Free math worksheets

- Interpret “\(4\) times as much” as \(4\) equal groups. - Use the small tank’s capacity as the size of each group. - Keep liters as the unit in the answer.

1. Four times as much means \(4\) equal groups of \(3\) liters. 2. Multiply: \(4 \times 3 = 12\).

The larger tank holds \(12\,\text{L}\).

- Which operation shows how many times a smaller price fits into a larger price? - Use a related multiplication fact or break apart the larger number.

1. Divide the bicycle price by the light price: \(360 \div 6 = 60\). 2. Therefore, the bicycle costs \(60\) times as much as the light.

The bicycle costs \(60\) times as much as the light.

- Use the small tree as the reference height in both comparisons. - Ask what factor changes \(40\,\text{cm}\) into each greater height. - Check each factor by multiplying it by the small tree’s height.

1. For the large tree, find the factor that makes \(40\) equal \(320\): \(40 \times 8=320\). 2. For the utility pole, find the factor that makes \(40\) equal \(800\): \(40 \times 20=800\).

a) The large tree is \(8\) times as tall. b) The utility pole is \(20\) times as tall.

- Your question must ask for a quantity that is not already stated. - Make sure the four-times relationship is necessary to answer your question. - Check that your calculation matches the comparison in the story.

1. One valid question is, “How many tickets were sold on Saturday?” 2. The Saturday amount is \(62 \times 4=248\).

One valid response is: Question: “How many tickets were sold on Saturday?” Answer: The theater sold \(248\) tickets on Saturday.

- Do not ask for the lock price because it is already given. - Your question must require the three-times relationship. - Check that your calculation matches the comparison.

1. One valid question is, “How much does the helmet cost?” 2. The helmet costs \(\$14 \times 3=\$42\).

One valid response is: Question: “How much does the helmet cost?” Answer: The helmet costs \(\$42\).

- Convert both measurements in a pair to the same unit. - Recall the conversion factors for yards, feet, inches, pounds, ounces, and tons. - Divide the larger numerical value by the smaller one.

1. a) Since \(1\,\text{yd}=3\,\text{ft}\), the first measurement is \(3\) times as large. 2. b) Since \(1\,\text{ft}=12\,\text{in.}\), the first measurement is \(12\) times as large. 3. c) Since \(1\,\text{lb}=16\,\text{oz}\), the first measurement is \(16\) times as large. 4. d) Since \(1\,\text{ton}=2000\,\text{lb}\), calculate \(2000 \div 200=10\). The first measurement is \(10\) times as large.

a) \(3\) times b) \(12\) times c) \(16\) times d) \(10\) times

- Think about how many bucketfuls fit into the aquarium. - Which operation separates a total into equal-size groups? - Use a related multiplication fact to check the division.

1. Divide the aquarium's capacity by the bucket's capacity: \(200\,\text{L} \div 5\,\text{L} = 40\). Therefore, \(40\) full buckets are needed. 2. Since \(40\) bucketfuls fit in the aquarium, the aquarium holds \(40\) times as much water as the bucket.

a) \(40\) full buckets b) \(40\) times as much water

- One comparison should describe how much greater one capacity is. - The other should describe how many equal copies of the smaller capacity make the larger capacity. - Make sure your two comparison sentences express different kinds of relationships.

1. The additive difference is \(360-9=351\), so the tank holds \(351\,\text{L}\) more than the watering can. 2. For the multiplicative comparison, find the factor that changes \(9\) to \(360\): \(9\times40=360\). Thus the tank holds \(40\) times as much as the watering can. 3. The additive comparison describes a difference of \(351\,\text{L}\); the multiplicative comparison describes a factor of \(40\).

a) \(360-9=351\). The tank holds \(351\,\text{L}\) more than the watering can. b) \(9\times40=360\). The tank holds \(40\) times as much as the watering can. c) The additive comparison gives the difference between the capacities, while the multiplicative comparison gives the factor relating them.

- Use the rows and columns in each panel to determine the number of plants efficiently. - Compare the two totals by thinking about equal groups. - The multiplication and division equations should use the same three numbers.

1. Panel a) shows \(60\) tomato plants, and panel b) shows \(20\) pepper plants. 2. There are three groups of \(20\) in \(60\), so \(3 \times 20=60\). 3. The related division equation is \(60 \div 20=3\).

There are three times as many tomato plants as pepper plants: \(3 \times 20=60\) and \(60 \div 20=3\).

- Read the comics value first. - Think about what quantity would be one-half of the comics value. - Find the category whose bar has that value.

1. Comics were read \(20\) times. 2. Half of \(20\) is \(10\). 3. The mysteries bar shows \(10\), so mysteries were read half as often as comics.

