Free math worksheets

- Can you write the starting and ending charge with the same denominator? - Which operation finds the increase from the starting charge to the ending charge? - Simplify your final fraction.

1. The starting charge is \(\frac{1}{5}\), and the ending charge is \(\frac{13}{15}\). 2. Rewrite \(\frac{1}{5}\) as \(\frac{3}{15}\). 3. Find the increase: \(\frac{13}{15} - \frac{3}{15} = \frac{10}{15} = \frac{2}{3}\).

\(\frac{2}{3}\) of the battery's total capacity was added.

- Rewrite the fractions with a common denominator before adding. - Compare the total with one whole cup. - An improper fraction can be rewritten as a mixed number.

1. Use denominator \(8\): \(\frac{1}{4}=\frac{2}{8}\) and \(\frac{1}{2}=\frac{4}{8}\). 2. Add the amounts: \(\frac{3}{8}+\frac{2}{8}+\frac{4}{8}=\frac{9}{8}=1 \frac{1}{8}\) cups. 3. Since \(1 \frac{1}{8}>1\), the drink will not fit in a \(1\)-cup container.

Mia makes \(1 \frac{1}{8}\) cups. No, it will not all fit in a \(1\)-cup container.

- Read one factor from the vertical partition and the other from the horizontal partition. - The overlap is the part that satisfies both fractional conditions. - Compare the number of overlap cells with the total number of equal cells in the whole grid.

1. The vertical region covers \(5\) of \(8\) equal columns, so \(\frac{5}{8}\) of the animals are dogs. 2. The horizontal region covers \(2\) of \(5\) equal rows, so the model takes \(\frac{2}{5}\) of the dog region. 3. The overlap covers \(10\) of the \(40\) equal cells, so \(\frac{2}{5}\times\frac{5}{8}=\frac{10}{40}=\frac{1}{4}\).

\(\frac{2}{5}\times\frac{5}{8}=\frac{1}{4}\). Puppies make up \(\frac{1}{4}\) of all the animals.

- Both bars represent fractions of the same full tank. - Rewrite fifteenths and twentieths as equal-sized parts. - Subtract the amount used from the amount initially in the tank.

1. Subtract the fraction used from the fraction initially in the tank: \(\frac{13}{15}-\frac{7}{20}\). 2. The least common denominator is \(60\). 3. Rewrite: \(\frac{13}{15}=\frac{52}{60}\) and \(\frac{7}{20}=\frac{21}{60}\). 4. Subtract: \(\frac{52}{60}-\frac{21}{60}=\frac{31}{60}\).

\(\frac{31}{60}\) of the full tank remains.

- Find the complement of the first fraction used. - Find the complement of the fraction used from the remainder. - The final amount is a fraction of a fraction of the original whole.

1. After the first use, \(1-\frac{4}{9}=\frac{5}{9}\) of the original sugar remains. 2. After the second use, \(1-\frac{7}{10}=\frac{3}{10}\) of that remainder is left. 3. Therefore the final fraction of the original amount is \(\frac{3}{10}\times\frac{5}{9}=\frac{15}{90}=\frac{1}{6}\).

After the first use, \(\frac{5}{9}\) remains. Keeping \(\frac{3}{10}\) of that gives \(\frac{3}{10}\times\frac{5}{9}=\frac{1}{6}\) of the original amount.

- A whole is eight times a part that represents \(\frac{1}{8}\). - When the whole is known, divide it by \(8\) to find one eighth. - Convert units before or after calculating, as needed.

1. For a), the whole is \(8\times125\,\text{mL}=1000\,\text{mL}=1\,\text{L}\), so \(x=1\). 2. For b), \(2\,\text{m}=200\,\text{cm}\). One eighth of \(200\,\text{cm}\) is \(200\div8=25\,\text{cm}\), so \(x=25\). 3. For c), the whole is \(8\times45\,\text{s}=360\,\text{s}=6\,\text{min}\), so \(x=6\).

a) \(x=1\) b) \(x=25\) c) \(x=6\)

- The whole is the combined mass of the sand and gravel. - Convert both masses to the same unit before adding. - Write the sand mass over the total mass and simplify.

