Free math worksheets

- What does one-eighth mean as an operation? - What number do you divide by to find one-eighth of a set? - Could \(32 \div 8\) help you calculate \(320 \div 8\)?

1. Find one-eighth of the total: \(320 \div 8 = 40\). 2. Therefore, \(40\) guinea pigs have completely white fur.

There are \(40\) white guinea pigs.

- Read the fraction represented by the shaded bar. - Divide the total number of tickets into that many equal groups. - Use the number of shaded groups to determine the number of tickets used.

1. The bar has \(1\) of \(4\) equal parts shaded, so the used fraction is \(\frac{1}{4}\). 2. Divide \(20\) tickets into \(4\) equal groups: \(20\div4=5\). 3. One fourth of \(20\) tickets is \(5\) tickets.

\(5\) tickets

- How can you find one-fourth of the total number of marbles? - After finding how many marbles were given away, how can you find how many remain? - What number do you divide by to find one-fourth?

1. Find one-fourth of \(36\): \(\frac{1}{4} \times 36 = 9\). 2. Subtract the marbles he gives away: \(36 - 9 = 27\).

Leon keeps \(27\) marbles.

- What does it mean to hike one-eighth of a trail? - First find the distance the family hiked before stopping. - How can you use the total distance and the distance already hiked to find the distance remaining?

1. Find one-eighth of the trail: \(\frac{1}{8} \times 8\,\text{mi} = 1\,\text{mi}\). 2. Subtract the distance already hiked: \(8\,\text{mi} - 1\,\text{mi} = 7\,\text{mi}\).

The family has \(7\,\text{mi}\) left to hike.

- Can you divide the total number of flowers into four equal groups? - How many flowers are in one of those groups? - After finding the number of tulips, how can you find the number of flowers that remain?

1. Find the number of tulips: \(\frac{1}{4} \times 32 = 8\). 2. Subtract the tulips from the total number of flowers: \(32 - 8 = 24\).

There are \(24\) daffodils in the bouquet.

- First find one-fifth of \(40\). - What does one-fifth mean when the seats are divided into equal groups? - After finding the number of occupied seats, how can you find the number available?

1. Find the number of occupied seats: \(\frac{1}{5} \times 40 = 8\). 2. Subtract the occupied seats from the total: \(40 - 8 = 32\).

There are \(32\) seats still available.

- First find the weight of the packaging alone. - What operation can you use to find one-eighth of a quantity? - What do you get when you subtract the packaging weight from the total weight?

1. Find the weight of the packaging: \(\frac{1}{8} \times 32\,\text{lb} = 4\,\text{lb}\). 2. Subtract the packaging weight from the total weight: \(32\,\text{lb} - 4\,\text{lb} = 28\,\text{lb}\).

The contents weigh \(28\,\text{lb}\).

- What does one-sixth mean in this calculation? - First find how many bulbs have already been planted. - What operation can you use to find the number remaining?

1. Find the number of bulbs already planted: \(\frac{1}{6} \times 480 = 80\). 2. Subtract the planted bulbs from the total: \(480 - 80 = 400\).

There are \(400\) flower bulbs left to plant.

- First find how many meters Paul has run. - Then find how many meters Anna has run. - What operation finds how much farther one distance is than another?

1. Find Paul's distance: \(\frac{3}{4} \times 400\,\text{m} = 300\,\text{m}\). 2. Find Anna's distance: \(\frac{1}{2} \times 400\,\text{m} = 200\,\text{m}\). 3. Find the difference: \(300\,\text{m} - 200\,\text{m} = 100\,\text{m}\).

Paul has run \(100\,\text{m}\) farther than Anna.

- First find the weight of one-fifth of the flour. - After finding one part, how can you find two parts? - Which operation divides the total into equal groups?

1. Find one-fifth of the flour: \(600\,\text{lb} \div 5 = 120\,\text{lb}\). 2. Find two-fifths: \(2 \times 120\,\text{lb} = 240\,\text{lb}\).

The bakery uses \(240\,\text{lb}\) of flour during the morning.

- First find one-fourth of the total amount of flour. - Check whether the question asks for the amount used or the amount remaining. - Imagine dividing the flour into four equal groups.

1. Find the flour used: \(\frac{1}{4} \times 32\,\text{oz} = 8\,\text{oz}\). 2. Subtract from the full bag: \(32\,\text{oz} - 8\,\text{oz} = 24\,\text{oz}\).