Mysteries were read half as often as comics.

- Interpret “half the actual number” as a relationship between the recorded and actual honey quantities. - Decide how the actual honey amount must compare with the recorded \(15\). - Change only the honey entry after finding the actual amount.

1. If \(15\) is half the actual honey amount, the actual amount is twice \(15\): \(2 \times 15=30\). 2. The other product values do not change.

Actual honey sales: \(30\) units Eggs: \(40\) Honey: \(30\) Cheese: \(25\) Juice: \(20\)

- Read both quantities from the horizontal scale. - Ask how many equal copies of the smaller amount make the larger amount. - Your multiplication and division equations should describe the same comparison.

1. Read the bars: Amina read \(9\) books and Mateo read \(45\) books. 2. Since \(9\times5=45\), the multiplication equation is \(5\times9=45\). 3. The related division equation is \(45\div9=5\). 4. Mateo read \(5\) times as many books as Amina.

\(5\times9=45\) and \(45\div9=5\). Mateo read \(5\) times as many books as Amina.

- Decide which person's amount is the unknown smaller quantity. - The phrase “\(5\) times as many” tells how the larger amount is built from the smaller one. - Check your value by substituting it into the equation.

1. Priya's amount is \(5\) times Noah's amount, so \(5\times n=45\). 2. Since \(5\times9=45\), \(n=9\).

\(5\times n=45\), so \(n=9\).

- Count the equal sections in the bar. - One section is the comparison unit represented by Leo's amount. - Relate the number of equal sections to the phrase “times as many.”

1. The bar has \(4\) equal sections. 2. Jada has \(4\times7=28\) beads. 3. Leo has one section, or \(7\) beads, so Jada has \(4\) times as many beads as Leo.

Jada has \(28\) beads. She has \(4\) times as many as Leo, and the model matches \(4\times7=28\).

- Find Sarah’s score before trying to find Tim’s score. - Think about how Sarah’s score is related to Lucas’s score. - Then use the same relationship between Tim’s score and Sarah’s score.

1. Sarah’s score is twice Lucas’s score: \(135{,}000 \times 2=270{,}000\). 2. Tim’s score is twice Sarah’s score: \(270{,}000 \times 2=540{,}000\).

Sarah scored \(270{,}000\) points. Tim scored \(540{,}000\) points.

- Compare the first number of bushes with the number Mr. Schmidt wants. - Think about how an equal-item total changes when the number of items changes by that factor. - You can use a unit-price method as a check after finding a comparison-based method.

1. One method is to find the unit price: \(\$54 \div 6=\$9\), then calculate \(12 \times \$9=\$108\). 2. Another method uses the multiplicative relationship between the quantities. Since \(12\) is twice \(6\), the cost is twice \(\$54\). 3. Doubling \(\$54\) gives \(\$108\).

a) The \(12\) rose bushes cost \(\$108\) altogether. b) Yes. Since \(12\) is twice \(6\), double \(\$54\) to get \(\$108\).

- Ask how many times the smaller weight fits into the greater weight. - Use a question that includes “times as heavy.” - Find a related multiplication fact.

1. One possible question is, “How many times as heavy is the large dog as the small dog?” 2. Divide the greater weight by the smaller weight: \(42 \div 6 = 7\).

Question: “How many times as heavy is the large dog as the small dog?” Answer: The large dog is \(7\) times as heavy as the small dog.

- Find the price of one flower in each package. - Then compare the two unit prices using division. - Check whether the larger unit price is a whole-number multiple of the smaller one.

1. One rose costs \(\$32 \div 8 = \$4\). 2. One tulip costs \(\$10 \div 5 = \$2\). 3. Since \(\$4 \div \$2 = 2\), one rose costs twice as much as one tulip.

One rose costs \(2\) times as much as one tulip.

- How many hours are in a full day and night? - Subtract the nighttime hours to find the remaining daylight hours. - Which operation tells how many times one amount fits into another?

1. Find the daylight period: \(24\,\text{hr} - 18\,\text{hr} = 6\,\text{hr}\). 2. Compare the durations: \(18\,\text{hr} \div 6\,\text{hr} = 3\). The night is \(3\) times as long as the daylight period.

There are \(6\) hours of daylight. The night is \(3\) times as long as the daylight period.

- What operation finds half of an amount? - First find the number of daylight hours in December. - Subtract the daylight hours from the full \(24\)-hour day.

1. Find the December daylight period: \(12\,\text{hr} \div 2 = 6\,\text{hr}\). 2. Find the nighttime period: \(24\,\text{hr} - 6\,\text{hr} = 18\,\text{hr}\).

The night lasts \(18\) hours.

- First find the cost of one crate of pansies. - How can you find how many times the smaller price fits into the larger price? - Which operation compares prices multiplicatively?

