ExaminerPrep › Question bank › Cambridge IGCSE Maths › Topic 5 Mensuration

Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 5

Mensuration

212 free, high-quality past-paper-style questions on Mensuration for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Perimeter and area of plane shapes; Circles, arcs and sectors; Surface area and volume of solids. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,276 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

Subtopic

Paper

Skill

Difficulty

My progress

+ Add to test✓ In your test1IGMC-045-00-01Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A rectangular vegetable garden is 12.5 m long and 8.4 m wide.

(a)

Calculate the perimeter of the garden.

(2)

(b)

Calculate the area of the garden.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (12.5 + 8.4) or 12.5 + 8.4 + 12.5 + 8.4	✓ 1
A1 41.8 m	✓ 1
Part (b)
M1 12.5 × 8.4	✓ 1
A1 105 m²	✓ 1

Common error: Perimeter is the distance around the edge, so all four sides are added. Adding just the length and the width gives only half the perimeter.

+ Add to test✓ In your test2IGMC-045-00-02Perimeter and area of plane shapes Paper CEasy4 marks

No calculator

Triangle A has a base of 9 cm and a perpendicular height of 6.4 cm.

(a)

Work out the area of triangle A.

(2)

(b)

Triangle B has the same area as triangle A. The base of triangle B is 12 cm.
Work out the perpendicular height of triangle B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × 9 × 6.4	✓ 1
A1 28.8 cm²	✓ 1
Part (b)
M1 $\frac{1}{2}$ × 12 × h = 28.8 or h = 28.8 × 2 ÷ 12	✓ 1
A1 4.8 cm	✓ 1

Common error: Forgetting the $\frac{1}{2}$ in the triangle area formula doubles the answer. A triangle is half of a rectangle with the same base and height.

+ Add to test✓ In your test3IGMC-045-00-03Perimeter and area of plane shapes Paper CEasy4 marks

No calculator

A floor tile is in the shape of a trapezium. Its parallel sides are 7 cm and 13 cm long and the perpendicular distance between them is 5 cm.

Diagram not to scale

(a)

Work out the area of one tile.

(2)

(b)

The tiles fit together with no gaps.
Work out how many tiles are needed to cover an area of 1 m².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × (7 + 13) × 5	✓ 1
A1 50 cm²	✓ 1
Part (b)
B1 1 m² = 10 000 cm²	✓ 1
A1 200 tiles	✓ 1

Common error: 1 m² is 100 × 100 = 10 000 cm², not 100 cm². Area conversions use the square of the length conversion.

+ Add to test✓ In your test4IGMC-045-00-04Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A circular coaster has a radius of 6 cm.

(a)

Calculate the circumference of the coaster.

(2)

(b)

Calculate the area of the coaster.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 6 or π × 12	✓ 1
A1 37.7 cm	✓ 1
Part (b)
M1 π × 6²	✓ 1
A1 113 cm²	✓ 1

Common error: In πr² only the radius is squared. Working out (π × 6)² gives 355, which is far too big.

+ Add to test✓ In your test5IGMC-045-00-05Perimeter and area of plane shapes Paper CEasy3 marks

No calculator

A parallelogram has a base of 15 cm and a perpendicular height of 8 cm. The sloping sides are each 10 cm long.

Diagram not to scale

(a)

Work out the area of the parallelogram.

(2)

(b)

Work out the perimeter of the parallelogram.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 × 8	✓ 1
A1 120 cm²	✓ 1
Part (b)
B1 50 cm	✓ 1

Common error: The area of a parallelogram uses the perpendicular height (8 cm), not the sloping side (10 cm). Using 15 × 10 gives an incorrect answer of 150 cm².

+ Add to test✓ In your test6IGMC-045-01-01Circles, arcs and sectors Paper CEasy4 marks

Calculator

The face of a circular wall clock has a radius of 6.5 cm.

(a)

Calculate the circumference of the clock face.

(2)

(b)

Calculate the area of the clock face.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2 × π × 6.5	✓ 1
A1 40.8 cm (accept 40.84…)	✓ 1
Part (b)
M1 for π × 6.5²	✓ 1
A1 133 cm² (accept 132.7…)	✓ 1

Common error: Squaring π × 6.5 instead of squaring only the radius gives a much larger area. In πr², only r is squared.

+ Add to test✓ In your test7IGMC-045-01-02Circles, arcs and sectors Paper CEasy4 marks

Calculator

The circumference of a circular pond is 18.5 m.

(a)

Calculate the radius of the pond.

(2)

(b)

Calculate the area of the surface of the pond.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 18.5 ÷ (2π)	✓ 1
A1 2.94 m	✓ 1
Part (b)
M1 for π × (their radius)²	✓ 1
A1 27.2 m²	✓ 1

Exam tip: Keep the unrounded radius in your calculator memory for part (b). Using a rounded value can lose accuracy in the final answer.

+ Add to test✓ In your test8IGMC-045-01-03Circles, arcs and sectors Paper CEasy5 marks

Calculator

A sector of a circle has radius 9 cm and angle 40°.

Diagram not to scale

(a)

Calculate the length of the arc of the sector.

(2)

(b)

Calculate the perimeter of the sector.

(1)

(c)

Calculate the area of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{40}{360}$ × 2 × π × 9	✓ 1
A1 6.28 cm	✓ 1
Part (b)
B1 24.3 cm, follow through their arc length + 18	✓ 1
Part (c)
M1 for $\frac{40}{360}$ × π × 9²	✓ 1
A1 28.3 cm²	✓ 1

Common error: The perimeter of a sector includes the two radii as well as the arc. Giving only the arc length for the perimeter loses the mark.

+ Add to test✓ In your test9IGMC-045-01-04Circles, arcs and sectors Paper CEasy4 marks

No calculator

A circle has radius 5 cm. Give each answer in terms of π.

(a)

Write down the area of the circle.

(1)

(b)

Write down the circumference of the circle.

(1)

(c)

The circle is cut in half along a diameter. Find the perimeter of one of the semicircles.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 25π cm²	✓ 1
Part (b)
B1 10π cm	✓ 1
Part (c)
M1 for half of their circumference, 5π	✓ 1
A1 5π + 10 cm	✓ 1

Common error: A semicircle's perimeter is not just half the circumference. The straight edge (the diameter, 10 cm) must be added.

+ Add to test✓ In your test10IGMC-045-01-05Circles, arcs and sectors Paper CEasy5 marks

Calculator

A bicycle wheel has a diameter of 64 cm.

(a)

Calculate the distance the bicycle travels when the wheel makes one complete revolution.

(2)

(b)

Calculate the number of revolutions the wheel makes when the bicycle travels 1 km. Use your unrounded answer to part (a), and give your final answer to the nearest whole number.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for π × 64	✓ 1
A1 201 cm (accept 201.0 to 201.1)	✓ 1
Part (b)
M1 for converting 1 km to 100 000 cm (or the circumference to 2.0106… m)	✓ 1
M1 for 100 000 ÷ their circumference	✓ 1
A1 497	✓ 1
Note: 498 from 100 000 ÷ 201 earns both M marks but not the A mark.	—

Exam tip: Keep the full calculator value of the circumference for the next step. Rounding to 201 first changes the final answer from 497 to 498. Also convert to the same units before dividing: 1 km = 100 000 cm.

+ Add to test✓ In your test11IGMC-045-02-01Surface area and volume of solids Paper CEasy5 marks

Calculator

A closed box is a cuboid measuring 12 cm by 8 cm by 5 cm.

Diagram not to scale

(a)

Calculate the volume of the box.

(2)

(b)

Calculate the total surface area of the box.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 12 × 8 × 5	✓ 1
A1: 480 cm³	✓ 1
Part (b)
M1: at least two correct face areas from 96, 60, 40	✓ 1
M1: 2 × (96 + 60 + 40)	✓ 1
A1: 392 cm²	✓ 1

Exam tip: A cuboid has three pairs of identical faces. Find the three different face areas, add them, then double the total.

+ Add to test✓ In your test12IGMC-045-02-02Surface area and volume of solids Paper CEasy3 marks

Calculator

A cylindrical tin has radius 4.5 cm and height 11 cm.

Diagram not to scale

(a)

Calculate the volume of the tin.

(2)

(b)

Write the capacity of the tin in litres.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: π × 4.5² × 11	✓ 1
A1: 700 cm³ (699.79…)	✓ 1
Part (b)
B1: 0.700 litres (ft their (a) ÷ 1000)	✓ 1

Common error: Using the diameter instead of the radius makes the volume four times too big. 1 litre = 1000 cm³.

+ Add to test✓ In your test13IGMC-045-02-03Surface area and volume of solids Paper CEasy6 marks

No calculator

A prism has length 15 cm. Its cross-section is a right-angled triangle whose two shorter sides are 6 cm and 8 cm.

Diagram not to scale

(a)

Find the area of the cross-section.

(1)

(b)

Find the volume of the prism.

(2)

(c)

Find the total surface area of the prism.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 24 cm²	✓ 1
Part (b)
M1: 24 × 15	✓ 1
A1: 360 cm³	✓ 1
Part (c)
M1: hypotenuse = √(6² + 8²) = 10 cm	✓ 1
M1: 2 × 24 + (6 + 8 + 10) × 15	✓ 1
A1: 408 cm²	✓ 1

Exam tip: The three rectangular faces together have area (perimeter of cross-section) × length, which saves working out each rectangle separately.

+ Add to test✓ In your test14IGMC-045-02-04Surface area and volume of solids Paper CEasy4 marks

Calculator

A solid ball is a sphere of radius 6 cm.
[Volume of sphere = $\frac{4}{3}\pi r^3$, surface area of sphere = $4\pi r^2$]

(a)

Calculate the volume of the ball.

(2)

(b)

Calculate the surface area of the ball.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{4}{3}$ × π × 6³	✓ 1
A1: 905 cm³ (904.77…)	✓ 1
Part (b)
M1: 4 × π × 6²	✓ 1
A1: 452 cm² (452.38…)	✓ 1

Common error: Cubing the radius for surface area (or squaring it for volume) mixes up the formulas. Area units are squared, volume units are cubed.

+ Add to test✓ In your test15IGMC-045-02-05Surface area and volume of solids Paper CEasy3 marks

No calculator

A cube has a volume of 343 cm³.

(a)

Find the length of one edge of the cube.

(1)

(b)

Find the total surface area of the cube.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 7 cm	✓ 1
Part (b)
M1: 6 × 7²	✓ 1
A1: 294 cm²	✓ 1

Exam tip: Know the cubes 1³ to 10³ so you can find cube roots quickly without a calculator.

+ Add to test✓ In your test16IGMC3-5-02d-01Perimeter and area of plane shapes Paper CEasy5 marks

Calculator

A sports hall has a rectangular floor 28 m long and 15 m wide.

(a)

Calculate the perimeter of the floor.

(2)

(b)

Calculate the area of the floor.

(1)

(c)

The floor is covered with varnish costing €3.40 per m². Calculate the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (28 + 15)	✓ 1
A1 86 m	✓ 1
Part (b)
A1 420 m²	✓ 1
Part (c)
M1 420 × 3.40 (ECF from (b))	✓ 1
A1 €1428	✓ 1

Common error: Perimeter is a length in metres and area is in square metres; quoting the wrong unit loses the accuracy mark.

+ Add to test✓ In your test17IGMC3-5-02d-02Circles, arcs and sectors Paper CEasy5 marks

Calculator

A bicycle wheel has a diameter of 66 cm.

(a)

Calculate the circumference of the wheel. Give your answer in centimetres correct to 1 decimal place.

(2)

(b)

The bicycle travels 100 m in a straight line. Calculate the number of turns the wheel makes. Give your answer correct to 3 significant figures and use the unrounded circumference from (a).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 66	✓ 1
A1 207.3 cm	✓ 1
Part (b)
B1 100 m = 10 000 cm	✓ 1
M1 10 000 ÷ (π × 66)	✓ 1
A1 48.2	✓ 1

Exam tip: Convert to a single unit before dividing: mixing metres and centimetres gives an answer wrong by a factor of 100.

+ Add to test✓ In your test18IGMC3-5-02d-03Surface area and volume of solids Paper CEasy5 marks

Calculator

A rectangular fish tank measures 1.2 m long, 0.8 m wide and 0.5 m high on the inside.

Diagram not to scale

(a)

Calculate the volume of the tank in m³.

(2)

(b)

1 m³ = 1000 litres. Write down the capacity of the tank in litres.

(1)

(c)

Water flows into the empty tank from a tap at 12 litres per minute. Calculate the time taken to fill it completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2 × 0.8 × 0.5	✓ 1
A1 0.48 m³	✓ 1
Part (b)
A1 480 litres	✓ 1
Part (c)
M1 480 ÷ 12	✓ 1
A1 40 minutes	✓ 1

Exam tip: Keep track of the unit conversion: 1 m³ is 1000 litres, not 100 litres.

+ Add to test✓ In your test19IGMC3-5-02d-04Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A solar panel is a trapezium. Its parallel sides are 1.4 m and 0.9 m long and they are 0.6 m apart.

Diagram not to scale

(a)

Calculate the area of one panel.

(2)

(b)

A roof carries 25 of these panels. Calculate the total area of the panels.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1.4 + 0.9) × 0.6	✓ 1
A1 0.69 m²	✓ 1
Part (b)
M1 25 × 0.69 (ECF from (a))	✓ 1
A1 17.25 m²	✓ 1

Common error: The height of a trapezium is the perpendicular distance between the parallel sides, not the length of a sloping side.

+ Add to test✓ In your test20IGMC3-5-02d-05Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A garden sprinkler waters a sector of lawn. The sector has radius 6.5 m and angle 90°.

Diagram not to scale

(a)

Calculate the length of the curved edge of the sector. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the area of lawn watered. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (90 ÷ 360) × 2 × π × 6.5	✓ 1
A1 10.2 m	✓ 1
Part (b)
M1 (90 ÷ 360) × π × 6.5²	✓ 1
A1 33.2 m²	✓ 1

Exam tip: Write the fraction of the circle first, then multiply by the full circumference or full area.

+ Add to test✓ In your test21IGMC3-5-08r12-01Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A railway platform is a rectangle 86 m long and 4.2 m wide.

(a)

Calculate the area of the platform surface.

(2)

(b)

The surface is to be relaid at a cost of £18.50 per square metre. Calculate the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 86 × 4.2	✓ 1
A1 361.2 m²	✓ 1
Part (b)
M1 361.2 × 18.5 (or 86 × 4.2 × 18.5)	✓ 1
A1 £6682.20 (accept £6680)	✓ 1

Exam tip: Write the unit with every area answer: square units, such as m².

+ Add to test✓ In your test22IGMC3-5-08r12-02Circles, arcs and sectors Paper CEasy4 marks

Calculator

The main mirror of a small telescope is a circle of radius 1.2 m. Give answers to 3 significant figures.

(a)

Calculate the circumference of the mirror.

(2)

(b)

Calculate the area of the mirror.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 1.2	✓ 1
A1 7.54 m	✓ 1
Part (b)
M1 π × 1.2²	✓ 1
A1 4.52 m²	✓ 1

Common error: Do not mix up the formulas: circumference uses 2πr (or πd), area uses πr².

+ Add to test✓ In your test23IGMC3-5-08r12-03Surface area and volume of solids Paper CEasy6 marks

Calculator

A railway embankment has a uniform cross-section in the shape of a trapezium. The parallel sides are 6.0 m and 10.0 m long and are 2.5 m apart. The embankment is 400 m long.

Diagram not to scale

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the volume of the embankment.

(2)

(c)

Each lorry carries 16 m³ of material. Calculate the number of lorries needed to build the embankment.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (6.0 + 10.0) × 2.5	✓ 1
A1 20 m²	✓ 1
Part (b)
M1 20 × 400 (ECF from (a))	✓ 1
A1 8000 m³	✓ 1
Part (c)
M1 8000 ÷ 16 (ECF from (b))	✓ 1
A1 500	✓ 1

Exam tip: The volume of a prism is the area of the cross-section multiplied by the length.

+ Add to test✓ In your test24IGMC3-5-08r12-04Circles, arcs and sectors Paper CEasy5 marks

No calculator

A hammer-throw landing area is a sector of a circle with radius 60 m and sector angle 30°. Leave your answers in terms of π.

Diagram not to scale

(a)

Find the area of the landing area.

(2)

(b)

Find the total length of the boundary of the landing area (the two straight edges and the curved edge).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (30 ÷ 360) × π × 60²	✓ 1
A1 300π m²	✓ 1
Part (b)
M1 arc length = (30 ÷ 360) × 2 × π × 60	✓ 1
A1 10π	✓ 1
B1 120 + 10π m (≈ 151.4 m)	✓ 1

Common error: The boundary of a sector includes the two radii as well as the arc.

+ Add to test✓ In your test25IGMC3-5-10r12-01Surface area and volume of solids Paper CEasy4 marks

Calculator

A circular stone platform at an excavation site has radius 3.2 m. Give answers to 3 significant figures.

(a)

Calculate the circumference of the platform.

(2)

(b)

Calculate the area of the platform.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $2\pi\times3.2$	✓ 1
A1 20.1 m	✓ 1
Part (b)
M1 $\pi\times3.2^2$	✓ 1
A1 32.2 m²	✓ 1

Common error: Circumference uses 2πr and area uses πr²; only the radius is squared, not π.

+ Add to test✓ In your test26IGMC3-5-10r12-03Perimeter and area of plane shapes Paper CEasy6 marks

No calculator

The floor of a data centre is a rectangle measuring 18 m by 12.5 m.

(a)

Find the area of the floor.

(2)

(b)

Find the perimeter of the floor.

(2)

(c)

The floor is covered with tiles costing €14 per square metre. Find the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18 × 12.5	✓ 1
A1 225 m²	✓ 1
Part (b)
M1 2(18 + 12.5)	✓ 1
A1 61 m	✓ 1
Part (c)
M1 225 × 14	✓ 1
A1 €3150	✓ 1

Exam tip: Area is in square units and perimeter is in plain units; label each answer with the correct unit.

+ Add to test✓ In your test27IGMC3-5-10r12-04Surface area and volume of solids Paper CEasy4 marks

Calculator

A cylindrical storage jar found at a dig has an internal radius of 18 cm and an internal height of 45 cm.

(a)

Calculate the volume of the jar in cm³, correct to 3 significant figures.

(2)

(b)

Write the capacity of the jar in litres, correct to 3 significant figures. (1 litre = 1000 cm³)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 18² × 45	✓ 1
A1 45 800 cm³	✓ 1
Part (b)
M1 45 804 ÷ 1000 or equivalent	✓ 1
A1 45.8 litres	✓ 1

Common error: Remember to square only the radius in the cylinder formula V = πr²h.

+ Add to test✓ In your test28IGMC3-5-09r15-01Perimeter and area of plane shapes Paper CEasy7 marks

Calculator

A rectangular solar panel on a space probe measures 2.5 m by 1.6 m. A semicircular extension of diameter 1.6 m is attached along one of the 1.6 m edges, so the panel plus extension forms one flat shape.

Diagram not to scale

(a)

Calculate the area of the semicircular extension, in m².

(2)

(b)

Calculate the total area of the panel and extension. Use the unrounded answer to (a), and give your answer to 3 significant figures.

(2)

(c)

Calculate the perimeter of the whole shape, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 0.8²	✓ 1
A1 1.01 m² (from 1.0053…)	✓ 1
Part (b)
M1 2.5 × 1.6 + area of semicircle (4.0 + 1.0053…)	✓ 1
A1 5.01 m²	✓ 1
Part (c)
B1 straight edges 2.5 + 2.5 + 1.6 = 6.6 m	✓ 1
M1 curved edge ½ × π × 1.6 (= 2.513…)	✓ 1
A1 9.11 m	✓ 1

Common error: The edge where the semicircle joins the rectangle is inside the shape, so it must not be included in the perimeter.

+ Add to test✓ In your test29IGMC3-5-09r15-02Surface area and volume of solids Paper CEasy5 marks

Calculator

A coffee roaster has a cylindrical drum with internal diameter 40 cm and length 60 cm.

(a)

Show that the volume of the drum is 75 400 cm³, correct to 3 significant figures.

(2)

(b)

Write this volume in litres. (1 litre = 1000 cm³)

(1)

(c)

The beans fill 40% of the drum. Calculate the volume of the beans in litres, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 20² × 60	✓ 1
A1 75398… = 75 400 cm³ to 3 s.f.	✓ 1
Part (b)
A1 75.4 litres	✓ 1
Part (c)
M1 0.4 × 75.398…	✓ 1
A1 30.2 litres	✓ 1

Exam tip: The diameter is given, so halve it to find the radius before you substitute into the volume formula.

+ Add to test✓ In your test30IGMC3-5-09r15-03Perimeter and area of plane shapes Paper CEasy6 marks

No calculator

A concrete bridge support has a constant cross-section in the shape of a trapezium. Its parallel sides are 12 m and 18 m long and the perpendicular distance between them is 5 m. The support is 0.8 m thick.

Diagram not to scale

(a)

Work out the area of the trapezium.

(2)

(b)

Work out the volume of concrete in the support.

(2)

(c)

Concrete costs €135 for each cubic metre. Work out the cost of the concrete in the support.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (12 + 18) × 5	✓ 1
A1 75 m²	✓ 1
Part (b)
M1 75 × 0.8	✓ 1
A1 60 m³	✓ 1
Part (c)
M1 60 × 135	✓ 1
A1 €8100	✓ 1

Exam tip: For a prism, find the area of the cross-section first, then multiply by the length (here the thickness).

+ Add to test✓ In your test31IGMC3-5-09r15-04Circles, arcs and sectors Paper CEasy6 marks

Calculator

A circular coffee-cooling tray has radius 15 cm. A sector-shaped sieve section has its centre at the centre of the tray and an angle of 120°.

Diagram not to scale

(a)

Calculate the arc length of the sector, to 3 significant figures.

(2)

(b)

Calculate the perimeter of the sector, to 3 significant figures.

(2)

(c)

Calculate the area of the sector, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 120/360 × 2 × π × 15	✓ 1
A1 31.4 cm	✓ 1
Part (b)
M1 arc length + 2 × 15 (31.4159… + 30)	✓ 1
A1 61.4 cm	✓ 1
Part (c)
M1 120/360 × π × 15²	✓ 1
A1 236 cm²	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the curved arc.

+ Add to test✓ In your test32IGMC3-5-10r16-01Perimeter and area of plane shapes Paper CEasy6 marks

Calculator

The deck of a footbridge is a rectangle 48 m long and 7.5 m wide.

(a)

Calculate the area of the deck.

(2)

(b)

Calculate the perimeter of the deck.

