 320 = 770\), and \(770 + 180 = 950\). The items weigh \(950\,\text{g}\). 3. Convert \(1\,\text{kg} = 1000\,\text{g}\), then find the difference: \(1000\,\text{g} - 950\,\text{g} = 50\,\text{g}\).

The items weigh \(950\,\text{g}\) altogether. Another \(50\,\text{g}\) would make exactly \(1\,\text{kg}\).

- How many grams are in \(1\) kilogram? - Express every weight in grams before adding. - Convert the final total back to kilograms and grams.

1. Convert the empty backpack's weight to grams: \(1\,\text{kg}\ 200\,\text{g} = 1200\,\text{g}\). 2. Add all the weights: \(1200\,\text{g} + 1500\,\text{g} + 350\,\text{g} + 750\,\text{g} = 3800\,\text{g}\). 3. Convert the total to kilograms and grams: \(3800\,\text{g} = 3\,\text{kg}\ 800\,\text{g}\).

The packed backpack weighs \(3\,\text{kg}\ 800\,\text{g}\).

- What does “too heavy” mean when you compare two weights? - Express the \(3\,\text{kg}\) limit in grams. - Add the weights of the backpack and all its contents first.

1. Find the packed weight: \(1100\,\text{g} + 1600\,\text{g} + 800\,\text{g} = 3500\,\text{g}\). 2. Convert the limit to grams: \(3\,\text{kg} = 3000\,\text{g}\). 3. Compare the weights: \(3500\,\text{g} > 3000\,\text{g}\). 4. Find how far the backpack is over the limit: \(3500\,\text{g} - 3000\,\text{g} = 500\,\text{g}\).

Yes. The backpack weighs \(3500\,\text{g}\), or \(3\,\text{kg}\ 500\,\text{g}\), so it is \(500\,\text{g}\) over the limit.

- Convert the full backpack weight to grams first. - Add the weights of all the items whose weights are known. - What operation finds a missing part when you know the whole and the other parts?

1. Convert the total weight to grams: \(4\,\text{kg} = 4000\,\text{g}\). 2. Convert the textbook weight to grams: \(1\,\text{kg}\ 300\,\text{g} = 1300\,\text{g}\). 3. Add the known parts: \(1250\,\text{g} + 1300\,\text{g} + 300\,\text{g} = 2850\,\text{g}\). 4. Subtract the known weight from the total: \(4000\,\text{g} - 2850\,\text{g} = 1150\,\text{g}\).

The water bottle weighs \(1150\,\text{g}\).

- Convert all masses to grams. - One kilogram equals \(1000\) grams. - After ordering the masses, subtract to find the difference.

1. Convert each mass to grams: Max has \(3250\,\text{g}\), Mia has \(2900\,\text{g}\), and Leo has \(3040\,\text{g}\). 2. Order the masses: \(2900 < 3040 < 3250\). 3. Find the difference between Max's and Mia's backpacks: \(3250 - 2900 = 350\,\text{g}\).

a) Mia, Leo, Max b) Max's backpack is \(350\,\text{g}\) heavier than Mia's.

- First find the new total weight. - How many grams are in \(4\,\text{kg}\)? - Compare the new total with the limit.

1. Find the new weight: \(3580\,\text{g} + 470\,\text{g} = 4050\,\text{g}\). 2. Convert the limit: \(4\,\text{kg} = 4000\,\text{g}\). Since \(4050\,\text{g} > 4000\,\text{g}\), the bag is above the limit. 3. Find the difference: \(4050\,\text{g} - 4000\,\text{g} = 50\,\text{g}\).

a) The bag weighs \(4050\,\text{g}\). b) It is above the limit by \(50\,\text{g}\).

- Convert the full bag weight to grams first. - Remember that Sarah removes two books. - Convert the final answer back to kilograms and grams.

1. Convert the bag's weight to grams: \(4\,\text{kg}\ 250\,\text{g} = 4250\,\text{g}\). 2. Find the combined weight of the books: \(850\,\text{g} + 850\,\text{g} = 1700\,\text{g}\). 3. Subtract the books' weight: \(4250\,\text{g} - 1700\,\text{g} = 2550\,\text{g}\). 4. Convert back to kilograms and grams: \(2550\,\text{g} = 2\,\text{kg}\ 550\,\text{g}\).

The school bag weighs \(2\,\text{kg}\ 550\,\text{g}\).

- Add the backpack and all its contents first. - Remember that \(16\,\text{oz} = 1\,\text{lb}\). - Then add the packed backpack's weight to Lucas's weight.

1. Add the backpack and its contents: \(2\,\text{lb}\,10\,\text{oz} + 6\,\text{lb}\,3\,\text{oz} + 1\,\text{lb}\,3\,\text{oz} = 9\,\text{lb}\,16\,\text{oz}\). 2. Convert \(16\,\text{oz}\) to \(1\,\text{lb}\), so the packed backpack weighs \(10\,\text{lb}\). 3. Add Lucas's weight: \(75\,\text{lb} + 10\,\text{lb} = 85\,\text{lb}\).

Lucas and the packed backpack weigh \(85\,\text{lb}\) altogether.

- Determine how many one-way trips each child walks per day. - Convert all distances to meters. - Find each child's distance for one day and then for five days.

1. Emma walks \(750 + 750 = 1500\,\text{m}\) each day. 2. In five days, Emma walks \(5 \times 1500 = 7500\,\text{m}\). 3. Convert Leo's one-way distance: \(1\,\text{km}\,200\,\text{m} = 1200\,\text{m}\). 4. In five days, Leo walks \(5 \times 1200 = 6000\,\text{m}\). 5. Since \(7500 > 6000\), Emma walks farther.

Emma walks farther. Emma walks \(7500\,\text{m}\), and Leo walks \(6000\,\text{m}\).

- Find half of the bottle's starting amount. - Keep track of both the amount consumed and the amount remaining. - Convert \(1\,\text{L}\) to milliliters before comparing.

1. Find the amount Sofia drinks during the first break: \(400\,\text{mL} \div 2 = 200\,\text{mL}\). 2. Find the amount remaining after the first break: \(400\,\text{mL} - 200\,\text{mL} = 200\,\text{mL}\). 3. Find the amount remaining after the second break: \(200\,\text{mL} - 150\,\text{mL} = 50\,\text{mL}\). 4. Find the amount she has consumed: \(200\,\text{mL} + 150\,\text{mL} = 350\,\text{mL}\). 5. Convert the daily goal: \(1\,\text{L} = 1000\,\text{mL}\). 6. Find the additional amount needed: \(1000\,\text{mL} - 350\,\text{mL} = 650\,\text{mL}\).

1. \(50\,\text{mL}\) remain in the bottle. 2. Sofia must drink \(650\,\text{mL}\) more.

- Convert the mixed kilogram-and-gram amount to grams. - Look for two item weights that add to the target amount. - There is more than one correct combination.

1. Convert the target reduction: \(1\,\text{kg}\ 500\,\text{g} = 1500\,\text{g}\). 2. One combination is \(1200\,\text{g} + 300\,\text{g} = 1500\,\text{g}\), so Lucas can remove the atlas and pencil case. 3. Another combination is \(800\,\text{g} + 700\,\text{g} = 1500\,\text{g}\), so he can remove the book and water bottle.

Solution 1: Remove the atlas and pencil case. Solution 2: Remove the book and water bottle.

- Convert \(2\,\text{kg}\) to grams and compare it with the total weight of all the groceries. - Find how much capacity remains when only the potatoes are in the bag. - Find how many grams over the limit the bag would be with every item included.

1. Find the current weight: \(1500\,\text{g} + 150\,\text{g} = 1650\,\text{g}\). 2. With the carrots, the total would be \(1650\,\text{g} + 500\,\text{g} = 2150\,\text{g}\), which is \(150\,\text{g}\) more than \(2000\,\text{g}\). 3. If Annie removes the apple, the potatoes and carrots weigh \(1500\,\text{g} + 500\,\text{g} = 2000\,\text{g} = 2\,\text{kg}\).

With the carrots, the bag would weigh \(2\,\text{kg}\ 150\,\text{g}\), so it would be \(150\,\text{g}\) too heavy. Annie must remove the apple; then the potatoes and carrots weigh exactly \(2\,\text{kg}\).

- Express every weight in grams before comparing or adding. - Remember how many grams are in \(1\) kilogram.

1. Convert the flour's weight: \(1\,\text{kg} = 1000\,\text{g}\). 2. Order the weights: \(20\,\text{g} < 100\,\text{g} < 350\,\text{g} < 400\,\text{g} < 1000\,\text{g}\). 3. Add all the weights: \(20\,\text{g} + 100\,\text{g} + 350\,\text{g} + 400\,\text{g} + 1000\,\text{g} = 1870\,\text{g}\). 4. Convert the total: \(1870\,\text{g} = 1\,\text{kg}\ 870\,\text{g}\).

a) Eraser, chocolate bar, bicycle helmet, pencil case, bag of flour b) \(1\,\text{kg}\ 870\,\text{g}\)

- How many meters are in \(1\) kilometer? - Subtract the distance already walked from the whole distance. - A round trip includes the distance there and the same distance back.

1. Since \(1\,\text{km} = 1000\,\text{m}\), Elias takes \(1000\) strides when each stride is \(1\,\text{m}\). 2. After \(400\) strides, he has walked \(400\,\text{m}\). The remaining distance is \(1000\,\text{m} - 400\,\text{m} = 600\,\text{m}\). 3. The trip to the playground and back is \(1000\,\text{m} + 1000\,\text{m} = 2000\,\text{m}\).

a) \(1000\) strides b) \(600\,\text{m}\) c) \(2000\,\text{m}\)

- How many groups of \(100\,\text{m}\) make \(1000\,\text{m}\)? - Subtract the completed distance from the full distance. - How many times can you add \(250\,\text{m}\) to reach \(1000\,\text{m}\)?

1. Convert \(1\,\text{km}=1000\,\text{m}\). Since \(10\times100\,\text{m}=1000\,\text{m}\), the course has \(10\) one-hundred-meter intervals after the starting line, so \(10\) cones are needed at \(100\,\text{m}\) through \(1000\,\text{m}\). 2. The remaining distance is \(1000\,\text{m}-750\,\text{m}=250\,\text{m}\). 3. Since \(4\times250\,\text{m}=1000\,\text{m}\), students need \(4\) laps.

a) \(10\) cones b) \(250\,\text{m}\) c) \(4\) laps

- Remember that \(1\,\text{km}=1000\,\text{m}\). - Use the number of lengths for \(100\,\text{m}\) to think about the number for \(1000\,\text{m}\). - Look for multiplication facts that make the target distance exactly.

1. a) Since \(10\times10\,\text{m}=100\,\text{m}\), the swimmer needs \(10\) lengths. 2. b) Convert \(1\,\text{km}=1000\,\text{m}\). Since \(100\times10\,\text{m}=1000\,\text{m}\), the swimmer needs \(100\) lengths. 3. c) Since \(20\times50\,\text{m}=1000\,\text{m}\), the swimmer needs \(20\) lengths.

a) \(10\) lengths b) \(100\) lengths c) \(20\) lengths

- How many times does Leon walk the route each day? - How many school days are in the week? - Find the distance for one day before finding the weekly distance. - Convert the final distance from meters to kilometers.

1. Leon walks the route twice each day: \(600\,\text{m} \times 2 = 1200\,\text{m}\). 2. For \(5\) days, he walks \(1200\,\text{m} \times 5 = 6000\,\text{m}\). 3. Convert meters to kilometers: \(6000\,\text{m} = 6\,\text{km}\).

Leon walks \(6\,\text{km}\) during one school week.

- How many meters are in \(1\,\text{km}\)? - Add the two lengths already completed. - Subtract that amount from the total length.

1. Add the lengths already completed: \(420\,\text{m} + 390\,\text{m} = 810\,\text{m}\). 2. Convert the total length: \(1\,\text{km} = 1000\,\text{m}\). 3. Subtract: \(1000\,\text{m} - 810\,\text{m} = 190\,\text{m}\).

\(190\,\text{m}\) still need barriers.

- Read carefully to identify what each part asks. - Part a) asks for the distance still remaining. - Part b) asks for the total distance already traveled. - Convert the kilometer to meters before calculating.

1. Convert the total distance: \(1\,\text{km}=1000\,\text{m}\). 2. The remaining distance after the first break is \(1000\,\text{m}-480\,\text{m}=520\,\text{m}\). 3. The total distance ridden after the next segment is \(480\,\text{m}+220\,\text{m}=700\,\text{m}\).

a) \(520\,\text{m}\) remain. b) The bicyclist has ridden \(700\,\text{m}\) in all.

- How many meters make \(1\) kilometer? - Add all three distances for one day first. - Determine how many times the route is walked in one school week. - Use the weekly distance to find the longer distances.

1. Add the three parts of the daily route: \(450\,\text{m} + 250\,\text{m} + 300\,\text{m} = 1000\,\text{m}\). 2. Convert: \(1000\,\text{m} = 1\,\text{km}\). 3. One school week: \(5 \times 1\,\text{km} = 5\,\text{km}\). 4. Four weeks: \(4 \times 5\,\text{km} = 20\,\text{km}\). 5. Twenty weeks: \(20 \times 5\,\text{km} = 100\,\text{km}\).

One day: \(1\,\text{km}\) One school week: \(5\,\text{km}\) Four weeks: \(20\,\text{km}\) Twenty weeks: \(100\,\text{km}\)

- Convert the daily distance to meters first. - How many times do you need the daily distance for each part? - Multiplying by \(10\) shifts each digit one place to the left.

