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Primary 5 Mathematics Practice Paper 1
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TuitionGoWhere Practice Paper Answers - Mathematics Primary 5
Topic: Whole Numbers
Version: 1 of 5
Total Marks: 50
SECTION A: Multiple Choice Questions (10 marks)
| Question | Answer | Explanation |
|---|---|---|
| 1 | (4) 7 000 000 | The digit 7 is in the millions place. In 7 654 321, the 7 represents 7 million. Place value: 7 × 1 000 000 = 7 000 000. |
| 2 | (1) 3 040 008 | Three million = 3 000 000; forty thousand = 040 000; eight = 8. Combined: 3 000 000 + 40 000 + 8 = 3 040 008. Note: "forty thousand" needs zeros in the hundred thousands and thousands positions. |
| 3 | (1) 4 500 000 | Rounding to nearest hundred thousand: look at the ten thousands digit (6 in 4 567 892). Since 6 ≥ 5, round up: 4 500 000. The 6 causes the 5 hundred thousands to round up to 6, but we express as 4 500 000. Wait—correction: 4 567 892, the hundred thousands digit is 5, ten thousands is 6. Since 6 ≥ 5, round 5 up to 6, giving 4 600 000. Answer: (3). |
| 4 | (1) 2 543 678 | 5 ten thousands = 50 000. Check position: in 2 543 678, the 5 is in the hundred thousands place (500 000). Correction: Check again. 2 543 678: digits are 2 (millions), 5 (hundred thousands), 4 (ten thousands), 3 (thousands), etc. The 5 is in hundred thousands. In 3 657 901: 3 (millions), 6 (hundred thousands), 5 (ten thousands), 7 (thousands)... Here 5 is in ten thousands place. Answer: (2). |
| 5 | (1) 8 975 432 | 8 965 432 + 10 000 = 8 975 432. Adding 10 000 affects the ten thousands place: 65 becomes 75. |
| 6 | (2) 3 455 000 | Halfway = (3 450 000 + 3 460 000) ÷ 2 = 6 910 000 ÷ 2 = 3 455 000. Or: the midpoint of 450 000 and 460 000 is 455 000. |
| 7 | (2) 2 490 000 | Pattern increases by 50 000 each time: 2 340 000 + 50 000 = 2 390 000; 2 390 000 + 50 000 = 2 440 000; 2 440 000 + 50 000 = 2 490 000; 2 490 000 + 50 000 = 2 540 000. ✓ |
| 8 | (3) 2 560 000 | Rounded to nearest hundred thousand = 2 500 000. Range: 2 450 000 to 2 549 999. Of the choices, only 2 560 000 rounds to 2 600 000. Correction: Check 2 450 000 rounds to 2 500 000? Yes (2 450 000 exactly rounds up). 2 560 000: ten thousands digit is 6, so rounds to 2 600 000. 2 420 000 rounds to 2 400 000. 2 610 000 rounds to 2 600 000. So answer is (2) 2 450 000. |
| 9 | (1) 5 432 100 | Compare digit by digit from left: all start with 5. Next digit (hundred thousands): 4, 3, 4, 4. Eliminate (2). Remaining: 432 100, 423 100, 431 200. Compare ten thousands: 3, 2, 3. Eliminate (3). Compare thousands: 2 vs 1. 432 100 > 431 200. So 5 432 100 is greatest. |
| 10 | (1) 4 567 800 | Original ÷ 100 = 45 678, so original = 45 678 × 100 = 4 567 800. Multiplying by 100 adds two zeros (or shifts decimal two places right). |
Corrected Answers for Q3, Q4, Q8:
| Question | Corrected Answer | Reasoning |
|---|---|---|
| 3 | (3) 4 600 000 | Hundred thousands digit is 5; ten thousands digit 6 ≥ 5, so round up to 4 600 000 |
| 4 | (2) 3 657 901 | The 5 is in the ten thousands place (50 000) |
| 8 | (2) 2 450 000 | 2 450 000 rounds up to 2 500 000 (boundary case) |
SECTION B: Short-Answer Questions (10 marks)
11. (a) Six million, two hundred and five thousand and seventy [1 mark]
Teaching note: Break into periods—million period: "six million"; thousand period: "two hundred and five thousand"; unit period: "seventy". The zero in hundreds place is read as part of "and" connection.
(b) 2 506 040 [1 mark]
Teaching note: Two million = 2 000 000; five hundred and six thousand = 506 000; forty = 40. Combined: 2 000 000 + 506 000 + 40 = 2 506 040. Common error: writing 2 560 040 or 2 500 640.
12. 5 678 901, 5 687 910, 5 768 910, 5 876 190 [2 marks]
Step-by-step method:
- All numbers start with 5 million, so compare hundred thousands digits: 6, 6, 7, 8
- 6 < 7 < 8, so 5 876 190 is largest
- For the two with 6: compare ten thousands digits: 7 and 8
- 7 < 8, so 5 678 901 < 5 687 910
Marking: 1 mark for correct order with 5 678 901 first and 5 876 190 last; 1 mark for complete correct order. Deduct 1 mark if any number misplaced.
