This exam tips covers Exam Tips within Volume of Cuboids & Prisms for GCSE Mathematics. Revise Volume of Cuboids & Prisms in Geometry & Measures for GCSE Mathematics with 20 exam-style questions and 12 flashcards. This topic shows up very often in GCSE exams, so students should be able to explain it clearly, not just recognise the term. It is section 8 of 8 in this topic. Treat this as a marking guide for what examiners are looking for, not just a fact list.
Topic position
Section 8 of 8
Practice
20 questions
Recall
12 flashcards
Exam Tips
- A cylinder is just a circular prism - same formula principle applies!
- If the cross-section is compound (L-shape, T-shape), split it into rectangles first
- Check units carefully - if given in different units, convert first
Practice Questions
Q1 Find the volume of a cuboid: length 5 cm, width 4 cm, height 3 cm [1 mark]
[1] 5 × 4 × 3 = 60 cm³
Q2 A triangular prism has cross-section area 15 cm² and length 6 cm. Find the volume. [1 mark]
[1] 15 × 6 = 90 cm³
Q3 Find the volume of a cylinder with radius 4 cm and height 6 cm (1 dp) [2 marks]
[1] V = πr²h = π × 4² × 6 = π × 96
[1] V = 301.6 cm³
Q4 A triangular prism: base 4 cm, height 3 cm, length 8 cm. Find volume. [3 marks]
[1] Area of triangle = ½ × 4 × 3 = 6 cm²
[1] Volume = 6 × 8
[1] V = 48 cm³
Q5 Calculate the volume of a cuboid with length 10 cm, width 6 cm, and height 4 cm. [2 marks]
[1] V = l × w × h = 10 × 6 × 4
[1] = 240 cm³
Examiner says: Volume is ALWAYS in cubic units (cm³, m³, mm³) - never forget the ³!
Q6 Find the volume of a cylinder with radius 5 cm and height 8 cm. Give your answer to 1 decimal place. [3 marks]
[1] V = πr²h
[1] V = π × 5² × 8 = π × 25 × 8 = 200π
[1] = 628.3 cm³ (1 dp)
Q7 A triangular prism has a triangle cross-section with base 8 cm and height 5 cm. The prism is 12 cm long. Find its volume. [3 marks]
[1] Cross-sectional area = ½ × 8 × 5 = 20 cm²
[1] Volume = area × length = 20 × 12
[1] = 240 cm³
Q8 A water tank is a cuboid measuring 2 m by 1.5 m by 0.8 m. Find its volume in: (a) m³, (b) litres (1 m³ = 1000 litres). [4 marks]
[1] (a) V = 2 × 1.5 × 0.8
[1] = 2.4 m³
[1] (b) Volume in litres = 2.4 × 1000
[1] = 2400 litres
Q9 A cylindrical can has diameter 10 cm and height 15 cm. Find its volume to the nearest cm³. [4 marks]
[1] Radius = 10 ÷ 2 = 5 cm
[1] V = πr²h = π × 5² × 15
[1] = π × 25 × 15 = 375π
[1] = 1178 cm³ (nearest cm³)
Examiner says: When given diameter, HALVE IT to get radius before using V = πr²h!
Q10 A cuboid has volume 360 cm³. Its length is 10 cm and width is 6 cm. Find the height. [4 marks]
[1] V = l × w × h, so 360 = 10 × 6 × h
[1] 360 = 60h
[1] h = 360 ÷ 60
[1] h = 6 cm
Q11 Find the volume of a cylinder with radius 6 cm and height 10 cm. Leave your answer in terms of π. [3 marks]
[1] V = πr²h
[1] V = π × 6² × 10 = π × 36 × 10
[1] = 360π cm³
Q12 A prism has a trapezoidal cross-section with parallel sides 7 cm and 11 cm, and height 4 cm. The prism is 15 cm long. Find its volume. [5 marks]
[2] Cross-sectional area = ½ × (7 + 11) × 4 = ½ × 18 × 4 = 36 cm²
[2] Volume = area × length = 36 × 15
[1] = 540 cm³
Q13 A cylinder has volume 500 cm³ and height 8 cm. Find its radius. Give your answer to 2 decimal places. [5 marks]
[1] V = πr²h, so 500 = πr² × 8
[1] πr² = 500 ÷ 8 = 62.5
[1] r² = 62.5 ÷ π = 19.894...