1. Convert the daily distance to meters: \(1\,\text{km}\ 500\,\text{m} = 1500\,\text{m}\). 2. For \(5\) days: \(5 \times 1500\,\text{m} = 7500\,\text{m} = 7\,\text{km}\ 500\,\text{m}\). 3. For \(10\) days: \(10 \times 1500\,\text{m} = 15{,}000\,\text{m} = 15\,\text{km}\).

a) \(7\,\text{km}\ 500\,\text{m}\) b) \(15\,\text{km}\)

- Remember that she makes two one-way trips each workday. - Find the total miles and total yards separately. - Use \(1760\,\text{yd} = 1\,\text{mi}\) to convert the yards.

1. In \(10\) days, Ms. Miller rides the \(4\)-mile part \(20\) times: \(20 \times 4\,\text{mi} = 80\,\text{mi}\). 2. She also rides the \(440\)-yard part \(20\) times: \(20 \times 440\,\text{yd} = 8800\,\text{yd}\). 3. Since \(1760\,\text{yd} = 1\,\text{mi}\), \(8800\,\text{yd} = 5\,\text{mi}\). 4. Add: \(80\,\text{mi} + 5\,\text{mi} = 85\,\text{mi}\).

Ms. Miller bikes \(85\,\text{mi}\) in \(10\) workdays.

- How many meters are in \(1\) kilometer? - First find how far Jonah and Emma have already walked. - Subtract the distance already walked from the whole route.

1. Convert the total distance: \(1\,\text{km} = 1000\,\text{m}\). 2. Find the distance already walked: \(400\,\text{m} + 250\,\text{m} = 650\,\text{m}\). 3. Find the remaining distance: \(1000\,\text{m} - 650\,\text{m} = 350\,\text{m}\).

They still need to walk \(350\,\text{m}\).

- Convert each day's distance to meters first. - Pay attention to the place value of the \(75\) meters on the second day. - Then add the two distances.

1. Convert the first day's distance: \(23\,\text{km}\ 500\,\text{m} = 23{,}500\,\text{m}\). 2. Convert the second day's distance: \(18\,\text{km}\ 75\,\text{m} = 18{,}075\,\text{m}\). 3. Add the distances: \(23{,}500\,\text{m} + 18{,}075\,\text{m} = 41{,}575\,\text{m}\).

Diego walks \(41{,}575\,\text{m}\) altogether.

- Convert meters to centimeters first. - Find how many \(10\)-centimeter side lengths make \(100\) centimeters. - Use the number of rows and the number in each row to find one quilt's total. - Use place value to find how many groups of \(100\) make \(1{,}000{,}000\).

1. Convert \(1\,\text{m}=100\,\text{cm}\). Since \(10\times10\,\text{cm}=100\,\text{cm}\), one row contains \(10\) squares. 2. The quilt has \(10\) rows of \(10\) squares, so it contains \(10\times10=100\) squares. 3. Each quilt contains \(100\) squares. Since \(10{,}000\times100=1{,}000{,}000\), \(10{,}000\) quilts contain \(1{,}000{,}000\) squares.

a) \(10\) fabric squares b) \(100\) fabric squares c) \(10{,}000\) quilts

- How many seconds are in \(1\) minute? - Express both times in seconds before comparing. - Subtract to find the difference.

1. Convert Lucas's time to seconds: \(1\,\text{min} = 60\,\text{s}\), so \(60\,\text{s} + 15\,\text{s} = 75\,\text{s}\). 2. Compare the times: \(85\,\text{s} > 75\,\text{s}\), so Emma jumps longer. 3. Find the difference: \(85\,\text{s} - 75\,\text{s} = 10\,\text{s}\).

Emma jumps longer by \(10\,\text{s}\).

- In the start time, identify the hours, minutes, and seconds. - Add the minutes and seconds separately. - Regroup if the number of seconds reaches \(60\) or more.

1. Add the minutes: \(30\,\text{min} + 3\,\text{min} = 33\,\text{min}\). 2. Add the seconds: \(50\,\text{s} + 20\,\text{s} = 70\,\text{s}\). 3. Regroup \(70\,\text{s}\) as \(1\) minute \(10\) seconds. Add the extra minute: \(33\,\text{min} + 1\,\text{min} = 34\,\text{min}\). 4. The ending time is \(2{:}34{:}10\) p.m.

The song ends at \(2{:}34{:}10\) p.m.

- Express every time in seconds first. - Add the numbers of seconds. - How many full minutes are in the total?

1. Express all the times in seconds: \(25\,\text{s}\), \(70\,\text{s}\), and \(45\,\text{s}\). 2. Add the times: \(25\,\text{s} + 70\,\text{s} + 45\,\text{s} = 140\,\text{s}\). 3. Convert the total: \(140\,\text{s} = 120\,\text{s} + 20\,\text{s} = 2\,\text{min}\ 20\,\text{s}\).

The activities take \(2\) minutes \(20\) seconds altogether.

- A loaded truck's mass includes both the truck and its cargo. - Convert metric tons to kilograms before calculating. - Subtract the cargo or empty mass from the loaded mass, as appropriate.

1. For part a, subtract the empty mass from the loaded mass: \(26{,}450\,\text{kg} - 11{,}820\,\text{kg} = 14{,}630\,\text{kg}\). 2. For part b, convert the masses to kilograms: \(14\) metric tons \(500\,\text{kg} = 14{,}500\,\text{kg}\), and \(8\) metric tons \(900\,\text{kg} = 8900\,\text{kg}\). 3. Start with \(14{,}500\,\text{kg}\). Subtract \(8900\,\text{kg}\). The empty truck has a mass of \(5600\,\text{kg}\).

a) The truck delivered \(14{,}630\,\text{kg}\) of oil. b) The empty truck has a mass of \(5600\,\text{kg}\), or \(5\) metric tons \(600\,\text{kg}\).

- Keep kilograms and grams aligned, or convert everything to grams. - Remember that \(1000\,\text{g} = 1\,\text{kg}\). - Decide whether each part requires subtraction or addition.

1. Find the apples' mass: \(22\,\text{kg}\,450\,\text{g} - 2\,\text{kg}\,150\,\text{g} = 20\,\text{kg}\,300\,\text{g}\). 2. Find the filled pear crate's mass: \(1\,\text{kg}\,850\,\text{g} + 15\,\text{kg}\,400\,\text{g} = 16\,\text{kg}\,1250\,\text{g}\). 3. Regroup \(1000\,\text{g}\) as \(1\,\text{kg}\): \(16\,\text{kg}\,1250\,\text{g} = 17\,\text{kg}\,250\,\text{g}\).

a) The apples have a mass of \(20\,\text{kg}\,300\,\text{g}\). b) The filled pear crate has a mass of \(17\,\text{kg}\,250\,\text{g}\).

- Find the sand load in each car by subtracting the empty mass from the loaded mass. - Compare the two sand loads. - Subtract to find the difference.

1. Find Car 1's sand load: \(78{,}500\,\text{kg} - 22{,}800\,\text{kg} = 55{,}700\,\text{kg}\). 2. Find Car 2's sand load: \(81{,}200\,\text{kg} - 25{,}600\,\text{kg} = 55{,}600\,\text{kg}\). 3. Car 1 carries more because \(55{,}700\,\text{kg} > 55{,}600\,\text{kg}\). 4. Subtract \(55{,}600\,\text{kg}\) from \(55{,}700\,\text{kg}\). The difference is \(100\,\text{kg}\).

Car 1 carries \(100\,\text{kg}\) more sand than Car 2.

- Convert the starting amount to milliliters. - Find the total amount taken by the five children. - Subtract the amount taken from the starting amount.

1. Convert the starting volume: \(1\,\text{L}\,500\,\text{mL} = 1500\,\text{mL}\). 2. Find the amount taken: \(5 \times 120\,\text{mL} = 600\,\text{mL}\). 3. Find the amount remaining: \(1500\,\text{mL} - 600\,\text{mL} = 900\,\text{mL}\).

\(900\,\text{mL}\) of punch remain in the bowl.

- Convert metric tons to kilograms. - The payload is the difference between the maximum total mass and the empty mass. - Put both measurements in the same unit before subtracting.

1. Convert both masses to kilograms: \(3\) metric tons \(500\,\text{kg} = 3500\,\text{kg}\), and \(2\) metric tons \(180\,\text{kg} = 2180\,\text{kg}\). 2. Subtract the empty mass from the maximum total mass: \(3500\,\text{kg} - 2180\,\text{kg} = 1320\,\text{kg}\).

The maximum payload is \(1320\,\text{kg}\).

- Convert the maximum total mass to kilograms. - Subtract the empty trailer's mass to find the maximum payload. - Compare the grain's mass with that payload.

1. Convert the maximum total mass: \(5\) metric tons \(= 5000\,\text{kg}\). 2. Find the maximum payload: \(5000\,\text{kg} - 1650\,\text{kg} = 3350\,\text{kg}\). 3. Since \(3400\,\text{kg} > 3350\,\text{kg}\), the planned load is too heavy. 4. The trailer would exceed its limit by \(3400\,\text{kg} - 3350\,\text{kg} = 50\,\text{kg}\).

No. The trailer would be overloaded by \(50\,\text{kg}\).

- Find the full payload capacity by subtracting the empty mass from the maximum total mass. - Find the combined mass of the three machines. - Subtract the machines' mass from the payload capacity.

1. Convert the maximum total mass to \(12{,}000\,\text{kg}\) and the empty mass to \(5200\,\text{kg}\). Subtract the empty mass from the maximum total mass. The truck's total payload capacity is \(6800\,\text{kg}\). 2. Find the mass of the three machines: \(3 \times 1500\,\text{kg} = 4500\,\text{kg}\). 3. Find the remaining capacity: \(6800\,\text{kg} - 4500\,\text{kg} = 2300\,\text{kg}\).

The truck can carry \(2300\,\text{kg}\) more.

- Convert the metric tons to kilograms first. - For each truck, subtract the empty mass from the maximum total mass. - Compare the payload capacities and find their difference.

1. Find Truck A's payload capacity: \(7500\,\text{kg} - 4230\,\text{kg} = 3270\,\text{kg}\). 2. Convert Truck B's maximum mass: \(8\) metric tons \(= 8000\,\text{kg}\). 3. Find Truck B's payload capacity: \(8000\,\text{kg} - 4850\,\text{kg} = 3150\,\text{kg}\). 4. Compare and subtract: \(3270\,\text{kg} - 3150\,\text{kg} = 120\,\text{kg}\).

Truck A has the greater payload capacity. Truck A can carry \(3270\,\text{kg}\), Truck B can carry \(3150\,\text{kg}\), and the difference is \(120\,\text{kg}\).

- How many milliliters are in \(1\) liter? - Find the total amount in the five small boxes. - Compare prices only after confirming that the quantities are equal.

1. Find the total amount in the juice boxes: \(5 \times 200\,\text{mL} = 1000\,\text{mL}\). 2. Since \(1000\,\text{mL} = 1\,\text{L}\), both options contain the same amount. 3. Find the difference in price: \(\$2.50 - \$1.90 = \$0.60\). 4. The bottle costs \(\$0.60\) less.

The \(1\,\text{L}\) bottle costs less by \(\$0.60\).

- How many \(250\,\text{mL}\) dispensers equal \(1\,\text{L}\)? - Find the cost of that many small dispensers. - Subtract the refill price from the small-dispenser cost.

1. Convert the refill amount: \(1\,\text{L} = 1000\,\text{mL}\). 2. Since \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), four small dispensers contain the same amount as the refill. 3. Find the cost of four small dispensers: \(4 \times \$1.60 = \$6.40\). 4. Find the savings: \(\$6.40 - \$5.20 = \$1.20\).

Buying the refill saves \(\$1.20\).

- Each US customary ton has \(2000\,\text{lb}\). - Put the slab and the table limits into the same unit. - Which is the longest arm length that can hold the slab?

1. At \(60\,\text{ft}\), \(9700\,\text{lb}=4\) tons \(1700\,\text{lb}\). 2. At \(120\,\text{ft}\), \(4300\,\text{lb}=2\) tons \(300\,\text{lb}\). 3. The slab's mass is \(3\times2000+200=6200\,\text{lb}\). 4. The limit at \(90\,\text{ft}\) is \(6800\,\text{lb}\), enough for the slab; the limit at the next reach is too small. Thus \(90\,\text{ft}\) is greatest.

a) At \(60\,\text{ft}\): \(4\) tons \(1700\,\text{lb}\); at \(120\,\text{ft}\): \(2\) tons \(300\,\text{lb}\) b) \(90\,\text{ft}\)

- Convert all masses to kilograms. - Subtract the empty mass from the maximum total mass. - Determine how many \(500\)-kilogram machines fit within the remaining capacity.

1. Convert the maximum mass: \(12\) metric tons \(= 12{,}000\,\text{kg}\). 2. Convert the empty mass: \(7\) metric tons \(500\,\text{kg} = 7500\,\text{kg}\). 3. Start with the maximum mass of \(12{,}000\,\text{kg}\) and subtract the empty mass of \(7500\,\text{kg}\). The remaining payload capacity is \(4500\,\text{kg}\). 4. Nine machines have a mass of \(9 \times 500\,\text{kg} = 4500\,\text{kg}\), which uses the full remaining capacity.

The truck can carry at most \(9\) machines.

- Find the mass of all four beams. - How many kilograms make a metric ton? - Compare the combined mass with the maximum allowed mass.

1. Four beams have a mass of \(4\times600=2400\,\text{kg}\). 2. The limit is \(2\times1000+200=2200\,\text{kg}\). 3. Because \(2400>2200\), the load exceeds the limit by \(2400-2200=200\,\text{kg}\).

Yes. The combined mass is \(200\,\text{kg}\) above the limit.

