Free math worksheets

Build your own math worksheets from 30,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Two-step word problems

Click problems to add them to your worksheet.

Difficulty: 1 – 5

5381393

Six more rings are added to the equipment room. Which other kind of equipment will then have the same number as the rings?

Figure for problem 538139

Hints

- Increase only the number of rings. - Find the new number of rings. - Compare that number with the other bars.

Solution

1. The number of rings changes from \(12\) to \(12 + 6 = 18\). 2. The graph shows \(18\) balls. 3. Therefore, there will be the same number of rings and balls.

Answer

There will be the same number of rings and balls.

5381823

In Graph b), the four values from Graph a) were rearranged. 1) Which value moved from A to C? 2) Which value moved from B to D? 3) Which value moved from C to A? 4) Which value moved from D to B?

Figure for problem 538182

Hints

- Match equal values across the two graphs. - Follow each starting bar named in the question. - Record all four value moves.

Solution

1. A in a) and C in b) both have \(12\). 2. B in a) and D in b) both have \(18\). 3. C in a) and A in b) both have \(24\). 4. D in a) and B in b) both have \(30\).

Answer

1) A to C: \(12\) 2) B to D: \(18\) 3) C to A: \(24\) 4) D to B: \(30\)

5383513

On Monday and Friday, club attendance was as follows: Drumming had \(8\) and \(11\) students, Drama had \(12\) and \(10\), and Chess had \(9\) and \(9\). Which statement is true? A: Fewer students attend Drumming on Friday than on Monday. B: Two more students attend Drama on Monday than on Friday. C: One more student attends Chess on Friday than on Monday.

Hints

- Match each statement to the two numbers for that club. - Check all three statements before choosing.

Solution

1. Statement A is false because \(11 > 8\). 2. Statement B is true because \(12 - 10 = 2\). 3. Statement C is false because both values are \(9\).

Answer

Statement B is true.

5383963

A music group practiced \(12\) flute sections on Monday and \(9\) on Wednesday, \(8\) guitar sections on Monday and \(13\) on Wednesday, and \(10\) drum sections on each day. Which instrument has the greatest difference between the two days?

Hints

- Find the difference between the two days for each instrument. - Compare the three differences.

Solution

1. The flute difference is \(12 - 9 = 3\). 2. The guitar difference is \(13 - 8 = 5\). 3. The drum difference is \(10 - 10 = 0\). 4. The greatest difference is \(5\), for guitar.

Answer

Guitar

5383983

A craft workshop has \(25\) pieces of cardboard and uses \(8\), has \(18\) balls of yarn and uses \(6\), and has \(30\) craft sticks and uses \(12\). Which material has the least amount remaining?

Hints

- Subtract the amount used from the amount available for each material. - Compare the three remaining amounts.

Solution

1. Cardboard remaining: \(25 - 8 = 17\). 2. Yarn remaining: \(18 - 6 = 12\). 3. Craft sticks remaining: \(30 - 12 = 18\). 4. The least remaining amount is \(12\), for yarn.

Answer

Yarn has the least amount remaining.

5384023

On Monday, Tuesday, and Wednesday, students checked out \(6\), \(8\), and \(5\) balls; \(4\), \(3\), and \(6\) jump ropes; and \(5\), \(7\), and \(4\) hula hoops. Which kind of equipment was checked out most often during the three days?

Hints

- Find the three-day total for each kind of equipment. - Compare the three totals.

Solution

1. Balls: \(6 + 8 + 5 = 19\). 2. Jump ropes: \(4 + 3 + 6 = 13\). 3. Hula hoops: \(5 + 7 + 4 = 16\). 4. The greatest total is \(19\).

Answer

Balls were checked out most often, with \(19\) checkouts.

5550863

A school collects \(135\) cans on Monday and \(120\) cans on Tuesday. Its goal is \(300\) cans. How many more cans must the school collect to reach the goal?

Hints

- First find how many cans have already been collected altogether. - Then compare that subtotal with the goal. - Check that the collected amount plus your answer equals \(300\).

Solution

1. Combine the two collection days: \(135+120=255\) cans. 2. Find the amount still needed: \(300-255=45\) cans.

Answer

The school must collect \(45\) more cans.

5550873

A classroom has \(4\) boxes with \(6\) markers in each box and \(9\) loose markers. How many markers are there altogether?

Hints

- First find the number of markers in the boxes. - Then add the loose markers. - Keep the two steps separate until you have found the product.

Solution

1. Find the markers in the boxes: \(4 \times 6=24\). 2. Add the loose markers: \(24+9=33\).

Answer

There are \(33\) markers altogether.

5550883

Ms. Rivera has \(50\) stickers. She sets aside \(14\) stickers, then shares the rest equally among \(6\) students. How many stickers does each student receive?

Hints

- The stickers set aside are not shared. - Find the amount that remains before forming equal shares. - The final answer is the size of one equal share.

Solution

1. Find the number left to share: \(50-14=36\). 2. Share equally among six students: \(36 \div 6=6\).

Answer

Each student receives \(6\) stickers.

5550893

Mr. Kim shares \(30\) pencils equally among \(5\) tables. Then he places \(2\) extra pencils on each table. How many pencils does each table have now?

Hints

- First find the equal share before any extra pencils are added. - The same number of extra pencils is then added to each table. - The question asks for the new amount on one table, not the total number of extra pencils.

Solution

1. The equal share is \(30 \div 5=6\) pencils per table. 2. Add the extra pencils: \(6+2=8\) pencils per table.

Answer

Each table has \(8\) pencils.

5550903

Camila has \(5\) packs with \(8\) stickers in each pack. Camila uses \(6\) stickers. Write one equation using \(n\) for the number of stickers left, then find \(n\).

Hints

- Decide what \(n\) represents before writing the equation. - Represent the equal packs first, then show what happens to that total. - Check that your equation follows the order of events in the story.

Solution

1. Camila starts with \(5\times8\) stickers, and then \(6\) are used. 2. An equation is \(n=5\times8-6\). 3. \(5\times8=40\), and \(40-6=34\), so \(n=34\).

Answer

\(n=5 \times 8-6\), so \(n=34\).

5156623

Lucas buys \(8\) trading cards each week for \(4\) weeks. He then receives \(15\) more cards for his birthday. How many cards are in his new collection altogether?

Hints

- First find how many cards Lucas buys over the four weeks. - Equal groups suggest multiplication. - Then add the cards he receives for his birthday.

Solution

1. Find the number of cards Lucas buys: \(4 \times 8 = 32\). 2. Add the birthday cards: \(32 + 15 = 47\).

Answer

Lucas has \(47\) cards altogether.

5156873

A school event begins with \(120\) bottles of apple juice and \(215\) bottles of water. By the end of the event, \(217\) bottles have been used. How many full bottles remain?

Hints

- First combine the two starting groups of bottles. - Then decide how the total changes when bottles are used. - Keep the two steps separate and check that the final amount is less than the starting amount.

Solution

1. Find the starting total: \(120 + 215 = 335\) bottles. 2. Subtract the bottles used: \(335 - 217 = 118\) bottles.

Answer

There are \(118\) full bottles remaining.

5156883

A bird park has \(42\) penguins. It has \(15\) more flamingos than penguins. The park also has \(124\) parakeets. How many flamingos and parakeets are there altogether?

Hints

- Use the penguin count to find the number of flamingos first. - Then focus only on the two bird groups named in the question. - Check that the flamingo count is greater than \(42\) before finding the final total.

Solution

1. Find the number of flamingos: \(42 + 15 = 57\). 2. Combine the flamingos and parakeets: \(57 + 124 = 181\).

Answer

There are \(181\) flamingos and parakeets altogether.

5157413

Students are logging miles in a school bike challenge. Class 3A rides \(267\) miles and Class 3B rides \(284\) miles. a) How many miles do the two third-grade classes ride altogether? b) The fourth-grade classes ride \(686\) miles altogether. How many more miles do they ride than the two third-grade classes?

Hints

- Combine the two third-grade distances first. - For “how many more,” compare the larger total with the smaller total. - Check that your difference added to the third-grade total reaches the fourth-grade total.

Solution

1. Add the two third-grade distances: \(267 + 284 = 551\) miles. 2. Compare the totals: \(686 - 551 = 135\) miles.

Answer

a) The two third-grade classes ride \(551\) miles altogether. b) The fourth-grade classes ride \(135\) more miles.

5161573

Paul wants to buy an electronic keyboard and a stand. The prices are: <table> <tr><td>Electronic keyboard</td><td>\(\$215\)</td></tr> <tr><td>Keyboard stand</td><td>\(\$38\)</td></tr> </table> Paul has \(\$265\). How much money will he have left after the purchase?

Hints

- First combine the two purchase prices. - Then compare the total cost with the amount Paul has. - The money left should be less than \(\$265\).

Solution

1. Find the total cost: \(\$215 + \$38 = \$253\). 2. Subtract the cost from the money Paul has: \(\$265 - \$253 = \$12\).

Answer

Paul will have \(\$12\) left.

5161703

Ms. Brooks has \(\$75\) for a class party. Drinks cost \(\$36\) and snacks cost \(\$45\). Is the budget enough? If not, how much more is needed?

Hints

- Combine the two costs first. - Compare the total cost with the budget before deciding whether there is enough money. - If the cost is larger, the amount needed is the difference between the two amounts.

Solution

1. Find the total cost: \(\$36 + \$45 = \$81\). 2. Since \(\$81 > \$75\), find the shortfall: \(\$81 - \$75 = \$6\).

Answer

No. Ms. Brooks needs \(\$6\) more.

5162084

Lucas lives \(70\,\text{m}\) from school. Each school day, he walks to school in the morning and walks home in the afternoon. How many meters does Lucas walk to and from school during a \(5\)-day school week?

Hints

- Determine how many times Lucas walks the route each school day. - Find his total walking distance for one day. - Use the daily distance to find the distance for all five days.

Solution

1. The distance Lucas walks each day is \(70\,\text{m} \times 2=140\,\text{m}\). 2. The distance for five days is \(140\,\text{m} \times 5=700\,\text{m}\).

Answer

Lucas walks \(700\,\text{m}\) during the school week.

5162094

Sophie goes to music lessons twice each week. Her home is \(110\,\text{m}\) from the music school, and she walks both ways each time. How many meters does Sophie walk for music lessons over two weeks?

Hints

- Find the walking distance for one round trip. - Account for both lessons in one week. - Then extend the weekly distance across two weeks.

Solution

1. One round trip is \(110\,\text{m} \times 2=220\,\text{m}\). 2. Two lessons in one week require \(220\,\text{m} \times 2=440\,\text{m}\) of walking. 3. Over two weeks, Sophie walks \(440\,\text{m} \times 2=880\,\text{m}\).

Answer

Sophie walks \(880\,\text{m}\) over two weeks.

5162103

Tim and Sarah walk to a playground on \(5\) days in one week. Tim walks \(120\,\text{m}\) to the playground and back each day. Sarah walks \(160\,\text{m}\) to the playground and back each day. How many more meters does Sarah walk than Tim during the week?

Hints

- Compare Tim's and Sarah's distances for one day before thinking about the whole week. - The daily difference is repeated on each of the \(5\) days. - The question asks for the difference between their weekly distances, not either weekly total.

Solution

1. Find the difference in their daily distances: \(160\,\text{m} - 120\,\text{m} = 40\,\text{m}\). 2. The same \(40\,\text{m}\) difference occurs on each of \(5\) days, so \(5 \times 40\,\text{m} = 200\,\text{m}\).

Answer

Sarah walks \(200\,\text{m}\) more than Tim during the week.

5162763

During recess, Mia runs \(320\,\text{m}\). She then walks for \(9\) minutes at \(60\,\text{m}\) per minute. How many meters does Mia travel altogether?

Hints

- Find the walking distance from the minutes and the distance per minute. - Then combine that distance with the running distance already given. - Check that the final distance is greater than each part.

Solution

1. Find the walking distance: \(9 \times 60\,\text{m} = 540\,\text{m}\). 2. Add the running and walking distances: \(320\,\text{m} + 540\,\text{m} = 860\,\text{m}\).

Answer

Mia travels \(860\,\text{m}\) altogether.

5162833

The Weber family plans a three-day hike that is \(78\) miles long. They hike \(24\) miles on the first day and \(29\) miles on the second day. They hike the rest on the third day. On which day do they hike the greatest distance?

Hints

- Add the distances from the first two days. - Subtract that sum from the total distance. - Compare all three daily distances.

Solution

1. Add the distances from the first two days: \(24 + 29 = 53\) miles. 2. Subtract from the total to find the third-day distance: \(78 - 53 = 25\) miles. 3. Compare the three distances: \(24\), \(29\), and \(25\). The greatest is \(29\).

Answer

The family hikes the greatest distance on the second day: \(29\) miles.

5162843

A class takes a three-day bicycle trip. The total distance is \(167\) miles. <table> <tr> <th colspan="3">Total distance: \(167\) miles</th> </tr> <tr> <td>Day 1: \(52\) miles</td> <td>Day 2: \(64\) miles</td> <td>Day 3: \(?\) miles</td> </tr> </table> How many miles does the class ride on Day 3? Which day has the shortest distance?

Hints

- Combine the two known daily distances first. - The three daily distances must add to the trip total. - After finding Day 3, compare the three daily distances without doing another calculation.

Solution

1. Add the two known distances: \(52 + 64 = 116\) miles. 2. Subtract from the trip total: \(167 - 116 = 51\) miles on Day 3. 3. Compare \(52\), \(64\), and \(51\). Day 3 is the shortest.

Answer

The class rides \(51\) miles on Day 3. Day 3 has the shortest distance.

5174233

A gardener harvests \(32\,\text{kg}\) of apples and divides them equally among \(4\) crates. a) How many kilograms of apples are in one crate? b) How many kilograms are in \(3\) crates?

Hints

- First find the weight in one crate when all four crates hold equal amounts. - Once you know the weight in one crate, how can you find the weight in three crates?

Solution

1. Find the weight in one crate: \(32\,\text{kg} \div 4 = 8\,\text{kg}\). 2. Find the weight in three crates: \(8\,\text{kg} \times 3 = 24\,\text{kg}\).

Answer

a) One crate contains \(8\,\text{kg}\) of apples. b) Three crates contain \(24\,\text{kg}\).

5174243

A farm packs \(45\,\text{kg}\) of potatoes equally into \(9\) bags. A family buys \(4\) bags. How many kilograms of potatoes does the family buy?

Hints

- First find the equal amount in one bag. - Then use the number of bags the family buys. - Check that the amount for \(4\) bags is less than the full \(45\,\text{kg}\).

Solution

1. Find the weight of one bag: \(45\,\text{kg} \div 9 = 5\,\text{kg}\). 2. Find the weight of \(4\) bags: \(4 \times 5\,\text{kg} = 20\,\text{kg}\).

Answer

The family buys \(20\,\text{kg}\) of potatoes.

5174333

At a school cafeteria, \(4\) servings of fruit salad cost \(\$12\) altogether. a) How much does one serving cost? b) How much do \(7\) servings cost?

