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Higher June 2025 Paper 2 Q21
21 The diagram shows a cuboid.
Diagram NOT accurately drawn
The cuboid measures \(3x\) cm by \(2x\) cm by \(y\) cm
The volume of the cuboid is 1014 cm3 The total surface area of the cuboid is \(A\) cm2
Show that \(A = 12x^2 + \dfrac{1690}{x}\)
You must show all the stages of your working.
(3)
Mark scheme
Mark scheme Scheme Marks \(3x \times 2x \times y = 1014\) oe or \(6x^2y = 1014\) oe or \(x^2y = 169\) oe M1 \(2 \times 3x \times y + 2 \times 2x \times y + 2 \times 3x \times 2x\) oe or \(2 \times (3xy + 2xy + 6x^2)\) oe or \(6xy + 4xy + 12x^2\) oe or \(10xy + 12x^2\) oe M1 Using \(y = \dfrac{1014}{6x^2}\left(= \dfrac{169}{x^2}\right)\) in formula for surface area to obtain correct expression
eg
(SA =) \(2 \times 5x \times \dfrac{169}{x^2} + 2 \times 6x^2 = 12x^2 + \dfrac{1690}{x}\)
or equating their surface area equations
eg \(10xy + 12x^2 = 12x^2 + \dfrac{1690}{x}\) leading to \(x^2y = 169\) oe
Working required
Answer: Shown
A1 (3) (3 marks)
Notes M1: for an equation for volume in terms of \(x\) and \(y\)
M1: (indep) for a correct expression for the surface area
NB May not explicitly see the surface area expression in terms of \(y\) eg
\(2 \times 3x \times \dfrac{169}{x^2} + 2 \times 2x \times \dfrac{169}{x^2} + 2 \times 3x \times 2x\) oe
May be fully substituted using \(y = \dfrac{169}{x^2}\) oe
A1: dep on M2 For completing the ‘show that’ by clearly showing the stages that lead to the given expression for the surface area.
Higher June 2025 Paper 1R Q10
10 The diagram shows two water containers. One is a cuboid and one is a cylinder.
Diagram NOT accurately drawn
The cuboid measures 35 cm by 28 cm by 20 cm The surface of the water in the cuboid is 9 cm above the base of the cuboid.
The cylinder has a radius of 10 cm and a height of 33 cm The cylinder is completely full of water.
Izzy is going to pour all the water from the cylinder into the cuboid.
Show that the cuboid will not be completely full of water.
(3)
Mark scheme
Mark scheme Scheme Marks (volume of water =) \(9 \times 35 \times 28 (= 8820)\)or (total volume of cuboid =) \(20 \times 35 \times 28 (= 19600)\)or (volume of space =) \((20 - 9) \times 35 \times 28 (= 10780)\) M1 \(\pi \times 10^2 \times 33\) (= \(3300\pi\) or 10367(.25...)) oe M1 (total volume of water =) “8820” + “10367(.25...)” (= 19187(.25...)) (difference between volumes of both solids =) “19600” − “10367(.25…)”(= 9232(.74…)) (volume not filled =) “19600” – “8820” – “10367(.25…)” (=412(.74…))Working required Answer: Shown A1 (3) (3 marks)
Notes M1: for a method to find a relevant volume for the cuboid
M1: (indep) for a method to find the volume of the cylinder, accept a volume in the range 10362 to 10368.6
Allow 3.14… or \(\dfrac{22}{7}\) for \(\pi\)
A1: correct workings with accurate figures eg
Value 1 Value 2 10780 10367(.25…) accept 10362 to 10372 19600 19187(.25…) accept 19182 to 19192 8820 9232(.74…) accept 9228 to 9238 412(.74…) or 413 accept 408 to 418 none needed
Higher June 2025 Paper 1 Q7
7 Here is a solid cylinder.
Diagram NOT accurately drawn
The radius of the cylinder is 8 cm The height of the cylinder is \(h\) cm
The volume of the cylinder is 3892 cm3
Work out the value of \(h\) Give your answer correct to one decimal place.
(3)
Mark scheme
Mark scheme Scheme Marks \(3892 = \pi \times 8^2 \times h\) or \(\pi \times 8^2 \left(= 64\pi = 201\ldots\right)\) M1 \((h =)\dfrac{3892}{\pi \times 8^2}\) oe eg 3892 ÷ 64 = 60.8… and 60.8… ÷ π M1 Correct answer scores full marks (unless from obvious incorrect working) Answer: 19.4A1 (3) (3 marks)
Notes M1: allow use of 3.14… or \(\dfrac{22}{7}\) for \(\pi\)
M1: allow use of 3.14… or \(\dfrac{22}{7}\) for \(\pi\)
A1: allow 19.3 – 19.4
Higher November 2024 Paper 2 Q6
6 The diagram shows a solid triangular prism.
Diagram NOT accurately drawn
Work out the total surface area of the triangular prism.
(3)
Mark scheme
Mark scheme Scheme Marks 8 × 6 (= 48) 0.5 × 8 × 6 (= 24 ) 15 × 8 (= 120) 15 × 6 (= 90) 15 × 10 (= 150) M1 0.5 × 8 × 6 (= 24 ) (× 2 (= 48) ) oe 15 × 8 (= 120) 15 × 6 (= 90) 15 × 10 (= 150) (measurements with intention to add for the 2nd M mark) Surface area = “120” + “90” + “150” + “24” + “24” [allow “120” + “90” + “150” + “48” + “48”] M1 Correct answer scores full marks (unless from obvious incorrect working) Answer: 408A1 (3) (3 marks)
Notes M1: For a correct method to find the areas of 2 different faces (ie not 2 triangles) allow 8 × 6 as one area (allow with incorrect areas for this mark)
M1: for adding together 4 or 5 values for area (condone 48 as 1 or 2 areas) at least 3 of which are from a correct method NB: (6 + 8 + 10) × 15 (sum must be seen) is 3 faces but only award this if clearly not intended to be the volume – eg by the addition of the area of a triangular end.
A1: cao SCB2 for an answer of 456 if no other marks awarded
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