- Convert all amounts to milliliters before adding. - Convert the total back to liters for part a. - Determine how many \(250\)-milliliter glasses make \(1\) liter, then use the \(2\)-liter total.

1. Convert the apple juice: \(1\,\text{L}\,500\,\text{mL} = 1500\,\text{mL}\). 2. Find the total volume: \(1500\,\text{mL} + 500\,\text{mL} = 2000\,\text{mL}\). 3. Convert for part a: \(2000\,\text{mL} = 2\,\text{L}\). 4. Four \(250\,\text{mL}\) glasses hold \(1000\,\text{mL}\), or \(1\,\text{L}\). Therefore, \(2 \times 4 = 8\) full glasses can be poured.

a) The container holds \(2\,\text{L}\) altogether. b) Jordan can pour \(8\) full glasses.

- Convert the total volume to milliliters. - Divide by the number of milliliters in one bottle. - You can divide both numbers by \(100\) to simplify the calculation.

1. Convert the total volume: \(72\,\text{L} = 72{,}000\,\text{mL}\). 2. Measure both volumes in units of \(100\,\text{mL}\). The total is \(720\) such units, and each bottle holds \(3\) such units. 3. Divide by the one-digit number of units per bottle: \(720 \div 3 = 240\).

The orchard can fill \(240\) bottles completely.

- First find how much water is used by \(10\) watering cans. - Determine how many groups of \(10\) cans are in \(250\) cans. - Subtract the total amount used from the starting volume.

1. Ten watering cans use \(10 \times 400\,\text{mL} = 4000\,\text{mL}\), or \(4\,\text{L}\). 2. There are \(25\) groups of \(10\) cans in \(250\) cans. Find the total used: \(25 \times 4\,\text{L} = 100\,\text{L}\). 3. Subtract from the starting amount: \(150\,\text{L} - 100\,\text{L} = 50\,\text{L}\).

\(50\,\text{L}\) of water remain in the rain barrel.

- Determine how many \(250\)-milliliter cups make \(1\) liter. - Find how many equal groups of that size are in \(80\) cups. - Each group represents \(1\) liter.

1. Four \(250\,\text{mL}\) cups hold \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), or \(1\,\text{L}\). 2. Find how many groups of \(4\) cups are in \(80\) cups: \(80 \div 4 = 20\). Therefore, the total volume is \(20\,\text{L}\).

\(20\,\text{L}\) of punch are prepared altogether.

- How many seconds are in one minute? - Convert both lap times to the same unit before comparing them. - For a race time, what does “faster” mean about the number of seconds?

1. Convert the second-lap time to seconds: \(1\,\text{minute} = 60\,\text{seconds}\), so \(60\,\text{seconds} + 15\,\text{seconds} = 75\,\text{seconds}\). 2. Compare the times. Since \(75\,\text{seconds} < 85\,\text{seconds}\), the second lap is faster. 3. Find the difference: \(85\,\text{seconds} - 75\,\text{seconds} = 10\,\text{seconds}\).

a) The second lap takes \(75\) seconds. b) The second lap is faster by \(10\) seconds.

- Find complete groups of \(60\) minutes inside \(230\) minutes. - Use the difference from the greatest multiple of \(60\) for the leftover minutes. - To compare the two times, express them in the same unit.

1. Find the greatest multiple of \(60\) below \(230\): \(3\times60=180\) and \(4\times60=240\). So there are \(3\) full hours, with \(230-180=50\) minutes remaining. Thus \(230\) minutes is \(3\) hours \(50\) minutes. 2. Convert the second train's time to minutes: \(4\times60+10=250\) minutes. 3. Compare the times: \(230\,\text{minutes}<250\,\text{minutes}\), so the express train is faster.

a) \(230\) minutes is \(3\) hours \(50\) minutes. b) The express train is faster.

- How many minutes are in one hour? - Convert the full hiking time to minutes. - Break the number of minutes into friendly place-value parts before multiplying.

1. Convert \(4\) hours to minutes: \(4 \times 60 = 240\) minutes. 2. Decompose \(240 = 200 + 40\). Then \(55 \times 200 = 11{,}000\) and \(55 \times 40 = 2200\). 3. Add: \(11{,}000 + 2200 = 13{,}200\).

Omar takes about \(13{,}200\) steps.

- How many minutes does the machine run altogether? - Break \(480\) into hundreds and tens before multiplying by \(125\). - Use one-digit multiplication facts and place value for each partial product.

1. Convert the shift length to minutes: \(8\times60=480\) minutes. 2. Decompose \(480=400+80\). Then \(125\times400=(125\times4)\times100=500\times100=50{,}000\). 3. Also, \(125\times80=(125\times8)\times10=1000\times10=10{,}000\). 4. Add: \(50{,}000+10{,}000=60{,}000\) parts.

The machine makes \(60{,}000\) parts during the shift.

- How many days are in \(1\) week? - First find the elephant's age in days. - How many hours are in each day? - What final operation converts the total days to hours?

1. Convert weeks to days: \(3 \times 7=21\) days. 2. Convert days to hours: \(21 \times 24=504\) hours.

The baby elephant is \(504\,\text{hours}\) old.

- Start by finding the cost of a smaller \(4\)-ounce portion. - Use \(1\,\text{lb} = 16\,\text{oz}\). - For part c), determine how many \(4\)-ounce portions fit the budget.

1. Four ounces is half of \(8\,\text{oz}\), so \(\$1.80 \div 2 = \$0.90\). 2. Since \(1\,\text{lb} = 16\,\text{oz}\), one pound is two \(8\)-ounce bags. The cost is \(2 \times \$1.80 = \$3.60\). 3. Each \(4\,\text{oz}\) costs \(\$0.90\). Since \(5 \times \$0.90 = \$4.50\), the budget buys five \(4\)-ounce portions: \(5 \times 4\,\text{oz} = 20\,\text{oz}\).

a) \(\$0.90\) b) \(\$3.60\) c) \(20\,\text{oz}\)

- Convert half a kilogram to grams. - Compare the masses of \(8\) eggs and \(9\) eggs. - Choose the total that is closer to the target mass.

1. Convert the target mass: \(\frac{1}{2}\,\text{kg} = 500\,\text{g}\). 2. Compare nearby multiples of \(60\,\text{g}\): \(8 \times 60\,\text{g} = 480\,\text{g}\), and \(9 \times 60\,\text{g} = 540\,\text{g}\). 3. Since \(480\,\text{g}\) is closer to \(500\,\text{g}\) than \(540\,\text{g}\), about \(8\) eggs are needed.

About \(8\) eggs.

- Convert the carton's volume to milliliters. - Find the total volume poured into the glasses. - Subtract the poured volume from the starting volume.

1. Convert the starting volume: \(1\,\text{L}\,500\,\text{mL} = 1500\,\text{mL}\). 2. Find the volume poured: \(6 \times 200\,\text{mL} = 1200\,\text{mL}\). 3. Find the remaining volume: \(1500\,\text{mL} - 1200\,\text{mL} = 300\,\text{mL}\).

\(300\,\text{mL}\) of juice remain in the carton.

- How many \(8\)-ounce containers make \(1\) pound? - How many \(4\)-ounce containers make \(1\) pound? - Multiply each container price by the number of containers needed.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), one pound of strawberries requires \(16 \div 8 = 2\) containers. The cost is \(2 \times \$2.90 = \$5.80\). 2. One pound of raspberries requires \(16 \div 4 = 4\) containers. The cost is \(4 \times \$2.10 = \$8.40\).

Strawberries cost \(\$5.80\) per pound. Raspberries cost \(\$8.40\) per pound.

- Find the cost of the grapes and apples separately. - Add the two costs to find the purchase total. - Subtract the total from \(\$10.00\).

1. Eight ounces is half a pound, so the grapes cost \(\$3.60 \div 2 = \$1.80\). 2. The apples cost \(2 \times \$1.90 = \$3.80\). 3. The total cost is \(\$1.80 + \$3.80 = \$5.60\). 4. The change is \(\$10.00 - \$5.60 = \$4.40\).

Ms. Bauer receives \(\$4.40\) in change.

- Start by finding the cost of \(4\,\text{oz}\). - Combine the prices for \(8\,\text{oz}\) and \(4\,\text{oz}\) to find the price of \(12\,\text{oz}\). - How many \(8\)-ounce portions are in \(1\) pound?

1. Four ounces is half of \(8\,\text{oz}\), so \(\$3.60 \div 2 = \$1.80\). 2. Twelve ounces is \(8\,\text{oz} + 4\,\text{oz}\), so \(\$3.60 + \$1.80 = \$5.40\). 3. Since \(1\,\text{lb} = 16\,\text{oz}\), one pound is two \(8\)-ounce portions. The cost is \(2 \times \$3.60 = \$7.20\).

a) \(\$1.80\) b) \(\$5.40\) c) \(\$7.20\)

- What separate time intervals are given? - Add the hours and minutes. - Remember that \(60\) minutes make one hour. - Make sure the total includes the time before the flight and after landing.

1. Add the preflight time and flight time: \(2\,\text{hours} + 1\,\text{hour}\,45\,\text{minutes} = 3\,\text{hours}\,45\,\text{minutes}\). 2. Add the baggage time: \(3\,\text{hours}\,45\,\text{minutes} + 45\,\text{minutes} = 3\,\text{hours}\,90\,\text{minutes}\). 3. Regroup \(90\) minutes as \(1\) hour \(30\) minutes. The total time is \(4\,\text{hours}\,30\,\text{minutes}\).

The total time is \(4\) hours \(30\) minutes.

- How much time do the two breaks take altogether? - Combine the break time and the construction delay. - Add that total to the driving time. - Regroup the minutes if they total more than \(60\).

1. Find the total break time: \(2 \times 25\,\text{minutes} = 50\,\text{minutes}\). 2. Add the break time and delay to the driving time: \(5\,\text{hours}\,15\,\text{minutes} + 50\,\text{minutes} + 40\,\text{minutes}\). 3. Add the minutes: \(15 + 50 + 40 = 105\) minutes, which is \(1\) hour \(45\) minutes. 4. Add the hours: \(5\,\text{hours} + 1\,\text{hour}\,45\,\text{minutes} = 6\,\text{hours}\,45\,\text{minutes}\).

The trip takes \(6\) hours \(45\) minutes.

- Find the complete trip time for each train. - What time must be added to Train B’s travel time? - Compare the two totals. - Subtract the shorter total from the longer total.

1. Find Train B’s total time: \(2\,\text{hours}\,50\,\text{minutes} + 40\,\text{minutes} = 3\,\text{hours}\,30\,\text{minutes}\). 2. Compare the totals: Train A takes \(3\) hours \(20\) minutes, and Train B takes \(3\) hours \(30\) minutes. Train A is faster. 3. Find the difference: \(3\,\text{hours}\,30\,\text{minutes} - 3\,\text{hours}\,20\,\text{minutes} = 10\,\text{minutes}\).

Train A is faster by \(10\) minutes.

- How many days are in \(3\) months using the model's \(30\)-day months? - First find the calf's total modeled weight gain. - Express both weights in pounds before comparing them.

1. The first \(3\) months contain \(3 \times 30 = 90\) days. 2. The calf gains \(90 \times 5\,\text{lb} = 450\,\text{lb}\). 3. Its modeled weight is \(100\,\text{lb} + 450\,\text{lb} = 550\,\text{lb}\). 4. Convert the model adult weight: \(2\) tons \(= 4000\,\text{lb}\). 5. The difference is \(4000\,\text{lb} - 550\,\text{lb} = 3450\,\text{lb}\).

a) The calf weighs \(550\,\text{lb}\) in the model. b) It is \(3450\,\text{lb}\) short of the model adult weight.

- Convert the starting amount to pounds. - Find the total delivered during the \(12\) days. - Add the delivery to the starting amount, then regroup the final number of pounds into tons and pounds.

1. Convert the starting amount: \(8\) tons \(= 16{,}000\,\text{lb}\). 2. Find the total delivered: \(12 \times 1200\,\text{lb} = 14{,}400\,\text{lb}\). 3. Add the amounts: \(16{,}000\,\text{lb} + 14{,}400\,\text{lb} = 30{,}400\,\text{lb}\). 4. Since \(15 \times 2000\,\text{lb} = 30{,}000\,\text{lb}\), \(30{,}400\,\text{lb} = 15\) tons \(400\,\text{lb}\).

After \(12\) days, the warehouse contains \(30{,}400\,\text{lb}\), or \(15\) tons \(400\,\text{lb}\), of iron.

- Find the total mass of each kind of rhinoceros separately. - Add the two totals. - Convert the final number of kilograms to metric tons and kilograms.

1. Estimate the mass of the white rhinoceroses: \(3 \times 2300\,\text{kg} \approx 6900\,\text{kg}\). 2. Estimate the mass of the black rhinoceroses: \(5 \times 1100\,\text{kg} \approx 5500\,\text{kg}\). 3. Add \(6900\,\text{kg}\) and \(5500\,\text{kg}\). The combined mass is about \(12{,}400\,\text{kg}\). 4. Convert the estimate: \(12{,}400\,\text{kg}\) is \(12\) metric tons \(400\,\text{kg}\).

The animals have a combined mass of about \(12\) metric tons \(400\,\text{kg}\).

- Recall how many quarts equal one gallon. - Multiply to convert the tank's capacity to quarts. - Divide the number of gallons in the tank by the number of gallons in one watering can.

1. Convert gallons to quarts: \(75\,\text{gallons} \times 4 = 300\,\text{quarts}\). 2. Divide the tank's capacity by the watering can's capacity: \(75\,\text{gallons} \div 3\,\text{gallons} = 25\).

a) The full tank contains \(300\,\text{quarts}\) of water. b) The watering can can be filled \(25\) times.

