Primary 6 PSLE Mathematics Semestral Assessment 2 (End of Year) Paper 4

Answer Key and Marking Scheme

Subject: Mathematics Primary 6
Topic: Whole Numbers
Paper: SA2 Practice Paper (Version 4)

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Section A: Multiple Choice Questions (1 mark each)

1. (4)
Reasoning: The number is 4,702,195. The digit 7 is in the hundred thousands place. Value = $7 \times 100,000 = 700,000$.

2. (2)
Reasoning:

• Divisibility by 4: Last two digits must be divisible by 4.
  • 3,216 (16 is div by 4) - Yes
  • 4,536 (36 is div by 4) - Yes
  • 5,124 (24 is div by 4) - Yes
  • 6,018 (18 is not div by 4) - No
• Divisibility by 9: Sum of digits must be divisible by 9.
  • 3,216: $3+2+1+6=12$ (No)
  • 4,536: $4+5+3+6=18$ (Yes)
  • 5,124: $5+1+2+4=12$ (No)
    Only 4,536 is divisible by both.

3. (3)
Reasoning: To round 8,459,321 to the nearest ten thousand, look at the thousands digit (9). Since $9 \ge 5$, round up the ten thousands digit (5 becomes 6). Result: 8,460,000.

4. (2)
Reasoning: $205 \times 38$.
$205 \times 30 = 6,150$
$205 \times 8 = 1,640$
$6,150 + 1,640 = 7,790$.

5. (2)
Reasoning:
(1) $12 \times 12 \div 2 = 144 \div 2 = 72$
(2) $96 + 2 = 98$
(3) $(96 + 4) \div 2 = 100 \div 2 = 50$
(4) $12 + 32 \div 2 = 12 + 16 = 28$
Largest value is 98.

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Section B: Short Answer Questions (2 marks each)

6. 3,045,006
Working:
Millions: 3
Hundred Thousands: 0
Ten Thousands: 4
Thousands: 5
Hundreds: 0
Tens: 0
Ones: 6
Note: Ensure zeros are placed correctly for empty positions.

7. 58,583
Working:
45,678

• 12,905

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58,583

8. 1
Working:
$5,432 \div 11$
$54 \div 11 = 4$ rem 10
$103 \div 11 = 9$ rem 4
$42 \div 11 = 3$ rem 9? Wait.
Let's re-calculate:
$11 \times 493 = 5,423$.
$5,432 - 5,423 = 9$.
Correction in mental check:
$5432 / 11$:
$54/11 = 4$ (rem 10) -> 103
$103/11 = 9$ (rem 4) -> 42
$42/11 = 3$ (rem 9).
Answer is 9.
Self-Correction for Answer Key: The question asks for remainder.
$11 \times 493 = 5423$.
$5432 - 5423 = 9$.
Answer: 9

9. $2^3 \times 3^2$
Working:
$72 = 8 \times 9$
$8 = 2 \times 2 \times 2 = 2^3$
$9 = 3 \times 3 = 3^2$
So, $72 = 2^3 \times 3^2$.

10. 12
Working:
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Common Factors: 1, 2, 3, 4, 6, 12
Highest Common Factor is 12.

11. 24
Working:
Multiples of 8: 8, 16, 24, 32...
Multiples of 12: 12, 24, 36...
Lowest Common Multiple is 24.

12. 60
Working:
Order of operations (BODMAS): Multiplication first.
$25 \times 4 = 100$
Expression becomes: $150 - 100 + 10$
Subtraction and Addition from left to right:
$150 - 100 = 50$
$50 + 10 = 60$

13. 35,000
Working:
$1,250 \times 28$
$1,250 \times 20 = 25,000$
$1,250 \times 8 = 10,000$
$25,000 + 10,000 = 35,000$

**14. $3,200** *Working:* Cost of chairs:$15 \times 120$$15 \times 12 = 180$, so$15 \times 120 = 1,800$. Money left:$5,000 - 1,800 = 3,200$.

15. 50
Working:
$4,500 \div \Box = 90$
$\Box = 4,500 \div 90$
$\Box = 450 \div 9$
$\Box = 50$

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Section C: Long Answer Questions (4 marks each)

16.
(a) 1,800 adults
(b) 1,800 seniors

Working:
Total spectators = 4,500
(a) Adults = $\frac{2}{5}$ of 4,500
$4,500 \div 5 = 900$
$900 \times 2 = 1,800$ adults.

