Primary 6 PSLE Mathematics Measurement Quiz

Primary 6 PSLE Mathematics Quiz - Measurement: Answer Key

Version 1 of 5 | Total Marks: 70

Teaching notes are included so a student new to the topic can follow the method, not just the final answer.

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Section A (Questions 1–5)

Answer 1.

Circumference = $44$ cm [2]

• Key idea: Circumference of a circle is the distance around it: $C = \pi d$.
• Working: $C = \frac{22}{7} \times 14 = 22 \times 2 = 44$ cm.
• Marking: M1 for using $C = \pi d$ with correct substitution; A1 for $44$ cm.
• Common mistake: Using $C = \pi r^2$ (that is the area formula) or halving the diameter wrongly. Always check whether the question gives diameter or radius.

Answer 2.

Area = $314 \text{ m}^2$ [2]

• Key idea: Area of a circle is $A = \pi r^2$, where $r$ is the radius.
• Working: $A = 3.14 \times 10^2 = 3.14 \times 100 = 314 \text{ m}^2$.
• Marking: M1 for $A = \pi r^2$ with $r = 10$; A1 for $314 \text{ m}^2$.
• Common mistake: Squaring the diameter instead of the radius, giving $1256$. Halve the diameter first if you are given the diameter.

Answer 3.

(a) $350$ (b) $2.75$ [2]

• Key idea: Unit conversion uses multiplication or division by the link factor.
• (a) $1$ m = $100$ cm, so $3.5 \times 100 = 350$ cm. [1]
• (b) $1000$ mL = $1$ L, so $2750 \div 1000 = 2.75$ L. [1]
• Common mistake: Dividing instead of multiplying when changing to a smaller unit. Smaller units need more of them, so multiply going from m to cm.

Answer 4.

Volume = $160 \text{ cm}^3$ [2]

• Key idea: Volume of a cuboid is $V = l \times w \times h$.
• Working: $V = 8 \times 5 \times 4 = 160 \text{ cm}^3$.
• Marking: M1 for correct formula/substitution; A1 for $160 \text{ cm}^3$ with units.
• Common mistake: Omitting units, or adding the sides instead of multiplying.

Answer 5.

Area = $113.04 \text{ cm}^2$ [2]

• Key idea: A quarter circle is $\frac{1}{4}$ of a full circle, so $A = \frac{1}{4}\pi r^2$.
• Working: $A = \frac{1}{4} \times 3.14 \times 12^2 = \frac{1}{4} \times 3.14 \times 144 = 36 \times 3.14 = 113.04 \text{ cm}^2$.
• Marking: M1 for $\frac{1}{4}\pi r^2$; A1 for $113.04 \text{ cm}^2$.
• Common mistake: Using the diameter in place of the radius, or forgetting the $\frac{1}{4}$.

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Section B (Questions 6–10)

Answer 6.

Radius = $10$ cm [3]

• Key idea: Work backwards from $C = 2\pi r$: divide the circumference by $2\pi$ to get the radius.
• Working: $2\pi r = 62.8$, so $r = 62.8 \div (2 \times 3.14) = 62.8 \div 6.28 = 10$ cm.
• Marking: M1 for $C = 2\pi r$; M1 for dividing $62.8$ by $6.28$; A1 for $10$ cm.
• Common mistake: Dividing by $\pi$ only once (forgetting the factor of $2$), which gives $20$ cm.

Answer 7.

Height = $10$ cm [3]

• Key idea: Since $V = l \times w \times h$, the missing dimension is $h = V \div (l \times w)$.
• Working: Base area $= 12 \times 6 = 72 \text{ cm}^2$. Height $= 720 \div 72 = 10$ cm.
• Marking: M1 for base area $72 \text{ cm}^2$; M1 for $720 \div 72$; A1 for $10$ cm.
• Common mistake: Dividing by only one dimension (e.g. $720 \div 12$). Both known dimensions must be divided out.

Answer 8.