Hints

- First find the cost of one serving. - Use that price for one serving for the new number of servings. - Check that buying more than \(4\) servings costs more than \(\$12\).

Solution

1. Find the cost of one serving: \(\$12 \div 4 = \$3\). 2. Find the cost of seven servings: \(7 \times \$3 = \$21\).

Answer

a) One serving costs \(\$3\). b) Seven servings cost \(\$21\).

5174343

A gardener plants \(27\) tulips equally in \(3\) rows. a) How many tulips are in one row? b) The gardener has \(45\) more tulips. How many additional rows of the same size can be planted?

Hints

- Find the size of one equal row first. - Use that row size to group the additional tulips. - The second answer is a number of rows, not a number of tulips.

Solution

1. Find the number of tulips in one row: \(27 \div 3 = 9\). 2. Find how many rows of \(9\) fit into \(45\): \(45 \div 9 = 5\).

Answer

a) One row contains \(9\) tulips. b) The gardener can plant \(5\) additional rows.

5174373

Lucas buys a notebook for \(\$7\) and \(4\) identical pencils. He pays \(\$15\) altogether. How much does one pencil cost?

Hints

- First find how much Lucas spends on the pencils altogether. - Then divide that amount equally among the four pencils. - Use the total cost and the notebook price.

Solution

1. Find the cost of all four pencils: \(\$15 - \$7 = \$8\). 2. Divide by the number of pencils: \(\$8 \div 4 = \$2\).

Answer

One pencil costs \(\$2\).

5174453

Lucas buys a poster for \(\$14\). Two identical picture frames together cost \(\$6\) more than the poster. How much does one picture frame cost?

Hints

- First find the total cost of the two frames. - Then divide that cost equally between the two frames. - The two frames cost more than the poster.

Solution

1. Find the cost of both frames: \(\$14 + \$6 = \$20\). 2. Divide by two: \(\$20 \div 2 = \$10\).

Answer

One picture frame costs \(\$10\).

5174463

Ms. Okafor buys \(8\) identical packages of markers. She pays with \(\$50\) and receives \(\$18\) in change. How much does one package cost?

Hints

- First find how much Ms. Okafor spends altogether. - Then divide the total cost among the eight equal packages. - Change is subtracted from the amount paid.

Solution

1. Find the total amount spent: \(\$50 - \$18 = \$32\). 2. Divide by the number of packages: \(\$32 \div 8 = \$4\).

Answer

One package of markers costs \(\$4\).

5174713

A gym has \(14\) boys and \(16\) girls. They form groups of \(5\) students for a game. How many groups are formed?

Hints

- First find the total number of students. - Then divide the students into equal groups. - Solve the problem in two steps.

Solution

1. Find the total number of students: \(14 + 16 = 30\). 2. Divide into groups of five: \(30 \div 5 = 6\).

Answer

The students form \(6\) groups.

5174724

A nursery has \(160\) tulips and \(240\) daffodils. The flowers are arranged into bouquets with exactly \(8\) flowers in each bouquet. How many bouquets can be made altogether?

Hints

- The bouquets can use flowers from both types, so determine how many flowers are available altogether. - Relate the complete flower total to the fixed bouquet size. - Check that your number of bouquets accounts for all the flowers.

Solution

1. There are \(160+240=400\) flowers altogether. 2. Grouping \(400\) flowers into bouquets of \(8\) gives \(400 \div 8=50\) bouquets.

Answer

The nursery can make \(50\) bouquets.

5174813

Lucas receives \(\$15\) from his grandmother and \(\$13\) from his grandfather. He wants to buy toy cars that cost \(\$4\) each. How many toy cars can Lucas buy with all of the money?

Hints

- How much money does Lucas receive altogether? - Think about how many times the price of one car fits into the total amount. - Which operation divides a total into equal-size amounts?

Solution

1. Find the total amount of money: \(\$15 + \$13 = \$28\). 2. Divide by the cost of one toy car: \(\$28 \div \$4 = 7\).

Answer

Lucas can buy \(7\) toy cars.

5174823

Ms. Weber pays \(\$25\) for some notebooks and receives \(\$4\) in change. Each notebook costs \(\$3\). How many notebooks does Ms. Weber buy?

Hints

- First find how much money was actually spent after the change was returned. - Then use the price of one notebook to form equal groups. - Check that the number of notebooks times \(\$3\) gives the amount spent.

Solution

1. Find the amount spent: \(\$25 - \$4 = \$21\). 2. Divide by the price of one notebook: \(\$21 \div \$3 = 7\).

Answer

Ms. Weber buys \(7\) notebooks.

5175333

A baker needs \(24\) apples to make \(4\) identical apple pies. How many apples are needed to make \(9\) of these pies?

Hints

- First find the number of apples needed for one pie. - Which operation divides a total equally? - Once you know the amount for one pie, how can you find the amount for nine pies?

Solution

1. Find the number of apples needed for one pie: \(24 \div 4 = 6\). 2. Find the number needed for nine pies: \(6 \times 9 = 54\).

Answer

The baker needs \(54\) apples.

5175393

Five classes set up booths for a school fair. At each booth, \(2\) adults and \(6\) students help. How many people help at all five booths altogether?

Hints

- Find the number of people at one booth. - Then use equal groups to find the total for five booths. - You can also find the total numbers of adults and students separately.

Solution

1. Find the number of helpers at one booth: \(2 + 6 = 8\). 2. Find the number at five booths: \(5 \times 8 = 40\).

Answer

There are \(40\) helpers altogether.

5175403

A store has \(9\) school-supply sets. Each set originally contains \(6\) items. The store adds \(1\) highlighter to each set. How many items are in the \(9\) sets now?

Hints

- First find the new number of items in one set. - Then treat the \(9\) sets as equal groups. - Check that adding one item to every set makes the final total larger than \(9 \times 6\).

Solution

1. Each changed set has \(6 + 1 = 7\) items. 2. Nine equal sets contain \(9 \times 7 = 63\) items.

Answer

The \(9\) sets contain \(63\) items.

5175633

A school receives \(100\) wide-ruled notebooks. It also orders \(3\) packages with \(50\) graph-paper notebooks in each package. How many notebooks does the school receive altogether?

Hints

- Find the number of graph-paper notebooks from the equal packages. - Then combine that amount with the wide-ruled notebooks already given. - The final total should be greater than both separate amounts.

Solution

1. Find the graph-paper notebooks: \(3 \times 50 = 150\). 2. Add the two amounts: \(100 + 150 = 250\).

Answer

The school receives \(250\) notebooks altogether.

5175663

A garden has \(18\) tulips planted equally in \(3\) flower beds. a) How many tulips are in each flower bed? b) How many tulips are needed for \(8\) flower beds with the same number in each bed?

Hints

- First find the equal amount in one flower bed. - Use that bed size for the new number of beds. - Check that \(8\) beds require more tulips than \(3\) beds.

Solution

1. Find the number in each flower bed: \(18 \div 3 = 6\) tulips. 2. For \(8\) flower beds, \(8 \times 6 = 48\) tulips are needed.

Answer

a) \(6\) tulips b) \(48\) tulips

5175673

Six boxes of apples have the same weight. Together they weigh \(30\,\text{kg}\). A garden cart can carry at most \(48\,\text{kg}\). Can the cart carry \(10\) boxes at once? Show how you know.

Hints

- Find the weight of one equal box first. - Use that weight to find the weight of \(10\) boxes. - Compare the result with the cart’s maximum load.

Solution

1. Find the weight of one box: \(30\,\text{kg} \div 6 = 5\,\text{kg}\). 2. Ten boxes weigh \(10 \times 5\,\text{kg} = 50\,\text{kg}\). 3. Since \(50\,\text{kg} > 48\,\text{kg}\), the cart cannot carry \(10\) boxes at once.

Answer

No. Ten boxes weigh \(50\,\text{kg}\), which is more than the \(48\,\text{kg}\) limit.

5175803

Luke uses \(8\) ounces of flour to bake \(4\) muffins. How many ounces of flour does he need for \(6\) muffins?

Hints

- First find the amount of flour for one muffin. - Use that amount for the new number of muffins. - Check that the flour amount increases because the batch has more muffins.

Solution

1. Find the amount for one muffin: \(8\,\text{oz} \div 4 = 2\,\text{oz}\). 2. For \(6\) muffins: \(6 \times 2\,\text{oz} = 12\,\text{oz}\).

Answer

Luke needs \(12\,\text{oz}\) of flour for \(6\) muffins.

5175813

Three equal packages contain \(24\) colored pencils altogether. How many colored pencils are in \(7\) packages of the same size?

Hints

- First find the number of colored pencils in one package. - Use that package size for the new number of packages. - Check that \(7\) packages contain more pencils than \(3\) packages.

Solution

1. Find the number in one package: \(24 \div 3 = 8\). 2. Seven packages contain \(7 \times 8 = 56\) colored pencils.

Answer

Seven packages contain \(56\) colored pencils.

5175924

A nursery delivers \(120\) seedlings for a school garden. The teachers plant \(30\) seedlings first. The remaining seedlings are shared equally among \(6\) classes for their own garden beds. How many seedlings does each class receive?

Hints

- First determine how many seedlings remain after the teachers plant some. - The remaining seedlings are split into \(6\) equal shares. - Use division after finding the remainder.

Solution

1. Find the number of seedlings remaining: \(120 - 30 = 90\). 2. Divide the remaining seedlings equally among the classes: \(90 \div 6 = 15\).

Answer

Each class receives \(15\) seedlings.

5175934

A tennis club buys \(100\) new tennis balls. After the first practice, \(28\) balls are missing. The remaining balls are stored in cans that hold exactly \(4\) balls each. How many cans are needed?

Hints

- Find how many balls remain after the missing balls are removed from the total. - Each can holds the same number of balls. - Divide the remaining number of balls by the number that fits in one can.

Solution

1. Find the number of tennis balls remaining: \(100 - 28 = 72\). 2. Divide by the number of balls in each can: \(72 \div 4 = 18\).

Answer

The club needs \(18\) cans.

5176183

During physical education class, Coach Ramirez divides \(40\) jump ropes equally among \(8\) groups. How many jump ropes do \(5\) groups receive altogether?

Hints

- First find how many jump ropes one group receives. - Once you know the amount for one group, how can you find the amount for five groups? - Which operation helps with equal sharing?

Solution

1. Find the number of jump ropes for one group: \(40 \div 8 = 5\). 2. Find the number for five groups: \(5 \times 5 = 25\).

Answer

Five groups receive \(25\) jump ropes altogether.

5176193

For a school event, Class 3A buys \(4\) equal bags with \(32\) oranges altogether. Class 3B buys \(3\) more bags of the same size. How many oranges are in the \(3\) extra bags?

Hints

- First use Class 3A’s total and bag count to find how many oranges are in one bag. - The question asks only about the additional bags, so no subtraction of bag counts is needed. - After finding the amount in one bag, scale it to the stated number of additional bags.

Solution

1. Find the number of oranges in one bag: \(32 \div 4=8\). 2. Multiply the \(3\) additional bags by \(8\) oranges per bag: \(3 \times 8=24\). 3. The additional bags contain \(24\) oranges.

Answer

The \(3\) additional bags contain \(24\) oranges.

5176293

Five identical museum tickets for a class trip cost \(\$35\) altogether. How much do \(9\) of these tickets cost?

Hints

- First find the cost of one ticket. - Which operation divides the total cost equally among the tickets? - Once you know the unit price, how can you find the cost of nine tickets?

Solution

1. Find the cost of one ticket: \(\$35 \div 5 = \$7\). 2. Find the cost of nine tickets: \(9 \times \$7 = \$63\).

Answer

Nine museum tickets cost \(\$63\) altogether.

5176304

At a nursery, \(6\) identical rose bushes cost \(\$54\) altogether. Mr. Schmidt wants to buy \(12\) of these rose bushes. a) How much do the \(12\) rose bushes cost altogether? b) Can you find the answer without first finding the cost of one rose bush? Explain.

Hints

- 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.

Solution

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 \(\$108\).

Answer

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

5176973

Four equal packages of colored pencils cost \(\$32\) altogether. How much do \(9\) packages cost?

Hints

- Find the cost of one package first. - Use that price for one package for \(9\) equal packages. - Check that \(9\) packages cost more than \(4\) packages.

Solution

1. Find the cost of one package: \(\$32 \div 4 = \$8\). 2. Nine packages cost \(9 \times \$8 = \$72\).

Answer

Nine packages cost \(\$72\).

5176983

Finn pays \(\$12\) for \(4\) equal packages of trading cards. He has saved \(\$21\). How many of these packages can he buy with all of his savings?

Hints

- Find the cost of one package first. - Then determine how many one-package prices fit into \(\$21\). - A table can help you match packages with prices.

Solution

1. Find the cost of one package: \(\$12 \div 4 = \$3\). 2. Find the number of packages Finn can buy: \(\$21 \div \$3 = 7\).

Answer

Finn can buy \(7\) packages of trading cards.

5177853

A school cafeteria has \(85\) apples in a crate. There are \(25\) fewer pears than apples. The number of bananas equals the total number of apples and pears. How many bananas are in the crate?

Hints

- Find the number of pears first. - Use “fewer than” to choose the first operation. - Then add the apples and pears.

Solution

1. Find the number of pears: \(85 - 25 = 60\). 2. Add the apples and pears to find the number of bananas: \(85 + 60 = 145\).

Answer

There are \(145\) bananas in the crate.

5177873

A school garden has a vegetable bed that is \(15\,\text{m}\) long. Students extend it by \(6\,\text{m}\) on the left for strawberries and by \(8\,\text{m}\) on the right for herbs. How long is the garden bed after both extensions?

Hints

- Draw the original bed and the two extensions. - Add both new lengths to the original length. - Extending a length means adding, not subtracting.

Solution

1. Add the first extension: \(15 + 6 = 21\,\text{m}\). 2. Add the second extension: \(21 + 8 = 29\,\text{m}\).

Answer

The garden bed is \(29\,\text{m}\) long after both extensions.

5178503

The diagram shows apple trees planted in one straight row. Each marker represents one tree. Neighboring trees are \(4\,\text{m}\) apart. What is the distance from the first tree to the last tree?

Figure for problem 517850

Hints

- Count the spaces between neighboring tree markers rather than counting the markers as spaces. - How many spaces are made by \(8\) objects in one row? - Use the same distance for every space.

Solution

1. The diagram shows \(8\) trees, so there are \(8 - 1 = 7\) spaces between neighboring trees. 2. Multiply the number of spaces by the length of each space: \(7 \times 4\,\text{m} = 28\,\text{m}\).

Answer

The distance from the first tree to the last tree is \(28\,\text{m}\).

5178513

A \(12\,\text{m}\)-long string of pennant flags has a flag at each end. The diagram shows all the equally spaced flags. What is the distance between each pair of neighboring flags?

Figure for problem 517851

Hints

- Count the spaces between the flag markers, not just the flags. - The full \(12\,\text{m}\) is shared equally among those spaces. - Choose the operation that splits one total into equal parts.