1. Convert the gravel mass: \(1.05\,\text{kg}=1050\,\text{g}\). 2. Find the total mass: \(350\,\text{g}+1050\,\text{g}=1400\,\text{g}\). 3. Write the sand mass over the total mass: \(\frac{350}{1400}\). 4. Simplify: \(\frac{350}{1400}=\frac{1}{4}\).

Sand makes up \(\frac{1}{4}\) of the mixture's total mass.

- Rewrite the fractional amounts with a common denominator. - Find how many cups were added altogether. - Subtract the amount added from the bowl's total capacity.

1. Add the amounts. Using denominator \(8\), \(2\frac{1}{4}=\frac{18}{8}\), \(1\frac{3}{8}=\frac{11}{8}\), \(3\frac{1}{2}=\frac{28}{8}\), and \(1\frac{3}{4}=\frac{14}{8}\). 2. The total added is \(\frac{18+11+28+14}{8}=\frac{71}{8}=8\frac{7}{8}\) cups. 3. Subtract from the capacity: \(12-8\frac{7}{8}=3\frac{1}{8}\) cups.

\(3\frac{1}{8}\) cups of space are left.

- Include both pieces of the repeated length. - Add the mixed numbers using a common denominator. - Multiply the total number of yards by the price per yard.

1. Find the total length: \(2\times2\frac{1}{2}+1\frac{1}{4}+2\frac{1}{5}=5+1\frac{1}{4}+2\frac{1}{5}=8\frac{9}{20}\,\text{yd}\). 2. Convert to a decimal: \(8\frac{9}{20}=8.45\). 3. Multiply by the price per yard: \(8.45\times\$2.00=\$16.90\).

The total cost is \(\$16.90\).

- Add the lengths of the pieces removed. - Subtract that sum from the original length. - Multiply the remaining length by the price per foot.

1. Add the lengths removed: \(1\frac{1}{3}+\frac{5}{6}=\frac{4}{3}+\frac{5}{6}=\frac{13}{6}=2\frac{1}{6}\,\text{ft}\). 2. Find the remaining length: \(4-2\frac{1}{6}=1\frac{5}{6}=\frac{11}{6}\,\text{ft}\). 3. Find its value: \(\frac{11}{6}\times\$6.00=\$11.00\).

The remaining piece is worth \(\$11.00\).

- What fraction represents the whole fence? - First find the total fraction painted during the first three days. - Subtract the painted part from one whole.

1. Use denominator \(20\): \(\frac{1}{5}=\frac{4}{20}\), \(\frac{2}{10}=\frac{4}{20}\), and \(\frac{1}{4}=\frac{5}{20}\). 2. The group has painted \(\frac{4}{20}+\frac{4}{20}+\frac{5}{20}=\frac{13}{20}\) of the fence. 3. Subtract from one whole: \(1-\frac{13}{20}=\frac{20}{20}-\frac{13}{20}=\frac{7}{20}\).

\(\frac{7}{20}\) of the fence remains to be painted.

- Use the stated value of a quarter note before applying the dot rule. - Add the values of the two known notes. - Subtract their total from the value of a full measure.

1. A dotted quarter note has value \(\frac{1}{4}+\frac{1}{2}\times\frac{1}{4}=\frac{3}{8}\). 2. Two dotted quarter notes have total value \(2\times\frac{3}{8}=\frac{6}{8}\). 3. The full measure has value \(\frac{4}{4}=\frac{8}{8}\). 4. The missing value is \(\frac{8}{8}-\frac{6}{8}=\frac{2}{8}=\frac{1}{4}\), which is an undotted quarter note.

The third note must be an undotted quarter note with value \(\frac{1}{4}\).

- Use the stated base fraction for each note before applying the dot rule. - Compare repeated copies of the resulting fractional durations. - Rewrite the dotted half note's value in eighths for part b).

