Primary 6 PSLE Mathematics Whole Numbers Quiz

Primary 6 PSLE Mathematics Quiz - Whole Numbers - Answer Key (Version 1)

Total Marks: 50 | Section A: 5 marks | Section B: 10 marks | Section C: 15 marks | Section D: 20 marks

Teaching notes are written for students new to the topic. Method marks and common mistakes are flagged where useful.

Answer 1.

Answer: 2,705,040 (1 mark)

• Key idea: Read large numbers in groups of three digits from the right: millions, thousands, ones.
• "Two million" gives 2,000,000. "Seven hundred and five thousand" gives 705,000. "Forty" gives 040 in the last three places.
• Combine: 2,000,000 + 705,000 + 40 = 2,705,040.
• Common mistake: writing 2,705,400 (mixing up tens and hundreds in the last group). Always check the final three digits carefully.

Answer 2.

Answer: 40,000 (1 mark)

• Key idea: The value of a digit depends on its place, not just the digit itself.
• Label the places of 8,346,215 from the right: 5 ones, 1 tens, 2 hundreds, 6 thousands, 4 ten-thousands, 3 hundred-thousands, 8 millions.
• The digit 4 sits in the ten-thousands place, so its value is 4 × 10,000 = 40,000.
• Common mistake: answering "4" or "ten thousands" instead of the value 40,000.

Answer 3

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Primary 6 PSLE Mathematics Quiz - Whole Numbers - Answer Key (Version 1)

Total Marks: 50 | Section A: 5 marks | Section B: 10 marks | Section C: 15 marks | Section D: 20 marks

Teaching notes are written for students new to the topic. Method marks and common mistakes are flagged where useful.

Answer 1.

Answer: 2,705,040 (1 mark)

• Key idea: Read large numbers in groups of three digits from the right: millions, thousands, ones.
• "Two million" gives 2,000,000. "Seven hundred and five thousand" gives 705,000. "Forty" gives 040 in the last three places.
• Combine: 2,000,000 + 705,000 + 40 = 2,705,040.
• Common mistake: writing 2,705,400 (mixing up tens and hundreds in the last group). Always check the final three digits carefully.

Answer 2.

Answer: 40,000 (1 mark)

• Key idea: The value of a digit depends on its place, not just the digit itself.
• Label the places of 8,346,215 from the right: 5 ones, 1 tens, 2 hundreds, 6 thousands, 4 ten-thousands, 3 hundred-thousands, 8 millions.
• The digit 4 sits in the ten-thousands place, so its value is 4 × 10,000 = 40,000.
• Common mistake: answering "4" or "ten thousands" instead of the value 40,000.

Answer 3.

Answer: 5,847,000 (1 mark)

• Key idea: To round to the nearest thousand, look only at the hundreds digit.
• 5,847,296 has 7 thousands and a hundreds digit of 2. Since 2 < 5, round down.
• 5,847,296 ≈ 5,847,000.
• Common mistake: rounding using the thousands digit instead of the hundreds digit.

Answer 4.

Answer: 24 (1 mark)

• Key idea: The LCM is the smallest number that appears in the times tables of both numbers.
• Multiples of 8: 8, 16, 24, 32, ...
• Multiples of 12: 12, 24, 36, ...
• The first common multiple is 24.
• Quick check: 24 ÷ 8 = 3 and 24 ÷ 12 = 2, both exact.

Answer 5.

Answer: 28 (1 mark)

• Key idea: Follow the order of operations - division and multiplication first, left to right, then subtraction and addition.
• 36 − 18 ÷ 3 × 2 + 4
• = 36 − 6 × 2 + 4
• = 36 − 12 + 4
• = 24 + 4 = 28
• Common mistake: doing addition before subtraction (getting 36 − 12 + 4 wrong as 36 − 16 = 20). Work strictly left to right for − and +.

Answer 6.

Answer: 66 (2 marks)

• Key idea: If a number leaves remainder 3 when divided by 7 and by 9, then (number − 3) is divisible by both 7 and 9.
• LCM of 7 and 9 = 63 (they share no common factors).
• So the number is of the form 63 × k + 3.
• k = 1 gives 63 + 3 = 66, which is already a two-digit number.
• Check: 66 ÷ 7 = 9 remainder 3; 66 ÷ 9 = 7 remainder 3. ✓
• Method marks: 1 mark for recognising the LCM idea, 1 mark for the correct answer with checks.

