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Fraction word problems

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5106135
A phone battery is \(\frac{1}{5}\) charged in the morning. During a train ride, it is charged until it is \(\frac{13}{15}\) full. What fraction of the battery's total capacity was added during the ride?

Hints

- 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.

Solution

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: \(\frac{13}{15} - \frac{3}{15} = \frac{10}{15} = \frac{2}{3}\).

Answer

\(\frac{2}{3}\) of the battery's total capacity was added.
5107035
Mia mixes \(\frac{3}{8}\) cup of apple juice, \(\frac{1}{4}\) cup of cranberry juice, and \(\frac{1}{2}\) cup of sparkling water. How many cups of drink does she make altogether? Will it all fit in a \(1\)-cup container? Explain.

Hints

- 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.

Solution

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.

Answer

Mia makes \(1 \frac{1}{8}\) cups. No, it will not all fit in a \(1\)-cup container.
5113955
At an animal shelter, \(\frac{5}{8}\) of the animals are dogs. Of those dogs, \(\frac{2}{5}\) are puppies. What fraction of all the animals at the shelter are puppies?

Hints

- Think about what it means to find a fraction of another fraction. - Which operation represents “\(\frac{2}{5}\) of \(\frac{5}{8}\)”? - Simplify your product at the end.

Solution

1. The puppies are \(\frac{2}{5}\) of the \(\frac{5}{8}\) of the animals that are dogs. 2. Multiply the fractions: \(\frac{2}{5}\times\frac{5}{8}=\frac{10}{40}=\frac{1}{4}\). 3. Therefore, puppies make up \(\frac{1}{4}\) of all the animals.

Answer

\(\frac{1}{4}\) of all the animals are puppies.
5102875
Find the missing number \(x\) so that the part is exactly \(\frac{1}{8}\) of the whole. a) \(125\,\text{mL}\) is \(\frac{1}{8}\) of \(x\,\text{L}\). b) \(x\,\text{cm}\) is \(\frac{1}{8}\) of \(2\,\text{m}\). c) \(45\,\text{s}\) is \(\frac{1}{8}\) of \(x\,\text{min}\).

Hints

- 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.

Solution

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 \(x=6\).

Answer

a) \(x=1\) b) \(x=25\) c) \(x=6\)
5102885
A soil mixture contains \(350\,\text{g}\) of sand and \(1.05\,\text{kg}\) of gravel. What fraction of the mixture's total mass is sand? Write the fraction in simplest form.

Hints

- 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.

Solution

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: \(\frac{350}{1400}=\frac{1}{4}\).

Answer

Sand makes up \(\frac{1}{4}\) of the mixture's total mass.
5106345
A punch bowl holds \(12\) cups. A student adds \(2 \frac{1}{4}\) cups of apple juice, \(1 \frac{3}{8}\) cups of cranberry juice, \(3 \frac{1}{2}\) cups of sparkling water, and \(1 \frac{3}{4}\) cups of orange juice. How many cups of space are left in the bowl?

Hints

- 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.

Solution

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.

Answer

\(3 \frac{1}{8}\) cups of space are left.
5106405
Ms. Meyer buys ribbon for wrapping gifts: two pieces that are each \(2\frac{1}{2}\,\text{yd}\), one piece that is \(1\frac{1}{4}\,\text{yd}\), and one piece that is \(2\frac{1}{5}\,\text{yd}\). The ribbon costs \(\$2.00\) per yard. What is the total cost?

Hints

- 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.

Solution

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: \(8.45\times\$2.00=\$16.90\).

Answer

The total cost is \(\$16.90\).
5106415
A carpenter has a \(4\,\text{ft}\) board. The carpenter cuts off pieces measuring \(1\frac{1}{3}\,\text{ft}\) and \(\frac{5}{6}\,\text{ft}\). The remaining wood is worth \(\$6.00\) per foot. What is the value of the remaining piece?

Hints

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

Solution

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: \(\frac{11}{6}\times\$6.00=\$11.00\).

Answer

The remaining piece is worth \(\$11.00\).
5107045
A school garden group is painting a fence. They paint \(\frac{1}{5}\) of the fence on Monday, \(\frac{2}{10}\) on Tuesday, and \(\frac{1}{4}\) on Wednesday. What fraction of the fence remains to be painted on Thursday?

