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Primary 6 PSLE Mathematics Semestral Assessment 2 (End of Year) Paper 2
Free P6 PSLE Maths SA2 Paper 2, LongCat Exam version, with questions, answers, and PSLE-focused practice for Singapore students.
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TuitionGoWhere Practice Paper — Mathematics Primary 6 PSLE
SA2 Practice — Version 2 of 5: Answer Key
Section A — Short Answer Questions [20 marks]
1. Write 7,030,506 in words. [2 marks]
Answer: Seven million, thirty thousand, five hundred and six.
Working:
- Millions period: 7 → "Seven million"
- Thousands period: 030 → "thirty thousand"
- Ones period: 506 → "five hundred and six"
Marking notes: Award 2 marks for the correct answer. Award 1 mark if the student writes the correct words but makes a minor error in punctuation or hyphenation. Accept "seven million thirty thousand five hundred six" (without "and") as acceptable in some marking conventions, but the standard Singapore convention includes "and" before the last part.
2. What is the value of the digit 8 in 3,842,157? [2 marks]
Answer: 800,000 (eight hundred thousand)
Working:
- The digit 8 is in the hundred-thousands place.
- 8 × 100,000 = 800,000
Marking notes: Award 2 marks for "800,000". Award 1 mark if the student writes "8 hundred thousand" or identifies the correct place value but gives an incorrect numerical value.
3. Round 4,678,293 to the nearest hundred thousand. [2 marks]
Answer: 4,700,000
Working:
- The hundred-thousands digit is 6 (4,678,293).
- The ten-thousands digit is 7, which is ≥ 5, so we round up.
- 4,600,000 → 4,700,000
Marking notes: Award 2 marks for the correct answer. Award 1 mark if the student identifies the correct rounding digit but rounds incorrectly (e.g., 4,600,000).
4. Find the HCF of 36 and 54. [2 marks]
Answer: 18
Working (prime factorisation method):
- 36 = 2² × 3²
- 54 = 2 × 3³
- HCF = 2¹ × 3² = 2 × 9 = 18
Alternative (listing method):
- Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
- Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54
- Highest common factor = 18
Marking notes: Award 2 marks for the correct answer with working. Award 1 mark for correct method with arithmetic error.
5. Find the LCM of 8 and 14. [2 marks]
Answer: 56
Working:
- 8 = 2³
- 14 = 2 × 7
- LCM = 2³ × 7 = 8 × 7 = 56
Alternative (listing multiples):
- Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, ...
- Multiples of 14: 14, 28, 42, 56, 70, ...
- LCM = 56
Marking notes: Award 2 marks for the correct answer with working. Award 1 mark for correct method with arithmetic error.
6. Express 72 as a product of its prime factors in index notation. [2 marks]
Answer: 2³ × 3²
Working (factor tree):
72
/ \
8 9
/ \ / \
2 4 3 3
/ \
2 2
- 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
Marking notes: Award 2 marks for the correct answer. Award 1 mark if the student lists all prime factors but does not use index notation (e.g., 2 × 2 × 2 × 3 × 3). Do not accept 8 × 9 as these are not prime factors.
7. List all the factors of 60. [2 marks]
Answer: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Working (systematic pairing):
- 1 × 60 = 60
- 2 × 30 = 60
- 3 × 20 = 60
- 4 × 15 = 60
- 5 × 12 = 60
- 6 × 10 = 60
Factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 (12 factors)
Marking notes: Award 2 marks for all 12 factors listed correctly. Award 1 mark if the student lists at least 8 correct factors with no incorrect ones. Deduct ½ mark for each missing or incorrect factor at the marker's discretion.
8. Find the value of: 240 ÷ 8 × 5 + 120 [2 marks]
Answer: 270
Working (order of operations — BODMAS/BIDMAS):
- Step 1: Division first → 240 ÷ 8 = 30
- Step 2: Multiplication → 30 × 5 = 150
- Step 3: Addition → 150 + 120 = 270
Marking notes: Award 2 marks for the correct answer. Award 1 mark if the student shows correct order of operations but makes an arithmetic error. Common mistake: adding 5 + 120 first, then multiplying — this is incorrect.
