Elementary School Math Quiz: Decompose Fractions Multiple Ways

11 questions · exam conditions

0:00

Decompose Fractions Multiple WaysQuestion 1 of 11

Emma wrote three different decompositions for $\frac{8}{10}$ : Decomposition 1: $\frac{8}{10} = \frac{5}{10} + \frac{3}{10}$ Decomposition 2: $\frac{8}{10} = \frac{4}{10} + \frac{4}{10}$ Decomposition 3: $\frac{8}{10} = \frac{2}{10} + \frac{2}{10} + \frac{4}{10}$ Which statement about her decompositions is true?

Only Decomposition 1 is correct

All three decompositions are correct

Only Decomposition 2 is correct

Only Decomposition 3 is correct

← Back to quizzes

Elementary School Math Quiz

Elementary School Math Quiz: Decompose Fractions Multiple Ways

Practice Decompose Fractions Multiple Ways in Elementary School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Decompose Fractions Multiple Ways, giving you a quick way to practice the rules, question types, and explanations that matter most for Elementary School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Only Decomposition 1 is correct
All three decompositions are correct (correct answer)
Only Decomposition 2 is correct
Only Decomposition 3 is correct

Explanation: The correct answer is B because each decomposition breaks

\frac{8}{10}

into parts with the same denominator that add up to 8 tenths, whether split into two parts or three. Choice A wrongly excludes the other two valid decompositions. Choice C makes the same mistake, picking only one valid decomposition as if the others were wrong. Choice D does the same for the three-part decomposition, when in fact all three ways of splitting the eighths are equally valid.

Question 2

Write $6/12$ as a sum of fractions with denominator 12 in two different ways. Choose the answer that shows two different correct decompositions.

$6/12=2/12+3/12$ and $6/12=4/12+2/12$
$6/12=2/12+4/12$ and $6/12=4/12+2/12$
$6/12=3/12+3/12$ and $6/12=1/12+5/12$ (correct answer)
$6/12=1/12+4/12$ and $6/12=2/12+4/12$

Explanation: Both parts of Choice C add correctly to

6/12

, and the two decompositions use genuinely different pairs of twelfths rather than just reordering the same pair. Choice A has one part that adds to only

5/12

, so that decomposition is not correct. Choice B uses the same two fractions in both decompositions, just written in a different order, so it isn't really two different ways. Choice D has one part that adds to only

5/12

, so it isn't fully correct either. Choice C is the only option with two decompositions that are both correct and genuinely different.

Question 3

Chen wants to break apart $\frac{6}{8}$ into a sum of fractions with denominator 8. Choose the option that shows two different correct decompositions.

$\frac{6}{8}=\frac{3}{8}+\frac{3}{8}$ and $\frac{6}{8}=\frac{2}{8}+\frac{4}{8}$ (correct answer)
$\frac{6}{8}=\frac{1}{8}+\frac{4}{8}$ and $\frac{6}{8}=\frac{2}{8}+\frac{5}{8}$
$\frac{6}{8}=\frac{5}{8}+\frac{2}{8}$ and $\frac{6}{8}=\frac{4}{8}+\frac{3}{8}$
$\frac{6}{8}=\frac{4}{8}+\frac{4}{8}$ and $\frac{6}{8}=\frac{5}{8}+\frac{3}{8}$

Explanation: 6/8 breaks apart correctly into 3/8 + 3/8 and 2/8 + 4/8 because both pairs of eighths add up to 6/8. Choice B's pairs add to 5/8 and 7/8 instead of 6/8. Choice C's pairs both add to 7/8. Choice D's pairs both add to 8/8, one whole.

Question 4

Decompose $\frac{4}{5}$ into a sum of fractions with the same denominator. Show two different ways. Choose the correct pair of equations.