Solution

1. The diagram shows \(5\) flags, so there are \(5 - 1 = 4\) equal spaces. 2. Divide the total length among the spaces: \(12\,\text{m} \div 4 = 3\,\text{m}\).

Answer

The distance between neighboring flags is \(3\,\text{m}\).

5178723

Ms. Miller already has \(56\) glitter stickers. She buys \(9\) packs with \(5\) star stickers in each pack. How many stickers does she have altogether?

Hints

- Find the number of star stickers in the equal packs. - Then combine that amount with the glitter stickers already given. - Check that the final total is greater than each separate amount.

Solution

1. Find the number of star stickers: \(9 \times 5 = 45\). 2. Add the two amounts: \(56 + 45 = 101\).

Answer

Ms. Miller has \(101\) stickers altogether.

5178813

A school receives \(100\) new notebooks. It gives out \(60\) of them. The notebooks left are shared equally among \(5\) study groups. How many notebooks does each group receive?

Hints

- First find how many notebooks remain after the ones already given out. - Then share the remaining notebooks equally among the groups. - Check that \(5\) equal groups of your answer use all \(40\) remaining notebooks.

Solution

1. Find the number of notebooks remaining: \(100 - 60 = 40\). 2. Share the \(40\) notebooks equally: \(40 \div 5 = 8\).

Answer

Each study group receives \(8\) notebooks.

5179163

Emma buys \(4\) notebooks for \(\$3\) each. After paying, she has \(\$24\) left. How much money did Emma have before shopping?

Hints

- First find how much Emma spends on the equal-price notebooks. - Her starting amount is the money spent together with the money left. - Check by subtracting the purchase cost from your starting amount.

Solution

1. Find the cost of the notebooks: \(4 \times \$3 = \$12\). 2. Add the amount spent and the money left: \(\$12 + \$24 = \$36\).

Answer

Emma had \(\$36\) before shopping.

5179173

A class buys \(6\) cases of drinks for \(\$9\) each. After paying, \(\$18\) remains in the class fund. How much money was in the fund before the purchase?

Hints

- First find the total cost of the equal-price cases. - The starting fund includes both the amount spent and the money left. - Check by subtracting the purchase cost from your starting amount.

Solution

1. Find the cost of the drinks: \(6 \times \$9 = \$54\). 2. Add the amount spent and the money left: \(\$54 + \$18 = \$72\).

Answer

The class fund originally contained \(\$72\).

5179584

Two swim teams practice three times each week. The Minnows practice for \(45\) minutes each time, and the Sharks practice for \(60\) minutes each time. Over \(4\) weeks, how many more minutes do the Sharks practice than the Minnows?

Hints

- Compare the teams’ practice times for one practice first. - Determine how that difference accumulates over all practices in one week. - Extend the weekly difference across four weeks.

Solution

1. The difference for one practice is \(60-45=15\) minutes. 2. The weekly difference is \(3 \times 15=45\) minutes. 3. Over four weeks, the difference is \(4 \times 45=180\) minutes.

Answer

The Sharks practice \(180\) more minutes over four weeks.

5179593

Three child tickets for a school play cost \(\$5\) each. A group ticket for three children costs \(\$12\). How much is saved by buying the group ticket instead of three individual tickets?

Hints

- Find the cost of the three individual tickets first. - Savings is the difference between the larger price and the smaller price. - Check that the group ticket costs less than the three individual tickets together.

Solution

1. Find the cost of three individual tickets: \(3 \times \$5 = \$15\). 2. Compare that cost with the group ticket: \(\$15 - \$12 = \$3\).

Answer

Buying the group ticket saves \(\$3\).

5180103

Three packages contain \(24\) collectible stickers altogether, with the same number in each package. Jonas wants to collect \(56\) stickers. How many packages must he buy?

Hints

- How many stickers are in one package? - Once you know the package size, how many groups of that size make \(56\)?

Solution

1. Find the number of stickers in one package: \(24 \div 3 = 8\). 2. Find the number of packages needed: \(56 \div 8 = 7\).

Answer

Jonas must buy \(7\) packages.

5180113

A sprinkler uses exactly \(45\,\text{L}\) of water in \(5\) minutes. It uses the same amount of water each minute. How much water does it use in \(8\) minutes?

Hints

- First find how much water is used in one minute. - Use that constant amount for \(8\) minutes. - Check that the amount for \(8\) minutes is greater than the amount for \(5\) minutes.

Solution

1. Find the amount used per minute: \(45\,\text{L} \div 5 = 9\,\text{L}\). 2. Find the amount used in eight minutes: \(8 \times 9\,\text{L} = 72\,\text{L}\).

Answer

The sprinkler uses \(72\,\text{L}\) in \(8\) minutes.

5180153

A school copier makes \(10\) copies in \(20\) seconds. It keeps working at the same speed. How many copies can it make in one minute, or \(60\) seconds?

Hints

- Compare \(60\) seconds with the given \(20\)-second interval. - Find the number of equal \(20\)-second intervals in one minute. - The copier makes the same number of copies during each equal interval.

Solution

1. Find how many \(20\)-second intervals fit in \(60\) seconds: \(60 \div 20 = 3\). 2. The copier makes \(10\) copies in each interval, so it makes \(3 \times 10 = 30\) copies.

Answer

The copier can make \(30\) copies in one minute.

5180204

At a grocery store, \(4\) packages of juice cost \(\$12\) altogether. Ms. Weber has \(\$45\) to buy juice for a school event. What is the greatest number of packages she can buy?

Hints

- First determine the cost of one package. - Then determine how many equal \(\$3\) amounts fit in the budget. - Check that the total cost does not exceed \(\$45\).

Solution

1. Find the cost of one package: \(\$12 \div 4 = \$3\). 2. Divide the budget by the cost per package: \(\$45 \div \$3 = 15\).

Answer

Ms. Weber can buy at most \(15\) packages of juice.

5180323

Five bags of potting soil weigh \(40\,\text{kg}\) altogether. How much do \(8\) identical bags weigh altogether?

Hints

- First find the weight of one bag. - Once you know the weight of one bag, how can you find the weight of eight bags? - A table may help you organize the quantities.

Solution

1. Find the weight of one bag: \(40\,\text{kg} \div 5 = 8\,\text{kg}\). 2. Find the weight of eight bags: \(8 \times 8\,\text{kg} = 64\,\text{kg}\).

Answer

Eight bags weigh \(64\,\text{kg}\) altogether.

5180383

Jonas plants \(7\) flowers each minute. Mia plants \(3\) more flowers each minute than Jonas. How many flowers does Mia plant in \(5\) minutes?

Hints

- First find how many flowers Mia plants in one minute. - Then use that amount for \(5\) equal minutes. - Check that the final total is five times Mia’s one-minute amount.

Solution

1. Mia plants \(7+3=10\) flowers each minute. 2. In \(5\) minutes, she plants \(10 \times 5 = 50\) flowers.

Answer

Mia plants \(50\) flowers in \(5\) minutes.

5180483

Students in Class 3A use \(45\,\text{L}\) of water to irrigate the school garden over \(5\) days. Class 3B uses \(4\,\text{L}\) more each day than Class 3A. How many liters of water does Class 3B use in one day?

Hints

- First find how much water Class 3A uses in one day. - Which operation divides a total amount equally among several days? - Once you know Class 3A's daily use, how can you find Class 3B's?

Solution

1. Find Class 3A's daily water use: \(45\,\text{L} \div 5 = 9\,\text{L}\). 2. Add Class 3B's additional daily use: \(9\,\text{L} + 4\,\text{L} = 13\,\text{L}\).

Answer

Class 3B uses \(13\,\text{L}\) of water in one day.

5181004

Three children have \(80\) marbles altogether. After Tom loses \(8\) marbles, all three children have the same number of marbles. a) How many marbles does each child have now? b) How many marbles did Tom have at first? c) At first, did each of the other two children have more or fewer marbles than Tom?

Hints

- First determine the total after Tom loses the marbles. - Equal amounts for three children call for division by \(3\). - Work backward from Tom's final amount to find his starting amount.

Solution

1. Find the total number of marbles remaining: \(80 - 8 = 72\). 2. Find the number each child has now: \(72 \div 3 = 24\). 3. Find Tom's original number of marbles: \(24 + 8 = 32\). 4. The other two children did not lose any marbles, so each had \(24\) at first. Since \(24 < 32\), each had fewer marbles than Tom.

Answer

a) Each child has \(24\) marbles now. b) Tom had \(32\) marbles at first. c) Each of the other children had fewer marbles than Tom.

5181294

Students plant flowers in a school garden on three Fridays. They plant \(15\) flowers on the first Friday. On the second Friday, they plant three times as many as on the first. On the third Friday, they plant \(12\) more than on the second. 1) How many flowers do they plant on the third Friday? 2) Do they plant more flowers on the third Friday than on the first two Fridays combined? Justify your answer.

Hints

- Determine the Friday amounts in order because each later amount depends on an earlier one. - For part 2, compare the third-Friday result with the combined amount from the first two Fridays. - Check that the comparison uses the correct totals.

Solution

1. On the second Friday, they plant \(15 \times 3=45\) flowers. 2. On the third Friday, they plant \(45+12=57\) flowers. 3. The first two Fridays total \(15+45=60\) flowers. 4. Since \(57<60\), the third-Friday amount is smaller.

Answer

1) They plant \(57\) flowers on the third Friday. 2) No. They plant \(60\) flowers on the first two Fridays combined, and \(57<60\).

5182143

A gardener pays \(\$35\) for a new garden hose. The hose costs \(\$5\) per foot. At home, the gardener cuts off a \(2\)-foot piece to water a small garden bed. How many feet of hose remain?

Hints

- First find how many feet of hose the gardener bought. - Which operation uses the total price and the price per foot? - When a piece is cut off, does the remaining length increase or decrease?

Solution

1. Find the original length of the hose: \(\$35 \div \$5 = 7\) feet. 2. Subtract the piece that was cut off: \(7 - 2 = 5\) feet.

Answer

The gardener has \(5\) feet of hose remaining.

5182153

A class has a \(9\)-yard ribbon. The students cut off \(3\) yards for a large poster. Then they use exactly half of the remaining ribbon to wrap gifts. How many yards of ribbon remain at the end?

Hints

- First find how much ribbon remains after the poster piece is cut off. - The students use exactly half of that remaining amount. - If half is used, what fraction of that same amount is left?

Solution

1. After the first cut, \(9-3=6\) yards remain. 2. The students use half of those \(6\) yards, so the other half remains: \(6 \div 2 = 3\) yards.

Answer

\(3\) yards of ribbon remain.

5182233

A large paint set costs \(\$18\). One paintbrush costs one-sixth as much as the paint set. How much do the paint set and paintbrush cost altogether?

Hints

- First find the cost of the paintbrush. - What operation finds one-sixth of an amount? - Remember that the question asks for the cost of both items together.

Solution

1. Find the cost of the paintbrush: \(\$18 \div 6 = \$3\). 2. Add the two prices: \(\$18 + \$3 = \$21\).

Answer

The paint set and paintbrush cost \(\$21\) altogether.

5183143

Leon has \(48\) stickers. Sophie has \(7\) packs with \(7\) stickers in each pack. Sophie says, “I have more stickers than Leon.” Is she correct? How many more stickers does she have, if any?

Hints

- Find Sophie’s total from the equal packs. - Compare Sophie’s total with Leon’s given total. - The difference tells how many more stickers the larger total has.

Solution

1. Find Sophie’s total: \(7 \times 7 = 49\) stickers. 2. Compare and find the difference: \(49 - 48 = 1\).

Answer

Yes. Sophie has \(1\) more sticker than Leon.

5183153

Ms. Miller wants to bake at least \(60\) muffins for a school fair. She already has \(36\) muffins. Then she bakes \(3\) trays with \(7\) muffins on each tray. Does she reach her goal? Explain.

Hints

- Find how many muffins are in the new trays. - Combine that amount with the muffins already baked. - Compare the final total with the goal; no extra calculation is needed for the comparison.

Solution

1. Find the number in the new trays: \(3 \times 7 = 21\) muffins. 2. Add them to the muffins already baked: \(36 + 21 = 57\). 3. Since \(57 < 60\), she does not reach the goal.

Answer

No. Ms. Miller has \(57\) muffins, which is fewer than \(60\).

5185453

At a wildlife park, \(5\) small goats receive \(40\) apple slices altogether. Each goat receives the same number. How many apple slices are needed for \(8\) goats?

Hints

- Find the number of apple slices for one goat. - Use that number to find the amount for \(8\) goats. - The total should increase because more goats are being fed.

Solution

1. Find the number of apple slices for one goat: \(40 \div 5 = 8\). 2. Find the number for \(8\) goats: \(8 \times 8 = 64\).

Answer

The park needs \(64\) apple slices.

5185803

A game figure costs \(\$7\). Noah wants to buy \(3\) figures but has only \(\$15\). How much more money does he need?

Hints

- First find the total cost of the three equal-price figures. - Compare that cost with the money Noah already has. - The difference is the additional amount he needs.

Solution

1. Find the cost of three figures: \(3 \times \$7 = \$21\). 2. Find the shortfall: \(\$21 - \$15 = \$6\).

Answer

Noah needs \(\$6\) more.

5185984

At a flea market, a child sells packages of trading cards. Each package contains \(10\) cards and costs \(\$2\). By the end of the day, the child has collected \(\$48\). How many trading cards were sold altogether?

Hints

- Use the money collected and the price per package to determine how many packages were sold. - Then relate the number of packages to the number of cards in each package. - Check that your final answer counts individual cards rather than packages.

Solution

1. The child sold \(\$48 \div \$2=24\) packages. 2. The number of cards sold is \(24 \times 10=240\).

Answer

The child sold \(240\) trading cards altogether.

5186723

A zoo has \(4\) groups of penguins with \(8\) penguins in each group. Each penguin eats \(3\) fish per day. How many fish are needed each day for all the penguins?

Hints

- First find the total number of penguins. - Each penguin receives the same number of fish. - Use the total number of penguins to find the daily amount of fish.

Solution

1. Find the total number of penguins: \(4 \times 8 = 32\). 2. Find the number of fish needed: \(32 \times 3 = 96\).

Answer

The penguins need \(96\) fish each day.

5186783

A rain barrel can hold \(400\,\text{L}\). It starts with \(280\,\text{L}\) of water. After some water is used, it takes \(230\,\text{L}\) to fill the barrel again. How many liters of water were used?

Hints

- Find the empty space after the first filling. - Compare that amount with the amount later needed to refill the barrel. - The extra refill amount equals the water that was removed.

Solution

1. After the first filling, the empty space is \(400\,\text{L}-280\,\text{L}=120\,\text{L}\). 2. If no water had been removed, only \(120\,\text{L}\) would be needed to fill the barrel. 3. The actual refill is \(230\,\text{L}\), so the amount removed is \(230\,\text{L}-120\,\text{L}=110\,\text{L}\).

