Word Problems of fractions worksheets - Math Worksheets ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

Question: An aquarium has exhibits that feature different marine animals. \( \frac{5}{8} \) of the staff are male. What fraction of the staff is female?

Solution:
- The total fraction of the staff is 1 (or \( \frac{8}{8} \)).
- If \( \frac{5}{8} \) of the staff are male, then the fraction of the staff that is female is:
\[
1 - \frac{5}{8} = \frac{8}{8} - \frac{5}{8} = \frac{3}{8}
\]

Answer: \( \boxed{\frac{3}{8}} \)

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Problem 2:

Question: The sharks are fed three times a day. During the morning feeding, \( \frac{2}{15} \) of a ton of fish is fed. During the afternoon feeding, the weight of fish fed will be \( \frac{1}{15} \) of a ton more than the fish fed during the morning. If the total weight of fish fed in a day is \( \frac{1}{2} \) of a ton, how much is fed during the feeding at night?

Solution:
1. Morning feeding: \( \frac{2}{15} \) tons.
2. Afternoon feeding: \( \frac{2}{15} + \frac{1}{15} = \frac{3}{15} = \frac{1}{5} \) tons.
3. Total feeding for the day: \( \frac{1}{2} \) tons.
4. Let \( x \) be the amount of fish fed at night.
5. The total amount of fish fed in a day is the sum of the morning, afternoon, and night feedings:
\[
\frac{2}{15} + \frac{1}{5} + x = \frac{1}{2}
\]
6. Convert \( \frac{1}{5} \) to a fraction with a denominator of 15:
\[
\frac{1}{5} = \frac{3}{15}
\]
7. Substitute back into the equation:
\[
\frac{2}{15} + \frac{3}{15} + x = \frac{1}{2}
\]
8. Combine the fractions on the left side:
\[
\frac{5}{15} + x = \frac{1}{2}
\]
9. Simplify \( \frac{5}{15} \):
\[
\frac{5}{15} = \frac{1}{3}
\]
10. So the equation becomes:
\[
\frac{1}{3} + x = \frac{1}{2}
\]
11. Subtract \( \frac{1}{3} \) from both sides:
\[
x = \frac{1}{2} - \frac{1}{3}
\]
12. Find a common denominator (6) for \( \frac{1}{2} \) and \( \frac{1}{3} \):
\[
\frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6}
\]
13. Subtract the fractions:
\[
x = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
\]

Answer: \( \boxed{\frac{1}{6}} \)

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Problem 3:

Question: A baby otter was born \( \frac{3}{4} \) of a month early. At birth, its weight was \( \frac{7}{8} \) kilograms, which is \( \frac{9}{10} \) kilogram less than the average weight of newborn otters in the aquarium. What is the average weight of newborn otters?

Solution:
1. Let the average weight of newborn otters be \( x \) kilograms.
2. According to the problem, the baby otter's weight is \( \frac{9}{10} \) kilogram less than the average weight:
\[
x - \frac{9}{10} = \frac{7}{8}
\]
3. Solve for \( x \):
\[
x = \frac{7}{8} + \frac{9}{10}
\]
4. Find a common denominator for \( \frac{7}{8} \) and \( \frac{9}{10} \). The least common denominator is 40:
\[
\frac{7}{8} = \frac{7 \times 5}{8 \times 5} = \frac{35}{40}, \quad \frac{9}{10} = \frac{9 \times 4}{10 \times 4} = \frac{36}{40}
\]
5. Add the fractions:
\[
x = \frac{35}{40} + \frac{36}{40} = \frac{71}{40}
\]

Answer: \( \boxed{\frac{71}{40}} \)

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Problem 4:

Question: The penguin nursery is open two times a day: \( \frac{2}{3} \) hour at noon and \( \frac{5}{12} \) hour in the afternoon. How much time is the penguin nursery open every day?

Solution:
1. The total time the nursery is open is the sum of the time it is open at noon and in the afternoon:
\[
\frac{2}{3} + \frac{5}{12}
\]
2. Find a common denominator for \( \frac{2}{3} \) and \( \frac{5}{12} \). The least common denominator is 12:
\[
\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}
\]
3. Add the fractions:
\[
\frac{8}{12} + \frac{5}{12} = \frac{13}{12}
\]

Answer: \( \boxed{\frac{13}{12}} \)

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Problem 5:

Question: Two kinds of fish can be found in a small tank that is \( 5 \frac{1}{7} \) feet long. The blue fish is \( \frac{2}{15} \) feet long and the orange fish is \( \frac{7}{10} \) feet long. How much longer is the orange fish?

Solution:
1. Convert \( 5 \frac{1}{7} \) to an improper fraction:
\[
5 \frac{1}{7} = \frac{5 \times 7 + 1}{7} = \frac{35 + 1}{7} = \frac{36}{7}
\]
However, this information is not needed to solve the problem.
2. The length of the blue fish is \( \frac{2}{15} \) feet.
3. The length of the orange fish is \( \frac{7}{10} \) feet.
4. To find how much longer the orange fish is, subtract the length of the blue fish from the length of the orange fish:
\[
\frac{7}{10} - \frac{2}{15}
\]
5. Find a common denominator for \( \frac{7}{10} \) and \( \frac{2}{15} \). The least common denominator is 30:
\[
\frac{7}{10} = \frac{7 \times 3}{10 \times 3} = \frac{21}{30}, \quad \frac{2}{15} = \frac{2 \times 2}{15 \times 2} = \frac{4}{30}
\]
6. Subtract the fractions:
\[
\frac{21}{30} - \frac{4}{30} = \frac{17}{30}
\]

Answer: \( \boxed{\frac{17}{30}} \)

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Problem 6:

Question: An octopus weighed \( \frac{5}{6} \) kilograms. After two weeks, its weight was increased by \( \frac{3}{10} \) kilograms. But afterwards, it lost \( \frac{1}{5} \) kilograms in weight as it was sick. What is its current weight?

Solution:
1. Initial weight of the octopus: \( \frac{5}{6} \) kilograms.
2. Weight increase after two weeks: \( \frac{3}{10} \) kilograms.
3. Weight after the increase:
\[
\frac{5}{6} + \frac{3}{10}
\]
4. Find a common denominator for \( \frac{5}{6} \) and \( \frac{3}{10} \). The least common denominator is 30:
\[
\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}, \quad \frac{3}{10} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30}
\]
5. Add the fractions:
\[
\frac{25}{30} + \frac{9}{30} = \frac{34}{30} = \frac{17}{15}
\]
6. Weight loss due to sickness: \( \frac{1}{5} \) kilograms.
7. Convert \( \frac{1}{5} \) to a fraction with a denominator of 15:
\[
\frac{1}{5} = \frac{1 \times 3}{5 \times 3} = \frac{3}{15}
\]
8. Subtract the weight loss from the current weight:
\[
\frac{17}{15} - \frac{3}{15} = \frac{14}{15}
\]

Answer: \( \boxed{\frac{14}{15}} \)

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Final Answers:

1. \( \boxed{\frac{3}{8}} \)
2. \( \boxed{\frac{1}{6}} \)
3. \( \boxed{\frac{71}{40}} \)
4. \( \boxed{\frac{13}{12}} \)
5. \( \boxed{\frac{17}{30}} \)
6. \( \boxed{\frac{14}{15}} \)