5th Grade Adding and Subtracting Fractions with Unlike Denominators Test Solutions and Scoring (PDF)

5th Grade math test solutions: Adding and Subtracting Fractions with Unlike Denominators – Maths PDF


This answer key gives worked solutions and a scoring guide for each of the 9 test questions, worth 20 points in all. Use the scoring notes to give partial credit when a student sets up the problem correctly but makes an arithmetic slip.

Grade 5 · Common Core State Standards: 5.NF.A.1, 5.NF.A.2

Question 1 – Add unlike fractions [2 pts]

(a)

The least common denominator of 3 and 4 is 12.

Rename the fractions: \(\frac{1}{3} = \frac{4}{12}\) and \(\frac{1}{4} = \frac{3}{12}\).

Add: \(\frac{4}{12} + \frac{3}{12} = \frac{7}{12}\).

\(\frac{7}{12}\)

(b)

The least common denominator of 6 and 8 is 24.

Rename the fractions: \(\frac{5}{6} = \frac{20}{24}\) and \(\frac{1}{8} = \frac{3}{24}\).

Add: \(\frac{20}{24} + \frac{3}{24} = \frac{23}{24}\).

\(\frac{23}{24}\)

Scoring: 1 point for each correct sum in simplest form (renaming correctly with a common denominator).

Question 2 – Subtract unlike fractions [2 pts]

(a)

The least common denominator of 8 and 2 is 8.

Rename the fractions: \(\frac{7}{8} = \frac{7}{8}\) and \(\frac{1}{2} = \frac{4}{8}\).

Subtract: \(\frac{7}{8} – \frac{4}{8} = \frac{3}{8}\).

\(\frac{3}{8}\)

(b)

The least common denominator of 3 and 5 is 15.

Rename the fractions: \(\frac{2}{3} = \frac{10}{15}\) and \(\frac{1}{5} = \frac{3}{15}\).

Subtract: \(\frac{10}{15} – \frac{3}{15} = \frac{7}{15}\).

\(\frac{7}{15}\)

Scoring: 1 point for each correct difference.

Question 3 – Add from a model [2 pts]

2/31/411/12
Twelfths: 8/12 + 3/12 = 11/12.

(a) \(\frac{2}{3} + \frac{1}{4}\).

(b) \(\frac{8}{12} + \frac{3}{12} = \frac{11}{12}\): \(\frac{11}{12}\).

Scoring: 1 point for the correct sentence (either order); 1 point for the sum.

Question 4 – Mixed numbers [3 pts]

(a)

The least common denominator of 5 and 10 is 10.

Rename the fractions: \(3\frac{2}{5} = 3\frac{4}{10}\) and \(2\frac{7}{10} = 2\frac{7}{10}\).

Add: \(3\frac{4}{10} + 2\frac{7}{10} = 5\frac{11}{10}\).

Simplify: \(5\frac{11}{10} = 6\frac{1}{10}\).

\(6\frac{1}{10}\)

(b)

The least common denominator of 4 and 6 is 12.

Rename the fractions: \(6\frac{1}{4} = 6\frac{3}{12}\) and \(2\frac{5}{6} = 2\frac{10}{12}\).

The first fraction is smaller than the second, so regroup 1 whole as \(\frac{12}{12}\): \(6\frac{3}{12} = 5\frac{15}{12}\).

Subtract: \(5\frac{15}{12} – 2\frac{10}{12} = 3\frac{5}{12}\).

\(3\frac{5}{12}\)

Scoring: (a) 1 point for the correct sum in simplest form. (b) 1 point for regrouping correctly, 1 point for the correct difference.

Question 5 – Is it reasonable? [2 pts]

(a) \(\frac{3}{8}\) is a little less than \(\frac{1}{2}\) and \(\frac{5}{9}\) is a little more than \(\frac{1}{2}\), so the sum is close to 1. But \(\frac{8}{17}\) is less than \(\frac{1}{2}\), so it is too small. (Adding the denominators is the error.)

(b) \(\frac{27}{72} + \frac{40}{72} = \frac{67}{72}\): \(\frac{67}{72}\).

Scoring: 1 point for a valid benchmark explanation; 1 point for the correct sum.

Question 6 – Two ropes [2 pts]

4 3/6– 2 4/6regroup3 9/6– 2 4/6subtract1 5/6
Regroup, then subtract.

The least common denominator of 2 and 3 is 6.

Rename the fractions: \(4\frac{1}{2} = 4\frac{3}{6}\) and \(2\frac{2}{3} = 2\frac{4}{6}\).

The first fraction is smaller than the second, so regroup 1 whole as \(\frac{6}{6}\): \(4\frac{3}{6} = 3\frac{9}{6}\).

Subtract: \(3\frac{9}{6} – 2\frac{4}{6} = 1\frac{5}{6}\).

\(1\frac{5}{6}\)

The first rope is \(1\frac{5}{6}\) feet longer.

Scoring: 1 point for a correct setup with a common denominator; 1 point for the correct difference with units.

Question 7 – Trail mix [3 pts]

(a) The LCD of 4, 3, and 2 is 12: \(\frac{9}{12} + \frac{8}{12} + \frac{6}{12} = \frac{23}{12} = 1\frac{11}{12}\): \(1\frac{11}{12}\) cups.

(b) \(2 – 1\frac{11}{12} = \frac{1}{12}\): Yes, it fits, with \(\frac{1}{12}\) cup of room left.

Scoring: (a) 1 point for renaming all three fractions with a common denominator, 1 point for the total. (b) 1 point for a correct answer to both parts.

Question 8 – Practice goal [2 pts]

(a) \(1\frac{3}{12} + 2\frac{6}{12} + 1\frac{8}{12} = 4\frac{17}{12} = 5\frac{5}{12}\): \(5\frac{5}{12}\) hours.

(b) \(6 – 5\frac{5}{12} = \frac{7}{12}\): \(\frac{7}{12}\) hour.

Scoring: 1 point for each correct answer.

Question 9 – Missing numbers [2 pts]

(a) \(? = \frac{5}{6} – \frac{1}{4} = \frac{10}{12} – \frac{3}{12} = \frac{7}{12}\): \(\frac{7}{12}\).

(b) \(? = 5\frac{3}{6} – 3\frac{2}{6} = 2\frac{1}{6}\): \(2\frac{1}{6}\).

Scoring: 1 point for each correct missing number.

Total

20 points

All Grade 5 chapters: Grade 5 math.

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