- Recall how many quarts equal one gallon. - Recall how many cups equal one quart. - Convert in two steps: gallons to quarts, then quarts to cups.

1. Convert gallons to quarts: \(50\,\text{gallons} \times 4 = 200\,\text{quarts}\). 2. Each quart contains \(4\) cups, so \(200\,\text{quarts} \times 4 = 800\,\text{cups}\).

a) The event receives \(200\,\text{quarts}\) of cider. b) The event can serve \(800\) one-cup portions.

- Round each amount to a convenient hundred milliliters. - Use the same daily estimate for each later part. - Keep track of when the result should be expressed in liters instead of milliliters.

1. Round the daily amounts to convenient hundreds: \(800\,\text{mL} + 200\,\text{mL} + 200\,\text{mL} + 200\,\text{mL} + 200\,\text{mL} = 1600\,\text{mL}\). 2. For \(7\) days, \(1600\,\text{mL} \times 7 = 11{,}200\,\text{mL} = 11.2\,\text{L}\), or about \(11\,\text{L}\). 3. For \(30\) days, \(1600\,\text{mL} \times 30 = 48{,}000\,\text{mL} = 48\,\text{L}\).

a) About \(1600\,\text{mL}\) b) About \(11\,\text{L}\) c) About \(48\,\text{L}\)

- Round each daily amount to a nearby hundred for the estimate. - Align place values when adding the exact amounts. - Remember that \(1000\,\text{g} = 1\,\text{kg}\).

1. Estimate by rounding: \(1300\,\text{g} + 800\,\text{g} + 600\,\text{g} + 500\,\text{g} = 3200\,\text{g}\). 2. Add the exact amounts: \(1250\,\text{g} + 810\,\text{g} + 590\,\text{g} + 450\,\text{g} = 3100\,\text{g}\). 3. Find the weekly amount: \(3100\,\text{g} \times 7 = 21{,}700\,\text{g}\). 4. Convert to kilograms: \(21{,}700\,\text{g} = 21.7\,\text{kg}\), which is about \(22\,\text{kg}\).

a) About \(3200\,\text{g}\) b) \(3100\,\text{g}\) c) About \(22\,\text{kg}\)

- Count how many increases of ten reach the required mark. - Each increase has the same physical length. - Change the total centimeters into meters.

1. a) From \(0\) to \(1000\) there are \(1000\div10=100\) intervals. Their total length is \(100\times5=500\,\text{cm}=5\,\text{m}\). 2. b) From \(0\) to \(20{,}000\) there are \(20{,}000\div10=2000\) intervals. Their total length is \(2000\times5=10{,}000\,\text{cm}=100\,\text{m}\).

a) \(5\,\text{m}\) b) \(100\,\text{m}\)

- How many unit intervals run from zero to the final number? - Find the total length in millimeters first. - For part b), convert the line's length to centimeters before dividing.

1. There are \(4500\) equal spaces, each \(2\,\text{mm}\) long, so the line is \(4500\times2=9000\,\text{mm}\). 2. a) Convert \(9000\,\text{mm}=9\,\text{m}\). 3. b) The length is \(900\,\text{cm}\), so \(900\div20=45\) notebooks would span it.

a) \(9\,\text{m}\) b) \(45\) notebooks

- Convert the kilogram amounts to grams first. - Find the total weight in Lucas's backpack. - Find the total weight of Sophie's pasta packages. - Compare the two totals.

1. Find Lucas's total: \(1000\,\text{g} + 1000\,\text{g} + 500\,\text{g} = 2500\,\text{g}\). 2. Find Sophie's total: \(5 \times 500\,\text{g} = 2500\,\text{g}\). 3. Compare: \(2500\,\text{g} = 2500\,\text{g}\), so they carry the same weight.

They carry the same weight: \(2500\,\text{g}\), or \(2\,\text{kg}\ 500\,\text{g}\), each.

- Express the starting amount in ounces. - Find how much dough was used for all \(12\) rolls. - Subtract the amount used from the starting amount.

1. Convert the total amount of dough: \(4\,\text{lb} = 64\,\text{oz}\). 2. Find the dough used: \(12 \times 4\,\text{oz} = 48\,\text{oz}\). 3. Subtract the amount used: \(64\,\text{oz} - 48\,\text{oz} = 16\,\text{oz}\).

\(16\,\text{oz}\) of dough remain.

- Convert the distance between the houses to meters. - Add the distances Julia and Tom have walked. - Compare their combined distance with the distance between the houses.

1. Convert the distance between the houses: \(1\,\text{km} = 1000\,\text{m}\). 2. Add the distances Julia and Tom have walked: \(460 + 550 = 1010\,\text{m}\). 3. Since \(1010 > 1000\), their combined distance is greater than the distance between the houses. They have already met and passed each other.

Yes. Together they have walked \(1010\,\text{m}\), which is \(10\,\text{m}\) more than the \(1000\,\text{m}\) between their houses, so they have already passed each other.

- Express the cruise ship's entire stay in hours. - How many hours are in \(2\) full days? - Use the same unit before comparing the durations. - Subtract to find the difference.

1. Convert \(2\) days to hours: \(2 \times 24\,\text{hr} = 48\,\text{hr}\). 2. Find the cruise ship's total time: \(48\,\text{hr} + 10\,\text{hr} = 58\,\text{hr}\). 3. Compare: \(60\,\text{hr} > 58\,\text{hr}\), so the cargo ship stays longer. 4. Find the difference: \(60\,\text{hr} - 58\,\text{hr} = 2\,\text{hr}\).

The cargo ship stays longer by \(2\) hours.

- How many minutes are in half an hour? - How many more bottles does Machine B fill in one minute? - Multiply the one-minute difference by the total number of minutes.

1. Convert half an hour to minutes: \(30\) minutes. 2. Find the difference in the machines’ rates: \(55 - 42 = 13\) bottles per minute. 3. Find the total difference over \(30\) minutes: \(13 \times 30 = 390\) bottles.

Machine B fills \(390\) more bottles than Machine A.

- Find how much elastic is needed for each type of item. - Express the needed and available lengths in the same unit. - Add the two needed amounts before comparing with the roll.

1. Find the elastic needed for the masks: \(6 \times 12\,\text{in} = 72\,\text{in}\). 2. Find the elastic needed for the hats: \(4 \times 20\,\text{in} = 80\,\text{in}\). 3. Find the total needed: \(72\,\text{in} + 80\,\text{in} = 152\,\text{in}\). 4. Convert the available length: \(15\,\text{ft} = 180\,\text{in}\). 5. Since \(152\,\text{in} < 180\,\text{in}\), there is enough elastic.

Yes, there is enough elastic.

- Express all the lengths in centimeters first. - Find the combined length of the two pieces removed. - Subtract the amount removed from the original length.

1. Convert the full beam length to centimeters: \(3\,\text{m} = 300\,\text{cm}\). 2. Find the total length cut off: \(110\,\text{cm} + 95\,\text{cm} = 205\,\text{cm}\). 3. Subtract the amount removed: \(300\,\text{cm} - 205\,\text{cm} = 95\,\text{cm}\).

\(95\,\text{cm}\) of the beam remain.

- How many fluid ounces are in \(1\) quart? - Find how many fluid ounces are in one glass. - Find the total volume poured into three glasses. - Subtract the poured amount from the full pitcher.

1. Convert the pitcher's volume: \(1\,\text{qt} = 32\,\text{fl oz}\). 2. Find the volume of one glass: \(32\,\text{fl oz} \div 8 = 4\,\text{fl oz}\). 3. Find the volume that was drunk: \(3 \times 4\,\text{fl oz} = 12\,\text{fl oz}\). 4. Find the remaining volume: \(32\,\text{fl oz} - 12\,\text{fl oz} = 20\,\text{fl oz}\).

\(20\,\text{fl oz}\) of juice remain.

- Convert the total weight to ounces. - Subtract the weight of the empty bag. - Find the total flour removed, then subtract it from the starting flour mass.

1. Convert the total weight: \(2\,\text{lb}\ 4\,\text{oz} = 36\,\text{oz}\). 2. Subtract the packaging: \(36\,\text{oz} - 1\,\text{oz} = 35\,\text{oz}\) of flour at the start. 3. Find the amount removed: \(4 \times 8\,\text{oz} = 32\,\text{oz}\). 4. Find the amount remaining: \(35\,\text{oz} - 32\,\text{oz} = 3\,\text{oz}\).

\(3\,\text{oz}\) of flour remain.

- Convert the full path length to meters. - Find how many meters were completed on Tuesday. - Add the first two days and subtract from the full length.

1. Convert the total length: \(1\,\text{km} = 1000\,\text{m}\). 2. Find Tuesday's distance: \(345 + 120 = 465\,\text{m}\). 3. Find the total finished on Monday and Tuesday: \(345 + 465 = 810\,\text{m}\). 4. Find the distance remaining: \(1000 - 810 = 190\,\text{m}\). 5. Since \(190 < 200\), the workers are correct.

Yes. The workers have \(190\,\text{m}\) left, which is less than \(200\,\text{m}\).

- Convert the pound weight to ounces. - Add the known weights on the left side. - Because the scale is balanced, subtract those known weights from the right side.

1. Convert the weight on the right: \(1\,\text{lb} = 16\,\text{oz}\). 2. Add the metal weights on the left: \(2 \times 2\,\text{oz} + 1\,\text{oz} = 5\,\text{oz}\). 3. Subtract to find the nuts' weight: \(16\,\text{oz} - 5\,\text{oz} = 11\,\text{oz}\).

The nuts weigh \(11\,\text{oz}\).

- Convert pounds to ounces. - Subtract the extra weight placed with the jars. - Divide the remaining weight equally among the three jars.

1. Convert the right side: \(4\,\text{lb} = 64\,\text{oz}\). 2. Subtract the added \(1\)-pound weight from the total: \(64\,\text{oz} - 16\,\text{oz} = 48\,\text{oz}\) for the three jars. 3. Divide equally: \(48\,\text{oz} \div 3 = 16\,\text{oz}\) per jar.

One honey jar weighs \(16\,\text{oz}\).

- How many hours are in a full week? - Find the library’s open hours for one weekday and then for all five weekdays. - Add the Saturday hours. - Subtract the total open time from the total time in a week.

1. Find the total hours in a week: \(7 \times 24 = 168\) hours. 2. The library is open \(8\) hours each weekday, so the weekday total is \(5 \times 8 = 40\) hours. 3. On Saturday, the library is open \(5\) hours. 4. Find the total open time: \(40 + 5 = 45\) hours. 5. Subtract from the full week: \(168 - 45 = 123\) hours.

The library is closed for \(123\) hours during a full week.

- First find the total volume of apple juice. - Put the juice and water in the same unit before adding. - After finding the total punch volume, compare it with the amount one cup holds.

1. The apple juice volume is \(12 \times 750\,\text{mL} = 9000\,\text{mL} = 9\,\text{L}\). 2. Add the water: \(9\,\text{L} + 6\,\text{L} = 15\,\text{L}\). 3. Convert the punch to milliliters: \(15\,\text{L} = 15{,}000\,\text{mL}\). 4. Divide by the cup capacity: \(15{,}000 \div 300 = 50\).

\(50\) cups can be filled completely.

- Multiply the pool's length by the number of lengths in each part. - Remember that \(1000\,\text{m} = 1\,\text{km}\). - For part c, find one day's distance before finding the weekly distance.

1. For part a, multiply: \(25 \times 12 = 300\,\text{m}\). 2. For part b, multiply: \(25 \times 60 = 1500\,\text{m}\). Convert: \(1500\,\text{m} = 1.5\,\text{km}\). 3. For part c, find the distance per day: \(25 \times 40 = 1000\,\text{m}\). Then multiply by \(5\): \(1000 \times 5 = 5000\,\text{m}\).

a) \(300\,\text{m}\) b) \(1.5\,\text{km}\) c) \(5000\,\text{m}\)

- Try starting with the largest available weight. - For \(300\,\text{g}\), look for one combination with two weights and another with more weights. - Remember how many grams are in \(1\) kilogram. - In part b, use each weight no more times than it appears in the list.

1. Two combinations for \(300\,\text{g}\) are \(200\,\text{g} + 100\,\text{g} = 300\,\text{g}\) and \(200\,\text{g} + 50\,\text{g} + 50\,\text{g} = 300\,\text{g}\). 2. One combination for \(850\,\text{g}\) is \(500\,\text{g} + 200\,\text{g} + 100\,\text{g} + 50\,\text{g} = 850\,\text{g}\). 3. The total of all six weights is \(500\,\text{g} + 200\,\text{g} + 100\,\text{g} + 100\,\text{g} + 50\,\text{g} + 50\,\text{g} = 1000\,\text{g}\). Since \(1000\,\text{g} = 1\,\text{kg}\), they weigh \(1\,\text{kg}\) altogether.

a) For example, \(200\,\text{g} + 100\,\text{g}\) and \(200\,\text{g} + 50\,\text{g} + 50\,\text{g}\) b) \(500\,\text{g}\), \(200\,\text{g}\), \(100\,\text{g}\), and \(50\,\text{g}\) c) \(1\,\text{kg}\)

- Break \(387\) into hundreds, tens, and ones to find suitable weights. - Look closely at the ones digit of \(387\). - For part b, add all the weights carefully. - How many grams would the total need to equal \(1\,\text{kg}\)?

