Adding and Subtracting Fractions: practice solutions, Grade 5 – download the PDF

1 Missing numbers in equivalent fractions ★★★

1. \(5 \times 4 = 20\), so \(3 \times 4 = 12\): \(\dfrac{3}{5} = \dfrac{12}{20}\).
2. \(7 \times 3 = 21\), so \(8 \times 3 = 24\): \(\dfrac{7}{8} = \dfrac{21}{24}\).
3. \(9 \times 5 = 45\), so \(4 \times 5 = 20\): \(\dfrac{4}{9} = \dfrac{20}{45}\).
4. \(10 \div 5 = 2\), so \(15 \div 5 = 3\): \(\dfrac{10}{15} = \dfrac{2}{3}\).

2 Simplify the fractions ★★★

Divide the numerator and denominator by their greatest common factor.

1. GCF is \(6\): \(\dfrac{12}{18} = \dfrac{2}{3}\).
2. GCF is \(5\): \(\dfrac{15}{25} = \dfrac{3}{5}\).
3. GCF is \(9\): \(\dfrac{18}{45} = \dfrac{2}{5}\).
4. GCF is \(4\): \(\dfrac{28}{16} = \dfrac{7}{4} = 1\dfrac{3}{4}\).

3 Least common multiples ★★★

1. Multiples of \(6\): \(6, 12\). \(12\) is divisible by \(4\). LCM \(= 12\).
2. Multiples of \(8\): \(8, 16, 24\). Only \(24\) is divisible by \(3\). LCM \(= 24\).
3. Multiples of \(9\): \(9, 18\). \(18\) is divisible by \(6\). LCM \(= 18\).
4. \(10\) is a multiple of \(5\). LCM \(= 10\).
5. Multiples of \(12\): \(12, 24\). \(24\) is divisible by \(8\). LCM \(= 24\).

4 Easy sums with one common denominator ★★★

1. \(\dfrac{1}{2} = \dfrac{2}{4}\), so \(\dfrac{1}{4} + \dfrac{2}{4} = \dfrac{3}{4}\).
2. \(\dfrac{2}{3} = \dfrac{4}{6}\), so \(\dfrac{4}{6} + \dfrac{1}{6} = \dfrac{5}{6}\).
3. \(\dfrac{1}{5} = \dfrac{2}{10}\), so \(\dfrac{3}{10} + \dfrac{2}{10} = \dfrac{5}{10} = \dfrac{1}{2}\).

5 Easy differences ★★★

1. Five sixths are shaded and two are crossed out: \(\dfrac{5}{6} - \dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}\). This is the same as \(\dfrac{5}{6} - \dfrac{1}{3}\).
2. \(\dfrac{1}{4} = \dfrac{2}{8}\), so \(\dfrac{7}{8} - \dfrac{2}{8} = \dfrac{5}{8}\).
3. \(\dfrac{2}{5} = \dfrac{4}{10}\), so \(\dfrac{9}{10} - \dfrac{4}{10} = \dfrac{5}{10} = \dfrac{1}{2}\).

6 True or false? ★★★

1. True: multiply the numerator and denominator of \(\dfrac{2}{3}\) by \(3\) to get \(\dfrac{6}{9}\).
2. False: \(\dfrac{3}{4} = \dfrac{15}{20}\) but \(\dfrac{4}{5} = \dfrac{16}{20}\).
3. True: divide the numerator and denominator of \(\dfrac{5}{10}\) by \(5\).
4. False: \(\dfrac{1}{3} + \dfrac{1}{4} = \dfrac{4}{12} + \dfrac{3}{12} = \dfrac{7}{12}\). Adding numerators and denominators is not allowed.

7 Juice in the pitcher ★★★

We add: \(\dfrac{1}{2} + \dfrac{1}{4} = \dfrac{2}{4} + \dfrac{1}{4} = \dfrac{3}{4}\).

Answer: there are \(\dfrac{3}{4}\) liter (750 milliliters) of juice in the pitcher.

8 Which is greater? ★★★

1. \(\dfrac{3}{4} = \dfrac{6}{8}\) and \(6 > 5\), so \(\dfrac{3}{4} > \dfrac{5}{8}\).
2. The LCM of \(10\) and \(3\) is \(30\): \(\dfrac{7}{10} = \dfrac{21}{30}\) and \(\dfrac{2}{3} = \dfrac{20}{30}\). Since \(21 > 20\), \(\dfrac{7}{10} > \dfrac{2}{3}\).

9 Sums with unlike denominators ★★★

1. LCM \(= 20\): \(\dfrac{8}{20} + \dfrac{15}{20} = \dfrac{23}{20} = 1\dfrac{3}{20}\).
2. LCM \(= 24\): \(\dfrac{20}{24} + \dfrac{9}{24} = \dfrac{29}{24} = 1\dfrac{5}{24}\).

10 Differences with unlike denominators ★★★

1. LCM \(= 18\): \(\dfrac{14}{18} - \dfrac{3}{18} = \dfrac{11}{18}\).
2. LCM \(= 24\): \(\dfrac{22}{24} - \dfrac{9}{24} = \dfrac{13}{24}\).

11 Adding mixed numbers ★★★

1. Wholes: \(2 + 1 = 3\). Fractions: \(\dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6}\). Total: \(3\dfrac{5}{6}\).
2. Wholes: \(3 + 2 = 5\). Fractions: \(\dfrac{9}{12} + \dfrac{8}{12} = \dfrac{17}{12} = 1\dfrac{5}{12}\). Total: \(5 + 1\dfrac{5}{12} = 6\dfrac{5}{12}\).