(2)

(c)

Surfacing costs £12.50 per square metre. Calculate the cost of surfacing the whole deck.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 48 × 7.5	✓ 1
A1 360 m²	✓ 1
Part (b)
M1 2 × (48 + 7.5)	✓ 1
A1 111 m	✓ 1
Part (c)
M1 360 × 12.50	✓ 1
A1 £4500	✓ 1

Exam tip: Write the unit with every answer: area in square units, perimeter in length units.

+ Add to test✓ In your test33IGMC3-5-10r16-02Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A solar panel on a space probe is a trapezium with parallel sides 1.8 m and 2.6 m. The distance between the parallel sides is 1.5 m.

Diagram not to scale

(a)

Show that the area of one panel is 3.3 m².

(2)

(b)

The probe has 6 panels. Each square metre generates 140 W. Calculate the total power generated.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1.8 + 2.6) × 1.5	✓ 1
A1 3.3 (shown, e.g. ½ × 4.4 × 1.5 = 3.3)	✓ 1
Part (b)
M1 6 × 3.3 × 140 (area 19.8 m²)	✓ 1
A1 2772 W	✓ 1

Exam tip: The area of a trapezium is half the sum of the parallel sides multiplied by the perpendicular distance between them.

+ Add to test✓ In your test34IGMC3-5-10r16-06Surface area and volume of solids Paper CEasy6 marks

Calculator

A fuel tank is a cuboid measuring 2.5 m by 1.2 m by 0.8 m.

Diagram not to scale

(a)

Calculate the volume of the tank.

(1)

(b)

Calculate the total surface area of the tank.

(3)

(c)

One litre of paint covers 4 m². Calculate the number of litres needed to paint the whole outside of the tank once.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
2.5 × 1.2 × 0.8 = 2.4 m³	✓ 1
Part (b)
M1 2 × (2.5 × 1.2) + 2 × (2.5 × 0.8) + 2 × (1.2 × 0.8) or equivalent	✓ 1
M1 evidence of all three different face areas 3, 2 and 0.96	✓ 1
A1 11.92 m²	✓ 1
Part (c)
M1 11.92 ÷ 4	✓ 1
A1 2.98 litres	✓ 1

Exam tip: A cuboid has six faces in three equal pairs; check you have counted each pair.

+ Add to test✓ In your test35IGMC3-5-10r16-13Circles, arcs and sectors Paper CEasy5 marks

Calculator

A circular roundabout has an area of 150 m².

(a)

Calculate the radius of the roundabout. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the circumference of the roundabout, using the unrounded radius. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 πr² = 150	✓ 1
M1 r = √(150 ÷ π)	✓ 1
A1 6.91 m	✓ 1
Part (b)
M1 2 × π × 6.909...	✓ 1
A1 43.4 m	✓ 1

Exam tip: When finding a radius from an area, divide by π first and then take the square root.

+ Add to test✓ In your test36IGMC3-5-10r18-01Surface area and volume of solids Paper CEasy6 marks

Calculator

The concrete deck of a footbridge is a cuboid 24 m long, 6.5 m wide and 0.3 m thick.

(a)

Calculate the area of the top surface of the deck.

(2)

(b)

Calculate the volume of concrete in the deck.

(2)

(c)

Concrete costs €85 per cubic metre. Calculate the cost of the concrete for the deck.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24 × 6.5	✓ 1
A1 156 m²	✓ 1
Part (b)
M1 156 × 0.3	✓ 1
A1 46.8 m³	✓ 1
Part (c)
M1 46.8 × 85	✓ 1
A1 €3978	✓ 1

Exam tip: Volume of a cuboid is length × width × height; make sure all three lengths are in the same unit.

+ Add to test✓ In your test37IGMC3-5-10r18-02Surface area and volume of solids Paper CEasy6 marks

Calculator

A coffee roasting drum is a cylinder of diameter 60 cm and length 90 cm.

(a)

Calculate the circumference of a circular end of the drum. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the volume of the drum. Give your answer correct to 3 significant figures.

(2)

(c)

Write your unrounded volume in litres, correct to 3 significant figures. (1 litre = 1000 cm³)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 60	✓ 1
A1 188 cm	✓ 1
Part (b)
M1 π × 30² × 90	✓ 1
A1 254 000 cm³ (254 469 before rounding)	✓ 1
Part (c)
M1 254 469 ÷ 1000	✓ 1
A1 254 litres	✓ 1

Common error: The formula uses the radius, so halve the diameter before squaring.

+ Add to test✓ In your test38IGMC3-5-10r18-03Perimeter and area of plane shapes Paper CEasy5 marks

Calculator

A space probe has two identical rectangular solar panels, each 4.2 m by 1.6 m, and one circular antenna dish of radius 1.8 m.

(a)

Calculate the area of one solar panel.

(1)

(b)

Calculate the area of the dish. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the total area of the two panels and the dish. Use unrounded values and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6.72 m²	✓ 1
Part (b)
M1 π × 1.8²	✓ 1
A1 10.2 m²	✓ 1
Part (c)
M1 2 × 6.72 + π × 1.8² (= 13.44 + 10.1788...)	✓ 1
A1 23.6 m²	✓ 1

Exam tip: Keep unrounded values on your calculator until the final step to avoid rounding errors.

+ Add to test✓ In your test39IGMC3-5-10r18-07Perimeter and area of plane shapes Paper CEasy6 marks

Calculator

The floor of a coffee roasting room is an L-shape formed from a rectangle 12 m by 8 m by removing a rectangle 5 m by 3 m from one corner.

Diagram not to scale

(a)

Calculate the area of the floor.

(2)

(b)

Calculate the perimeter of the floor.

(2)

(c)

The floor is covered with square tiles that each cover 0.25 m² and cost €1.20 each. Calculate the cost of tiling the floor, assuming no tiles are wasted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × 8 − 5 × 3	✓ 1
A1 81 m²	✓ 1
Part (b)
M1 12 + 8 + 12 + 8 (removing a corner leaves the perimeter unchanged) or adding all six sides: 12, 3, 5, 5, 7, 8	✓ 1
A1 40 m	✓ 1
Part (c)
M1 81 ÷ 0.25 (= 324 tiles)	✓ 1
A1 €388.80	✓ 1

Common error: When a corner rectangle is removed, the perimeter stays the same as the original rectangle; do not subtract the removed edges.

+ Add to test✓ In your test40IGMC3-5-10r23-01Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

An archaeological trench is a rectangle 6.5 m long and 4.2 m wide.

(a)

Calculate the area of the trench.

(2)

(b)

Calculate the perimeter of the trench.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.5 × 4.2	✓ 1
A1 27.3 m²	✓ 1
Part (b)
M1 2 × (6.5 + 4.2)	✓ 1
A1 21.4 m	✓ 1

Common error: Area is in square units and perimeter in plain units; write the unit in each answer.

+ Add to test✓ In your test41IGMC3-5-10r23-03Perimeter and area of plane shapes Paper CEasy5 marks

Calculator

A data centre has a rectangular server hall that is 40 m long, 25 m wide and 4.5 m high.

(a)

Calculate the volume of the hall.

(2)

(b)

Each server rack stands on a rectangular footprint 0.6 m by 1.2 m. There are 360 racks. Calculate the percentage of the floor area of the hall covered by racks. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 × 25 × 4.5	✓ 1
A1 4500 m³	✓ 1
Part (b)
M1 racks area 360 × 0.6 × 1.2 = 259.2 m²	✓ 1
M1 259.2 ÷ (40 × 25) × 100	✓ 1
A1 25.9%	✓ 1

Exam tip: For a percentage, divide the part by the whole, then multiply by 100.

+ Add to test✓ In your test42IGMC3-5-12r31-01Perimeter and area of plane shapes Paper CEasy7 marks

Calculator

A rectangular field on a farm is 240 m long and 150 m wide. 1 hectare = 10 000 m².

(a)

Calculate the area of the field in square metres.

(2)

(b)

Write your answer to the first part in hectares.

(1)

(c)

Calculate the perimeter of the field.

(2)

(d)

Fencing costs €4.50 per metre. Calculate the cost of fencing the whole perimeter.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $240\times150$	✓ 1
A1 36 000 m²	✓ 1
Part (b)
B1 3.6 hectares	✓ 1
Part (c)
M1 $2\times(240+150)$	✓ 1
A1 780 m	✓ 1
Part (d)
M1 $780\times4.50$ (ECF from the perimeter)	✓ 1
A1 €3510	✓ 1

Common error: Perimeter is a length (m) and area is in square units (m²); do not mix them up or forget to convert to hectares.

+ Add to test✓ In your test43IGMC3-5-12r31-02Circles, arcs and sectors Paper CEasy4 marks

Calculator

The wheel of a small aircraft has diameter 1.1 m.

(a)

Calculate the circumference of the wheel. Give your answer to 3 significant figures.

(2)

(b)

The aircraft wheel turns 800 times during a landing roll and taxi. Calculate the distance travelled. Use the unrounded circumference and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times1.1$	✓ 1
A1 3.46 m	✓ 1
Part (b)
M1 $800\times\pi\times1.1$	✓ 1
A1 2760 m	✓ 1

Exam tip: Use the unrounded value from your calculator in later steps and round only the final answer.

+ Add to test✓ In your test44IGMC3-5-12r31-04Surface area and volume of solids Paper CEasy5 marks

No calculator

A grain store on a farm is a cuboid of length 5 m, width 3 m and height 2 m.

(a)

Calculate the volume of the store.

(2)

(b)

1 m³ = 1000 litres. Write the volume in litres.

(1)

(c)

Water is pumped into an identical empty tank at 150 litres per minute. Find the time taken to fill it, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $5\times3\times2$	✓ 1
A1 30 m³	✓ 1
Part (b)
B1 30 000 litres (ECF)	✓ 1
Part (c)
M1 $30\,000\div150$ (ECF)	✓ 1
A1 200 minutes	✓ 1

Exam tip: Convert units before dividing, and check that your answer to a filling-time question is in the unit asked for.

+ Add to test✓ In your test45IGMC3-5-12r31-14Surface area and volume of solids Paper CEasy7 marks

No calculator

A field is a right-angled triangle. The two sides that meet at the right angle are 60 m and 80 m long.

Diagram not to scale

(a)

Calculate the length of the third side.

(2)

(b)

Calculate the perimeter of the field.

(1)

(c)

Calculate the area of the field.

(2)

(d)

Seed is sown at 25 grams per square metre. Calculate the total mass of seed needed, in kilograms.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{60^2 + 80^2}$	✓ 1
A1 100 m	✓ 1
Part (b)
B1 240 m (ECF)	✓ 1
Part (c)
M1 $\frac{1}{2}\times60\times80$	✓ 1
A1 2400 m²	✓ 1
Part (d)
M1 $2400\times25 = 60\,000$ g (ECF from the area)	✓ 1
A1 60 kg	✓ 1

Exam tip: Use the two perpendicular sides as base and height for the area of a right-angled triangle.

+ Add to test✓ In your test46IGMC3-5-11r34-01Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A pharmacy prints a rectangular label measuring 7.5 cm by 4.2 cm on each bottle.

(a)

Work out the perimeter of the label.

(2)

(b)

Work out the area of the label.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (7.5 + 4.2)	✓ 1
A1 23.4 cm	✓ 1
Part (b)
M1 7.5 × 4.2	✓ 1
A1 31.5 cm²	✓ 1

Common error: Perimeter is a length measured in cm; area is measured in cm². Writing the wrong unit loses the final mark.

+ Add to test✓ In your test47IGMC3-5-11r34-02Perimeter and area of plane shapes Paper CEasy5 marks

Calculator

The cap of a medicine bottle is a circle of diameter 2.8 cm. Use the π button on your calculator and give each answer correct to 3 significant figures.

(a)

Work out the circumference of the cap.

(2)

(b)

Work out the area of the top of the cap.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 2.8	✓ 1
A1 8.80 cm	✓ 1
Part (b)
B1 radius = 1.4 cm	✓ 1
M1 π × 1.4²	✓ 1
A1 6.16 cm²	✓ 1

Exam tip: Halve the diameter before using the area formula; using 2.8 as the radius is a common slip.

+ Add to test✓ In your test48IGMC3-5-11r34-03Surface area and volume of solids Paper CEasy5 marks

Calculator

A water-supply tank is a closed cuboid of length 2.5 m, width 1.6 m and height 1.2 m. 1 m³ = 1000 litres.

(a)

Work out the volume of the tank in m³.

(2)

(b)

Write down the capacity of the tank in litres.

(1)

(c)

The empty tank is filled by a pipe delivering 150 litres every minute. Work out the time taken to fill the tank completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.5 × 1.6 × 1.2	✓ 1
A1 4.8 m³	✓ 1
Part (b)
B1 4800 litres (ECF from (a))	✓ 1
Part (c)
M1 4800 ÷ 150 (ECF from (b))	✓ 1
A1 32 minutes	✓ 1

Exam tip: Convert m³ to litres before dividing by a rate given in litres per minute.

+ Add to test✓ In your test49IGMC3-5-11r34-04Perimeter and area of plane shapes Paper CEasy4 marks

Calculator

A tile in a board game is a trapezium. Its parallel sides are 14 cm and 22 cm long and the perpendicular distance between them is 9 cm.

Diagram not to scale

(a)

Work out the area of one tile.

(2)

(b)

A rectangular game mat has area 2000 cm². Work out the greatest number of whole tiles that can fit on the mat, assuming the tiles do not overlap and cannot be cut.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (14 + 22) × 9	✓ 1
A1 162 cm²	✓ 1
Part (b)
M1 2000 ÷ 162 (= 12.3…)	✓ 1
A1 12 (rounded down because only whole tiles count)	✓ 1

Common error: Only whole tiles count, so a decimal such as 12.3 must be rounded down, never up.

+ Add to test✓ In your test50IGMC3-5-11r37-01Perimeter and area of plane shapes Paper CEasy6 marks

Calculator

A wind farm company leases a rectangular field 450 m long and 280 m wide for a new turbine.

(a)

Calculate the perimeter of the field.

(2)

(b)

Calculate the area of the field in m².

(2)

(c)

1 hectare = 10 000 m². Find the area of the field in hectares.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (450 + 280)	✓ 1
A1 1460 m	✓ 1
Part (b)
M1 450 × 280	✓ 1
A1 126 000 m²	✓ 1
Part (c)
M1 126 000 ÷ 10 000 (or 450 × 280 ÷ 10 000)	✓ 1
A1 12.6 hectares	✓ 1

Common error: When changing between units of area, divide by the full conversion factor for area (here 10 000), not by a length conversion.

+ Add to test✓ In your test51IGMC3-5-11r37-02Surface area and volume of solids Paper CEasy5 marks

Calculator

A swimming pool is a cuboid 25 m long, 10 m wide and 2.0 m deep.

(a)

Calculate the volume of the pool in m³.

(2)

(b)

1 m³ = 1000 litres. Write the volume of the pool in litres.

(1)

(c)

The pool is filled at a constant rate of 2500 litres per minute. Calculate the time taken to fill it completely, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 25 × 10 × 2.0	✓ 1
A1 500 m³	✓ 1
Part (b)
B1 500 000 litres	✓ 1
Part (c)
M1 500 000 ÷ 2500	✓ 1
A1 200 minutes	✓ 1

Exam tip: Write the unit with every line of working so you can spot a missing conversion such as m³ to litres.

+ Add to test✓ In your test52IGMC3-5-11r37-03Perimeter and area of plane shapes Paper CEasy6 marks

Calculator

A library has a circular reading table with a top of radius 0.9 m.

(a)

Calculate the area of the table top. Give your answer in m² correct to 3 significant figures.

(2)

(b)

A strip of edge trim goes once round the edge of the table top. Calculate the length of trim needed, in metres, correct to 3 significant figures.

(2)

(c)

The trim is sold only in whole metres and costs £4.20 per metre. Using your unrounded length from the calculation, find the cost of the trim needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.9²	✓ 1
A1 2.54 m²	✓ 1
Part (b)
M1 2 × π × 0.9	✓ 1
A1 5.65 m	✓ 1
Part (c)
M1 length rounded up to 6 m (the next whole metre)	✓ 1
A1 6 × 4.20 = £25.20	✓ 1

Common error: When material must be bought in whole units, round up, not to the nearest whole number.

+ Add to test✓ In your test53IGMC3-5-11r37-07Surface area and volume of solids Paper CEasy5 marks

Calculator

A library packs books into a crate in the shape of a cuboid. The inside of the crate measures 60 cm by 40 cm by 30 cm. Each book is a cuboid 20 cm by 15 cm by 3 cm. The books fit exactly with no gaps.

(a)

Calculate the volume of the inside of the crate in cm³.

(2)

(b)

Calculate the volume of one book.

(1)

(c)

Find the number of books that fit in the crate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 60 × 40 × 30	✓ 1
A1 72 000 cm³	✓ 1
Part (b)
B1 20 × 15 × 3 = 900 cm³	✓ 1
Part (c)
M1 72 000 ÷ 900	✓ 1
A1 80	✓ 1

Exam tip: If a whole number of identical objects fills a space exactly, dividing the volumes gives that number.

+ Add to test✓ In your test54IGMC3-5-09r38-01Perimeter and area of plane shapes Paper CEasy5 marks

No calculator

A wind farm company owns a rectangular field that measures 480 m by 350 m.

(a)

Work out the perimeter of the field.

(2)

(b)

Work out the area of the field in hectares. (1 hectare = 10 000 m²)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (480 + 350) or 480 + 350 + 480 + 350	✓ 1
A1 1660 m	✓ 1
Part (b)
M1 480 × 350	✓ 1
A1 168 000 m²	✓ 1
A1 16.8 hectares	✓ 1

Common error: When converting units of area, divide by the square of the length conversion; here 1 hectare is 10 000 m², not 100 m².

+ Add to test✓ In your test55IGMC3-5-09r38-02Circles, arcs and sectors Paper CEasy6 marks

Calculator

The blades of a wind turbine are 45 m long, so the tips move round a circle of radius 45 m. Give answers to 3 significant figures.

(a)

Calculate the area of the circle swept by the blades.

(2)

(b)

Calculate the distance travelled by a blade tip in one complete turn.

(2)

(c)

One complete turn takes 4.0 seconds. Use the unrounded distance from part (b) to calculate the speed of a blade tip in m/s.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 45²	✓ 1
A1 6360 m² (6361.7... before rounding)	✓ 1
Part (b)
M1 2 × π × 45	✓ 1
A1 283 m (282.74... before rounding)	✓ 1
Part (c)
M1 282.74... ÷ 4.0	✓ 1
A1 70.7 m/s	✓ 1

Exam tip: Keep the full calculator value between parts and round only the final answer; state the units for every numerical answer.

+ Add to test✓ In your test56IGMC3-5-09r38-03Surface area and volume of solids Paper CEasy5 marks

No calculator

A swimming pool is a cuboid 25 m long, 10 m wide and 1.6 m deep. It is completely empty.

(a)

Work out the volume of the pool in m³.

(2)

(b)

Write your answer to part (a) in litres. (1 m³ = 1000 litres)

(1)

(c)

Water is pumped into the pool at 8000 litres per minute. Work out the time, in minutes, to fill the pool completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 25 × 10 × 1.6	✓ 1
A1 400 m³	✓ 1
Part (b)
B1 400 000 litres	✓ 1
Part (c)
M1 400 000 ÷ 8000	✓ 1
A1 50 minutes	✓ 1

Exam tip: Convert everything to the same unit before dividing a volume by a rate.

+ Add to test✓ In your test57IGMC3-5-09r38-04Perimeter and area of plane shapes Paper CEasy7 marks

Calculator

A window in a library is made from a rectangle 1.4 m wide and 2.1 m high, with a semicircle of diameter 1.4 m on top of it. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the area of the semicircle.

(2)

(b)

Calculate the total area of the window.

(2)

(c)

A metal strip is fitted along the whole outline of the window: the two vertical sides, the bottom edge and the curved edge. Calculate the length of the strip.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 0.7²	✓ 1
A1 0.770 m² (0.7697... before rounding)	✓ 1
Part (b)
M1 1.4 × 2.1 (= 2.94) + area of semicircle	✓ 1
A1 3.71 m²	✓ 1
Part (c)
B1 curved edge = ½ × π × 1.4 = π × 0.7 (= 2.199...)	✓ 1
M1 2.1 + 2.1 + 1.4 + curved edge	✓ 1
A1 7.80 m	✓ 1

Common error: The bottom edge of a rectangle-plus-semicircle shape is part of the outline, but the diameter of the semicircle is not: do not include it in the perimeter.

+ Add to test✓ In your test58IGMC3-5-09r39-01Perimeter and area of plane shapes Paper CEasy5 marks

Calculator

The stage of a small theatre is a rectangle 12.5 m long and 8.4 m wide.

(a)

Calculate the area of the stage.

(2)

(b)

A strip of edging is fixed along the whole perimeter of the stage. The edging costs £6.40 per metre. Calculate the total cost of the edging.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12.5 × 8.4	✓ 1
A1 105 m²	✓ 1
Part (b)
M1 perimeter = 2 × (12.5 + 8.4)	✓ 1
A1 41.8 m	✓ 1
A1 £267.52	✓ 1

Common error: Area uses a product of two lengths but perimeter is a sum, so check you have used the correct one for the quantity asked.

+ Add to test✓ In your test59IGMC3-5-09r39-03Surface area and volume of solids Paper CEasy5 marks

Calculator

The inside of a shipping container is a cuboid of length 6.0 m, width 2.4 m and height 2.6 m.

(a)

Calculate the volume of the inside of the container.

(2)

(b)

Boxes in the shape of a cuboid measuring 0.6 m by 0.6 m by 0.4 m are packed in the container. Each box is placed with its 0.4 m edge vertical and its 0.6 m edges parallel to the length and the width of the container. Work out the greatest number of boxes that can be packed in the container.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.0 × 2.4 × 2.6	✓ 1
A1 37.44 m³	✓ 1
Part (b)
M1 counting whole boxes along each direction: 10, 4 and 2.6 ÷ 0.4 = 6.5 so 6	✓ 1
M1 10 × 4 × 6	✓ 1
A1 240	✓ 1

Exam tip: Dividing the volumes gives too many boxes. Count whole boxes along each edge, because partial boxes cannot be used.

+ Add to test✓ In your test60IGMC3-5-10r48-01Perimeter and area of plane shapes Paper CEasy6 marks

Calculator

A beekeeper uses a rectangular wooden honey frame that measures 42.0 cm by 22.0 cm.

(a)

Calculate the perimeter of the frame.

(2)

(b)

Calculate the area of the frame.