Hints

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

Solution

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: \(1-\frac{13}{20}=\frac{20}{20}-\frac{13}{20}=\frac{7}{20}\).

Answer

\(\frac{7}{20}\) of the fence remains to be painted.
5107075
A full measure in \(\frac{4}{4}\) time has a total value of one whole note. A dot after a musical note adds one half of that note’s original value. The measure must contain exactly three notes, and two of them are dotted quarter notes. What value must the third note have to fill the measure exactly? Is it dotted or undotted?

Hints

- A dot adds half of a note’s original value. - Add the values of the two known notes. - Subtract their total from the value of a full measure.

Solution

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.

Answer

The third note must be an undotted quarter note with value \(\frac{1}{4}\).
5107085
A dot after a musical note increases its duration by one half of its original value. a) Show mathematically that two dotted quarter notes have the same duration as three undotted quarter notes. b) How many eighth notes have the same total duration as one dotted half note?

Hints

- A dot adds one half of the note’s original value. - Express both totals using a common denominator. - Rewrite the dotted half note’s value in eighths.

Solution

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.

Answer

a) Both durations equal \(\frac{3}{4}\). b) \(6\) eighth notes
5107365
Anya buys \(8\) bottles of sparkling water. Each bottle contains \(\frac{3}{4}\,\text{qt}\). a) How many quarts of water did she buy altogether? b) She pours exactly \(\frac{1}{4}\,\text{qt}\) from each bottle into a large bowl. How many quarts remain in the bottles altogether?

Hints

- 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.

Solution

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 \(6-2=4\,\text{qt}\).

Answer

a) \(6\,\text{qt}\) b) \(4\,\text{qt}\)
5107375
A painter has \(10\) gallons of white paint. The painter fills \(6\) small containers with \(\frac{3}{8}\) gallon each and \(4\) larger containers with \(\frac{3}{4}\) gallon each. How many gallons remain in the original bucket?

Hints

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

Solution

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.

Answer

\(4\frac{3}{4}\) gallons remain.
5107775
Class A has \(24\) students, and \(\frac{2}{3}\) of them have a pet. Class B has \(30\) students, and \(\frac{1}{2}\) of them have a pet. a) Which class has more students with pets? Show your calculations. b) What fraction of all students in the two classes have a pet?

Hints

- 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.

Solution

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

Answer

a) Class A, with \(16\) students compared with \(15\) in Class B b) \(\frac{31}{54}\)
5111705
A juice bottle is being emptied. First, \(\frac{1}{4}\) of the juice is poured into a glass. Then \(\frac{2}{3}\) of the remaining juice is poured into a pitcher. Exactly \(150\,\text{mL}\) remains in the bottle. How much juice was in the bottle at first?

Hints

- 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.

Solution

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 \(150\times4=600\,\text{mL}\).

Answer

\(600\,\text{mL}\)
5111715
A rain barrel is partly filled. First, \(\frac{3}{10}\) of the water is used for flower beds. Then half of the remaining water is used for a small pond. Afterward, \(21\,\text{L}\) remains. Find the original amount of water in liters and milliliters.

Hints

- 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.

Solution

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 \(1\,\text{L}=1000\,\text{mL}\), \(60\,\text{L}=60{,}000\,\text{mL}\).

Answer

\(60\,\text{L}\), or \(60{,}000\,\text{mL}\)
5111725
A bucket of paint is used for a renovation. The first wall uses \(\frac{2}{9}\) of the paint. The second wall uses \(\frac{3}{7}\) of the paint that remains. After both walls are painted, \(1200\,\text{mL}\) remains. How many liters of paint were in the bucket at first?

Hints

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

Solution

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 \(2700\,\text{mL}=2.7\,\text{L}\).

Answer

\(2.7\,\text{L}\)
5112705
An inflatable boat can safely carry at most \(550\,\text{lb}\). Jonah weighs \(120\,\text{lb}\), and his gear weighs \(90\,\text{lb}\). His sister Mia weighs \(\frac{4}{5}\) as much as Jonah. Their father weighs twice as much as Mia. Can all three people and the gear ride safely without exceeding the limit?

Hints

- 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.

Solution

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.