Answer 7.

Answer: Common factors 1, 2, 3, 4, 6, 12; HCF = 12 (2 marks)

• Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
• Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
• Common factors: 1, 2, 3, 4, 6, 12.
• Highest common factor: 12.
• Common mistake: forgetting 1 as a common factor, or stopping the list early and missing 12.

Answer 8.

Rule: multiply the previous term by 3, then subtract 1. Next term: 284 (2 marks)

• Test differences first: 11 − 4 = 7, 32 − 11 = 21, 95 − 32 = 63. Differences are not constant, so try a multiply rule.
• 4 × 3 − 1 = 11 ✓; 11 × 3 − 1 = 32 ✓; 32 × 3 − 1 = 95 ✓.
• Rule: multiply by 3, then subtract 1.
• Next term: 95 × 3 − 1 = 285 − 1 = 284.
• Award 1 mark for a correct stated rule and 1 mark for 284.

Answer 9.

Answer: 344,999 (2 marks)

• Key idea: Numbers rounding to 340,000 (nearest ten thousand) run from 335,000 up to 344,999.
• The largest such number is 344,999, because 344,999 is still less than 345,000 (the halfway point).
• Common mistake: writing 349,999 (which rounds to 350,000) or 345,000 (which rounds up to 350,000).
• Tip: always locate the halfway point between the two neighbouring rounded values.

Answer 10.

Answer: 7532 (2 marks)

• Key idea: An even number must end in an even digit. The only even digit available is 2, so 2 must be the last digit.
• Place the remaining digits 7, 5, 3 in descending order in front: 7, then 5, then 3.
• Largest possible 4-digit even number: 7532.
• Common mistake: forming 7532's near-rivals like 7352 or 5732 - always fix the units digit first, then maximise the thousands digit.

Answer 11.

Answer: There are 17 cars in the car park. (3 marks)

• Suppose all 25 vehicles were motorcycles: 25 × 2 = 50 wheels.
• Extra wheels counted: 84 − 50 = 34 wheels.
• Each car adds 4 − 2 = 2 extra wheels compared with a motorcycle.
• Number of cars: 34 ÷ 2 = 17 cars (and 25 − 17 = 8 motorcycles).
• Check: 17 × 4 + 8 × 2 = 68 + 16 = 84 ✓
• Marks: 1 for setting up the assumption, 1 for correct working, 1 for the final statement.

Answer 12.

Answer: The printer finishes printing the report at 11:00 a.m. (3 marks)

• Time needed: 2,850 ÷ 38 = 75 minutes.
• Convert: 75 minutes = 1 hour 15 minutes.
• Start 9:45 a.m. + 1 hour = 10:45 a.m.; 10:45 a.m. + 15 minutes = 11:00 a.m.
• Common mistake: dividing the wrong way (38 ÷ 2,850) or misreading 75 minutes as 1 hour 75 minutes.

Answer 13.

Answer: There are 11 pupils and Mrs Nadia has 67 stickers. (3 marks)

• Compare the two situations: giving 7 instead of 5 uses 2 more stickers per pupil.
• The shortfall changes from "12 left over" to "short of 10", a swing of 12 + 10 = 22 stickers.
• Number of pupils: 22 ÷ 2 = 11 pupils.
• Stickers: 11 × 5 + 12 = 55 + 12 = 67 stickers.
• Check: 11 × 7 = 77, and 77 − 67 = 10 short ✓
• Marks: 1 for the total difference of 22, 1 for 11 pupils, 1 for 67 stickers.

Answer 14.

Answer: He gets 90 packets of pens. (3 marks)

• Total pens: 45 × 16 = 720 pens.
• Packets: 720 ÷ 8 = 90 packets.
• Alternative: each box of 16 pens makes exactly 2 packets, so 45 × 2 = 90 packets.
• Marks: 1 for total pens, 1 for division, 1 for the final statement with units.

Answer 15.