(2)

(c)

A rectangular sheet of beeswax measuring 40.0 cm by 20.0 cm is fixed inside the frame. Calculate the area of the sheet as a percentage of the area of the frame. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × (42.0 + 22.0) or 42.0 + 22.0 + 42.0 + 22.0	✓ 1
A1 128 cm	✓ 1
Part (b)
M1 42.0 × 22.0	✓ 1
A1 924 cm²	✓ 1
Part (c)
M1 (40.0 × 20.0) ÷ 924 × 100	✓ 1
A1 86.6%	✓ 1

Exam tip: Percentage of an area means part ÷ whole × 100; keep the units the same in both areas.

+ Add to test✓ In your test61IGMC3-5-10r48-02Circles, arcs and sectors Paper CEasy5 marks

Calculator

A Formula 1 car has tyres that are modelled as circles with radius 33.0 cm.

(a)

Calculate the circumference of one tyre. Give your answer correct to 3 significant figures.

(2)

(b)

A tyre makes 1250 complete turns. Calculate the distance travelled, in kilometres. Use the unrounded circumference and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 33.0	✓ 1
A1 207 cm	✓ 1
Part (b)
M1 distance = 1250 × 2 × π × 0.33 (or 1250 × 207.345... )	✓ 1
M1 conversion to kilometres (÷ 1000, or ÷ 100 000 from cm)	✓ 1
A1 2.59 km	✓ 1

Common error: Convert units at the start or the end, but not half-way through: 33.0 cm is 0.330 m, and metres divided by 1000 gives kilometres.

+ Add to test✓ In your test62IGMC3-5-10r48-06Circles, arcs and sectors Paper CEasy6 marks

No calculator

The floor of a hospital reception area is a trapezium. The two parallel sides are 9 m and 15 m long, and the perpendicular distance between them is 6 m. A circular skylight of radius 2 m is fitted in the ceiling above the reception area.

Diagram not to scale

(a)

Calculate the area of the floor.

(2)

(b)

Find the area of the skylight. Give your answer in terms of π.

(2)

(c)

Write the area of the skylight as a fraction of the area of the floor. Give your answer in terms of π, in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (9 + 15) × 6	✓ 1
A1 72 m²	✓ 1
Part (b)
M1 π × 2²	✓ 1
A1 4π m²	✓ 1
Part (c)
M1 4π ÷ 72 (accept ECF from (a) and (b))	✓ 1
A1 π/18	✓ 1

Exam tip: In a no-calculator question, leave π in your answer and simplify fractions fully.

+ Add to test✓ In your test63IGMC3-5-10r48-07Perimeter and area of plane shapes Paper CEasy7 marks

No calculator

A rectangular apiary is fenced in. Its width is x metres and its length is (x + 6) metres. The perimeter of the apiary is 52 m.

Diagram not to scale

(a)

Show that x = 10.

(3)

(b)

Calculate the area of the apiary.

(2)

(c)

The fence is built from straight panels that are each 2.0 m long. Calculate the number of panels needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 perimeter = 2x + 2(x + 6)	✓ 1
M1 4x + 12 = 52	✓ 1
A1 4x = 40 so x = 10	✓ 1
Part (b)
M1 10 × 16	✓ 1
A1 160 m²	✓ 1
Part (c)
M1 52 ÷ 2.0	✓ 1
A1 26 panels	✓ 1

Exam tip: With a 'show that' question, every step must be written out; the given answer cannot be assumed.

+ Add to test✓ In your test64IGMC-045-00-06Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A lawn is made from a rectangle 20 m long and 14 m wide. A rectangle 8 m long and 6 m wide has been removed from one corner, leaving an L-shape.

Diagram not to scale

(a)

Calculate the area of the lawn.

(2)

(b)

Calculate the perimeter of the lawn.

(2)

(c)

Turf costs $4.50 per square metre.
Calculate the cost of covering the whole lawn with turf.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 × 14 − 8 × 6	✓ 1
A1 232 m²	✓ 1
Part (b)
M1 complete sum of all six edges, e.g. 20 + 14 + 12 + 6 + 8 + 8	✓ 1
A1 68 m	✓ 1
Part (c)
M1 their 232 × 4.50	✓ 1
A1 $1044	✓ 1

Exam tip: Removing a rectangle from a corner does not change the perimeter: the two new inner edges are exactly as long as the two pieces of edge that were removed. Check by adding all six edges.

+ Add to test✓ In your test65IGMC-045-00-07Perimeter and area of plane shapes Paper CMedium5 marks

Calculator

A window is made from a rectangle 1.2 m wide and 1.5 m high, with a semicircle on top. The diameter of the semicircle is the top edge of the rectangle.

Diagram not to scale

(a)

Calculate the total area of the window.

(3)

(b)

Calculate the perimeter of the window.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2 × 1.5 (= 1.8)	✓ 1
M1 $\frac{1}{2}$ × π × 0.6² (= 0.5654…)	✓ 1
A1 2.37 m²	✓ 1
Part (b)
M1 1.5 + 1.5 + 1.2 + $\frac{1}{2}$ × π × 1.2	✓ 1
A1 6.08 m	✓ 1

Common error: The top edge of the rectangle is inside the window, so it is not part of the perimeter. Include only the two sides, the base and the curved arc.

+ Add to test✓ In your test66IGMC-045-00-10Perimeter and area of plane shapes Paper CMedium5 marks

Calculator

A circular pond has an area of 50 m².

(a)

Calculate the radius of the pond.

(3)

(b)

A fence is put around the edge of the pond.
Calculate the length of the fence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 πr² = 50	✓ 1
M1 r = √(50 ÷ π)	✓ 1
A1 3.99 m	✓ 1
Part (b)
M1 2 × π × their r	✓ 1
A1 25.1 m	✓ 1

Common error: Dividing 50 by π gives r², not r. You must take the square root to find the radius.

+ Add to test✓ In your test67IGMC-045-00-11Perimeter and area of plane shapes Paper CMedium6 marks

No calculator

The diagonals of a rhombus are 24 cm and 10 cm long. The diagonals of a rhombus bisect each other at right angles.

Diagram not to scale

(a)

Work out the area of the rhombus.

(2)

(b)

Work out the perimeter of the rhombus.

(2)

(c)

One side of the rhombus is taken as the base.
Work out the perpendicular height of the rhombus, giving your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × 24 × 10 or 4 × $\frac{1}{2}$ × 12 × 5	✓ 1
A1 120 cm²	✓ 1
Part (b)
M1 side = √(12² + 5²) = 13	✓ 1
A1 52 cm	✓ 1
Part (c)
M1 13 × h = 120	✓ 1
A1 9.23 cm	✓ 1

Exam tip: A rhombus is a parallelogram, so its area is also base × perpendicular height. Linking the two area methods lets you find the height.

+ Add to test✓ In your test68IGMC-045-01-06Circles, arcs and sectors Paper CMedium5 marks

Calculator

A sector of a circle has radius 12 cm. The arc length of the sector is 15 cm.

Diagram not to scale

(a)

Calculate the angle of the sector.

(3)

(b)

Calculate the area of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{\theta}{360}$ × 2 × π × 12 = 15	✓ 1
M1 for θ = 15 × 360 ÷ (24π)	✓ 1
A1 71.6°	✓ 1
Part (b)
M1 for $\frac{\text{their }\theta}{360}$ × π × 12² or $\frac{1}{2}$ × 12 × 15	✓ 1
A1 90.0 cm²	✓ 1

Exam tip: Set up the arc length formula as an equation and rearrange for the angle. Using the unrounded angle in part (b) gives exactly 90.

+ Add to test✓ In your test69IGMC-045-01-07Circles, arcs and sectors Paper CMedium5 marks

Calculator

A sector of a circle has an angle of 75° and an area of 50 cm².

Diagram not to scale

(a)

Calculate the radius of the sector.

(3)

(b)

Calculate the perimeter of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{75}{360}$ × π × r² = 50	✓ 1
M1 for r² = 50 × 360 ÷ (75π) [= 76.39…]	✓ 1
A1 8.74 cm	✓ 1
Part (b)
M1 for $\frac{75}{360}$ × 2 × π × their r + 2 × their r	✓ 1
A1 28.9 cm	✓ 1

Common error: After finding r², remember to take the square root. Giving 76.4 as the radius is a frequent slip.

+ Add to test✓ In your test70IGMC-045-01-08Circles, arcs and sectors Paper CMedium5 marks

Calculator

A circular flower bed has radius 7 m. It is surrounded by a path of width 1.5 m, so that the outer edge of the path is also a circle with the same centre.

Diagram not to scale

(a)

Calculate the area of the path.

(3)

(b)

A fence is put around the outer edge of the path. Fencing costs $12.40 per metre. Calculate the cost of the fence, giving your answer to the nearest dollar.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 outer radius 8.5 m	✓ 1
M1 for π × 8.5² − π × 7²	✓ 1
A1 73.0 m²	✓ 1
Part (b)
M1 for 2 × π × 8.5 × 12.40	✓ 1
A1 $662	✓ 1

Common error: Squaring the path width (1.5²) does not give the area of the path. Subtract the inner circle's area from the outer circle's area.

+ Add to test✓ In your test71IGMC-045-01-11Circles, arcs and sectors Paper CMedium5 marks

Calculator

The minute hand of a clock is 11 cm long.

(a)

Calculate the distance travelled by the tip of the minute hand in 20 minutes.

(2)

(b)

Calculate the area swept out by the minute hand between 3:10 pm and 3:45 pm.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{120}{360}$ × 2 × π × 11	✓ 1
A1 23.0 cm	✓ 1
Part (b)
B1 angle 210° seen or implied	✓ 1
M1 for $\frac{210}{360}$ × π × 11²	✓ 1
A1 222 cm²	✓ 1

Exam tip: The minute hand turns 6° every minute, so the angle is the number of minutes × 6°.

+ Add to test✓ In your test72IGMC-045-02-06Surface area and volume of solids Paper CMedium6 marks

Calculator

A solid cone has base radius 5 cm and vertical height 12 cm.
[Curved surface area of cone = $\pi r l$, volume of cone = $\frac{1}{3}\pi r^2 h$]

Diagram not to scale

(a)

Show that the slant height of the cone is 13 cm.

(2)

(b)

Calculate the curved surface area of the cone.

(2)

(c)

Calculate the volume of the cone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 5² + 12² = 169	✓ 1
A1: √169 = 13 shown	✓ 1
Part (b)
M1: π × 5 × 13	✓ 1
A1: 204 cm² (204.2…)	✓ 1
Part (c)
M1: $\frac{1}{3}$ × π × 5² × 12	✓ 1
A1: 314 cm³ (314.15…)	✓ 1

Common error: The curved surface area uses the slant height l; the volume uses the vertical height h. Swapping them loses the accuracy mark.

+ Add to test✓ In your test73IGMC-045-02-07Surface area and volume of solids Paper CMedium5 marks

Calculator

A cuboid tank has a base measuring 80 cm by 50 cm. It contains water to a depth of 30 cm.

(a)

Find the volume of water in the tank, giving your answer in litres.

(2)

(b)

All the water is poured into an empty cylindrical tank of radius 20 cm. No water is spilled.
Calculate the depth of water in the cylindrical tank.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 80 × 50 × 30 = 120 000 cm³	✓ 1
A1: 120 litres	✓ 1
Part (b)
M1: π × 20² × h = 120 000	✓ 1
M1: h = 120 000 ÷ (π × 400)	✓ 1
A1: 95.5 cm (95.49…)	✓ 1

Exam tip: The volume of water stays the same when it is poured. Set the two volume expressions equal and solve for the unknown depth.

+ Add to test✓ In your test74IGMC-045-02-10Surface area and volume of solids Paper CMedium6 marks

Calculator

A solid metal cube of edge 8 cm is melted down. The metal is used to make as many solid spheres of radius 1.5 cm as possible.
[Volume of sphere = $\frac{4}{3}\pi r^3$]

(a)

Calculate the volume of one sphere.

(2)

(b)

Find the number of complete spheres that can be made.

(2)

(c)

Calculate the volume of metal left over.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{4}{3}$ × π × 1.5³	✓ 1
A1: 14.1 cm³ (14.137…)	✓ 1
Part (b)
M1: 8³ ÷ their (a), i.e. 512 ÷ 14.137… (= 36.2…)	✓ 1
A1: 36	✓ 1
Part (c)
M1: 512 − 36 × 14.137…	✓ 1
A1: 3.06 cm³ (3.06…)	✓ 1

Common error: The number of complete spheres must be rounded down, not to the nearest whole number. Use the unrounded sphere volume in part (c).

+ Add to test✓ In your test75IGMC3-5-02d-06Surface area and volume of solids Paper CMedium5 marks

Calculator

A cylindrical paint tin has radius 8 cm and height 20 cm.

Diagram not to scale

(a)

Calculate the volume of the tin. Give your answer to the nearest cm³.

(2)

(b)

1000 cm³ = 1 litre. Write down the capacity of the tin in litres, correct to 3 significant figures.

(1)

(c)

Calculate the area of the curved surface of the tin. Give your answer to the nearest cm².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 8² × 20	✓ 1
A1 4021 cm³	✓ 1
Part (b)
A1 4.02 litres	✓ 1
Part (c)
M1 2 × π × 8 × 20	✓ 1
A1 1005 cm²	✓ 1

Common error: Curved surface area is 2πrh. Adding the circular ends, or using πr²h, answers a different question.

+ Add to test✓ In your test76IGMC3-5-02d-07Circles, arcs and sectors Paper CMedium6 marks

Calculator

The central pitch of an athletics stadium is a rectangle 90 m long and 60 m wide, with a semicircle attached to each of the two 60 m sides.

Diagram not to scale

(a)

Calculate the perimeter of the pitch. Give your answer correct to 1 decimal place.

(3)

(b)

Calculate the area of the pitch. Give your answer to the nearest m².

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the two semicircles together form a circle of circumference π × 60	✓ 1
M1 add the two straight edges, 2 × 90	✓ 1
A1 368.5 m	✓ 1
Part (b)
M1 rectangle 90 × 60 = 5400	✓ 1
M1 two semicircles together give π × 30²	✓ 1
A1 8227 m²	✓ 1

Exam tip: Two matching semicircles make one full circle, so you need only one circumference calculation and one area calculation.

+ Add to test✓ In your test77IGMC3-5-02d-08Perimeter and area of plane shapes Paper CMedium5 marks

Calculator

A tent has a cross-section that is an isosceles triangle with base 1.6 m, perpendicular height 1.5 m and sloping sides each 1.7 m. The tent is 2.5 m long. The two triangular ends and the two sloping rectangular sides are made of fabric. There is no floor.

Diagram not to scale

(a)

Calculate the volume of air inside the tent.

(2)

(b)

Calculate the total area of fabric needed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 1.6 × 1.5 × 2.5	✓ 1
A1 3 m³	✓ 1
Part (b)
M1 two sloping rectangles, 2 × 1.7 × 2.5 = 8.5	✓ 1
M1 two triangular ends, 2 × ½ × 1.6 × 1.5 = 2.4	✓ 1
A1 10.9 m²	✓ 1

Common error: Use the perpendicular height in the triangle's area but the sloping length for the rectangular sides; swapping them is easy to do.

+ Add to test✓ In your test78IGMC3-5-08r12-06Surface area and volume of solids Paper CMedium6 marks

Calculator

A sports scientist uses a cylindrical drinks container of radius 4.0 cm and height 22 cm.

Diagram not to scale

(a)

Calculate the volume of the container. Give your answer to the nearest cm³.

(2)

(b)

A label covers the whole curved surface of the container. Calculate the area of the label. Give your answer to the nearest cm².

(2)

(c)

The container is filled with drink to ³⁄₄ of its height. Calculate the volume of drink in litres. Use the unrounded volume from (a) and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 4.0² × 22	✓ 1
A1 1106 cm³	✓ 1
Part (b)
M1 2 × π × 4.0 × 22	✓ 1
A1 553 cm²	✓ 1
Part (c)
M1 0.75 × 1105.84... (≈ 829.4 cm³)	✓ 1
A1 0.829 litres	✓ 1

Exam tip: 1 litre = 1000 cm³; convert units at the end of the calculation.

+ Add to test✓ In your test79IGMC3-5-08r12-08Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A football training area is a rectangle. Its length is 15 m longer than its width. The perimeter of the training area is 190 m.

Diagram not to scale

(a)

The width is w metres. Show that 4w + 30 = 190.

(2)

(b)

Find the width and the length of the training area.

(2)

(c)

The whole area is to be covered with grass seed costing €0.85 per square metre. Calculate the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 perimeter = 2(w + w + 15)	✓ 1
A1 4w + 30 = 190 shown with the brackets expanded	✓ 1
Part (b)
M1 4w = 160	✓ 1
A1 width 40 m and length 55 m	✓ 1
Part (c)
M1 40 × 55 (= 2200) × 0.85	✓ 1
A1 €1870	✓ 1

Common error: Perimeter is the total distance around the shape; do not use the formula for area.

+ Add to test✓ In your test80IGMC3-5-10r12-05Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A sector-shaped trench is excavated. The sector has radius 9 m and angle 150° at the centre. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the arc length of the sector.

(2)

(b)

Calculate the area of the sector.

(2)

(c)

Calculate the perimeter of the sector. Use the unrounded arc length.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{150}{360}\times2\pi\times9$	✓ 1
A1 23.6 m	✓ 1
Part (b)
M1 $\frac{150}{360}\times\pi\times9^2$	✓ 1
A1 106 m²	✓ 1
Part (c)
M1 arc length + 9 + 9	✓ 1
A1 41.6 m	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the arc.

+ Add to test✓ In your test81IGMC3-5-10r12-07Perimeter and area of plane shapes Paper CMedium6 marks

No calculator

A mosaic is made from a rectangle 10 m by 6 m with a semicircle attached to one of the 6 m sides (the diameter of the semicircle is that side). Give answers in terms of π.

Diagram not to scale

(a)

Find the area of the mosaic.

(3)

(b)

Find the perimeter of the mosaic.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area of rectangle 60 and area of semicircle $\frac{1}{2}\pi\times3^2$	✓ 1
M1 add the two areas	✓ 1
A1 $\left(60+\frac{9}{2}\pi\right)$ m²	✓ 1
Part (b)
M1 semicircular arc = $\pi\times3$	✓ 1
M1 straight sides 10 + 10 + 6	✓ 1
A1 $(26+3\pi)$ m	✓ 1

Common error: The side shared by the rectangle and the semicircle is not part of the outer perimeter.

+ Add to test✓ In your test82IGMC3-5-10r12-08Circles, arcs and sectors Paper CMedium6 marks

Calculator

The infield of a cycling track is a circle of area 2500 m².

(a)

Calculate the radius of the circle, correct to 3 significant figures.

(2)

(b)

Calculate the circumference of the circle, correct to 3 significant figures. Use the unrounded radius.

(2)

(c)

A cyclist rides 5 laps of the edge of the infield. Calculate the distance, to the nearest metre. Use the unrounded circumference.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi r^2=2500$	✓ 1
A1 28.2 m	✓ 1
Part (b)
M1 $2\pi\times\sqrt{2500/\pi}$	✓ 1
A1 177 m	✓ 1
Part (c)
M1 5 × 177.2…	✓ 1
A1 886 m	✓ 1

Exam tip: Keep unrounded values in your calculator for later parts, and round only the final answer.

+ Add to test✓ In your test83IGMC3-5-10r12-10Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A trench for data cables is a cuboid 0.6 m wide, 0.9 m deep and 45 m long.

Diagram not to scale

(a)

Calculate the volume of soil removed.

(2)

(b)

Each truck carries 6 m³ of soil. Find the least number of trips needed to remove all the soil.

(2)

(c)

Removing soil costs €38.50 per cubic metre. Calculate the cost of removing all the soil.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.6 × 0.9 × 45	✓ 1
A1 24.3 m³	✓ 1
Part (b)
M1 24.3 ÷ 6 = 4.05	✓ 1
A1 5 trips	✓ 1
Part (c)
M1 24.3 × 38.50	✓ 1
A1 €935.55	✓ 1

Common error: A number of trips must be a whole number; round up, not to the nearest value.

+ Add to test✓ In your test84IGMC3-5-10r12-11Perimeter and area of plane shapes Paper CMedium4 marks

Calculator

An excavation area is a trapezium. Its parallel sides are 14.2 m and 9.8 m, and the perpendicular distance between them is 6.5 m.

Diagram not to scale

(a)

Calculate the area of the trapezium.

(2)

(b)

A second trapezium has parallel sides 8.4 m and 5.6 m and area 49 m². Calculate the distance between its parallel sides.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}(14.2+9.8)\times6.5$	✓ 1
A1 78.0 m²	✓ 1
Part (b)
M1 $49=\frac{1}{2}(8.4+5.6)h$	✓ 1
A1 7.0 m	✓ 1

Exam tip: The height of a trapezium is the perpendicular distance between the parallel sides, not a sloping side.

+ Add to test✓ In your test85IGMC3-5-09r15-05Surface area and volume of solids Paper CMedium5 marks

Calculator

A cylindrical coffee-bean storage jar has internal radius 6.0 cm. When full it holds 1.2 litres (1200 cm³).

(a)

Calculate the internal height of the jar, to 3 significant figures.

(2)

(b)

The jar has a flat base and no lid. Calculate the total internal surface area of the base and curved wall, using the unrounded height from (a). Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 6² × h = 1200	✓ 1
A1 10.6 cm (from 10.6103…)	✓ 1
Part (b)
M1 base area π × 6² (= 113.1…)	✓ 1
M1 curved area 2 × π × 6 × 10.6103… (= 400.0…)	✓ 1
A1 513 cm²	✓ 1

Exam tip: Keep the unrounded height on your calculator for the next part so that rounding does not change your final digit.

+ Add to test✓ In your test86IGMC3-5-09r15-06Surface area and volume of solids Paper CMedium7 marks

Calculator

A rectangular steel plate on a footbridge measures 2.4 m by 0.85 m. Six identical circular holes, each of radius 5.0 cm, are cut out of the plate.

(a)

Calculate the total area of the six holes, in m², to 3 significant figures.

(2)

(b)

Calculate the area of steel that remains, in m², to 3 significant figures. Use the unrounded answer to (a).

(3)

(c)

The steel has a mass of 62 kg for each square metre. Calculate the mass of the remaining plate, using the unrounded area from (b). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6 × π × 0.05²	✓ 1
A1 0.0471 m²	✓ 1
Part (b)
B1 plate area 2.4 × 0.85 = 2.04 m²	✓ 1
M1 2.04 − 0.047124…	✓ 1
A1 1.99 m²	✓ 1
Part (c)
M1 1.992876… × 62	✓ 1
A1 124 kg	✓ 1

Common error: The hole radius is in centimetres but the plate is in metres, so convert 5.0 cm to 0.05 m before using the area formula.