Answer

Yes. The total load is \(498\,\text{lb}\), which is below the \(550\,\text{lb}\) limit.
5113105
Two pirate crews are comparing plans for dividing treasure. Plan A gives \(\frac{1}{2}\) to the captain, \(\frac{1}{4}\) to the officers, and \(\frac{1}{10}\) to the sailors. Plan B gives \(\frac{1}{3}\) to the captain, \(\frac{1}{6}\) to the officers, and \(\frac{1}{4}\) to the sailors. Which plan distributes a greater fraction of the treasure altogether? Show your calculation.

Hints

- 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.

Solution

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.

Answer

Plan A distributes more: \(\frac{17}{20}\) of the treasure, compared with \(\frac{3}{4}=\frac{15}{20}\) for Plan B.
5113925
A school garden has a total area of \(6000\,\text{ft}^2\). Flowers are planted on \(\frac{1}{4}\) of the area. Of the remaining area, \(\frac{2}{3}\) is used for vegetables. The rest is lawn. What is the area of the lawn?

Hints

- 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.

Solution

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 \(4500-3000=1500\,\text{ft}^2\).

Answer

The lawn has an area of \(1500\,\text{ft}^2\).
5113975
A pitcher is filled to \(\frac{4}{5}\) of its capacity. Then \(\frac{3}{8}\) of the juice in the pitcher is poured into a large glass. a) What fraction of the pitcher's total capacity is now in the glass? b) Compare this with a second situation: the pitcher is filled to \(\frac{3}{5}\) of its capacity, and exactly \(\frac{1}{2}\) of that juice is poured into the glass. In which situation is there more juice in the glass?

Hints

- For part a), multiply the fraction poured by the fraction of the pitcher that is full. - Set up the same kind of calculation for part b). - Simplify both results before comparing them. - Compare the fractions after simplifying.

Solution

1. For a), multiply the fraction poured by the fraction of the pitcher that was full: \(\frac{3}{8}\times\frac{4}{5}=\frac{12}{40}=\frac{3}{10}\). 2. For b), the second situation gives \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\). 3. Since both results are \(\frac{3}{10}\), the glass contains the same amount of juice in both situations.

Answer

a) \(\frac{3}{10}\) of the pitcher's total capacity b) The amount is the same in both situations; each gives \(\frac{3}{10}\) of the pitcher's capacity.
5116345
Solve each problem. a) Find \(\frac{2}{3}\) of \(1 \frac{1}{2}\) pounds. b) One jug holds \(\frac{3}{4}\) gallon. How many gallons do \(5\) such jugs hold altogether? Write the result as a mixed number.

Hints

- 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.

Solution

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.

Answer

a) \(1\) pound b) \(3 \frac{3}{4}\) gallons
5118015
A school garden is being redesigned. One third of the entire garden will be a pond. Of the area that remains, \(\frac{3}{5}\) will be flower beds. The rest will be grass. What fraction of the original garden will be grass?

Hints

- First find the fraction of the garden left after the pond is planned. - Next find what fraction of that remaining area is not used for flowers. - Pay attention to whether each fraction refers to the whole garden or only to the remaining area.

Solution

1. After the pond is set aside, \(1-\frac{1}{3}=\frac{2}{3}\) of the garden remains. 2. Because \(\frac{3}{5}\) of the remaining area becomes flower beds, \(1-\frac{3}{5}=\frac{2}{5}\) of that remaining area becomes grass. 3. Find \(\frac{2}{5}\) of \(\frac{2}{3}\): \(\frac{2}{5}\times\frac{2}{3}=\frac{4}{15}\). 4. Therefore, grass covers \(\frac{4}{15}\) of the original garden.

Answer

\(\frac{4}{15}\) of the original garden
5170505
A fruit punch is mixed in a large pitcher. One-eighth of the punch is berry concentrate, one-half is water, and the rest is \(12\,\text{fl oz}\) of apple juice. How many fluid ounces of punch are in the pitcher altogether?

Hints

- 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.

Solution

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 \(8 \times 4\,\text{fl oz} = 32\,\text{fl oz}\).

Answer

The pitcher contains \(32\,\text{fl oz}\) of punch.
5213575
A family bicycles \(14.6\,\text{mi}\) to a rest stop and then another \(7.9\,\text{mi}\) to an old bridge. At the bridge, they have completed three-fourths of the entire route. What is the total length of the bicycle route?