1. Decompose \(387\,\text{g}\) as \(300\,\text{g} + 80\,\text{g} + 7\,\text{g}\). Use \(200\,\text{g} + 100\,\text{g}\), \(50\,\text{g} + 20\,\text{g} + 10\,\text{g}\), and \(5\,\text{g} + 2\,\text{g}\). 2. Add all ten weights: \(1 + 2 + 2 + 5 + 10 + 20 + 50 + 100 + 200 + 500 = 890\), so the total is \(890\,\text{g}\). 3. Since \(1\,\text{kg} = 1000\,\text{g}\) and \(890\,\text{g} < 1000\,\text{g}\), Lucas is not correct. The set is \(110\,\text{g}\) short of \(1\,\text{kg}\).

a) Use \(200\,\text{g}\), \(100\,\text{g}\), \(50\,\text{g}\), \(20\,\text{g}\), \(10\,\text{g}\), \(5\,\text{g}\), and one \(2\,\text{g}\) weight. b) No. All ten weights total \(890\,\text{g}\), which is \(110\,\text{g}\) less than \(1\,\text{kg}\).

- Find how much the scale reading increased when the flour was added. - Convert the final weight from pounds to ounces. - Find the increase from the flour-only reading to the final reading.

1. Find the flour's weight: \(19\,\text{oz} - 7\,\text{oz} = 12\,\text{oz}\). 2. Convert the final scale reading: \(2\,\text{lb} = 32\,\text{oz}\). 3. Find the sugar's weight: \(32\,\text{oz} - 19\,\text{oz} = 13\,\text{oz}\).

Lucas added \(12\,\text{oz}\) of flour and \(13\,\text{oz}\) of sugar.

- Convert the length in feet to inches first. - Add the ribbon used for all three gifts. - Subtract the amount used from the original amount.

1. Convert the full ribbon length to inches: \(12\,\text{ft} = 144\,\text{in}\). 2. Find the total amount used: \(45\,\text{in} + 33\,\text{in} + 42\,\text{in} = 120\,\text{in}\). 3. Find the amount left: \(144\,\text{in} - 120\,\text{in} = 24\,\text{in}\).

\(24\,\text{in}\) of ribbon remain.

- Convert the maximum mass to kilograms. - Find the combined mass of the pallets. - Add the empty vehicle and cargo, then subtract from the limit.

1. Convert the maximum mass: \(3\) metric tons \(500\,\text{kg} = 3500\,\text{kg}\). 2. Find the pallets' mass: \(4 \times 300\,\text{kg} = 1200\,\text{kg}\). 3. Find the current total mass: \(2100\,\text{kg} + 1200\,\text{kg} = 3300\,\text{kg}\). 4. Find the remaining capacity: \(3500\,\text{kg} - 3300\,\text{kg} = 200\,\text{kg}\).

The van can carry \(200\,\text{kg}\) more.

- Convert both lengths to centimeters. - Compare the two measurements. - Subtract to find the difference.

1. Convert the blue ribbon's length: \(2\,\text{m}\ 40\,\text{cm} = 240\,\text{cm}\). 2. Compare: \(270 > 240\), so the red ribbon is longer. 3. Find the difference: \(270 - 240 = 30\,\text{cm}\).

The red ribbon is \(30\,\text{cm}\) longer.

- How many centimeters make one meter? - Separate \(305\) centimeters into hundreds and leftover centimeters. - Convert \(4\,\text{m}\) to centimeters before finding the difference.

1. Decompose the measurement: \(305\,\text{cm} = 300\,\text{cm} + 5\,\text{cm}\). 2. Since \(300\,\text{cm} = 3\,\text{m}\), the jump is \(3\,\text{m}\ 5\,\text{cm}\). 3. Convert the target: \(4\,\text{m} = 400\,\text{cm}\). 4. Subtract: \(400\,\text{cm} - 305\,\text{cm} = 95\,\text{cm}\).

a) \(3\,\text{m}\ 5\,\text{cm}\) b) \(95\,\text{cm}\)

- Convert both lengths to centimeters. - How many centimeters are in \(1\) meter? In \(1\) decimeter? - Subtract the shorter length from the longer length.

1. Convert the first beam to centimeters: \(1\,\text{m}\,45\,\text{cm} = 100 + 45 = 145\,\text{cm}\). 2. Convert the second beam to centimeters: \(16\,\text{dm} = 160\,\text{cm}\). 3. Since \(160 > 145\), the second beam is longer. 4. Find the difference: \(160 - 145 = 15\,\text{cm}\).

The second beam is longer by \(15\,\text{cm}\).

- Convert every length to centimeters. - Add the three pieces. - Compare the total with \(200\,\text{cm}\) and find the difference.

1. Convert the target length: \(2\,\text{m} = 200\,\text{cm}\). 2. Convert the pieces to centimeters: \(95\,\text{cm}\), \(6\,\text{dm} = 60\,\text{cm}\), and \(300\,\text{mm} = 30\,\text{cm}\). 3. Add the pieces: \(95 + 60 + 30 = 185\,\text{cm}\). 4. Since \(185 < 200\), the pieces are not long enough. 5. Find the missing length: \(200 - 185 = 15\,\text{cm}\).

No. The pieces are \(15\,\text{cm}\) too short.

- Add the masses of all the crates. - Put the forklift's limit into kilograms. - Compare the total load with the allowed mass.

1. The crates have a total mass of \(240+310+180+280=1010\,\text{kg}\). 2. One metric ton equals \(1000\,\text{kg}\). 3. The total is \(1010-1000=10\,\text{kg}\) over the limit.

No. The crates have a total mass of \(1010\,\text{kg}\), which is \(10\,\text{kg}\) over the forklift's limit.

- Convert both measurements to kilograms first. - Add the mass of the empty truck and the mass of the sand. - Check that your answer includes the correct unit.

1. Convert the empty truck's mass: \(2\) metric tons \(350\,\text{kg} = 2350\,\text{kg}\). 2. Convert the sand's mass: \(1\) metric ton \(800\,\text{kg} = 1800\,\text{kg}\). 3. Add the masses: \(2350\,\text{kg} + 1800\,\text{kg} = 4150\,\text{kg}\).

The loaded truck has a total mass of \(4150\,\text{kg}\).

- Convert both weights to grams before adding. - Add the backpack's weight and the books' weight. - Convert the total back to kilograms and grams.

1. Convert the books' weight to grams: \(2\,\text{kg}\ 350\,\text{g} = 2350\,\text{g}\). 2. Add the weights: \(1200\,\text{g} + 2350\,\text{g} = 3550\,\text{g}\). 3. Convert the total to mixed-unit form: \(3550\,\text{g} = 3\,\text{kg}\ 550\,\text{g}\).

The filled backpack weighs \(3\,\text{kg}\ 550\,\text{g}\).

- Convert both ages to the same unit. - Change Frieda’s years to months and add the extra months. - Compare the two totals and subtract.

1. Convert Frieda’s age to months: \(6 \times 12 + 10 = 72 + 10 = 82\) months. 2. Compare the ages: \(82 > 80\), so Frieda is older. 3. Find the difference: \(82 - 80 = 2\) months.

Frieda is older by \(2\) months.

- Convert each time unit to days. - Use the stated value of \(30\) days per month. - Convert the weeks using \(7\) days per week. - Add all three amounts.

1. Convert the months to days: \(2 \times 30 = 60\) days. 2. Convert the weeks to days: \(2 \times 7 = 14\) days. 3. Add all the days: \(60 + 14 + 5 = 79\) days.

The ship is at sea for \(79\) days.

- Convert the weeks to days first. - Add the remaining days. - How many hours are in one day? - Multiply the total number of days by \(24\).

1. Convert the weeks to days: \(3 \times 7 = 21\) days. 2. Add the extra days: \(21 + 4 = 25\) days. 3. Convert days to hours: \(25 \times 24 = 600\) hours.

The expedition lasts \(600\) hours.

- Convert both times to the same unit. - How many minutes are in one hour? - Compare the totals and subtract.

1. Convert Group B’s time to minutes: \(8 \times 60 + 45 = 525\) minutes. 2. Compare the times: \(525 > 510\), so Group B hikes longer. 3. Find the difference: \(525 - 510 = 15\) minutes.

Group B hikes longer by \(15\) minutes.

- How many hours are in one day? - Convert both times to hours. - In a race, the smaller time is faster.

1. Convert Team Blue’s time to hours: \(4 \times 24 + 5 = 101\) hours. 2. Since \(101 < 105\), Team Blue is faster. 3. Find the difference: \(105 - 101 = 4\) hours.

Team Blue is faster by \(4\) hours.

- Use multiples of \(24\) to find the greatest number of full days within \(300\) hours. - After finding the full days, subtract their hours from \(300\) to get the remainder. - Convert two weeks to days and then to hours before comparing.

1. Find the greatest whole number of days within \(300\) hours. Since \(12 \times 24 = 288\), \(12\) full days use \(288\) hours. 2. The remaining time is \(300-288=12\) hours, so the battery lasts \(12\) days \(12\) hours. 3. Two weeks is \(14\) days, or \(14 \times 24 = 336\) hours. 4. The additional time needed is \(336-300=36\) hours.

a) The battery lasts \(12\) days \(12\) hours. b) It would need to last \(36\) additional hours.

- Add all three weights in grams first. - One kilogram equals \(1000\) grams. - Separate the total into complete thousands and the remainder.

1. Add the three weights: \(12{,}400\,\text{g} + 11{,}950\,\text{g} + 12{,}050\,\text{g} = 36{,}400\,\text{g}\). 2. Since \(1000\,\text{g} = 1\,\text{kg}\), \(36{,}400\,\text{g}\) contains \(36\) kilograms with \(400\) grams remaining. 3. The total weight is \(36\,\text{kg}\ 400\,\text{g}\).

The three crates weigh \(36\,\text{kg}\ 400\,\text{g}\) in all.

- Convert every load to kilograms. - Add the three loads for part a. - Convert \(10\) metric tons to kilograms, then subtract the total load.

1. Convert each load to kilograms: Truck A has \(2450\,\text{kg}\), Truck B has \(1800\,\text{kg}\), and Truck C has \(3050\,\text{kg}\). 2. Add the loads: \(2450\,\text{kg} + 1800\,\text{kg} + 3050\,\text{kg} = 7300\,\text{kg}\). 3. Convert the target mass: \(10\) metric tons \(= 10{,}000\,\text{kg}\). 4. Start with the target of \(10{,}000\,\text{kg}\) and subtract the current total of \(7300\,\text{kg}\). Another \(2700\,\text{kg}\) is needed.

a) The total mass is \(7300\,\text{kg}\), or \(7\) metric tons \(300\,\text{kg}\). b) Another \(2700\,\text{kg}\), or \(2\) metric tons \(700\,\text{kg}\), is needed.

- Convert the packed bag's mass to grams. - Which part of the total is not the books? - After subtracting, can you write the mass using kilograms and grams?

1. Change the total mass into grams: \(2\,\text{kg}\ 100\,\text{g}=2100\,\text{g}\). 2. Subtract the empty bag's mass: \(2100-850=1250\,\text{g}\). 3. Express the mass as \(1\,\text{kg}\ 250\,\text{g}\).

The books weigh \(1\,\text{kg}\ 250\,\text{g}\).

- Find the weight of only the marbles in each package. - Remember that \(1\,\text{kg} = 1000\,\text{g}\). - Compare the two marble weights. - Subtract to find how much greater one weight is.

1. Find the weight of the marbles in Package A: \(1400\,\text{g} - 250\,\text{g} = 1150\,\text{g}\). 2. Find the weight of the marbles in Package B: \(1200\,\text{g} - 100\,\text{g} = 1100\,\text{g}\). 3. Compare: \(1150\,\text{g} > 1100\,\text{g}\), so Package A contains the greater weight of marbles. 4. Find the difference: \(1150\,\text{g} - 1100\,\text{g} = 50\,\text{g}\).

Package A contains the greater weight of marbles. The difference is \(50\,\text{g}\).

- One pound equals \(16\) ounces. - Find the number of complete groups of \(16\) in \(214\). - Convert \(14\) pounds to ounces before finding the difference.

1. Since \(1\,\text{lb} = 16\,\text{oz}\), \(214\,\text{oz}\) contains \(13\) complete pounds because \(13 \times 16 = 208\). 2. The remainder is \(214 - 208 = 6\), so the dog weighs \(13\,\text{lb}\ 6\,\text{oz}\). 3. Fourteen pounds equals \(14 \times 16 = 224\,\text{oz}\). 4. The dog needs \(224\,\text{oz} - 214\,\text{oz} = 10\,\text{oz}\) more.

The dog weighs \(13\,\text{lb}\ 6\,\text{oz}\), and it needs \(10\,\text{oz}\) more to weigh \(14\,\text{lb}\).

- Express the first jump in meters and centimeters. - Put both jumps in centimeters before comparing. - Find the difference between the longer and shorter jump.

1. Convert Ava's jump: \(305\,\text{cm} = 3\,\text{m}\ 5\,\text{cm}\). 2. Convert Mateo's jump to centimeters: \(3\,\text{m}\ 50\,\text{cm} = 350\,\text{cm}\). 3. Since \(350 > 305\), Mateo jumps farther. 4. The difference is \(350\,\text{cm} - 305\,\text{cm} = 45\,\text{cm}\).

Ava's jump is \(3\,\text{m}\ 5\,\text{cm}\). Mateo jumps farther by \(45\,\text{cm}\).

- Add the two parts of Train A’s trip. - Regroup the minutes if their sum is at least \(60\). - Compare Train A’s total with Train B’s time.

1. Add Train A’s times: \(2\,\text{hours}\,45\,\text{minutes} + 3\,\text{hours}\,35\,\text{minutes}\). 2. Add the minutes: \(45 + 35 = 80\) minutes, which is \(1\) hour \(20\) minutes. 3. Train A’s total time is \(6\) hours \(20\) minutes. 4. Compare with Train B’s \(6\) hours \(10\) minutes. Train B is faster. 5. Find the difference: \(6\,\text{hours}\,20\,\text{minutes} - 6\,\text{hours}\,10\,\text{minutes} = 10\,\text{minutes}\).