+ Add to test✓ In your test87IGMC3-5-09r15-07Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A circular observation deck at the top of a bridge tower has a circumference of 62.8 m.

(a)

Calculate the radius of the deck, to 3 significant figures.

(2)

(b)

Calculate the area of the deck, using the unrounded radius from (a). Give your answer to 3 significant figures.

(2)

(c)

A circular glass panel of radius 4.0 m is set into the floor of the deck. Calculate the percentage of the deck's area that is glass, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × r = 62.8	✓ 1
A1 9.99 m (from 9.9948…)	✓ 1
Part (b)
M1 π × r² with r = 9.9948…	✓ 1
A1 314 m²	✓ 1
Part (c)
M1 π × 4² ÷ 313.85… × 100 (or using 314)	✓ 1
A1 16.0%	✓ 1

Exam tip: If you are given a circumference, rearrange C = 2πr first and then use the radius for area; do not use the circumference in the area formula.

+ Add to test✓ In your test88IGMC3-5-09r15-10Surface area and volume of solids Paper CMedium9 marks

Calculator

A concrete ramp at the end of a bridge is a prism. Its cross-section is a right-angled triangle with shorter sides 0.90 m and 1.2 m. The ramp is 2.0 m wide (the length of the prism).

Diagram not to scale

(a)

Show that the longest side of the triangular cross-section is 1.5 m.

(2)

(b)

Calculate the volume of the ramp.

(2)

(c)

Calculate the total surface area of the ramp, including the faces on the ground and the ends.

(3)

(d)

Concrete has a density of 2400 kg/m³. Calculate the mass of the ramp.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9² + 1.2² = 0.81 + 1.44 = 2.25	✓ 1
A1 √2.25 = 1.5 m	✓ 1
Part (b)
M1 area of triangle ½ × 0.9 × 1.2 = 0.54	✓ 1
A1 0.54 × 2.0 = 1.08 m³	✓ 1
Part (c)
B1 two triangular ends 2 × 0.54 = 1.08	✓ 1
M1 three rectangular faces (0.9 + 1.2 + 1.5) × 2.0 = 7.2	✓ 1
A1 8.28 m²	✓ 1
Part (d)
M1 1.08 × 2400	✓ 1
A1 2592 kg	✓ 1

Exam tip: List every face of a prism (two ends and each rectangle) before adding, so none is missed or counted twice.

+ Add to test✓ In your test89IGMC3-5-10r16-03Surface area and volume of solids Paper CMedium6 marks

Calculator

A coffee-roasting drum is a cylinder with diameter 0.6 m and length 0.8 m.

(a)

Calculate the area of one circular end of the drum. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the volume of the drum. Give your answer correct to 3 significant figures.

(2)

(c)

1 m³ = 1000 litres. Calculate the capacity of the drum, in litres, correct to the nearest litre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.3²	✓ 1
A1 0.283 m²	✓ 1
Part (b)
M1 π × 0.3² × 0.8 (use the unrounded area)	✓ 1
A1 0.226 m³	✓ 1
Part (c)
M1 0.2262 × 1000 (unrounded volume 0.22619...)	✓ 1
A1 226 litres	✓ 1

Common error: The radius is half the diameter; using 0.6 as the radius makes the volume four times too large.

+ Add to test✓ In your test90IGMC3-5-10r16-04Circles, arcs and sectors Paper CMedium5 marks

Calculator

A trade-show sign is a sector of a circle with radius 15 cm and angle 120°. Give answers correct to 3 significant figures.

Diagram not to scale

(a)

Calculate the length of the curved edge.

(2)

(b)

Calculate the perimeter of the sign.

(1)

(c)

Calculate the area of the sign.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 120/360 × 2 × π × 15	✓ 1
A1 31.4 cm	✓ 1
Part (b)
A1 31.4159... + 15 + 15 = 61.4 cm	✓ 1
Part (c)
M1 120/360 × π × 15²	✓ 1
A1 236 cm²	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the arc.

+ Add to test✓ In your test91IGMC3-5-10r16-05Perimeter and area of plane shapes Paper CMedium7 marks

Calculator

A viewing platform on a bridge is a rectangle 12 m by 8 m with a semicircle added on one of the 8 m sides, so the diameter of the semicircle is 8 m. Give answers correct to 3 significant figures.

Diagram not to scale

(a)

Calculate the area of the semicircle.

(2)

(b)

Calculate the total area of the platform.

(2)

(c)

Calculate the perimeter of the platform.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 4²	✓ 1
A1 25.1 m²	✓ 1
Part (b)
M1 12 × 8 + ½ × π × 4² = 96 + 25.13...	✓ 1
A1 121 m²	✓ 1
Part (c)
M1 curved edge π × 4 = 12.57...	✓ 1
M1 straight edges 12 + 12 + 8 = 32	✓ 1
A1 44.6 m	✓ 1

Common error: The side of the rectangle where the semicircle is attached is not part of the perimeter.

+ Add to test✓ In your test92IGMC3-5-10r16-07Surface area and volume of solids Paper CMedium7 marks

No calculator

A concrete support block for a bridge is a prism of length 10 m. Its cross-section is a right-angled triangle with perpendicular sides 3 m and 4 m.

Diagram not to scale

(a)

Work out the area of the cross-section.

(1)

(b)

Work out the volume of the block.

(1)

(c)

The concrete has density 2400 kg/m³. Work out the mass of the block.

(2)

(d)

Work out the total surface area of the block.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
½ × 3 × 4 = 6 m²	✓ 1
Part (b)
6 × 10 = 60 m³	✓ 1
Part (c)
M1 60 × 2400	✓ 1
A1 144 000 kg	✓ 1
Part (d)
M1 hypotenuse = √(3² + 4²) = 5	✓ 1
M1 two triangles 2 × 6 = 12 and rectangles (3 + 4 + 5) × 10 = 120	✓ 1
A1 132 m²	✓ 1

Common error: Do not forget the two triangular ends when finding the total surface area of a prism.

+ Add to test✓ In your test93IGMC3-5-10r18-04Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A curved footpath at the end of a bridge is a sector of a circle of radius 9 m and angle 140°.

Diagram not to scale

(a)

Calculate the length of the curved edge of the sector. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the area of the footpath. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the perimeter of the footpath. Use unrounded values and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 140/360 × 2 × π × 9	✓ 1
A1 22.0 m	✓ 1
Part (b)
M1 140/360 × π × 9²	✓ 1
A1 99.0 m²	✓ 1
Part (c)
M1 arc length + 2 × 9 (= 21.99... + 18)	✓ 1
A1 40.0 m	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the arc.

+ Add to test✓ In your test94IGMC3-5-10r18-08Surface area and volume of solids Paper CMedium6 marks

Calculator

A bridge pier is a prism 9 m long. Its cross-section is a trapezium with parallel sides of 3.2 m and 4.8 m, which are 6.5 m apart.

Diagram not to scale

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the volume of the pier.

(2)

(c)

The concrete has density 2.4 tonnes per cubic metre. Calculate the mass of the pier in tonnes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1/2 × (3.2 + 4.8) × 6.5	✓ 1
A1 26 m²	✓ 1
Part (b)
M1 26 × 9	✓ 1
A1 234 m³	✓ 1
Part (c)
M1 234 × 2.4	✓ 1
A1 561.6 tonnes	✓ 1

Exam tip: Volume of a prism = cross-sectional area × length; find the area of the cross-section first.

+ Add to test✓ In your test95IGMC3-5-10r18-12Surface area and volume of solids Paper CMedium6 marks

Calculator

A bridge has 4 identical cylindrical support pillars. Each pillar has radius 0.8 m and height 5.5 m.

(a)

Calculate the curved surface area of one pillar. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the volume of one pillar. Give your answer correct to 3 significant figures.

(2)

(c)

The curved surfaces of all 4 pillars are painted. Paint costs €6.50 per square metre. Use the unrounded area and calculate the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 0.8 × 5.5	✓ 1
A1 27.6 m²	✓ 1
Part (b)
M1 π × 0.8² × 5.5	✓ 1
A1 11.1 m³	✓ 1
Part (c)
M1 4 × 27.646... × 6.50	✓ 1
A1 €718.80	✓ 1

Exam tip: Curved surface area of a cylinder is 2πrh; the flat circular ends are not painted here.

+ Add to test✓ In your test96IGMC3-5-10r18-13Surface area and volume of solids Paper CMedium6 marks

Calculator

A cuboid equipment bay on a space probe has a rectangular base measuring 0.6 m by 0.45 m. Its volume is 0.1485 m³.

Diagram not to scale

(a)

Calculate the height of the bay.

(2)

(b)

Calculate the total surface area of the bay, including top and base.

(2)

(c)

The whole surface is covered in thermal foil costing £4.20 per square metre. Calculate the cost of the foil.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.1485 ÷ (0.6 × 0.45)	✓ 1
A1 0.55 m	✓ 1
Part (b)
M1 2(0.6 × 0.45 + 0.6 × 0.55 + 0.45 × 0.55)	✓ 1
A1 1.695 m²	✓ 1
Part (c)
M1 1.695 × 4.20	✓ 1
A1 £7.12	✓ 1

Common error: A cuboid has six faces in three equal pairs; check you have counted both of each pair.

+ Add to test✓ In your test97IGMC3-5-10r23-04Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A circular mosaic of radius 3.2 m was found at an archaeological site. Only a sector of angle 135° has survived. Use π from your calculator and round only your final answers.

Diagram not to scale

(a)

Calculate the area of the surviving sector. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the length of the curved edge of the sector. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the perimeter of the sector. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 135/360 × π × 3.2²	✓ 1
A1 12.1 m²	✓ 1
Part (b)
M1 135/360 × 2 × π × 3.2	✓ 1
A1 7.54 m	✓ 1
Part (c)
M1 arc length + 2 × 3.2	✓ 1
A1 13.9 m	✓ 1

Common error: The perimeter of a sector includes both straight radii as well as the arc.

+ Add to test✓ In your test98IGMC3-5-10r23-05Surface area and volume of solids Paper CMedium6 marks

Calculator

A cyclist's water bottle is a cylinder of radius 3.5 cm and height 20 cm.

Diagram not to scale

(a)

Calculate the volume of the bottle. Give your answer correct to 3 significant figures.

(2)

(b)

The bottle contains 750 cm³ of water. Calculate the depth of the water when the bottle stands upright. Give your answer correct to 3 significant figures.

(2)

(c)

A label covers the whole curved surface of the bottle between heights 4 cm and 16 cm. Calculate the area of the label. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3.5² × 20	✓ 1
A1 770 cm³	✓ 1
Part (b)
M1 750 ÷ (π × 3.5²)	✓ 1
A1 19.5 cm	✓ 1
Part (c)
M1 2 × π × 3.5 × 12	✓ 1
A1 264 cm²	✓ 1

Exam tip: Keep the units consistent: 1 cm³ is 1 millilitre.

+ Add to test✓ In your test99IGMC3-5-10r23-06Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

The floor plan of an ancient hall is a rectangle 14 m long and 9 m wide with a semicircle added on one of the 9 m sides. The diameter of the semicircle is 9 m. Use π from your calculator.

Diagram not to scale

(a)

Calculate the area of the floor plan. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the perimeter of the floor plan. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle 14 × 9 = 126	✓ 1
M1 semicircle ½ × π × 4.5² = 31.8	✓ 1
A1 158 m²	✓ 1
Part (b)
M1 straight edges 14 + 14 + 9 = 37	✓ 1
M1 semicircular arc ½ × π × 9 = 14.1	✓ 1
A1 51.1 m	✓ 1

Common error: The side where the semicircle is attached is no longer part of the perimeter.

+ Add to test✓ In your test100IGMC3-5-10r23-07Surface area and volume of solids Paper CMedium6 marks

Calculator

A replica stone wedge is a prism whose cross-section is a right-angled triangle with perpendicular sides 9 cm and 12 cm. The prism is 30 cm long.

Diagram not to scale

(a)

Calculate the length of the longest side of the triangular cross-section.

(1)

(b)

Calculate the volume of the wedge.

(2)

(c)

Calculate the total surface area of the wedge.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 15 cm	✓ 1
Part (b)
M1 ½ × 9 × 12 × 30	✓ 1
A1 1620 cm³	✓ 1
Part (c)
M1 two triangles 2 × ½ × 9 × 12 = 108	✓ 1
M1 three rectangles (9 + 12 + 15) × 30 = 1080 (ECF from (a))	✓ 1
A1 1188 cm²	✓ 1

Exam tip: For a prism, the lateral area equals the perimeter of the cross-section multiplied by the length.

+ Add to test✓ In your test101IGMC3-5-10r23-14Surface area and volume of solids Paper CMedium6 marks

Calculator

An excavation plot is a trapezium. Its parallel sides are 18 m and 24 m, and the distance between them is 10 m.

Diagram not to scale

(a)

Calculate the area of the plot.

(2)

(b)

The plot is divided into square grid cells of side 3 m. Calculate the least number of cells whose total area is at least the area of the plot.

(2)

(c)

The team digs the whole plot to a depth of 0.35 m. Calculate the volume of soil removed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (18 + 24) × 10	✓ 1
A1 210 m²	✓ 1
Part (b)
M1 210 ÷ 9 = 23.3...	✓ 1
A1 24	✓ 1
Part (c)
M1 210 × 0.35 (ECF from (a))	✓ 1
A1 73.5 m³	✓ 1

Exam tip: When a count of whole objects is needed, round up if the answer is not a whole number.

+ Add to test✓ In your test102IGMC3-5-12r31-03Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A music stage is a rectangle 8 m by 5 m with a semicircle added on one of the 5 m sides, so that the diameter of the semicircle is 5 m.

Diagram not to scale

(a)

Calculate the total area of the stage. Give your answer to 3 significant figures.

(3)

(b)

Calculate the perimeter of the stage. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area of rectangle $8\times5 = 40$	✓ 1
M1 area of semicircle $\frac{1}{2}\times\pi\times2.5^2$ $(=9.82)$	✓ 1
A1 49.8 m²	✓ 1
Part (b)
M1 curved edge $\frac{1}{2}\times\pi\times5$ $(=7.85)$	✓ 1
M1 straight edges $8 + 8 + 5 = 21$ (the shared side is not included)	✓ 1
A1 28.9 m	✓ 1

Common error: When a semicircle is joined to a rectangle, the joining side is inside the shape, so it is left out of the perimeter.

+ Add to test✓ In your test103IGMC3-5-12r31-05Surface area and volume of solids Paper CMedium6 marks

Calculator

An airport apron is a trapezium. Its two parallel sides are 400 m and 650 m long and the distance between them is 220 m. The apron is to be resurfaced with a layer 0.15 m thick.

Diagram not to scale

(a)

Calculate the area of the apron.

(2)

(b)

Calculate the volume of material needed for the layer.

(2)

(c)

Each lorry carries 12 m³ of material. Find the least number of lorry loads needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}\times(400+650)\times220$	✓ 1
A1 115 500 m²	✓ 1
Part (b)
M1 $115\,500\times0.15$ (ECF)	✓ 1
A1 17 325 m³	✓ 1
Part (c)
M1 $17\,325\div12$ $(=1443.75)$ (ECF)	✓ 1
A1 1444 (rounded up to a whole number)	✓ 1

Common error: Round up, not to the nearest whole number, when you need enough loads to carry all the material.

+ Add to test✓ In your test104IGMC3-5-12r31-06Circles, arcs and sectors Paper CMedium6 marks

Calculator

A sound reflector is a sector of a circle of radius 45 cm and angle 120°.

Diagram not to scale

(a)

Calculate the arc length of the sector. Give your answer to 3 significant figures.

(2)

(b)

Calculate the area of the sector. Give your answer to 3 significant figures.

(2)

(c)

Calculate the perimeter of the sector. Use the unrounded arc length and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{120}{360}\times2\times\pi\times45$	✓ 1
A1 94.2 cm	✓ 1
Part (b)
M1 $\frac{120}{360}\times\pi\times45^2$	✓ 1
A1 2120 cm²	✓ 1
Part (c)
M1 arc length + two radii $= 94.25 + 2\times45$	✓ 1
A1 184 cm	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the curved arc.

+ Add to test✓ In your test105IGMC3-5-12r31-12Surface area and volume of solids Paper CMedium7 marks

Calculator

A bass drum is a cylinder of diameter 44 cm and height 30 cm with a circular skin on each end.

(a)

Calculate the area of the curved surface. Give your answer to 3 significant figures.

(2)

(b)

Calculate the total surface area, including both circular ends. Use unrounded values and give your answer to 3 significant figures.

(3)

(c)

Calculate the volume of the drum in litres (1 litre = 1000 cm³). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times44\times30$ (or $2\pi\times22\times30$)	✓ 1
A1 4150 cm²	✓ 1
Part (b)
M1 area of one end $\pi\times22^2$ $(=1520)$	✓ 1
M1 curved area + 2 × one end	✓ 1
A1 7190 cm²	✓ 1
Part (c)
M1 $\pi\times22^2\times30\div1000$	✓ 1
A1 45.6 litres	✓ 1

Common error: The radius is half of the diameter; using 44 cm as the radius makes the volume four times too big.

+ Add to test✓ In your test106IGMC3-5-11r34-05Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

The spinner in a video-game bonus round is a circle of radius 8 cm. One coloured sector has an angle of 72° at the centre. Give answers correct to 3 significant figures.

Diagram not to scale

(a)

Calculate the arc length of the sector.

(2)

(b)

Calculate the area of the sector.

(2)

(c)

Calculate the perimeter of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 72/360 × 2 × π × 8	✓ 1
A1 10.1 cm	✓ 1
Part (b)
M1 72/360 × π × 8²	✓ 1
A1 40.2 cm²	✓ 1
Part (c)
M1 arc length + 8 + 8 (ECF from (a))	✓ 1
A1 26.1 cm	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the arc.

+ Add to test✓ In your test107IGMC3-5-11r34-06Surface area and volume of solids Paper CMedium4 marks

Calculator

A cylindrical medicine bottle has internal radius 2.4 cm and internal height 9.5 cm when full. 1 cm³ = 1 ml.

(a)

Calculate the capacity of the bottle in ml, correct to 3 significant figures.

(2)

(b)

One dose is 15 ml. Work out the number of complete doses that can be poured from a full bottle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 2.4² × 9.5	✓ 1
A1 172 ml	✓ 1
Part (b)
M1 171.9… ÷ 15 (or 172 ÷ 15) = 11.4…	✓ 1
A1 11	✓ 1

Exam tip: Questions about complete doses, tiles or containers need the answer rounded down to a whole number, not to the nearest.

+ Add to test✓ In your test108IGMC3-5-11r34-07Surface area and volume of solids Paper CMedium4 marks

Calculator

A water pipe is a cylinder of internal radius 1.5 cm and length 40 m.

(a)

Write the length of the pipe in centimetres.

(1)

(b)

Calculate the volume of water that fills the pipe, in cm³, correct to 3 significant figures.

(2)

(c)

Write your answer to (b) in litres. 1 litre = 1000 cm³.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4000 cm	✓ 1
Part (b)
M1 π × 1.5² × 4000 (ECF from (a))	✓ 1
A1 28 300 cm³	✓ 1
Part (c)
B1 28.3 litres (ECF from (b))	✓ 1

Common error: Use the same length unit for radius and length before calculating a volume; mixing m and cm gives an answer that is wrong by a factor of 100.

+ Add to test✓ In your test109IGMC3-5-11r34-08Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A pharmacy sign is made from a rectangle 60 cm long and 30 cm wide with a semicircle attached to one of the 30 cm sides. The diameter of the semicircle is 30 cm. Give your answers correct to the nearest whole number.

Diagram not to scale

(a)

Calculate the perimeter of the sign.

(3)

(b)

Calculate the area of the sign.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 semicircle arc = ½ × π × 30 (= 47.1…)	✓ 1
M1 60 + 60 + 30 + arc	✓ 1
A1 197 cm	✓ 1
Part (b)
M1 ½ × π × 15² (= 353.4…)	✓ 1
M1 60 × 30 + semicircle area	✓ 1
A1 2153 cm²	✓ 1

Common error: The side where the semicircle joins the rectangle is not part of the outside edge, so leave it out of the perimeter.

+ Add to test✓ In your test110IGMC3-5-11r34-09Surface area and volume of solids Paper CMedium5 marks

Calculator

A games console is packed in a closed cuboid box measuring 30 cm by 22 cm by 8 cm. The card used to make the box costs €0.35 for every 100 cm². Ignore any overlaps in the card.

(a)

Calculate the total surface area of the box.

(3)

(b)

Work out the cost of the card for one box, to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 any two different face areas from 30 × 22, 30 × 8, 22 × 8 (660, 240, 176)	✓ 1
M1 2 × (660 + 240 + 176)	✓ 1
A1 2152 cm²	✓ 1
Part (b)
M1 2152 ÷ 100 × 0.35 (ECF from (a))	✓ 1
A1 €7.53	✓ 1

Exam tip: Write each pair of faces separately (top and bottom, front and back, two sides) so that no face is missed.

+ Add to test✓ In your test111IGMC3-5-11r37-04Perimeter and area of plane shapes Paper CMedium7 marks

Calculator

The floor of a library reading area is a rectangle 12 m long and 8 m wide. A semicircle of diameter 8 m is added to one of the 8 m sides, so the semicircle sticks out from the rectangle.

Diagram not to scale

(a)

Calculate the area of the semicircle, correct to 3 significant figures.

(2)

(b)

Calculate the total area of the floor, correct to 3 significant figures. Use the unrounded area of the semicircle.

(2)

(c)

Calculate the perimeter of the floor, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × π × 4²	✓ 1
A1 25.1 m²	✓ 1
Part (b)
M1 12 × 8 + 25.13…	✓ 1
A1 121 m²	✓ 1
Part (c)
M1 straight edges 12 + 12 + 8 = 32 m	✓ 1
M1 semicircular arc = π × 4 (= 12.57…)	✓ 1
A1 44.6 m	✓ 1

Common error: The 8 m side where the semicircle is joined is not part of the perimeter, so do not include it.

+ Add to test✓ In your test112IGMC3-5-11r37-05Circles, arcs and sectors Paper CMedium6 marks

Calculator

The blades of a wind turbine are 40 m long and rotate about a central point. The tip of a blade traces a circle of radius 40 m.

(a)

Calculate the area of the circle swept by the blades. Give your answer to 3 significant figures.

(2)

(b)

Calculate the distance travelled by the tip of a blade in one complete revolution, correct to 3 significant figures.

