Primary 5 Mathematics Fractions Quiz

Primary 5 Mathematics Quiz - Fractions: Answer Key

Total Marks: 40 marks

Section A: Multiple Choice (Questions 1–5) — 5 marks

1. B) $\frac{9}{15}$ ✓ (1 mark)

Explanation: To find an equivalent fraction, multiply or divide both numerator and denominator by the same number.

• $\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$ ✓
• Check: A) $\frac{6}{15} = \frac{2}{5}$ ✗; C) $\frac{12}{20} = \frac{3}{5}$ ✓ Wait — actually $\frac{12}{20} = \frac{3}{5}$ also. Let me recheck: $\frac{12 \div 4}{20 \div 4} = \frac{3}{5}$.

Correction: Both B and C are equivalent to $\frac{3}{5}$. In a real exam, only one option would be correct. Corrected answer: C) $\frac{12}{20}$ is also equivalent, so this question has an error. Intended single correct answer: D) $\frac{15}{25} = \frac{3}{5}$ — all of B, C, D are equivalent.

Marking note for teacher: This question has a flaw. Accept any of B, C, or D as correct, or use it as a teaching moment about multiple equivalent fractions.

2. B) 0.875 ✓ (1 mark)

Explanation: To convert fraction to decimal, divide numerator by denominator: $7 \div 8 = 0.875$

Alternative: Convert to denominator of 1000: $\frac{7}{8} = \frac{7 \times 125}{8 \times 125} = \frac{875}{1000} = 0.875$

3. A) 8 ✓ (1 mark)

Working:

• Fraction given away: $\frac{5}{6}$
• Fraction kept: $1 - \frac{5}{6} = \frac{1}{6}$
• Cookies kept: $\frac{1}{6} \times 48 = 48 \div 6 = 8$

4. A) $\frac{3}{4}$ ✓ (1 mark)

Explanation: Simplify by finding HCF of 36 and 48, which is 12.

• $\frac{36 \div 12}{48 \div 12} = \frac{3}{4}$

Other options are equivalent but not in simplest form: B) $\frac{6}{8} = \frac{3}{4}$; C) $\frac{9}{12} = \frac{3}{4}$; D) $\frac{18}{24} = \frac{3}{4}$

5. A) $\frac{1}{6}$ ✓ (1 mark)

Working:

• Ravi drank $\frac{2}{3}$, so left: $1 - \frac{2}{3} = \frac{1}{3}$
• Sister drank $\frac{1}{2}$ of remainder: $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Common mistake: Taking $\frac{1}{2}$ of the whole bottle instead of $\frac{1}{2}$ of the remainder.

Section B: Short Answer (Questions 6–15) — 20 marks

6. $\frac{13}{20}$ ✓ (2 marks)

Working:

• Find LCM of 5 and 4: LCM = 20
• $\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$
• $\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$
• $\frac{8}{20} + \frac{5}{20} = \frac{13}{20}$

Already in simplest form since HCF(13, 20) = 1.

7. $\frac{11}{24}$ ✓ (2 marks)

Working:

• Find LCM of 6 and 8: LCM = 24
• $\frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}$
• $\frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}$
• $\frac{20}{24} - \frac{9}{24} = \frac{11}{24}$

Already in simplest form.

8. $\frac{2}{3}$ ✓ (2 marks)

Working:

• $\frac{3}{7} \times \frac{14}{9} = \frac{3 \times 14}{7 \times 9}$
• Simplify before multiplying: 3 and 9 have common factor 3; 14 and 7 have common factor 7
• $= \frac{1 \times 2}{1 \times 3} = \frac{2}{3}$

Or: $\frac{42}{63} = \frac{42 \div 21}{63 \div 21} = \frac{2}{3}$

9. $\frac{1}{10}$ ✓ (2 marks)

Working:

• $\frac{4}{5} \div 8 = \frac{4}{5} \times \frac{1}{8}$ (dividing by 8 = multiplying by $\frac{1}{8}$)
• $= \frac{4 \times 1}{5 \times 8} = \frac{4}{40}$
• Simplify: $\frac{4 \div 4}{40 \div 4} = \frac{1}{10}$

Or simplify first: $\frac{4}{5} \times \frac{1}{8} = \frac{1}{5} \times \frac{1}{2} = \frac{1}{10}$

10. $\frac{3}{20}$ m ✓ (2 marks)

Working:

• $\frac{9}{10} \div 6 = \frac{9}{10} \times \frac{1}{6}$
• $= \frac{9 \times 1}{10 \times 6} = \frac{9}{60}$
• Simplify: $\frac{9 \div 3}{60 \div 3} = \frac{3}{20}$ m

11. $\frac{7}{16}$ kg ✓ (2 marks)

Working:

• Used $\frac{1}{2}$ of $\frac{7}{8}$: $\frac{1}{2} \times \frac{7}{8} = \frac{7}{16}$ kg

Concept: "Of" means multiply when working with fractions.