Answer

\(110\,\text{L}\)

5187474

A gardener buys \(15\) red flowers for \(\$4\) each and then has \(\$20\) remaining. The gardener wonders how many yellow flowers could have been bought with all the original money if each yellow flower cost \(\$8\). How many yellow flowers could the gardener have bought?

Hints

- Reconstruct the gardener’s original amount of money from what was spent and what remained. - Then compare that original amount with the price of one yellow flower. - Check the number of yellow flowers with a multiplication equation.

Solution

1. The red flowers cost \(15 \times \$4=\$60\). 2. The gardener originally had \(\$60+\$20=\$80\). 3. Since \(\$8 \times 10=\$80\), the gardener could have bought \(10\) yellow flowers.

Answer

The gardener could have bought \(10\) yellow flowers.

5189273

For a class trip, Mr. Miller buys juice. Eight identical cases contain \(48\) bottles altogether. a) How many bottles are in one case? b) Mr. Miller wants each of the \(65\) children on the trip to have one bottle. Are \(10\) cases enough? Justify your answer with a calculation.

Hints

- First find how many bottles are in one case. - How many bottles are in ten cases? - Compare that amount with the number of children.

Solution

1. Find the number of bottles in one case: \(48 \div 8 = 6\). 2. Find the number of bottles in ten cases: \(10 \times 6 = 60\). 3. Compare the available bottles with the number needed: \(60 < 65\). 4. Ten cases are not enough.

Answer

a) One case contains \(6\) bottles. b) No. Ten cases contain only \(60\) bottles, but \(65\) bottles are needed.

5189333

At the Star Theater, \(5\) tickets cost \(\$45\). At the Moon Theater, \(4\) tickets cost \(\$40\). At which theater does one ticket cost less? Show the price of one ticket at each theater.

Hints

- Find the price of one ticket at each theater. - Compare the two prices for one ticket. - The smaller one-ticket price tells which theater costs less.

Solution

1. Find the price of one ticket at the Star Theater: \(\$45 \div 5 = \$9\). 2. Find the price of one ticket at the Moon Theater: \(\$40 \div 4 = \$10\). 3. Since \(\$9 < \$10\), one ticket is less expensive at the Star Theater.

Answer

One ticket is less expensive at the Star Theater because it costs \(\$9\), compared with \(\$10\) at the Moon Theater.

5190013

Grandpa Henry is \(65\) years old today. His grandson Ben is \(56\) years younger. How old will Ben be \(7\) years from now?

Hints

- First use the age difference to find Ben’s age today. - Then move Ben’s age forward by the stated number of years. - Check that both people would be \(7\) years older after \(7\) years.

Solution

1. Find Ben’s age today: \(65-56=9\). 2. Seven years from now, Ben will be \(9+7=16\) years old.

Answer

Ben will be \(16\) years old.

5190284

Ms. Berger is \(42\) years old. Her son Lucas is \(7\), and her daughter Mia is twice as old as Lucas. a) How old was Ms. Berger when Mia was born? b) How old was Ms. Berger when Lucas was born? c) At which child’s birth was she older?

Hints

- Determine Mia’s current age from the comparison with Lucas. - A parent’s age at a child’s birth can be found from their current age difference. - Compare the two ages you obtain for Ms. Berger.

Solution

1. Mia is \(2 \times 7=14\) years old. 2. At Mia’s birth, Ms. Berger was \(42-14=28\) years old. 3. At Lucas’s birth, Ms. Berger was \(42-7=35\) years old. 4. Since \(35>28\), she was older when Lucas was born.

Answer

a) Ms. Berger was \(28\) years old when Mia was born. b) She was \(35\) years old when Lucas was born. c) She was older when Lucas was born.

5191463

A school event starts with \(450\) muffins. Students sell \(200\) muffins in the morning and \(150\) muffins in the afternoon. How many muffins remain at the end of the day?

Hints

- First find the total number of muffins sold. - Subtract the number sold from the starting amount. - You can also subtract the morning and afternoon sales one at a time.

Solution

1. Add the muffins sold: \(200 + 150 = 350\). 2. Subtract from the starting amount: \(450 - 350 = 100\).

Answer

\(100\) muffins remain at the end of the day.

5191473

Two classes collect paper for a recycling contest. Class 3A collects \(340\,\text{kg}\), and Class 3B collects \(420\,\text{kg}\). The school's goal is \(900\,\text{kg}\). How many more kilograms are needed to reach the goal?

Hints

- Add the amounts collected by both classes. - Subtract that total from the goal. - Check that the missing amount and collected amount add to \(900\,\text{kg}\).

Solution

1. Add the amounts collected by the two classes: \(340 + 420 = 760\,\text{kg}\). 2. Subtract the amount collected from the goal: \(900 - 760 = 140\,\text{kg}\).

Answer

The classes need \(140\,\text{kg}\) more to reach the goal.

5191643

Two classes collect empty bottles for a recycling project. Class 3A collects \(370\) bottles. Together, Classes 3A and 3B collect \(720\) bottles. Which class collects more bottles, and how many more?

Hints

- Use the combined total and Class 3A's amount to find Class 3B's amount. - Compare the two class amounts. - Subtract to find how many more.

Solution

1. Find the number collected by Class 3B: \(720 - 370 = 350\). 2. Compare the two amounts: \(370 > 350\), so Class 3A collects more. 3. Find the difference: \(370 - 350 = 20\).

Answer

Class 3A collects more bottles—\(20\) more than Class 3B.

5191653

A book has \(560\) pages. Max has read \(240\) pages. Klara is reading the same book and has \(310\) pages left. Who has read more pages? Justify your answer with a calculation.

Hints

- First find how many pages Klara has already read. - Subtract the pages she has left from the total number of pages. - Compare Klara's result with Max's \(240\) pages.

Solution

1. Find the number of pages Klara has read: \(560 - 310 = 250\). 2. Compare the amounts read: \(250 > 240\), so Klara has read more pages.

Answer

Klara has read more. She has read \(250\) pages, while Max has read \(240\) pages.

5191663

A trail is \(720\,\text{m}\) long and has two sections: a forest section and a field section. The forest section is \(450\,\text{m}\) long. How many meters longer is the forest section than the field section?

Hints

- First find the length of the field section. - Then compare the two section lengths. - Subtract the shorter length from the longer length.

Solution

1. Find the length of the field section: \(720 - 450 = 270\,\text{m}\). 2. Find the difference between the two sections: \(450 - 270 = 180\,\text{m}\).

Answer

The forest section is \(180\,\text{m}\) longer than the field section.

5192963

A hot-air balloon pilot plans a trip of \(900\) miles. The balloon travels \(342\) miles on the first day and \(285\) miles on the second day. How many miles must it travel on the third day to reach the destination?

Hints

- Find the total distance already traveled. - Subtract that total from \(900\) miles. - Check that all three daily distances add to \(900\).

Solution

1. Add the distances traveled on the first two days: \(342 + 285 = 627\) miles. 2. Subtract from the total distance: \(900 - 627 = 273\) miles.

Answer

The balloon must travel \(273\) miles on the third day.

5193063

A baker buys \(245\,\text{kg}\) of wheat flour and \(180\,\text{kg}\) of rye flour. During the day, the baker uses \(315\,\text{kg}\) of flour. How many kilograms of flour remain?

Hints

- Add the two kinds of flour first. - Subtract the amount used from the total. - Include the unit in your answer.

Solution

1. Find the total amount of flour: \(245 + 180 = 425\,\text{kg}\). 2. Subtract the amount used: \(425 - 315 = 110\,\text{kg}\).

Answer

The baker has \(110\,\text{kg}\) of flour remaining.

5193073

A school supply store begins the day with \(356\) blue notebooks and \(289\) red notebooks. At the end of the day, \(218\) notebooks remain. How many notebooks were sold?

Hints

- Combine the two starting notebook amounts first. - Then compare the starting total with the number left at the end of the day. - Check that sold notebooks plus remaining notebooks equal the starting total.

Solution

1. Find the starting total: \(356+289=645\). 2. Subtract the number remaining: \(645-218=427\).

Answer

The store sold \(427\) notebooks.

5193083

A bakery plans to make \(850\) pretzels for a community festival. It makes \(412\) pretzels on Friday and \(465\) on Saturday. How many pretzels does the bakery make altogether, and by how many does it exceed its goal?

Hints

- Add the amounts from Friday and Saturday. - Compare the total with the goal. - Subtract the goal from the total to find how far it was exceeded.

Solution

1. Add the pretzels made on both days: \(412 + 465 = 877\). 2. Subtract the goal from the total: \(877 - 850 = 27\).

Answer

The bakery makes \(877\) pretzels, exceeding its goal by \(27\) pretzels.

5193093

Lucas is saving for a mountain bike that costs \(\$450\). He saves \(\$215\) during the first year and \(\$278\) during the second year. Lucas says, “I have saved more than the bike costs.” Is Lucas correct? How much more has he saved than he needs?

Hints

- Add the two amounts Lucas saved. - Compare the total savings with the bike price. - Subtract to find the extra amount.

Solution

1. Add Lucas's savings: \(\$215 + \$278 = \$493\). 2. Compare with the bike price: \(\$493 > \$450\), so Lucas is correct. 3. Find the extra amount: \(\$493 - \$450 = \$43\).

Answer

Lucas is correct. He has saved \(\$493\), which is \(\$43\) more than the bike costs.

5193113

A produce seller has \(312\) apples. There are \(125\) fewer pears than apples. How many apples and pears are at the stand altogether?

Hints

- First find the number of pears from the “fewer than” relationship. - Then combine the two fruit amounts. - Check that the pear count is smaller than the apple count by \(125\).

Solution

1. Find the number of pears: \(312-125=187\). 2. Add the apples and pears: \(312+187=499\).

Answer

There are \(499\) apples and pears altogether.

5194324

A school buys \(2\) packages containing \(250\) balloons each for a celebration. After decorating, \(112\) balloons remain. How many balloons were used?

Hints

- First determine the total number of balloons purchased. - The balloons used and the balloons remaining make the starting total. - Compare the starting total with the amount left after decorating.

Solution

1. The school bought \(2 \times 250=500\) balloons. 2. The number used is \(500-112=388\).

Answer

The school used \(388\) balloons.

5194824

A truck travels \(50\) miles in one hour. An express train travels twice that distance in one hour. How many miles does the train travel in \(5\) hours?

Hints

- Use the multiplicative comparison to determine the train’s one-hour distance. - Keep the one-hour comparison separate from the repeated five-hour distance. - Check that the five-hour result is five times the train’s one-hour distance.

Solution

1. The train travels \(2 \times 50=100\) miles in one hour. 2. In five hours, the train travels \(5 \times 100=500\) miles.

Answer

The train travels \(500\) miles in five hours.

5196723

Start with \(340\). a) What number is \(60\) greater than \(340\)? b) What number is \(400\) greater than your answer to part a)?

Hints

- Which operation matches the phrase “greater than”? - Find the first new number before finding the second one. - Add hundreds to hundreds and tens to tens.

Solution

1. Find the first number: \(340 + 60 = 400\). 2. Use the first number to find the second number: \(400 + 400 = 800\).

Answer

a) \(400\) b) \(800\)

5197193

A school library tracked the number of books borrowed in September and October. Find the increase for each category. Which category had the greatest increase? <table> <tr><th>Category</th><th>September</th><th>October</th><th>Increase</th></tr> <tr><td>Nonfiction</td><td>145</td><td>210</td><td></td></tr> <tr><td>Fiction</td><td>238</td><td>312</td><td></td></tr> <tr><td>Comics</td><td>189</td><td>254</td><td></td></tr> <tr><td>Picture books</td><td>92</td><td>167</td><td></td></tr> </table>

Hints

- Look at how each category changes from September to October. - Subtract the September number from the October number in each row. - Compare the four increases after calculating them.

Solution

1. Nonfiction increased by \(210 - 145 = 65\). 2. Fiction increased by \(312 - 238 = 74\). 3. Comics increased by \(254 - 189 = 65\). 4. Picture books increased by \(167 - 92 = 75\). 5. Since \(75\) is greatest, picture books had the greatest increase.

Answer

Nonfiction: \(65\) Fiction: \(74\) Comics: \(65\) Picture books: \(75\) Picture books had the greatest increase.

5197203

A school garden compares this year’s harvest with last year’s harvest. <table> <tr><th>Fruit</th><th>Last year</th><th>This year</th><th>Increase</th></tr> <tr><td>Apples</td><td>\(345\,\text{kg}\)</td><td>\(412\,\text{kg}\)</td><td></td></tr> <tr><td>Pears</td><td></td><td></td><td>\(77\,\text{kg}\)</td></tr> </table> How much did the apple harvest increase? How much did the apple and pear harvests increase altogether?

Hints

- For apples, compare this year’s amount with last year’s amount. - The pear increase is already given in the table. - Combine the two increases for the overall change.

Solution

1. Apples increased by \(412-345=67\,\text{kg}\). 2. Add the two increases: \(67+77=144\,\text{kg}\).

Answer

Apples increased by \(67\,\text{kg}\). Altogether, the apple and pear harvests increased by \(144\,\text{kg}\).

5199193

A bakery has \(8\) trays of muffins. On each tray, the muffins are arranged in \(4\) rows of \(5\). How many muffins are there altogether?

Hints

- First find how many muffins are on one tray. - Then use the number of trays to find the total. - Think of each tray as an array.

Solution

1. Find the number on one tray: \(4 \times 5 = 20\). 2. Find the number on eight trays: \(8 \times 20 = 160\).

Answer

There are \(160\) muffins altogether.

5199203

A craft group has \(6\) students. Each student needs \(8\) white beads and \(2\) gold beads for a necklace. How many beads are needed altogether?

Hints

- First find how many beads one student needs. - Then use the number of students as equal groups. - Check that the final total is six times the amount for one student.

Solution

1. Each student needs \(8+2=10\) beads. 2. Six students need \(6 \times 10=60\) beads altogether.

Answer

The group needs \(60\) beads altogether.

5200814

A stationery store packs pens into boxes. <table> <tr><th>Pen color</th><th>Pens per box</th><th>Total pens</th></tr> <tr><td>Blue pens</td><td>\(8\)</td><td>\(480\)</td></tr> <tr><td>Red pens</td><td>\(6\)</td><td>?</td></tr> </table> The store has the same number of boxes of each color. How many red pens are there altogether?

Hints

- Use the blue-pen row to determine the shared number of boxes. - Apply that same box count to the red-pen row. - Check that your result matches \(6\) pens in each of the equal number of boxes.

Solution

1. The number of blue-pen boxes is \(480\div8=60\). 2. There are also \(60\) boxes of red pens. 3. The total number of red pens is \(60 \times 6=360\).

Answer

There are \(360\) red pens altogether.

5203193

Students sell raffle tickets at a school event. They sell \(465\) tickets in the morning, which is \(130\) more than they sell in the afternoon. a) How many tickets do they sell in the afternoon? b) How many tickets do they sell during the entire day?