Train B is faster by \(10\) minutes.

- Find each age difference by subtraction. - Remember that \(1\) year equals \(12\) months. - Regroup one year as \(12\) months when the top number of months is too small.

1. For Lucas and Mia, regroup Lucas’s age as \(10\) years \(15\) months. Then \(10\,\text{years}\,15\,\text{months} - 8\,\text{years}\,7\,\text{months} = 2\,\text{years}\,8\,\text{months}\). 2. For Mia and Ben, regroup Mia’s age as \(7\) years \(19\) months. Then \(7\,\text{years}\,19\,\text{months} - 5\,\text{years}\,10\,\text{months} = 2\,\text{years}\,9\,\text{months}\).

a) Lucas is \(2\) years \(8\) months older than Mia. b) Mia is \(2\) years \(9\) months older than Ben.

- For part a, find the difference in grams for one loaf. - Multiply each per-loaf difference by \(100\). - Convert the final amounts from grams to kilograms.

1. Find the flour saved per loaf: \(450\,\text{g} - 380\,\text{g} = 70\,\text{g}\). 2. Find the savings for \(100\) loaves: \(70\,\text{g} \times 100 = 7000\,\text{g}\). 3. Convert the savings: \(7000\,\text{g} = 7\,\text{kg}\). 4. Find the difference per loaf for part b: \(380\,\text{g} - 200\,\text{g} = 180\,\text{g}\). 5. Find the difference for \(100\) loaves: \(180\,\text{g} \times 100 = 18{,}000\,\text{g}\). 6. Convert the difference: \(18{,}000\,\text{g} = 18\,\text{kg}\).

a) The bakery saves \(7\,\text{kg}\) of flour. b) The new recipe uses \(18\,\text{kg}\) more flour than the small-loaf recipe.

- One meter equals \(100\) centimeters. - Find each unit fraction of \(100\) separately. - Compare the three centimeter measurements.

1. One meter equals \(100\) centimeters. 2. Lucas cuts \(100\,\text{cm} \div 2 = 50\,\text{cm}\). 3. Sarah cuts \(100\,\text{cm} \div 4 = 25\,\text{cm}\). 4. Tyler cuts \(100\,\text{cm} \div 5 = 20\,\text{cm}\). 5. Since \(20 < 25 < 50\), Tyler cuts the shortest piece.

Lucas: \(50\,\text{cm}\) Sarah: \(25\,\text{cm}\) Tyler: \(20\,\text{cm}\) Tyler cuts the shortest piece.

- Convert the full route to meters. - How many times does Leonie walk \(75\,\text{m}\) in \(8\) minutes? - Subtract the distance already walked from the full route.

1. Find the approximate distance walked in \(8\) minutes: \(8 \times 75 \approx 600\,\text{m}\). 2. Convert the full route to meters: \(1\,\text{km} = 1000\,\text{m}\). 3. Find the approximate remaining distance: \(1000 - 600 \approx 400\,\text{m}\).

a) Leonie walks about \(600\,\text{m}\). b) She still needs to walk about \(400\,\text{m}\).

- Find how many one-second steps occur in one minute. - Find the one-minute distance in centimeters before converting. - Use the one-minute result to scale to five minutes.

1. One minute has \(60\) seconds. 2. Find the distance in one minute: \(60 \times 80\,\text{cm} = 4800\,\text{cm}\). 3. Convert: \(4800\,\text{cm} = 48\,\text{m}\). 4. Find the distance in \(5\) minutes: \(5 \times 48\,\text{m} = 240\,\text{m}\).

a) \(48\,\text{m}\) b) \(240\,\text{m}\)

- Find the combined mass of the books. - Find the combined mass of the board games. - Convert the box's limit to grams. - Subtract the current mass from the limit.

1. Find the total mass of the books: \(3 \times 450\,\text{g} = 1350\,\text{g}\). 2. Find the total mass of the board games: \(2 \times 850\,\text{g} = 1700\,\text{g}\). 3. Find the current mass: \(1350\,\text{g} + 1700\,\text{g} = 3050\,\text{g}\). 4. Convert the limit: \(5\,\text{kg} = 5000\,\text{g}\). 5. Find the remaining mass allowed: \(5000\,\text{g} - 3050\,\text{g} = 1950\,\text{g}\).

The remaining contents may have a mass of at most \(1950\,\text{g}\).

- Choose a reasonable step length and use it to estimate how many steps span \(30\,\text{ft}\). - For the yardstick check, express the repeated yard lengths in feet. - Compare that result with the floor-plan length.

1. One reasonable estimate is a step length of about \(2\,\text{ft}\). Fifteen such steps make \(15 \times 2\,\text{ft} = 30\,\text{ft}\), so about \(15\) steps is reasonable. Other reasonable step-length estimates can give nearby answers. 2. One yard is \(3\,\text{ft}\), so \(6\) yardstick lengths measure \(6 \times 3\,\text{ft} = 18\,\text{ft}\). 3. Since \(18\,\text{ft}\) is much less than \(30\,\text{ft}\), Jordan's result is not reasonable.

a) About \(15\) steps using a \(2\)-foot step estimate; other reasonable estimates may give a nearby answer. b) No. Six yardstick lengths equal \(18\,\text{ft}\), not \(30\,\text{ft}\).

- Divide the total volume by the capacity of each larger container. - Convert \(1\,\text{L}\) to milliliters for the cups. - Use multiplication to determine how many \(250\)-milliliter cups make \(1\) liter, then multiply by \(60\).

1. For part a, divide: \(60\,\text{L} \div 2\,\text{L} = 30\) pitchers. 2. For part b, divide: \(60\,\text{L} \div 5\,\text{L} = 12\) containers. 3. For part c, convert \(1\,\text{L} = 1000\,\text{mL}\). Four cups hold \(4 \times 250\,\text{mL} = 1000\,\text{mL}\), so there are \(4\) cups per liter. 4. Scale to \(60\,\text{L}\): \(60 \times 4 = 240\) cups.

a) \(30\) pitchers b) \(12\) containers c) \(4\) cups per liter and \(240\) cups from \(60\,\text{L}\)

- One kilometer equals \(1000\) meters. - Use each denominator to find the corresponding unit fraction of a kilometer in meters. - Subtract the shorter length from the longer length.

1. One kilometer equals \(1000\) meters. 2. Half a kilometer is \(1000\,\text{m} \div 2 = 500\,\text{m}\). 3. One-fifth of a kilometer is \(1000\,\text{m} \div 5 = 200\,\text{m}\). 4. Since \(500 > 200\), the \(\frac{1}{2}\)-kilometer section is longer. The difference is \(500\,\text{m} - 200\,\text{m} = 300\,\text{m}\).

a) \(\frac{1}{2}\,\text{km} = 500\,\text{m}\), and \(\frac{1}{5}\,\text{km} = 200\,\text{m}\). b) The \(\frac{1}{2}\)-kilometer section is longer by \(300\,\text{m}\).

- Write every package mass in the same unit. - Add the packages in each delivery separately. - Compare the two totals.

1. Convert Delivery A's masses to grams and add: \(850\,\text{g} + 1400\,\text{g} + 750\,\text{g} = 3000\,\text{g}\). 2. Convert Delivery B's masses to grams and add: \(1900\,\text{g} + 1100\,\text{g} = 3000\,\text{g}\). 3. Since both totals are \(3000\,\text{g} = 3\,\text{kg}\), the deliveries have the same mass.

The two deliveries have the same mass. Each has a mass of \(3\,\text{kg}\).

- Write the length planted on each day in order. - Each day after the first is longer than the previous day by the same amount. - Use one common unit if that makes the additions easier to organize.

1. Find the second day's length: \(12\,\text{m}\ 40\,\text{cm} + 1\,\text{m}\ 50\,\text{cm} = 13\,\text{m}\ 90\,\text{cm}\). 2. Find the third day's length: \(13\,\text{m}\ 90\,\text{cm} + 1\,\text{m}\ 50\,\text{cm} = 15\,\text{m}\ 40\,\text{cm}\). 3. Add all three days: \(12\,\text{m}\ 40\,\text{cm} + 13\,\text{m}\ 90\,\text{cm} + 15\,\text{m}\ 40\,\text{cm} = 41\,\text{m}\ 70\,\text{cm}\).

Sofia plants \(41\,\text{m}\ 70\,\text{cm}\) of hedge altogether.

- First find the combined length of the North and South sections. - Remember that \(1000\,\text{m} = 1\,\text{km}\). - Then add the amount by which the West section is longer.

1. Add the North and South sections: \(5\,\text{km}\,350\,\text{m} + 4\,\text{km}\,800\,\text{m} = 10\,\text{km}\,150\,\text{m}\). 2. Add the extra length: \(10\,\text{km}\,150\,\text{m} + 1\,\text{km}\,150\,\text{m} = 11\,\text{km}\,300\,\text{m}\).

The West section is \(11\,\text{km}\,300\,\text{m}\) long.

- Add the kilogram parts and gram parts separately. - Decide whether the gram total contains a full kilogram. - Regroup before writing the final mixed-unit amount.

1. Add the kilogram parts: \(3\,\text{kg} + 2\,\text{kg} + 4\,\text{kg} = 9\,\text{kg}\). 2. Add the gram parts: \(450\,\text{g} + 800\,\text{g} + 50\,\text{g} = 1300\,\text{g}\). 3. Regroup \(1300\,\text{g}\) as \(1\,\text{kg}\ 300\,\text{g}\). 4. Combine the amounts: \(9\,\text{kg} + 1\,\text{kg}\ 300\,\text{g} = 10\,\text{kg}\ 300\,\text{g}\).

Amina measures \(10\,\text{kg}\ 300\,\text{g}\) of flour altogether.

- Identify the two parts of the post that are not visible. - Convert the full length to centimeters. - Subtract the hidden parts from the total.

1. Convert the full length: \(4\,\text{m} = 400\,\text{cm}\). 2. Add the hidden parts: \(85\,\text{cm} + 12\,\text{cm} = 97\,\text{cm}\). 3. Subtract the hidden length: \(400\,\text{cm} - 97\,\text{cm} = 303\,\text{cm}\). 4. Convert: \(303\,\text{cm} = 3\,\text{m}\,3\,\text{cm}\).

The visible part is \(3\,\text{m}\,3\,\text{cm}\) long.

- Find the second stage first. - Add the lengths of the first two stages. - Subtract that sum from the total trail length.

1. Find the second stage: \(4\,\text{km}\,250\,\text{m} - 2\,\text{km}\,100\,\text{m} = 2\,\text{km}\,150\,\text{m}\). 2. Add the first two stages: \(4\,\text{km}\,250\,\text{m} + 2\,\text{km}\,150\,\text{m} = 6\,\text{km}\,400\,\text{m}\). 3. Subtract from the full trail: \(15\,\text{km} - 6\,\text{km}\,400\,\text{m} = 8\,\text{km}\,600\,\text{m}\).

The third stage is \(8\,\text{km}\,600\,\text{m}\) long.

- Find the afternoon distance first. - The afternoon distance is shorter, so use the stated difference accordingly. - Then combine the morning and afternoon distances.

1. Find the afternoon distance: \(15\,\text{km}\ 650\,\text{m} - 4\,\text{km}\ 800\,\text{m} = 10\,\text{km}\ 850\,\text{m}\). 2. Add the two distances: \(15\,\text{km}\ 650\,\text{m} + 10\,\text{km}\ 850\,\text{m} = 26\,\text{km}\ 500\,\text{m}\).

Leila walks \(26\,\text{km}\ 500\,\text{m}\) altogether.

- Find the mass of the second load first. - Regroup \(1000\,\text{kg}\) as \(1\) metric ton when needed. - Add the masses of both loads.

1. Find the second load's mass: \(3\) metric tons \(450\,\text{kg} + 1\) metric ton \(600\,\text{kg} = 5\) metric tons \(50\,\text{kg}\). 2. Add both loads: \(3\) metric tons \(450\,\text{kg} + 5\) metric tons \(50\,\text{kg} = 8\) metric tons \(500\,\text{kg}\).

The two loads have a combined mass of \(8\) metric tons \(500\,\text{kg}\).

- Convert the total load to kilograms. - Find the mass of the pears. - Add the apples and pears, then subtract their mass from the total.

1. Convert the total load: \(3\) metric tons \(= 3000\,\text{kg}\). 2. Find the mass of the pears: \(1250\,\text{kg} + 420\,\text{kg} = 1670\,\text{kg}\). 3. Find the combined mass of the apples and pears: \(1250\,\text{kg} + 1670\,\text{kg} = 2920\,\text{kg}\). 4. Find the mass of the plums: \(3000\,\text{kg} - 2920\,\text{kg} = 80\,\text{kg}\).

The truck carries \(80\,\text{kg}\) of plums.

- Find the total weight of all the bags. - Express the trailer capacity in pounds. - Compare the two weights in the same unit.

1. The \(120\) bags weigh \(120 \times 50\,\text{lb} = 6000\,\text{lb}\). 2. Convert the trailer capacity: \(1.5\) tons \(= 3000\,\text{lb}\). 3. Since \(6000\,\text{lb} > 3000\,\text{lb}\), the potatoes are too heavy for one trip.

The statement is not reasonable. The potatoes weigh \(6000\,\text{lb}\), or \(3\) tons, while the trailer capacity is \(3000\,\text{lb}\), or \(1.5\) tons.

- Convert \(1\,\text{km}\) to meters. - Find each student's total distance. - Subtract each distance from \(1000\,\text{m}\) and compare the results.

