Primary 6 PSLE Mathematics Semestral Assessment 2 (End of Year) Paper 4

TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

Answer Key and Marking Scheme

Paper: SA2 End-of-Year Practice Paper - Whole Numbers (Version 4 of 5) Total Marks: 60

Marking codes used below: M = method mark, A = accuracy mark. Award M marks for correct method even if the final answer is wrong due to a later slip.

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Section A (Questions 1-10, 20 marks)

Question 1 (2 marks) Answer: $7{,}306{,}405$

• M1: Correctly identifies millions period (7 million).
• A1: Full numeral $7{,}306{,}405$ correct. Teaching note: Split the words into periods: seven million | three hundred six thousand | four hundred five. Each period is written as a 3-digit group. Common mistake: writing $7{,}360{,}405$ (swapping hundreds and tens within the thousand period).

Question 2 (2 marks) Answer: $40{,}000$

• M1: Identifies the digit 4 as being in the ten thousands place.
• A1: States the value as $40{,}000$ (not just "ten thousands"). Teaching note: The value of a digit = digit $\times$ its place value. Here $4 \times 10{,}000 = 40{,}000$. Common mistake: answering "ten thousands", which is the place, not the value.

Question 3 (2 marks) Answer: $6{,}300{,}000$

• M1: Looks at the digit after the hundred-thousands place ($7$).
• A1: Rounds up correctly to $6{,}300{,}000$. Teaching note: To round to the nearest hundred thousand, check the ten-thousands digit. Since $7 \geq 5$, round the $2$ up to $3$. Common mistake: rounding to the nearest million ($6{,}000{,}000$) instead.

Question 4 (2 marks) Answer: $5{,}260{,}490 > 5{,}260{,}094 > 5{,}206{,}409 > 5{,}206{,}049$

• M1: Compares digits place by place from the left (all have 5 millions; compare hundred thousands: 2 vs 2; compare ten thousands: 6 vs 6 vs 0 vs 0; continue to smaller places).
• A1: All four numbers in fully correct descending order. Teaching note: Line up place values in a table mentally. Numbers sharing the first three digits differ first at the ten-thousands place: $6$ beats $0$, so the $5{,}260{,}\ldots$ pair comes first.

Question 5 (2 marks) Answer: $129$

• M1: Applies order of operations: $32 \times 3 = 96$ and $60 \div 4 = 15$ before adding/subtracting: $48 + 96 - 15$.
• A1: $129$. Teaching note: Multiplication and division come before addition and subtraction, worked left to right. Common mistake: computing left to right without priority, e.g. $(48+32)\times3$, giving $378$.

Question 6 (2 marks) Answer: $1, 2, 3, 4, 6, 12$

• M1: Lists factors of both numbers correctly (24: 1, 2, 3, 4, 6, 8, 12, 24; 36: 1, 2, 3, 4, 6, 9, 12, 18, 36).
• A1: Identifies all six common factors. Teaching note: A common factor divides both numbers exactly. List factors in pairs ($1\times24$, $2\times12$, $3\times8$, $4\times6$) so none are missed. Common mistake: stopping at $6$ and forgetting $12$.

Question 7 (2 marks) Answer: $36$

• M1: Lists multiples of 9 (9, 18, 27, 36, ...) and multiples of 12 (12, 24, 36, ...) OR uses prime factors.
• A1: Smallest common multiple $36$. Teaching note: The smallest common multiple is the first number appearing in both lists. Checking multiples of the larger number (12, 24, 36...) against 9's times table speeds this up.

Question 8 (2 marks) Answer: Quotient = $55$, Remainder = $25$

• M1: Finds $45 \times 55 = 2{,}475$ and subtracts: $2{,}500 - 2{,}475 = 25$.
• A1: Quotient $55$, remainder $25$. Teaching note: The remainder must be smaller than the divisor ($25 < 45$). Check: $45 \times 55 + 25 = 2{,}500$. Common mistake: quotient 56 (then remainder would be negative, showing 56 is too big).

Question 9 (2 marks) Answer: $6{,}000 \times 30 = 180{,}000$

• M1: Rounds $5{,}982 \to 6{,}000$ and $31 \to 30$.
• A1: Estimated product $180{,}000$. Teaching note: Estimation replaces numbers with close "round" values to make mental calculation possible. The estimate is near the exact value $185{,}442$. Common mistake: rounding $31$ to $31$ or $0$ instead of the nearest ten.

