Primary 6 PSLE Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

Answer Key with Marking Scheme - Whole Numbers (Version 1 of 5)

TuitionGoWhere Practice Paper (AI)

Marking guide: M1 = method mark, A1 = accuracy mark. Award marks for correct method even if the final answer is wrong due to a slip.

Section A (2 marks each)

1. Seven million = 7,000,000; two hundred and five thousand = 205,000; forty = 40. 7,000,000 + 205,000 + 40 = 7,205,040 (2) Teaching note: Line up the periods (millions, thousands, ones). "Two hundred and five thousand" is 205,000, not 200,005. Common mistake: writing 7,205,040 as 7,250,040.

2. Read the places from the right: ones, tens, hundreds, thousands, ten thousands, hundred thousands, millions. In 8,046,391 the digit 4 is in the ten thousands place, so its value is 40,000 (2). Common mistake: answering "ten thousands" (the place name) instead of the value, or 4,000.

3. To round to the nearest thousand, look at the hundreds digit. 5,648,295: the hundreds digit is 2 (less than 5), so round down. 5,648,295 ≈ 5,648,000 (2) Teaching note: Rounding never changes the digits to the left of the rounding place; everything after becomes zeros.

4. Compare digit by digit from the left. 4,503,216 and 4,503,612 match through the thousands place; compare hundreds: 2 < 6, so 4,503,216 comes first. Both are smaller than 4,530,126 because their ten-thousands digit is 0 while 4,530,126 has 3. Increasing order: 4,503,216, 4,503,612, 4,530,126 (2)

5. Follow the order of operations: multiply before subtracting. 25 × 400 = 10,000; 10,000 − 2,000 = 8,000 (2) Common mistake: working left to right (25 × 400 = 10,000 is the same here, but subtracting first would wrongly give 23 × 400 = 9,200).

Section B (3 marks each)

6. (a) Factors of 48 (pairs: 1×48, 2×24, 3×16, 4×12, 6×8): 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 (1) (b) Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Numbers appearing in both lists: 1, 2, 3, 4, 6, 12 (2) Teaching note: A factor divides a number exactly (no remainder). List factors in pairs so none are missed. Common mistake: forgetting 1 or the number itself.

7. The lights blink together again after the first common multiple of 8 and 12, i.e. the LCM. Multiples of 8: 8, 16, 24, 32, ... Multiples of 12: 12, 24, 36, ... LCM = 24. They will blink together again after 24 seconds (3: M1 listing/multiples idea, M1 LCM identified, A1 answer). Teaching note: "Together again" signals a common multiple, not a difference.

8. 5,894 rounds to 6,000 (nearest thousand); 62 rounds to 60 (nearest ten). Estimate: 6,000 ÷ 60 = 100 (3: M1 both roundings, M1 division, A1). Teaching note: Estimation replaces numbers with friendly rounded values to check roughly how big an answer should be. The exact answer (about 95) is near 100, so the estimate is reasonable.

9. 185 ÷ 40 = 4 remainder 25. Four buses seat 160 pupils, leaving 25 pupils without seats, so one more bus is required. Smallest number of buses =

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TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

Answer Key with Marking Scheme - Whole Numbers (Version 1 of 5)

TuitionGoWhere Practice Paper (AI)

Marking guide: M = method mark, A = accuracy mark. Award method marks even if the final answer is wrong due to a slip.

Section A (2 marks each)

1. Seven million = 7,000,000; two hundred and five thousand = 205,000; forty = 40. 7,000,000 + 205,000 + 40 = 7,205,040 (2) Teaching note: "Two hundred and five thousand" is 205,000, not 200,005.

2. In 8,046,391 the digit 4 is in the ten thousands place, so its value is 40,000 (2). Common mistake: giving the place name "ten thousands" or 4,000 instead of the value.

3. Look at the hundreds digit: 5,648,295 → 2 < 5, round down. 5,648,295 ≈ 5,648,000 (2)

4. Compare digit by digit. 4,503,216 vs 4,503,612 match to the thousands place; hundreds: 2 < 6. Both are less than 4,530,126 (ten-thousands digit 0 < 3). Increasing order: 4,503,216, 4,503,612, 4,530,126 (2)

5. Multiply first: 25 × 400 = 10,000; 10,000 − 2,000 = 8,000 (2) Common mistake: subtracting before multiplying.

Section B (3 marks each)

6. (a) Factors of 48 (pairs 1×48, 2×24, 3×16, 4×12, 6×8): 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 (1) (b) Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Common factors: 1, 2, 3, 4, 6, 12 (2) Teaching note: List factors in pairs so none are missed; don't forget 1 and the number itself.