[1] r = √19.894
[1] r = 4.46 cm (2 dp)
Q14 An L-shaped prism has a cross-section made of two rectangles: one 10 cm × 3 cm and one 4 cm × 5 cm. The prism is 20 cm long. Find its volume. [6 marks]
[2] Rectangle 1 area = 10 × 3 = 30 cm²
[1] Rectangle 2 area = 4 × 5 = 20 cm²
[1] Total cross-sectional area = 30 + 20 = 50 cm²
[1] Volume = 50 × 20
[1] = 1000 cm³
Q15 The radius of a cylinder is increased by 10% and the height decreased by 10%. Find the percentage change in volume. [6 marks]
[1] Let original r = r, h = h. V = πr²h
[1] New radius = 1.1r, new height = 0.9h
[2] New V = π(1.1r)²(0.9h) = π × 1.21r² × 0.9h = 1.089πr²h
[1] Change = (1.089πr²h - πr²h)/(πr²h) × 100
[1] = 0.089 × 100 = 8.9% increase
Examiner says: Radius is squared, so 10% increase → 1.1² = 1.21 factor!
Q16 A metal cuboid measures 50 cm × 30 cm × 20 cm. The metal has density 7.8 g/cm³. Find the mass of the cuboid in kg. [6 marks]
[1] Volume = 50 × 30 × 20 = 30,000 cm³
[2] Mass = density × volume = 7.8 × 30,000
[1] = 234,000 g
[1] Convert to kg: 234,000 ÷ 1000
[1] = 234 kg
Q17 A cylindrical water tank has diameter 1.2 m and height 2 m. Water flows in at 50 litres per minute. How long does it take to fill the tank? (1 m³ = 1000 litres) [7 marks]
[1] Radius = 1.2 ÷ 2 = 0.6 m
[2] V = πr²h = π × 0.6² × 2 = 0.72π ≈ 2.262 m³
[1] Capacity in litres = 2.262 × 1000 = 2262 litres
[2] Time = 2262 ÷ 50 = 45.24 minutes
[1] = 45 minutes 14 seconds (or 45.2 minutes)
Q18 A cuboid has length (x + 3) cm, width x cm, and height (x - 1) cm. Its volume is 60 cm³. Form and solve an equation to find x. [7 marks]
[1] V = (x+3) × x × (x-1) = 60
[2] x(x+3)(x-1) = 60, expand: x(x² + 3x - x - 3) = x(x² + 2x - 3)
[2] x³ + 2x² - 3x = 60, so x³ + 2x² - 3x - 60 = 0
[1] By trial: x = 3 works (27 + 18 - 9 - 60 = -24, try x=4: 64+32-12-60=24, try x=3.5...)
[1] By systematic trial or calculator: x ≈ 3.39 cm (or exact: solve cubic)
Examiner says: Cubic equations are tricky - use calculator graph or systematic trial!
Q19 A hollow cylindrical pipe has outer radius 5 cm, inner radius 4 cm, and length 50 cm. Find the volume of metal needed to make the pipe. Give your answer to the nearest cm³. [8 marks]
[2] Outer cylinder volume = π × 5² × 50 = 1250π cm³
[2] Inner cylinder volume (hollow) = π × 4² × 50 = 800π cm³
[2] Volume of metal = 1250π - 800π = 450π
[2] = 1414 cm³ (nearest cm³)
Examiner says: Hollow shapes: subtract the inner volume from the outer volume!
Q20 A cuboid box has a square base of side x cm and height h cm. Its volume must be 1000 cm³. Express h in terms of x, and find the value of x that minimises the surface area. [8 marks]
[1] V = x²h = 1000, so h = 1000/x²
[2] Surface area S = 2x² + 4xh = 2x² + 4x(1000/x²) = 2x² + 4000/x
[2] For minimum, dS/dx = 0: 4x - 4000/x² = 0
[2] 4x = 4000/x², x³ = 1000, x = ∛1000
[1] x = 10 cm
Examiner says: This is calculus optimization - A-level territory! The cube minimizes surface area for given volume.
Q21 Two similar cylinders have volumes in the ratio 8:27. The smaller cylinder has radius 4 cm. Find: (a) the radius of the larger cylinder, (b) the ratio of their surface areas. [9 marks]
[2] (a) Volume ratio = 8:27, so linear scale factor = ∛(8/27) = 2/3 or ∛(27/8) = 3/2
[2] Since larger volume, SF = 3/2
[1] Larger radius = 4 × (3/2) = 6 cm
[2] (b) Surface area ratio = (linear SF)² = (3/2)² = 9/4
[2] Ratio = 4:9
Examiner says: For similar shapes: length ratio = ∛(volume ratio), area ratio = (length ratio)²!
Q22 A cylindrical bucket has top radius 15 cm, bottom radius 10 cm, and height 20 cm (frustum of a cone). Estimate its volume by treating it as a prism with average radius. Give answer to nearest 10 cm³. [9 marks]
[2] Average radius = (15 + 10) ÷ 2 = 12.5 cm
[2] Average cross-sectional area = π × 12.5² = 156.25π cm²
[2] Volume ≈ area × height = 156.25π × 20
[2] = 3125π ≈ 9817.48
[1] ≈ 9820 cm³ (nearest 10)
Examiner says: This is an approximation method. Exact frustum formula is more complex!