12 Subtracting mixed numbers ★★★

1. \(5\dfrac{3}{6} - 2\dfrac{2}{6} = 3\dfrac{1}{6}\).
2. \(4\dfrac{3}{12} - 1\dfrac{8}{12}\). Since \(3\) twelfths is less than \(8\) twelfths, regroup: \(4\dfrac{3}{12} = 3\dfrac{15}{12}\). Then \(3\dfrac{15}{12} - 1\dfrac{8}{12} = 2\dfrac{7}{12}\).

13 A walk in the park ★★★

1. \(\dfrac{3}{4} + \dfrac{2}{3} = \dfrac{9}{12} + \dfrac{8}{12} = \dfrac{17}{12} = 1\dfrac{5}{12}\). Leo walks \(1\dfrac{5}{12}\) miles in all.
2. \(\dfrac{9}{12} - \dfrac{8}{12} = \dfrac{1}{12}\). He walked \(\dfrac{1}{12}\) mile farther in the morning.

14 Estimate first ★★★

1. Estimate: \(\dfrac{5}{8} \approx \dfrac{1}{2}\) and \(\dfrac{7}{12} \approx \dfrac{1}{2}\), so the sum is about \(1\). Exact: \(\dfrac{15}{24} + \dfrac{14}{24} = \dfrac{29}{24} = 1\dfrac{5}{24}\), which is close to \(1\).
2. Estimate: \(\dfrac{13}{15} \approx 1\) and \(\dfrac{4}{9} \approx \dfrac{1}{2}\), so the difference is about \(\dfrac{1}{2}\). Exact: \(\dfrac{39}{45} - \dfrac{20}{45} = \dfrac{19}{45}\), which is close to \(\dfrac{1}{2}\).

15 Three fractions ★★★

1. LCM \(= 12\): \(\dfrac{6}{12} + \dfrac{4}{12} + \dfrac{3}{12} = \dfrac{13}{12} = 1\dfrac{1}{12}\).
2. LCM \(= 12\): \(\dfrac{9}{12} + \dfrac{10}{12} - \dfrac{4}{12} = \dfrac{15}{12} = \dfrac{5}{4} = 1\dfrac{1}{4}\).

16 Find the error ★★★

1. \(\dfrac{3}{4}\) alone is already more than \(\dfrac{1}{2}\), and \(\dfrac{2}{3}\) is also more than \(\dfrac{1}{2}\), so the sum must be more than \(1\). But \(\dfrac{5}{7}\) is less than \(1\) (it is even less than \(\dfrac{3}{4}\)), so it cannot be right.
2. Sam added the numerators and added the denominators. The pieces (thirds and fourths) are not the same size. Correct: \(\dfrac{8}{12} + \dfrac{9}{12} = \dfrac{17}{12} = 1\dfrac{5}{12}\).

17 Ribbon cutting ★★★

Cut pieces: \(1\dfrac{2}{3} + 2\dfrac{1}{2} = 1\dfrac{4}{6} + 2\dfrac{3}{6} = 3\dfrac{7}{6} = 4\dfrac{1}{6}\) feet.

Left: \(5\dfrac{1}{4} - 4\dfrac{1}{6} = 5\dfrac{3}{12} - 4\dfrac{2}{12} = 1\dfrac{1}{12}\) feet.

In inches: \(1\dfrac{1}{12} \times 12 = 13\) inches.

Answer: \(1\dfrac{1}{12}\) feet, or \(13\) inches, of ribbon are left.

18 Missing fractions ★★★

1. \(\square = \dfrac{11}{12} - \dfrac{3}{8} = \dfrac{22}{24} - \dfrac{9}{24} = \dfrac{13}{24}\).
2. \(\square = \dfrac{1}{3} + \dfrac{2}{5} = \dfrac{5}{15} + \dfrac{6}{15} = \dfrac{11}{15}\).

19 The community garden ★★★

1. Tomatoes and peppers: \(\dfrac{2}{5} + \dfrac{1}{4} = \dfrac{8}{20} + \dfrac{5}{20} = \dfrac{13}{20}\). Herbs: \(1 - \dfrac{13}{20} = \dfrac{20}{20} - \dfrac{13}{20} = \dfrac{7}{20}\).
2. \(\dfrac{7}{20} \times 40 = 7 \times 2 = 14\). The herbs cover \(14\) square feet.

20 Training plan ★★★

Monday and Tuesday: \(2\dfrac{4}{12} + 1\dfrac{9}{12} = 3\dfrac{13}{12} = 4\dfrac{1}{12}\) miles.

Wednesday: \(7 - 4\dfrac{1}{12} = 6\dfrac{12}{12} - 4\dfrac{1}{12} = 2\dfrac{11}{12}\) miles.

Answer: she must run \(2\dfrac{11}{12}\) miles on Wednesday.

21 A pattern with differences ★★★

1. \(\dfrac{5}{20} - \dfrac{4}{20} = \dfrac{1}{20}\).
2. \(\dfrac{6}{30} - \dfrac{5}{30} = \dfrac{1}{30}\).
3. \(\dfrac{1}{20} + \dfrac{1}{30} = \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + \left(\dfrac{1}{5} - \dfrac{1}{6}\right) = \dfrac{1}{4} - \dfrac{1}{6}\), because the \(\dfrac{1}{5}\) cancels. That is \(\dfrac{3}{12} - \dfrac{2}{12} = \dfrac{1}{12}\). Check: \(\dfrac{3}{60} + \dfrac{2}{60} = \dfrac{5}{60} = \dfrac{1}{12}\). It matches.