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Primary 6 PSLE Mathematics Whole Numbers Quiz
Free P6 PSLE Maths Whole Numbers quiz, LongCat Exam version, with questions, answers, and PSLE-focused practice for Singapore students.
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Primary 6 PSLE Mathematics Quiz - Whole Numbers
Answer Key
Section A: Short Answer
1. 5 206 043 [2]
Working: 5 000 000 + 206 000 + 43 = 5 206 043
Marking note: Award 2 marks for correct numeral. Award 1 mark if the student writes the correct digits but with a minor place-value error (e.g., 5 260 043).
2. Hundred thousands (or 700 000) [2]
Working: In 3 728 415, the digit 7 is in the 6th position from the right → hundred thousands place.
Marking note: Accept "hundred thousands" or "700 000".
3. 4 900 000 [2]
Working: The digit in the hundred thousands place is 8. The digit to its right (ten thousands) is 7, which is ≥ 5, so we round up. 8 becomes 9, and all digits to the right become 0.
Marking note: Award 1 mark if student writes 4 800 000 (rounded down incorrectly).
4. 9 000 [2]
Working:
250
× 36
-----
1500 (250 × 6)
+7500 (250 × 30)
-----
9000
Alternative: 250 × 36 = 250 × 4 × 9 = 1 000 × 9 = 9 000
Marking note: Award 1 mark for correct method with arithmetic error. Award 0 if no working shown and answer is wrong.
5. 1 [2]
Working: 9 876 ÷ 25 = 395 remainder 1
Check: 25 × 395 = 9 875; 9 876 − 9 875 = 1
Marking note: Award 1 mark for correct method (long division shown) with minor arithmetic slip.
Section B: Short Answer
6. 6 025 130 , 6 052 301 , 6 205 013 , 6 250 031 [2]
Working: Compare digit by digit from the left. All start with 6. Compare the hundred-thousands digit: 0 < 2, so the first two are smaller. Between 6 025 130 and 6 052 301, compare the ten-thousands digit: 2 < 5, so 6 025 130 < 6 052 301. Similarly, 6 205 013 < 6 250 031.
Marking note: Award 2 marks for fully correct order. Award 1 mark if two adjacent numbers are swapped.
7. 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 [2]
Working: Systematic listing:
- 1 × 84 = 84
- 2 × 42 = 84
- 3 × 28 = 84
- 4 × 21 = 84
- 6 × 14 = 84
- 7 × 12 = 84
Factors: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Marking note: Award 2 marks for all 12 factors listed. Award 1 mark if 8 or more factors are listed correctly.
8. 24 [2]
Working: Prime factorisation method:
- 48 = 2⁴ × 3
- 72 = 2³ × 3²
- HCF = 2³ × 3 = 8 × 3 = 24
Alternative (listing):
- Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
- Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
- Highest common factor = 24
Marking note: Award 1 mark for correct method with minor error.
9. 90 [2]
Working:
- 18 = 2 × 3²
- 30 = 2 × 3 × 5
- LCM = 2 × 3² × 5 = 2 × 9 × 5 = 90
Alternative: Multiples of 18: 18, 36, 54, 72, 90, …
Multiples of 30: 30, 60, 90, …
LCM = 90
Marking note: Award 1 mark for correct method with minor error.
10. 3.875 × 10⁵ (or 387 500) [2]
Working: 12 500 × 31 = 387 500
In standard form: 387 500 = 3.875 × 10⁵
Marking note: If the student gives 387 500 without standard form, award 1 mark. Standard form is required for full marks.
11. 15 [2]
Working: 1 000 000 ÷ 17 = 58 823 remainder r
17 × 58 823 = 999 991
1 000 000 − 999 991 = 9
Wait — recheck: 17 × 58 824 = 1 000 008 (too high).
17 × 58 823 = 999 991. Remainder = 1 000 000 − 999 991 = 9.
The smallest number to subtract = 9.
Correction: Answer is 9, not 15.
Answer: 9 [2]
Marking note: Award 1 mark for correct long division setup with minor arithmetic error.
12. 2⁵ × 3³ [2]
Working:
864 ÷ 2 = 432
432 ÷ 2 = 216
216 ÷ 2 = 108
108 ÷ 2 = 54
54 ÷ 2 = 27
27 ÷ 3 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1
864 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2⁵ × 3³
Marking note: Award 1 mark if the student lists all prime factors but does not write in index notation.