\(24\,\text{oz}\) of flour remain in the bag.

- Find each student's cut length separately. - How can you find one-eighth of a number? - Compare the two final lengths.

1. Find the length Lucas cuts: \(\frac{1}{8} \times 16\,\text{ft} = 2\,\text{ft}\). 2. Find the length Sarah cuts: \(\frac{1}{2} \times 6\,\text{ft} = 3\,\text{ft}\). 3. Since \(3\,\text{ft} > 2\,\text{ft}\), Sarah cuts the longer piece.

Sarah cuts the longer piece: \(3\,\text{ft}\) compared with Lucas's \(2\,\text{ft}\).

- Count the total number of pieces and the number of orange-shaded pieces. - Write and simplify the fraction represented by the shaded pieces. - To find a fraction of a quantity, divide by the denominator and multiply by the numerator.

1. The array has \(4\) rows and \(6\) columns, so there are \(4\times6=24\) pieces in all. 2. There are \(9\) orange-shaded pieces, so the fraction eaten is \(\frac{9}{24}=\frac{3}{8}\). 3. Find \(\frac{3}{8}\) of \(160\,\text{g}\): \(160\div8=20\), and \(3\times20\,\text{g}=60\,\text{g}\).

a) \(\frac{3}{8}\) b) \(60\,\text{g}\)

- The largest sector represents the most popular snack. - Add fractions with the same denominator by adding their numerators. - Multiply the class size by the pretzel fraction.

1. The largest sector is apple, representing \(\frac{1}{2}\) of the class. 2. Add the banana and granola-bar fractions: \(\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}\). 3. Pretzel represents \(\frac{1}{4}\) of the class. Compute \(\frac{1}{4} \times 24 = 6\) students.

1. Apple 2. \(\frac{1}{4}\) 3. \(6\) students

- Multiply \(32\) by the fraction for each fruit. - Look for two sectors whose fractions add to \(\frac{1}{4}\). - Two eighths are equivalent to one-fourth.

1. Apples represent \(\frac{1}{2}\) of the class: \(\frac{1}{2} \times 32 = 16\) students. 2. Pears represent \(\frac{1}{8}\) of the class: \(\frac{1}{8} \times 32 = 4\) students. 3. Pears and strawberries each represent \(\frac{1}{8}\). Together, \(\frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}\).

a) \(16\) students b) \(4\) students c) Pears and strawberries

- The whole circle represents all \(100\) students. - Multiply \(100\) by each fraction shown by the circle sectors.

1. Soccer represents \(\frac{1}{2}\) of the students: \(\frac{1}{2} \times 100 = 50\). 2. Swimming represents \(\frac{1}{4}\) of the students: \(\frac{1}{4} \times 100 = 25\). 3. Tennis also represents \(\frac{1}{4}\) of the students: \(\frac{1}{4} \times 100 = 25\).

Soccer: \(50\) students; swimming: \(25\) students; tennis: \(25\) students

- Use the unshaded sections, not the shaded sections, for the fraction that remains. - For part b), the fraction from part a) should be one factor and the full trail length should be the other. - To evaluate the product, first find one eighth of the full length.

1. The diagram shows \(3\) of the \(8\) equal sections unshaded, so the remaining fraction is \(\frac{3}{8}\). 2. A multiplication equation for the remaining distance is \(\frac{3}{8}\times24\,\text{km}\). 3. One eighth of \(24\,\text{km}\) is \(24\div8=3\,\text{km}\). Three eighths is \(3\times3\,\text{km}=9\,\text{km}\).

a) \(\frac{3}{8}\) b) \(\frac{3}{8}\times24\,\text{km}=9\,\text{km}\) c) \(9\,\text{km}\)

- Compare the sector sizes for part a). - Add the fractions for comics and nonfiction. - Multiply the total number of books by the combined fraction.

1. Adventure is the largest sector, representing \(\frac{1}{2}\) of the books. 2. Comics and nonfiction each represent \(\frac{1}{4}\). Together they represent \(\frac{1}{2}\) of the box, so \(\frac{1}{2} \times 60 = 30\) books.

a) Adventure b) \(30\) books

- Compare each sector with the whole circle. - Find one-half of \(24\) for part a). - For part b), write the campground votes over the total votes and simplify.