9. What is the smallest number that must be added to 43,567 to make it divisible by 100? [2 marks]
Answer: 33
Working:
- A number divisible by 100 must end in 00.
- 43,567 → the last two digits are 67.
- To reach the next hundred: 43,567 + ? = 43,600
- 43,600 − 43,567 = 33
Marking notes: Award 2 marks for the correct answer with working. Award 1 mark for correct method with arithmetic error.
10. A number is divisible by both 6 and 8. It lies between 200 and 300. What is the largest possible value? [2 marks]
Answer: 288
Working:
- LCM of 6 and 8 = 24
- Multiples of 24 between 200 and 300:
- 24 × 8 = 192 (too small)
- 24 × 9 = 216
- 24 × 10 = 240
- 24 × 11 = 264
- 24 × 12 = 288
- 24 × 13 = 312 (too large)
- Largest value = 288
Marking notes: Award 2 marks for the correct answer with working. Award 1 mark for finding the LCM correctly but listing multiples incorrectly, or for finding a correct multiple that is not the largest.
Section B — Structured Short Answer [20 marks]
11. Town populations. [4 marks total — 1 mark for (a), 1 mark for (b), 2 marks for (c)]
(a) Arrange from smallest to largest.
Answer: C (2,405,830), A (2,450,380), D (2,504,308), B (2,540,038)
Working:
- Compare digit by digit from the left:
- All start with 2 million.
- Hundred-thousands: C=4, A=4, D=5, B=5 → C and A are smaller than D and B.
- Between C and A: ten-thousands digit C=0, A=5 → C < A.
- Between D and B: ten-thousands digit D=0, B=4 → D < B.
- Order: C, A, D, B
(b) Difference between largest and smallest.
Answer: 134,208
Working:
- Largest: B = 2,540,038
- Smallest: C = 2,405,830
- 2,540,038 − 2,405,830 = 134,208
(c) How many people must move to Town C to equal Town B?
Answer: 134,208 people
Working:
- Same calculation as (b): 2,540,038 − 2,405,830 = 134,208
Marking notes: (a) Award 1 mark for correct order. (b) Award 1 mark for correct answer. (c) Award 2 marks — 1 for correct method, 1 for correct answer. If (c) is answered correctly but (b) was wrong, award full marks for (c) if the working is correct (error carried forward).
12. Factory toys. [4 marks total — 1 mark for (a), 1 mark for (b), 2 marks for (c)]
(a) How many boxes are needed?
Answer: 525 boxes
Working:
- 12,600 ÷ 24 = 525
(b) Toys produced each day?
Answer: 2,100 toys per day
Working:
- 12,600 ÷ 6 = 2,100
(c) Money earned from one week's production?
Answer: $7,875
Working:
- Number of boxes = 525 (from part a)
- 525 × 7,875**
Alternative: 12,600 ÷ 24 × 15 = 525 × 15 = $7,875
Marking notes: (a) 1 mark for correct answer. (b) 1 mark for correct answer. (c) Award 2 marks — 1 for correct method, 1 for correct answer. Error carried forward from (a) accepted.
13. Powers and roots. [4 marks total — 1 mark each for (a), (b), (c), (d)]
(a) 3⁴
Answer: 81
Working: 3 × 3 × 3 × 3 = 9 × 9 = 81
(b) √144
Answer: 12
Working: 12 × 12 = 144, so √144 = 12
(c) Cube of 7
Answer: 343
Working: 7³ = 7 × 7 × 7 = 49 × 7 = 343
(d) √2,025
Answer: 45
Working: 45 × 45 = (40 + 5)² = 1,600 + 400 + 25 = 2,025, so √2,025 = 45
Marking notes: Award 1 mark per correct part. No working required for these recall questions, but working shown should be credited if correct.
14. School students. [4 marks total — 1 mark for (a), 1 mark for (b), 2 marks for (c)]
(a) Groups of 16?