1. Find the cost of one crate of pansies: \(\$20 \div 4 = \$5\). 2. Compare the prices by division: \(\$15 \div \$5 = 3\).

A crate of roses costs \(3\) times as much as a crate of pansies.

- Can you compare the group's \(48\) muffins directly with the teacher's \(24\)? - What must you know about one child to make a fair comparison? - Imagine each child's muffins shown separately beside the teacher's muffins.

1. Find the number of muffins baked by one child: \(48 \div 6 = 8\). 2. Compare the teacher's amount with one child's amount: \(24 \div 8 = 3\). 3. The total of \(48\) muffins represents six children together, so it cannot be compared directly with the work of one teacher. A per-person amount is needed for a fair comparison.

a) The teacher bakes \(3\) times as many muffins as one child. b) You must first find one child's amount so that the comparison is between two individual people.

- First find the number of pages in the reference section. - Once you know both page counts, determine how many times the smaller number fits into the larger one. - Removing a factor of ten from both numbers may make the division easier.

1. Find the number of pages in the reference section: \(240 - 210 = 30\). 2. Compare the page counts by division: \(210 \div 30 = 7\). The main text has seven times as many pages as the reference section.

The main text has \(7\) times as many pages as the reference section.

- How much does the paint set cost without the paintbrush? - Compare the cost of the paint set with the cost of the paintbrush. - Which operation finds how many times one amount fits into another?

1. Find the cost of the paint set: \(\$45 - \$5 = \$40\). 2. Compare the costs by division: \(\$40 \div \$5 = 8\). The paint set costs eight times as much as the paintbrush.

The paint set costs \(8\) times as much as the paintbrush.

- First find the total number of yellow tulips. - Compare the number of yellow tulips with the number of red tulips. - How many times does the number of red tulips fit into the number of yellow tulips?

1. Find the number of yellow tulips: \(9 + 18 = 27\). 2. Compare the numbers by division: \(27 \div 9 = 3\). There are three times as many yellow tulips as red tulips.

There are \(3\) times as many yellow tulips as red tulips.

- First find Sophie’s exact score. - Use place value to simplify the multiplicative comparison. - How many copies of Lucas’s score make Sophie’s score?

1. Find Sophie's score: \(150 + 450 = 600\). 2. Compare the scores by division: \(600 \div 150 = 4\). Sophie scores four times as many points as Lucas.

Sophie scores \(4\) times as many points as Lucas.

- Use each multiplicative comparison to find the amount raised by each class. - Add the three amounts. - Compare the total with the equipment cost.

1. Class 4B raises \(3 \times \$135 = \$405\). 2. Class 4C raises \(2 \times \$405 = \$810\). 3. Together, the classes raise \(\$135 + \$405 + \$810 = \$1350\). 4. Since \(\$1350 < \$1500\), they did not raise enough. 5. They are short by \(\$1500 - \$1350 = \$150\).

No. The classes raised \(\$1350\), so they are \(\$150\) short.

- First find the total after the number of apples becomes five times as large. - The question asks only for the added apples. - Subtract the original amount from the new total.

1. Find the new total: \(24 \times 5 = 120\). 2. Subtract the original apples: \(120 - 24 = 96\).

The vendor adds \(96\) apples.

- First find the current number of streetlights. - The question asks for the increase, not the new total. - Subtract the old number from the new number.

1. Find the current number: \(8 \times 6 = 48\). 2. Find the increase: \(48 - 8 = 40\).

The number of streetlights increased by \(40\).

- First find the amount of orange juice. - “Four times as many” indicates multiplication. - Then add both amounts.

1. Find the amount of orange juice: \(15\,\text{L} \times 4 = 60\,\text{L}\). 2. Add both kinds of juice: \(15\,\text{L} + 60\,\text{L} = 75\,\text{L}\).

There are \(75\,\text{L}\) of juice altogether.

- Use division to find how many times one price fits into the other. - For part b), find the total cost of two soccer balls. - Divide that total by the price of one playground ball.

1. Divide the prices: \(27 \div 3 = 9\). The soccer ball costs \(9\) times as much. 2. Two soccer balls cost \(2 \times 27 = 54\) dollars. 3. Divide by the price of a playground ball: \(54 \div 3 = 18\).

a) \(9\) times as much b) \(18\) playground balls

- First find the total cost of all the hot dog packages. - Compare that total with the amount spent on mustard. - Which operation finds how many times one amount fits into another?

1. Find the total cost of the hot dogs: \(15 \times \$4 = \$60\). 2. Compare the hot dog cost with the mustard cost: \(\$60 \div \$6 = 10\). Mr. Weber spends ten times as much on hot dogs as on mustard.