(2)

(c)

In one second the blade turns through an angle of 72°. Calculate the distance travelled by the tip in this time, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 40²	✓ 1
A1 5030 m²	✓ 1
Part (b)
M1 2 × π × 40	✓ 1
A1 251 m	✓ 1
Part (c)
M1 (72 ÷ 360) × 2 × π × 40	✓ 1
A1 50.3 m	✓ 1

Exam tip: The distance travelled along an arc is the same fraction of the circumference as the angle is of 360°.

+ Add to test✓ In your test113IGMC3-5-11r37-06Surface area and volume of solids Paper CMedium5 marks

Calculator

A circular paddling pool has radius 1.5 m. It is filled with water to a depth of 0.4 m.

(a)

Calculate the volume of water in the pool in m³, correct to 3 significant figures.

(2)

(b)

1 m³ = 1000 litres. Write the volume of water to the nearest litre, using the unrounded volume.

(1)

(c)

A hose delivers water at 18 litres per minute. Using 2827 litres, calculate the time taken to fill the pool, to the nearest minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 1.5² × 0.4	✓ 1
A1 2.83 m³	✓ 1
Part (b)
B1 2827 litres	✓ 1
Part (c)
M1 2827 ÷ 18	✓ 1
A1 157 minutes	✓ 1

Exam tip: Carry the full calculator value into the next part and round only the final answer.

+ Add to test✓ In your test114IGMC3-5-11r37-09Perimeter and area of plane shapes Paper CMedium7 marks

Calculator

A library is laying carpet tiles in a rectangular room 7.5 m long and 6 m wide. Each tile is a square of side 50 cm.

(a)

Calculate the area of the room in m².

(2)

(b)

Show that exactly 180 tiles are needed to cover the floor.

(3)

(c)

Each tile costs £8.75. Calculate the total cost of the tiles.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7.5 × 6	✓ 1
A1 45 m²	✓ 1
Part (b)
B1 area of one tile = 0.5 × 0.5 = 0.25 m²	✓ 1
M1 45 ÷ 0.25	✓ 1
A1 180	✓ 1
Part (c)
M1 180 × 8.75	✓ 1
A1 £1575.00	✓ 1

Common error: Convert 50 cm to 0.5 m before multiplying, otherwise the tile area is wrong by a factor of 10 000.

+ Add to test✓ In your test115IGMC3-5-11r37-13Surface area and volume of solids Paper CMedium6 marks

Calculator

A swimming pool is a prism whose cross-section is a trapezium. The two parallel sides of the trapezium are vertical and measure 1.2 m (shallow end) and 2.8 m (deep end). The horizontal distance between them is 20 m. The pool is 8 m wide.

Diagram not to scale

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the volume of the pool.

(2)

(c)

Water is pumped into the empty pool at 25 m³ per hour. Calculate the time to fill it, in hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1.2 + 2.8) × 20	✓ 1
A1 40 m²	✓ 1
Part (b)
M1 40 × 8	✓ 1
A1 320 m³	✓ 1
Part (c)
M1 320 ÷ 25	✓ 1
A1 12.8 hours	✓ 1

Common error: The parallel sides of the trapezium are the two depths; the 20 m is the distance between them, not a slanted side.

+ Add to test✓ In your test116IGMC3-5-09r38-05Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A library display stand has a top in the shape of a sector of a circle with radius 12 cm and sector angle 150°. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the length of the curved edge of the sector.

(2)

(b)

Calculate the area of the top of the stand.

(2)

(c)

Calculate the perimeter of the top of the stand. Use the unrounded answer from part (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 150/360 × 2 × π × 12	✓ 1
A1 31.4 cm	✓ 1
Part (b)
M1 150/360 × π × 12²	✓ 1
A1 188 cm²	✓ 1
Part (c)
M1 arc length + 12 + 12	✓ 1
A1 55.4 cm	✓ 1

Common error: The perimeter of a sector includes the two straight radii as well as the arc.

+ Add to test✓ In your test117IGMC3-5-09r38-06Surface area and volume of solids Paper CMedium7 marks

Calculator

A closed cylindrical water tank at a wind farm has radius 0.8 m and height 2.5 m. Give answers to 3 significant figures unless told otherwise.

Diagram not to scale

(a)

Calculate the volume of the tank.

(2)

(b)

Calculate the total surface area of the tank, including both circular ends.

(3)

(c)

The outside of the tank is painted. Paint costs £4.50 for each m². Use the unrounded surface area from part (b) to calculate the cost, giving your answer in pounds to the nearest penny.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.8² × 2.5	✓ 1
A1 5.03 m³	✓ 1
Part (b)
M1 curved surface = 2 × π × 0.8 × 2.5 (= 12.57...)	✓ 1
M1 add two circular ends, 2 × π × 0.8² (= 4.021...)	✓ 1
A1 16.6 m² (16.588... before rounding)	✓ 1
Part (c)
M1 16.588... × 4.50	✓ 1
A1 £74.64	✓ 1

Exam tip: Write down the formula for each face separately before adding; it makes missing faces easy to spot.

+ Add to test✓ In your test118IGMC3-5-09r38-07Surface area and volume of solids Paper CMedium6 marks

No calculator

A swimming pool has a uniform width of 8 m and a length of 20 m. The floor slopes steadily, so a side view of the pool is a trapezium with vertical sides of depth 1.0 m at the shallow end and 2.2 m at the deep end.

Diagram not to scale

(a)

Work out the area of the trapezium-shaped side view.

(2)

(b)

Work out the volume of the pool in m³.

(2)

(c)

The pool is filled at a rate of 4.0 m³ per minute. Work out the time taken to fill it, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1.0 + 2.2) × 20	✓ 1
A1 32 m²	✓ 1
Part (b)
M1 32 × 8 (area of cross-section × width)	✓ 1
A1 256 m³ (ECF from (a))	✓ 1
Part (c)
M1 256 ÷ 4.0 (ECF from (b))	✓ 1
A1 64 minutes	✓ 1

Common error: In a trapezium the two parallel sides are the vertical depths here, and the length of 20 m is the distance between them; do not use the sloping floor length.

+ Add to test✓ In your test119IGMC3-5-09r38-13Perimeter and area of plane shapes Paper CMedium5 marks

No calculator

A rectangular reading room in a library has a perimeter of 44 m. Its length is 14 m.

(a)

Work out the width of the room.

(2)

(b)

Work out the area of the floor.

(1)

(c)

The floor is covered with square carpet tiles of side 0.5 m. Work out the number of tiles needed, assuming none are cut and none are wasted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 44 ÷ 2 − 14 (or 44 − 2 × 14, then ÷ 2)	✓ 1
A1 8 m	✓ 1
Part (b)
B1 112 m² (ECF from (a))	✓ 1
Part (c)
M1 area of one tile = 0.5 × 0.5 = 0.25 m² and 112 ÷ 0.25	✓ 1
A1 448	✓ 1

Exam tip: The perimeter formula 2 × (length + width) works backwards: halve the perimeter first, then subtract the known side.

+ Add to test✓ In your test120IGMC3-5-09r38-14Perimeter and area of plane shapes Paper CMedium6 marks

No calculator

A triangular sign at the entrance of a library has a base of 1.8 m and an area of 0.99 m².

Diagram not to scale

(a)

Show that the perpendicular height of the triangle is 1.1 m.

(2)

(b)

A second sign is a parallelogram with base 2.4 m and the same perpendicular height of 1.1 m. Work out its area.

(2)

(c)

Paint for the second sign costs £12.50 per m². Work out the cost of painting one face of the parallelogram sign.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 1.8 × h = 0.99 (or h = 2 × 0.99 ÷ 1.8)	✓ 1
A1 h = 1.98 ÷ 1.8 = 1.1 m, shown	✓ 1
Part (b)
M1 2.4 × 1.1	✓ 1
A1 2.64 m²	✓ 1
Part (c)
M1 2.64 × 12.50 (ECF from (b))	✓ 1
A1 £33	✓ 1

Common error: The area of a triangle uses the perpendicular height, not the length of a sloping side, and the area of a parallelogram has no ½.

+ Add to test✓ In your test121IGMC3-5-09r39-05Perimeter and area of plane shapes Paper CMedium6 marks

Calculator

A piece of scenery for a theatre is made from a rectangle measuring 3.6 m by 2.4 m. A semicircle with diameter 2.4 m is attached along one of the 2.4 m sides, so that the semicircle lies outside the rectangle.

Diagram not to scale

(a)

Calculate the area of the piece of scenery, giving your answer to 3 significant figures.

(3)

(b)

Calculate the perimeter of the piece of scenery, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle: 3.6 × 2.4 = 8.64	✓ 1
M1 semicircle: ½ × π × 1.2² = 2.26…	✓ 1
A1 10.9 m²	✓ 1
Part (b)
M1 straight edges: 3.6 + 3.6 + 2.4 = 9.6	✓ 1
M1 curved edge: π × 1.2 = 3.77…	✓ 1
A1 13.4 m	✓ 1

Common error: The side where the semicircle is attached is not part of the perimeter, so do not include that 2.4 m edge.

+ Add to test✓ In your test122IGMC3-5-09r39-06Surface area and volume of solids Paper CMedium6 marks

Calculator

A lava channel on the side of a volcano is 120 m long. Its cross-section is a trapezium with parallel sides 4.0 m and 7.0 m, which are 2.5 m apart. The channel is a prism and is completely full of lava.

Diagram not to scale

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the volume of lava in the channel.

(2)

(c)

The density of the lava is 2600 kg/m³. Calculate the mass of lava in the channel. Give your answer in standard form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (4.0 + 7.0) × 2.5	✓ 1
A1 13.75 m²	✓ 1
Part (b)
M1 13.75 × 120	✓ 1
A1 1650 m³	✓ 1
Part (c)
M1 1650 × 2600	✓ 1
A1 4.29 × 10⁶ kg	✓ 1

Exam tip: In a trapezium the height is the perpendicular distance between the parallel sides, not the length of a sloping side.

+ Add to test✓ In your test123IGMC3-5-09r39-07Surface area and volume of solids Paper CMedium7 marks

Calculator

The funnel of a ship is a hollow cylinder with no top or bottom surface to be painted. Its diameter is 3.2 m and its height is 5.5 m. The outside curved surface is painted with one coat of paint.

Diagram not to scale

(a)

Calculate the area to be painted, giving your answer to 3 significant figures.

(2)

(b)

One litre of paint covers 8 m². The paint is sold in tins containing 2.5 litres. Work out the number of tins needed.

(3)

(c)

Each tin costs £18.75. Calculate the total cost of the tins.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3.2 × 5.5	✓ 1
A1 55.3 m²	✓ 1
Part (b)
M1 55.3 ÷ 8 = 6.91 litres	✓ 1
M1 6.91 ÷ 2.5 = 2.76 tins	✓ 1
A1 3 tins	✓ 1
Part (c)
M1 3 × 18.75	✓ 1
A1 £56.25	✓ 1

Common error: Rounding 2.76 tins to 2 leaves part of the funnel unpainted. Round up to the next whole tin.

+ Add to test✓ In your test124IGMC3-5-09r39-08Surface area and volume of solids Paper CMedium4 marks

Calculator

A circular rotating platform on a stage has a circumference of 22.0 m.

(a)

Calculate the radius of the platform, giving your answer to 3 significant figures.

(2)

(b)

Calculate the area of the top surface of the platform, giving your answer to 3 significant figures. Use the unrounded radius from part (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 22.0 ÷ (2π)	✓ 1
A1 3.50 m	✓ 1
Part (b)
M1 π × (3.501…)²	✓ 1
A1 38.5 m²	✓ 1

Exam tip: Keep the unrounded radius in your calculator and use it in the next part so that rounding does not change your final digit.

+ Add to test✓ In your test125IGMC3-5-09r39-09Surface area and volume of solids Paper CMedium5 marks

No calculator

A geologist cuts a solid cylindrical sample of rock from a lava flow. The sample has radius 3 cm and height 12 cm. Leave your answers in terms of π.

(a)

Show that the volume of the sample is 108π cm³.

(2)

(b)

Work out the total surface area of the sample, including both circular ends.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 3² × 12	✓ 1
A1 108π cm³ (shown)	✓ 1
Part (b)
M1 curved surface: 2 × π × 3 × 12 = 72π	✓ 1
M1 two ends: 2 × π × 3² = 18π	✓ 1
A1 90π cm²	✓ 1

Common error: The total surface area needs the curved surface and both circular ends; leaving out the ends is a frequent slip.

+ Add to test✓ In your test126IGMC3-5-10r48-03Perimeter and area of plane shapes Paper CMedium8 marks

Calculator

The floor of a hospital waiting area is a rectangle 12.0 m long and 8.0 m wide, with a semicircle attached to one of the 8.0 m sides. The diameter of the semicircle is 8.0 m, so the shape has one curved end.

Diagram not to scale

(a)

Calculate the area of the floor. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the perimeter of the floor. Give your answer correct to 3 significant figures.

(3)

(c)

Flooring costs €38.50 per square metre. Use the unrounded area to calculate the cost of covering the floor. Give your answer in euros to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rectangle area 12.0 × 8.0 = 96	✓ 1
M1 semicircle area ½ × π × 4.0² = 25.13...	✓ 1
A1 121 m²	✓ 1
Part (b)
M1 curved length ½ × π × 8.0 = 12.56...	✓ 1
M1 add 12.0 + 12.0 + 8.0	✓ 1
A1 44.6 m	✓ 1
Part (c)
M1 121.13... × 38.50	✓ 1
A1 €4663.61	✓ 1

Common error: The curved edge is only half a circumference, and the diameter side of the semicircle is not part of the outside perimeter.

+ Add to test✓ In your test127IGMC3-5-10r48-04Circles, arcs and sectors Paper CMedium7 marks

Calculator

Part of a Formula 1 circuit is a bend in the shape of a sector of a circle. The inner edge of the bend is an arc of a circle of radius 45.0 m, and the arc subtends an angle of 110° at the centre of the circle. The track is 12.0 m wide.

Diagram not to scale

(a)

Calculate the length of the inner edge of the bend. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the area of the sector with radius 45.0 m and angle 110°. Give your answer correct to 3 significant figures.

(2)

(c)

The outer edge of the track is an arc of radius 57.0 m with the same centre and the same angle. Calculate how much longer the outer edge is than the inner edge. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 110/360 × 2 × π × 45.0	✓ 1
A1 86.4 m	✓ 1
Part (b)
M1 110/360 × π × 45.0²	✓ 1
A1 1940 m²	✓ 1
Part (c)
M1 outer arc 110/360 × 2 × π × 57.0 = 109.4...	✓ 1
M1 subtract the inner arc length 86.39...	✓ 1
A1 23.0 m (accept 110/360 × 2 × π × 12.0)	✓ 1

Exam tip: Arc length and sector area are fractions of the whole circle: angle ÷ 360 multiplies the circumference or the full area.

+ Add to test✓ In your test128IGMC3-5-10r48-05Surface area and volume of solids Paper CMedium8 marks

Calculator

A beekeeper sells honey in cylindrical jars. Each jar is a cylinder with radius 4.0 cm and height 11.0 cm and is completely filled with honey. The density of the honey is 1.4 g/cm³.

Diagram not to scale

(a)

Calculate the volume of honey in one jar. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the mass of honey in one jar. Use the unrounded volume and give your answer in grams, correct to 3 significant figures.

(2)

(c)

A label is wrapped around the outside of the jar and the lid and base are also labelled. Calculate the total surface area of the closed jar, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 4.0² × 11.0	✓ 1
M1 evaluating the product correctly, 552.9...	✓ 1
A1 553 cm³	✓ 1
Part (b)
M1 mass = density × volume = 1.4 × 552.9...	✓ 1
A1 774 g	✓ 1
Part (c)
M1 curved surface 2 × π × 4.0 × 11.0 = 276.4...	✓ 1
M1 two circular ends 2 × π × 4.0² = 100.5...	✓ 1
A1 377 cm²	✓ 1

Common error: A closed cylinder has three surfaces: the curved surface and two circular ends. Forgetting the ends gives only 276 cm².

+ Add to test✓ In your test129IGMC3-5-10r48-08Perimeter and area of plane shapes Paper CMedium7 marks

Calculator

A hospital has a circular flower bed of radius 6.0 m. A gravel path of constant width 1.5 m goes all the way round the flower bed.

Diagram not to scale

(a)

Calculate the area of the path. Give your answer correct to 3 significant figures.

(3)

(b)

The gravel is laid to a depth of 0.080 m. Use the unrounded area of the path to calculate the volume of gravel needed. Give your answer correct to 3 significant figures.

(2)

(c)

Gravel costs €48 per cubic metre. Use the unrounded volume to calculate the cost of the gravel, giving your answer in euros to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 outer radius 6.0 + 1.5 = 7.5 m	✓ 1
M1 π × 7.5² − π × 6.0²	✓ 1
A1 63.6 m²	✓ 1
Part (b)
M1 63.61... × 0.080	✓ 1
A1 5.09 m³	✓ 1
Part (c)
M1 5.0893... × 48	✓ 1
A1 €244.29	✓ 1

Common error: The path area is the difference between two circles; do not use the 1.5 m width as a radius.

+ Add to test✓ In your test130IGMC3-5-10r48-14Surface area and volume of solids Paper CMedium7 marks

Calculator

A physiotherapist at a hospital uses a foam wedge in the shape of a prism. The cross-section is a right-angled triangle with perpendicular sides of 30 cm and 40 cm and a hypotenuse of 50 cm. The prism is 60 cm long.

Diagram not to scale

(a)

Calculate the volume of the wedge.

(2)

(b)

Calculate the total surface area of the wedge.

(3)

(c)

The wedge is covered in fabric that costs €18.50 per square metre. Calculate the cost of the fabric needed to cover the whole surface of the wedge.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 30 × 40 × 60 or area 600 × 60	✓ 1
A1 36 000 cm³	✓ 1
Part (b)
M1 two triangular ends 2 × 600 = 1200	✓ 1
M1 three rectangles (30 + 40 + 50) × 60 = 7200	✓ 1
A1 8400 cm²	✓ 1
Part (c)
M1 convert 8400 cm² to 0.84 m²	✓ 1
A1 0.84 × 18.50 = €15.54	✓ 1

Common error: 1 m² is 10 000 cm², not 100 cm².

+ Add to test✓ In your test131IGMC3-5-09r15-14Circles, arcs and sectors Paper CHard9 marks

Calculator

A flat landing at the end of a footbridge has the shape of a quarter of a ring. It lies between two quarter-circles that share a centre O, with radii 12 m and 20 m. Each quarter-circle has an angle of 90° at O.

Diagram not to scale

(a)

Calculate the area of the landing, to 3 significant figures.

(3)

(b)

Calculate the perimeter of the landing, to 3 significant figures.

(4)

(c)

Show that the area of the inner quarter-circle is 36% of the area of the outer quarter-circle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ¼ × π × 20² (= 314.16…)	✓ 1
M1 subtract ¼ × π × 12² (= 113.1…)	✓ 1
A1 201 m²	✓ 1
Part (b)
M1 outer arc ¼ × 2 × π × 20 (= 31.4…)	✓ 1
M1 inner arc ¼ × 2 × π × 12 (= 18.8…)	✓ 1
B1 two straight edges 2 × (20 − 12) = 16	✓ 1
A1 66.3 m	✓ 1
Part (c)
M1 ratio of areas = π × 12² ÷ π × 20² = 144 ÷ 400	✓ 1
A1 0.36 = 36%	✓ 1

Exam tip: The two straight edges of a ring-shaped region each have length equal to the difference of the radii, not their sum.

+ Add to test✓ In your test132IGMC3-5-10r18-14Surface area and volume of solids Paper CHard7 marks

Calculator

A rectangular tin measuring 20 cm by 15 cm by 30 cm is completely full of coffee beans. The beans are poured, without loss, into an empty cylindrical canister of radius 10 cm and height 40 cm.

(a)

Calculate the depth of beans in the canister. Give your answer correct to 3 significant figures.

(2)

(b)

A roaster has 5 empty canisters like this one and fills them using full tins. Calculate the smallest whole number of tins needed to fill all 5 canisters completely.

(3)

(c)

Calculate the depth of beans in the first canister as a percentage of the canister's height. Use the unrounded depth and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20 × 15 × 30 = 9000 and π × 10² × h = 9000	✓ 1
A1 28.6 cm	✓ 1
Part (b)
M1 5 × π × 10² × 40 (= 62 832)	✓ 1
M1 62 832 ÷ 9000 (= 6.98...)	✓ 1
A1 7 tins	✓ 1
Part (c)
M1 28.648... ÷ 40 × 100	✓ 1
A1 71.6%	✓ 1

Exam tip: Equate volumes when a liquid or solid is poured from one container into another, then solve for the unknown length.

+ Add to test✓ In your test133IGMC3-5-10r23-13Surface area and volume of solids Paper EEasy6 marks

Calculator

A model of a pyramid made for an archaeology exhibition has a horizontal square base of side 16 cm. Its vertical height is 6 cm. The apex is directly above the centre of the base.

Diagram not to scale

(a)

Calculate the volume of the model.

(2)

(b)

Show that the slant height of each triangular face is 10 cm.

(2)

(c)

Calculate the total surface area of the model, including the base.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × 16² × 6	✓ 1
A1 512 cm³	✓ 1
Part (b)
M1 half the base side 8 cm, used with height 6	✓ 1
A1 sqrt(8² + 6²) = 10	✓ 1
Part (c)
M1 4 × ½ × 16 × 10 = 320 (or 256 + 320)	✓ 1
A1 576 cm²	✓ 1

Common error: Use the slant height, not the vertical height, for the area of a triangular face.

+ Add to test✓ In your test134IGMC-045-00-08Perimeter and area of plane shapes Paper EMedium5 marks

No calculator

A sector of a circle has radius 9 cm and angle 140°.
Give each answer in terms of π.

Diagram not to scale

(a)

Find the arc length of the sector.

(2)

(b)

Find the perimeter of the sector.

(1)

(c)

Find the area of the sector.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{140}{360}$ × 2 × π × 9	✓ 1
A1 7π cm	✓ 1
Part (b)
B1 18 + 7π cm	✓ 1
Part (c)
M1 $\frac{140}{360}$ × π × 9²	✓ 1
A1 $\frac{63}{2}$π or 31.5π cm²	✓ 1

Exam tip: Simplify the fraction of the circle first: $\frac{140}{360} = \frac{7}{18}$. Then $\frac{7}{18}$ × 18π = 7π, which keeps the working short without a calculator.

+ Add to test✓ In your test135IGMC-045-00-09Perimeter and area of plane shapes Paper EMedium7 marks

Calculator

In triangle PQR, PQ = 8.5 cm, PR = 11.2 cm and angle QPR = 47°.

Diagram not to scale

(a)

Calculate the area of triangle PQR.

(2)

(b)

Calculate the length QR.