1. The farm sector represents one-half of the class. Compute \(\frac{1}{2} \times 24 = 12\) students. 2. The campground received \(3\) of the \(24\) votes. The fraction is \(\frac{3}{24} = \frac{1}{8}\).

a) \(12\) students b) \(\frac{1}{8}\)

- Count all equal grid squares and the blue squares, then simplify that fraction. - Use the simplified fraction as one factor and the total poster area as the other factor. - Find one twelfth of the total area before finding five twelfths.

1. The grid has \(4\times6=24\) equal squares, and \(10\) are blue, so the blue fraction is \(\frac{10}{24}=\frac{5}{12}\). 2. A multiplication equation for the blue area is \(\frac{5}{12}\times72\,\text{cm}^2\). 3. One twelfth of \(72\,\text{cm}^2\) is \(72\div12=6\,\text{cm}^2\). Five twelfths is \(5\times6\,\text{cm}^2=30\,\text{cm}^2\).

a) \(\frac{5}{12}\) b) \(\frac{5}{12}\times72\,\text{cm}^2=30\,\text{cm}^2\) c) \(30\,\text{cm}^2\)

- First find the total in the unit given for one pennant. - Use the customary-unit relationship between feet and yards. - The unit changes, but the physical length does not.

1. The total width is \(10\times\frac{3}{10}=3\,\text{ft}\). 2. Since \(3\,\text{ft}=1\,\text{yd}\), the display is \(1\,\text{yd}\) wide.

\(3\,\text{ft}=1\,\text{yd}\)

- The boxes are equal groups with the same fractional amount in each. - Count the total number of fourth-sized parts across all eight groups. - Regroup the fourths into complete wholes.

1. The total is \(8\times\frac{3}{4}\) pounds. 2. Eight groups of three fourths contain \(24\) fourths: \(\frac{24}{4}\). 3. \(\frac{24}{4}=6\), so the boxes contain \(6\) pounds of clay.

\(6\,\text{lb}\)

- Multiply the amount in one bottle by the number of bottles. - Find the total amount poured out. - Subtract the amount poured out from the original total.

1. The total amount purchased is \(8\times\frac{3}{4}=6\,\text{qt}\). 2. The amount poured out is \(8\times\frac{1}{4}=2\,\text{qt}\). 3. The amount remaining is \(6-2=4\,\text{qt}\).

a) \(6\,\text{qt}\) b) \(4\,\text{qt}\)

- Find the number of students with pets in each class. - A larger fraction does not always mean a larger number; consider each class size. - Combine the numbers of students and pet owners for part b.

1. In Class A, \(\frac{2}{3}\times24=16\) students have a pet. 2. In Class B, \(\frac{1}{2}\times30=15\) students have a pet. 3. Therefore, Class A has more students with pets. 4. There are \(24+30=54\) students altogether and \(16+15=31\) students with pets. The fraction is \(\frac{31}{54}\).

a) Class A, with \(16\) students compared with \(15\) in Class B b) \(\frac{31}{54}\)

- How many times do you need to find half of the original amount? - Write down the result of the first step before finding half again. - Include the unit in your answer.

1. Find the forest area: \(\frac{1}{2} \times 960\,\text{acres} = 480\,\text{acres}\). 2. Find half of the forest area: \(\frac{1}{2} \times 480\,\text{acres} = 240\,\text{acres}\).

The protected oak grove covers \(240\,\text{acres}\).

- First find how many marbles each person gives away. - For the same total, is one-third greater or less than one-fourth? - If one person gives away more marbles, will that person have more or fewer left?

1. Lucas gives away \(\frac{1}{4} \times 24 = 6\) marbles, so he has \(24 - 6 = 18\) marbles left. 2. Sarah gives away \(\frac{1}{3} \times 24 = 8\) marbles, so she has \(24 - 8 = 16\) marbles left. 3. Since \(18 > 16\), Lucas has more marbles left.

Lucas has more marbles left. Lucas has \(18\) marbles, and Sarah has \(16\) marbles.

- Divide the total weight into eight equal parts for part a). - Use the weight of one eighth to build several eighths. - For part c), calculate the two fractional weights separately before comparing them.

1. One-eighth of the bar weighs \(\frac{1}{8}\times16\,\text{oz}=2\,\text{oz}\). 2. Three-eighths weigh \(3\times2\,\text{oz}=6\,\text{oz}\). 3. Two-eighths weigh \(2\times2\,\text{oz}=4\,\text{oz}\). One-fourth weighs \(\frac{1}{4}\times16\,\text{oz}=4\,\text{oz}\). 4. The two fractional amounts have the same weight, so Noah is correct.

a) \(2\,\text{oz}\) b) \(6\,\text{oz}\) c) Yes. Both two-eighths and one-fourth weigh \(4\,\text{oz}\).