He spends \(10\) times as much on hot dogs as on mustard.

- First determine how much of the \(\$100\) remains after the soccer-ball purchase. - Compare the soccer-ball amount with that remaining amount. - Think of a multiplication fact that connects the smaller amount to the larger amount.

1. The amount spent on jump ropes is \(\$100-\$80=\$20\). 2. Since \(\$20 \times 4=\$80\), the soccer-ball amount is \(4\) times the jump-rope amount.

The class spends \(4\) times as much on soccer balls as on jump ropes.

- First find the second-year harvest. - The question asks for the difference, not the new total. - Subtract the first-year harvest from the second-year harvest.

1. Find the second-year harvest: \(15\,\text{kg} \times 6 = 90\,\text{kg}\). 2. Find the increase: \(90\,\text{kg} - 15\,\text{kg} = 75\,\text{kg}\).

The second-year harvest is \(75\,\text{kg}\) greater.

- First find the cost of one chair. - Then find the cost of one table. - How many times does the smaller unit price fit into the larger one?

1. Find the cost of one chair: \(\$400 \div 8 = \$50\). 2. Find the cost of one table: \(\$300 \div 2 = \$150\). 3. Compare the unit prices: \(\$150 \div \$50 = 3\).

One table costs \(3\) times as much as one chair.

- First find the cost of one basketball. - Then find the cost of one medicine ball. - Check whether the medicine-ball price is exactly ten times the basketball price.

1. Find the cost of one basketball: \(\$54 \div 6 = \$9\). 2. Find the cost of one medicine ball: \(\$270 \div 3 = \$90\). 3. Compare the prices: \(\$90 \div \$9 = 10\). The teacher is correct.

Yes. One basketball costs \(\$9\), and one medicine ball costs \(\$90\). Since \(90\) is ten times \(9\), the teacher is correct.

- Your question must ask for a quantity that is not stated directly. - The comparison statement must be necessary to answer it. - Check your result by comparing it with this year’s \(135\,\text{kg}\).

1. One valid question is, “How many kilograms of honey did the beekeeper harvest last year?” 2. The unknown amount must satisfy \(3 \times ?=135\), so it is \(45\,\text{kg}\).

One valid response is: Question: “How many kilograms of honey did the beekeeper harvest last year?” Answer: The beekeeper harvested \(45\,\text{kg}\) last year.

- First use the relationship between the red and green apple counts. - The two questions ask for different kinds of comparison, so keep their meanings separate. - After finding both counts, determine what “how many more” asks you to compare.

1. The number of green apples is \(32 \times 4=128\). 2. The difference is \(128-32=96\).

There are \(128\) green apples, which is \(96\) more than the number of red apples.

- First find the total number of storybooks. - Once you know both numbers, determine how many times the smaller fits into the larger. - What operation does “more than” suggest for the first step?

1. Find the number of storybooks: \(15 + 45 = 60\). 2. Compare the numbers by division: \(60 \div 15 = 4\). There are four times as many storybooks as nonfiction books.

There are \(4\) times as many storybooks as nonfiction books.

- How many liters does the large tank hold altogether? - How many groups of \(25\,\text{L}\) make the large tank's capacity? - Counting by twenty-fives may help.

1. Find the capacity of the large tank: \(25\,\text{L} + 175\,\text{L} = 200\,\text{L}\). 2. Compare the capacities by division: \(200 \div 25 = 8\). The large tank holds eight times as much water as the small tank.

The large tank holds \(8\) times as much water as the small tank.

- First find the elephant's current age. - Once you know both ages, find their difference. - Does the age difference between two living things change over time?

1. Find the elephant's current age: \(60 \div 5 = 12\) years. 2. Find the difference in their ages: \(60 - 12 = 48\) years. 3. Their age difference stays constant, so the turtle was \(48\) years old when the elephant was born.

The turtle was \(48\) years old when the elephant was born.

- Find the total number of chairs in each section first. - Keep track of which section is the reference quantity in the comparison. - Look for a multiplication fact that connects the two section totals.

1. The front section has \(4 \times 18=72\) chairs. 2. The back section has \(3 \times 8=24\) chairs. 3. Since \(24 \times 3=72\), the front section has \(3\) times as many chairs.

The front section has \(3\) times as many chairs as the back section.

- Find the total number of each kind of roll before comparing them. - Decide which total is the reference quantity in “times as many.” - Look for a multiplication fact that connects the two totals.

1. The baker makes \(5 \times 36=180\) plain rolls. 2. The baker makes \(3 \times 15=45\) multigrain rolls. 3. Since \(45 \times 4=180\), the baker makes \(4\) times as many plain rolls.