Hints

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

Solution

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 \(7.5\times 4=30\,\text{mi}\).

Answer

The bicycle route is \(30\,\text{mi}\) long.
5223285
An aquarium has a capacity of \(V\) gallons and is \(\frac{4}{5}\) full. To clean it, \(\frac{1}{4}\) of the aquarium's total capacity is drained. a) What fraction of the total capacity remains in the aquarium? b) How many gallons remain when \(V=80\) gallons?

Hints

- 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.

Solution

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 \(80\times\frac{11}{20}=44\).

Answer

a) \(\frac{11}{20}\) b) \(44\) gallons
5319815
A rectangular community garden is divided into \(15\) equal plots, as shown. The shaded green plots are planted with strawberries. Together, the strawberry plots have an area of \(12\,\text{m}^2\). What is the total area of the garden?
Figure for problem 531981

Hints

- Count the total number of plots and the number of shaded plots. - Use the known area of the shaded plots to find the area of one plot. - Multiply the area of one plot by the total number of plots.

Solution

1. The diagram shows \(6\) shaded plots out of \(15\), so the shaded fraction is \(\frac{6}{15}=\frac{2}{5}\). 2. Since \(6\) plots have a total area of \(12\,\text{m}^2\), one plot has area \(12\div6=2\,\text{m}^2\). 3. The area of all \(15\) plots is \(15\times2\,\text{m}^2=30\,\text{m}^2\).

Answer

The total area of the garden is \(30\,\text{m}^2\).
5355055
Part of a large chocolate bar has been eaten. The blue-shaded pieces in the diagram are the pieces that remain. These pieces weigh \(105\,\text{g}\) altogether. What was the original mass of the whole chocolate bar?
Figure for problem 535505

Hints

- Count all the pieces in the whole bar and the blue-shaded pieces that remain. - Divide the remaining mass by the number of remaining pieces to find the mass of one piece. - Multiply the mass of one piece by the total number of pieces.

Solution

1. The array has \(4\) rows and \(6\) columns, so the whole bar had \(4\times6=24\) pieces. 2. There are \(14\) blue-shaded pieces. Since those pieces weigh \(105\,\text{g}\), one piece weighs \(105\div14=7.5\,\text{g}\). 3. The whole bar weighed \(24\times7.5\,\text{g}=180\,\text{g}\).

Answer

The whole chocolate bar originally weighed \(180\,\text{g}\).
5105795
A person spends \(\frac{3}{8}\) of an amount of money on a gift. The gift costs \(\$45\). a) What was the original amount of money? b) How much money remains after the purchase?

Hints

- 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.

Solution

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 \(5\times\$15=\$75\).

Answer

a) \(\$120\) b) \(\$75\)
5107385
A juice bar sells two cup sizes: Standard, which holds \(\frac{1}{3}\,\text{qt}\), and Large, which holds \(\frac{1}{2}\,\text{qt}\). During the morning, the shop sells \(15\) Standard cups and \(12\) Large cups. The juice comes in \(5\)-quart containers. a) How many quarts of juice were sold? b) What is the minimum number of containers that had to be opened? c) How much juice remains in the last container opened?

Hints

- 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.

Solution

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.

Answer

a) \(11\) quarts b) \(3\) containers c) \(4\) quarts
5358085
A trail runs from a mountain town to a cabin. The entire trail is exactly \(6\) miles longer than the part that passes through the shaded forest. The remaining sunny-meadow section is \(\frac{1}{4}\) of the entire trail. How long is the entire trail? The shaded part of the model represents the forest section.
Figure for problem 535808

Hints

- The part not in the forest is the difference between the whole trail and the forest section. - Use the given difference to find the meadow length. - If one fourth has a known length, find all four fourths. - Use the bar model to count the equal parts in the whole.

Solution

1. The difference between the entire trail and the forest section is the meadow section, so the meadow section is \(6\) miles long. 2. The meadow section is \(\frac{1}{4}\) of the entire trail. 3. If one fourth is \(6\) miles, the entire trail is \(4\times6=24\) miles.

Answer

The entire trail is \(24\) miles long.

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