- How many pears are one-fourth of \(40\)? - After finding the number of pears, how can you find the number of apples? - Is one-eighth of a whole greater or less than one-fourth? - What happens to the number of apples if there are fewer pears?

1. Find the number of pears: \(\frac{1}{4} \times 40 = 10\). 2. Subtract to find the number of apples: \(40 - 10 = 30\). 3. One-eighth is less than one-fourth. With the same total number of fruit, a smaller fraction of pears means a greater number of apples.

a) There are \(30\) apples. b) The number of apples would be greater because one-eighth is less than one-fourth, so there would be fewer pears.

- How many squares are one-fourth of \(16\)? - How many squares are one-eighth of \(16\)? - After accounting for the tulips and daffodils, how many squares remain? - Write the rose squares as a fraction of all \(16\) squares.

1. Find the tulip squares: \(\frac{1}{4} \times 16 = 4\). 2. Find the daffodil squares: \(\frac{1}{8} \times 16 = 2\). 3. Find the number of planted squares already used: \(4 + 2 = 6\). 4. Find the rose squares: \(16 - 6 = 10\). 5. The rose fraction is \(\frac{10}{16}\), which simplifies to \(\frac{5}{8}\).

There are \(10\) squares planted with roses. The roses cover \(\frac{5}{8}\) of the bed (or \(\frac{10}{16}\)).

- First find exactly how much Tim spends. - Then find exactly how much Sarah spends. - Compare the two amounts.

1. Find the amount Tim spends: \(\frac{1}{2} \times \$40 = \$20\). 2. Find the amount Sarah spends: \(\frac{1}{4} \times \$60 = \$15\). 3. Since \(\$20 > \$15\), Tim spends more.

Tim spends more. He spends \(\$20\), while Sarah spends \(\$15\).

- Imagine dividing the rope into equal-length pieces. - What does the denominator tell you about the number of equal parts? - What does the numerator tell you about the number of parts being used? - How are \(\frac{1}{6}\) and \(\frac{5}{6}\) related?

1. Find one-sixth of the rope: \(\frac{1}{6} \times 24\,\text{ft} = 4\,\text{ft}\). 2. Find five-sixths of the rope: \(5 \times 4\,\text{ft} = 20\,\text{ft}\). 3. Compare the lengths: \(20\,\text{ft} \div 4\,\text{ft} = 5\). The shorter piece fits into the longer piece \(5\) times because \(\frac{5}{6}\) is five times \(\frac{1}{6}\).

a) \(4\,\text{ft}\) b) \(20\,\text{ft}\) c) \(5\) times, because \(\frac{5}{6}\) is five times \(\frac{1}{6}\).

- First find the length of one-fourth of the cord. - After finding one-fourth, how can you find three-fourths? - What operation finds the amount left from the original length?

1. Find one-fourth of the cord: \(24\,\text{in} \div 4 = 6\,\text{in}\). 2. Find three-fourths: \(3 \times 6\,\text{in} = 18\,\text{in}\). 3. The sister is correct because \(\frac{3}{4}\) of \(24\,\text{in}\) is \(18\,\text{in}\). 4. Find the remaining length: \(24\,\text{in} - 18\,\text{in} = 6\,\text{in}\).

Yes. \(\frac{3}{4}\) of \(24\,\text{in}\) is \(18\,\text{in}\), and \(6\,\text{in}\) remain.

- How can you find a fraction of a number? - First find how many miles each rider has completed. - How can you find the distance remaining from the total route? - Compare the two remaining distances.

1. Find Leon's completed distance: \(\frac{1}{4} \times 80\,\text{mi} = 20\,\text{mi}\). 2. Find Leon's remaining distance: \(80\,\text{mi} - 20\,\text{mi} = 60\,\text{mi}\). 3. Find Mia's completed distance: \(\frac{3}{8} \times 80\,\text{mi} = 30\,\text{mi}\). 4. Find Mia's remaining distance: \(80\,\text{mi} - 30\,\text{mi} = 50\,\text{mi}\). 5. Since \(60\,\text{mi} > 50\,\text{mi}\), Leon has the longer distance left.

Leon has \(60\,\text{mi}\) left, and Mia has \(50\,\text{mi}\) left. Leon has the longer distance remaining.