1. A dotted quarter note has value \(\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\). Two have value \(2\times\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\). 2. Three undotted quarter notes have value \(3\times\frac{1}{4}=\frac{3}{4}\), so the durations are equal. 3. A dotted half note has value \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}=\frac{6}{8}\). Therefore, it has the duration of six eighth notes.

a) Both durations equal \(\frac{3}{4}\). b) \(6\) eighth notes

- 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) \(4\,\text{qt}\)

- The long jumpers are a fraction of a group that is itself a fraction of all students. - For one method, combine the two fractions before using the total number of students. - For the second method, find the track-and-field group first and then take half of it.

1. The long jumpers are \(\frac{1}{2}\) of the \(\frac{2}{5}\) who choose track and field, so \(\frac{1}{2}\times\frac{2}{5}=\frac{1}{5}\) of all students are long jumpers. 2. One method is to find \(\frac{1}{5}\) of \(60\), which is \(12\). 3. Another method is to find \(\frac{2}{5}\) of \(60\), which is \(24\), and then take half of \(24\), which is \(12\).

a) \(\frac{1}{2}\times\frac{2}{5}=\frac{1}{5}\) of all students b) \(12\) students. Either find \(\frac{1}{5}\) of \(60\), or find \(\frac{2}{5}\) of \(60\) and then take one half.

- 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) \(\frac{31}{54}\)

- Find the total fraction distributed under each plan separately. - Rewrite the fractions with common denominators before adding. - Finally, rewrite the two totals with a common denominator so you can compare them. - The plan with the greater total gives away the larger share of the treasure.

1. For Plan A, use denominator \(20\): \(\frac{1}{2}+\frac{1}{4}+\frac{1}{10}=\frac{10}{20}+\frac{5}{20}+\frac{2}{20}=\frac{17}{20}\). 2. For Plan B, use denominator \(12\): \(\frac{1}{3}+\frac{1}{6}+\frac{1}{4}=\frac{4}{12}+\frac{2}{12}+\frac{3}{12}=\frac{9}{12}=\frac{3}{4}\). 3. Rewrite \(\frac{3}{4}\) as \(\frac{15}{20}\). Since \(\frac{17}{20}>\frac{15}{20}\), Plan A distributes the greater fraction.

Plan A distributes more: \(\frac{17}{20}\) of the treasure, compared with \(\frac{3}{4}=\frac{15}{20}\) for Plan B.

- Read one factor from the vertical partition and the other from the horizontal partition in each model. - The overlap represents the fraction that satisfies both conditions. - Simplify both overlap fractions before comparing them.

1. In Model A, the vertical region is \(\frac{4}{5}\) of the pitcher and the horizontal region is \(\frac{3}{8}\) of that amount. The overlap is \(12\) of \(40\) cells, so \(\frac{3}{8}\times\frac{4}{5}=\frac{12}{40}=\frac{3}{10}\). 2. In Model B, the vertical region is \(\frac{3}{5}\) of the pitcher and the horizontal region is \(\frac{1}{2}\) of that amount. The overlap is \(3\) of \(10\) cells, so \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\). 3. Both overlaps represent \(\frac{3}{10}\) of a full pitcher, so the amounts are equal.

a) Model A: \(\frac{3}{8}\times\frac{4}{5}=\frac{3}{10}\) Model B: \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\) b) The amounts are equal.

- In part a), interpret “of” as multiplication. - In part b), multiply the amount in one jug by the number of jugs. - Keep the unit with each final answer.

1. For a), rewrite \(1 \frac{1}{2}=\frac{3}{2}\). Then \(\frac{2}{3}\times\frac{3}{2}=1\), so the amount is \(1\) pound. 2. For b), multiply the capacity of one jug by \(5\): \(5\times\frac{3}{4}=\frac{15}{4}=3 \frac{3}{4}\). The jugs hold \(3 \frac{3}{4}\) gallons altogether.

a) \(1\) pound b) \(3 \frac{3}{4}\) gallons

- Use the colored regions in the grid; the row and column counts are not stated in the text. - First compare the non-pond columns with all columns. - Then compare the grass rows within the non-pond region before writing the fraction-of-a-fraction product.