Question 10 (2 marks) Answer: $26{,}820$ cookies

• M1: Total packets: $1{,}250 + 985 = 2{,}235$.
• A1: $2{,}235 \times 12 = 26{,}820$ cookies. Teaching note: Two-step problem: first combine (total packets), then multiply (packets $\times$ cookies per packet). Always end with the unit asked for: cookies.

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Section B (Questions 11-15, 15 marks)

Question 11 (3 marks) Answer: (a) $109$ and $221$ (b) Multiply the previous term by 2 and add 3.

• M1: Detects the pattern, e.g. differences $7, 14, 28$ (doubling) or $4 \times 2 + 3 = 11$, $11 \times 2 + 3 = 25$, $25 \times 2 + 3 = 53$.
• A1: $53 \times 2 + 3 = 109$; $109 \times 2 + 3 = 221$.
• A1(b): Rule stated correctly in words. Teaching note: Test whether differences form their own pattern. Here differences double each time, which matches the rule "$\times 2$ then $+3$". Either description of the rule is acceptable if mathematically correct.

Question 12 (3 marks) Answer: (a) $17{,}325$ flyers (b) $17{,}000$

• M1: Recognises total = rate $\times$ time: $385 \times 45$.
• A1: $385 \times 45 = 15{,}400 + 1{,}925 = 17{,}325$ flyers.
• A1(b): Rounds to nearest thousand: $17{,}000$ (since $325 < 500$). Teaching note: "Per minute" means each minute gives 385 flyers, so multiply. For rounding, look only at the hundreds digit ($3$), so round down.

Question 13 (3 marks) Answer: (a) $32$ cm (b) $7$ pieces

• M1: Recognises the longest equal length = highest common factor of 96 and 128.
• A1: $\text{HCF}(96, 128) = 32$, so each piece is $32$ cm.
• A1(b): $96 \div 32 = 3$ pieces and $128 \div 32 = 4$ pieces; total $3 + 4 = 7$ pieces. Teaching note: "Longest possible equal pieces with nothing left over" signals HCF. Check: $32$ divides both $96$ and $128$ exactly, and no larger number does ($128 = 4 \times 32$; $96 = 3 \times 32$). Common mistake: using LCM instead of HCF.

Question 14 (3 marks) Answer: $16$

• M1: Works backwards with inverse operations: $25 \times 4 = 100$.
• M1: $100 + 28 = 128$.
• A1: $128 \div 8 = 16$. Teaching note: Undo each step in reverse order using the opposite operation: divide became multiply ($\times 4$), subtract became add ($+28$), multiply became divide ($\div 8$). Check forwards: $16 \times 8 = 128$; $128 - 28 = 100$; $100 \div 4 = 25$. ✓

Question 15 (3 marks) Answer: $1{,}080$ children

• M1: Draws units: children = $1$ unit, adults = $3$ units, so total $= 4$ units.
• M1: $4$ units $= 4{,}320$, so $1$ unit $= 4{,}320 \div 4 = 1{,}080$.
• A1: Children $= 1{,}080$. *Teaching

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TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

Answer Key and Marking Scheme

Paper: SA2 End-of-Year Practice Paper - Whole Numbers (Version 4 of 5) Total Marks: 60

Marking codes used below: M = method mark, A = accuracy mark. Award M marks for correct method even if the final answer is wrong due to a later slip.

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Section A (Questions 1-10, 20 marks)

Question 1 (2 marks) Answer: $7{,}306{,}405$

• M1: Correctly identifies millions period (7 million).
• A1: Full numeral $7{,}306{,}405$ correct. Teaching note: Split the words into periods: seven million | three hundred six thousand | four hundred five. Each period is written as a 3-digit group. Common mistake: writing $7{,}360{,}405$ (swapping hundreds and tens within the thousand period).

Question 2 (2 marks) Answer: $40{,}000$

• M1: Identifies the digit 4 as being in the ten thousands place.
• A1: States the value as $40{,}000$ (not just "ten thousands"). Teaching note: The value of a digit = digit $\times$ its place value. Here $4 \times 10{,}000 = 40{,}000$. Common mistake: answering "ten thousands", which is the place, not the value.

Question 3 (2 marks) Answer: $6{,}300{,}000$

• M1: Looks at the digit after the hundred-thousands place ($7$).
• A1: Rounds up correctly to $6{,}300{,}000$. Teaching note: To round to the nearest hundred thousand, check the ten-thousands digit. Since $7 \geq 5$, round the $2$ up to $3$. Common mistake: rounding to the nearest million ($6{,}000{,}000$) instead.