Hints

- Use “more than” to find the afternoon amount. - Then add the morning and afternoon amounts. - Add the two parts to check the daily total.

Solution

1. Find the afternoon sales: \(465 - 130 = 335\). 2. Add the morning and afternoon sales: \(465 + 335 = 800\).

Answer

a) Students sell \(335\) tickets in the afternoon. b) Students sell \(800\) tickets during the entire day.

5203763

An aquarium has \(340\) goldfish in a large tank and \(280\) silver fish in a small tank. a) How many fish are in both tanks altogether? b) The aquarium adds \(40\) more goldfish to the large tank. How many fish are there altogether now? Use your answer from part a.

Hints

- First add the fish in the two tanks. - Then add the fish that are introduced later. - Use the result from part a in part b.

Solution

1. Add the fish in both tanks: \(340 + 280 = 620\). 2. Add the new goldfish: \(620 + 40 = 660\).

Answer

a) There are \(620\) fish altogether. b) There are now \(660\) fish altogether.

5203973

Anton has \(160\) marbles, and Bea has \(125\) marbles. a) How many marbles do they have altogether? b) They give \(35\) marbles to their younger brother. How many marbles do they have left altogether?

Hints

- Add the two starting amounts. - Then subtract the marbles that are given away. - Use the answer from part a in part b.

Solution

1. Add to find the starting total: \(160 + 125 = 285\). 2. Subtract the marbles they give away: \(285 - 35 = 250\).

Answer

a) They have \(285\) marbles altogether. b) They have \(250\) marbles left.

5203983

A wildlife park has \(24\) meerkats in one habitat and \(19\) in another. The park sends \(7\) meerkats to another zoo. How many meerkats remain in the park?

Hints

- Combine the two starting habitats first. - Then subtract the animals that leave the park. - Check that the final number is smaller than the starting total.

Solution

1. Add the two habitats: \(24+19=43\). 2. Subtract the meerkats that leave: \(43-7=36\).

Answer

\(36\) meerkats remain in the park.

5204233

Maria adds \(55\) apples to a basket while her brother removes \(20\) pears. By how much does the total number of pieces of fruit change?

Hints

- Decide how adding apples changes the total. - Decide how removing pears changes the total. - Combine the increase and decrease.

Solution

1. Adding the apples changes the total by \(+55\). 2. Removing the pears changes the total by \(-20\). 3. Combine the changes: \(55 - 20 = 35\). The positive result means the total increases.

Answer

The total number of pieces of fruit increases by \(35\).

5205613

Lucas has \(345\) trading cards, and his sister Marie has \(415\) trading cards. Their cousin Tom has \(230\) fewer cards than Lucas and Marie have altogether. How many cards does Tom have?

Hints

- First find how many cards Lucas and Marie have altogether. - Use “fewer than” to choose the next operation.

Solution

1. Add Lucas's and Marie's cards: \(345 + 415 = 760\). 2. Subtract \(230\): \(760 - 230 = 530\).

Answer

Tom has \(530\) trading cards.

5208823

A school supply store has \(340\) pencils on a shelf. A worker adds \(270\) new pencils, and later the store sells \(180\) pencils. How many pencils are on the shelf at the end of the day?

Hints

- Add the pencils that are placed on the shelf. - Then subtract the pencils that are sold. - Follow the events in order.

Solution

1. Add the new pencils: \(340 + 270 = 610\). 2. Subtract the pencils sold: \(610 - 180 = 430\).

Answer

There are \(430\) pencils on the shelf at the end of the day.

5209983

Students sell \(350\) raffle tickets on Friday. On Saturday, they sell \(125\) more tickets than on Friday. Jordan says, “We sold more than \(800\) tickets over the two days.” Is Jordan correct? Justify your answer with a calculation.

Hints

- Find the number of tickets sold on Saturday. - Add the sales from both days. - Compare the total with \(800\).

Solution

1. Find the Saturday sales: \(350 + 125 = 475\). 2. Add the sales from both days: \(350 + 475 = 825\). 3. Compare: \(825 > 800\), so the statement is correct.

Answer

Yes. The students sold \(825\) tickets over the two days, and \(825>800\).

5210013

Lucas has \(350\) soccer stickers. He has \(120\) fewer animal stickers than soccer stickers. How many stickers does Lucas have altogether?

Hints

- First find the number of animal stickers. - Then add the two sticker collections. - Check that the animal-sticker count is smaller than \(350\).

Solution

1. Find the number of animal stickers: \(350 - 120 = 230\). 2. Add both kinds of stickers: \(350 + 230 = 580\).

Answer

Lucas has \(580\) stickers altogether.

5210023

A bakery makes \(240\) plain rolls in the morning. It makes \(70\) more whole-wheat rolls than plain rolls. How many plain and whole-wheat rolls does the bakery make altogether?

Hints

- First use the comparison to find the number of whole-wheat rolls. - Then combine the two kinds of rolls. - Check that the whole-wheat count is \(70\) greater than the plain-roll count.

Solution

1. Find the number of whole-wheat rolls: \(240+70=310\). 2. Add the two kinds: \(240+310=550\).

Answer

The bakery makes \(550\) plain and whole-wheat rolls altogether.

5210083

A school library has \(450\) books. Of these, \(180\) are fiction books and \(130\) are nonfiction books. All the remaining books are picture books. How many picture books are in the library?

Hints

- Combine the two categories whose counts are given. - The picture books are what remains from the full library total. - Check that all three categories add to \(450\).

Solution

1. Find the number of fiction and nonfiction books together: \(180+130=310\). 2. Subtract from the library total: \(450-310=140\).

Answer

There are \(140\) picture books in the library.

5210093

Three third-grade classes collect \(820\,\text{kg}\) of paper altogether. Ms. Lee’s class collects \(245\,\text{kg}\), and Mr. Grant’s class collects \(275\,\text{kg}\). How many kilograms does Ms. Patel’s class collect?

Hints

- Combine the two class amounts that are known. - The third class collected the part of the total that is left. - Check that all three class amounts add to \(820\,\text{kg}\).

Solution

1. Add the first two class amounts: \(245+275=520\,\text{kg}\). 2. Subtract from the total: \(820-520=300\,\text{kg}\).

Answer

Ms. Patel’s class collects \(300\,\text{kg}\) of paper.

5210153

A mail carrier drives \(18\) miles during the first hour of a route. During the second hour, the carrier drives \(4\) miles farther than during the first hour. How far does the carrier drive during the two hours altogether?

Hints

- First use the comparison to find the second-hour distance. - Then combine the distances from the two hours. - Check that the second-hour distance is \(4\) miles greater than the first-hour distance.

Solution

1. Find the second-hour distance: \(18+4=22\) miles. 2. Add the two hours: \(18+22=40\) miles.

Answer

The carrier drives \(40\) miles altogether.

5210173

Lucas is saving for a bicycle that costs \(\$450\). He saves \(\$135\) in May and \(\$185\) in June. How much more must he save to have enough money for the bicycle?

Hints

- Combine the two savings amounts first. - Then compare the saved amount with the bicycle price. - Check that the amount saved plus the amount still needed equals \(\$450\).

Solution

1. Find the amount already saved: \(\$135+\$185=\$320\). 2. Subtract from the bicycle price: \(\$450-\$320=\$130\).

Answer

Lucas must save \(\$130\) more.

5210183

At a field day, three third-grade classes complete \(840\) laps altogether. Ms. Lee’s class completes \(260\) laps, and Mr. Grant’s class completes \(220\) laps. Ms. Patel’s class completes the remaining laps. How many laps does Ms. Patel’s class complete, and which class completes the most laps?

Hints

- Combine the two known class totals. - Subtract their combined amount from the overall total. - After finding the remaining class amount, compare the three numbers without another calculation.

Solution

1. Add the first two class totals: \(260+220=480\) laps. 2. Find the remaining laps: \(840-480=360\) laps. 3. Compare \(360\), \(260\), and \(220\). Ms. Patel’s class completes the most laps.

Answer

Ms. Patel’s class completes \(360\) laps and completes the most laps.

5211263

A baker receives \(100\) eggs. In the morning, the baker uses \(37\) eggs for cakes. The remaining eggs will be used for waffles, with exactly \(7\) eggs in each batch. How many batches of waffles can the baker make?

Hints

- Identify the important starting quantity. - How many eggs remain after the cakes are made? - Calculate the subtraction carefully. - Use a multiplication fact to find how many groups of seven fit into the remainder.

Solution

1. Find the number of eggs remaining: \(100 - 37 = 63\). 2. Divide by the number of eggs in each batch: \(63 \div 7 = 9\).

Answer

The baker can make \(9\) batches of waffles.

5211324

A toy store receives a shipment of \(84\) card games. The games are divided equally among \(7\) shelf sections. How many card games are in \(3\) of the sections altogether?

Hints

- Determine one equal shelf section before combining several sections. - The same section size applies to all seven sections. - Check that your per-section amount would reproduce the original shipment across seven sections.

Solution

1. Each shelf section receives \(84\div7=12\) card games. 2. Three sections contain \(12 \times 3=36\) card games.

Answer

There are \(36\) card games in \(3\) shelf sections altogether.

5211413

A school library needs to shelve \(850\) books. Volunteers shelve \(260\) books on Monday and \(215\) books on Tuesday. How many books still need to be shelved after Tuesday?

Hints

- Combine the numbers shelved on Monday and Tuesday. - Then compare that subtotal with the full number of books. - Check that shelved books plus remaining books equal \(850\).

Solution

1. Find the total shelved on both days: \(260+215=475\). 2. Subtract from the full number of books: \(850-475=375\).

Answer

There are \(375\) books left to shelve.

5211423

A walking trail is \(1000\,\text{m}\) long. The first section is \(340\,\text{m}\) long, and the second section is \(220\,\text{m}\) long. How long is the third section?

Hints

- Combine the lengths of the two known sections first. - The three sections together make the full trail length. - Check that the three section lengths add to \(1000\,\text{m}\).

Solution

1. Find the combined length of the first two sections: \(340+220=560\,\text{m}\). 2. Find the remaining length: \(1000-560=440\,\text{m}\).

Answer

The third section is \(440\,\text{m}\) long.

5211873

A gardener has \(54\) tulip bulbs. He wants to plant \(10\) full rows with exactly \(6\) bulbs in each row. How many more bulbs does he need?

Hints

- First find how many bulbs are needed for all \(10\) rows. - Then compare that total with the \(54\) bulbs already available. - Check that adding your answer to \(54\) reaches the needed total.

Solution

1. Ten rows require \(10 \times 6=60\) bulbs. 2. Compare the required number with the bulbs he has: \(60-54=6\).

Answer

The gardener needs \(6\) more bulbs.

5212703

A school garden has a water tank holding \(850\,\text{L}\). Students use \(186\,\text{L}\) on Monday and another \(275\,\text{L}\) on Tuesday. a) How many liters of water remain in the tank? b) Did the students use more water over the two days than remains in the tank? Explain.

Hints

- First find the total amount used on both days. - Subtract the amount used from the starting amount. - Compare the amount used with the amount remaining.

Solution

1. Find the total amount of water used: \(186\,\text{L} + 275\,\text{L} = 461\,\text{L}\). 2. Find the amount remaining: \(850\,\text{L} - 461\,\text{L} = 389\,\text{L}\). 3. Compare the amounts: \(461\,\text{L} > 389\,\text{L}\), so more water was used than remains.

Answer

a) \(389\,\text{L}\) remain in the tank. b) Yes. The students used \(461\,\text{L}\), and \(461\,\text{L} > 389\,\text{L}\).

5213433

A bakery starts the morning with \(650\) rolls. By noon, it sells \(385\) rolls. The baker then makes \(120\) more rolls. How many rolls are there now? Is this more or fewer than the bakery had at the start of the morning?

Hints

- Subtract the rolls that were sold. - Then add the newly baked rolls. - Compare the result with \(650\).

Solution

1. Find the number left after the sales: \(650 - 385 = 265\). 2. Add the newly baked rolls: \(265 + 120 = 385\). 3. Compare with the starting amount: \(385 < 650\), so there are fewer rolls now.

Answer

There are \(385\) rolls now, which is fewer than the \(650\) rolls at the start of the morning.

5213523

A store receives \(6\) large cartons of pencils. Each carton contains \(4\) packages, and each package contains \(10\) pencils. 1) How many pencils are in one large carton? 2) How many pencils are delivered altogether?

Hints

- First find the number of pencils in one carton. - Then find the total for all six cartons. - Equal groups suggest multiplication.

Solution

1. Find the number in one carton: \(4 \times 10 = 40\). 2. Find the total in six cartons: \(6 \times 40 = 240\).

Answer

1) One large carton contains \(40\) pencils. 2) The delivery contains \(240\) pencils altogether.

5214843

A bakery has \(480\) rolls in two baskets altogether. The baker removes \(65\) rolls from one basket for packaging and adds \(65\) freshly baked rolls to the other basket. a) How many rolls are now in the two baskets altogether? b) What would happen to the total if the baker removed \(70\) rolls and added \(80\) rolls instead?

Hints

- In part a, compare the number removed with the number added. - In part b, find the difference between the number added and the number removed. - Use that change to find the new total.

Solution

1. For part a, the baker removes and adds the same number of rolls. These changes cancel, so the total remains \(480\). 2. For part b, compare the number added with the number removed: \(80 - 70 = 10\). The total increases by \(10\). 3. Find the new total: \(480 + 10 = 490\).

Answer

a) There are still \(480\) rolls altogether. b) The total increases by \(10\), so there would be \(490\) rolls.

5215493

A large aquarium has \(260\) fish. A keeper moves \(45\) fish to another tank and later adds \(72\) young fish. How many fish are in the aquarium now? Is this more or fewer than the aquarium had at first?

Hints

- First subtract the fish that are moved. - Then add the young fish. - Compare the final number with \(260\).

Solution

1. Find the number after \(45\) fish are moved: \(260 - 45 = 215\). 2. Add the young fish: \(215 + 72 = 287\). 3. Compare with the starting number: \(287 > 260\), so there are more fish now.

Answer

There are \(287\) fish now, which is more than the starting number of \(260\).

5373693

The picture shows trees planted in equal rows. After a storm, all the trees in the shaded bottom row fell. How many trees are still standing? Show two ways to calculate.

Figure for problem 537369

Hints

- Use the picture to find the number of rows and the number of trees in one row. - One complete bottom row is affected. - You can subtract the fallen row or multiply using only the rows that remain.

Solution

1. The picture shows \(6\) rows with \(8\) trees in each row, so the original total is \(6 \times 8=48\) trees. Subtract the fallen row: \(48-8=40\). 2. Alternatively, \(5\) rows remain standing, so \(5 \times 8=40\).

Answer

\(40\) trees are still standing. Two methods are \(6 \times 8-8=40\) and \(5 \times 8=40\).

5373713

The picture shows all the seats in a small movie theater. The shaded seats are already reserved. Without counting the available seats one by one: a) write a multiplication equation for the total number of seats; b) find the number of reserved seats from the picture; c) write the subtraction that gives the number of available seats.