The baker makes \(4\) times as many plain rolls as multigrain rolls.

- Find Lucas’s number and Sophie’s number separately first. - Keep track of which number is the reference quantity in the final comparison. - Look for a multiplication fact that connects the smaller number to the larger number.

1. Lucas’s number is \(48 \div 4=12\). 2. Sophie’s number is \(100-4=96\). 3. Since \(12 \times 8=96\), Sophie’s number is \(8\) times as large as Lucas’s number.

Sophie’s number is \(8\) times as large as Lucas’s number.

- Try expressing all of the ticket costs in terms of just one ticket type. - Use the “twice as much” relationship to connect the two ticket prices. - Check that your two prices make the total for \(3\) adults and \(4\) children equal \(\$70\).

1. One adult ticket has the same cost as two child tickets, so the \(3\) adult tickets have the same cost as \(3 \times 2=6\) child tickets. 2. Together with the \(4\) child tickets, the total cost is equivalent to \(6+4=10\) child-ticket costs. 3. One child ticket costs \(\$70 \div 10=\$7\). 4. One adult ticket costs \(2 \times \$7=\$14\).

A child ticket costs \(\$7\), and an adult ticket costs \(\$14\).

- Try expressing the fountain pen’s weight using only one of the smaller object types. - Use both comparison statements before assigning a numerical weight to one object. - Check that your two weights reproduce the stated \(200\,\text{g}\) fountain-pen weight.

1. Two packages of pencils weigh as much as \(2 \times 4=8\) erasers. 2. Including the two additional erasers, the fountain pen weighs as much as \(8+2=10\) erasers. 3. One eraser weighs \(200\,\text{g} \div 10=20\,\text{g}\). 4. One package of pencils weighs \(4 \times 20\,\text{g}=80\,\text{g}\).

One eraser weighs \(20\,\text{g}\), and one package of pencils weighs \(80\,\text{g}\).

- Represent the three amounts in a way that preserves both comparison statements. - Look for a common equal-sized unit that can describe all three shares. - Check that your three amounts total \(120\) and satisfy both relationships.

1. Represent Ava’s amount as one equal part. Ben receives two such parts, and Chloe receives the same amount as Ava and Ben together, or three parts. 2. Altogether, there are \(1+2+3=6\) equal parts. 3. Each part is \(120 \div 6=20\) marbles. 4. Ava receives \(20\), Ben receives \(2 \times 20=40\), and Chloe receives \(3 \times 20=60\).

Ava receives \(20\) marbles, Ben receives \(40\), and Chloe receives \(60\).

- Use the total and the “four times as many” relationship together to determine the original two groups. - In parts b) and c), remember which bird count changes and which count stays fixed. - Recheck the comparison after the new geese arrive rather than assuming the original factor stays the same.

1. Geese represent one equal part and ducks represent four such parts, for five equal parts altogether. 2. The number of geese is \(150 \div 5=30\), and the number of ducks is \(4 \times 30=120\). 3. After \(10\) geese arrive, there are \(150+10=160\) birds. 4. There are now \(30+10=40\) geese. Since \(40 \times 3=120\), there are now only \(3\) times as many ducks as geese.

a) There are \(120\) ducks and \(30\) geese. b) There are \(160\) birds. c) No. There are now \(40\) geese, and \(40 \times 3=120\), so there are \(3\) times as many ducks as geese.

- The red-package count must be determined before the overall total can be found. - Use the stated relationship between the red and blue package counts. - After finding both color counts, combine them to answer the question asked.

1. The number of red packages is \(135 \times 4=540\). 2. The total number of packages is \(135+540=675\).

The store has \(675\) packages altogether.

- First find the number of relay-race awards. - Add the awards for both events. - Compare the total needed with \(1000\).

1. Find the number of relay-race awards: \(124 \times 7 = 868\). 2. Find the total needed: \(124 + 868 = 992\). 3. Since \(992 < 1000\), there are enough awards.

Yes. The school needs \(992\) awards, so the \(1000\) printed awards are enough.

- Neither question may ask only for the given bus count. - The four-times relationship must be necessary for each answer. - Make your two questions ask for different unknown quantities.

1. One valid first question is, “How many passengers are in the train car?” The train car has \(15 \times 4=60\) passengers. 2. One valid second question is, “How many passengers are in the bus and train car altogether?” The total is \(15+60=75\) passengers.

One valid pair is: 1. “How many passengers are in the train car?” \(60\) passengers. 2. “How many passengers are there altogether?” \(75\) passengers.

- First convert both measurements to the same smaller unit. - Divide to determine how many equal parts fit into the whole. - For numbers ending in zeros, use place value to simplify the division.