Edge = $7$ cm [3]

• Key idea: For a cube, $V = e \times e \times e = e^3$, so the edge is the cube root of the volume: $e = \sqrt[3]{V}$.
• Working: Test perfect cubes: $7 \times 7 \times 7 = 343$. So $e = \sqrt[3]{343} = 7$ cm.
• Marking: M1 for recognising $e^3 = 343$; M1 for attempting $7 \times 7 \times 7$; A1 for $7$ cm.
• Common mistake: Taking the square root instead of the cube root ($\sqrt{343} \

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Primary 6 PSLE Mathematics Quiz - Measurement: Answer Key

Version 1 of 5 | Total Marks: 70

Teaching notes are included so a student new to the topic can follow the method, not just the final answer.

---

Section A (Questions 1–5)

Answer 1.

Circumference = $44$ cm [2]

• Key idea: Circumference is the distance around a circle: $C = \pi d$.
• Working: $C = \frac{22}{7} \times 14 = 22 \times 2 = 44$ cm.
• Marking: M1 for $C = \pi d$ with correct substitution; A1 for $44$ cm.
• Common mistake: Using the area formula $\pi r^2$, or halving the diameter wrongly. Check whether diameter or radius is given.

Answer 2.

Area = $314 \text{ m}^2$ [2]

• Key idea: Area of a circle is $A = \pi r^2$.
• Working: $A = 3.14 \times 10^2 = 3.14 \times 100 = 314 \text{ m}^2$.
• Marking: M1 for $A = \pi r^2$ with $r = 10$; A1 for $314 \text{ m}^2$.
• Common mistake: Squaring the diameter instead of the radius (gives $1256$).

Answer 3.

(a) $350$ (b) $2.75$ [2]

• Key idea: Unit conversion multiplies or divides by the link factor.
• (a) $1$ m = $100$ cm, so $3.5 \times 100 = 350$ cm. [1]
• (b) $1000$ mL = $1$ L, so $2750 \div 1000 = 2.75$ L. [1]
• Common mistake: Dividing when changing to a smaller unit. Smaller units need more of them, so multiply going from m to cm.

Answer 4.

Volume = $160 \text{ cm}^3$ [2]

• Key idea: Volume of a cuboid is $V = l \times w \times h$.
• Working: $V = 8 \times 5 \times 4 = 160 \text{ cm}^3$.
• Marking: M1 for correct formula/substitution; A1 for $160 \text{ cm}^3$ with units.
• Common mistake: Omitting units, or adding sides instead of multiplying.

Answer 5.

Area = $113.04 \text{ cm}^2$ [2]

• Key idea: A quarter circle is $\frac{1}{4}$ of a full circle: $A = \frac{1}{4}\pi r^2$.
• Working: $A = \frac{1}{4} \times 3.14 \times 12^2 = \frac{1}{4} \times 3.14 \times 144 = 113.04 \text{ cm}^2$.
• Marking: M1 for $\frac{1}{4}\pi r^2$; A1 for $113.04 \text{ cm}^2$.
• Common mistake: Forgetting the $\frac{1}{4}$, or using the diameter as the radius.

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Section B (Questions 6–10)

Answer 6.

Radius = $10$ cm [3]

• Key idea: Work backwards from $C = 2\pi r$: divide the circumference by $2\pi$.
• Working: $r = 62.8 \div (2 \times 3.14) = 62.8 \div 6.28 = 10$ cm.
• Marking: M1 for $C = 2\pi r$; M1 for dividing by $6.28$; A1 for $10$ cm.
• Common mistake: Dividing by $\pi$ only once (forgetting the factor of $2$), giving $20$ cm.

Answer 7.

Height = $10$ cm [3]

• Key idea: Since $V = l \times w \times h$, then $h = V \div (l \times w)$.
• Working: Base area $= 12 \times 6 = 72 \text{ cm}^2$. Height $= 720 \div 72 = 10$ cm.
• Marking: M1 for base area $72 \text{ cm}^2$; M1 for $720 \div 72$; A1 for $10$ cm.
• Common mistake: Dividing by only one dimension (e.g. $720 \div 12$).

Answer 8.