1. The pond occupies one of three equal columns, so \(\frac{2}{3}\) of the garden remains. 2. Within the two remaining columns, grass occupies two of five equal rows, so grass is \(\frac{2}{5}\) of the remaining region. 3. Therefore \(\frac{2}{5}\times\frac{2}{3}=\frac{4}{15}\). The four green cells out of fifteen total cells confirm the product.

a) \(\frac{2}{3}\) b) \(\frac{2}{5}\) c) \(\frac{2}{5}\times\frac{2}{3}=\frac{4}{15}\)

- How many eighths are equal to one-half? - What fraction of the punch is already accounted for by the berry concentrate and water? - What fraction remains for the apple juice, and how can that help you find one-eighth? - The whole mixture contains eight equal eighths.

1. Write one-half in eighths: \(\frac{1}{2} = \frac{4}{8}\). 2. Add the known fractions: \(\frac{1}{8} + \frac{4}{8} = \frac{5}{8}\). 3. Find the fraction that is apple juice: \(1 - \frac{5}{8} = \frac{3}{8}\). 4. Since \(\frac{3}{8}\) of the punch is \(12\,\text{fl oz}\), one-eighth is \(12\,\text{fl oz} \div 3 = 4\,\text{fl oz}\). 5. Eight eighths equal \(8 \times 4\,\text{fl oz} = 32\,\text{fl oz}\).

The pitcher contains \(32\,\text{fl oz}\) of punch.

- Add the distances already traveled. - The result represents three equal parts of the whole. - Divide by \(3\) to find one part, then multiply by \(4\).

1. Add the completed sections: \(14.6+7.9=22.5\,\text{mi}\). 2. The \(22.5\,\text{mi}\) represents \(\frac{3}{4}\) of the route. One-fourth is \(22.5\div 3=7.5\,\text{mi}\). 3. The whole route is \(7.5\times 4=30\,\text{mi}\).

The bicycle route is \(30\,\text{mi}\) long.

- Identify the fraction of the capacity that was filled at the start. - Use a common denominator before subtracting the fractions. - The fraction drained is based on the aquarium's total capacity. - Substitute the given value of \(V\) after finding the remaining fraction.

1. Subtract the drained fraction from the starting fraction: \(\frac{4}{5}-\frac{1}{4}=\frac{16}{20}-\frac{5}{20}=\frac{11}{20}\). 2. For \(V=80\), calculate \(80\times\frac{11}{20}=44\).

a) \(\frac{11}{20}\) b) \(44\) gallons

- Use the image to compare the number of shaded equal plots with the total number of equal plots. - Express the shaded area as a fraction of an unknown whole area. - Solve the fraction equation by reasoning about equal fifths of the whole.

1. The diagram shows \(6\) shaded plots out of \(15\), so the shaded fraction is \(\frac{6}{15}=\frac{2}{5}\). 2. The shaded area is \(\frac{2}{5}\) of the whole garden, so \(\frac{2}{5}\times A=12\). 3. Five equal fifths together have area \(12\times\frac{5}{2}=30\,\text{m}^2\), so \(A=30\,\text{m}^2\).

a) \(\frac{2}{5}\) b) \(\frac{2}{5}\times A=12\), so \(A=30\,\text{m}^2\).

- Count the remaining equal pieces and all equal pieces to write a fraction of the whole. - Use that fraction as a multiplier of the unknown original mass. - Reason from seven twelfths to one twelfth, then to twelve twelfths.

