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

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Primary 6 PSLE Mathematics From Real Exams Generated by Qwen3.7 Plus Updated 2026-06-04

Questions

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TuitionGoWhere Exam Practice (AI) - Mathematics Primary 6 PSLE

Subject: Mathematics
Level: Primary 6
Paper: WA3 Practice Paper (Version 2 of 5)
Topic Focus: Whole Numbers
Duration: 1 hour 15 minutes
Total Marks: 40

Name: __________________________
Class: __________________________
Date: __________________________


Instructions to Candidates

  1. This paper consists of 20 questions.
  2. Answer all questions.
  3. Write your answers in the spaces provided.
  4. For questions requiring working, show your working clearly. Marks may be awarded for method marks even if the final answer is incorrect.
  5. Unless otherwise stated, give your answers in the simplest form.
  6. The use of calculators is not allowed for this specific topic practice to reinforce mental arithmetic and heuristic strategies, unless specified.

Section A (10 marks)

Questions 1 to 10 carry 1 mark each. Write your answer in the space provided.

1. Write the number four million, sixty thousand, and five in numerals.
Answer: __________________________

2. Round off 8,456,721 to the nearest ten thousand.
Answer: __________________________

3. What is the value of the digit 7 in the number 12,704,390?
Answer: __________________________

4. Find the product of 245 and 38.
Answer: __________________________

5. Find the quotient of 4,536 divided by 12.
Answer: __________________________

6. What is the remainder when 5,000 is divided by 13?
Answer: __________________________

7. Evaluate: 15+25×41015 + 25 \times 4 - 10.
Answer: __________________________

8. Find the Highest Common Factor (HCF) of 18 and 24.
Answer: __________________________

9. Find the Lowest Common Multiple (LCM) of 6 and 8.
Answer: __________________________

10. A number is divisible by both 4 and 9. What is the smallest possible value of this number?
Answer: __________________________


Section B (20 marks)

Questions 11 to 15 carry 2 marks each. Show your working.

11. Mr. Tan bought 15 boxes of pencils. Each box contained 24 pencils. He repacked them into smaller packets of 8 pencils each. How many packets did he make?
<br> <br> <br> Answer: __________________________

12. The sum of two numbers is 450. One number is twice the other. Find the larger number.
<br> <br> <br> Answer: __________________________

13. A factory produces 1,200 toys in 5 days. At this rate, how many toys will it produce in 12 days?
<br> <br> <br> Answer: __________________________

14. Find the value of AA in the following equation:
3×(A12)=453 \times (A - 12) = 45
<br> <br> <br> Answer: A=A = __________________________

15. Sarah has some stickers. If she gives 5 stickers to each of her friends, she will have 3 stickers left. If she gives 6 stickers to each of her friends, she will be short of 4 stickers. How many friends does Sarah have?
<br> <br> <br> Answer: __________________________


Section C (10 marks)

Questions 16 to 20 carry 2 marks each. Show your working.

16. The product of three consecutive whole numbers is 210. Find the sum of these three numbers.
<br> <br> <br> Answer: __________________________

17. A number when divided by 8 gives a quotient of 125 and a remainder of 3. What is the number?
<br> <br> <br> Answer: __________________________

18. Study the pattern below:
Figure 1: 1 dot
Figure 2: 4 dots
Figure 3: 9 dots
Figure 4: 16 dots

How many dots are there in Figure 10?
<br> <br> <br> Answer: __________________________

19. Box A contains 3 times as many marbles as Box B. If 20 marbles are transferred from Box A to Box B, both boxes will have the same number of marbles. How many marbles were there in Box A initially?
<br> <br> <br> Answer: __________________________

20. The total cost of 3 pens and 2 notebooks is \14.Thetotalcostof2pensand3notebooksis. The total cost of **2** pens and **3** notebooks is $16.Findthecostof1pen.<br><br><br>Answer:. Find the cost of **1** pen. <br> <br> <br> Answer: $ $ __________________________


End of Paper

Answers

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Answer Key & Marking Scheme - Mathematics Primary 6 PSLE

Paper: WA3 Practice Paper (Version 2 of 5)
Topic: Whole Numbers


Section A (1 mark each)

1. 4,060,005
Teaching Note: Break down the place values. Millions: 4. Thousands: 060 (sixty thousand). Ones: 005 (five). Ensure zeros are placed correctly in the hundred-thousands, ten-thousands, hundreds, and tens columns.