Figure for problem 537371

Hints

- Use the rectangular arrangement to find the total rather than counting individual seats. - The shaded block gives the number that must be removed from the total. - Show both the multiplication equation and the subtraction equation.

Solution

1. The picture has \(7\) rows with \(10\) seats in each row, so \(7 \times 10=70\) seats are shown. 2. There are \(13\) shaded reserved seats. 3. Subtract the reserved seats: \(70-13=57\). 4. There are \(57\) available seats.

Answer

a) \(7 \times 10=70\) b) \(13\) reserved seats c) \(70-13=57\) available seats

5373724

The picture shows a rectangular sheet of collectible stamps. Each blue stamp is worth \(2\) cents, and each green stamp is worth \(3\) cents. What is the total value of all the stamps on the sheet?

Figure for problem 537372

Hints

- Use the picture to count the blue and green stamps separately. - Find the value of each color group using its price per stamp. - Add the two color-group values.

Solution

1. The picture shows \(30\) blue stamps and \(24\) green stamps. 2. The blue stamps are worth \(30\times2=60\) cents. 3. The green stamps are worth \(24\times3=72\) cents. 4. The total value is \(60+72=132\) cents, or \(\$1.32\).

Answer

The sheet is worth \(132\) cents, or \(\$1.32\).

5373883

A market stand begins the day with \(72\) apples arranged in rows of \(8\). During the morning, it sells \(3\) complete rows. How many apples remain?

Hints

- First find how many apples are in the three complete rows that were sold. - Then subtract the sold apples from the starting amount. - Check that the number sold and the number remaining add to \(72\).

Solution

1. Find the number sold in three rows: \(3 \times 8=24\). 2. Subtract from the starting amount: \(72-24=48\).

Answer

\(48\) apples remain.

5381233

Teams collect nature cards during a forest scavenger hunt. How many more cards does Team Fox need, at minimum, to have more cards than Team Badger?

Figure for problem 538123

Hints

- Read the values for Team Fox and Team Badger. - Decide the smallest total that is greater than Team Badger's total. - Subtract Team Fox's current total from that target.

Solution

1. Team Fox has \(15\) cards, and Team Badger has \(20\) cards. 2. To have more than \(20\), Team Fox needs at least \(21\) cards. 3. The number still needed is \(21 - 15 = 6\) cards.

Answer

Team Fox needs at least \(6\) more cards.

5381273

Which two statements match the graph? A: Dogs and rabbits received the same number of votes. B: Horses received the most votes. C: Cats received \(4\) more votes than dogs. D: There were fewer than \(40\) votes altogether.

Figure for problem 538127

Hints

- Check each statement separately against the graph. - Pay attention to words such as “same,” “more,” and “most.” - For the total, add all four bar values.

Solution

1. Dogs and rabbits each received \(12\) votes, so A is true. 2. Cats received \(16 - 12 = 4\) more votes than dogs, so C is true. 3. B is false because cats received the most votes. 4. D is false because \(12 + 16 + 8 + 12 = 48\), which is not fewer than \(40\).

Answer

Statements A and C are true.

5381284

The graph shows glass containers collected for recycling. Which statement cannot be true? A: In February, \(15\) more containers were collected than in January. B: In March, half as many containers were collected as in February. C: In April, \(5\) more containers were collected than in March.

Figure for problem 538128

Hints

- Distinguish statements that compare by a difference from statements that compare by a factor. - Check each statement against the values shown in the graph. - A statement is ruled out if its stated comparison does not match the two quantities.

Solution

1. A is true because \(40 - 25 = 15\). 2. B is false because half of \(40\) is \(20\), not \(30\). 3. C is true because \(35 - 30 = 5\).

Answer

Statement B cannot be true.

5381293

During a field game, Team Green later receives \(6\) more points. Which team is then in the lead, and with how many points?

Figure for problem 538129

Hints

- Update Team Green's score first. - Keep the other teams' scores unchanged. - Compare all four scores after the change.

Solution

1. Team Green's score changes from \(30\) to \(30 + 6 = 36\). 2. Compare \(36\) with the other scores: \(25\), \(35\), and \(20\). 3. Since \(36\) is greatest, Team Green is in the lead.

Answer

Team Green is in the lead with \(36\) points.

5381303

Because of an error in a quiz, Team Star loses \(8\) points. What place is Team Star in after the deduction?

Figure for problem 538130

Hints

- Change only Team Star's score. - Compare the new score with all three other scores. - Order the four scores from greatest to least.

Solution

1. Team Star's new score is \(40 - 8 = 32\) points. 2. The other teams have \(35\), \(40\), and \(45\) points. 3. Since \(32\) is the least score, Team Star is in fourth place.

Answer

Team Star is in fourth place.

5381323

The graph shows how students travel to a sports field. Do more or fewer students travel by bike and on foot together than by car? How many more or fewer?

Figure for problem 538132

Hints

- Add the bike and walking bars first. - Compare that sum with the car bar. - Subtract to find how much greater or smaller it is.

Solution

1. By bike and on foot together, \(12 + 8 = 20\) students travel to the field. 2. By car, \(16\) students travel to the field. 3. The difference is \(20 - 16 = 4\), so the bike-and-walking total is greater.

Answer

\(4\) more students travel by bike and on foot together than by car.

5381363

The graph shows soft pretzels baked during one week. How many pretzels were baked on all days except Wednesday?

Figure for problem 538136

Hints

- Leave out the Wednesday bar. - Add the other four bar values. - Pair numbers to make the addition easier.

Solution

1. The values for Monday, Tuesday, Thursday, and Friday are \(15\), \(20\), \(25\), and \(30\). 2. Add them: \(15 + 20 + 25 + 30 = 90\).

Answer

\(90\) pretzels were baked on all days except Wednesday.

5381373

Four groups estimate the number of steps along a short path. Which group's estimate is closest to \(55\) steps?

Figure for problem 538137

Hints

- Find how far each estimate is from \(55\). - Use a positive difference whether the estimate is above or below \(55\). - Choose the least difference.

Solution

1. Find how far each estimate is from \(55\). 2. Group A: \(55 - 40 = 15\); Group B: \(55 - 50 = 5\); Group C: \(55 - 30 = 25\); Group D: \(70 - 55 = 15\). 3. The least difference is \(5\), for Group B.

Answer

Group B is closest to \(55\) steps.

5381383

Lina and Mina want to have at least \(45\) stickers altogether. How many more stickers do they need?

Figure for problem 538138

Hints

- Add Lina's and Mina's amounts first. - Compare their total with \(45\). - Subtract to find how many are still needed.

Solution

1. Lina and Mina have \(20 + 15 = 35\) stickers altogether. 2. The number still needed is \(45 - 35 = 10\) stickers.

Answer

They need \(10\) more stickers.

5381423

A class used \(60\) paper strips for a craft. The graph shows the three colors. How many strips were purple? Confirm the value using the other two bars.

Figure for problem 538142

Hints

- Add the orange and green values. - Subtract their sum from the total of \(60\). - Compare your result with the purple bar.

Solution

1. The orange and green strips total \(20 + 15 = 35\). 2. The remaining number is \(60 - 35 = 25\). 3. This matches the purple bar, which has a value of \(25\).

Answer

There were \(25\) purple paper strips.

5381433

Are the two shortest bars together greater or less than the tallest bar? By how much?

Figure for problem 538143

Hints

- Identify the two shortest bars and add their values. - Identify the tallest bar. - Compare the sum with the tallest bar by subtraction.

Solution

1. The two least values are \(10\) and \(15\), with a sum of \(10 + 15 = 25\). 2. The greatest value is \(30\). 3. Since \(30 - 25 = 5\), the sum of the two least values is less by \(5\).

Answer

The two shortest bars together are \(5\) less than the tallest bar.

5381443

Teams collect keys during a puzzle challenge. Which team is in second place, and how far behind first place is it?

Figure for problem 538144

Hints

- Find the greatest and second-greatest bar values. - The second-greatest value is second place. - Subtract the two values to find the gap.

Solution

1. The two greatest values are Team Dune with \(40\) keys and Team Meadow with \(35\) keys. 2. Team Meadow is in second place. 3. The gap is \(40 - 35 = 5\) keys.

Answer

Team Meadow is in second place, \(5\) keys behind Team Dune.

5381473

For a nature collage, the number of stones is doubled. The numbers of wood pieces and leaves stay the same. How many pieces will the collage have altogether?

Figure for problem 538147

Hints

- Double only the stone value. - Keep the wood and leaf values unchanged. - Add the three new values.

Solution

1. The number of stones changes from \(10\) to \(2 \times 10 = 20\). 2. The new total is \(15 + 20 + 20 = 55\) pieces.

Answer

The collage will have \(55\) pieces altogether.

5381483

Some birds leave the largest flock. How many birds must leave so that this flock has the same number of birds as the second-largest flock?

Figure for problem 538148

Hints

- Identify the largest and second-largest flocks. - Find the difference between their sizes. - Check that removing that number makes the flock sizes equal.

Solution

1. The largest flock is the cranes with \(40\) birds. The second-largest flock is the geese with \(35\) birds. 2. The difference is \(40 - 35 = 5\) birds.

Answer

\(5\) cranes must leave the flock.

5381523

Four collection boxes contain the amounts shown in the graph. More items will be added to Box S4 so that the boxes contain \(70\) items altogether. How many items must be added?

Figure for problem 538152

Hints

- Add the amounts in all four boxes first. - Compare that total with \(70\). - Subtract to find how many items must be added.

Solution

1. The current total is \(10 + 15 + 20 + 20 = 65\) items. 2. The number needed is \(70 - 65 = 5\) items.

Answer

\(5\) items must be added to Box S4.

5381543

Each group should have at least \(20\) points. How many points must be distributed altogether if points are added only to groups below \(20\)?

Figure for problem 538154

Hints

- Find how many points each group below \(20\) needs. - Do not add points to groups already at or above \(20\). - Add the needed amounts.

Solution

1. Group A needs \(20 - 15 = 5\) points. 2. Group C needs \(20 - 10 = 10\) points. 3. Groups B and D already meet the goal. 4. Altogether, \(5 + 10 = 15\) points must be distributed.

Answer

\(15\) points must be distributed altogether.

5381563

The graph shows red and blue beads. Check both claims: “There are \(40\) beads altogether” and “There are \(8\) more red beads than blue beads.” Do both claims match the graph?

Figure for problem 538156

Hints

- Check the total claim by addition. - Check the comparison claim by subtraction. - Decide whether each claim is true.

Solution

1. The total is \(24 + 16 = 40\), so the first claim is true. 2. The difference is \(24 - 16 = 8\), so the second claim is true. 3. Both claims match the graph.

Answer

Yes. Both claims match the graph.

5381573

Group A gives \(5\) cards to Group B. Which groups then have the same number of cards?

Figure for problem 538157

Hints

- Subtract \(5\) from Group A and add \(5\) to Group B. - Keep Group C's value unchanged. - Compare the three new values.

Solution

1. Group A will have \(30 - 5 = 25\) cards. 2. Group B will have \(20 + 5 = 25\) cards. 3. Group C already has \(25\) cards.

Answer

All three groups will have \(25\) cards each.

5381633

Nuri describes Bar D this way: “It is \(3\) greater than C and \(3\) less than B.” Is the description correct?

Figure for problem 538163

Hints

- Check each part of the description separately. - Find the difference between D and C. - Find the difference between B and D.

Solution

1. Bar C has a value of \(18\), Bar D has \(21\), and Bar B has \(24\). 2. Since \(21 - 18 = 3\) and \(24 - 21 = 3\), both parts of the description are correct.

Answer

Yes. Nuri's description is correct.

5381683

The day with the greatest value is removed from the data. What is the sum of the other three days?

Figure for problem 538168

Hints

- Identify and leave out the greatest bar. - Add the other three values. - Check that you used exactly three days.

Solution

1. Tuesday has the greatest value, \(25\). 2. The remaining values are \(15\), \(10\), and \(20\). 3. Their sum is \(15 + 10 + 20 = 45\).

Answer

The remaining three days have a sum of \(45\).

5381693

Which individual bars are taller than Bars A and B combined?

Figure for problem 538169

Hints

- Add the values of A and B. - Compare each other bar with that sum. - Remember that equal to is not the same as greater than.

Solution

1. Bars A and B have a combined value of \(10 + 15 = 25\). 2. Bar C has a value of \(20\), and Bar D has a value of \(25\). 3. Neither value is greater than \(25\).

Answer

No individual bar is taller than Bars A and B combined.

5381713

Compare Bars A and B combined with Bars C and D combined. Which pair has the greater total, and by how much?

Figure for problem 538171

Hints

- Find each pair total separately. - Compare the two totals. - Subtract to find the difference.

Solution

1. Bars A and B total \(10 + 20 = 30\). 2. Bars C and D total \(25 + 10 = 35\). 3. The difference is \(35 - 30 = 5\).

Answer

Bars C and D have the greater total, by \(5\).

5381733

Graph a) shows the points in Round \(1\), and Graph b) shows the points in Round \(2\). Which group increased its score the most?

Figure for problem 538173

Hints

- Compare the same group in both graphs. - Find each change from a) to b). - Choose the greatest positive change.

Solution

1. Group A increased by \(15 - 10 = 5\) points. 2. Group B changed by \(20 - 20 = 0\) points. 3. Group C increased by \(25 - 15 = 10\) points. 4. Group D decreased by \(25 - 20 = 5\) points. The greatest increase is \(10\), for Group C.

Answer

Group C increased its score the most, by \(10\) points.

5381743

The graphs show visitor counts in two different weeks. On which weekday did the number decrease the most from a) to b)?

Figure for problem 538174

Hints

- Compare the same weekday in both graphs. - Find each decrease from a) to b). - Choose the greatest decrease.

Solution

1. The decreases are Monday: \(30 - 25 = 5\), Tuesday: \(35 - 20 = 15\), Wednesday: \(25 - 25 = 0\), and Thursday: \(30 - 20 = 10\). 2. The greatest decrease is \(15\), on Tuesday.

Answer

The number decreased the most on Tuesday, by \(15\) visitors.

5381763

Graph a) shows bags of recyclables collected by three teams in Week \(1\), and Graph b) shows the bags collected by the same teams in Week \(2\). In which week were more bags collected altogether, and how many more?

Figure for problem 538176

Hints

- Find the total for each graph separately. - Compare the two totals. - Subtract to find how many more.

Solution

1. Week \(1\) total: \(15 + 25 + 20 = 60\) bags. 2. Week \(2\) total: \(20 + 15 + 30 = 65\) bags. 3. The difference is \(65 - 60 = 5\) bags.

Answer

Week \(2\) had \(5\) more bags collected.

5381773

Do the two graphs have the same total? Justify your answer using the values shown.

Figure for problem 538177

Hints

- Add all four values in Graph a). - Add all four values in Graph b). - Compare the totals.

Solution

1. Graph a) has a total of \(10 + 20 + 15 + 25 = 70\). 2. Graph b) has a total of \(15 + 10 + 25 + 20 = 70\). 3. The totals are equal.