1. The model has \(24\) equal pieces, with \(14\) remaining, so the remaining fraction is \(\frac{14}{24}=\frac{7}{12}\). 2. Therefore \(\frac{7}{12}\times M=105\). 3. One twelfth of the whole has mass \(105\div7=15\,\text{g}\). Twelve twelfths have mass \(12\times15=180\,\text{g}\).

a) \(\frac{7}{12}\) remains. b) \(\frac{7}{12}\times M=105\), so \(M=180\,\text{g}\).

- 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.

- Subtract the cut length from the original board length. - Rewrite the fractional parts using equal-sized pieces. - Simplify the fractional part and keep the length unit.

1. Subtract the cut length from the original length: \(4\frac{5}{6}-2\frac{3}{10}\). 2. Use denominator \(30\): \(\frac{5}{6}=\frac{25}{30}\) and \(\frac{3}{10}=\frac{9}{30}\). 3. Subtract: \(4\frac{25}{30}-2\frac{9}{30}=2\frac{16}{30}=2\frac{8}{15}\).

\(2\frac{8}{15}\,\text{ft}\) of the board remains.

- Part a) explicitly asks for the fraction of the same whole book, so combine the two reading fractions first. - Subtract the fraction read from one whole. - Only after finding the unread fraction should you apply it to the page count.

1. Add the fractions read: \(\frac{3}{8}+\frac{5}{12}=\frac{9}{24}+\frac{10}{24}=\frac{19}{24}\). 2. Subtract from one whole: \(1-\frac{19}{24}=\frac{5}{24}\), so \(\frac{5}{24}\) of the book is unread. 3. Find \(\frac{5}{24}\) of \(96\): \(96\div24=4\), then \(5\times4=20\).

a) \(\frac{5}{24}\) of the book is unread. b) \(20\) pages are unread.

- First combine the two fractions that refer to the same whole novel. - Then compare the total read with the target fraction. - If the target is larger, subtract the amount read from the target.

1. Add the portions read: \(\frac{2}{7}+\frac{3}{10}=\frac{20}{70}+\frac{21}{70}=\frac{41}{70}\). 2. Compare with the goal: \(\frac{3}{4}=\frac{105}{140}\) and \(\frac{41}{70}=\frac{82}{140}\). 3. The shortfall is \(\frac{105}{140}-\frac{82}{140}=\frac{23}{140}\). 4. Since \(\frac{41}{70}<\frac{3}{4}\), Eli does not meet the goal.

No. Eli is \(\frac{23}{140}\) of the novel short of his goal.

- The distance between two positions on the same whole can be found by subtracting their fractional locations. - Use a common denominator for the two checkpoint fractions. - Compare the resulting distance with an equivalent form of \(\frac{1}{4}\).

1. Subtract the checkpoint positions: \(\frac{7}{10}-\frac{5}{12}\). 2. Use denominator \(60\): \(\frac{7}{10}=\frac{42}{60}\) and \(\frac{5}{12}=\frac{25}{60}\). 3. The distance between them is \(\frac{42}{60}-\frac{25}{60}=\frac{17}{60}\). 4. Since \(\frac{1}{4}=\frac{15}{60}\), the distance \(\frac{17}{60}\) is greater than \(\frac{1}{4}\).

The checkpoints are \(\frac{17}{60}\) of the course apart, which is greater than \(\frac{1}{4}\).

- Both fractions refer to the same total number of seats. - First find how much of the original empty portion remains after more seats are filled. - The occupied fraction is the part of one whole not left empty.

1. The newly filled seats reduce the empty fraction: \(\frac{2}{9}-\frac{1}{6}\). 2. Use denominator \(18\): \(\frac{2}{9}=\frac{4}{18}\) and \(\frac{1}{6}=\frac{3}{18}\). 3. The empty fraction is \(\frac{4}{18}-\frac{3}{18}=\frac{1}{18}\). 4. The occupied fraction is \(1-\frac{1}{18}=\frac{17}{18}\).

\(\frac{1}{18}\) of the seats are empty and \(\frac{17}{18}\) are occupied.

