Primary 6 PSLE Mathematics Weighted Assessment 2 (Term 3) Paper 5

Answer Key and Marking Scheme - Mathematics Primary 6 PSLE (WA2 Version 5)

Section A: Multiple Choice Questions (10 marks)

1. (4)

• Reasoning: The digit 7 is in the hundred-thousands place. Value = $7 \times 100,000 = 700,000$.
• Common Mistake: Confusing place value positions (e.g., choosing 70,000).

2. (2)

• Reasoning:
  • Divisible by 4: Last two digits must be divisible by 4.
    • 1,236 (36 $\div$ 4 = 9) - Yes
    • 2,304 (04 $\div$ 4 = 1) - Yes
    • 3,412 (12 $\div$ 4 = 3) - Yes
    • 4,518 (18 $\div$ 4 = 4.5) - No
  • Divisible by 9: Sum of digits must be divisible by 9.
    • 1,236: $1+2+3+6=12$ (No)
    • 2,304: $2+3+0+4=9$ (Yes)
    • 3,412: $3+4+1+2=10$ (No)
  • Only 2,304 is divisible by both.

3. (1)

• Reasoning: Look at the hundreds digit (4). Since $4 < 5$, round down. 58,492 becomes 58,000.

4. (1)

• Reasoning: Numbers divisible by 5 end in 0 or 5. The next number ending in 0 or 5 after 1,234 is 1,235.
  • $1,235 - 1,234 = 1$.

5. (3)

• Reasoning:
  • 51 = $3 \times 17$ (Not prime)
  • 57 = $3 \times 19$ (Not prime)
  • 61 has only factors 1 and 61 (Prime)
  • 63 = $9 \times 7$ (Not prime)

6. (2)

• Reasoning:
  • Factors of 18: 1, 2, 3, 6, 9, 18
  • Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
  • Common factors: 1, 2, 3, 6. Highest is 6.

7. (2)

• Reasoning:
  • Multiples of 6: 6, 12, 18, 24, 30...
  • Multiples of 8: 8, 16, 24, 32...
  • Lowest Common Multiple is 24.

8. (1)

• Reasoning: $$\begin{array}{r} 45,000 \\ - 12,345 \\ \hline 32,655 \end{array}$$

9. (4)

• Reasoning: $1,000 \div 7 = 142$ remainder $6$.
  • Check: $142 \times 7 = 994$. $1,000 - 994 = 6$.

10. (2)

• Reasoning: "Product of 3 and $n$" is $3n$. "5 more than" means add 5. Result: $3n + 5$.

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Section B: Short Answer Questions (20 marks)

11. Two million, forty thousand and five.

• Marking: 1 mark for "Two million", 1 mark for "forty thousand and five". Spelling must be correct.

12. $2^3 \times 3^2 \times 5$

• Working:
  • $360 \div 2 = 180$
  • $180 \div 2 = 90$
  • $90 \div 2 = 45$
  • $45 \div 3 = 15$
  • $15 \div 3 = 5$
  • $5 \div 5 = 1$
  • Prime factors: $2, 2, 2, 3, 3, 5$.
  • Index notation: $2^3 \times 3^2 \times 5$.

13. 47

• Working: Follow BODMAS/PEMDAS.
  • Multiplication/Division first: $8 \times 5 = 40$ and $10 \div 2 = 5$.
  • Expression becomes: $12 + 40 - 5$.
  • Addition/Subtraction: $52 - 5 = 47$.

14. 150

• Working:
  • Let smaller number be $u$. Larger number is $2u$.
  • $u + 2u = 450$
  • $3u = 450$
  • $u = 150$.

15. 45

• Working:
  • Divisible by 3 and 5 means divisible by 15.
  • Multiples of 15: 15, 30, 45, 60...
  • Between 40 and 60: Only 45 fits.

16. 13

• Working:
  • $2^3 = 8$
  • $3^2 = 9$
  • $4^1 = 4$
  • $8 + 9 - 4 = 17 - 4 = 13$.

17. 15 boxes

• Working:
  • $120 \div 8 = 15$.
  • No remainder, so exactly 15 boxes.

18. 5

• Working:
  • $3n + 7 = 22$
  • $3n = 22 - 7$
  • $3n = 15$
  • $n = 15 \div 3 = 5$.

19. 89,999

• Working:
  • Largest 5-digit odd number: 99,999.
  • Smallest 5-digit even number: 10,000.
  • Difference: $99,999 - 10,000 = 89,999$.