12. $\frac{7}{12}$, $\frac{5}{8}$, $\frac{2}{3}$, $\frac{5}{6}$ ✓ (2 marks)

Working: Find LCM of denominators: 3, 8, 12, 6. LCM = 24

• $\frac{2}{3} = \frac{16}{24}$
• $\frac{5}{8} = \frac{15}{24}$
• $\frac{7}{12} = \frac{14}{24}$
• $\frac{5}{6} = \frac{20}{24}$

Ascending order (smallest to largest): $\frac{14}{24}$, $\frac{15}{24}$, $\frac{16}{24}$, $\frac{20}{24}$

So: $\frac{7}{12}$, $\frac{5}{8}$, $\frac{2}{3}$, $\frac{5}{6}$

13. $\frac{1}{12}$ km ✓ (2 marks)

Working:

• $\frac{5}{6} - \frac{3}{4}$
• LCM of 6 and 4 is 12
• $\frac{5}{6} = \frac{10}{12}$, $\frac{3}{4} = \frac{9}{12}$
• $\frac{10}{12} - \frac{9}{12} = \frac{1}{12}$ km

14. 120 litres ✓ (2 marks)

Working:

• Water added: $\frac{7}{10} - \frac{3}{5} = \frac{7}{10} - \frac{6}{10} = \frac{1}{10}$ of tank
• $\frac{1}{10}$ of tank = 12 litres
• Full capacity: $12 \times 10 = 120$ litres

Or: Let capacity be C. Then $\frac{3}{5}C + 12 = \frac{7}{10}C$

• $12 = \frac{7}{10}C - \frac{6}{10}C = \frac{1}{10}C$
• $C = 120$ litres

15. $\frac{2}{5}$ ✓ (2 marks)

Working:

• Spent on book: $\frac{2}{5}$, remaining: $1 - \frac{2}{5} = \frac{3}{5}$
• Spent on pen: $\frac{1}{3} \times \frac{3}{5} = \frac{1}{5}$
• Total spent: $\frac{2}{5} + \frac{1}{5} = \frac{3}{5}$
• Left: $1 - \frac{3}{5} = \frac{2}{5}$

Common mistake: Taking $\frac{1}{3}$ of the original amount instead of $\frac{1}{3}$ of the remainder.

Section C: Problem Solving (Questions 16–20) — 15 marks

16. (a) $\frac{1}{2}$ m$^3$ ✓ (2 marks)

Working:

• Volume = base area $\times$ height $= \frac{3}{4} \times \frac{2}{3}$
• $= \frac{3 \times 2}{4 \times 3} = \frac{6}{12} = \frac{1}{2}$ m$^3$

(b) $\frac{2}{9}$ m$^3$ ✓ (2 marks)

Working:

• Remaining water: $1 - \frac{5}{9} = \frac{4}{9}$ of original volume
• Volume remaining: $\frac{4}{9} \times \frac{1}{2} = \frac{4}{18} = \frac{2}{9}$ m$^3$

Note: The question asks for the amount that remains, not the fraction. Correction to question phrasing: "What volume of water remains" would be clearer than "What fraction of a cubic metre" — but as written, $\frac{2}{9}$ m$^3$ equals $\frac{2}{9}$ of a cubic metre, so the answer is $\frac{2}{9}$.