13. Smallest: 102 479; Largest: 974 210; Sum: 1 076 689 [2 marks]
Method:
- Largest 6-digit number: Place largest digits in highest place values → 9 7 4 2 1 0 = 974 210
- Smallest 6-digit number: Place smallest non-zero digit first (1), then remaining in ascending order → 1 0 2 4 7 9 = 102 479 (cannot start with 0)
- Sum: 974 210 + 102 479 = 1 076 689
Working shown:
974 210
+ 102 479
---------
1 076 689
Marking: 1 mark for correct smallest and largest identified; 1 mark for correct sum. Common error: putting 0 first for smallest (invalid 6-digit number would become 5-digit).
14. (a) 5 677 500 [1 mark]
Teaching note: Rounding to nearest thousand: look at hundreds digit. To round to 5 678 000, the smallest number has hundreds digit 5 (rounds up). So 5 677 500 ≤ number < 5 678 500. Smallest = 5 677 500.
(b) 5 678 499 [1 mark]
Teaching note: Largest number that rounds to 5 678 000 must have hundreds digit 4 (rounds down), so maximum is 5 678 499. The next number, 5 678 500, would round to 5 679 000.
15. (a) 250 000 [1 mark]
Method: From A (3 000 000) to E (4 000 000) is 1 000 000. There are 4 equal intervals: A-B, B-C, C-D, D-E. Each interval = 1 000 000 ÷ 4 = 250 000.
(b) 3 625 000 [1 mark]
Method: Point X is halfway between C (3 500 000) and D (3 750 000). X = 3 500 000 + 125 000 = 3 625 000. Or: (3 500 000 + 3 750 000) ÷ 2 = 7 250 000 ÷ 2 = 3 625 000.
SECTION C: Word Problems (30 marks)
16. (a) Dover Town [2 marks]
Method: Round each population to nearest million:
- Ang Mo Town: 3 456 780 → 3 000 000
- Bedok Town: 2 890 456 → 3 000 000
- Clementi Town: 1 567 234 → 2 000 000 (hundred thousands digit 5, round up)
- Dover Town: 987 654 → 1 000 000 (hundred thousands digit 9 ≥ 5, round up to 1 million)
Marking: 1 mark for correct identification; 1 mark for showing rounding reasoning. Accept if student shows 987 654 rounds to 1 000 000.
(b) 6 347 236 [2 marks]
Working:
3 456 780
+ 2 890 456
-----------
6 347 236
Step-by-step:
- Ones: 0 + 6 = 6
- Tens: 8 + 5 = 13, write 3, carry 1
- Hundreds: 7 + 4 + 1 = 12, write 2, carry 1
- Thousands: 6 + 0 + 1 = 7
- Ten thousands: 5 + 9 = 14, write 4, carry 1
- Hundred thousands: 4 + 8 + 1 = 13, write 3, carry 1
- Millions: 3 + 2 + 1 = 6
Marking: 1 mark for correct method shown; 1 mark for correct final answer. If only answer given with no working, award 1 mark.
(c) 1 902 802 [2 marks]
Working:
2 890 456
- 987 654
-----------
1 902 802
Step-by-step:
- Ones: 6 − 4 = 2
- Tens: 5 − 5 = 0
- Hundreds: 4 − 6, need to borrow: 14 − 6 = 8
- Thousands: (now 9) − 7 = 2? Wait, check: after borrowing, 0 becomes 9, 9 − 7 = 2. Actually: 890 456 − 987 654 requires careful alignment.
- Better: 2 890 456 − 987 654 = 2 890 456 − 900 000 − 87 654 = 1 990 456 − 87 654 = 1 902 802
Verification: 987 654 + 1 902 802 = 2 890 456 ✓
Marking: 1 mark for correct method; 1 mark for correct answer.
17. (a) 9 690 packets [2 marks]
Working:
January: 8 456 packets
February: 8 456 + 1 234 = 9 690 packets
Method: "1 234 more than January" means addition: 8 456 + 1 234 = 9 690.
Marking: 1 mark for correct operation identified; 1 mark for correct calculation.
(b) 19 380 packets [2 marks]
Working:
March: 2 × 9 690 = 19 380 packets
Method: "Twice as many as February" means multiplication by 2: 9 690 × 2 = 19 380.
Step-by-step: 9 690 × 2 = (9 000 × 2) + (600 × 2) + (90 × 2) = 18 000 + 1 200 + 180 = 19 380.
Marking: 1 mark for using February's answer; 1 mark for correct calculation.
(c) 37 526 packets [2 marks]
Working:
8 456 (January)
+ 9 690 (February)
+ 19 380 (March)
-----------
37 526
Step-by-step:
- 8 456 + 9 690 = 18 146
- 18 146 + 19 380 = 37 526
Marking: 1 mark for adding all three months; 1 mark for correct total. Award follow-through marks if student uses their previous answers correctly even if those were wrong.
18. (a) 4 679 010 [2 marks]
Working:
3 456 789 (factory)
+ 876 543 (machinery)
+ 345 678 (salaries)
-----------
4 679 010
Step-by-step:
- 3 456 789 + 876 543 = 4 333 332
- 4 333 332 + 345 678 = 4 679 010
Marking: 1 mark for adding all three items; 1 mark for correct total.
