
Test solutions with the detailed point scale. Add up your points and spot what to review.
1 Equivalent fractions / 3 pts
- \(7 \times 4 = 28\), so \(5 \times 4 = 20\): \(\dfrac{20}{28}\). (1 pt)
- GCF is \(6\): \(\dfrac{18}{30} = \dfrac{3}{5}\). (1 pt)
- \(\dfrac{8}{12} = \dfrac{2}{3}\) and \(\dfrac{10}{15} = \dfrac{2}{3}\), so yes, they are equivalent. (1 pt)
2 Least common multiple / 3 pts
- Multiples of \(12\): \(12, 24, 36\). LCM \(= 36\). (1 pt)
- No common factor, so LCM \(= 5 \times 7 = 35\). (1 pt)
- Multiples of \(20\): \(20, 40\). LCM \(= 40\). (1 pt)
3 Adding and subtracting fractions / 4 pts
- \(\dfrac{12}{20} + \dfrac{5}{20} = \dfrac{17}{20}\). (1 pt)
- LCM \(= 36\): \(\dfrac{15}{36} + \dfrac{14}{36} = \dfrac{29}{36}\). (1.5 pts: 0.5 for the denominator, 0.5 for the conversions, 0.5 for the sum)
- LCM \(= 45\): \(\dfrac{33}{45} - \dfrac{10}{45} = \dfrac{23}{45}\). (1.5 pts: 0.5 for the denominator, 0.5 for the conversions, 0.5 for the difference)
4 Mixed numbers / 3 pts
- \(3\dfrac{4}{10} + 1\dfrac{7}{10} = 4\dfrac{11}{10} = 5\dfrac{1}{10}\). (1.5 pts: 0.5 common denominator, 0.5 sum, 0.5 regrouping the improper fraction)
- \(6\dfrac{2}{12} - 2\dfrac{9}{12} = 5\dfrac{14}{12} - 2\dfrac{9}{12} = 3\dfrac{5}{12}\). (1.5 pts: 0.5 common denominator, 0.5 regrouping, 0.5 final difference)
5 Estimating / 3 pts
- \(\dfrac{4}{9} \approx \dfrac{1}{2}\) and \(\dfrac{11}{12} \approx 1\), so the sum is about \(1\dfrac{1}{2}\). (1.5 pts)
- LCM \(= 36\): \(\dfrac{16}{36} + \dfrac{33}{36} = \dfrac{49}{36} = 1\dfrac{13}{36}\). (1 pt) It is close to the estimate \(1\dfrac{1}{2}\), so it is reasonable. (0.5 pt)
6 The bike trip / 4 pts
- \(12\dfrac{9}{12} + 9\dfrac{10}{12} = 21\dfrac{19}{12} = 22\dfrac{7}{12}\) miles. (2 pts)
- \(25 - 22\dfrac{7}{12} = 24\dfrac{12}{12} - 22\dfrac{7}{12} = 2\dfrac{5}{12}\) miles. (1 pt)
- \(12\dfrac{9}{12} - 9\dfrac{10}{12} = 11\dfrac{21}{12} - 9\dfrac{10}{12} = 2\dfrac{11}{12}\) miles. (1 pt)
Test yourself: quick challenge for Grade 5
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