$\frac{4}{5}=\frac{1}{5}+\frac{3}{5}$ and $\frac{4}{5}=\frac{2}{5}+\frac{2}{5}$ (correct answer)
$\frac{4}{5}=\frac{1}{10}+\frac{7}{10}$ and $\frac{4}{5}=\frac{2}{5}+\frac{2}{5}$
$\frac{4}{5}=\frac{1}{5}+\frac{2}{5}$ and $\frac{4}{5}=\frac{3}{5}+\frac{2}{5}$
$\frac{4}{5}=\frac{1}{5}+\frac{3}{5}$ and $\frac{4}{5}=\frac{2}{5}+\frac{3}{5}$

Explanation: The first pair is correct because both equations use fifths and both sums equal

\frac{4}{5}

. The second pair is incorrect because it mixes tenths with fifths. The third pair is incorrect because

\frac{1}{5}+\frac{2}{5}

only equals

\frac{3}{5}

, not

\frac{4}{5}

. The fourth pair is incorrect because

\frac{2}{5}+\frac{3}{5}

equals

\frac{5}{5}

, not

\frac{4}{5}

Question 5

Ava found that $\frac{12}{16}$ can be decomposed as $\frac{12}{16} = \frac{8}{16} + \frac{4}{16}$ . She wants to create a new decomposition by breaking apart the $\frac{8}{16}$ term into two equal parts while keeping the $\frac{4}{16}$ term unchanged. What new equation will she write?

$\frac{12}{16} = \frac{2}{16} + \frac{6}{16} + \frac{4}{16}$
$\frac{12}{16} = \frac{4}{16} + \frac{4}{16} + \frac{3}{16}$
$\frac{12}{16} = \frac{3}{16} + \frac{5}{16} + \frac{4}{16}$
$\frac{12}{16} = \frac{4}{16} + \frac{4}{16} + \frac{4}{16}$ (correct answer)

Explanation: The correct answer is D because splitting

\frac{8}{16}

into two equal parts gives

\frac{4}{16} + \frac{4}{16}

, so the full decomposition is

\frac{4}{16} + \frac{4}{16} + \frac{4}{16}

. Choice A splits

\frac{8}{16}

unevenly into

\frac{2}{16}

and

\frac{6}{16}

. Choice B changes the total, since

\frac{4}{16} + \frac{4}{16} + \frac{3}{16}

adds up to

\frac{11}{16}

, not

\frac{12}{16}

. Choice C also splits

\frac{8}{16}

unevenly, into

\frac{3}{16}

and

\frac{5}{16}

Question 6

Write $\tfrac{5}{8}$ as a sum of fractions with denominator 8 in two different ways. Choose the answer that shows two correct and different decompositions.

Decomposition 1: $\tfrac{5}{8}=\tfrac{1}{8}+\tfrac{4}{8}$ ; Decomposition 2: $\tfrac{5}{8}=\tfrac{4}{8}+\tfrac{1}{8}$
Decomposition 1: $\tfrac{5}{8}=\tfrac{2}{8}+\tfrac{2}{8}$ ; Decomposition 2: $\tfrac{5}{8}=\tfrac{1}{8}+\tfrac{2}{8}$
Decomposition 1: $\tfrac{5}{8}=\tfrac{1}{8}+\tfrac{3}{8}$ ; Decomposition 2: $\tfrac{5}{8}=\tfrac{2}{8}+\tfrac{3}{8}$
Decomposition 1: $\tfrac{5}{8}=\tfrac{1}{8}+\tfrac{4}{8}$ ; Decomposition 2: $\tfrac{5}{8}=\tfrac{2}{8}+\tfrac{3}{8}$ (correct answer)

Explanation: Both parts of Choice D add correctly to

\tfrac{5}{8}

, and the two decompositions use genuinely different pairs of fractions rather than just reordering the same pair. Choice A uses the same two fractions in both decompositions, just written in a different order, so it isn't really two different ways. Choice B has both parts add up to less than

\tfrac{5}{8}

, so neither decomposition is correct. Choice C has one part that adds to

\tfrac{5}{8}

correctly, but the other part adds to only

\tfrac{4}{8}

, so it isn't fully correct. Choice D is the only option with two decompositions that are both correct and genuinely different.

Question 7

Use the number line to determine which decomposition equation correctly represents the fraction shown by the marked segments.

$\frac{7}{9} = \frac{2}{9} + \frac{2}{9} + \frac{3}{9}$
$\frac{7}{9} = \frac{3}{9} + \frac{3}{9} + \frac{1}{9}$
$\frac{7}{9} = \frac{4}{9} + \frac{2}{9} + \frac{1}{9}$ (correct answer)
$\frac{7}{9} = \frac{5}{9} + \frac{1}{9} + \frac{1}{9}$

Explanation: The number line shows three marked segments representing

\frac{7}{9}

total: the first segment spans

\frac{4}{9}

, the second segment spans

\frac{2}{9}

, and the third segment spans

\frac{1}{9}

, giving

\frac{4}{9} + \frac{2}{9} + \frac{1}{9} = \frac{7}{9}

. The other choices represent different groupings that don't match the marked segments shown on the number line.