1. Convert the goal: \(1\,\text{km} = 1000\,\text{m}\). 2. Lucas runs \(4 \times 215\,\text{m} = 860\,\text{m}\), so he needs \(1000\,\text{m} - 860\,\text{m} = 140\,\text{m}\) more. 3. Sarah runs \(6 \times 145\,\text{m} = 870\,\text{m}\), so she needs \(1000\,\text{m} - 870\,\text{m} = 130\,\text{m}\) more. 4. Since \(130\,\text{m} < 140\,\text{m}\), Sarah is closer to \(1\,\text{km}\).

Lucas needs \(140\,\text{m}\) more, and Sarah needs \(130\,\text{m}\) more. Sarah is closer to \(1\,\text{km}\).

- Put the needed amount and package amounts in the same unit. - Find the amount in each group of packages. - Compare the total available with the target and find any shortage.

1. Convert the amount needed: \(1\,\text{kg} = 1000\,\text{g}\). 2. Find the flour in the first group: \(3 \times 185\,\text{g} = 555\,\text{g}\). 3. Find the flour in the second group: \(2 \times 210\,\text{g} = 420\,\text{g}\). 4. Find the total available: \(555\,\text{g} + 420\,\text{g} = 975\,\text{g}\). 5. Since \(975\,\text{g} < 1000\,\text{g}\), there is not enough flour. 6. The shortage is \(1000\,\text{g} - 975\,\text{g} = 25\,\text{g}\).

No. Hana needs \(25\,\text{g}\) more flour.

- Put all three times in the same form before comparing them. - Use multiples of \(60\) to find complete hours. - Compare the hours first and then the minutes.

1. For Leo, \(3\times60=180\) and \(4\times60=240\). So \(215\) minutes is \(3\) hours with \(215-180=35\) minutes remaining: \(3\) hours \(35\) minutes. 2. For Noah, \(3\times60=180\), so \(220\) minutes is \(3\) hours with \(220-180=40\) minutes remaining: \(3\) hours \(40\) minutes. 3. Compare: \(3\,\text{hours}\ 35\,\text{minutes}<3\,\text{hours}\ 40\,\text{minutes}<3\,\text{hours}\ 45\,\text{minutes}\). 4. The order is Leo, Noah, Mia.

Leo: \(3\) hours \(35\) minutes Noah: \(3\) hours \(40\) minutes Mia: \(3\) hours \(45\) minutes Order: Leo, Noah, Mia.

- Subtract the empty backpack's mass from the total to find the contents' mass. - Compare the contents' mass with the empty backpack's mass. - Convert between kilograms and grams as needed.

1. Convert the total mass: \(4\,\text{kg}\,150\,\text{g} = 4150\,\text{g}\). 2. Find the mass of the contents: \(4150\,\text{g} - 950\,\text{g} = 3200\,\text{g}\). 3. Compare the contents with the empty backpack: \(3200\,\text{g} - 950\,\text{g} = 2250\,\text{g}\). 4. Convert the difference: \(2250\,\text{g} = 2\,\text{kg}\,250\,\text{g}\).

The contents have a mass that is \(2\,\text{kg}\,250\,\text{g}\) greater than the mass of the empty backpack.

- Separate the pears' mass from the crate's mass first. - Remove half of the pears, not half of the crate-and-pears total. - Add the empty crate back to the remaining pears.

1. Convert the filled mass to \(14{,}200\,\text{g}\) and the empty crate mass to \(1350\,\text{g}\). The pears have a mass of \(14{,}200\,\text{g} - 1350\,\text{g} = 12{,}850\,\text{g}\). 2. Half of the pears have a mass of \(12{,}850\,\text{g} \div 2 = 6425\,\text{g}\). 3. Add the empty crate mass: \(6425\,\text{g} + 1350\,\text{g} = 7775\,\text{g}\). 4. Convert the total: \(7775\,\text{g} = 7\,\text{kg}\ 775\,\text{g}\).

The crate and remaining pears have a total mass of \(7\,\text{kg}\ 775\,\text{g}\).

- When strips are end to end, their lengths add. - Repeated equal lengths can be found with multiplication. - For part b, regroup the inch total into feet and leftover inches.

1. Fifteen strips have total length \(15 \times 22\,\text{in} = 330\,\text{in}\). 2. Eighty strips have total length \(80 \times 22\,\text{in} = 1760\,\text{in}\). 3. Since \(1760 = 146 \times 12 + 8\), \(1760\,\text{in} = 146\,\text{ft}\ 8\,\text{in}\).

a) \(330\,\text{in}\) b) \(1760\,\text{in}\), or \(146\,\text{ft}\ 8\,\text{in}\)

- Convert all four times to seconds first. - Add the four values. - Use a nearby multiple of \(60\) to convert the total back to minutes and seconds.

1. Convert each time to seconds: \(58\), \(62\), \(65\), and \(57\) seconds. 2. Add the times: \(58+62+65+57=242\) seconds. 3. Since \(4 \times 60=240\), \(242\) seconds is \(4\) minutes with \(2\) seconds remaining. 4. The total time was \(4\) minutes \(2\) seconds.

The relay team's total time was \(4\) minutes \(2\) seconds.

- Find the total time for all class periods. - Add the total break time. - Use a multiple of \(60\) to express the total minutes as hours and minutes before adding it to the start time.

1. Find the total class time: \(5 \times 45=225\) minutes. 2. Find the total break time: \(2 \times 15=30\) minutes. 3. Add the times: \(225+30=255\) minutes. 4. Since \(4 \times 60=240\), \(255\) minutes is \(4\) hours \(15\) minutes. 5. Count forward from \(8{:}15\,\text{a.m.}\): \(8{:}15\,\text{a.m.}+4\,\text{hours}\,15\,\text{minutes}=12{:}30\,\text{p.m.}\).

The program ends at \(12{:}30\,\text{p.m.}\)

- What unit should all three masses use before subtraction? - Evaluate the grouped addition on the right branch before subtracting it from the starting mass.

1. Convert the starting mass: \(1\,\text{kg} = 1000\,\text{g}\). 2. Add the masses removed: \(350\,\text{g} + 250\,\text{g} = 600\,\text{g}\). 3. Subtract: \(1000\,\text{g} - 600\,\text{g} = 400\,\text{g}\).

\(400\,\text{g}\) remain.

- Use the arrangement of dots to identify equal groups of bottles. - Find the volume represented by one row before finding the total volume. - Determine how many \(250\)-milliliter cups make \(1\) liter, then scale to the total.

1. The diagram has \(4\) rows of \(8\) bottles. One row contains \(8 \times 125\,\text{mL}=1000\,\text{mL}\), or \(1\,\text{L}\). 2. Four rows therefore contain \(4\,\text{L}\) altogether. 3. Four \(250\)-milliliter cups hold \(1\,\text{L}\). Therefore, \(4\,\text{L}\) fills \(4 \times 4=16\) cups.

There are \(4\,\text{L}\) of juice, enough to fill \(16\) cups completely.

- What does half an hour tell you about the length of one practice? - Could a time line help compare the practices together with the game?

1. Each half-hour practice is \(30\) minutes. Two practices last \(60\) minutes. 2. Compare with the game: \(60-50=10\) minutes.

Yes. The practices last \(10\) minutes longer.

- Draw the two sides of the scale and label what sits on each. - What would have to be true for the scale to stay level?

1. Three bags weigh \(3\times200=600\,\text{g}\). 2. The second side has \(400+600=1000\,\text{g}\), equal to \(1\,\text{kg}\).

Yes. Both sides weigh \(1000\,\text{g}\).

- Cross the next whole-hour mark when finding the end time. - Find the gap from that end time to the stated later time.

1. Ten minutes brings the time to \(3{:}00\) p.m. Another thirty minutes brings it to \(3{:}30\) p.m. 2. From \(3{:}30\) p.m. to \(4{:}00\) p.m. is \(30\) minutes.

Class ends at \(3{:}30\) p.m.; \(4{:}00\) p.m. is \(30\) minutes later.

- Which lengths make up the whole route? - What unit would let you compare the known parts with the full route?

1. Three sections total \(3\times500=1500\,\text{m}\). 2. The whole bike route is \(2000\,\text{m}\), so \(2000-1500=500\,\text{m}\) remains.

\(500\,\text{m}\)

- Subtract what was already drunk before counting cups. - Use milliliters throughout, then think of equal groups.

1. Three liters contain \(3000\,\text{mL}\). The juice left is \(3000-750=2250\,\text{mL}\). 2. Divide into full cups: \(2250\div250=9\).

\(9\) full cups

- Which single unit works for both the banner and the strips? - Try sketching the banner and the repeated strips as two lengths to compare.

1. The banner measures \(150\,\text{cm}\). 2. The strips total \(3\times40=120\,\text{cm}\), so the banner is \(150-120=30\,\text{cm}\) longer.

\(30\,\text{cm}\)

- You can draw a strip and mark the two cuts without calculating yet. - How will you compare the cut lengths with the original ribbon?

1. Two feet equal \(24\,\text{in}\), so the two cut pieces use \(24+10=34\,\text{in}\). 2. The ribbon left is \(36-34=2\,\text{in}\).

\(2\,\text{in}\)

- Compare both amounts using milliliters. - A quarter of a liter helps you find the benchmark.

1. The jug now has \(350+500=850\,\text{mL}\). 2. Three quarters of a liter is \(750\,\text{mL}\). The jug has \(850-750=100\,\text{mL}\) more.

More, by \(100\,\text{mL}\).

- The scale measures the basket and carrots together. - Which part of the measured mass comes from the basket rather than the food?

1. The carrots weigh \(2\times350=700\,\text{g}\) altogether. 2. The basket alone weighs \(1000-700=300\,\text{g}\).

\(300\,\text{g}\)

- First find the total attendance. - Treat the stated waste amount per spectator as the model for this exercise. - Track the units carefully before comparing the total with the amount removed per shift.

1. Multiply the spectators per event by the number of events: \(2500 \times 4 = 10{,}000\). 2. Find the waste in grams: \(10{,}000 \times 20\,\text{g} = 200{,}000\,\text{g}\). 3. Convert: \(200{,}000\,\text{g} = 200\,\text{kg}\). 4. Since \(4 \times 50\,\text{kg} = 200\,\text{kg}\), the crew needs \(4\) shifts.

a) \(10{,}000\) spectators b) \(200\,\text{kg}\) c) \(4\) shifts

- How many grams are in \(3\,\text{kg}\)? - Find the weight of everything already packed. - Find how much weight remains before the bag reaches the limit. - Share the remaining amount equally between the two books.

1. Convert the maximum weight: \(3\,\text{kg} = 3000\,\text{g}\). 2. Find the current weight: \(1150\,\text{g} + 300\,\text{g} + 650\,\text{g} = 2100\,\text{g}\). 3. Find the weight available for both books: \(3000\,\text{g} - 2100\,\text{g} = 900\,\text{g}\). 4. Divide the available weight equally: \(900\,\text{g} \div 2 = 450\,\text{g}\).

Each book can weigh at most \(450\,\text{g}\).

- Convert both trip lengths to minutes before finding the difference. - How many minutes are in one hour? - For the departure time, work backward by the full hours and then the remaining minutes.

1. Convert the bus time to minutes: \(7 \times 60 + 15 = 435\) minutes. 2. Convert the train time to minutes: \(3 \times 60 + 55 = 235\) minutes. 3. Find the difference: \(435 - 235 = 200\) minutes. 4. Work backward \(7\) hours from \(4{:}30\,\text{p.m.}\) to get \(9{:}30\,\text{a.m.}\). 5. Subtract the remaining \(15\) minutes: \(9{:}30\,\text{a.m.} - 15\,\text{minutes} = 9{:}15\,\text{a.m.}\).

a) The train trip is \(200\) minutes shorter. b) The bus left at \(9{:}15\,\text{a.m.}\).

- How many cuts are needed to make \(4\) pieces? - Convert all lengths to millimeters before calculating. - Subtract the material lost in every cut before dividing the remaining length.

1. Cutting one post into \(4\) pieces requires \(3\) cuts. 2. Find the total wood lost: \(3 \times 4 = 12\,\text{mm}\). 3. Convert the original length: \(2\,\text{m} = 2000\,\text{mm}\). 4. Subtract the wood lost: \(2000 - 12 = 1988\,\text{mm}\). 5. Divide the remaining length into \(4\) equal pieces: \(1988 \div 4 = 497\,\text{mm}\).

Each finished piece is \(497\,\text{mm}\) long.

- Find the total length needed for the \(6\) pieces. - Find the total waste from all the cuts and convert it to centimeters. - Subtract both the pieces and the waste from the original ribbon length.

1. Find the total length of the six pieces: \(6 \times 80 = 480\,\text{cm}\). 2. Cutting off \(6\) separate pieces requires \(6\) cuts. 3. Find the total waste: \(6 \times 5 = 30\,\text{mm}\). 4. Convert the waste: \(30\,\text{mm} = 3\,\text{cm}\). 5. Find the total ribbon used: \(480 + 3 = 483\,\text{cm}\). 6. Convert the original length: \(5\,\text{m} = 500\,\text{cm}\). 7. Subtract: \(500 - 483 = 17\,\text{cm}\).

\(17\,\text{cm}\) of ribbon remain.

- First find the number of kernels on one square meter. - Scale that amount to the full \(5\,\text{m}^2\). - How many groups of \(1000\) kernels are in the total? - Convert the final weight from grams to kilograms.

1. Find the approximate number of kernels per square meter: \(400 \times 40 \approx 16{,}000\). 2. Find the approximate number of kernels in \(5\,\text{m}^2\): \(5 \times 16{,}000 \approx 80{,}000\). 3. Find the approximate number of \(1000\)-kernel groups: \(80{,}000 \div 1000 \approx 80\). 4. Find the approximate total weight: \(80 \times 50\,\text{g} \approx 4000\,\text{g}\). 5. Convert to kilograms: \(4000\,\text{g} = 4\,\text{kg}\), so the harvest is about \(4\,\text{kg}\).