- Convert the total goal to meters. - Find the distance of half a lap. - Subtract the distance already completed from the goal.

1. Convert the goal distance: \(2\,\text{km} = 2000\,\text{m}\). 2. Seven full laps cover \(7 \times 200\,\text{m} = 1400\,\text{m}\). 3. Half a lap covers \(\frac{1}{2} \times 200\,\text{m} = 100\,\text{m}\). 4. Sarah has run \(1400\,\text{m} + 100\,\text{m} = 1500\,\text{m}\). 5. She has \(2000\,\text{m} - 1500\,\text{m} = 500\,\text{m}\) left.

Sarah still needs to run \(500\,\text{m}\).

- What fraction of the whole roll is the \(20\,\text{yd}\) piece? - How much does that fraction of the total ribbon weigh? - How can you find the weight of the ribbon that remains?

1. The cut piece is \(\frac{20}{80} = \frac{1}{4}\) of the total length. 2. Find the weight of the cut piece: \(\frac{1}{4} \times 24\,\text{oz} = 6\,\text{oz}\). 3. Subtract the cut piece's weight: \(24\,\text{oz} - 6\,\text{oz} = 18\,\text{oz}\).

The ribbon remaining on the roll weighs \(18\,\text{oz}\).

- Look at the angles in the circle. Which familiar fractions, such as one-half or one-fourth, do you recognize? - After identifying a fraction, use it to find that part of the total weight. - Compare the size of the cherry slice with the pear slice.

1. The apple slice is one-half of the circle, so apples are \(\frac{1}{2}\) of the harvest. 2. The pear slice is one-fourth of the circle, so pears are \(\frac{1}{4}\) of the harvest. 3. Find the pear weight: \(\frac{1}{4} \times 80\,\text{lb} = 20\,\text{lb}\). 4. The cherry slice is one-half the size of the pear slice, so cherries are \(\frac{1}{8}\) of the harvest. 5. Find the cherry weight: \(\frac{1}{8} \times 80\,\text{lb} = 10\,\text{lb}\).

a) \(\frac{1}{2}\) b) \(20\,\text{lb}\) c) \(10\,\text{lb}\)

- For each field, count the shaded dots and the total dots shown. - Compare the two fractions separately from comparing the two shaded counts. - Ask whether the two fractions refer to wholes of the same size.

1. Field A has \(30\) shaded dots out of \(40\), so its shaded fraction is \(\frac{30}{40}=\frac{3}{4}\). 2. Field B has \(40\) shaded dots out of \(60\), so its shaded fraction is \(\frac{40}{60}=\frac{2}{3}\). 3. Compare the fractions: \(\frac{3}{4}=\frac{9}{12}\) and \(\frac{2}{3}=\frac{8}{12}\), so \(\frac{3}{4}>\frac{2}{3}\). 4. Compare the actual shaded counts: Field B has \(40\) shaded dots and Field A has \(30\), so Field B has \(10\) more shaded dots. 5. The fractions describe portions of different-sized wholes. A greater fraction of a smaller whole can still represent fewer objects than a smaller fraction of a larger whole.

a) A: \(\frac{3}{4}\); B: \(\frac{2}{3}\) b) Field A has the greater shaded fraction. c) Field B has \(10\) more shaded dots. d) The fields have different total numbers of dots, so comparing the fractions alone does not compare the actual shaded counts.

- First find how many of the \(84\) groups are not working. - Subtract to find the number of working groups. - Each working group contributes the same number of lights.

1. One-sixth of \(84\) groups is \(\frac{1}{6}\times84=14\) groups, so \(14\) groups are not working. 2. The number of working groups is \(84-14=70\). 3. Each working group has \(5\) lights, so \(70\times5=350\) lights are working.

\(350\) lights

- Express the mixed-number total entirely in fifths. - Compare the numerator contributed by one packet with the total numerator. - Find the whole-number factor that changes \(4\) fifths into \(16\) fifths.

1. Rewrite the total mass: \(3\frac{1}{5}=\frac{16}{5}\) pounds. 2. The equation becomes \(p\times\frac{4}{5}=\frac{16}{5}\). 3. Each group contributes four fifth-sized parts. Since \(4\times4=16\), \(p=4\).

\(p=4\), so there are \(4\) packets.

- Find the total ribbon required by all seven projects. - Express the required and available amounts using the same fractional unit. - Subtract the available amount from the required amount.