Question 4 (2 marks) Answer: $5{,}260{,}490 > 5{,}260{,}094 > 5{,}206{,}409 > 5{,}206{,}049$

• M1: Compares digits place by place from the left (all have 5 millions; compare hundred thousands, then ten thousands, then smaller places).
• A1: All four numbers in fully correct descending order. Teaching note: Numbers sharing the first three digits differ first at the ten-thousands place: $6$ beats $0$, so the $5{,}260{,}\ldots$ pair comes first.

Question 5 (2 marks) Answer: $129$

• M1: Applies order of operations: $32 \times 3 = 96$ and $60 \div 4 = 15$ before adding/subtracting: $48 + 96 - 15$.
• A1: $129$. Teaching note: Multiplication and division come before addition and subtraction. Common mistake: computing left to right without priority, e.g. $(48+32)\times3$, giving $378$.

Question 6 (2 marks) Answer: $1, 2, 3, 4, 6, 12$

• M1: Lists factors of both numbers correctly (24: 1, 2, 3, 4, 6, 8, 12, 24; 36: 1, 2, 3, 4, 6, 9, 12, 18, 36).
• A1: Identifies all six common factors. Teaching note: List factors in pairs ($1\times24$, $2\times12$, $3\times8$, $4\times6$) so none are missed. Common mistake: stopping at $6$ and forgetting $12$.

Question 7 (2 marks) Answer: $36$

• M1: Lists multiples of 9 (9, 18, 27, 36, ...) and multiples of 12 (12, 24, 36, ...).
• A1: Smallest common multiple $36$. Teaching note: Checking multiples of the larger number (12, 24, 36...) against 9's times table speeds this up.

Question 8 (2 marks) Answer: Quotient = $55$, Remainder = $25$

• M1: Finds $45 \times 55 = 2{,}475$ and subtracts: $2{,}500 - 2{,}475 = 25$.
• A1: Quotient $55$, remainder $25$. Teaching note: The remainder must be smaller than the divisor ($25 < 45$). Check: $45 \times 55 + 25 = 2{,}500$. Common mistake: quotient 56 (then the remainder would be negative).

Question 9 (2 marks) Answer: $6{,}000 \times 30 = 180{,}000$

• M1: Rounds $5{,}982 \to 6{,}000$ and $31 \to 30$.
• A1: Estimated product $180{,}000$. Teaching note: The estimate is near the exact value $185{,}442$. Common mistake: rounding $31$ to something other than the nearest ten.

Question 10 (2 marks) Answer: $26{,}820$ cookies

• M1: Total packets: $1{,}250 + 985 = 2{,}235$.
• A1: $2{,}235 \times 12 = 26{,}820$ cookies. Teaching note: Two-step problem: combine packets first, then multiply by cookies per packet. End with the unit asked for: cookies.

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Section B (Questions 11-15, 15 marks)

Question 11 (3 marks) Answer: (a) $109$ and $221$ (b) Multiply the previous term by 2 and add 3.

• M1: Detects the pattern, e.g. differences $7, 14, 28$ (doubling), or $4 \times 2 + 3 = 11$, $11 \times 2 + 3 = 25$, $25 \times 2 + 3 = 53$.
• A1: $53 \times 2 + 3 = 109$; $109 \times 2 + 3 = 221$.
• A1(b): Rule stated correctly in words. Teaching note: Test whether differences form their own pattern. Either mathematically correct description of the rule is acceptable.

Question 12 (3 marks) Answer: (a) $17{,}325$ flyers (b) $17{,}000$

• M1: Recognises total = rate $\times$ time: $385 \times 45$.
• A1: $385 \times 45 = 17{,}325$ flyers.
• A1(b): Rounds to nearest thousand: $17{,}000$ (since $325 < 500$). Teaching note: "Per minute" means multiply. For rounding, look only at the hundreds digit ($3$), so round down.

Question 13 (3 marks) Answer: (a) $32$ cm (b) $7$ pieces

• M1: Recognises the longest equal length = highest common factor of 96 and 128.
• A1: $\text{HCF}(96, 128) = 32$, so each piece is $32$ cm.
• A1(b): $96 \div 32 = 3$ pieces and $128 \div 32 = 4$ pieces; total $3 + 4 = 7$ pieces. Teaching note: "Longest possible equal pieces with nothing left over" signals HCF. Common mistake: using LCM instead of HCF.