The garden can produce about \(4\,\text{kg}\) of wheat.

- Find the number of kernels per square meter, then for the entire field. - Use the weight of \(1000\) kernels to find the total weight. - Divide the total weight by the amount in one bag.

1. Find the approximate number of kernels per square meter: \(400 \times 50 \approx 20{,}000\). 2. Find the approximate total number of kernels: \(10 \times 20{,}000 \approx 200{,}000\). 3. Find the approximate number of \(1000\)-kernel groups: \(200{,}000 \div 1000 \approx 200\). 4. Find the approximate harvest weight: \(200 \times 40\,\text{g} \approx 8000\,\text{g}\). 5. Estimate the number of bags: \(8000\,\text{g} \div 100\,\text{g} \approx 80\).

The harvest could fill about \(80\) bags.

- Express the filled crate and empty crate in ounces. - Remove the crate's own weight before accounting for the filled bags. - Compare the remaining apple mass with the fifth bag's target.

1. Convert the filled crate's weight: \(11\,\text{lb} = 176\,\text{oz}\). 2. Convert the empty crate's weight: \(1\,\text{lb}\ 8\,\text{oz} = 24\,\text{oz}\). 3. Find the apples' starting weight: \(176\,\text{oz} - 24\,\text{oz} = 152\,\text{oz}\). 4. The first four bags use \(4 \times 32\,\text{oz} = 128\,\text{oz}\). 5. The remaining apples weigh \(152\,\text{oz} - 128\,\text{oz} = 24\,\text{oz}\). 6. The fifth bag needs \(1\,\text{lb}\ 12\,\text{oz} = 28\,\text{oz}\), so the remaining amount is not enough. 7. The shortage is \(28\,\text{oz} - 24\,\text{oz} = 4\,\text{oz}\).

No. Only \(1\,\text{lb}\ 8\,\text{oz}\) remain, so Tariq is \(4\,\text{oz}\) short.

- Convert the total length to meters. - Find half of the total length. - Find the second week's amount, then add both weeks before subtracting from the halfway point.

1. Convert the total length: \(2\,\text{km} = 2000\,\text{m}\). 2. Find half of the total length: \(2000 \div 2 = 1000\,\text{m}\). 3. Find the second week's distance: \(320 - 45 = 275\,\text{m}\). 4. Find the total installed after two weeks: \(320 + 275 = 595\,\text{m}\). 5. Find the amount still needed to reach halfway: \(1000 - 595 = 405\,\text{m}\).

The workers must install \(405\,\text{m}\) more to reach half of the total length.

- For each site, convert all measurements to kilograms. - Add the deliveries for each site. - Subtract each delivered total from that site's target. - Compare the two missing amounts.

1. For Oak Street, find the delivered mass: \(4650\,\text{kg} + 3800\,\text{kg} = 8450\,\text{kg}\). 2. Start with Oak Street's target of \(12{,}000\,\text{kg}\) and subtract the delivered mass of \(8450\,\text{kg}\). Oak Street still needs \(3550\,\text{kg}\). 3. For Robin Street, find the delivered mass: \(5200\,\text{kg} + 2950\,\text{kg} = 8150\,\text{kg}\). 4. Start with Robin Street's target of \(10{,}000\,\text{kg}\) and subtract the delivered mass of \(8150\,\text{kg}\). Robin Street still needs \(1850\,\text{kg}\). 5. Compare: \(3550\,\text{kg} > 1850\,\text{kg}\), so Oak Street still needs more sand.

The Oak Street site still needs the greater amount. Oak Street needs \(3550\,\text{kg}\), while Robin Street needs \(1850\,\text{kg}\).

- Subtract the age difference to find Bertha’s age. - Regroup one year as \(12\) months when subtracting. - Add the two ages for the second question. - Regroup any total of \(12\) or more months.

1. Convert one year to \(12\) months before subtracting: \(14\,\text{years}\,8\,\text{months} = 13\,\text{years}\,20\,\text{months}\). 2. Subtract the age difference: \(13\,\text{years}\,20\,\text{months} - 5\,\text{years}\,9\,\text{months} = 8\,\text{years}\,11\,\text{months}\). 3. Add the two ages: \(14\,\text{years}\,8\,\text{months} + 8\,\text{years}\,11\,\text{months} = 22\,\text{years}\,19\,\text{months}\). 4. Regroup \(19\) months as \(1\) year \(7\) months. The total is \(23\) years \(7\) months.

1. Bertha is \(8\) years \(11\) months old. 2. The sum of their ages is \(23\) years \(7\) months.

- Find the total seed-potato mass used in both beds. - Find the second bed's harvest from the stated difference. - Add both harvests, then compare that total with the seed-potato mass.

1. Find the total mass of seed potatoes: \(2 \times 12\,\text{kg}\ 500\,\text{g} = 25\,\text{kg}\). 2. Find the harvest from the second bed: \(110\,\text{kg}\ 400\,\text{g} - 15\,\text{kg}\ 250\,\text{g} = 95\,\text{kg}\ 150\,\text{g}\). 3. Find the total harvest: \(110\,\text{kg}\ 400\,\text{g} + 95\,\text{kg}\ 150\,\text{g} = 205\,\text{kg}\ 550\,\text{g}\). 4. Compare the harvest with the seed potatoes: \(205\,\text{kg}\ 550\,\text{g} - 25\,\text{kg} = 180\,\text{kg}\ 550\,\text{g}\).

The total harvest is \(180\,\text{kg}\ 550\,\text{g}\) greater than the mass of the seed potatoes.

- Find the total mass of the two wax blocks. - Find the mass of the candles made from the second block. - Add the masses of all the candles. - Subtract the candle mass from the original wax mass.

1. Find the total mass of the wax blocks: \(2 \times 5\,\text{kg}\,200\,\text{g} = 10\,\text{kg}\,400\,\text{g}\). 2. Find the mass of the candles from the second block: \(4\,\text{kg}\,850\,\text{g} - 420\,\text{g} = 4\,\text{kg}\,430\,\text{g}\). 3. Find the total mass of the candles: \(4\,\text{kg}\,850\,\text{g} + 4\,\text{kg}\,430\,\text{g} = 9\,\text{kg}\,280\,\text{g}\). 4. Find the waste: \(10\,\text{kg}\,400\,\text{g} - 9\,\text{kg}\,280\,\text{g} = 1\,\text{kg}\,120\,\text{g}\).

The total mass of the wax waste is \(1\,\text{kg}\,120\,\text{g}\).

- Convert weeks to days first. - One day equals \(24\) hours. - Regroup one day as \(24\) hours when subtracting. - Compare Beta with each longer mission separately.

1. Convert Mission Gamma to days: \(2 \times 7 + 3 = 17\) days. 2. Compare Alpha and Beta. Regroup Alpha as \(13\) days \(29\) hours: \(13\,\text{days}\,29\,\text{hours} - 9\,\text{days}\,18\,\text{hours} = 4\,\text{days}\,11\,\text{hours}\). 3. Compare Gamma and Beta. Regroup Gamma as \(16\) days \(24\) hours: \(16\,\text{days}\,24\,\text{hours} - 9\,\text{days}\,18\,\text{hours} = 7\,\text{days}\,6\,\text{hours}\).

Mission Beta is \(4\) days \(11\) hours shorter than Mission Alpha and \(7\) days \(6\) hours shorter than Mission Gamma.

- Find the daylight duration first. - How much of the full twenty-four-hour period is left over? - Compare the two durations in hours and minutes.

1. From \(8{:}15\) a.m. to \(4{:}05\) p.m. is \(7\,\text{hr}\ 50\,\text{min}\) of daylight. 2. Subtract from a full day: \(24\,\text{hr}-7\,\text{hr}\ 50\,\text{min}=16\,\text{hr}\ 10\,\text{min}\) of night. 3. The difference is \(16\,\text{hr}\ 10\,\text{min}-7\,\text{hr}\ 50\,\text{min}=8\,\text{hr}\ 20\,\text{min}\).

The dark time is \(8\,\text{hr}\ 20\,\text{min}\) longer.

- Determine the second and third day's distances in order. - Pay attention to whether each comparison says farther or less. - A common unit can make the final total easier to calculate.

1. Convert the first day's distance: \(14\,\text{km}\ 600\,\text{m} = 14{,}600\,\text{m}\). 2. The second-day distance is \(14{,}600\,\text{m} + 3850\,\text{m} = 18{,}450\,\text{m}\). 3. The third-day distance is \(18{,}450\,\text{m} - 2500\,\text{m} = 15{,}950\,\text{m}\). 4. Add the three days: \(14{,}600\,\text{m} + 18{,}450\,\text{m} + 15{,}950\,\text{m} = 49{,}000\,\text{m}\). 5. Convert the total: \(49{,}000\,\text{m} = 49\,\text{km}\).

Ibrahim walks \(49\,\text{km}\) altogether.

- Find Trail C before testing the claim. - Find twice Trail B using the same units. - Compare the two resulting lengths.

1. Add Trails A and B: \(12\,\text{km}\ 250\,\text{m} + 9\,\text{km}\ 900\,\text{m} = 22\,\text{km}\ 150\,\text{m}\). 2. Find Trail C: \(22\,\text{km}\ 150\,\text{m} - 4\,\text{km}\ 500\,\text{m} = 17\,\text{km}\ 650\,\text{m}\). 3. Find twice the length of Trail B: \(2 \times 9\,\text{km}\ 900\,\text{m} = 19\,\text{km}\ 800\,\text{m}\). 4. Since \(17\,\text{km}\ 650\,\text{m} < 19\,\text{km}\ 800\,\text{m}\), Marcus's statement is not correct.

No. Trail C is \(17\,\text{km}\ 650\,\text{m}\), while twice Trail B is \(19\,\text{km}\ 800\,\text{m}\).

- Both the total mass and the difference between the masses must be satisfied. - Try the same structure with two small whole numbers to see how the total and difference constrain the lighter value. - Check a candidate answer by finding the heavier mass and verifying both given facts.

1. Convert both given masses to grams: the combined mass is \(24{,}200\,\text{g}\), and the difference is \(5600\,\text{g}\). 2. Subtract the difference from the combined mass: \(24{,}200-5600=18{,}600\,\text{g}\). This is twice the lighter mass. 3. Divide by \(2\): \(18{,}600\div2=9300\,\text{g}\). 4. Regroup: \(9300\,\text{g}=9\,\text{kg}\ 300\,\text{g}\).

\(9\,\text{kg}\ 300\,\text{g}\)

- Find the red roll's original length first. - For each roll, subtract the amount remaining from the original amount. - Compare the amounts used and find their difference.

1. Find the red roll's original length: \(12\,\text{m}\,40\,\text{cm} + 1\,\text{m}\,50\,\text{cm} = 13\,\text{m}\,90\,\text{cm}\). 2. Find the amount used from the blue roll: \(12\,\text{m}\,40\,\text{cm} - 3\,\text{m}\,80\,\text{cm} = 8\,\text{m}\,60\,\text{cm}\). 3. Find the amount used from the red roll: \(13\,\text{m}\,90\,\text{cm} - 4\,\text{m}\,25\,\text{cm} = 9\,\text{m}\,65\,\text{cm}\). 4. Compare and subtract: \(9\,\text{m}\,65\,\text{cm} - 8\,\text{m}\,60\,\text{cm} = 1\,\text{m}\,5\,\text{cm}\).

More ribbon was used from the red roll, by \(1\,\text{m}\,5\,\text{cm}\).

- Picture two copies of the shorter trail plus the extra length. - What remains if you remove the extra part from their combined length? - Use the two equal parts to find the shorter trail, then add back the extra part.

1. The two trails total \(12{,}400\,\text{m}\), and their difference is \(2200\,\text{m}\). 2. Remove the longer trail's extra \(2200\,\text{m}\) from the total: \(12{,}400-2200=10{,}200\,\text{m}\). This leaves two equal shorter-trail lengths. 3. One shorter trail is \(10{,}200\div2=5100\,\text{m}\). 4. The longer trail is \(5100+2200=7300\,\text{m}=7\,\text{km}\ 300\,\text{m}\).

The longer trail is \(7\,\text{km}\ 300\,\text{m}\) long.

- Convert all masses to grams. - Find the mass of the second package. - Add the first two packages, then subtract their mass from the total. - Convert the result back to kilograms and grams.

1. Convert the total and first package to grams: \(12\,\text{kg} = 12{,}000\,\text{g}\) and \(4\,\text{kg}\,250\,\text{g} = 4250\,\text{g}\). 2. Find the second package's mass: \(4250\,\text{g} - 850\,\text{g} = 3400\,\text{g}\). 3. Find the combined mass of the first two packages: \(4250\,\text{g} + 3400\,\text{g} = 7650\,\text{g}\). 4. Start with the total of \(12{,}000\,\text{g}\) and subtract \(7650\,\text{g}\). The third package has a mass of \(4350\,\text{g}\). 5. Convert: \(4350\,\text{g} = 4\,\text{kg}\,350\,\text{g}\).

The third package has a mass of \(4\,\text{kg}\,350\,\text{g}\).

- Convert the full length of the break from weeks and days to days. - Remember that July has 31 days. - Find the final day of summer break before counting the days that remain.

1. Convert the length of the break to days: \(6 \times 7 + 2 = 44\) days. 2. From July 12 through July 31, there are \(20\) days. Therefore, \(44 - 20 = 24\) days of the break are in August, so August 24 is the last day of break. 3. After August 20, the remaining days are August 21, 22, 23, and 24. That is \(4\) days.

There are \(4\) days of summer break remaining after the family returns.