1. a) Convert \(1\,\text{m}\) to \(100\,\text{cm}\). Then \(100 \div 20=5\), so \(20\,\text{cm}\) fits \(5\) times. 2. b) Convert \(1\,\text{kg}\) to \(1000\,\text{g}\). Then \(1000 \div 250=4\), so \(250\,\text{g}\) fits \(4\) times. 3. c) Convert \(1\,\text{km}\) to \(1000\,\text{m}\). Then \(1000 \div 50=20\), so \(50\,\text{m}\) fits \(20\) times.

a) \(5\) times b) \(4\) times c) \(20\) times

- The “four times” statement describes the two-week total. - Subtract the first-week pages from that total. - Compare the second-week amount with \(3 \times 112\).

1. Find the total after two weeks: \(112 \times 4 = 448\). 2. Find the second-week amount: \(448 - 112 = 336\). 3. Check the claim: \(112 \times 3 = 336\), so the student is correct.

a) The class reads \(336\) pages during the second week. b) Yes. Since \(112 \times 3 = 336\), the second-week amount is three times the first-week amount.

- The question asks for a difference, but one of the two card counts must be found first. - Use the relationship between Finn’s amount and Lucas’s amount to determine both counts. - Check that your final result answers “how many more,” not Finn’s total.

1. Finn has \(48 \times 4=192\) cards. 2. The difference is \(192-48=144\).

Finn has \(144\) more cards than Lucas.

- Determine both drink amounts before answering the comparison question. - Use the relationship between the water amount and the apple-juice amount. - Make sure the final number represents “how many more,” not the total amount of water.

1. The fair prepares \(115 \times 3=345\) cups of water. 2. The difference is \(345-115=230\) cups.

The fair prepares \(230\) more cups of water than apple juice.

- Find both shelf counts before determining the requested comparison. - Use the stated relationship between the mystery and nonfiction counts. - Check that the final value is a difference rather than the larger shelf total.

1. The mystery shelf has \(45 \times 3=135\) books. 2. The difference is \(135-45=90\) books.

There are \(90\) more mystery books than nonfiction books.

- First find Marie's total savings. - The question asks for the difference, not Marie's total. - Subtract Lucas's amount from Marie's amount.

1. Find Marie's savings: \(\$160 \times 4 = \$640\). 2. Find the difference: \(\$640 - \$160 = \$480\).

Marie has saved \(\$480\) more than Lucas.

- Use both the total cost and the “six times as much” relationship together. - Represent the two prices with a common equal-sized unit before assigning dollar values. - Check that the two prices add to \(\$42\) and have the stated multiplicative relationship.

1. The pump represents one equal part and the soccer ball represents six such parts, for seven equal parts altogether. 2. One part is \(\$42 \div 7=\$6\). 3. The pump costs \(\$6\), and the soccer ball costs \(6 \times \$6=\$36\).

The pump costs \(\$6\), and the soccer ball costs \(\$36\).

- Use equal parts to represent “three times as many.” - Removing \(4\) fruits and adding \(4\) others keeps the total unchanged. - Compare the two new amounts at the end.

1. Pears represent \(1\) equal part and apples represent \(3\) parts, for \(4\) parts altogether. 2. Each part is \(48\div 4=12\), so there are \(12\) pears and \(3\times 12=36\) apples. 3. After the change, there are \(36-4=32\) apples and \(12+4=16\) pears. 4. The difference is \(32-16=16\).

a) \(36\) apples and \(12\) pears b) There are \(16\) more apples than pears.

- Find today’s complete device total before making the comparison. - Use the original \(14\) devices as the reference quantity. - Look for a multiplication fact that connects the original amount to today’s total.

1. The computer labs contain \(3 \times 28=84\) computers. 2. The school now has \(84+42=126\) digital devices. 3. Since \(14 \times 9=126\), the school now owns \(9\) times as many devices as before.

The school owns \(9\) times as many digital devices today.

- Represent the youngest person's age with one equal part. - How many of those parts represent each of the other ages? - How many equal parts make the total of \(66\) years?

1. Let Leo's age be represented by \(1\) equal part. 2. Maya's age is \(2\) parts. 3. Their father's age is \(4 \times 2=8\) parts. 4. Altogether, their ages make \(1+2+8=11\) equal parts. 5. One part is \(66\div11=6\) years. 6. Their father is \(8 \times 6=48\) years old.

Their father is \(48\) years old.

- Express the two longer pieces in terms of the shortest piece. - How many equal parts make the full \(135\,\text{cm}\) length? - After finding the shortest length, use the comparisons to find the other lengths.