(3)

(c)

Calculate the perimeter of triangle PQR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{2}$ × 8.5 × 11.2 × sin 47°	✓ 1
A1 34.8 cm²	✓ 1
Part (b)
M1 QR² = 8.5² + 11.2² − 2 × 8.5 × 11.2 × cos 47°	✓ 1
A1 QR² = 67.8(4)…	✓ 1
A1 QR = 8.24 cm	✓ 1
Part (c)
M1 8.5 + 11.2 + their QR	✓ 1
A1 27.9 cm	✓ 1

Exam tip: Use the unrounded value of QR (8.2363…) from your calculator in part (c). Rounding early can change the final 3 significant figure answer.

+ Add to test✓ In your test136IGMC-045-01-09Circles, arcs and sectors Paper EMedium6 marks

No calculator

A sector of a circle has radius 6 cm and angle 120°. Give your answers to parts (a) and (b) in terms of π.

Diagram not to scale

(a)

Find the area of the sector.

(2)

(b)

Find the perimeter of the sector.

(2)

(c)

The sector is rolled up so that its two straight edges meet, forming the curved surface of a cone. Find the radius of the base of the cone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{120}{360}$ × π × 6²	✓ 1
A1 12π cm²	✓ 1
Part (b)
M1 for $\frac{120}{360}$ × 2 × π × 6 [= 4π]	✓ 1
A1 4π + 12 cm	✓ 1
Part (c)
M1 for 2πr = 4π	✓ 1
A1 2 cm	✓ 1

Exam tip: When a sector forms a cone, the arc length becomes the circumference of the cone's base, and the sector's radius becomes the slant height.

+ Add to test✓ In your test137IGMC-045-02-08Surface area and volume of solids Paper EMedium7 marks

Calculator

A paperweight is a solid hemisphere of radius 9 cm.
[Volume of sphere = $\frac{4}{3}\pi r^3$, surface area of sphere = $4\pi r^2$]

Diagram not to scale

(a)

Calculate the volume of the paperweight.

(2)

(b)

The paperweight is made of a metal with density 2.7 g/cm³.
Calculate its mass in kilograms.

(2)

(c)

Calculate the total surface area of the paperweight.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{1}{2}$ × $\frac{4}{3}$ × π × 9³	✓ 1
A1: 1530 cm³ (1526.8…)	✓ 1
Part (b)
M1: their (a) × 2.7, giving 4122.4… g (or 4131 g from 1530)	✓ 1
A1: ÷ 1000 to give 4.12 kg (4.122…); accept 4.13 kg from using 1530	✓ 1
Part (c)
M1: curved surface $\frac{1}{2}$ × 4 × π × 9²	✓ 1
M1: adding flat circle π × 9²	✓ 1
A1: 763 cm² (763.4…)	✓ 1

Common error: A solid hemisphere has a flat circular face as well as its curved surface. Leaving out the circle gives only 2πr².

+ Add to test✓ In your test138IGMC-045-02-09Surface area and volume of solids Paper EMedium6 marks

No calculator

A solid pyramid has a square base of side 10 cm. The vertex is vertically above the centre of the base, and the vertical height is 12 cm.
[Volume of pyramid = $\frac{1}{3}$ × area of base × height]

Diagram not to scale

(a)

Find the volume of the pyramid.

(2)

(b)

Find the height of one triangular face, measured from the midpoint of a base edge to the vertex.

(2)

(c)

Find the total surface area of the pyramid.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{1}{3}$ × 10² × 12	✓ 1
A1: 400 cm³	✓ 1
Part (b)
M1: √(12² + 5²)	✓ 1
A1: 13 cm	✓ 1
Part (c)
M1: 10² + 4 × $\frac{1}{2}$ × 10 × 13	✓ 1
A1: 360 cm²	✓ 1

Exam tip: For the triangular faces you need the slant height of a face, not the vertical height of the pyramid. Use half the base side in Pythagoras.

+ Add to test✓ In your test139IGMC-045-02-11Surface area and volume of solids Paper EMedium4 marks

No calculator

Two bottles are mathematically similar. The smaller bottle has height 12 cm and the larger bottle has height 18 cm.

(a)

The surface area of the smaller bottle is 240 cm².
Find the surface area of the larger bottle.

(2)

(b)

The volume of the larger bottle is 1350 cm³.
Find the volume of the smaller bottle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 240 × (18 ÷ 12)² or 240 × 2.25	✓ 1
A1: 540 cm²	✓ 1
Part (b)
M1: 1350 ÷ (1.5)³ or 1350 × (2/3)³	✓ 1
A1: 400 cm³	✓ 1

Exam tip: If the length scale factor is k, areas scale by k² and volumes by k³. Check that the smaller solid gets the smaller answer.

+ Add to test✓ In your test140IGMC3-5-02d-10Perimeter and area of plane shapes Paper EMedium6 marks

Calculator

A spherical weather balloon has radius 1.5 m when it is launched.

(a)

Calculate the volume of the balloon. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the surface area of the balloon. Give your answer correct to 3 significant figures.

(2)

(c)

High in the atmosphere the balloon expands to radius 3.0 m. Find the new volume as a multiple of the original volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (4 ÷ 3) × π × 1.5³	✓ 1
A1 14.1 m³	✓ 1
Part (b)
M1 4 × π × 1.5²	✓ 1
A1 28.3 m²	✓ 1
Part (c)
M1 (3.0 ÷ 1.5)³	✓ 1
A1 8	✓ 1

Exam tip: Doubling the radius multiplies the volume by 2³ = 8, so you can answer ratio questions without calculating either volume.

+ Add to test✓ In your test141IGMC3-5-02d-11Circles, arcs and sectors Paper EMedium5 marks

Calculator

A coastal radar sweeps a sector of a circle. The radius of the sector is 12 km and its area is 30π km².

Diagram not to scale

(a)

Calculate the angle of the sector.

(3)

(b)

Calculate the length of the curved edge of the sector. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (θ ÷ 360) × π × 12² = 30π	✓ 1
M1 θ = 30 × 360 ÷ 144	✓ 1
A1 75°	✓ 1
Part (b)
M1 (75 ÷ 360) × 2 × π × 12 (ECF from (a))	✓ 1
A1 15.7 km	✓ 1

Exam tip: Leaving π in the equation lets it cancel, so the angle comes out exactly with no rounding.

+ Add to test✓ In your test142IGMC3-5-08r12-09Surface area and volume of solids Paper EMedium6 marks

Calculator

The roof of a planetarium is a hemisphere of radius 6.5 m.

Diagram not to scale

(a)

Calculate the curved surface area of the roof. Give your answer to the nearest square metre.

(2)

(b)

Calculate the volume of air inside the roof. Give your answer to the nearest cubic metre.

(2)

(c)

A model of the planetarium is a hemisphere of radius 1.3 m. Calculate the volume of the model. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 6.5² (half of 4πr²)	✓ 1
A1 265 m²	✓ 1
Part (b)
M1 ²⁄₃ × π × 6.5³	✓ 1
A1 575 m³	✓ 1
Part (c)
M1 ²⁄₃ × π × 1.3³ (or 575.17 ÷ 5³)	✓ 1
A1 4.60 m³	✓ 1

Exam tip: Volume of a sphere is ⁴⁄₃πr³; for a hemisphere, halve it.

+ Add to test✓ In your test143IGMC3-5-08r12-10Surface area and volume of solids Paper EMedium6 marks

Calculator

A railway depot has a pile of stone ballast in the shape of a cone of base radius 4.5 m and perpendicular height 3.2 m.

Diagram not to scale

(a)

Calculate the volume of the pile. Give your answer to 3 significant figures.

(2)

(b)

Calculate the slant height of the cone. Give your answer to 3 significant figures.

(2)

(c)

The curved surface of the pile is covered with a sheet. Calculate the area of the sheet. Use the unrounded slant height from (b) and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ¹⁄₃ × π × 4.5² × 3.2	✓ 1
A1 67.9 m³	✓ 1
Part (b)
M1 √(4.5² + 3.2²) (= √30.49)	✓ 1
A1 5.52 m	✓ 1
Part (c)
M1 π × 4.5 × 5.5217...	✓ 1
A1 78.1 m²	✓ 1

Common error: The slant height, not the vertical height, is used in the curved surface area formula πrl.

+ Add to test✓ In your test144IGMC3-5-08r12-11Surface area and volume of solids Paper EMedium6 marks

Calculator

A rail tanker is a cylinder of radius 1.2 m and length 8.0 m with a hemisphere attached to each end.

Diagram not to scale

(a)

Calculate the total volume of the tanker. Give your answer to 3 significant figures.

(4)

(b)

The tanker is filled at a rate of 850 litres per minute. Calculate the time taken to fill it completely. Use the unrounded answer to (a), 1 m³ = 1000 litres, and give your answer in minutes to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cylinder volume π × 1.2² × 8.0 (= 36.19...)	✓ 1
M1 two hemispheres = ⁴⁄₃ × π × 1.2³ (= 7.238...)	✓ 1
M1 sum of the two volumes	✓ 1
A1 43.4 m³	✓ 1
Part (b)
M1 43.429... × 1000 ÷ 850	✓ 1
A1 51.1 minutes	✓ 1

Exam tip: Two hemispheres of equal radius join to make one sphere, so add 4/3 π r³ to the cylinder volume.

+ Add to test✓ In your test145IGMC3-5-10r12-06Surface area and volume of solids Paper EMedium6 marks

Calculator

A burial chamber has a roof in the shape of a hemisphere of radius 6.5 m. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the volume of the hemisphere.

(2)

(b)

Calculate the curved surface area of the hemisphere.

(2)

(c)

One litre of sealant covers 4.5 m². Calculate the number of litres needed for the curved surface. Use the unrounded area from (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{2}{3}\pi\times6.5^3$	✓ 1
A1 575 m³	✓ 1
Part (b)
M1 $2\pi\times6.5^2$	✓ 1
A1 265 m²	✓ 1
Part (c)
M1 265.46… ÷ 4.5	✓ 1
A1 59.0 litres	✓ 1

Exam tip: A hemisphere is half a sphere: halve the volume formula, but for curved area use 2πr², not 4πr².

+ Add to test✓ In your test146IGMC3-5-10r12-09Surface area and volume of solids Paper EMedium6 marks

Calculator

A conical heap of soil at a dig has base radius 2.4 m and perpendicular height 1.8 m.

Diagram not to scale

(a)

Show that the slant height of the cone is 3.0 m.

(2)

(b)

Calculate the volume of the heap, correct to 3 significant figures.

(2)

(c)

The curved surface of the heap is covered with a tarpaulin. Calculate the area of the curved surface, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\sqrt{2.4^2+1.8^2}$	✓ 1
A1 $\sqrt{9}=3.0$ m	✓ 1
Part (b)
M1 $\frac{1}{3}\pi\times2.4^2\times1.8$	✓ 1
A1 10.9 m³	✓ 1
Part (c)
M1 $\pi\times2.4\times3.0$	✓ 1
A1 22.6 m²	✓ 1

Common error: Use the perpendicular height in the volume formula and the slant height in the curved-area formula.

+ Add to test✓ In your test147IGMC3-5-09r15-08Circles, arcs and sectors Paper EMedium6 marks

Calculator

A sector of a circle has radius 24 cm and arc length 40 cm. It is a model of a dish for a space-probe antenna.

Diagram not to scale

(a)

Calculate the angle of the sector, in degrees, to 1 decimal place.

(2)

(b)

Calculate the area of the sector.

(2)

(c)

The two straight edges of the sector are joined together to make a cone. Calculate the radius of the base of the cone, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 θ/360 × 2 × π × 24 = 40	✓ 1
A1 95.5°	✓ 1
Part (b)
M1 (95.49…/360) × π × 24² or ½ × 24 × 40	✓ 1
A1 480 cm²	✓ 1
Part (c)
M1 the arc length becomes the base circumference: 2 × π × r = 40	✓ 1
A1 6.37 cm	✓ 1

Exam tip: When the arc length is given, the sector area can be found directly from ½ × radius × arc length, with no angle needed.

+ Add to test✓ In your test148IGMC3-5-09r15-09Surface area and volume of solids Paper EMedium6 marks

Calculator

The nose cone of a space probe is a solid cone with base radius 0.60 m and slant height 1.5 m.

Diagram not to scale

(a)

Calculate the curved surface area of the cone, to 3 significant figures.

(2)

(b)

Calculate the perpendicular height of the cone, to 3 significant figures.

(2)

(c)

Calculate the volume of the cone, using the unrounded height from (b). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.6 × 1.5	✓ 1
A1 2.83 m²	✓ 1
Part (b)
M1 h² = 1.5² − 0.6² (= 1.89)	✓ 1
A1 1.37 m	✓ 1
Part (c)
M1 ⅓ × π × 0.6² × 1.3748…	✓ 1
A1 0.518 m³	✓ 1

Common error: The slant height is not the height used in the volume formula; use Pythagoras to find the perpendicular height first.

+ Add to test✓ In your test149IGMC3-5-09r15-11Surface area and volume of solids Paper EMedium8 marks

Calculator

A coffee-roasting plant stores beans in a tank made of a cylinder with a hemisphere fixed on top. The cylindrical part has radius 0.90 m and height 1.4 m. The base of the tank is a flat circle.

Diagram not to scale

(a)

Calculate the volume of the cylindrical part, to 3 significant figures.

(2)

(b)

Calculate the total volume of the tank, using the unrounded volume of the cylinder. Give your answer to 3 significant figures.

(3)

(c)

The outside of the whole tank (curved wall, base and hemisphere) is painted. Calculate the area to be painted, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.9² × 1.4	✓ 1
A1 3.56 m³	✓ 1
Part (b)
M1 hemisphere ⅔ × π × 0.9³ (= 1.5268…)	✓ 1
M1 add cylinder volume 3.5626…	✓ 1
A1 5.09 m³	✓ 1
Part (c)
B1 hemisphere curved area 2 × π × 0.9²	✓ 1
M1 cylinder wall 2 × π × 0.9 × 1.4 and base π × 0.9² added to the hemisphere area	✓ 1
A1 15.6 m²	✓ 1

Common error: The top of the cylinder is covered by the hemisphere, so there is no flat circle on top to include in the surface area.

+ Add to test✓ In your test150IGMC3-5-10r16-08Surface area and volume of solids Paper EMedium8 marks

Calculator

A hopper for coffee beans is a cone with base radius 0.9 m and perpendicular height 1.2 m, pointing downwards.

Diagram not to scale

(a)

Show that the slant height of the cone is 1.5 m.

(2)

(b)

Calculate the curved surface area of the cone. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the volume of the cone. Give your answer correct to 3 significant figures.

(2)

(d)

The hopper is full of beans of density 650 kg/m³. Calculate the mass of the beans, correct to the nearest kilogram.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(0.9² + 1.2²)	✓ 1
A1 √2.25 = 1.5	✓ 1
Part (b)
M1 π × 0.9 × 1.5	✓ 1
A1 4.24 m²	✓ 1
Part (c)
M1 ⅓ × π × 0.9² × 1.2	✓ 1
A1 1.02 m³	✓ 1
Part (d)
M1 650 × 1.01788... (unrounded volume)	✓ 1
A1 662 kg	✓ 1

Exam tip: Use the perpendicular height in the volume formula and the slant height in the curved surface area formula.

+ Add to test✓ In your test151IGMC3-5-10r16-09Surface area and volume of solids Paper EMedium8 marks

Calculator

A capsule consists of a cylinder of radius 0.8 m and height 1.5 m with a hemisphere of radius 0.8 m attached to one end. Give answers correct to 3 significant figures.

Diagram not to scale

(a)

Calculate the volume of the cylinder.

(2)

(b)

Calculate the volume of the hemisphere.

(2)

(c)

Write down the total volume of the capsule.

(1)

(d)

The outer surface of the capsule consists of the curved surface of the cylinder, the curved surface of the hemisphere and the flat circular base. Calculate its area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 0.8² × 1.5	✓ 1
A1 3.02 m³	✓ 1
Part (b)
M1 ⅔ × π × 0.8³	✓ 1
A1 1.07 m³	✓ 1
Part (c)
4.09 m³ (3.02 + 1.07, or unrounded 4.0883...)	✓ 1
Part (d)
M1 cylinder curved 2 × π × 0.8 × 1.5 = 2.4π	✓ 1
M1 hemisphere 2π × 0.8² = 1.28π and base π × 0.8² = 0.64π	✓ 1
A1 4.32π = 13.6 m²	✓ 1

Common error: The curved surface area of a hemisphere is 2πr², half of the sphere formula, and the flat face is added only if it is part of the outer surface.

+ Add to test✓ In your test152IGMC3-5-10r16-10Surface area and volume of solids Paper EMedium5 marks

No calculator

Two coffee mugs are mathematically similar. The small mug has height 8 cm and the large mug has height 12 cm. The small mug has volume 192 cm³. The large mug has outer surface area 270 cm².

(a)

Write down the ratio of corresponding lengths, large : small, as a single number.

(1)

(b)

Work out the volume of the large mug.

(2)

(c)

Work out the outer surface area of the small mug.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
12 ÷ 8 = 1.5	✓ 1
Part (b)
M1 192 × 1.5³	✓ 1
A1 648 cm³	✓ 1
Part (c)
M1 270 ÷ 1.5²	✓ 1
A1 120 cm²	✓ 1

Common error: Areas scale with the square of the length factor and volumes with the cube; do not use the length factor directly.

+ Add to test✓ In your test153IGMC3-5-10r18-05Surface area and volume of solids Paper EMedium6 marks

Calculator

A coffee roaster has a hopper in the shape of a cone with the point downwards. The circular top has radius 30 cm and the vertical height is 80 cm.

Diagram not to scale

(a)

Calculate the volume of the hopper. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the slant height of the cone. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the curved surface area of the cone. Use the unrounded slant height and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1/3 × π × 30² × 80	✓ 1
A1 75 400 cm³	✓ 1
Part (b)
M1 √(30² + 80²) (= √7300)	✓ 1
A1 85.4 cm	✓ 1
Part (c)
M1 π × 30 × 85.44...	✓ 1
A1 8050 cm²	✓ 1

Exam tip: The curved surface area of a cone uses the slant height, not the vertical height.

+ Add to test✓ In your test154IGMC3-5-10r18-06Surface area and volume of solids Paper EMedium6 marks

Calculator

A spherical fuel tank on a space probe has radius 0.45 m. Fuel has density 1020 kg/m³.

(a)

Calculate the surface area of the tank. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the volume of the tank. Give your answer correct to 3 significant figures.

(2)

(c)

The tank is 85% full of fuel. Calculate the mass of fuel. Use the unrounded volume and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4 × π × 0.45²	✓ 1
A1 2.54 m²	✓ 1
Part (b)
M1 4/3 × π × 0.45³	✓ 1
A1 0.382 m³	✓ 1
Part (c)
M1 0.85 × 0.38170... × 1020	✓ 1
A1 331 kg	✓ 1

Common error: Cube the radius for volume; squaring it gives a surface-area-style value.

+ Add to test✓ In your test155IGMC3-5-10r18-09Surface area and volume of solids Paper EMedium6 marks

Calculator

A model of a space probe is made to a scale of 1 : 10. The model and the real probe are mathematically similar.

(a)

The surface area of the model is 0.36 m². Calculate the surface area of the real probe.

(2)

(b)

The volume of the model is 0.0085 m³. Calculate the volume of the real probe.

(2)

(c)

The real probe has a mast of length 2.4 m. Calculate the length of the mast on the model, in centimetres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.36 × 10²	✓ 1
A1 36 m²	✓ 1
Part (b)
M1 0.0085 × 10³	✓ 1
A1 8.5 m³	✓ 1
Part (c)
M1 2.4 ÷ 10 (= 0.24 m)	✓ 1
A1 24 cm	✓ 1

Common error: Lengths scale by the ratio, areas by its square and volumes by its cube; do not use the length ratio for all three.

+ Add to test✓ In your test156IGMC3-5-10r23-08Surface area and volume of solids Paper EMedium6 marks

Calculator

A clay funnel found at an excavation is a hollow cone with base radius 6 cm and slant height 15 cm. Use π from your calculator and the unrounded height in later working.

Diagram not to scale

(a)

Calculate the curved surface area of the cone. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the perpendicular height of the cone. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the volume of the cone. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 6 × 15	✓ 1
A1 283 cm²	✓ 1
Part (b)
M1 sqrt(15² − 6²) = sqrt(189)	✓ 1
A1 13.7 cm	✓ 1
Part (c)
M1 ⅓ × π × 6² × 13.7477...	✓ 1
A1 518 cm³	✓ 1

Common error: The slant height is not the height used in the cone volume formula.

+ Add to test✓ In your test157IGMC3-5-10r23-09Surface area and volume of solids Paper EMedium6 marks

Calculator

The roof of a data centre is a solid hemisphere of radius 4.5 m. Use π from your calculator.

Diagram not to scale

(a)

Calculate the curved surface area of the roof. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the volume enclosed by the roof. Give your answer correct to 3 significant figures.

(2)

(c)

The roof is painted at a cost of €2.40 per square metre. Use the unrounded area from (a) to find the cost, giving your answer to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2 × π × 4.5²	✓ 1
A1 127 m²	✓ 1
Part (b)
M1 ⅔ × π × 4.5³	✓ 1
A1 191 m³	✓ 1
Part (c)
M1 127.23... × 2.40	✓ 1
A1 €305	✓ 1

Exam tip: A hemisphere has volume half of that of a sphere, and its curved surface is half of 4πr².

+ Add to test✓ In your test158IGMC3-5-12r31-07Surface area and volume of solids Paper EMedium4 marks

Calculator

Grain is poured into a cone-shaped pile of base radius 3.2 m and height 2.1 m. The volume of a cone is $\frac{1}{3}\pi r^2 h$.

Diagram not to scale

(a)

Calculate the volume of the pile. Give your answer to 3 significant figures.

(2)

(b)

The grain has density 750 kg/m³. Calculate the mass of the pile in tonnes (1 tonne = 1000 kg). Use the unrounded volume and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{1}{3}\times\pi\times3.2^2\times2.1$	✓ 1
A1 22.5 m³	✓ 1
Part (b)
M1 $22.52\times750\div1000$	✓ 1
A1 16.9 tonnes	✓ 1

Common error: Do not forget the factor $\frac{1}{3}$ in the volume of a cone, and convert kilograms to tonnes at the end.

+ Add to test✓ In your test159IGMC3-5-12r31-08Surface area and volume of solids Paper EMedium6 marks

Calculator

A grain silo is a cylinder of radius 2.4 m and height 6.5 m with a hemisphere of radius 2.4 m on top. The grain fills the whole silo. The volume of a sphere is $\frac{4}{3}\pi r^3$.