Question 14 (3 marks) Answer: $16$

• M1: Works backwards with inverse operations: $25 \times 4 = 100$.
• M1: $100 + 28 = 128$.
• A1: $128 \div 8 = 16$. Teaching note: Undo each step in reverse order using the opposite operation. Check forwards: $16 \times 8 = 128$; $128 - 28 = 100$; $100 \div 4 = 25$. ✓

Question 15 (3 marks) Answer: $1{,}080$ children

• M1: Draws units: children = $1$ unit, adults = $3$ units, so total $= 4$ units.
• M1: $4$ units $= 4{,}320$, so $1$ unit $= 4{,}320 \div 4 = 1{,}080$.
• A1: Children $= 1{,}080$. Teaching note: "3 times as many adults as children" means adults + children = 4 equal units, not 3. Common mistake: dividing by 3 instead of 4.

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Section C (Questions 16-20, 25 marks)

Question 16 (5 marks) Answer: $252$ stamps

• Working: At first, Ravi $= 3$ units, Ken $= 1$ unit. After giving away $84$: Ravi $= 3u - 84$, Ken $= u + 84$.
• Since they are equal: $3u - 84 = u + 84 \Rightarrow 2u = 168 \Rightarrow u = 84$.
• Ravi at first $= 3 \times 84 = 252$ stamps.
• M1: Sets up units/model correctly. M1: Forms the equality or uses the difference ($3u - u = 2u$ represents the $84 \times 2 = 168$ transferred difference). A1: Correct units found. A1: $252$ stamps with working shown. A1: Answer stated with units. Teaching note: When Ravi gives 84 stamps, he loses 84 while Ken gains 84, closing a gap of 168. Common mistake: setting $2u = 84$ instead of $2u = 168$.

Question 17 (5 marks) Answer: $38$ motorcycles

• Working: Let cars $= c$, motorcycles $= m$. Then $c + m = 120$ and $4c + 2m = 404$.
• Divide the wheel equation by 2: $2c + m = 202$. Subtract $c + m = 120$: $c = 82$ cars.
• Motorcycles $= 120 - 82 = 38$.
• Check: $82 \times 4 + 38 \times 2 = 328 + 76 = 404$. ✓
• M1: Forms both equations/relationships. M1: Solves the system correctly. A1: $82$ cars found. A1: $38$ motorcycles. A1: Verification shown. Teaching note: Alternative (assumption method): assume all 120 are motorcycles $= 240$ wheels; shortfall $= 404 - 240 = 164$; each car adds 2 wheels, so cars $= 164 \div 2 = 82$.

Question 18 (5 marks) Answer: (a) $62$ seats (b) Row $22$

• (a) Working: Row 12 $= 18 + (12 - 1) \times 4 = 18 + 44 = 62$ seats.
• (b) Working: $102 = 18 + (n - 1) \times 4 \Rightarrow 84 = (n - 1) \times 4 \Rightarrow n - 1 = 21 \Rightarrow n = 22$.
• M1: Recognises the pattern rule (start 18, add 4). A1: $62$ seats. M1: Sets up equation for part (b). A1: $n - 1 = 21$. A1: Row $22$. Teaching note: Row $n$ has $18 + (n-1) \times 4$ seats because Row 1 already has 18, so only $n - 1$ additions of 4 occur. Common mistake: using $n$ instead of $n - 1$, giving Row 23.

Question 19 (5 marks) Answer: (a) $8$ passengers (b) $15$ trips

• (a) Working: $638 \div 45 = 14$ remainder $8$, since $45 \times 14 = 630$ and $638 - 630 = 8$. Last trip carries $8$ passengers.
• (b) Working: $14$ full trips carry $630$ people; the remaining $8$ people need one more trip. Least number of trips $= 14 + 1 = 15$.
• M1: Divides correctly. A1: Remainder $8$ identified as last-trip load. M1: Interprets the remainder as requiring an extra trip. A1: $15$ trips. A1: Clear reasoning linking (a) and (b). Teaching note: Always round UP for "how many trips/vehicles needed" questions — you cannot leave people behind. Common mistake: answering 14 trips (rounding down) or 14.4 trips.

Question 20 (5 marks) Answer: $807$ boxes

• Working:
  • Total toys produced: $1{,}275 \times 16 = 20{,}400$ toys.
  • After removing defective toys: $20{,}400 - 225 = 20{,}175$ toys.
  • Number of boxes: $20{,}175 \div 25 = 807$ boxes.
  • Check: $807 \times 25 = 20{,}175$. ✓
• M1: Multiplies daily production by days. A1: $20{,}400$. M1: Subtracts defectives. A1: $20{,}175$. M1+A1: Divides by 25 correctly to get $807$ boxes. Teaching note: This is a chained multi-step problem — label each intermediate result so no step is skipped. Common mistake: subtracting 225 before multiplying, or dividing by 25 before removing defectives.

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End of Answer Key