Fractions Word Problems Worksheet for students, featuring real-life scenarios like coffee cups, milk pitchers, cookies, and cheesecakes.

Let's solve each of these fraction word problems step by step.

---

Problem 1:

The coffee cups can hold 7/9 of a pint of liquid. If Emily pours 2/3 of a pint of coffee into a cup, how much milk can a customer add?

We need to find the remaining space in the cup after pouring in 2/3 pint of coffee.

- Total capacity = $ \frac{7}{9} $
- Coffee poured = $ \frac{2}{3} $

We subtract:
$$
\frac{7}{9} - \frac{2}{3}
$$

Convert $ \frac{2}{3} $ to ninths:
$$
\frac{2}{3} = \frac{6}{9}
$$

Now subtract:
$$
\frac{7}{9} - \frac{6}{9} = \frac{1}{9}
$$

✔ Answer: $ \frac{1}{9} $ pint of milk can be added.

---

Problem 2:

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

We are asked to find the total amount of milk in both pitchers:
$$
\frac{3}{8} + \frac{5}{12}
$$

Find the least common denominator (LCD) of 8 and 12 → LCD = 24

Convert:
- $ \frac{3}{8} = \frac{9}{24} $
- $ \frac{5}{12} = \frac{10}{24} $

Add:
$$
\frac{9}{24} + \frac{10}{24} = \frac{19}{24}
$$

✔ Answer: $ \frac{19}{24} $ liters of milk altogether.

---

Problem 3:

Each large cookie is 5/6 oz and each small cookie is 4/9 oz. What is the total weight of 2 large cookies and 1 small cookie?

Calculate:
- 2 large cookies: $ 2 \times \frac{5}{6} = \frac{10}{6} $
- 1 small cookie: $ \frac{4}{9} $

Now add:
$$
\frac{10}{6} + \frac{4}{9}
$$

Simplify $ \frac{10}{6} = \frac{5}{3} $

Now find LCD of 3 and 9 → LCD = 9

- $ \frac{5}{3} = \frac{15}{9} $
- $ \frac{4}{9} = \frac{4}{9} $

Add:
$$
\frac{15}{9} + \frac{4}{9} = \frac{19}{9} = 2\frac{1}{9}
$$

✔ Answer: $ 2\frac{1}{9} $ ounces.

---

Problem 4:

The coffee shop sells bags of coffee beans: 4/5 kg and 7/12 kg. How much more coffee does the bigger bag have?

Compare $ \frac{4}{5} $ and $ \frac{7}{12} $

Find which is larger:

Convert both to common denominator. LCD of 5 and 12 is 60.

- $ \frac{4}{5} = \frac{48}{60} $
- $ \frac{7}{12} = \frac{35}{60} $

So $ \frac{4}{5} > \frac{7}{12} $

Now subtract:
$$
\frac{4}{5} - \frac{7}{12} = \frac{48}{60} - \frac{35}{60} = \frac{13}{60}
$$

✔ Answer: The bigger bag has $ \frac{13}{60} $ kg more coffee.

---

Problem 5:

Two cheesecakes: First cut into 16 slices, 3 left. Second cut into 10 slices, 4 sold. How much cake is left?

First cheesecake:
- Total: 16 slices
- Left: 3 slices → $ \frac{3}{16} $ of the cake

Second cheesecake:
- Total: 10 slices
- Sold: 4 → Left: $ 10 - 4 = 6 $ slices → $ \frac{6}{10} = \frac{3}{5} $ of the cake

Total cake left:
$$
\frac{3}{16} + \frac{3}{5}
$$

LCD of 16 and 5 is 80.

- $ \frac{3}{16} = \frac{15}{80} $
- $ \frac{3}{5} = \frac{48}{80} $

Add:
$$
\frac{15}{80} + \frac{48}{80} = \frac{63}{80}
$$

✔ Answer: $ \frac{63}{80} $ of a cheesecake is left in the fridge.

---

Problem 6:

A bag had 4/5 kg of sugar. Chef took out 1/10 kg for a can and then used 3/20 kg for a cake. How much sugar was left?

Start with $ \frac{4}{5} $ kg

Subtract:
- $ \frac{1}{10} $
- $ \frac{3}{20} $

Total removed:
$$
\frac{1}{10} + \frac{3}{20}
$$

LCD of 10 and 20 is 20:
- $ \frac{1}{10} = \frac{2}{20} $
- $ \frac{3}{20} = \frac{3}{20} $

Sum: $ \frac{2}{20} + \frac{3}{20} = \frac{5}{20} = \frac{1}{4} $

Now subtract from original:
$$
\frac{4}{5} - \frac{1}{4}
$$

LCD of 5 and 4 is 20:
- $ \frac{4}{5} = \frac{16}{20} $
- $ \frac{1}{4} = \frac{5}{20} $

Subtract:
$$
\frac{16}{20} - \frac{5}{20} = \frac{11}{20}
$$

✔ Answer: $ \frac{11}{20} $ kilograms of sugar left.

---

✔ Final Answers Summary:

1. $ \frac{1}{9} $ pint
2. $ \frac{19}{24} $ liters
3. $ 2\frac{1}{9} $ oz
4. $ \frac{13}{60} $ kg
5. $ \frac{63}{80} $ of a cheesecake
6. $ \frac{11}{20} $ kg

Let me know if you'd like these answers explained further or shown with diagrams!