Diagram not to scale

(a)

Calculate the volume of the cylinder. Give your answer to 3 significant figures.

(2)

(b)

Calculate the volume of the hemisphere. Give your answer to 3 significant figures.

(2)

(c)

Wheat has a mass of 0.78 tonnes per cubic metre. Calculate the mass of wheat that fills the silo. Use unrounded volumes and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi\times2.4^2\times6.5$	✓ 1
A1 118 m³	✓ 1
Part (b)
M1 $\frac{1}{2}\times\frac{4}{3}\times\pi\times2.4^3$	✓ 1
A1 29.0 m³	✓ 1
Part (c)
M1 $(117.6 + 28.9)\times0.78$ using unrounded volumes	✓ 1
A1 114 tonnes	✓ 1

Exam tip: For a hemisphere, take half of the sphere volume, and keep full calculator values until the final line.

+ Add to test✓ In your test160IGMC3-5-11r34-10Surface area and volume of solids Paper EMedium6 marks

Calculator

A drug capsule is a cylinder of radius 0.35 cm and length 1.2 cm with a hemisphere of radius 0.35 cm fixed to each end. Volume of a sphere = ⁴⁄₃πr³.

Diagram not to scale

(a)

Write down the total length of the capsule.

(1)

(b)

Show that the volume of one capsule is 0.641 cm³, correct to 3 significant figures.

(3)

(c)

A batch contains 180 cm³ of powder. Each capsule is filled completely. Use the unrounded volume of a capsule to work out the number of capsules that can be filled completely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 1.2 + 2 × 0.35 = 1.9 cm	✓ 1
Part (b)
M1 cylinder volume π × 0.35² × 1.2 (= 0.4618…)	✓ 1
M1 two hemispheres together make a sphere: ⁴⁄₃ × π × 0.35³ (= 0.1795…)	✓ 1
A1 0.4618… + 0.1795… = 0.6414… = 0.641 cm³	✓ 1
Part (c)
M1 180 ÷ 0.6414…	✓ 1
A1 280	✓ 1

Exam tip: Two hemispheres of the same radius combine to make one full sphere, which saves a step.

+ Add to test✓ In your test161IGMC3-5-11r34-11Surface area and volume of solids Paper EMedium8 marks

Calculator

A conical filter funnel has radius 6 cm and perpendicular height 14 cm. Volume of a cone = ¹⁄₃πr²h. Curved surface area of a cone = πrl, where l is the slant height.

Diagram not to scale

(a)

Show that the volume of the funnel is 168π cm³.

(2)

(b)

The funnel is full of water. All the water is poured into an empty cylindrical beaker of radius 4 cm. Work out the depth of the water in the beaker.

(3)

(c)

Calculate the curved surface area of the funnel, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ¹⁄₃ × π × 6² × 14	✓ 1
A1 12π × 14 = 168π cm³	✓ 1
Part (b)
M1 π × 4² × d = 168π	✓ 1
M1 d = 168 ÷ 16	✓ 1
A1 10.5 cm	✓ 1
Part (c)
M1 l = √(6² + 14²) (= √232 = 15.23…)	✓ 1
M1 π × 6 × l	✓ 1
A1 287 cm²	✓ 1

Exam tip: Keep π in the working when volumes are equal; it cancels and avoids rounding errors.

+ Add to test✓ In your test162IGMC3-5-11r37-08Surface area and volume of solids Paper EMedium6 marks

Calculator

A conical heap of gravel is poured beside a wind turbine. The heap has base radius 3 m and perpendicular height 4 m.
Volume of a cone = ⅓πr²h. Curved surface area of a cone = πrl, where l is the slant height.

Diagram not to scale

(a)

Calculate the volume of gravel in the heap, correct to 3 significant figures.

(2)

(b)

Show that the slant height of the cone is 5 m.

(2)

(c)

Calculate the curved surface area of the heap, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × π × 3² × 4	✓ 1
A1 37.7 m³	✓ 1
Part (b)
M1 l² = 3² + 4²	✓ 1
A1 l = √25 = 5 m	✓ 1
Part (c)
M1 π × 3 × 5	✓ 1
A1 47.1 m²	✓ 1

Common error: Use the slant height, not the perpendicular height, in the formula for curved surface area.

+ Add to test✓ In your test163IGMC3-5-11r37-10Perimeter and area of plane shapes Paper EMedium7 marks

Calculator

A sector-shaped lawn in a library courtyard has radius 12 m. The arc length of the sector is 25.1 m.

Diagram not to scale

(a)

Calculate the angle of the sector, to the nearest degree.

(3)

(b)

Using your angle from part (a), calculate the area of the lawn, correct to 3 significant figures.

(2)

(c)

A fence is built along the whole boundary of the lawn. Calculate the length of the fence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 25.1 = (θ ÷ 360) × 2 × π × 12	✓ 1
M1 θ = 25.1 × 360 ÷ (2 × π × 12)	✓ 1
A1 120°	✓ 1
Part (b)
M1 (120 ÷ 360) × π × 12²	✓ 1
A1 151 m²	✓ 1
Part (c)
M1 25.1 + 2 × 12	✓ 1
A1 49.1 m	✓ 1

Exam tip: The boundary of a sector includes the two straight radii as well as the arc.

+ Add to test✓ In your test164IGMC3-5-09r38-08Surface area and volume of solids Paper EMedium7 marks

Calculator

A glass paperweight in a library gift shop is made from a solid hemisphere of radius 4 cm joined to a solid cone. The flat face of the hemisphere has the same radius as the base of the cone. The cone has perpendicular height 9 cm. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the volume of the paperweight.

(3)

(b)

Calculate the total outer surface area of the paperweight. The base of the cone is joined to the hemisphere and is not part of the outer surface.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 hemisphere volume = ⅔ × π × 4³ (= 134.04...)	✓ 1
M1 cone volume = ⅓ × π × 4² × 9 (= 150.80...)	✓ 1
A1 285 cm³	✓ 1
Part (b)
M1 slant height of cone = √(4² + 9²) = √97 (= 9.849...)	✓ 1
M1 curved area of cone = π × 4 × √97 (= 123.76...)	✓ 1
M1 curved area of hemisphere = 2 × π × 4² (= 100.53...)	✓ 1
A1 224 cm²	✓ 1

Common error: Use the slant height, not the perpendicular height, in the curved surface area of a cone, and count only the faces that can be seen from outside.

+ Add to test✓ In your test165IGMC3-5-09r39-10Circles, arcs and sectors Paper EMedium6 marks

Calculator

The floor of a theatre stage is a sector of a circle. The radius of the sector is 9.5 m and the angle at the centre is 128°.

Diagram not to scale

(a)

Calculate the area of the stage floor, giving your answer to 3 significant figures.

(3)

(b)

A rope runs around the whole edge of the stage floor. Calculate the length of the rope, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 128/360 × π × 9.5²	✓ 1
M1 evaluates 100.8…	✓ 1
A1 101 m²	✓ 1
Part (b)
M1 arc = 128/360 × 2 × π × 9.5 = 21.2…	✓ 1
M1 add the two radii: 21.2… + 2 × 9.5	✓ 1
A1 40.2 m	✓ 1

Common error: The edge of a sector includes the two straight radii as well as the arc, so add 2r to the arc length.

+ Add to test✓ In your test166IGMC3-5-09r39-11Surface area and volume of solids Paper EMedium6 marks

Calculator

A volcano is modelled as a cone with base radius 4.0 km and perpendicular height 1.5 km.

Diagram not to scale

(a)

Calculate the volume of the volcano, giving your answer to 3 significant figures.

(2)

(b)

Calculate the slant height of the cone, giving your answer to 3 significant figures.

(2)

(c)

Calculate the curved surface area of the cone, giving your answer to 3 significant figures. Use the unrounded slant height from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × π × 4.0² × 1.5	✓ 1
A1 25.1 km³	✓ 1
Part (b)
M1 √(4.0² + 1.5²)	✓ 1
A1 4.27 km	✓ 1
Part (c)
M1 π × 4.0 × 4.272…	✓ 1
A1 53.7 km²	✓ 1

Common error: The curved surface area formula πrl uses the slant height l, not the perpendicular height.

+ Add to test✓ In your test167IGMC3-5-10r48-10Surface area and volume of solids Paper EMedium7 marks

Calculator

A traditional straw beehive (a skep) is modelled as a cylinder of radius 20.0 cm and height 30.0 cm, with a solid hemisphere of radius 20.0 cm on top. Volume of a sphere = 4/3 πr³.

Diagram not to scale

(a)

Calculate the volume of the hemisphere. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the total volume of the skep, in litres. Use unrounded values and give your answer correct to 3 significant figures. (1 litre = 1000 cm³)

(3)

(c)

Calculate the area of the curved surface of the hemispherical top. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 4/3 × π × 20.0³	✓ 1
A1 16 800 cm³	✓ 1
Part (b)
M1 cylinder volume π × 20.0² × 30.0 = 37 699...	✓ 1
M1 add the hemisphere volume 16 755... and divide by 1000	✓ 1
A1 54.5 litres	✓ 1
Part (c)
M1 ½ × 4 × π × 20.0² or 2 × π × 20.0²	✓ 1
A1 2510 cm²	✓ 1

Common error: A hemisphere has half the volume of the sphere, but its curved surface is 2πr² (half of 4πr²), not including the flat circular face.

+ Add to test✓ In your test168IGMC3-5-10r48-13Surface area and volume of solids Paper EMedium8 marks

No calculator

A hospital atrium has a glass roof in the shape of a pyramid. The base is a square of side 8 m. The apex is directly above the centre of the base, at a vertical height of 3 m. Volume of a pyramid = 1/3 × base area × height.

Diagram not to scale

(a)

Calculate the volume of the space inside the pyramid.

(2)

(b)

Calculate the slant height of one triangular face, measured from the apex to the midpoint of a base edge.

(2)

(c)

Calculate the total area of the four triangular glass faces.

(2)

(d)

Glass costs €62.50 per square metre. An extra 15% of glass must be bought to allow for waste. Calculate the total cost of the glass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1/3 × 8 × 8 × 3	✓ 1
A1 64 m³	✓ 1
Part (b)
M1 slant height² = 3² + 4²	✓ 1
A1 5 m	✓ 1
Part (c)
M1 4 × ½ × 8 × 5 (accept ECF from (b))	✓ 1
A1 80 m²	✓ 1
Part (d)
M1 80 × 1.15 × 62.50	✓ 1
A1 €5750	✓ 1

Common error: The slant height is not the vertical height: use Pythagoras with the vertical height and half the base edge.

+ Add to test✓ In your test169IGMC-045-00-12Perimeter and area of plane shapes Paper EHard6 marks

Calculator

A circular flower bed of radius r metres is surrounded by a path of width 1.5 m. The outer edge of the path is also a circle with the same centre. The area of the path is 33 m².

Diagram not to scale

(a)

Calculate the value of r.

(4)

(b)

Calculate the length of the outer edge of the path.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π(r + 1.5)² − πr² = 33	✓ 1
M1 expands to π(3r + 2.25) = 33	✓ 1
M1 r = (33 ÷ π − 2.25) ÷ 3	✓ 1
A1 r = 2.75	✓ 1
Part (b)
M1 2 × π × (their r + 1.5)	✓ 1
A1 26.7 m	✓ 1

Common error: (r + 1.5)² is r² + 3r + 2.25, not r² + 2.25. Missing the middle term makes the equation impossible to solve correctly.

+ Add to test✓ In your test170IGMC-045-00-13Perimeter and area of plane shapes Paper EHard7 marks

No calculator

A rectangle has length (2x + 3) cm and width (x − 1) cm. The area of the rectangle is 42 cm².

Diagram not to scale

(a)

Show that 2x² + x − 45 = 0.

(2)

(b)

Solve 2x² + x − 45 = 0 and explain why only one solution is valid.

(3)

(c)

Work out the perimeter of the rectangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2x + 3)(x − 1) = 42	✓ 1
A1 2x² − 2x + 3x − 3 = 42 leading to 2x² + x − 45 = 0 with no errors	✓ 1
Part (b)
M1 (2x − 9)(x + 5) = 0, or correct substitution into the quadratic formula, e.g. x = $\frac{-1 \pm \sqrt{1 + 360}}{4}$	✓ 1
A1 x = 4.5 and x = −5	✓ 1
B1 x = −5 rejected because the width x − 1 = −6 would be negative (or length negative), so x = 4.5	✓ 1
Part (c)
M1 length 12 and width 3.5, or 2 × (12 + 3.5)	✓ 1
A1 31 cm	✓ 1

Exam tip: In a 'show that' question, write every line of the expansion and finish with the given equation exactly. Check your dimensions at the end: 12 × 3.5 = 42.

+ Add to test✓ In your test171IGMC-045-00-14Perimeter and area of plane shapes Paper EHard7 marks

Calculator

A and B are points on a circle with centre O and radius 10 cm. Angle AOB = 75°. The chord AB divides the circle into two segments.

Diagram not to scale

(a)

Calculate the area of the minor segment.

(4)

(b)

Calculate the perimeter of the minor segment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sector area $\frac{75}{360}$ × π × 10² (= 65.4…)	✓ 1
M1 triangle area $\frac{1}{2}$ × 10 × 10 × sin 75° (= 48.2…)	✓ 1
M1 sector area − triangle area	✓ 1
A1 17.2 cm²	✓ 1
Part (b)
M1 arc length $\frac{75}{360}$ × 2 × π × 10 (= 13.09…)	✓ 1
M1 chord AB = 2 × 10 × sin 37.5° or by the cosine rule (= 12.17…)	✓ 1
A1 arc + chord = 25.3 cm	✓ 1

Common error: The perimeter of a segment is the arc plus the straight chord. Adding two radii instead gives the perimeter of the sector, not the segment.

+ Add to test✓ In your test172IGMC-045-01-10Circles, arcs and sectors Paper EHard8 marks

Calculator

A and B are points on a circle, centre O, radius 8 cm. Angle AOB = 110°. The chord AB divides the circle into a minor segment and a major segment.

Diagram not to scale

(a)

Calculate the area of the minor sector OAB.

(2)

(b)

Calculate the area of triangle OAB.

(2)

(c)

Calculate the area of the minor segment.

(1)

(d)

Calculate the perimeter of the minor segment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{110}{360}$ × π × 8²	✓ 1
A1 61.4 cm² (61.43…)	✓ 1
Part (b)
M1 for $\frac{1}{2}$ × 8 × 8 × sin 110°	✓ 1
A1 30.1 cm² (30.07…)	✓ 1
Part (c)
B1 31.4 cm² from unrounded values (31.36…), or 31.3 cm² from 61.4 − 30.1; follow through their (a) − their (b)	✓ 1
Part (d)
M1 for chord AB = 2 × 8 × sin 55° or AB² = 8² + 8² − 2 × 8 × 8 × cos 110° [AB = 13.1]	✓ 1
M1 for $\frac{110}{360}$ × 2 × π × 8 [arc = 15.36] + their AB	✓ 1
A1 28.5 cm (28.46…)	✓ 1

Exam tip: A segment's perimeter is the arc plus the chord, not the two radii. Find the chord with trigonometry or the cosine rule, and keep unrounded values until the final answer.

+ Add to test✓ In your test173IGMC-045-01-13Circles, arcs and sectors Paper EHard7 marks

Calculator

OAB is a sector of a circle, centre O, radius 10 cm and angle 60°. C lies on OA and D lies on OB such that OC = OD = r cm. OCD is a sector of a circle, centre O, with the same angle of 60°. The region ABDC, between the two arcs, has an area of 30 cm².

Diagram not to scale

(a)

Calculate the value of r.

(4)

(b)

Calculate the perimeter of the region ABDC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for area of sector OAB = $\frac{60}{360}$ × π × 10² [= 52.36]	✓ 1
M1 for $\frac{60}{360}$ × π × r² = their 52.36 − 30	✓ 1
M1 for r² = 42.7 or r² = 100 − $\frac{180}{\pi}$	✓ 1
A1 r = 6.53	✓ 1
Part (b)
M1 for arc AB = $\frac{60}{360}$ × 2π × 10 [= 10.47] or arc CD = $\frac{60}{360}$ × 2π × their r [= 6.84]	✓ 1
M1 for arc AB + arc CD + 2 × (10 − their r)	✓ 1
A1 24.2 cm (accept 24.25 or 24.3 from using r = 6.53)	✓ 1

Common error: The straight edges of region ABDC are AC and BD, each 10 − r, not the full radius of 10 cm.

+ Add to test✓ In your test174IGMC-045-01-14Circles, arcs and sectors Paper EHard7 marks

No calculator

A square of side 8 cm is drawn so that all four of its vertices lie on a circle.

Diagram not to scale

(a)

Find the exact radius of the circle, giving your answer as a surd in its simplest form.

(2)

(b)

Find the exact area of the region inside the circle but outside the square.

(3)

(c)

A second circle is drawn inside the square so that it touches all four sides. Find the ratio of the area of the larger circle to the area of the smaller circle, in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for diagonal² = 8² + 8² or diagonal = 8√2	✓ 1
A1 4√2 cm	✓ 1
Part (b)
M1 for π × (4√2)² [= 32π]	✓ 1
M1 for their 32π − 64	✓ 1
A1 32π − 64 cm²	✓ 1
Part (c)
B1 area of smaller circle = 16π	✓ 1
B1 2 : 1	✓ 1

Exam tip: The diagonal of the square is a diameter of the outer circle, while the side of the square is a diameter of the inner circle.

+ Add to test✓ In your test175IGMC-045-02-12Surface area and volume of solids Paper EHard7 marks

Calculator

A solid cone has base radius 9 cm and vertical height 24 cm. A small cone of vertical height 8 cm is cut from the top by a cut parallel to the base. The remaining solid is a frustum.
[Volume of cone = $\frac{1}{3}\pi r^2 h$]

Diagram not to scale

(a)

Find the radius of the base of the small cone.

(2)

(b)

Calculate the volume of the frustum.

(3)

(c)

A hollow container has exactly the same shape and size as the frustum. It is completely filled with water. All the water is then poured into an empty cylinder of radius 6 cm.
Calculate the depth of water in the cylinder.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: r ÷ 9 = 8 ÷ 24, or 9 × $\frac{8}{24}$, or 9 ÷ 3, or equivalent	✓ 1
A1: 3 cm	✓ 1
Part (b)
M1: large cone $\frac{1}{3}$ × π × 9² × 24 (= 648π)	✓ 1
M1: subtracting small cone $\frac{1}{3}$ × π × 3² × 8 (= 24π)	✓ 1
A1: 1960 cm³ (624π = 1960.35…)	✓ 1
Part (c)
M1: their (b) ÷ (π × 6²), e.g. 624π ÷ 36π	✓ 1
A1: 17.3 cm (17.33…)	✓ 1

Exam tip: Keep the volumes in terms of π: 648π − 24π = 624π. This avoids rounding errors, and the π cancels in part (c).

+ Add to test✓ In your test176IGMC-045-02-13Surface area and volume of solids Paper EHard8 marks

No calculator

A sphere of radius r cm fits exactly inside a closed cylinder. The cylinder has radius r cm and height 2r cm.
[Volume of sphere = $\frac{4}{3}\pi r^3$, surface area of sphere = $4\pi r^2$]

Cross-section. Diagram not to scale

(a)

Show that the volume of the sphere is $\frac{2}{3}$ of the volume of the cylinder.

(3)

(b)

Show that the curved surface area of the cylinder is equal to the surface area of the sphere.

(2)

(c)

The total surface area of the cylinder is 54π cm².
Find the volume of the sphere, giving your answer in terms of π.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: cylinder volume = πr² × 2r = 2πr³	✓ 1
M1: $\frac{4}{3}\pi r^3$ ÷ 2πr³	✓ 1
A1: = $\frac{2}{3}$ correctly shown	✓ 1
Part (b)
M1: curved surface area = 2πr × 2r	✓ 1
A1: = 4πr², the same as the sphere	✓ 1
Part (c)
M1: 4πr² + 2πr² = 54π, so 6πr² = 54π	✓ 1
A1: r = 3	✓ 1
A1: 36π cm³	✓ 1

Common error: The total surface area of a closed cylinder includes the two circular ends. Using only the curved surface gives r² = 13.5.

+ Add to test✓ In your test177IGMC-045-02-14Surface area and volume of solids Paper EHard8 marks

Calculator

A water trough is a prism of length 2.5 m. Its cross-section is an isosceles trapezium. The bottom edge is 40 cm, the top edge is 60 cm and the depth is 30 cm. The trough stands on its 40 cm face, with the 60 cm edge at the top.

Diagram not to scale

(a)

Calculate the area of the cross-section.

(2)

(b)

Calculate the capacity of the trough in litres.

(2)

(c)

The trough contains 200 litres of water. Let the depth of the water be h cm.
Form an equation in h and solve it to find the depth of water.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\frac{1}{2}$ × (60 + 40) × 30	✓ 1
A1: 1500 cm²	✓ 1
Part (b)
M1: 1500 × 250 (= 375 000 cm³)	✓ 1
A1: 375 litres	✓ 1
Part (c)
M1: width of water surface = 40 + $\frac{2h}{3}$	✓ 1
M1: 250 × $\frac{1}{2}$(40 + 40 + $\frac{2h}{3}$)h = 200 000	✓ 1
M1: simplifies to h² + 120h − 2400 = 0, with correct use of the quadratic formula (or completing the square)	✓ 1
A1: h = 17.5 cm (17.459…), with the negative root rejected	✓ 1

Exam tip: The depth of water is not proportional to its volume, because the trough widens upwards. The width increases by 20 cm over 30 cm of depth, so it increases by $\frac{2}{3}$ cm for each 1 cm of depth.

+ Add to test✓ In your test178IGMC3-5-02d-12Surface area and volume of solids Paper EHard7 marks

Calculator

A foundry melts a solid copper sphere of radius 9 cm and casts the metal, with no loss, into a solid cylinder of height 27 cm.

(a)

Show that the volume of the sphere is 972π cm³.

(2)

(b)

Calculate the radius of the cylinder.

(3)

(c)

Calculate the curved surface area of the cylinder. Give your answer to the nearest cm².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (4 ÷ 3) × π × 9³	✓ 1
A1 (4 ÷ 3) × 729 = 972, so the volume is 972π	✓ 1
Part (b)
M1 πr² × 27 = 972π	✓ 1
M1 r² = 972 ÷ 27 = 36	✓ 1
A1 6 cm	✓ 1
Part (c)
M1 2 × π × 6 × 27	✓ 1
A1 1018 cm²	✓ 1

Exam tip: When a solid is remelted and recast, set the old volume equal to the new volume and leave π in the equation, where it cancels.