- The midpoint between \(\frac{1}{2}\) and \(1\) is \(\frac{3}{4}\). - Compare \(\frac{2}{5}\) separately with \(\frac{1}{4}\) and \(\frac{1}{2}\); do not compute the exact sum yet. - Use those inequalities to bracket the total before doing exact addition.

1. Since \(\frac{2}{5}>\frac{1}{4}\), the total is greater than \(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\). 2. Since \(\frac{2}{5}<\frac{1}{2}\), the total is less than \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\). Therefore the total lies between \(\frac{1}{2}\) and the midpoint \(\frac{3}{4}\), so it is closer to \(\frac{1}{2}\) than to \(1\). 3. Exactly, \(\frac{2}{5}+\frac{1}{4}=\frac{8}{20}+\frac{5}{20}=\frac{13}{20}\).

a) \(\frac{1}{2}<\frac{2}{5}+\frac{1}{4}<\frac{3}{4}\), so the total is closer to \(\frac{1}{2}\) cup. b) The exact total is \(\frac{13}{20}\) cup.

- Compare \(\frac{3}{8}\) and \(\frac{2}{5}\) without adding them. - Replace both addends by the smaller one for a lower bound and by the larger one for an upper bound. - Only after the bound is established should you use a common denominator for the exact result.

1. Since \(\frac{2}{5}>\frac{3}{8}\), the total is greater than \(\frac{3}{8}+\frac{3}{8}=\frac{3}{4}\). 2. Since \(\frac{3}{8}<\frac{2}{5}\), the total is less than \(\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\). Thus \(\frac{3}{4}<\text{total}<\frac{4}{5}\), so \(\frac{4}{5}\) is a reasonable nearby upper estimate. 3. Exactly, \(\frac{3}{8}+\frac{2}{5}=\frac{15}{40}+\frac{16}{40}=\frac{31}{40}\). 4. The fraction remaining is \(1-\frac{31}{40}=\frac{9}{40}\).

a) \(\frac{3}{4}<\frac{3}{8}+\frac{2}{5}<\frac{4}{5}\), so Devin's estimate is reasonable. b) Exactly \(\frac{31}{40}\) was used and \(\frac{9}{40}\) remains.

- Bound each used length by nearby quarters, halves, or whole numbers before adding exactly. - Use those bounds to create an interval for the total amount used, then subtract the interval endpoints from \(6\). - Do the exact common-denominator calculation only after the reasonableness interval is established.

1. \(2\frac{1}{4}<2\frac{3}{8}<2\frac{1}{2}\), and \(1\frac{3}{4}<1\frac{5}{6}<2\). 2. Therefore the amount used is greater than \(2\frac{1}{4}+1\frac{3}{4}=4\) yards and less than \(2\frac{1}{2}+2=4\frac{1}{2}\) yards. 3. Subtracting those bounds from \(6\) shows the remainder is greater than \(1\frac{1}{2}\) and less than \(2\) yards, so “about \(2\) yards” is reasonable. 4. Exactly, \(2\frac{3}{8}+1\frac{5}{6}=\frac{101}{24}=4\frac{5}{24}\). Thus \(6-4\frac{5}{24}=1\frac{19}{24}\) yards remain.

a) The benchmark bounds give \(1\frac{1}{2}<\text{remainder}<2\), so Priya's estimate is reasonable. b) Exactly \(1\frac{19}{24}\) yards remain.

- Use the value of \(\frac{3}{8}\) to find the value of \(\frac{1}{8}\). - Once you know one eighth, find all eight eighths. - After spending three eighths, determine how many eighths remain.

1. Since \(\frac{3}{8}\) of the total is \(\$45\), one eighth is \(\$45\div3=\$15\). 2. The whole amount is \(8\times\$15=\$120\). 3. The amount left is \(\$120-\$45=\$75\). Equivalently, \(\frac{5}{8}\) remains, and \(5\times\$15=\$75\).

a) \(\$120\) b) \(\$75\)

- 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 the fraction remaining after the first pour. - Interpret \(\frac{2}{3}\) of the remainder as multiplication. - Use the final fraction and \(150\,\text{mL}\) to find the whole.