Answer

Yes. Both graphs have a total of \(70\).

5381783

Compare Graphs a) and b). 1) Which regions increased? 2) Which regions decreased?

Figure for problem 538178

Hints

- Compare the same region in both graphs. - Mark each change as an increase or a decrease. - Check all four regions.

Solution

1. South increased from \(25\) to \(30\), and West increased from \(20\) to \(25\). 2. North decreased from \(30\) to \(25\), and East decreased from \(35\) to \(30\).

Answer

1) South and West increased. 2) North and East decreased.

5381803

How does the gap between A and B change from Graph a) to Graph b)?

Figure for problem 538180

Hints

- Find the A–B gap in each graph. - Compare the two gaps. - State whether the gap became larger or smaller.

Solution

1. In a), the gap is \(40 - 30 = 10\) points. 2. In b), the gap is \(45 - 40 = 5\) points. 3. The gap becomes \(10 - 5 = 5\) points smaller.

Answer

The gap becomes \(5\) points smaller.

5381813

Compare both graphs. 1) In Graph a), do more, fewer, or the same number of students travel by bike and on foot together than by bus? 2) In Graph b), do more, fewer, or the same number of students travel by bike and on foot together than by bus?

Figure for problem 538181

Hints

- Add the bike and walking values in each graph. - Compare each sum with the bus value in the same graph. - State the result for both a) and b).

Solution

1. In a), bike and walking total \(15 + 10 = 25\), the same as the bus value of \(25\). 2. In b), bike and walking total \(20 + 15 = 35\), while the bus value is \(20\). 3. In b), the combined total is \(35 - 20 = 15\) greater.

Answer

1) In Graph a), the numbers are equal. 2) In Graph b), \(15\) more students travel by bike and on foot together than by bus.

5381833

The total increases from a) to b). Which product contributes the most to this increase?

Figure for problem 538183

Hints

- Compare each product in the two graphs. - Find each increase or decrease. - Choose the greatest positive change.

Solution

1. Apple crates increase by \(25 - 20 = 5\). 2. Pear crates increase by \(25 - 15 = 10\). 3. Nut crates decrease by \(25 - 20 = 5\). 4. The greatest increase is \(10\), for pears.

Answer

Pears contribute the most, with an increase of \(10\) crates.

5381853

For each group, add its value from a) to its value from b). Which groups have the same sum?

Figure for problem 538185

Hints

- Add the two values for one group at a time. - Record all four sums. - Compare the sums.

Solution

1. Group A: \(5 + 20 = 25\). 2. Group B: \(10 + 15 = 25\). 3. Group C: \(15 + 10 = 25\). 4. Group D: \(20 + 5 = 25\). All four sums are equal.

Answer

All four groups have the same sum: \(25\) each.

5381863

In which graph are the greatest and least values farther apart?

Figure for problem 538186

Hints

- Find the greatest and least value in each graph. - Subtract to find each difference. - Compare the two differences.

Solution

1. In a), the difference between the greatest and least values is \(25 - 10 = 15\). 2. In b), the difference between the greatest and least values is \(20 - 15 = 5\). 3. Since \(15 > 5\), the values are farther apart in a).

Answer

The greatest and least values are farther apart in Graph a).

5381893

How many objects do the two graphs show altogether?

Figure for problem 538189

Hints

- Find the total in each graph first. - Add the two graph totals. - Check that all six bars were included.

Solution

1. Graph a) has \(15 + 20 + 10 = 45\) objects. 2. Graph b) has \(20 + 15 + 15 = 50\) objects. 3. Together, the graphs show \(45 + 50 = 95\) objects.

Answer

The two graphs show \(95\) objects altogether.

5381903

Which group is below \(20\) in a) but reaches at least \(20\) in b)?

Figure for problem 538190

Hints

- Check the value in a) first. - Then check the same group in b). - Apply both conditions to every group.

Solution

1. Group A increases from \(15\) to \(20\), so it meets both conditions. 2. Group C remains below \(20\). 3. Groups B and D were already above \(20\) in a).

Answer

Only Group A meets both conditions.

5381913

Which group has the greatest combined total from the two graphs?

Figure for problem 538191

Hints

- Add each group's two values. - Record the four totals. - Choose the greatest total.

Solution

1. Group A: \(10 + 20 = 30\). 2. Group B: \(15 + 10 = 25\). 3. Group C: \(20 + 15 = 35\). 4. Group D: \(25 + 20 = 45\). The greatest total is \(45\).

Answer

Group D has the greatest combined total, \(45\).

5381923

In Graph b), only Bar C will be increased until the two graphs have the same total. By how much must C increase?

Figure for problem 538192

Hints

- Find the total of each graph. - Find the difference between the totals. - Only Bar C changes.

Solution

1. Graph a) has a total of \(20 + 15 + 25 = 60\). 2. Graph b) has a total of \(15 + 20 + 20 = 55\). 3. The difference is \(60 - 55 = 5\), so C must increase by \(5\).

Answer

Bar C in b) must increase by \(5\).

5382124

The bar graph shows values for Morning, Noon, and Evening. a) Check whether Noon’s value is twice Morning’s value. b) Check whether Evening’s value is three times Morning’s value. c) Find the total of the three values.

Figure for problem 538212

Hints

- 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.

Solution

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 \(6+12+18=36\).

Answer

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 \(36\).

5382943

Art studio: <table><tr><th>Main material</th><th>Number of artworks</th></tr><tr><td>Clay</td><td>\(13\)</td></tr><tr><td>Wood</td><td>\(19\)</td></tr><tr><td>Fabric</td><td>\(11\)</td></tr><tr><td>Paper</td><td>\(17\)</td></tr></table> For the final bar graph, \(4\) artworks are moved from the Wood category to the Paper category. What are the new bar lengths?

Hints

- One category loses \(4\) artworks and another category gains \(4\). - Check that the total number of artworks stays the same.

Solution

1. For Wood, subtract \(4\): \(19 - 4 = 15\). 2. For Paper, add \(4\): \(17 + 4 = 21\). 3. The Clay and Fabric values stay the same.

Answer

Clay: \(13\) Wood: \(15\) Fabric: \(11\) Paper: \(21\)

5383193

Three teams make paper airplanes. <table><thead><tr><th>Team</th><th>Round 1</th><th>Round 2</th></tr></thead><tbody><tr><td>Red</td><td>\(12\)</td><td>\(8\)</td></tr><tr><td>Blue</td><td>\(15\)</td><td>\(11\)</td></tr><tr><td>Green</td><td>\(9\)</td><td>\(14\)</td></tr></tbody></table> How many paper airplanes did the teams make in both rounds altogether?

Hints

- Use all six numbers in the table. - Find the total for each round first. - Add the two round totals.

Solution

1. In Round 1, the teams made \(12 + 15 + 9 = 36\) airplanes. 2. In Round 2, the teams made \(8 + 11 + 14 = 33\) airplanes. 3. Altogether, they made \(36 + 33 = 69\) airplanes.

Answer

The teams made \(69\) paper airplanes altogether.

5383463

Raffle tickets are counted by color. Monday has \(14\) red and \(9\) blue tickets. Tuesday has \(12\) red and \(11\) blue tickets. Wednesday has \(16\) red and \(8\) blue tickets. Thursday has \(10\) red and \(15\) blue tickets. Which two days have the same total number of tickets?

Hints

- Find the total for each day. - Compare the four totals.

Solution

1. Monday's total is \(14 + 9 = 23\). 2. Tuesday's total is \(12 + 11 = 23\). 3. Wednesday's total is \(16 + 8 = 24\). 4. Thursday's total is \(10 + 15 = 25\). 5. Monday and Tuesday have the same total.

Answer

Monday and Tuesday

5383603

A classroom book box is counted before and after a book drive. <table><thead><tr><th>Day</th><th>Before</th></tr></thead><tbody><tr><td>Monday</td><td>\(11\)</td></tr><tr><td>Tuesday</td><td>\(15\)</td></tr><tr><td>Wednesday</td><td>\(13\)</td></tr></tbody></table> <table><thead><tr><th>Day</th><th>After</th></tr></thead><tbody><tr><td>Monday</td><td>\(14\)</td></tr><tr><td>Tuesday</td><td>\(14\)</td></tr><tr><td>Wednesday</td><td>\(18\)</td></tr></tbody></table> On which day did the number increase the most?

Hints

- Compare the Before and After values for each day. - Find each increase, then compare the increases.

Solution

1. Monday's increase is \(14 - 11 = 3\). 2. Tuesday's value decreased from \(15\) to \(14\). 3. Wednesday's increase is \(18 - 13 = 5\). 4. The greatest increase is \(5\), on Wednesday.

Answer

Wednesday

5383623

A colored-pencil case has \(16\) pencils in the top section and \(13\) in the bottom section. Four pencils are added to the top section. How many pencils are in the case afterward?

Hints

- Update the section that changes first. - The bottom section keeps its original number of pencils. - Add the two section amounts after the change.

Solution

1. The top section then has \(16+4=20\) pencils. 2. Add the unchanged bottom section: \(20+13=33\) pencils.

Answer

There are \(33\) colored pencils in the case afterward.

5383643

A theater sells \(24\) child tickets in the morning and \(19\) child tickets in the afternoon. It sells \(38\) adult tickets altogether. Were more child tickets or adult tickets sold, and how many more?

Hints

- Combine the morning and afternoon child-ticket counts. - Compare that total with the adult-ticket total already given. - Subtract the smaller total from the larger total to find how many more.

Solution

1. Find the child-ticket total: \(24+19=43\). 2. Compare with the adult-ticket total: \(43-38=5\).

Answer

\(5\) more child tickets were sold.

5383673

Paul reads \(25\) pages on Monday and \(17\) pages on Tuesday. Rayan reads \(39\) pages over the same two days. Who reads more altogether, and by how many pages?

Hints

- Find Paul’s two-day total first. - Rayan’s two-day total is already given. - Compare the two totals by finding their difference.

Solution

1. Paul reads \(25+17=42\) pages altogether. 2. Compare the totals: \(42-39=3\).

Answer

Paul reads \(3\) more pages than Rayan.

5383923

A bin has \(7\) red cubes. It also has \(5\) small blue cubes and \(2\) large blue cubes. How many red and blue cubes are in the bin altogether?

Hints

- Combine the two sizes of blue cubes first. - The red-cube total is already given. - Add the two color totals.

Solution

1. Find the number of blue cubes: \(5+2=7\). 2. Add red and blue cubes: \(7+7=14\).

Answer

There are \(14\) red and blue cubes altogether.

5384033

Fill in the two missing numbers. <table><thead><tr><th>Group</th><th>Red cards</th><th>Blue cards</th><th>Row total</th></tr></thead><tbody><tr><td>Group A</td><td>\(8\)</td><td>?</td><td>\(15\)</td></tr><tr><td>Group B</td><td>?</td><td>\(9\)</td><td>\(15\)</td></tr><tr><td>Column total</td><td>\(14\)</td><td>\(16\)</td><td>\(30\)</td></tr></tbody></table>

Hints

- Start with a row or column that has only one missing value. - Check each result against the other totals.

Solution

1. Group A has \(15 - 8 = 7\) blue cards. 2. Group B has \(14 - 8 = 6\) red cards. 3. These values also satisfy the remaining row and column totals.

Answer

Group A: \(7\) blue cards Group B: \(6\) red cards

5384043

Fill in the three missing numbers. <table><thead><tr><th>Color</th><th>Small stars</th><th>Large stars</th><th>Combined</th></tr></thead><tbody><tr><td>Silver</td><td>\(9\)</td><td>?</td><td>\(14\)</td></tr><tr><td>Gold</td><td>?</td><td>?</td><td>\(15\)</td></tr><tr><td>Total</td><td>\(16\)</td><td>\(13\)</td><td>\(29\)</td></tr></tbody></table>

Hints

- Begin with a row or column that has only one missing value. - Enter each result before using another total. - Check the completed row and column totals against the table.

Solution

1. The number of large silver stars is \(14-9=5\). 2. The number of small gold stars is \(16-9=7\). 3. The number of large gold stars is \(15-7=8\).

Answer

Large silver stars: \(5\) Small gold stars: \(7\) Large gold stars: \(8\)

5384073

Team North has \(25\) points. During a review, it receives \(6\) more points and then loses \(2\) points because of a penalty. Team South has \(28\) points and its score does not change. What is Team North’s final score, and which team has more points afterward?

Hints

- Apply the score changes in the order they occur. - The second change acts on the score after the first change. - Compare the final North score with South’s unchanged score only after both operations.

Solution

1. Add the awarded points: \(25+6=31\). 2. Apply the penalty: \(31-2=29\). 3. Team North finishes with \(29\) points, which is greater than Team South’s \(28\) points.

Answer

Team North finishes with \(29\) points, so Team North has more points afterward.

5402854

A water station begins with \(12\,\text{L}\) of water. After \(3\) bottles are filled, \(5\,\text{L}\) remain. First find how many liters were used. Is that amount a multiple of \(3\)? Use your answer to decide whether every bottle could have received the same whole number of liters.

Hints

- Subtract the amount remaining from the starting amount. - Compare the amount used with nearby multiples of \(3\). - Equal whole-number shares among \(3\) bottles are possible only when the amount used is a multiple of \(3\).

Solution

1. The bottles received \(12\,\text{L}-5\,\text{L}=7\,\text{L}\) altogether. 2. Seven is not a multiple of \(3\): the nearby multiples are \(6=3\times2\) and \(9=3\times3\). 3. Therefore, \(7\,\text{L}\) cannot be split into \(3\) equal whole-number amounts.

Answer

\(7\,\text{L}\) were used. Since \(7\) is not a multiple of \(3\), the bottles could not all receive the same whole number of liters.

5503643

The diagram shows an upper seating section and a lower seating section. Every row has the same number of seats. Without counting seats one by one: a) How many more rows does the upper section have? b) Use that row difference to find how many more seats the upper section has. Show both steps.

Figure for problem 550364

Hints

- Compare the sections by rows before thinking about individual seats. - The diagram shows the same number of seats in every row. - Show the row subtraction first, then multiply the extra rows by the number of seats in each row.

Solution

1. The upper section has \(4\) rows and the lower section has \(2\) rows, so the upper section has \(4-2=2\) extra rows. 2. Each row has \(7\) seats, so \(2 \times 7=14\). 3. The upper section has \(14\) more seats.

Answer

a) The upper section has \(2\) more rows. b) \(2 \times 7=14\), so the upper section has \(14\) more seats.

5550913

A club has \(18\) red badges and \(12\) blue badges. It divides all the badges equally among \(5\) teams. Write one equation using \(n\) for the number of badges each team receives, then find \(n\).

Hints

- The teams share all badges, so the two badge amounts must be combined before sharing. - Let \(n\) represent the size of one equal share. - Use parentheses so the equation shows which operation happens first.

Solution

1. First combine the two badge colors: \(18+12=30\). 2. The equation is \(n=(18+12) \div 5\). 3. \(30 \div 5=6\), so \(n=6\).