Q23 A prism has an equilateral triangle cross-section with side s cm. The prism length is 3s cm. Find an expression for the volume in terms of s. [Area of equilateral triangle = (s²√3)/4] [10 marks]
[3] Cross-sectional area = (s²√3)/4
[2] Length = 3s
[3] Volume = area × length = (s²√3)/4 × 3s
[2] V = (3s³√3)/4 or 0.75s³√3
Examiner says: Equilateral triangle area uses √3 - don't try to simplify it to a decimal!
Q24 A semi-cylindrical tunnel has diameter 4 m and length 50 m. Find: (a) the volume of air inside the tunnel, (b) the area of material needed for the curved surface (not the ends). Give answers to 1 dp. [10 marks]
[1] Radius = 4 ÷ 2 = 2 m
[2] (a) Full cylinder V = πr²h = π × 2² × 50 = 200π m³
[2] Semi-cylinder V = ½ × 200π = 100π = 314.2 m³ (1 dp)
[2] (b) Circumference of full circle = 2πr = 4π m
[1] Semi-circle arc = ½ × 4π = 2π m
[2] Curved surface area = arc × length = 2π × 50 = 100π = 314.2 m² (1 dp)
Q25 A regular hexagonal prism has side length 6 cm and height 10 cm. Find its volume. [A regular hexagon of side s has area (3s²√3)/2] [11 marks]
[3] Cross-sectional area = (3 × 6² × √3)/2
[2] = (3 × 36 × √3)/2 = (108√3)/2 = 54√3 cm²
[2] Volume = area × height = 54√3 × 10
[2] = 540√3
[2] ≈ 935.3 cm³
Examiner says: Regular hexagon = 6 equilateral triangles arranged around a center!
Q26 A solid is formed by rotating the region bounded by y = x², y = 0, x = 0, and x = 2 about the x-axis. Find its volume using V = ∫π y² dx. Give answer to 1 dp. [11 marks]
[2] y = x², so y² = x⁴
[2] V = ∫₀² π x⁴ dx
[3] = π [x⁵/5]₀²
[2] = π × (32/5 - 0) = 32π/5
[2] = 20.1 cubic units (1 dp)
Examiner says: This is A-level calculus - volumes of revolution using integration!
Q27 A swimming pool is 25 m long and 10 m wide. The depth increases uniformly from 1 m at the shallow end to 3 m at the deep end. Find the volume of water needed to fill it. [12 marks]
[3] This is a prism with trapezoidal cross-section (viewed from the side)
[2] Cross-section trapezoid: parallel sides 1 m and 3 m, width 25 m
[2] Area = ½ × (1 + 3) × 25 = ½ × 4 × 25 = 50 m²
[2] Alternative: Average depth = (1+3)/2 = 2 m, Volume = 25 × 10 × 2 = 500 m³
[2] Volume = area × width = 50 × 10
[1] = 500 m³
Q28 Prove that if all dimensions of a cuboid are doubled, the volume is multiplied by 8. [12 marks]
[2] Let original cuboid have dimensions l, w, h
[2] Original volume V₁ = l × w × h
[2] After doubling: new dimensions are 2l, 2w, 2h
[2] New volume V₂ = (2l) × (2w) × (2h)
[2] V₂ = 2 × 2 × 2 × l × w × h = 8lwh
[2] V₂ = 8V₁ ✓
Examiner says: This applies to ALL 3D shapes - if linear scale factor = 2, volume scale factor = 2³ = 8!
Q29 A cylindrical tank with radius 2 m is being filled with water at 0.5 m³/min. At what rate is the water level rising? (Give answer in cm/min to 2 dp) [13 marks]
[2] V = πr²h, where r = 2 m (constant), h varies with time
[2] V = π × 2² × h = 4πh
[3] dV/dt = 4π × dh/dt (differentiating with respect to time)
[2] Given dV/dt = 0.5 m³/min
[2] 0.5 = 4π × dh/dt
[1] dh/dt = 0.5/(4π) = 0.0398 m/min
[1] = 3.98 cm/min (2 dp)
Examiner says: This is A-level calculus - related rates of change!
Q30 A cylindrical can must hold 500 ml (500 cm³). Find the radius and height that minimize the surface area (to save material). Give answers to 2 dp. [14 marks]
[2] V = πr²h = 500, so h = 500/(πr²)
[2] Surface area S = 2πr² + 2πrh (top + bottom + curved)
[2] S = 2πr² + 2πr × 500/(πr²) = 2πr² + 1000/r
[3] For minimum, dS/dr = 0: 4πr - 1000/r² = 0
[2] 4πr = 1000/r², so 4πr³ = 1000, r³ = 1000/(4π) = 79.577
[1] r = ∛79.577 = 4.30 cm (2 dp)
[1] h = 500/(π × 4.30²) = 8.60 cm (2 dp)
[1] (Note: h = 2r for minimum surface area!)
Examiner says: For a cylinder with fixed volume, minimum surface area when h = 2r!