7. "Together again" → first common multiple (LCM) of 8 and 12. Multiples of 8: 8, 16, 24, ...; multiples of 12: 12, 24, ... LCM = 24 → 24 seconds (M1 listing multiples, M1 LCM, A1 answer)

8. 5,894 ≈ 6,000 (nearest thousand); 62 ≈ 60 (nearest ten). Estimate: 6,000 ÷ 60 = 100 (M1 both roundings, M1 division, A1) Note: exact value ≈ 95, so the estimate is reasonable.

9. 185 ÷ 40 = 4 remainder 25. Four buses seat only 160 pupils, so one more bus is needed. Smallest number of buses = 5 (M1 division, M1 interpretation of remainder, A1) Common mistake: answering 4 and ignoring the leftover 25 pupils.

10. (a) 1,208 + 986 + 1,320 + 1,095 = 4,609 pupils (2: M1 addition, A1) (b) Largest: Hillcrest 1,320; smallest: Eastbrook 986. 1,320 − 986 = 334 pupils (1)

Section C (4 marks each)

11. Target: 500,000. Produced: 128,450 + 96,780 = 225,230. 500,000 − 225,230 = 274,770 buttons (M1 total produced, M1 subtraction setup, A1 answer, A1 units/statement)

12. Ravi : Sean = 4 : 1; 5 units = 175, so 1 unit = 35. (a) Ravi = 4 × 35 = 140 marbles (2) (b) After giving away 30: Ravi = 140 − 30 = 110; Sean = 35. Difference = 110 − 35 = 75 marbles (2)

13. Amy + Ben = 305; Amy − Ben = 53. Amy = (305 + 53) ÷ 2 = 179; Ben = 305 − 179 = 126. In the end: Amy = 179 − 42 = 137; Ben = 126 + 25 = 151. Ben has 151 − 137 = $14 more than Amy (M1 finding both amounts, M1 end amounts, A1 comparison, A1 units)

14. Muffins: 250 ÷ 6 = 41 boxes (remainder 4). Cookies: 300 ÷ 8 = 37 boxes (remainder 4). Money: 41 × $3 =$123; 37 × $4 =$148. Total = 123 + 148 = $271 (M1 boxes of each, M1 money per type, A1 total, A1 units) Common mistake: rounding remainders up — incomplete boxes are not sold.

15. Pattern: add 5 each time; nth number = 5n + 2. (a) 10th number = 5 × 10 + 2 = 52 (2) (b) 5n + 2 = 102 → 5n = 100 → n = 20. It is the 20th number (2)

Section D (5 marks each)

16. At first: Box A = 3 units, Box B = 1 unit. Moving 60 beads changes the difference by 120. Difference at first = 2 units; difference after = 20. 2 units = 20 + 60 + 60 = 140 → 1 unit = 70. Box B = 70 beads; Box A = 3 × 70 = 210 beads (M1 model/units, M1 difference change, A1 Box A, A1 Box B, A1 check/statement) Check: 210 − 60 = 150; 70 + 60 = 130; 150 − 130 = 20 ✓

17. Let the number of pupils be p. 5p + 8 = 7p − 24 → 8 + 24 = 7p − 5p → 2p = 32 → p = 16 pupils (M1 equation/model, M1 solving, A1 answer, plus check: 5 × 16 + 8 = 88 stickers; 7 × 16 = 112, short 24 ✓)

18. Let cars = c, motorcycles = m. c + m = 30; 4c + 2m = 96 → 2c + m = 48. Subtract: c = 18, so m = 30 − 18 = 12 motorcycles (M1 equations, M1 elimination/supposition, A1 answer, A1 check: 18×4 + 12×2 = 72 + 24 = 96 ✓)

19. Son now = s; Mr Lim = 4s. In 12 years: 4s + 12 = 2(s + 12) → 4s + 12 = 2s + 24 → 2s = 12 → s = 6 years old (M1 expressions, M1 equation, A1 answer, A1 check: Mr Lim 24; in 12 yrs: 36 and 18 ✓)

20. Finally all three have equal amounts. Total = 570, so each ends with 570 ÷ 3 = 190. Nina received $25 → at first: 190 − 25 = 165. Mei received$15 → at first: 190 − 15 = 175. (Check: 175 = 165 + 10 ✓) Oli gave away 25 + 15 = 40 → at first: 190 + 40 = $230 (M1 final equal share, M1 working back for Nina/Mei, M1 Oli's give-away, A1 answer, A1 check: 175 + 165 + 230 = 570 ✓)

END OF ANSWER KEY