1. The projects require \(7\times\frac{3}{8}=\frac{21}{8}=2\frac{5}{8}\) feet. 2. Rewrite the available ribbon as \(2\frac{1}{4}=2\frac{2}{8}\) feet. 3. The shortage is \(2\frac{5}{8}-2\frac{2}{8}=\frac{3}{8}\) foot.

\(\frac{3}{8}\,\text{ft}\)

- Find each design’s total from its number of equal strips. - Compare the two totals before answering part a). - For part b), subtract the smaller total from the larger one.

1. Design A uses \(5\times\frac{2}{3}=\frac{10}{3}=3\frac{1}{3}\) feet. 2. Design B uses \(8\times\frac{3}{8}=\frac{24}{8}=3\) feet. 3. Since \(3\frac{1}{3}>3\), Design A uses more. 4. The difference is \(3\frac{1}{3}-3=\frac{1}{3}\) foot.

a) Design A b) \(\frac{1}{3}\,\text{ft}\)

- Think of the situation as four equal groups of fifth-sized parts. - Ask whether adding more jars changes the size of each fractional part. - Compare Luis's value with the amount in a single jar as a reasonableness check.

1. Four jars represent four groups of \(\frac{3}{5}\) pound. 2. Multiplication by \(4\) combines four groups of fifth-sized parts: \(4\times\frac{3}{5}=\frac{12}{5}\). 3. \(\frac{12}{5}=2\frac{2}{5}\) pounds. 4. Luis created an equivalent fraction for the amount in one jar instead of combining four jars. The denominator stays \(5\) because the size of each fifth does not change.

Luis incorrectly changed the denominator. The correct total is \(2\frac{2}{5}\,\text{lb}\).

- First find the duration of all seven equal parts. - Express three minutes using twelfths for an exact comparison. - The difference tells how far the routine is from three minutes.

1. The total duration is \(7\times\frac{5}{12}=\frac{35}{12}=2\frac{11}{12}\) minutes. 2. Three minutes equals \(\frac{36}{12}\) minutes. 3. The routine is shorter by \(\frac{36}{12}-\frac{35}{12}=\frac{1}{12}\) minute.

The routine is \(\frac{1}{12}\) minute shorter than \(3\) minutes.

- Express the total number of yards in fifths. - Each loop contributes two fifth-sized parts. - Find the whole-number factor that changes \(2\) fifths into \(10\) fifths.

1. Rewrite \(2\) as \(\frac{10}{5}\). 2. The equation becomes \(n\times\frac{2}{5}=\frac{10}{5}\). 3. Each loop contributes two fifth-sized parts. Since \(5\times2=10\), \(n=5\).

\(n=5\), so Mia makes \(5\) loops.

- Determine the total number of eighth-yard parts in all eleven pieces. - Group those eighths into as many complete sets of eight as possible. - Report both the number of whole yards and the leftover fraction.

1. Eleven pieces contain \(11\times\frac{3}{8}=\frac{33}{8}\) yards. 2. Four complete yards use \(\frac{32}{8}\). 3. The remaining length is \(\frac{1}{8}\) yard.

\(4\) complete yards with \(\frac{1}{8}\) yard remaining

- Keep both the number of equal amounts and the size of each amount unchanged. - Make the meaning of the five equal amounts different in the two situations. - In each situation, identify what quantity the product measures.

1. One equal-groups example is five containers holding \(\frac{2}{5}\) liter each; the product represents total volume. 2. One movement example is five jumps of \(\frac{2}{5}\) meter each; the product represents total distance. 3. In either structure, \(5\times\frac{2}{5}=\frac{10}{5}=2\).

One valid pair of situations is: Equal groups: Five containers each hold \(\frac{2}{5}\,\text{L}\). Together they hold \(2\,\text{L}\). Equal moves: Elena makes five jumps of \(\frac{2}{5}\,\text{m}\) each. The total distance is \(2\,\text{m}\).

- Separate the first marker's location from the equal gaps after it. - Count spaces between marker numbers rather than counting the markers themselves. - Add the initial offset only after finding the repeated distance.

1. From marker \(1\) to marker \(5\), there are \(4\) equal spaces. 2. Those spaces total \(4\times\frac{3}{8}=\frac{12}{8}=1\frac{1}{2}\) miles. 3. The starting offset is \(\frac{1}{4}=\frac{2}{8}\) mile, so the total distance is \(\frac{2}{8}+\frac{12}{8}=\frac{14}{8}=1\frac{3}{4}\) miles.

