Fraction word problems worksheet for students, themed around a coffee shop, with colorful design and illustrations.

Let's solve each problem step by step.

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Problem 1:

The coffee cups can hold \( \frac{7}{9} \) of a pint of liquid. If Emily pours \( \frac{2}{3} \) of a pint of coffee into a cup, how much milk can a customer add?

#### Solution:
1. The total capacity of the cup is \( \frac{7}{9} \) of a pint.
2. Emily pours \( \frac{2}{3} \) of a pint of coffee into the cup.
3. To find out how much milk can be added, we need to subtract the amount of coffee from the total capacity of the cup:
\[
\text{Remaining space} = \text{Total capacity} - \text{Amount of coffee}
\]
\[
\text{Remaining space} = \frac{7}{9} - \frac{2}{3}
\]
4. To subtract these fractions, we need a common denominator. The least common denominator (LCD) of 9 and 3 is 9.
\[
\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}
\]
5. Now subtract:
\[
\frac{7}{9} - \frac{6}{9} = \frac{7 - 6}{9} = \frac{1}{9}
\]

#### Final Answer:
\[
\boxed{\frac{1}{9}}
\]

---

Problem 2:

The capacity of the two milk pitchers on the counter is \( \frac{9}{10} \) liter each. One pitcher has \( \frac{3}{8} \) of a liter of milk in it and the other pitcher has \( \frac{5}{12} \) of a liter milk in it. How much milk is there altogether?

#### Solution:
1. The first pitcher contains \( \frac{3}{8} \) liter of milk.
2. The second pitcher contains \( \frac{5}{12} \) liter of milk.
3. To find the total amount of milk, we add these two fractions:
\[
\text{Total milk} = \frac{3}{8} + \frac{5}{12}
\]
4. To add these fractions, we need a common denominator. The least common denominator (LCD) of 8 and 12 is 24.
\[
\frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}
\]
\[
\frac{5}{12} = \frac{5 \times 2}{12 \times 2} = \frac{10}{24}
\]
5. Now add:
\[
\frac{9}{24} + \frac{10}{24} = \frac{9 + 10}{24} = \frac{19}{24}
\]

#### Final Answer:
\[
\boxed{\frac{19}{24}}
\]

---

Problem 3:

Each large cookie is \( \frac{5}{6} \) oz and each small cookie is \( \frac{4}{9} \) oz. What is the total weight of 2 large cookies and 1 small cookie?

#### Solution:
1. The weight of one large cookie is \( \frac{5}{6} \) oz.
2. The weight of 2 large cookies is:
\[
2 \times \frac{5}{6} = \frac{2 \times 5}{6} = \frac{10}{6} = \frac{5}{3} \text{ oz}
\]
3. The weight of one small cookie is \( \frac{4}{9} \) oz.
4. To find the total weight, we add the weight of 2 large cookies and 1 small cookie:
\[
\text{Total weight} = \frac{5}{3} + \frac{4}{9}
\]
5. To add these fractions, we need a common denominator. The least common denominator (LCD) of 3 and 9 is 9.
\[
\frac{5}{3} = \frac{5 \times 3}{3 \times 3} = \frac{15}{9}
\]
6. Now add:
\[
\frac{15}{9} + \frac{4}{9} = \frac{15 + 4}{9} = \frac{19}{9}
\]

#### Final Answer:
\[
\boxed{\frac{19}{9}}
\]

---

Problem 4:

The coffee shop also sells small bags of coffee beans. The bags come in two sizes: \( \frac{4}{5} \) kilogram and \( \frac{7}{12} \) kilogram. How much more coffee does the bigger bag have?

