Decomposing Fractions into Unit Fractions Worksheets - Free Printable

Explanation:
We are asked to decompose fractions into sums of unit fractions — that is, fractions with numerator 1.

A unit fraction looks like:
 $\frac{1}{n}$, where $n$ is a whole number.

To decompose a fraction like $\frac{a}{b}$ (where $a < b$ or $a > b$), we write it as a sum of $a$ copies of $\frac{1}{b}$, *if* the denominator stays the same.
This works because:
$$
\frac{a}{b} = \underbrace{\frac{1}{b} + \frac{1}{b} + \cdots + \frac{1}{b}}_{a \text{ times}}
$$

Let’s go through each part:

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Part A: Express each fraction as a sum of unit fractions

1) $\frac{8}{9}$
→ 8 copies of $\frac{1}{9}$:
 $\frac{1}{9} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9} + \frac{1}{9}$
✔ Already given in the problem — just confirm it's correct.

2) $\frac{5}{6}$
→ 5 copies of $\frac{1}{6}$:
 $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}$

3) $\frac{2}{3}$
→ 2 copies of $\frac{1}{3}$:
 $\frac{1}{3} + \frac{1}{3}$

4) $\frac{4}{7}$
→ 4 copies of $\frac{1}{7}$:
 $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

5) $\frac{9}{10}$
→ 9 copies of $\frac{1}{10}$:
 $\frac{1}{10} + \frac{1}{10} + \cdots + \frac{1}{10}$ (9 times)

6) $\frac{3}{5}$
→ 3 copies of $\frac{1}{5}$:
 $\frac{1}{5} + \frac{1}{5} + \frac{1}{5}$

All of these follow the pattern: numerator tells you how many unit fractions to add, and denominator stays the same.

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Part B: Choose the correct expression

1) Which shows $\frac{7}{8}$ decomposed?

Let’s count how many $\frac{1}{8}$ terms are in each option:

a) $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}$ → that’s 7 terms → correct!
b) Has 8 terms → $\frac{8}{8} = 1$, too big.
c) Only 4 terms → $\frac{4}{8} = \frac{1}{2}$, too small.
d) Only 3 terms → $\frac{3}{8}$, too small.

So answer is a).

2) Which shows $\frac{6}{11}$ decomposed?

Count the $\frac{1}{11}$ terms:

a) 6 terms → $\frac{6}{11}$ ✔
b) 3 terms → $\frac{3}{11}$
c) 3 terms → $\frac{3}{11}$
d) 5 terms → $\frac{5}{11}$

So answer is a).

Double-check:
- For $\frac{6}{11}$, we need six $\frac{1}{11}$’s. Option a has exactly six:
 $\frac{1}{11} + \frac{1}{11} + \frac{1}{11} + \frac{1}{11} + \frac{1}{11} + \frac{1}{11}$

Yes.

Final Answer:
A)
2) $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}$
3) $\frac{1}{3} + \frac{1}{3}$
4) $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$
5) $\frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10}$
6) $\frac{1}{5} + \frac{1}{5} + \frac{1}{5}$

B)
1) a
2) a