Marker \(5\) is \(1\frac{3}{4}\) miles from the trailhead. The repeated distance is used four times because five markers create four spaces from marker \(1\) to marker \(5\).

- One method combines the two ribbon amounts within each bag before multiplying. - The other method keeps the blue and yellow ribbon in separate equal groups. - Check that both methods account for all five bags and both ribbon colors.

1. One bag uses \(\frac{3}{10}+\frac{1}{10}=\frac{4}{10}\) yard, so \(5\times\frac{4}{10}=2\) yards. 2. Separately, the blue ribbon uses \(5\times\frac{3}{10}=1\frac{1}{2}\) yards and the yellow ribbon uses \(5\times\frac{1}{10}=\frac{1}{2}\) yard. 3. The separate totals also add to \(2\) yards.

Per bag first: \(5\times\left(\frac{3}{10}+\frac{1}{10}\right)=2\) yards Colors separately: \(5\times\frac{3}{10}+5\times\frac{1}{10}=2\) yards

- Remove the fixed-width title panel first. - Represent the six equal picture panels with \(6\times w\). - Find a familiar fraction that makes the unknown-factor equation true, then rebuild the full width to check.

1. The picture panels use \(5-\frac{1}{2}=4\frac{1}{2}\,\text{ft}\). 2. Let \(w\) be the width of one picture panel. Then \(6\times w=4\frac{1}{2}\). 3. Since \(6\times\frac{3}{4}=\frac{18}{4}=4\frac{1}{2}\), each panel is \(\frac{3}{4}\,\text{ft}\) wide. 4. Check: \(6\times\frac{3}{4}+\frac{1}{2}=4\frac{1}{2}+\frac{1}{2}=5\,\text{ft}\).

\(6\times w=4\frac{1}{2}\), so \(w=\frac{3}{4}\,\text{ft}\)

- Use the trays and row positions as two different ways to group the same tiles. - Find a subtotal for one chosen group before using the number of such groups. - Both methods must include all \(12\) tiles exactly once.

1. One tray weighs \(3\times\frac{1}{6}=\frac{1}{2}\,\text{lb}\), so four trays weigh \(4\times\frac{1}{2}=2\,\text{lb}\). 2. For each of the three row positions, the four corresponding tiles weigh \(4\times\frac{1}{6}=\frac{2}{3}\,\text{lb}\). Three such groups weigh \(3\times\frac{2}{3}=2\,\text{lb}\).

Grouping by trays: \(4\times\left(3\times\frac{1}{6}\right)=2\,\text{lb}\) Grouping by row position: \(3\times\left(4\times\frac{1}{6}\right)=2\,\text{lb}\)

- Do not multiply by the originally planned number of sections. - Determine how many equal sections remain after the closure. - Convert the resulting improper fraction to a mixed number.

1. The number of active sections is \(9-2=7\). 2. Their total length is \(7\times\frac{5}{12}=\frac{35}{12}=2\frac{11}{12}\) miles.

\(2\frac{11}{12}\) miles

- Follow the stated decomposition of the number of bookmarks. - Find the ribbon subtotal for each part before recombining them. - Check that the partial group counts include all \(13\) bookmarks.

1. Ten bookmarks use \(10\times\frac{1}{4}=2\frac{1}{2}\,\text{yd}\) of ribbon. 2. Three bookmarks use \(3\times\frac{1}{4}=\frac{3}{4}\,\text{yd}\) of ribbon. 3. Rename \(2\frac{1}{2}=2\frac{2}{4}\), so the total is \(2\frac{2}{4}+\frac{3}{4}=3\frac{1}{4}\,\text{yd}\). 4. The split is convenient because multiplying one fourth by \(10\) and by \(3\) uses familiar facts and small partial products.

\(10\times\frac{1}{4}=2\frac{1}{2}\,\text{yd}\) and \(3\times\frac{1}{4}=\frac{3}{4}\,\text{yd}\), for a total of \(3\frac{1}{4}\,\text{yd}\). The split uses familiar, easy partial products.

- Treat drill time and pause time as two separate repeated quantities. - Count the spaces between consecutive drills. - Add the two subtotals only after each has the correct group count.

1. The drills take \(6\times\frac{5}{6}=5\) minutes. 2. Six drills have \(5\) spaces between them, so pauses take \(5\times\frac{1}{12}=\frac{5}{12}\) minute. 3. The complete time is \(5\frac{5}{12}\) minutes.