13. 3 or 12 [2]
Working: For a number to be divisible by 9, the sum of its digits must be divisible by 9.
Sum of known digits: 5 + 4 + 3 + 2 + 1 = 15
Total sum = 15 + A + B
For divisibility by 9: 15 + A + B must be divisible by 9.
Possible values: 15 + A + B = 18 or 27 (since A and B are single digits, max A + B = 18)
- If 15 + A + B = 18 → A + B = 3
- If 15 + A + B = 27 → A + B = 12
Answer: 3 or 12
Marking note: Award 2 marks for both values. Award 1 mark for one correct value with valid reasoning.
14. 40 048 [2]
Working: Dividend = Divisor × Quotient + Remainder
= 125 × 320 + 48
= 40 000 + 48
= 40 048
Marking note: Award 1 mark for correct formula applied with arithmetic error.
15. 18 [2]
Working: 360 = 2³ × 3² × 5¹
Total number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24
Odd factors (exclude all factors of 2): 3² × 5¹ → (2+1)(1+1) = 3 × 2 = 6
Even factors = Total − Odd = 24 − 6 = 18
Marking note: Award 1 mark for finding total factors or odd factors correctly. Award 0 if the student simply guesses.
Section C: Problem Solving
16. 96 people [2]
Working: Let the number of adults = 1 unit. Then children = 3 units.
Difference = 3 units − 1 unit = 2 units = 48
1 unit = 24
Total = 4 units = 4 × 24 = 96
Marking note: Award 1 mark for correct model/unit method setup. Award 0 if answer is guessed without working.
17. 8 000 apples [2]
Working: Total sold = 1 250 + 2 380 = 3 630
Original amount = 3 630 + 4 370 = 8 000
Marking note: Award 1 mark for finding total sold correctly.
18. 71 [2]
Working: Let the three consecutive odd numbers be: n − 2, n, n + 2
Sum = (n − 2) + n + (n + 2) = 3n = 207
n = 69
The three numbers are: 67, 69, 71
Largest = 71
Marking note: Award 1 mark for correct equation setup. Accept alternative methods (e.g., 207 ÷ 3 = 69, then 69 + 2 = 71).
19. 3 584 613 [2]
Working: The number is 3 _ _ _ _ _ _ (between 3 000 000 and 4 000 000).
Hundred-thousands digit = 5 → 3 5 _ _ _ _ _
Ten-thousands digit = 2 × thousands digit.
Ones digit = 3 → 3 5 _ _ _ _ 3
Sum of all digits = 29 → 3 + 5 + (ten-thousands) + (thousands) + (hundreds) + (tens) + 3 = 29
→ (ten-thousands) + (thousands) + (hundreds) + (tens) = 18
Let thousands digit = x, then ten-thousands digit = 2x
Possible values: x = 4 → 2x = 8 → hundreds + tens = 18 − 8 − 4 = 6
Try hundreds = 6, tens = 0: 3 5 8 4 6 0 3 → sum = 3+5+8+4+6+0+3 = 29 ✓
But also try hundreds = 4, tens = 2: 3 5 8 4 4 2 3 → sum = 29 ✓ — multiple solutions possible.
Re-examining: The problem states a unique answer is expected. With x = 4: ten-thousands = 8, thousands = 4, hundreds + tens = 6. The simplest (and intended) answer with hundreds = 6, tens = 0: 3 584 603. Rechecking: 3+5+8+4+6+0+3 = 29 ✓.
However, with x = 3: ten-thousands = 6, thousands = 3, hundreds + tens = 9. E.g., 3 5 6 3 9 0 3 → sum = 29 ✓. Multiple solutions exist.
For a unique answer, the intended solution is: 3 584 603 (assuming tens digit is 0 for the simplest case with x = 4).
Answer: 3 584 603 [2]
Marking note: Award 2 marks for correct answer with full reasoning. Award 1 mark for correct setup and partial deduction. Accept any valid number satisfying all conditions.
20. 59 [2]
Working: Let the number of stamps = N.
N ÷ 7 gives remainder 3 → N = 7k + 3
N ÷ 9 gives remainder 5 → N = 9m + 5
List numbers that are 3 more than a multiple of 7: 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, …
Check which gives remainder 5 when divided by 9:
59 ÷ 9 = 6 remainder 5 ✓
Answer: 59
Marking note: Award 1 mark for correct listing of one sequence. Award 0 if no systematic method is shown.