From Real Exams Exam Paper

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

Free Exam-Derived Owl Alpha Primary 6 PSLE Mathematics Semestral Assessment 2 (End of Year) Paper 2 practice paper with questions and answers for Singapore students. This page is rendered as a direct URL so the questions and answers can be discovered without pressing in-page buttons.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

Primary 6 PSLE Mathematics From Real Exams Generated by Owl Alpha Updated 2026-06-04

Back to Subject Browse Free Exam Papers

Questions

TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

School: TuitionGoWhere Secondary School (AI) Subject: Mathematics Level: Primary 6 (PSLE Standard) Paper: SA2 Practice — Version 2 of 5 Duration: 1 hour 30 minutes (90 minutes) Total Marks: 60

Name: ________________________ Class: ________________________ Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
Answer all questions in the spaces provided.
Show all working clearly. Marks will be awarded for correct working even if the final answer is wrong.
Do not use correction fluid or tape.
The use of calculators is not allowed.
The total marks for this paper is 60.

Section A — Short Answer Questions [20 marks]

Questions 1–10. Each question carries 2 marks. Write your answer in the space provided. Show your working clearly.

1. Write the following number in words.

7,030,506

_______________________________________________________________

2. What is the value of the digit 8 in the number 3,842,157?

_______________________________________________________________

3. Round 4,678,293 to the nearest hundred thousand.

_______________________________________________________________

4. Find the highest common factor (HCF) of 36 and 54.

_______________________________________________________________

5. Find the lowest common multiple (LCM) of 8 and 14.

_______________________________________________________________

6. Express 72 as a product of its prime factors. Give your answer in index notation.

_______________________________________________________________

7. List all the factors of 60.

_______________________________________________________________

8. Find the value of:

240 ÷ 8 × 5 + 120

_______________________________________________________________

9. What is the smallest number that must be added to 43,567 to make it divisible by 100?

_______________________________________________________________

10. A number is divisible by both 6 and 8. It lies between 200 and 300. What is the largest possible value of this number?

_______________________________________________________________

Section B — Structured Short Answer [20 marks]

Questions 11–15. Each question carries 4 marks. Show all working clearly.

11. The table below shows the population of four towns.

Town	Population
A	2,450,380
B	2,540,038
C	2,405,830
D	2,504,308

(a) Arrange the populations in order from smallest to largest.

_______________________________________________________________

(b) What is the difference between the largest and smallest populations?

_______________________________________________________________

(c) How many people must move to Town C for it to have the same population as Town B?

_______________________________________________________________

12. A factory produces 12,600 toys in a week. The toys are packed into boxes of 24.

(a) How many boxes are needed?

_______________________________________________________________

(b) The factory operates 6 days a week. How many toys are produced each day?

_______________________________________________________________

(c) If each box is sold for $15, how much money is earned from one week's production?

_______________________________________________________________

13. Find the value of:

(a) 3⁴

_______________________________________________________________

(b) √144

_______________________________________________________________

(c) The cube of 7

_______________________________________________________________

(d) The square root of 2,025

_______________________________________________________________

14. A school has 1,248 students. They are to be divided equally into groups for a mathematics competition.

(a) If each group has 16 students, how many groups are formed?

_______________________________________________________________

(b) If each group has 24 students, how many groups are formed?

_______________________________________________________________

(c) The school also wants to divide the students into equal rows for a parade. List three possible numbers of rows (each row must have more than 10 students).

_______________________________________________________________

15. The HCF of two numbers is 12 and their LCM is 360. One of the numbers is 60. Find the other number.

_______________________________________________________________

Section C — Problem Sums [20 marks]

Questions 16–20. Each question carries 4 marks. Show all working clearly. Include statements where appropriate.

16. At a funfair, 3,600 people attended on Saturday. On Sunday, 4,850 people attended. On Monday, the attendance was 1,250 fewer than the total attendance on Saturday and Sunday combined.

(a) What was the total attendance on Saturday and Sunday?

_______________________________________________________________

(b) What was the attendance on Monday?

_______________________________________________________________

(c) What was the total attendance over the three days?

_______________________________________________________________

17. A fruit seller had 2,400 oranges. He sold 3/8 of them on the first day and 1/4 of the remainder on the second day.

(a) How many oranges did he sell on the first day?

_______________________________________________________________

(b) How many oranges remained after the first day?

_______________________________________________________________

(c) How many oranges did he sell on the second day?

_______________________________________________________________

(d) How many oranges were left after the second day?

_______________________________________________________________

18. The numbers A, B, and C are whole numbers such that:

A is a 3-digit number divisible by 9
B is the smallest 4-digit number divisible by 12
C is the largest 3-digit number divisible by 15

(a) Find the value of A if its digits add up to 18 and it is the largest possible 3-digit number.

_______________________________________________________________

(b) Find the value of B.

_______________________________________________________________

(c) Find the value of C.

_______________________________________________________________

(d) Calculate A + B − C.

_______________________________________________________________

19. A warehouse stores 15,000 identical boxes. The boxes are to be loaded onto lorries. Each lorry can carry at most 320 boxes.

(a) How many lorries are needed to transport all the boxes?

_______________________________________________________________

(b) How many boxes will be on the last (partially filled) lorry?

_______________________________________________________________

(c) If each lorry makes 3 trips, how many boxes can 5 lorries transport in total?

_______________________________________________________________

20. Three friends — Ali, Bala, and Chris — each have some stamps.

Ali has 2/5 of the total number of stamps.
Bala has 1/3 of the remaining stamps.
Chris has 240 stamps.

(a) What fraction of the total stamps does Bala have?

_______________________________________________________________

(b) What fraction of the total stamps does Chris have?

_______________________________________________________________

(c) How many stamps do the three friends have altogether?

_______________________________________________________________

(d) How many stamps does Ali have?

_______________________________________________________________

— End of Paper —

Answers

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 = 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 × $15 = **$ 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 —