Answer: In 3 years' time, Mr Ravi will be exactly 4 times as old as his son. (3 marks)

• Age gap stays constant: 41 − 8 = 33 years.
• When Mr Ravi is 4 times as old, the gap equals 3 units of the son's age.
• Son's age then: 33 ÷ 3 = 11 years old.
• Years from now: 11 − 8 = 3 years.
• Check: in 3 years, Mr Ravi is 44 and his son is 11; 44 = 4 × 11 ✓
• Marks: 1 for using the constant age gap, 1 for finding the son's future age, 1 for the answer.

Answer 16.

(a) Row 31 has exactly 94 chairs. (b) The total is 1,519 chairs. (4 marks)

• Row n has 4 + 3 × (n − 1) chairs.
• Set 4 + 3 × (n − 1) = 94 → 3 × (n − 1) = 90 → n − 1 = 30 → n = 31.
• Total chairs from Row 1 to Row 31: add the first and last terms, then multiply by the number of rows and divide by 2:
• Total = (4 + 94) × 31 ÷ 2 = 98 × 31 ÷ 2 = 49 × 31 = 1,519 chairs.
• Marks: 1 for the pattern expression, 1 for Row 31, 1 for a valid summation method, 1 for 1,519.

Answer 17.

Answer: Mrs Goh bought 25 pens and 20 files. (4 marks)

• Let all 45 items be pens: cost = 45 × 6 = $270.
• Shortfall: $370 −$270 = $100.
• Each file costs $11 −$6 = $5 more than a pen.
• Files: 100 ÷ 5 = 20 files; pens: 45 − 20 = 25 pens.
• Check: 25 × 6 + 20 × 11 = 150 + 220 = $370 ✓
• Marks: 1 for the assumption method, 1 for the number of files, 1 for the number of pens, 1 for verification or clear statements.

Answer 18.

Answer: Each boy had $101 at first. (4 marks)

• Both boys started with the same amount, say $S.
• After spending: Siva has S − 74; Kumar has S − 20.
• Kumar has 3 times Siva's amount: S − 20 = 3 × (S − 74).
• Expand: S − 20 = 3S − 222 → 222 − 20 = 3S − S → 202 = 2S → S = $101.
• Check: Siva has 101 − 74 = 27; Kumar has 101 − 20 = 81; 81 = 3 × 27 ✓
• Alternative model method: the difference in spending is 74 − 20 = $54, which creates the extra 2 units; 54 ÷ 2 = 27 left for Siva; 27 + 74 =$101.
• Marks: 1 for modelling the relationship, 1 for solving, 1 for $101, 1 for checking or a clear model.

Answer 19.

(a) Point Q stands for 505,000. (b) 505,000 rounded to the nearest ten thousand is 510,000. (4 marks)

• (a) The minor ticks fall halfway between the labelled major ticks, so each small step is 5,000. Q sits one minor tick past 500,000: 500,000 + 5,000 = 505,000.
• (b) Halfway between 500,000 and 510,000 is 505,000. By the rounding rule, a number exactly at the halfway point rounds up, so 505,000 → 510,000.
• Common mistakes: reading Q as 502,500 (forgetting each minor step is 5,000), or rounding 505,000 down to 500,000.
• Marks: 2 marks for part (a), 2 marks for part (b).

Answer 20.

Answer: The smallest possible mystery number is 117. (4 marks)

• Numbers leaving remainder 3 when divided by 6: 3, 9, 15, 21, 27, ... (add 6 each time).
• Numbers leaving remainder 5 when divided by 8: 5, 13, 21, 29, ... (add 8 each time).
• First common value: 21 (check: 21 ÷ 6 = 3 r 3; 21 ÷ 8 = 2 r 5 ✓).
• The number must repeat every LCM of 6 and 8, which is 24: 21, 45, 69, 93, 117, ...
• Since the number must be greater than 100, the smallest valid value is 117.
• Check: 117 ÷ 6 = 19 remainder 3 ✓; 117 ÷ 8 = 14 remainder 5 ✓
• Marks: 1 for listing either pattern, 1 for finding 21 as the base solution, 1 for using the LCM of 24, 1 for 117 with checks.

End of Answer Key - Version 1