Answer: 78 groups
Working: 1,248 ÷ 16 = 78
(b) Groups of 24?
Answer: 52 groups
Working: 1,248 ÷ 24 = 52
(c) Three possible numbers of rows (each row > 10 students).
Answer (any three of the following):
Working: Find factors of 1,248 where the number of students per row > 10.
- 1,248 = 2⁴ × 3 × 13 = 16 × 78
- Factors of 1,248: 1, 2, 3, 4, 6, 8, 12, 13, 16, 24, 26, 39, 48, 52, 78, 104, 156, 208, 312, 416, 624, 1248
Possible rows (where students per row > 10):
- 12 rows × 104 students per row
- 13 rows × 96 students per row
- 16 rows × 78 students per row
- 24 rows × 52 students per row
- 26 rows × 48 students per row
- 39 rows × 32 students per row
- 48 rows × 26 students per row
- 52 rows × 24 students per row
Accept any three from the above.
Marking notes: (a) and (b): 1 mark each. (c): Award 2 marks for three correct answers. Award 1 mark for one or two correct answers. The number of students per row must be greater than 10 (so the number of rows must be less than 124.8/10 ≈ 104).
15. HCF and LCM problem. [4 marks]
Answer: 72
Working:
- Formula: HCF × LCM = Product of the two numbers
- 12 × 360 = 60 × (other number)
- 4,320 = 60 × (other number)
- Other number = 4,320 ÷ 60 = 72
Verification:
- HCF of 60 and 72: 60 = 2² × 3 × 5, 72 = 2³ × 3² → HCF = 2² × 3 = 12 ✓
- LCM of 60 and 72: 2³ × 3² × 5 = 8 × 9 × 5 = 360 ✓
Marking notes: Award 4 marks for correct answer with complete working. Award 3 marks for correct formula and method with minor arithmetic error. Award 2 marks for correct formula but incorrect calculation. Award 1 mark for attempting to use HCF/LCM relationship.
Section C — Problem Sums [20 marks]
16. Funfair attendance. [4 marks total — 1 mark for (a), 1 mark for (b), 2 marks for (c)]
(a) Total attendance on Saturday and Sunday.
Answer: 8,450
Working: 3,600 + 4,850 = 8,450
(b) Attendance on Monday.
Answer: 7,200
Working: Monday = (Saturday + Sunday) − 1,250 = 8,450 − 1,250 = 7,200
(c) Total attendance over three days.
Answer: 15,650
Working: 8,450 + 7,200 = 15,650
Marking notes: (a) 1 mark. (b) 1 mark. (c) 2 marks — 1 for method, 1 for answer. Error carried forward accepted.
17. Fruit seller's oranges. [4 marks total — 1 mark each for (a), (b), (c), (d)]
(a) Oranges sold on the first day.
Answer: 900
Working: 3/8 × 2,400 = 3 × 300 = 900
(b) Oranges remaining after the first day.
Answer: 1,500
Working: 2,400 − 900 = 1,500
(c) Oranges sold on the second day.
Answer: 375
Working: 1/4 × 1,500 = 375
(d) Oranges left after the second day.
Answer: 1,125
Working: 1,500 − 375 = 1,125
Marking notes: Award 1 mark per part. Error carried forward accepted throughout. Common mistake in (c): using 1/4 of 2,400 instead of 1/4 of the remainder (1,500).
18. Numbers A, B, and C. [4 marks total — 1 mark each for (a), (b), (c), (d)]
(a) Largest 3-digit number divisible by 9 with digit sum 18.
Answer: 981
Working:
- A number is divisible by 9 if its digits sum to a multiple of 9.
- Digit sum = 18 (which is a multiple of 9 ✓).
- To find the largest 3-digit number: start with the largest hundreds digit.
- 9 + 8 + 1 = 18 → 981
- Check: 981 ÷ 9 = 109 ✓
(b) Smallest 4-digit number divisible by 12.