20. 1, 2, 3, 4, 6, 8, 12, 24

• Marking: All 8 factors must be listed correctly. Missing one or including a non-factor results in 0 marks.

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Section C: Long Answer Questions (30 marks)

21. (a) $\frac{1}{2}$ (b) 240 beads

• Concept: Fraction of Remainder.
• Working:
  • (a)
    • Used for necklace: $\frac{1}{3}$. Remainder: $1 - \frac{1}{3} = \frac{2}{3}$.
    • Used for bracelet: $\frac{1}{4}$ of remainder = $\frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}$.
    • Total used: $\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.
    • Fraction left: $1 - \frac{1}{2} = \frac{1}{2}$.
  • (b)
    • $\frac{1}{2}$ of total = 120 beads.
    • Total = $120 \times 2 = 240$ beads.
• Marking: 2 marks for correct fraction in (a), 3 marks for correct calculation and answer in (b).

22. (a) 1,400 (b) 1,680

• Concept: Average and Percentage Increase.
• Working:
  • (a)
    • Sum = $1,200 + 1,500 + 1,100 + 1,800 + 1,400 = 7,000$.
    • Average = $7,000 \div 5 = 1,400$.
  • (b)
    • Friday visitors = 1,400.
    • Increase = $20\%$ of $1,400 = 0.2 \times 1,400 = 280$.
    • Saturday visitors = $1,400 + 280 = 1,680$.
• Marking: 2 marks for (a), 3 marks for (b).

23. (a) 20 (b) 50

• Concept: Constant Difference / Algebra / Model Method.
• Working:
  • (a)
    • If 10 moved from A to B makes them equal, A must have had 20 more than B initially.
    • Let $A$ be initial A, $B$ be initial B.
    • $A - 10 = B + 10 \Rightarrow A - B = 20$.
    • Difference is 20.
  • (b)
    • If 20 moved from B to A:
    • New A = $A + 20$. New B = $B - 20$.
    • New A = $3 \times$ New B.
    • Substitute $A = B + 20$:
    • $(B + 20) + 20 = 3(B - 20)$
    • $B + 40 = 3B - 60$
    • $100 = 2B$
    • $B = 50$.
    • $A = 50 + 20 = 70$.
    • Correction Check: If A=70, B=50. Move 20 from B to A: A=90, B=30. $90 = 3 \times 30$. Correct.
    • Question asks for Box A at first: 70.
    • Wait, let me re-read carefully. "If 20 marbles are moved from Box B to Box A... Box A will have 3 times as many as Box B."
    • My calculation: $B=50, A=70$.
    • Let's check the first condition: Move 10 from A to B. A=60, B=60. Equal. Correct.
    • Answer for (b) is 70.
• Marking: 2 marks for (a), 3 marks for (b).

24. (a) 950 (b) 3,500

• Concept: Arithmetic Progression.
• Working:
  • Mon: 500
  • Tue: $500 + 150 = 650$
  • Wed: $650 + 150 = 800$
  • Thu: $800 + 150 = 950$
  • Fri: $950 + 150 = 1,100$
  • (a) Thursday production: 950.
  • (b) Total: $500 + 650 + 800 + 950 + 1,100$.
    • Sum = 3,500.
• Marking: 2 marks for (a), 3 marks for (b).

25. (a) 4 m (b) 6 tiles

• Concept: HCF for tiling without cutting.
• Working:
  • (a) Find HCF of 12 and 8.
    • Factors of 12: 1, 2, 3, 4, 6, 12.
    • Factors of 8: 1, 2, 4, 8.
    • HCF = 4. Side length = 4 m.
  • (b)
    • Number of tiles along length: $12 \div 4 = 3$.
    • Number of tiles along width: $8 \div 4 = 2$.
    • Total tiles: $3 \times 2 = 6$.
• Marking: 2 marks for (a), 3 marks for (b).

26. (a) 12 rabbits (b) 18 chickens

• Concept: Assumption Method or Algebra.
• Working:
  • Let $R$ be rabbits, $C$ be chickens.
  • $R + C = 30$ (Heads)
  • $4R + 2C = 84$ (Legs)
  • From first eq: $C = 30 - R$.
  • Substitute into second eq:
    • $4R + 2(30 - R) = 84$
    • $4R + 60 - 2R = 84$
    • $2R = 24$
    • $R = 12$.
  • $C = 30 - 12 = 18$.
  • Check: $12 \times 4 + 18 \times 2 = 48 + 36 = 84$. Correct.
• Marking: 2.5 marks for (a), 2.5 marks for (b). Working must be shown.