2. 8,460,000
Teaching Note: Identify the ten-thousands digit (5). Look at the digit to its right (thousands digit is 6). Since 656 \ge 5, round up the ten-thousands digit. 565 \rightarrow 6. Replace digits to the right with zeros.

3. 700,000 (or 7 hundred thousands)
Teaching Note: The digit 7 is in the hundred-thousands place. Its value is 7×100,0007 \times 100,000.

4. 9,310
Teaching Note: Standard multiplication algorithm.
245×8=1960245 \times 8 = 1960
245×30=7350245 \times 30 = 7350
1960+7350=93101960 + 7350 = 9310.

5. 378
Teaching Note: Long division.
45÷12=345 \div 12 = 3 rem 99. Bring down 3 93\rightarrow 93.
93÷12=793 \div 12 = 7 rem 99. Bring down 6 96\rightarrow 96.
96÷12=896 \div 12 = 8.
Result: 378.

6. 11
Teaching Note: 5000÷135000 \div 13.
50÷13=350 \div 13 = 3 rem 1111.
110÷13=8110 \div 13 = 8 rem 66 (13×8=10413 \times 8 = 104).
60÷13=460 \div 13 = 4 rem 88 (13×4=5213 \times 4 = 52).
Wait, let's re-calculate carefully.
13×300=390013 \times 300 = 3900. Remainder 11001100.
13×80=104013 \times 80 = 1040. Remainder 6060.
13×4=5213 \times 4 = 52. Remainder 88.
Correction: 5000=13×384+85000 = 13 \times 384 + 8.
Let's check 13×38413 \times 384: 13×300=390013 \times 300 = 3900, 13×80=104013 \times 80 = 1040, 13×4=5213 \times 4 = 52. Sum =4992= 4992.
50004992=85000 - 4992 = 8.
Answer is 8.
(Self-Correction during generation: Previous mental check was flawed. Final answer is 8.)

7. 105
Teaching Note: Order of operations (BODMAS). Multiplication first: 25×4=10025 \times 4 = 100.
Then addition/subtraction from left to right: 15+10010=11510=10515 + 100 - 10 = 115 - 10 = 105.

8. 6
Teaching Note: 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.

9. 24
Teaching Note: Multiples of 6: 6, 12, 18, 24, 30...
Multiples of 8: 8, 16, 24, 32...
Lowest common multiple is 24.

10. 36
Teaching Note: A number divisible by 4 and 9 must be divisible by their LCM. Since 4 and 9 are coprime (HCF is 1), LCM(4,9)=4×9=36\text{LCM}(4, 9) = 4 \times 9 = 36. The smallest positive whole number is 36.


Section B (2 marks each)

11. 45 packets
Working:
Total pencils =15×24=360= 15 \times 24 = 360.
Number of packets =360÷8= 360 \div 8.
360÷8=45360 \div 8 = 45.
Alternative Method:
15×248=15×3=45\frac{15 \times 24}{8} = 15 \times 3 = 45.
Marking: 1 mark for total pencils (360), 1 mark for correct division (45).

12. 300
Working:
Let the smaller number be 11 unit.
Larger number =2= 2 units.
Total =3= 3 units =450= 450.
11 unit =450÷3=150= 450 \div 3 = 150.
Larger number =2×150=300= 2 \times 150 = 300.
Marking: 1 mark for finding 1 unit (150), 1 mark for larger number (300).

13. 2,880 toys
Working:
Rate per day =1200÷5=240= 1200 \div 5 = 240 toys/day.
Toys in 12 days =240×12= 240 \times 12.
240×10=2400240 \times 10 = 2400.
240×2=480240 \times 2 = 480.
2400+480=28802400 + 480 = 2880.
Marking: 1 mark for daily rate (240), 1 mark for final answer (2880).