1. Represent the shortest piece with \(1\) equal part. 2. The middle piece is \(2\) parts. 3. The longest piece is \(3 \times 2=6\) parts. 4. The entire rope is \(1+2+6=9\) equal parts. 5. One part is \(135\div9=15\,\text{cm}\). 6. The three lengths are \(15\,\text{cm}\), \(2 \times 15=30\,\text{cm}\), and \(6 \times 15=90\,\text{cm}\).

The shortest piece is \(15\,\text{cm}\), the middle piece is \(30\,\text{cm}\), and the longest piece is \(90\,\text{cm}\).

- First find the July harvest. - “Six times as large” indicates multiplication. - Subtract the June harvest to test the claim.

1. Find the July harvest: \(12\,\text{kg} \times 6 = 72\,\text{kg}\). 2. Find the difference: \(72\,\text{kg} - 12\,\text{kg} = 60\,\text{kg}\). 3. The statement is correct.

Yes. The July harvest is \(72\,\text{kg}\), which is exactly \(60\,\text{kg}\) more than the June harvest.

- One of the two geranium counts must be determined from the comparison statement. - Keep the park amount and the flower-bed amount separate until both are known. - The question asks for the combined amount, not only the larger amount.

1. The park needs \(24 \times 8=192\) geraniums. 2. The flower bed and park need \(192+24=216\) geraniums altogether.

The nursery needs \(216\) geraniums altogether.

- Use the employee's comparison to find the number of tickets still available. - Once you know the sold and unsold amounts, how can you find the total? - Breaking \(480\) into hundreds and tens may help with the division.

1. Find the number of tickets still available: \(480 \div 4 = 120\). 2. Add the tickets already sold and the tickets still available: \(480 + 120 = 600\).

There were \(600\) tickets available for the concert altogether.

- Express both prices in the same unit. - One dollar equals \(100\) cents. - Think about what happens to the comparison factor when the smaller price is cut in half.

1. Convert the art set price: \(\$8 = 800\) cents. 2. For part a), \(800 \div 40 = 20\). 3. For part b), \(800 \div 20 = 40\).

a) The art set costs \(20\) times as much. b) The art set costs \(40\) times as much.

- Recall the conversion from liters to milliliters. - For part b), find how many groups of \(25\,\text{g}\) make \(1000\,\text{g}\). - If the comparison unit doubles in length, what happens to the number of times it fits into the same whole?

1. a) Since \(1\,\text{L}=1000\,\text{mL}\), calculate \(1000 \div 200=5\). 2. b) Since \(1\,\text{kg}=1000\,\text{g}\), calculate \(1000 \div 25=40\). 3. c) Since \(1\,\text{m}=100\,\text{cm}\), calculate \(100\,\text{cm} \div 2\,\text{cm}=50\). A \(2\,\text{cm}\) segment is twice as long as a \(1\,\text{cm}\) segment, so it fits half as many times.

a) \(5\) portions b) \(40\) objects c) \(50\) times; doubling the smaller segment halves the number of segments that fit.

- What quantity remains if the red gummy bears are removed from the total? - Describe the relationship shown by the yellow and red parts of the tape diagram. - Use the diagram and the known difference together before finding the total.

1. The difference between the total number and the number of red gummy bears is the number of yellow gummy bears, so there are \(12\) yellow gummy bears. 2. The tape diagram shows one yellow part and three equal red parts, so there are \(3 \times 12=36\) red gummy bears. 3. The total is \(12+36=48\) gummy bears.

There are \(48\) gummy bears in the bowl.

- Read the number represented by each bar. - Look for two quantities where the larger is a whole-number multiple of the smaller. - State the comparison in words after you identify the matching pair.

1. The graph shows \(18\) oak trees and \(9\) beech trees. 2. Since \(18=2 \times 9\), the number of oak trees is twice the number of beech trees.

There are twice as many oak trees as beech trees because \(18=2 \times 9\).

- Read the value represented by each bar. - Look for a pair in which the larger value is three groups of the smaller value. - State the relationship using the two bar names.

1. Bar B has a value of \(24\), and Bar A has a value of \(8\). 2. Since \(24=3 \times 8\), Bar B represents three times the quantity represented by Bar A.

Bar B represents three times the quantity represented by Bar A.

- Read each category’s value from the bar lengths and scale. - Look for a pair in which the larger value is two equal groups of the smaller value. - Use two different pairs for the two comparisons.

1. Apple has \(12\) votes, and pear has \(6\) votes. Since \(12=2 \times 6\), apple has twice as many votes as pear. 2. Pear has \(6\) votes, and melon has \(3\) votes. Since \(6=2 \times 3\), pear has twice as many votes as melon.

1) Apple has twice as many votes as pear. 2) Pear has twice as many votes as melon.

- Read the three values from the bar graph before checking either claim. - For each claim, compare the larger bar with Morning as the reference quantity. - Add the three graph values only after checking the two comparisons.