The complete session lasts \(5\frac{5}{12}\) minutes. The pause length is multiplied by \(5\) because six drills have five spaces between neighboring drills.

- Find the fraction of the full pack used by each team. - Compare the two resulting sheet counts. - The question asks for both which team used more and the difference.

1. Team A uses \(\frac{5}{8}\times48=30\) sheets. 2. Team B uses \(\frac{2}{3}\times48=32\) sheets. 3. Team B uses \(32-30=2\) more sheets.

Team B uses \(2\) more sheets.

- Find the total amount placed in each type of container. - Add the amounts removed. - Subtract from the original amount.

1. The small containers use \(6\times\frac{3}{8}=\frac{18}{8}=2\frac{1}{4}\) gallons. 2. The larger containers use \(4\times\frac{3}{4}=3\) gallons. 3. The total removed is \(2\frac{1}{4}+3=5\frac{1}{4}\) gallons. 4. The amount remaining is \(10-5\frac{1}{4}=4\frac{3}{4}\) gallons.

\(4\frac{3}{4}\) gallons remain.

- Find the total sold in each cup size. - Add those amounts. - Determine how many \(5\)-quart containers are needed to reach or exceed the total.

1. Standard cups use \(15\times\frac{1}{3}=5\) quarts. 2. Large cups use \(12\times\frac{1}{2}=6\) quarts. 3. The total sold is \(5+6=11\) quarts. 4. Two containers hold only \(10\) quarts, so \(3\) containers must be opened. 5. Three containers hold \(15\) quarts. The amount remaining is \(15-11=4\) quarts.

a) \(11\) quarts b) \(3\) containers c) \(4\) quarts

- Find Mia’s weight first. - Use Mia’s weight to find the father’s weight. - Add all three people and the gear, then compare with the limit.

1. Mia weighs \(\frac{4}{5}\times120=96\,\text{lb}\). 2. Their father weighs \(2\times96=192\,\text{lb}\). 3. The total load is \(120+90+96+192=498\,\text{lb}\). 4. Since \(498\le550\), the load is within the limit.

Yes. The total load is \(498\,\text{lb}\), which is below the \(550\,\text{lb}\) limit.

- Find the flower area and subtract it from the total. - The vegetable fraction applies to the remaining area, not the original total. - Subtract the vegetable area from the remaining area.

1. The flower area is \(\frac{1}{4}\times6000=1500\,\text{ft}^2\). 2. The remaining area is \(6000-1500=4500\,\text{ft}^2\). 3. The vegetable area is \(\frac{2}{3}\times4500=3000\,\text{ft}^2\). 4. The lawn area is \(4500-3000=1500\,\text{ft}^2\).

The lawn has an area of \(1500\,\text{ft}^2\).

- Put both length measurements in the same unit before finding the fraction cut off. - Use the stated proportional model to connect the fraction of length with the fraction of weight. - Keep the weight in one unit while finding and subtracting a fraction of it.

1. Convert the full length: \(16\,\text{ft} = 192\,\text{in}\). 2. The cut piece is \(\frac{24}{192} = \frac{1}{8}\) of the beam's total length. 3. Because weight is proportional to length in the stated model, the cut piece has \(\frac{1}{8}\) of the beam's weight. 4. Convert the beam's weight: \(46\,\text{lb} = 736\,\text{oz}\). 5. The cut piece weighs \(\frac{1}{8} \times 736\,\text{oz} = 92\,\text{oz}\), leaving \(736\,\text{oz} - 92\,\text{oz} = 644\,\text{oz}\). 6. Convert: \(644\,\text{oz} = 40\,\text{lb}\ 4\,\text{oz}\).

a) \(\frac{1}{8}\) b) \(40\,\text{lb}\ 4\,\text{oz}\)

- Find the length before any pieces overlap. - Count joints between strips rather than counting strips again. - Subtract the total hidden overlap from the original total.

1. Before overlapping, the strips total \(7\times\frac{3}{8}=\frac{21}{8}\,\text{yd}\). 2. Seven strips in one line create \(6\) joints, so the total overlap is \(6\times\frac{1}{8}=\frac{6}{8}\,\text{yd}\). 3. The displayed length is \(\frac{21}{8}-\frac{6}{8}=\frac{15}{8}=1\frac{7}{8}\,\text{yd}\).

The displayed length is \(1\frac{7}{8}\,\text{yd}\). There are six overlaps because joining seven strips in one line creates six joints.