17. (a) 90 stickers ✓ (2 marks)

Working:

• Brother: $\frac{3}{8} \times 240 = 240 \div 8 \times 3 = 30 \times 3 = 90$ stickers

(b) 90 stickers ✓ (2 marks)

Working:

• Remainder after brother: $240 - 90 = 150$
• Or: $240 \times \frac{5}{8} = 150$ stickers
• Cousin: $\frac{2}{5} \times 150 = 60$ stickers
• Left: $150 - 60 = 90$ stickers
• Or: $\frac{3}{5} \times 150 = 90$ stickers

18. (a) $\frac{3}{10}$ ✓ (2 marks)

Working:

• Bala ate $\frac{3}{4}$ of what Ali ate: $\frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10}$

(b) $\frac{3}{10}$ ✓ (2 marks)

Working:

• Total eaten by Ali and Bala: $\frac{2}{5} + \frac{3}{10} = \frac{4}{10} + \frac{3}{10} = \frac{7}{10}$
• Charles ate: $1 - \frac{7}{10} = \frac{3}{10}$

19. (a) 1.5 kg ✓ (1 mark)

Working:

• $\frac{3}{10} \times 5 = \frac{15}{10} = \frac{3}{2} = 1\frac{1}{2} = 1.5$ kg

(b) 1 kg ✓ (1 mark)

Working:

• Remainder after party: $5 - 1.5 = 3.5$ kg or $\frac{7}{2}$ kg
• Or: $5 \times \frac{7}{10} = \frac{35}{10} = \frac{7}{2}$ kg
• Fried rice: $\frac{2}{7} \times \frac{7}{2} = 1$ kg

(c) $\frac{1}{2}$ ✓ (2 marks)

Working:

• Left after fried rice: $\frac{7}{2} - 1 = \frac{5}{2}$ kg
• Or: $\frac{5}{7} \times \frac{7}{2} = \frac{5}{2}$ kg (fraction of remainder used)
• Wait: Let me recompute.
• After party: $\frac{7}{10} \times 5 = \frac{35}{10} = \frac{7}{2}$ kg remainder
• Used for fried rice: $\frac{2}{7} \times \frac{7}{2} = 1$ kg
• Left: $\frac{7}{2} - 1 = \frac{7}{2} - \frac{2}{2} = \frac{5}{2}$ kg
• Fraction of original: $\frac{5}{2} \div 5 = \frac{5}{2} \times \frac{1}{5} = \frac{5}{10} = \frac{1}{2}$

Or step by step with fractions only:

• After party: $\frac{7}{10}$ of original remains
• Of this remainder, $\frac{5}{7}$ is left (since $\frac{2}{7}$ used)
• Fraction left: $\frac{7}{10} \times \frac{5}{7} = \frac{5}{10} = \frac{1}{2}$

20. (a) 150 storybooks ✓ (1 mark)

Working:

• Monday: $\frac{5}{12} \times 360 = 360 \div 12 \times 5 = 30 \times 5 = 150$ books

(b) 130 books ✓ (1 mark)

Working:

• Remaining after Monday: $360 - 150 = 210$
• Tuesday: $\frac{3}{8} \times 210 = 210 \div 8 \times 3 = 26.25 \times 3 = 78.75$

Problem! This gives a non-whole number. Let me re-examine.

Actually: $\frac{3}{8} \times 210 = \frac{630}{8} = \frac{315}{4} = 78.75$. This is not a whole number.

Marking note: The question has a calculation that doesn't yield whole numbers, which is atypical for Primary 5. The intended answer using the given numbers is 78.75, or if we assume rounding, 79. However, this reveals a question design flaw.

Revised working with exact values:

• After Monday: $360 \times \frac{7}{12} = 210$ books
• Tuesday sales: $\frac{3}{8} \times 210 = 78.75$ books
• Left after Tuesday: $210 \times \frac{5}{8} = 131.25$ books, or $210 - 78.75 = 131.25$ books

Since this is problematic, acceptable answers: 131 or 131.25 depending on whether student rounds or gives exact. For P5, the numbers should have been chosen better — e.g., 240 books instead of 360 would give cleaner numbers.

(c) 278 or 278.25 books ✓ (2 marks)

Working (with exact values):

• Wednesday: $\frac{2}{5} \times 131.25 = 52.5$ books sold
• Total: $150 + 78.75 + 52.5 = 281.25$? Let me recheck: $131.25 \times \frac{2}{5} = 52.5$
• Actually need to check: after Tuesday, remaining is 131.25. Wednesday sold $\frac{2}{5}$ of this, so left is $\frac{3}{5} \times 131.25 = 78.75$
• Total sold: $360 - 78.75 = 281.25$

This confirms the numbers are poorly chosen. Teacher note: This question should be revised for future use. Suggest changing to: start with 240 books, or use fractions $\frac{1}{3}$ and $\frac{1}{2}$ instead.

Summary Table

Known issues to fix in next version:

• Question 1: Multiple correct options — select only one equivalent fraction
• Question 20: Numbers chosen don't yield whole numbers, making it atypical for P5