Edge = $7$ cm [3]

• Key idea: For a cube, $V = e^3$, so the edge is the cube root of the volume.
• Working: $7 \times 7 \times 7 = 343$, so $e = 7$ cm.
• Marking: M1 for recognising $e^3 = 343$; M1 for testing $7 \times 7 \times 7$; A1 for $7$ cm.
• Common mistake: Taking the square root instead of the cube root ($\sqrt{343} \approx 18.5$ is wrong here).

Answer 9.

Perimeter = $70.84$ cm [3]

• Key idea: The outer boundary consists of three straight sides plus the curved arc; side $BC$ (the diameter) is inside the figure and is not counted.
• Working: Straight parts: $AB + AD + DC = 20 + 12 + 20 = 52$ cm. Arc $= \frac{1}{2} \times \pi \times d = \frac{1}{2} \times 3.14 \times 12 = 18.84$ cm. Perimeter $= 52 + 18.84 = 70.84$ cm.
• Marking: M1 for identifying the three straight sides ($52$ cm); M1 for the semicircle arc $18.84$ cm; A1 for $70.84$ cm.
• Common mistake: Including side $BC$ ($12$ cm) in the perimeter, which double-counts an internal edge.

Answer 10.

Shaded area = $86 \text{ cm}^2$ [3]

• Key idea: Shaded area = area of square − area of inscribed circle. The circle's diameter equals the square's side.
• Working: Square $= 20 \times 20 = 400 \text{ cm}^2$. Circle $= 3.14 \times 10^2 = 314 \text{ cm}^2$. Shaded $= 400 - 314 = 86 \text{ cm}^2$.
• Marking: M1 for square area $400$; M1 for circle area $314$; A1 for $86 \text{ cm}^2$.
• Common mistake: Using radius $20$ instead of $10$ for the circle.

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Section C (Questions 11–15)

Answer 11.

(a) $30000 \text{ cm}^3$ (b) $30$ litres [4]

• (a) $V = 40 \times 30 \times 25 = 30000 \text{ cm}^3$. [2]
• (b) $30000 \div 1000 = 30$ L. [2]
• Key idea: Divide cubic centimetres by $1000$ to convert to litres.
• Common mistake: Multiplying by $1000$ instead of dividing in part (b).

Answer 12.

$10$ minutes [4]

• Key idea: Find the volume of water needed, convert to litres, then divide by the rate.
• Working: Water volume $= 50 \times 40 \times 20 = 40000 \text{ cm}^3 = 40$ L. Time $= 40 \div 4 = 10$ minutes.
• Marking: M1 for volume $40000 \text{ cm}^3$; M1 for converting to $40$ L; M1 for dividing by rate; A1 for $10$ minutes.
• Common mistake: Using the tank's full height ($30$ cm) instead of the water depth ($20$ cm).

Answer 13.

Area = $78.5 \text{ m}^2$ [4]

• Key idea: First find the radius from the circumference, then use $A = \pi r^2$.
• Working: $r = 31.4 \div (2 \times 3.14) = 31.4 \div 6.28 = 5$ m. $A = 3.14 \times 5^2 = 3.14 \times 25 = 78.5 \text{ m}^2$.
• Marking: M1 for finding $r = 5$; M1 for $A = \pi r^2$; A2 for $78.5 \text{ m}^2$.
• Common mistake: Squaring $31.4$ directly instead of finding the radius first.

Answer 14.

Height = $3$ cm [4]

• Key idea: Melting preserves volume: volume of cube = volume of cuboid.
• Working: Cube volume $= 6 \times 6 \times 6 = 216 \text{ cm}^3$. Cuboid base area $= 9 \times 8 = 72 \text{ cm}^2$. Height $= 216 \div 72 = 3$ cm.
• Marking: M1 for cube volume $216$; M1 for base area $72$; M1 for dividing; A1 for $3$ cm.
• Common mistake: Adding edges or forgetting that melting/recasting keeps the volume unchanged.

Answer 15.