14. 27
Working:
3×(A12)=453 \times (A - 12) = 45
Divide both sides by 3:
A12=45÷3A - 12 = 45 \div 3
A12=15A - 12 = 15
A=15+12A = 15 + 12
A=27A = 27
Marking: 1 mark for intermediate step (A12=15A-12=15), 1 mark for final answer (27).

15. 7 friends
Working:
Let number of friends be uu.
Case 1: Total stickers =5u+3= 5u + 3.
Case 2: Total stickers =6u4= 6u - 4.
Equating the totals:
5u+3=6u45u + 3 = 6u - 4
3+4=6u5u3 + 4 = 6u - 5u
7=u7 = u
Check:
If 7 friends, Case 1: 5(7)+3=385(7)+3 = 38.
Case 2: 6(7)4=424=386(7)-4 = 42-4 = 38. Matches.
Marking: 1 mark for setting up equation or logical difference method, 1 mark for answer (7).
Note on Difference Method: Difference in stickers per friend is 65=16-5=1. Difference in total surplus/deficit is 3(4)=73 - (-4) = 7. Number of friends =7÷1=7= 7 \div 1 = 7.


Section C (2 marks each)

16. 18
Working:
Find three consecutive numbers whose product is 210.
Estimate cube root of 210. 53=1255^3 = 125, 63=2166^3 = 216. So numbers are around 5, 6, 7.
Check: 5×6×7=30×7=2105 \times 6 \times 7 = 30 \times 7 = 210.
The numbers are 5, 6, and 7.
Sum =5+6+7=18= 5 + 6 + 7 = 18.
Marking: 1 mark for identifying numbers (5, 6, 7), 1 mark for sum (18).

17. 1,003
Working:
Formula: Dividend=(Divisor×Quotient)+Remainder\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}.
Number =(8×125)+3= (8 \times 125) + 3.
8×125=10008 \times 125 = 1000.
1000+3=10031000 + 3 = 1003.
Marking: 1 mark for multiplication (10001000), 1 mark for adding remainder (10031003).

18. 100 dots
Working:
Pattern analysis:
Figure 1: 12=11^2 = 1
Figure 2: 22=42^2 = 4
Figure 3: 32=93^2 = 9
Figure 4: 42=164^2 = 16
Rule: Number of dots =n2= n^2 where nn is the figure number.
For Figure 10: 102=10010^2 = 100.
Marking: 1 mark for identifying square number pattern, 1 mark for correct calculation (100).

19. 60 marbles
Working:
Let Box B have 11 unit. Box A has 33 units.
Transfer 20 from A to B:
Box A becomes 3u203u - 20.
Box B becomes 1u+201u + 20.
They are equal:
3u20=1u+203u - 20 = 1u + 20
2u=402u = 40
1u=201u = 20.
Box A initially =3u=3×20=60= 3u = 3 \times 20 = 60.
Check: A=60, B=20. Transfer 20: A=40, B=40. Equal.
Marking: 1 mark for finding 1 unit (20), 1 mark for Box A initial (60).

20. **\2Working:Let** *Working:* Let pbecostofpen,be cost of pen,nbecostofnotebook.(1)be cost of notebook. (1)3p + 2n = 14(2) (2)2p + 3n = 16Multiply(1)by3: Multiply (1) by 3:9p + 6n = 42Multiply(2)by2: Multiply (2) by 2:4p + 6n = 32Subtractthesecondnewequationfromthefirst: Subtract the second new equation from the first: (9p + 6n) - (4p + 6n) = 42 - 32 5p = 10 p = 2Marking:1markforcorrectelimination/substitutionmethodsetup,1markforanswer( *Marking:* 1 mark for correct elimination/substitution method setup, 1 mark for answer (2).
Alternative: Add both equations: 5p+5n=30p+n=65p + 5n = 30 \rightarrow p+n=6.
Subtract original equations: (2p+3n)(3p+2n)=1614np=2(2p+3n)-(3p+2n) = 16-14 \rightarrow n-p=2.
Solve system: p+n=6p+n=6 and np=2n-p=2. Adding them: 2n=8,n=42n=8, n=4. Then p=2p=2.