Answer: 1,008
Working:
- Smallest 4-digit number = 1,000
- 1,000 ÷ 12 = 83.33... → next whole multiple: 84 × 12 = 1,008
- Check: 1,008 ÷ 12 = 84 ✓
(c) Largest 3-digit number divisible by 15.
Answer: 990
Working:
- Largest 3-digit number = 999
- 999 ÷ 15 = 66.6 → 66 × 15 = 990
- Check: 990 ÷ 15 = 66 ✓
- (Must end in 0 or 5 and be divisible by 3: 9 + 9 + 0 = 18, divisible by 3 ✓)
(d) Calculate A + B − C.
Answer: 999
Working: 981 + 1,008 − 990 = 1,989 − 990 = 999
Marking notes: Award 1 mark per part. Error carried forward accepted for (d).
19. Warehouse boxes. [4 marks total — 1 mark for (a), 1 mark for (b), 2 marks for (c)]
(a) Lorries needed.
Answer: 47 lorries
Working:
- 15,000 ÷ 320 = 46.875
- Since we cannot have a fraction of a lorry, we round up: 47 lorries
(b) Boxes on the last lorry.
Answer: 160 boxes
Working:
- 46 full lorries carry: 46 × 320 = 14,720 boxes
- Remaining: 15,000 − 14,720 = 280 boxes
Correction: 15,000 − 14,720 = 280 boxes
(c) Boxes transported by 5 lorries making 3 trips each.
Answer: 4,800 boxes
Working:
- Each lorry carries 320 boxes per trip.
- 5 lorries × 3 trips × 320 boxes = 15 × 320 = 4,800 boxes
Marking notes: (a) Award 1 mark. Common mistake: truncating to 46 instead of rounding up to 47. (b) Award 1 mark. Error carried forward from (a): if student used 46, then 15,000 − (46 × 320) = 15,000 − 14,720 = 280 — this is correct for 46 full lorries, but the question asks about the last lorry when 47 are used, so the answer is 280. (c) Award 2 marks — 1 for method, 1 for answer.
20. Stamps problem. [4 marks total — 1 mark for (a), 1 mark for (b), 1 mark for (c), 1 mark for (d)]
(a) Fraction of total stamps that Bala has.
Answer: 2/5 (of the total... wait, let me recalculate)
Working:
- Ali has 2/5 of total.
- Remainder = 1 − 2/5 = 3/5 of total.
- Bala has 1/3 of the remainder = 1/3 × 3/5 = 3/15 = 1/5 of total.
Answer: 1/5
(b) Fraction of total stamps that Chris has.
Answer: 4/15... let me recalculate.
Working:
- Ali: 2/5 = 6/15 of total
- Bala: 1/5 = 3/15 of total
- Remaining fraction = 1 − 6/15 − 3/15 = 1 − 9/15 = 6/15 = 2/5
Wait — let me redo this carefully.
- Ali has 2/5 of total.
- Remainder after Ali = 3/5 of total.
- Bala has 1/3 of remainder = 1/3 × 3/5 = 1/5 of total.
- Remainder after Ali and Bala = 3/5 − 1/5 = 2/5 of total.
- Chris has 2/5 of total.
So Chris has 2/5 of total.
(b) Answer: 2/5
(c) Total number of stamps.
Answer: 600
Working:
- Chris has 2/5 of total = 240 stamps.
- 2/5 × Total = 240
- Total = 240 ÷ 2/5 = 240 × 5/2 = 600 stamps
(d) Number of stamps Ali has.
Answer: 240
Working:
- Ali has 2/5 of total = 2/5 × 600 = 240 stamps
Verification:
- Ali: 240
- Remainder: 600 − 240 = 360
- Bala: 1/3 × 360 = 120
- Chris: 360 − 120 = 240 ✓
Marking notes: Award 1 mark per part. Error carried forward accepted. Common mistake in (a): students may say Bala has 1/3 of the total instead of 1/3 of the remainder.
Summary of Marks
| Section | Marks |
|---|---|
| A: Questions 1–10 | 20 |
| B: Questions 11–15 | 20 |
| C: Questions 16–20 | 20 |
| Total | 60 |
— End of Answer Key —