(a) Perimeter = $72$ cm (b) Area = $308 \text{ cm}^2$ [4]

• Key idea: Semicircle perimeter = half-circumference + diameter; semicircle area = half the full circle area.
• (a) Arc $= \frac{1}{2} \times \frac{22}{7} \times 28 = 44$ cm. Perimeter $= 44 + 28 = 72$ cm. [2]
• (b) $A = \frac{1}{2} \times \frac{22}{7} \times 14^2 = \frac{1}{2} \times \frac{22}{7} \times 196 = 308 \text{ cm}^2$. [2]
• Common mistake: Omitting the diameter in the perimeter (arc only gives $44$ cm), or using radius $28$.

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Section D (Questions 16–20)

Answer 16.

New depth = $9.5$ cm [5]

• Key idea: The fully submerged block pushes up water equal to the block's volume spread over the tank's base area.
• Working: Block volume $= 15 \times 10 \times 5 = 750 \text{ cm}^3$. Base area $= 25 \times 20 = 500 \text{ cm}^2$. Rise $= 750 \div 500 = 1.5$ cm. New depth $= 8 + 1.5 = 9.5$ cm.
• Marking: M1 block volume $750$; M1 base area $500$; M1 rise $1.5$; A1 new depth $9.5$; A1 units.
• Common mistake: Forgetting to add the rise to the original depth of $8$ cm.

Answer 17.

Shaded area = $126 \text{ cm}^2$ [5]

• Key idea: Shaded area = rectangle − quarter circle.
• Working: Rectangle $= 20 \times 14 = 280 \text{ cm}^2$. Quarter circle $= \frac{1}{4} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154 \text{ cm}^2$. Shaded $= 280 - 154 = 126 \text{ cm}^2$.
• Marking: M1 rectangle $280$; M1 quarter-circle setup; A1 quarter circle $154$; M1 subtraction; A1 $126 \text{ cm}^2$.
• Common mistake: Using $20$ as the radius; the radius is $AD = 14$ cm.

Answer 18.

(a) $192$ blocks (b) $0 \text{ cm}^3$ [5]

• Key idea: Fit whole blocks along each dimension, then compare volumes.
• (a) Along length: $24 \div 3 = 8$; width: $18 \div 3 = 6$; height: $12 \div 3 = 4$. Blocks $= 8 \times 6 \times 4 = 192$. [3]
• (b) Box volume $= 24 \times 18 \times 12 = 5184 \text{ cm}^3$. Blocks' volume $= 192 \times 27 = 5184 \text{ cm}^3$. Unfilled space $= 5184 - 5184 = 0 \text{ cm}^3$. [2]
• Common mistake: Dividing total volumes to get blocks — this works here only because every dimension is a multiple of $3$; always check whole blocks fit along each edge.

Answer 19.

(a) $240$ litres (b) $9.20$ a.m. [5]

• Key idea: Use the net rate (inflow minus outflow) and divide capacity by it.
• (a) Capacity $= 80 \times 60 \times 50 = 240000 \text{ cm}^3 = 240$ L. [2]
• (b) Net rate $= 9 - 6 = 3$ L/min. Time $= 240 \div 3 = 80$ minutes $= 1$ h $20$ min. From $8$ a.m., the tank fills at $9.20$ a.m. [3]
• Common mistake: Using only Tap A's rate ($240 \div 9$), which ignores the drain.

Answer 20.

Shaded area = $78.5 \text{ cm}^2$ [5]

• Key idea: Shaded area = large semicircle − two small semicircles. Two identical small semicircles make one full circle.
• Working: Large semicircle $= \frac{1}{2} \times 3.14 \times 10^2 = 157 \text{ cm}^2$. Two small semicircles $=$ one circle of radius $5$: $3.14 \times 5^2 = 78.5 \text{ cm}^2$. Shaded $= 157 - 78.5 = 78.5 \text{ cm}^2$.
• Marking: M1 large semicircle setup; A1 $157$; M1 combining both small semicircles into one circle; A1 $78.5$; A1 final $78.5 \text{ cm}^2$.
• Common mistake: Treating the two small semicircles as two full circles, subtracting $157$ twice.

END OF ANSWER KEY