#### Solution:
1. The weight of the first bag is \( \frac{4}{5} \) kg.
2. The weight of the second bag is \( \frac{7}{12} \) kg.
3. To find out how much more coffee the bigger bag has, we need to determine which bag is larger and then subtract the smaller weight from the larger weight.
4. Compare \( \frac{4}{5} \) and \( \frac{7}{12} \):
- Convert both fractions to a common denominator. The LCD of 5 and 12 is 60.
\[
\frac{4}{5} = \frac{4 \times 12}{5 \times 12} = \frac{48}{60}
\]
\[
\frac{7}{12} = \frac{7 \times 5}{12 \times 5} = \frac{35}{60}
\]
- Since \( \frac{48}{60} > \frac{35}{60} \), the first bag (\( \frac{4}{5} \)) is larger.
5. Subtract the smaller weight from the larger weight:
\[
\text{Difference} = \frac{4}{5} - \frac{7}{12}
\]
6. Using the common denominator of 60:
\[
\frac{4}{5} = \frac{48}{60}, \quad \frac{7}{12} = \frac{35}{60}
\]
\[
\frac{48}{60} - \frac{35}{60} = \frac{48 - 35}{60} = \frac{13}{60}
\]

#### Final Answer:
\[
\boxed{\frac{13}{60}}
\]

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Problem 5:

There were two cheesecakes in the fridge. The first cheesecake was cut into 16 slices and there are 3 slices left. The other cheesecake was cut into 10 slices and 4 slices were sold. How much cake is left in the fridge?

#### Solution:
1. First cheesecake:
- Total slices: 16
- Slices left: 3
- Fraction of the first cheesecake left:
\[
\frac{\text{Slices left}}{\text{Total slices}} = \frac{3}{16}
\]

2. Second cheesecake:
- Total slices: 10
- Slices sold: 4
- Slices left: \( 10 - 4 = 6 \)
- Fraction of the second cheesecake left:
\[
\frac{\text{Slices left}}{\text{Total slices}} = \frac{6}{10} = \frac{3}{5}
\]

3. To find the total amount of cake left, we add the fractions:
\[
\text{Total cake left} = \frac{3}{16} + \frac{3}{5}
\]
4. Find the least common denominator (LCD) of 16 and 5, which is 80.
\[
\frac{3}{16} = \frac{3 \times 5}{16 \times 5} = \frac{15}{80}
\]
\[
\frac{3}{5} = \frac{3 \times 16}{5 \times 16} = \frac{48}{80}
\]
5. Add the fractions:
\[
\frac{15}{80} + \frac{48}{80} = \frac{15 + 48}{80} = \frac{63}{80}
\]

#### Final Answer:
\[
\boxed{\frac{63}{80}}
\]

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Problem 6:

There is a bag of sugar in the storage room. The bag contained \( \frac{4}{5} \) kilograms of sugar. The chef filled up an empty can with \( \frac{1}{10} \) kilogram of sugar and then used \( \frac{3}{20} \) of a kilogram of sugar for a cake. How much sugar was left in the bag?

#### Solution:
1. The initial amount of sugar in the bag is \( \frac{4}{5} \) kg.
2. The chef uses \( \frac{1}{10} \) kg for the can and \( \frac{3}{20} \) kg for the cake.
3. Total sugar used:
\[
\text{Total used} = \frac{1}{10} + \frac{3}{20}
\]
4. Find the least common denominator (LCD) of 10 and 20, which is 20.
\[
\frac{1}{10} = \frac{1 \times 2}{10 \times 2} = \frac{2}{20}
\]
\[
\frac{3}{20} = \frac{3}{20}
\]
5. Add the fractions:
\[
\frac{2}{20} + \frac{3}{20} = \frac{2 + 3}{20} = \frac{5}{20} = \frac{1}{4}
\]
6. Subtract the total sugar used from the initial amount:
\[
\text{Sugar left} = \frac{4}{5} - \frac{1}{4}
\]
7. Find the least common denominator (LCD) of 5 and 4, which is 20.
\[
\frac{4}{5} = \frac{4 \times 4}{5 \times 4} = \frac{16}{20}
\]
\[
\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}
\]
8. Subtract the fractions:
\[
\frac{16}{20} - \frac{5}{20} = \frac{16 - 5}{20} = \frac{11}{20}
\]

#### Final Answer:
\[
\boxed{\frac{11}{20}}
\]

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Final Answers:

1. \(\boxed{\frac{1}{9}}\)
2. \(\boxed{\frac{19}{24}}\)
3. \(\boxed{\frac{19}{9}}\)
4. \(\boxed{\frac{13}{60}}\)
5. \(\boxed{\frac{63}{80}}\)
6. \(\boxed{\frac{11}{20}}\)