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

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

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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 of a pint of coffee.

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

To subtract, we need a common denominator:

- $ \frac{2}{3} = \frac{6}{9} $

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

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

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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 pitcher has 5/12 of a liter milk in it. How much milk is there altogether?

We are asked for the total amount of milk in both pitchers.

Add: $ \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} $

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

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

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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} = \frac{5}{3} $
- 1 small cookie: $ \frac{4}{9} $

Now add: $ \frac{5}{3} + \frac{4}{9} $

LCD of 3 and 9 is 9:
- $ \frac{5}{3} = \frac{15}{9} $

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

✔ Answer: $ \boxed{2\frac{1}{9}} $ ounces.

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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 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:
$$
\frac{48}{60} - \frac{35}{60} = \frac{13}{60}
$$

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

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Problem 5:

There were two cheesecakes in the fridge. The first was cut into 16 slices, 3 left. The second was cut into 10 slices, 4 sold. How much cake is left?

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

Second cheesecake:
- Total slices: 10
- Sold: 4 → so $ 10 - 4 = 6 $ left → $ \frac{6}{10} = \frac{3}{5} $

Now add $ \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: $ \boxed{\frac{63}{80}} $ of a cheesecake is left in the fridge.

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Problem 6:

A bag had 4/5 kg of sugar. Chef used 1/10 kg to fill a can and 3/20 kg for a cake. How much sugar is left?

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

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

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

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

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✔ 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 explanations in simpler terms or visual models!