Answer

\(n=(18+12) \div 5\), so \(n=6\).

5550923

A school has \(248\) books in one reading room and \(367\) books in another. It sends \(190\) books to classrooms. a) How many books remain in the two reading rooms altogether? b) Round each starting number and the number sent to the nearest hundred. Use the rounded numbers to check whether your exact answer is reasonable.

Hints

- Find the exact two-step answer first. - For the estimate, round each number to the nearest hundred before calculating. - Your estimate should be close to the exact answer, but it does not have to match it.

Solution

1. Combine the starting amounts: \(248+367=615\) books. 2. Subtract the books sent out: \(615-190=425\) books. 3. To the nearest hundred, \(248\approx200\), \(367\approx400\), and \(190\approx200\). 4. The estimate is \(200+400-200=400\), which is close to \(425\). The exact answer is reasonable.

Answer

a) \(425\) books b) The estimate is about \(400\) books, so \(425\) is reasonable.

5550933

A museum shop starts with \(325\) postcards. It sells \(178\) postcards, then receives \(92\) new postcards. a) How many postcards are in the shop now? b) Round each number to the nearest hundred and use the rounded calculation to decide whether your exact answer is reasonable.

Hints

- Follow the events in order for the exact calculation. - Round the three quantities separately for the estimate. - Compare the exact answer with the estimate by size, not by exact equality.

Solution

1. After the sale, \(325-178=147\) postcards remain. 2. After the delivery, \(147+92=239\) postcards are in the shop. 3. To the nearest hundred, \(325\approx300\), \(178\approx200\), and \(92\approx100\). 4. The estimate is \(300-200+100=200\). This is close enough to \(239\) to support the exact answer.

Answer

a) \(239\) postcards b) The estimate is about \(200\) postcards, so \(239\) is reasonable.

5550943

A baker puts \(5\) cookies on each of \(8\) trays. Then \(12\) cookies are eaten. Noah writes \((8+12) \times 5\) to find how many cookies remain. Explain Noah’s error and find the correct number of cookies remaining.

Hints

- What do the \(8\), \(5\), and \(12\) mean in the story? - Find how many cookies there are before any are eaten. - Then show what happens when \(12\) cookies are eaten.

Solution

1. Noah adds the cookies eaten to the number of trays before finding how many cookies there were at the start. Those numbers should not be added. 2. First find the starting number of cookies: \(8 \times 5=40\). 3. Then subtract the cookies eaten: \(40-12=28\).

Answer

Noah should multiply to find the starting number of cookies before subtracting the cookies that were eaten. The correct calculation is \(8\times5-12=28\), so \(28\) cookies remain.

5550953

After \(4\) packs with \(6\) stickers in each pack are added to a collection, the collection has \(38\) stickers. How many stickers were in the collection before the packs were added?

Hints

- First find how many stickers were added in the \(4\) packs. - The \(38\) stickers are the amount after the packs were added, so work backward. - Check by adding the pack total to your starting amount.

Solution

1. Find how many stickers were added: \(4 \times 6=24\). 2. Work backward from the final total: \(38-24=14\).

Answer

There were \(14\) stickers in the collection before the packs were added.

5161713

A sports club has \(\$250\). Soccer balls cost \(\$125\), and practice vests cost \(\$130\). Maya adds the two costs and compares the total with \(\$250\). Jordan subtracts \(\$125\) from \(\$250\), then compares what is left with \(\$130\). Use both methods. Do they give the same result? How much more money does the club need?

Hints

- In Maya’s method, compare the total purchase cost with the budget. - In Jordan’s method, compare the second purchase with what remains after the first purchase. - Think about why both methods measure the same gap between cost and available money.

Solution

1. Maya’s method: \(\$125 + \$130 = \$255\), then \(\$255 - \$250 = \$5\). The club is short by \(\$5\). 2. Jordan’s method: \(\$250 - \$125 = \$125\), then \(\$130 - \$125 = \$5\). The club is short by \(\$5\). 3. Both methods compare the same two purchase costs with the same budget, so they measure the same shortfall.

Answer

Both methods show that the club needs \(\$5\) more.

5178473

Leon has \(25\) trading cards. Ben says, “If I give you \(8\) of my cards, we will have the same number.” How many cards does Ben have before giving Leon the cards?

Hints

- First find how many cards Leon has after the gift. - At that time, Ben has the same number. - Work backward to find Ben's starting amount.

Solution

1. After receiving \(8\) cards, Leon has \(25+8=33\) cards. 2. Ben also has \(33\) cards after giving away the cards because their amounts are equal then. 3. Before giving away \(8\) cards, Ben had \(33+8=41\) cards.

Answer

Ben has \(41\) cards at the start.

5185903

A library box contains some nonfiction books at the start of the day. A class checks out \(15\) books and later returns \(11\) of them. At the end of the day, \(28\) books remain in the box. How many books were in the box at the start of the day?

Hints

- First find the number of books left that stayed checked out. - Work backward from the ending number of books. - Check by starting with your answer, removing \(15\), and returning \(11\).

Solution

1. The class keeps \(15-11=4\) books checked out. 2. Add those \(4\) books back to the \(28\) remaining books: \(28+4=32\).

Answer

There were \(32\) books in the box at the start of the day.

5204303

At the end of the day, a school library has \(650\) books. Earlier that day, students checked out \(45\) books, and later the library received \(25\) new books. How many books were in the library at the start of the day?

Hints

- The ending amount is known, so undo the events in reverse order. - What operation undoes receiving \(25\) books? - What operation undoes checking out \(45\) books?

Solution

1. Work backward through the last event. Before the \(25\) new books arrived, there were \(650 - 25 = 625\) books. 2. Before \(45\) books were checked out, there were \(625 + 45 = 670\) books. 3. Check forward: \(670 - 45 + 25 = 650\).

Answer

There were \(670\) books in the library at the start of the day.

5204503

A school library removes \(18\) old books on Monday and buys \(45\) new books on Tuesday. On Wednesday, a community group donates a box of books. After all three changes, the library has \(110\) more books than it had before Monday. How many books were in the donated box?

Hints

- Find the change after removing and buying books. - Compare that change with the final increase of \(110\). - The difference is the number of donated books.

Solution

1. Find the change after Monday and Tuesday: \(45 - 18 = 27\). The library has \(27\) more books at that point. 2. The final increase is \(110\) books. Subtract the known increase: \(110 - 27 = 83\). 3. The donated box contains \(83\) books.

Answer

The donated box contained \(83\) books.

5213483

A bakery sells \(145\) loaves in its store and delivers \(120\) loaves to a restaurant. After that, the bakery has \(265\) loaves left. How many loaves did the bakery have at the start of the morning?

Hints

- First find how many loaves left the bakery. - The number left is the same as that amount. - Add the loaves that left and the loaves that remained.

Solution

1. The bakery sold and delivered \(145+120=265\) loaves. 2. It also had \(265\) loaves left. 3. So it started with \(265+265=530\) loaves.

Answer

\(530\) loaves

5381243

The chart shows the numbers of marbles in Box A and Box B. Move marbles from Box A to Box B until the two boxes contain the same number. How many marbles must be moved?

Figure for problem 538124

Hints

- Find the difference between the two bars. - Think about how moving one marble changes both boxes at once. - Check that the two boxes have the same number after the move.

Solution

1. Box A contains \(24\) marbles and Box B contains \(16\), so the difference is \(24-16=8\). 2. Each marble moved lowers the difference by \(2\): one leaves Box A and one is added to Box B. 3. Therefore, \(8 \div 2=4\) marbles must be moved.

Answer

\(4\) marbles must be moved from Box A to Box B.

5381413

Which two animal teams differ by exactly \(8\) collected items? Name every pair that works.

Figure for problem 538141

Hints

- Write down the four values. - Find the difference for each possible pair in an organized way. - Make sure you list every pair with a difference of \(8\).

Solution

1. Hedgehog and Mouse differ by \(20 - 12 = 8\). 2. Frog and Owl differ by \(16 - 8 = 8\). 3. Checking the other pairs shows that none of their differences is \(8\).

Answer

Hedgehog and Mouse; Frog and Owl.

5381513

Which two groups have the closest scores? What is the difference between their scores?

Figure for problem 538151

Hints

- Write down all four scores. - Compare the differences between possible pairs. - Choose the least positive difference.

Solution

1. Compare the score differences for all pairs. 2. The least difference is between Group B with \(30\) points and Group C with \(25\) points. 3. Their difference is \(30 - 25 = 5\) points.

Answer

Groups B and C have the closest scores. Their scores differ by \(5\) points.

5381603

Divide the four groups into two pairs so that the pairs have equal point totals. Give the pairing.

Figure for problem 538160

Hints

- Write down the four point values. - Test ways to split the groups into two pairs. - Add both pair totals to check that they are equal.

Solution

1. Groups A and B have \(10 + 25 = 35\) points. 2. Groups C and D have \(15 + 20 = 35\) points. 3. The two pair totals are equal.

Answer

Pair A with B and pair C with D. Each pair has \(35\) points.

5381663

Only Group C may receive more points. What is the least number of points it needs to be alone in first place?

Figure for problem 538166

Hints

- Find the current greatest score. - Determine the least score that is greater than it. - Subtract Group C's current score from that target.

Solution

1. Group C has \(20\) points, and the current greatest score is \(25\). 2. To be alone in first place, Group C must have at least \(26\) points. 3. It needs \(26 - 20 = 6\) more points.

Answer

Group C needs at least \(6\) more points.

5381703

On which two days were exactly \(36\) admission tickets sold altogether? Name every possible pair.

Figure for problem 538170

Hints

- Write down all five day values. - Test pairs in an organized way. - List every pair with a sum of \(36\).

Solution

1. Monday and Friday give \(12 + 24 = 36\). 2. Tuesday and Thursday give \(15 + 21 = 36\). 3. Checking the other day pairs shows that none has a sum of \(36\).

Answer

Monday and Friday; Tuesday and Thursday.

5382043

The chart shows student votes for cocoa and tea. Some students switch their vote from cocoa to tea. How many students must switch so that cocoa and tea have equal numbers of votes?

Figure for problem 538204

Hints

- Compare the lengths and values of the two bars. - One switch changes both vote totals at the same time. - Check that the two new vote totals are equal.

Solution

1. Cocoa has \(14\) votes and tea has \(8\), so the difference is \(14-8=6\). 2. Each switch lowers the difference by \(2\), because cocoa loses one vote while tea gains one. 3. Therefore, \(6 \div 2=3\) students must switch.

Answer

\(3\) students must switch from cocoa to tea.

5384053

Exactly one value inside the table is incorrect. The totals along the edges are correct. <table><thead><tr><th>Team</th><th>Round 1</th><th>Round 2</th><th>Combined</th></tr></thead><tbody><tr><td>Team A</td><td>\(6\)</td><td>\(8\)</td><td>\(14\)</td></tr><tr><td>Team B</td><td>\(7\)</td><td>\(10\)</td><td>\(16\)</td></tr><tr><td>Total</td><td>\(13\)</td><td>\(17\)</td><td>\(30\)</td></tr></tbody></table> Find and correct the incorrect value.

Hints

- Check every row total and column total. - The corrected value must satisfy both its row total and its column total.

Solution

1. Team B's Round 2 value must combine with \(7\) to make the row total \(16\). 2. The correct value is \(16 - 7 = 9\). 3. This also makes the Round 2 column total correct because \(8 + 9 = 17\).

Answer

The Team B, Round 2 entry should be \(9\), not \(10\).

5503653

A classroom has \(9\) rows with \(6\) chairs in each row. Two chairs are removed from the room. Leo writes \(9 \times (6 - 2)\) to find how many chairs remain. Explain Leo's error and find the correct number of chairs remaining.

Hints

- Decide whether the \(2\) removed chairs come from every row or from the room altogether. - Compare what \(6 - 2\) would mean inside the multiplication expression with what the story says. - Find the original total before applying the removal.

Solution

1. Leo's expression removes \(2\) chairs from every row, which would remove \(18\) chairs altogether. 2. The room starts with \(9 \times 6 = 54\) chairs. 3. Only \(2\) chairs are removed from the whole room, so \(54 - 2 = 52\) chairs remain.

Answer

Leo's expression removes \(2\) chairs from each row instead of \(2\) chairs total. The correct answer is \(52\) chairs.

5503663

At the end of an art lesson, the supply cabinet has \(87\) markers. Earlier, Ms. Davis gave out \(48\) markers, and later \(15\) unused markers were returned. How many markers were in the cabinet before the lesson?

Hints

- Work backward from the ending amount in reverse event order. - First undo the markers that were returned. - Then restore the markers that had been given out.

Solution

1. Work backward by undoing the return: \(87-15=72\). 2. Undo the markers that were given out: \(72+48=120\).

Answer

There were \(120\) markers in the cabinet before the lesson.

5503673

A supply room has \(72\) pencils packed equally with \(8\) pencils in each bag. Five full bags are given to classrooms. How many pencils remain? Solve in two ways: first by finding the pencils given away, and second by reasoning with the number of full bags. Explain why the methods agree.

Hints

- One route can work with individual pencils; the other can work with whole bags. - How many pencils are in the five bags that leave? - How many full bags are represented by \(72\) pencils?

Solution

1. Method 1: five bags contain \(5 \times 8 = 40\) pencils, so \(72 - 40 = 32\) pencils remain. 2. Method 2: \(72 \div 8 = 9\) full bags. After giving away \(5\) bags, \(9 - 5 = 4\) bags remain. Then \(4 \times 8 = 32\) pencils remain. 3. Both methods describe removing the same five equal groups from the same starting amount.

Answer

Method 1: \(5\times8=40\), then \(72-40=32\). Method 2: \(72\div8=9\) bags, \(9-5=4\) bags, then \(4\times8=32\). Both methods remove the same \(5\) bags, so \(32\) pencils remain.

5503683

A class has \(63\) stickers. Mr. Lewis keeps \(14\) stickers for a prize box and shares the rest equally among \(7\) students. Sam subtracts first and then divides. Priya says she can first notice that \(14\) is \(2\) groups of \(7\), then reason about how many groups of \(7\) remain. Use both approaches and explain why they agree.

Hints

- In Sam's method, find how many stickers remain before sharing. - In Priya's method, write \(63\) and \(14\) as groups of \(7\). - Explain how both methods remove the same \(14\) stickers.

Solution

1. Sam's method: \(63 - 14 = 49\), then \(49 \div 7 = 7\). Each student gets \(7\) stickers. 2. Priya's method: \(63 = 9 \times 7\) and \(14 = 2 \times 7\). Removing \(2\) groups from \(9\) groups leaves \(7\) groups of \(7\), so each student gets \(7\) stickers. 3. Both methods remove the same \(14\) stickers before finding the equal share.

Answer

Each student gets \(7\) stickers. Sam removes \(14\) stickers first and then divides. Priya sees \(63\) as \(9\) groups of \(7\) and removes \(2\) groups of \(7\). Both methods remove the same \(14\) stickers before sharing, so they agree.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.