1. After the first pour, \(1-\frac{1}{4}=\frac{3}{4}\) of the original amount remains. 2. The pitcher receives \(\frac{2}{3}\times\frac{3}{4}=\frac{1}{2}\) of the original amount. 3. The bottle keeps \(\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\) of the original amount. 4. If \(\frac{1}{4}\) is \(150\,\text{mL}\), the whole amount is \(150\times4=600\,\text{mL}\).

\(600\,\text{mL}\)

- Find the fraction remaining after each use. - Half of a remainder means multiply that remainder by \(\frac{1}{2}\). - Use the remaining fraction and \(21\,\text{L}\) to find the whole.

1. After watering the flower beds, \(1-\frac{3}{10}=\frac{7}{10}\) of the original amount remains. 2. Half of that remainder is used, so the fraction still in the barrel is \(\frac{1}{2}\times\frac{7}{10}=\frac{7}{20}\). 3. Let \(V\) be the original volume. Then \(\frac{7}{20}V=21\), so \(V=21\times\frac{20}{7}=60\,\text{L}\). 4. Since \(1\,\text{L}=1000\,\text{mL}\), \(60\,\text{L}=60{,}000\,\text{mL}\).

\(60\,\text{L}\), or \(60{,}000\,\text{mL}\)

- Find the fraction remaining after each wall. - Multiply to find a fraction of a remainder. - Convert the final answer from milliliters to liters.

1. After the first wall, \(1-\frac{2}{9}=\frac{7}{9}\) of the original amount remains. 2. The second wall uses \(\frac{3}{7}\times\frac{7}{9}=\frac{1}{3}\) of the original amount. 3. The remaining fraction is \(\frac{7}{9}-\frac{1}{3}=\frac{4}{9}\). 4. If \(\frac{4}{9}\) is \(1200\,\text{mL}\), then \(\frac{1}{9}\) is \(300\,\text{mL}\), so the whole amount is \(2700\,\text{mL}=2.7\,\text{L}\).

\(2.7\,\text{L}\)

- 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 \(1500\,\text{ft}^2\).

- Read the unshaded fraction directly from the equal-part bar. - The stated \(6\)-mile difference is exactly the unshaded part. - Represent that part as a fraction of the unknown whole before finding the total length.

1. The bar is divided into \(4\) equal parts, and one part is unshaded, so the sunny-meadow section is \(\frac{1}{4}\) of the trail. 2. The difference between the whole trail and the forest section is the unshaded section, so that section is \(6\) miles long. 3. Thus \(\frac{1}{4}\times L=6\). Four fourths have length \(4\times6=24\) miles, so \(L=24\) miles.

a) \(\frac{1}{4}\) b) \(\frac{1}{4}\times L=6\), so \(L=24\) miles.

- First replace the mixed numbers with nearby benchmark amounts that are easy to combine mentally. - For the exact calculation, use one common denominator for all three fractional parts. - Compare the exact remainder with \(2\) gallons only after completing the calculation.

1. For a benchmark check, the used amounts are about \(3\), \(1\frac{1}{2}\), and \(1\frac{3}{4}\) gallons. Their total is a little more than \(6\) gallons, so less than \(2\) gallons remaining is plausible. 2. Rename the fractional parts in twelfths: \(2\frac{5}{6}=2\frac{10}{12}\), \(1\frac{7}{12}\), and \(1\frac{3}{4}=1\frac{9}{12}\). 3. Add: \(2\frac{10}{12}+1\frac{7}{12}+1\frac{9}{12}=4+\frac{26}{12}=6\frac{1}{6}\). 4. Subtract from the total: \(8-6\frac{1}{6}=1\frac{5}{6}\). 5. Since \(1\frac{5}{6}<2\), Asha's claim is correct.

Asha's claim is reasonable and correct. Exactly \(1\frac{5}{6}\) gallons remain.