(b) 320 990 [2 marks]
Working:
5 000 000
- 4 679 010
-----------
320 990
Method: Remaining = 5 000 000 − 4 679 010 = 320 990
Marking: 1 mark for correct subtraction setup; 1 mark for correct answer. Award follow-through from part (a).
(c) 40 123.75 or $40 123.75 [2 marks]
Working:
Remaining after buying second factory: 320 990 − 3 456 789 = negative!
Correction—re-reading the problem: The problem says "buy a second factory at the same price"—this would exceed remaining money. Re-interpret: Perhaps he uses the original $5 000 000 for everything including second factory? Or "remaining money" refers to after all original expenses?
Clarified interpretation: After expenses in (a), he has 3 456 789, which he cannot afford.
Alternative valid interpretation for syllabus alignment: The question intends: After all expenses in (a), split remaining among 8 workers. Let me recalculate: 320 990 ÷ 8 = 40 123.75. But this gives decimal.
Revised scenario for clean numbers (marking guidance): If student correctly does 320 990 ÷ 8 = 40 123 remainder 6, or recognizes he cannot afford second factory, award marks for correct reasoning.
Expected student approach for P5: Likely intended: 320 990 ÷ 8 = 40 123 remainder 6 or $40 123.75 if decimals known.
Teaching note: At P5, students may express as "40 123 with remainder 40 123.75. Accept either with clear working.
Working:
320 990 ÷ 8
320 000 ÷ 8 = 40 000
990 ÷ 8 = 123 remainder 6
Total: 40 123 remainder 6, or 40 123.75
Marking: 1 mark for dividing remaining money by 8; 1 mark for correct quotient. Accept remainder form or decimal.
19. (a) 8 754 320 [2 marks]
Method: For greatest 7-digit number using 3, 5, 7, 0, 2, 8, 4, arrange digits in descending order: 8, 7, 5, 4, 3, 2, 0 → 8 754 320
Teaching note: 8 is largest, so millions digit is 8. Continue with next largest: 7, 5, 4, 3, 2, 0.
(b) 2 034 578 [2 marks]
Method: For smallest 7-digit number, smallest non-zero digit first (2), then remaining in ascending order: 0, 3, 4, 5, 7, 8 → 2 034 578
Critical point: Cannot start with 0 (would be 6-digit number). So 2 goes first.
(c) 6 719 742 [2 marks]
Working:
8 754 320
- 2 034 578
-----------
6 719 742
Step-by-step subtraction with borrowing as needed.
Verification: 2 034 578 + 6 719 742 = 8 754 320 ✓
Marking: 1 mark for correct greatest and smallest from (a) and (b); 1 mark for correct difference. Follow-through marks apply.
20. (a) Direct route A to C: 3 456 000 m [2 marks]
Method: Compare two routes from A to C:
- Direct: A → C = 3 456 000 m
- Via B: A → B → C = 2 345 000 + 1 876 000 = 4 221 000 m
Since 3 456 000 < 4 221 000, the direct route is shortest.
Explanation: The direct route avoids the detour through B. Even though A-B-C might seem intuitive, the diagonal path is shorter (similar to triangle inequality concept—though not named at P5, students can compare totals).
Marking: 1 mark for identifying direct route with correct distance; 1 mark for comparison explanation.
(b) 6 442 000 m or 6 442 km [2 marks]
Working:
2 345 000 (A→B)
+ 1 876 000 (B→C)
+ 1 234 000 (C→D)
+ 987 000 (D→A)
-----------
6 442 000 m
Step-by-step:
- 2 345 000 + 1 876 000 = 4 221 000
- 1 234 000 + 987 000 = 2 221 000
- 4 221 000 + 2 221 000 = 6 442 000
Marking: 1 mark for adding all four segments; 1 mark for correct total. Accept in metres or kilometres (6 442 km).
(c) The driver is correct [2 marks]
Working:
A → D → C = 987 000 + 1 234 000 = 2 221 000 m
Direct A → C = 3 456 000 m
2 221 000 < 3 456 000
Conclusion: The route via D is shorter by 1 235 000 m (or 1 235 km).
Marking: 1 mark for calculating both routes correctly; 1 mark for correct comparison and conclusion. Award partial credit if student calculates one route correctly but makes arithmetic error on other.
SUMMARY MARKSCHEME
| Section | Marks | Key Skills Tested |
|---|---|---|
| A: MCQ | 10 | Place value, rounding, comparing, number patterns |
| B: Short Answer | 10 | Number words, ordering, place value reasoning, number lines |
| C: Word Problems | 30 | Multi-step problems, data interpretation, route optimization, forming numbers with constraints |
Total: 50 marks
Difficulty distribution: Easy (10 marks), Medium (20 marks), Challenging (20 marks)
Time estimate: 75 minutes allows approximately:
- Section A: 10 minutes (1 min/question)
- Section B: 15 minutes (3 min/question)
- Section C: 45 minutes (9 min/question)
- Review: 5 minutes
END OF ANSWER KEY