1. Morning has a value of \(6\) and Noon has a value of \(12\). Since \(12=2 \times 6\), the first comparison is correct. 2. Evening has a value of \(18\). Since \(18=3 \times 6\), the second comparison is correct. 3. The total is \(6+12+18=36\).

a) Yes. Noon’s value is twice Morning’s value. b) Yes. Evening’s value is three times Morning’s value. c) The total is \(36\).

- Identify which quantity is larger and which quantity the factor \(6\) describes. - Compare the meaning of “\(6\) times as many” with “\(6\) more.” - Test the equation you choose against the story before solving it.

1. Rosa's amount is made from \(6\) equal copies of Ben's amount, so the correct equation is \(6\times b=42\). 2. Since \(6\times7=42\), \(b=7\). 3. The addition equation treats \(6\) as an additive difference, and \(b=6\times42\) reverses which quantity is larger.

The correct equation is \(6\times b=42\), because Rosa's amount is made from \(6\) equal copies of Ben's amount. The equation \(b+6=42\) treats \(6\) as an additive difference, and \(b=6\times42\) reverses which quantity is larger. Since \(6\times7=42\), \(b=7\).

- Separate the guide cost from the ticket cost before comparing ticket prices. - Think about how the half-price statement lets you express both ticket types using one common price unit. - Check your final prices against the full \(\$80\) bill, including the guides.

1. The three guides cost \(3 \times \$8=\$24\). 2. The tickets therefore cost \(\$80-\$24=\$56\) altogether. 3. One adult ticket has the same cost as two child tickets, so the \(3\) adult tickets are equivalent in cost to \(6\) child tickets. Including the \(2\) actual child tickets gives \(8\) equal child-ticket costs. 4. A child ticket costs \(\$56 \div 8=\$7\), and an adult ticket costs \(2 \times \$7=\$14\).

An adult ticket costs \(\$14\), and a child ticket costs \(\$7\).

- Treat each situation independently; the relationship among the four shares changes each time. - In each situation, represent amounts that are equal with equal-sized parts before finding a dollar value. - Check that the four individual shares add back to \(\$1800\) in every situation.

1. With four equal shares, each person receives \(\$1800 \div 4=\$450\). 2. In the second situation, Lucas’s amount counts as two equal shares while each other friend’s amount counts as one, for \(5\) equal shares altogether. One share is \(\$1800 \div 5=\$360\), so Lucas receives \(2 \times \$360=\$720\) and each other friend receives \(\$360\). 3. In the third situation, Lucas receives the same amount as the other three friends combined, so the prize is split into two equal amounts of \(\$900\). The other three friends divide \(\$900\) equally, so each receives \(\$900 \div 3=\$300\).

a) Each person receives \(\$450\). b) Lucas receives \(\$720\); each other friend receives \(\$360\). c) Lucas receives \(\$900\); each other friend receives \(\$300\).

- Compare how the relationships among the three program amounts differ between parts a) and b). - For part b), represent the linked “twice as much” statements with equal-sized parts. - Check that the three program amounts total \(\$2100\) in each case.

1. If the three programs receive equal amounts, each receives \(\$2100 \div 3=\$700\). 2. In the second situation, use one equal share for the chess amount. The tennis amount is two such shares, and the soccer amount is four such shares, for \(1+2+4=7\) equal shares altogether. 3. One share is \(\$2100 \div 7=\$300\). 4. Chess receives \(\$300\), tennis receives \(2 \times \$300=\$600\), and soccer receives \(4 \times \$300=\$1200\).

a) Each program receives \(\$700\). b) Chess receives \(\$300\), tennis receives \(\$600\), and soccer receives \(\$1200\).

- Start with the relationship between Room 12 and the combined amount for the other two rooms. - After finding how many seedlings Rooms 14 and 16 plant together, use their multiplicative relationship to separate that combined amount. - Check that all three class amounts add to \(120\) and satisfy both comparison statements.

1. Let the combined number planted by Rooms 14 and 16 be one amount. Room 12 plants that amount plus \(40\). 2. Remove the extra \(40\) from the total: \(120-40=80\). The remaining \(80\) represents two equal combined amounts, so Rooms 14 and 16 together plant \(80 \div 2=40\) seedlings. 3. Room 12 plants \(40+40=80\) seedlings. 4. Room 14 plants three times as many as Room 16. Their combined \(40\) seedlings therefore consist of four equal shares, so one share is \(40 \div 4=10\). 5. Room 16 plants \(10\) seedlings, and Room 14 plants \(3 \times 10=30\) seedlings.

Room 12 plants \(80\) seedlings, Room 14 plants \(30\), and Room 16 plants \(10\).