Question 8

Look at the equation: $\frac{9}{14} = \frac{a}{14} + \frac{b}{14} + \frac{c}{14}$ , where $a$ , $b$ , and $c$ are three whole numbers that are all different from each other. Which set of values for $a$ , $b$ , and $c$ makes this equation true?

$a = 2, b = 2, c = 5$
$a = 1, b = 3, c = 6$
$a = 3, b = 3, c = 3$
$a = 2, b = 3, c = 4$ (correct answer)

Explanation: The correct answer is D because

2 + 3 + 4 = 9

and all three values are different from one another. Choice A also adds to 9 but repeats the value 2, so the numbers aren't all different. Choice B adds to 10, not 9. Choice C adds to 9 but uses the same value three times.

Question 9

Jamal decomposed $\frac{5}{6}$ in two different ways: $\frac{5}{6} = \frac{2}{6} + \frac{3}{6}$ and $\frac{5}{6} = \frac{1}{6} + \frac{4}{6}$ . He wants to create a third decomposition using exactly four addends. Which equation shows a valid third way?

$\frac{5}{6} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{2}{6}$ (correct answer)
$\frac{5}{6} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}$
$\frac{5}{6} = \frac{2}{6} + \frac{1}{6} + \frac{1}{6} + \frac{2}{6}$
$\frac{5}{6} = \frac{0}{6} + \frac{2}{6} + \frac{2}{6} + \frac{1}{6}$

Explanation: Choice A correctly decomposes

\frac{5}{6}

into four addends:

\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{5}{6}

. Choice B sums to

\frac{4}{6}

, not

\frac{5}{6}

. Choice C sums to

\frac{6}{6}

, which is greater than

\frac{5}{6}

. Choice D sums to

\frac{5}{6}

but includes

\frac{0}{6}

, which is not a meaningful part in a decomposition.

Question 10

Write $\frac{7}{12}$ as a sum of fractions with the same denominator in two different ways. Choose the answer that is correct.

$\frac{7}{12}=\frac{3}{12}+\frac{4}{12}$ and $\frac{7}{12}=\frac{2}{12}+\frac{5}{12}$ (correct answer)
$\frac{7}{12}=\frac{1}{6}+\frac{5}{12}$ and $\frac{7}{12}=\frac{2}{12}+\frac{5}{12}$
$\frac{7}{12}=\frac{1}{12}+\frac{4}{12}$ and $\frac{7}{12}=\frac{3}{12}+\frac{4}{12}$
$\frac{7}{12}=\frac{2}{12}+\frac{4}{12}$ and $\frac{7}{12}=\frac{3}{12}+\frac{3}{12}$

Explanation: The first pair is correct because both equations use twelfths and both sums equal

\frac{7}{12}

. The second pair is incorrect because it mixes sixths with twelfths. The third pair is incorrect because

\frac{1}{12}+\frac{4}{12}

equals

\frac{5}{12}

, not

\frac{7}{12}

. The fourth pair is incorrect because both of its equations sum to

\frac{6}{12}

, not

\frac{7}{12}

Question 11

A recipe calls for $\frac{11}{15}$ cup of flour. Ben wants to measure this using three different measuring steps. Which equation shows a way Ben could NOT decompose $\frac{11}{15}$ ?

$\frac{11}{15} = \frac{5}{15} + \frac{3}{15} + \frac{3}{15}$
$\frac{11}{15} = \frac{6}{15} + \frac{3}{15} + \frac{2}{15}$
$\frac{11}{15} = \frac{4}{15} + \frac{4}{15} + \frac{4}{15}$ (correct answer)
$\frac{11}{15} = \frac{7}{15} + \frac{2}{15} + \frac{2}{15}$

Explanation: Choice C is incorrect because

\frac{4}{15} + \frac{4}{15} + \frac{4}{15} = \frac{12}{15}

, which is greater than

\frac{11}{15}

. Choices A, B, and D all correctly sum to

\frac{11}{15}

: Choice A gives

\frac{5+3+3}{15} = \frac{11}{15}

, Choice B gives

\frac{6+3+2}{15} = \frac{11}{15}

, and Choice D gives

\frac{7+2+2}{15} = \frac{11}{15}