Remaining spectators = $4,500 - 1,800 = 2,700$.

(b) Children = $\frac{1}{3}$ of remainder
$\frac{1}{3}$ of 2,700 = $2,700 \div 3 = 900$ children.

Seniors = Remainder - Children
Seniors = $2,700 - 900 = 1,800$.
Alternatively: Seniors are $\frac{2}{3}$ of remainder. $\frac{2}{3} \times 2,700 = 1,800$.

Marking:
1 mark for correct number of adults.
1 mark for correct remainder.
1 mark for correct number of children or logic for seniors.
1 mark for final answer for seniors.

17.
(a) 6 : 10 : 7
(b) 84 green marbles

Working:
(a) Red : Blue = 3 : 5
Blue : Green = 10 : 7
To combine, make the 'Blue' part equal. LCM of 5 and 10 is 10.
Multiply first ratio by 2:
Red : Blue = $6 : 10$
Blue : Green = $10 : 7$
Combined Ratio Red : Blue : Green = $6 : 10 : 7$.

(b) Blue units = 10 units.
Given 120 blue marbles.
10 units = 120
1 unit = $120 \div 10 = 12$.
Green marbles = 7 units.
$7 \times 12 = 84$.

Marking:
1 mark for correct combined ratio.
1 mark for value of 1 unit.
1 mark for substitution.
1 mark for final answer.

18.
(a) $10 \times 11 + 10$
(b) 120
(c) Pattern 9

Working:
Pattern $n$: $n \times (n+1) + n$
Simplify: $n^2 + n + n = n^2 + 2n$? No, let's stick to the visual pattern.
Pattern 1: $1(2)+1 = 3$
Pattern 2: $2(3)+2 = 8$
Pattern 3: $3(4)+3 = 15$

(a) For Pattern 10 ($n=10$):
Expression: $10 \times 11 + 10$.

(b) Value:
$10 \times 11 = 110$
$110 + 10 = 120$.

(c) Value = 99.
$n \times (n+1) + n = 99$
$n^2 + n + n = 99$
$n^2 + 2n = 99$
$n(n+2) = 99$
We look for two numbers differing by 2 that multiply to 99.
$9 \times 11 = 99$.
So $n = 9$.
Check Pattern 9: $9 \times 10 + 9 = 90 + 9 = 99$.
Answer is Pattern 9.

Marking:
1 mark for correct expression in (a).
1 mark for correct value in (b).
1 mark for setting up equation or trial in (c).
1 mark for correct pattern number.

19.
(a) 20 oranges
(b) 80 fruits

Working:
Let number of oranges = $1u$.
Number of apples = $3u$ (3 times as many).
Cost of oranges = 1u \times \3 = $3u$. Cost of apples =$3u \times $2 = $6u$. Total cost =$$3u + $6u = $9u$.

Given Total cost = $180.$9u = 180$$1u = 180 \div 9 = 20$.

(a) Oranges = $1u = 20$.

(b) Total fruits = Apples + Oranges = $3u + 1u = 4u$.
$4 \times 20 = 80$ fruits.

Marking:
1 mark for defining units/ratio.
1 mark for correct cost equation.
1 mark for value of 1 unit.
1 mark for final answers (both parts).

20.
(a) 540 boxes
(b) 39 stacks
(c) $19.50

Working:
(a) Total boxes = $45 \text{ stacks} \times 12 \text{ boxes/stack}$.
$45 \times 12 = 45 \times (10 + 2) = 450 + 90 = 540$ boxes.

(b) Boxes removed = 150.
Remaining boxes = $540 - 150 = 390$ boxes.
New stack size = 10 boxes.
Number of new stacks = $390 \div 10 = 39$ stacks.

(c) Cost of labels = 39 \text{ stacks} \times \0.50$.$39 \times 0.5 = 19.5$. Total cost =$19.50.

Marking:
1 mark for total initial boxes.
1 mark for remaining boxes.
1 mark for number of new stacks.
1 mark for total cost.