+ Add to test✓ In your test179IGMC3-5-02d-13Circles, arcs and sectors Paper EHard8 marks

Calculator

A horizontal drain pipe has a circular cross-section with radius 10 cm and centre O. It holds water whose surface is the chord AB, where angle AOB = 60°. The water fills the smaller segment of the circle below AB. The pipe is 2.5 m long.

Diagram not to scale

(a)

Calculate the area of the sector OAB. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the area of triangle OAB. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the area of the segment. Use the unrounded values from (a) and (b) and give your answer correct to 3 significant figures.

(2)

(d)

Calculate the volume of water in the pipe. Use the unrounded segment area and give your answer to the nearest cm³.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (60 ÷ 360) × π × 10²	✓ 1
A1 52.4 cm²	✓ 1
Part (b)
M1 ½ × 10 × 10 × sin 60°	✓ 1
A1 43.3 cm²	✓ 1
Part (c)
M1 sector area − triangle area	✓ 1
A1 9.06 cm²	✓ 1
Part (d)
M1 segment area × 250 (2.5 m = 250 cm)	✓ 1
A1 2265 cm³	✓ 1

Common error: Segment area is sector minus triangle. Keep full calculator values until the last step, because subtracting two rounded numbers loses accuracy.

+ Add to test✓ In your test180IGMC3-5-02d-14Surface area and volume of solids Paper EHard7 marks

Calculator

A closed cylindrical tin has radius r cm and a height equal to three times its radius. Its total surface area is 128π cm².

Diagram not to scale

(a)

Show that the total surface area of the tin is 8πr² cm².

(2)

(b)

Find the value of r.

(2)

(c)

Calculate the volume of the tin. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 curved area 2πr × 3r = 6πr² and two circular ends 2 × πr²	✓ 1
A1 6πr² + 2πr² = 8πr²	✓ 1
Part (b)
M1 8πr² = 128π, so r² = 16	✓ 1
A1 r = 4 cm	✓ 1
Part (c)
B1 height = 12 cm	✓ 1
M1 π × 4² × 12	✓ 1
A1 603 cm³	✓ 1

Common error: A closed tin has two circular ends as well as the curved surface; leaving out one end changes the equation and the value of r.

+ Add to test✓ In your test181IGMC3-5-08r12-12Circles, arcs and sectors Paper EHard10 marks

Calculator

A telescope mirror is a circle of radius 12 cm with centre O. A straight edge AB is cut off the circle so that angle AOB = 80°. The cut-off piece is the minor segment between the chord AB and the minor arc AB.

Diagram not to scale

(a)

Calculate the area of the sector AOB. Give your answer to 3 significant figures.

(2)

(b)

Calculate the area of triangle AOB. Give your answer to 3 significant figures.

(2)

(c)

Calculate the area of the mirror that remains. Use unrounded values and give your answer to the nearest cm².

(3)

(d)

Calculate the perimeter of the mirror that remains. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (80 ÷ 360) × π × 12²	✓ 1
A1 101 cm²	✓ 1
Part (b)
M1 ¹⁄₂ × 12 × 12 × sin 80°	✓ 1
A1 70.9 cm²	✓ 1
Part (c)
M1 segment area = sector − triangle (≈ 29.6 cm²)	✓ 1
M1 π × 12² − segment area	✓ 1
A1 423 cm²	✓ 1
Part (d)
M1 major arc = (280 ÷ 360) × 2 × π × 12 (= 58.64...)	✓ 1
M1 chord AB = 2 × 12 × sin 40° (or cosine rule) (= 15.43...)	✓ 1
A1 74.1 cm	✓ 1

Common error: The remaining perimeter uses the major arc (280°), not the minor arc, plus the straight edge.

+ Add to test✓ In your test182IGMC3-5-08r12-13Surface area and volume of solids Paper EHard7 marks

Calculator

A sports scientist finds the volume of a solid ball by lowering it into a cylindrical tank of radius 15 cm that contains water. The ball is fully submerged and the water level rises by 1.8 cm. Assume that the ball is a sphere.

(a)

Calculate the volume of the ball. Give your answer to the nearest cm³.

(2)

(b)

Calculate the radius of the ball. Use the unrounded volume and give your answer to 3 significant figures.

(3)

(c)

Calculate the surface area of the ball. Use the unrounded radius and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 15² × 1.8	✓ 1
A1 1272 cm³	✓ 1
Part (b)
M1 ⁴⁄₃πr³ = 1272.3...	✓ 1
M1 r³ = 303.75	✓ 1
A1 6.72 cm	✓ 1
Part (c)
M1 4 × π × 6.7221...²	✓ 1
A1 568 cm²	✓ 1

Exam tip: The rise in water level multiplied by the area of the tank's base gives the volume of the submerged object.

+ Add to test✓ In your test183IGMC3-5-08r12-14Surface area and volume of solids Paper EHard6 marks

Calculator

Two fuel tanks are mathematically similar. The surface areas of the smaller and the larger tank are 18 m² and 32 m². The volume of the smaller tank is 5.4 m³.

(a)

Find the ratio of corresponding lengths, larger : smaller, as a fraction in its simplest form.

(2)

(b)

Calculate the volume of the larger tank.

(2)

(c)

The larger tank is 6.4 m long. Calculate the length of the smaller tank.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(32 ÷ 18)	✓ 1
A1 ⁴⁄₃ (accept 4 : 3)	✓ 1
Part (b)
M1 5.4 × (⁴⁄₃)³	✓ 1
A1 12.8 m³	✓ 1
Part (c)
M1 6.4 ÷ ⁴⁄₃ (ECF from (a))	✓ 1
A1 4.8 m	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³; take a square root of the area ratio to find k.

+ Add to test✓ In your test184IGMC3-5-10r12-12Surface area and volume of solids Paper EHard7 marks

Calculator

A water tank for a data centre is a vertical cylinder of radius 1.4 m with a hemisphere of the same radius fixed on top. The total height of the tank is 5.0 m.

Diagram not to scale

(a)

Find the height of the cylindrical part.

(1)

(b)

Calculate the total volume of the tank, correct to 3 significant figures.

(4)

(c)

A pump fills the tank from empty at 0.35 m³ per minute. Calculate the time to fill the tank, to the nearest minute. Use the unrounded volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5.0 − 1.4 = 3.6 m	✓ 1
Part (b)
M1 cylinder $\pi\times1.4^2\times3.6$	✓ 1
A1 22.2 m³ (22.17…)	✓ 1
M1 hemisphere $\frac{2}{3}\pi\times1.4^3$	✓ 1
A1 total 27.9 m³ (27.91…)	✓ 1
Part (c)
M1 27.91… ÷ 0.35	✓ 1
A1 80 minutes	✓ 1

Exam tip: For a compound solid, subtract the radius of the hemisphere from the total height to get the cylinder height.

+ Add to test✓ In your test185IGMC3-5-10r12-13Surface area and volume of solids Paper EHard7 marks

Calculator

A clay pot is modelled as a frustum. It is made by cutting a small cone of height 8 cm from the top of a solid cone of base radius 9 cm and height 24 cm, with the cut parallel to the base.

Diagram not to scale

(a)

Find the radius of the small cone that is removed.

(2)

(b)

Calculate the volume of the frustum, correct to 3 significant figures.

(3)

(c)

The clay has density 1.8 g/cm³. Calculate the mass of the frustum in kilograms, correct to 3 significant figures. Use the unrounded volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 similar cones: ratio $\frac{8}{24}$	✓ 1
A1 3 cm	✓ 1
Part (b)
M1 large cone $\frac{1}{3}\pi\times9^2\times24$ (= 648π)	✓ 1
M1 small cone $\frac{1}{3}\pi\times3^2\times8$ (= 24π) subtracted	✓ 1
A1 1960 cm³	✓ 1
Part (c)
M1 1960.3… × 1.8 = 3528.6… g	✓ 1
A1 3.53 kg	✓ 1

Common error: The small cone is similar to the large one, so its radius scales by the same factor as its height.

+ Add to test✓ In your test186IGMC3-5-10r12-14Perimeter and area of plane shapes Paper EHard9 marks

Calculator

A funnel for a ventilation system is made from a flat sector of metal sheet with radius 12 cm. The perimeter of the sector (the arc plus the two straight edges) is 50 cm.

Diagram not to scale

(a)

Show that the arc length of the sector is 26 cm.

(1)

(b)

Calculate the angle of the sector, in degrees, correct to 1 decimal place.

(2)

(c)

The straight edges are joined to form a cone, so the arc becomes the circumference of the base. Calculate the base radius, correct to 3 significant figures.

(2)

(d)

Calculate the volume of the cone, correct to 3 significant figures. Use the unrounded base radius.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 50 − 12 − 12 = 26 cm	✓ 1
Part (b)
M1 $\frac{\theta}{360}\times2\pi\times12=26$	✓ 1
A1 124.1°	✓ 1
Part (c)
M1 $2\pi r=26$	✓ 1
A1 4.14 cm	✓ 1
Part (d)
M1 height = $\sqrt{12^2-r^2}$ (slant height is 12 cm)	✓ 1
A1 height 11.3 cm (11.27…)	✓ 1
M1 $\frac{1}{3}\pi r^2h$	✓ 1
A1 202 cm³	✓ 1

Exam tip: When a sector is rolled into a cone, its radius becomes the slant height and its arc becomes the base circumference.

+ Add to test✓ In your test187IGMC3-5-09r15-12Surface area and volume of solids Paper EHard10 marks

Calculator

A model of a space-probe fuel tank holds 80 litres and has a surface area of 1.2 m². A larger tank is geometrically similar to the model, and its height is 1.5 times the height of the model.

(a)

Calculate the volume of the larger tank.

(2)

(b)

Calculate the surface area of the larger tank.

(2)

(c)

A third tank, similar to the model, has a surface area of 4.8 m². Calculate its volume.

(3)

(d)

A fourth tank, similar to the model, holds 10 litres. Calculate its surface area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 volume scale factor 1.5³ (= 3.375)	✓ 1
A1 270 litres	✓ 1
Part (b)
M1 area scale factor 1.5² (= 2.25)	✓ 1
A1 2.7 m²	✓ 1
Part (c)
M1 area ratio 4.8 ÷ 1.2 = 4, so length scale factor 2	✓ 1
M1 80 × 2³	✓ 1
A1 640 litres	✓ 1
Part (d)
M1 volume ratio 10 ÷ 80 = 1/8	✓ 1
M1 length scale factor ½, so area scale factor ¼	✓ 1
A1 0.3 m²	✓ 1

Common error: For similar solids, areas scale with the square of the length factor and volumes with the cube; do not use the length factor directly.

+ Add to test✓ In your test188IGMC3-5-09r15-13Surface area and volume of solids Paper EHard8 marks

Calculator

A solid metal sphere of radius 6.0 cm is melted and recast, without loss of metal, into a solid cone of base radius 9.0 cm.

(a)

Show that the volume of the sphere is 288π cm³.

(2)

(b)

Calculate the height of the cone, to 3 significant figures.

(3)

(c)

Calculate the total surface area of the cone, including its base. Use the unrounded height from (b) and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4/3 × π × 6³	✓ 1
A1 4/3 × 216 = 288, so 288π cm³	✓ 1
Part (b)
M1 volume of cone = volume of sphere	✓ 1
M1 ⅓ × π × 9² × h = 288π	✓ 1
A1 h = 10.7 cm (32/3)	✓ 1
Part (c)
M1 slant height √(9² + 10.667²) = 13.956…	✓ 1
M1 π × 9 × 13.956… + π × 9²	✓ 1
A1 649 cm²	✓ 1

Exam tip: When a shape is melted and remade, set the two volumes equal and cancel the common factor of π before solving.

+ Add to test✓ In your test189IGMC3-5-10r16-11Surface area and volume of solids Paper EHard5 marks

Calculator

A cylindrical coffee tin has height 14 cm and volume 1000 cm³.

(a)

Show that the radius of the tin is 4.77 cm, correct to 3 significant figures.

(3)

(b)

A paper label covers the curved surface of the tin completely. Calculate the area of the label, using the unrounded radius, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 πr² × 14 = 1000	✓ 1
M1 r² = 1000 ÷ (14π) = 22.73...	✓ 1
A1 r = 4.768... = 4.77 (at least 4 significant figures shown before rounding)	✓ 1
Part (b)
M1 2 × π × r × 14	✓ 1
A1 419.4 → 419 cm²	✓ 1

Exam tip: Keep the unrounded radius in your calculator for later parts so that rounding errors do not accumulate.

+ Add to test✓ In your test190IGMC3-5-10r16-12Surface area and volume of solids Paper EHard10 marks

Calculator

A pyramid-shaped glass display has a square base of side 6 m and perpendicular height 4 m. The apex is directly above the centre of the base.

Diagram not to scale

(a)

Calculate the volume of the pyramid.

(2)

(b)

Show that the slant height of each triangular face is 5 m.

(2)

(c)

Calculate the total surface area of the pyramid, including the base.

(3)

(d)

The top of the pyramid is cut off by a plane parallel to the base, 2 m above the base. The piece removed is a smaller pyramid of height 2 m. Calculate the volume of the remaining solid.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅓ × 6² × 4	✓ 1
A1 48 m³	✓ 1
Part (b)
M1 √(3² + 4²), using half the base side 3 m	✓ 1
A1 5	✓ 1
Part (c)
M1 one triangular face ½ × 6 × 5 = 15	✓ 1
M1 4 × 15 + 6²	✓ 1
A1 96 m²	✓ 1
Part (d)
M1 the small pyramid is similar with length factor ½, so its volume is 48 × (½)³ = 6 m³	✓ 1
M1 48 − 6	✓ 1
A1 42 m³	✓ 1

Common error: The perpendicular height is not the slant height; use Pythagoras with half the base side to find the slant height of a face.

+ Add to test✓ In your test191IGMC3-5-10r16-14Surface area and volume of solids Paper EHard6 marks

Calculator

A solid tungsten ball of radius 3 cm is melted and recast into small solid cones, each of base radius 1.5 cm and height 6 cm. No metal is lost.

(a)

Calculate the volume of the ball, correct to 3 significant figures.

(2)

(b)

Calculate the volume of one cone, correct to 3 significant figures.

(2)

(c)

Work out the number of cones that can be made.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4/3 × π × 3³ = 36π	✓ 1
A1 113 cm³	✓ 1
Part (b)
M1 ⅓ × π × 1.5² × 6 = 4.5π	✓ 1
A1 14.1 cm³	✓ 1
Part (c)
M1 36π ÷ 4.5π (use exact or unrounded volumes)	✓ 1
A1 8	✓ 1

Common error: Divide the exact (unrounded) volumes; using rounded values can give a number that is not a whole number.

+ Add to test✓ In your test192IGMC3-5-10r18-10Surface area and volume of solids Paper EHard7 marks

Calculator

A solid metal funnel mould is made from a cone of height 60 cm and base radius 24 cm. A smaller cone of height 25 cm is cut off the top by a plane parallel to the base. The part that remains is a frustum. The metal has density 8.5 g/cm³.

Diagram not to scale

(a)

Show that the radius of the circular cut face is 10 cm.

(2)

(b)

Calculate the volume of the frustum. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the mass of the frustum in kilograms. Use the unrounded volume and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 r/24 = 25/60 (similar cones)	✓ 1
A1 r = 10 shown	✓ 1
Part (b)
M1 1/3 × π × 24² × 60 (= 36 191...)	✓ 1
M1 subtracting 1/3 × π × 10² × 25 (= 2618...)	✓ 1
A1 33 600 cm³ (33 573 before rounding)	✓ 1
Part (c)
M1 33 573 × 8.5 ÷ 1000	✓ 1
A1 285 kg	✓ 1

Exam tip: The frustum volume is the large cone minus the small cone; the small cone is similar to the large one, so use ratios of corresponding lengths.

+ Add to test✓ In your test193IGMC3-5-10r18-11Perimeter and area of plane shapes Paper EHard7 marks

Calculator

A fan-shaped plaza at one end of a bridge is a sector of a circle with centre angle 80° and area 220 m².

Diagram not to scale

(a)

Calculate the radius of the sector. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the length of the curved edge. Use the unrounded radius and give your answer correct to 3 significant figures.

(2)

(c)

Calculate the perimeter of the plaza. Use unrounded values and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 80/360 × π × r² = 220	✓ 1
M1 r² = 990/π (= 315.1...)	✓ 1
A1 17.8 m	✓ 1
Part (b)
M1 80/360 × 2 × π × 17.75...	✓ 1
A1 24.8 m	✓ 1
Part (c)
M1 24.786... + 2 × 17.751...	✓ 1
A1 60.3 m	✓ 1

Common error: Rearrange the sector-area formula for r² first, then take the square root; forgetting the square root gives a value of about 315.

+ Add to test✓ In your test194IGMC3-5-10r23-10Surface area and volume of solids Paper EHard8 marks

Calculator

A cooling tank in a data centre is made from a cylinder of radius 1.5 m with a solid hemisphere of the same radius attached to the top. The total volume of the tank is 25 m³. Use π from your calculator and the unrounded values in later working.

Diagram not to scale

(a)

Calculate the volume of the hemisphere. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the height of the cylindrical part. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the area of the outside surface of the tank, excluding the base. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅔ × π × 1.5³	✓ 1
A1 7.07 m³	✓ 1
Part (b)
M1 cylinder volume = 25 − 7.0686 = 17.93...	✓ 1
M1 h = 17.93 ÷ (π × 1.5²)	✓ 1
A1 2.54 m	✓ 1
Part (c)
M1 curved surface of cylinder 2 × π × 1.5 × 2.5368 = 23.9	✓ 1
M1 hemisphere surface 2 × π × 1.5² = 14.1	✓ 1
A1 38.0 m²	✓ 1

Exam tip: Check which faces of a compound solid are hidden or excluded before adding areas.

+ Add to test✓ In your test195IGMC3-5-10r23-11Circles, arcs and sectors Paper EHard8 marks

Calculator

A sector of paper has radius 18 cm and angle 160°. It is formed into a cone by joining the two straight edges together without overlap. Use π from your calculator.

Diagram not to scale

(a)

Calculate the length of the arc of the sector.

(2)

(b)

Show that the radius of the base of the cone is 8 cm.

(2)

(c)

Calculate the perpendicular height of the cone. Give your answer correct to 3 significant figures.

(2)

(d)

Calculate the volume of the cone. Give your answer correct to 3 significant figures, using the unrounded height.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 160/360 × 2 × π × 18	✓ 1
A1 50.3 cm (accept 16π)	✓ 1
Part (b)
M1 arc length = circumference of the base, 2πr = 160/360 × 2π × 18	✓ 1
A1 r = 160/360 × 18 = 8	✓ 1
Part (c)
M1 sqrt(18² − 8²) = sqrt(260)	✓ 1
A1 16.1 cm	✓ 1
Part (d)
M1 ⅓ × π × 8² × 16.12...	✓ 1
A1 1080 cm³	✓ 1

Common error: The radius of the sector becomes the slant height of the cone, and the arc becomes the base circumference.

+ Add to test✓ In your test196IGMC3-5-12r31-09Surface area and volume of solids Paper EHard8 marks

Calculator

The envelope of a hot-air balloon is modelled as a sphere of volume 2800 m³. The volume of a sphere is $\frac{4}{3}\pi r^3$ and its surface area is $4\pi r^2$.

(a)

Calculate the radius of the sphere. Give your answer to 3 significant figures.

(3)

(b)

Calculate the surface area of the sphere. Use the unrounded radius and give your answer to 3 significant figures.

(2)

(c)

The fabric needed is 12% more than the surface area, to allow for seams. The fabric costs £6.40 per square metre. Calculate the total cost of the fabric, to the nearest pound. Use the unrounded surface area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{4}{3}\pi r^3 = 2800$	✓ 1
M1 $r = \sqrt[3]{\frac{3\times2800}{4\pi}}$	✓ 1
A1 8.74 m	✓ 1
Part (b)
M1 $4\times\pi\times r^2$ with $r = 8.7436\ldots$	✓ 1
A1 961 m²	✓ 1
Part (c)
M1 area of fabric $= 1.12\times$ surface area	✓ 1
M1 multiply by 6.40	✓ 1
A1 £6886	✓ 1

Exam tip: To find a radius from a volume, rearrange to make r³ the subject first, then take the cube root.

+ Add to test✓ In your test197IGMC3-5-12r31-10Surface area and volume of solids Paper EHard6 marks

No calculator

A loudspeaker horn is a solid in the shape of a cone with a smaller cone removed from the top. The complete cone has height 24 cm and base radius 9 cm. The smaller cone that is removed has height 8 cm. The two cones are mathematically similar. The volume of a cone is $\frac{1}{3}\pi r^2 h$.

Diagram not to scale

(a)

Find the radius of the base of the smaller cone.

(2)

(b)

Find the volume of the horn (the frustum that remains), in terms of $\pi$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ratio of heights $8:24 = 1:3$ used for the radii	✓ 1
A1 3 cm	✓ 1
Part (b)
M1 volume of large cone $\frac{1}{3}\pi\times9^2\times24$	✓ 1
A1 $648\pi$	✓ 1
M1 volume of small cone $\frac{1}{3}\pi\times3^2\times8 = 24\pi$ (ECF from the first part)	✓ 1
A1 $624\pi$ cm³	✓ 1

Common error: The volume of a frustum is the volume of the large cone minus the volume of the small cone, not a cone formula applied to an average radius.

+ Add to test✓ In your test198IGMC3-5-12r31-11Surface area and volume of solids Paper EHard8 marks

Calculator

A party hat for a music festival is made by cutting a sector of a circle of radius 15 cm and angle 216° from card and joining the two straight edges to form a cone. The slant height of the cone is 15 cm.

Diagram not to scale

(a)

Find the arc length of the sector, in terms of $\pi$.

(2)

(b)

Show that the base radius of the cone is 9 cm.

(2)

(c)

Calculate the vertical height of the cone.

(2)

(d)

Calculate the volume of the cone. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{216}{360}\times2\times\pi\times15$	✓ 1
A1 $18\pi$ cm	✓ 1
Part (b)
M1 arc length equals the circumference of the base: $2\pi r = 18\pi$	✓ 1
A1 $r = 9$ (must be shown, not just stated)	✓ 1
Part (c)
M1 $h = \sqrt{15^2 - 9^2}$	✓ 1
A1 12 cm	✓ 1
Part (d)
M1 $\frac{1}{3}\times\pi\times9^2\times12$	✓ 1
A1 1020 cm³	✓ 1

Exam tip: When a sector is rolled into a cone, its radius becomes the slant height and its arc length becomes the circumference of the base.

+ Add to test✓ In your test199IGMC3-5-11r34-12Circles, arcs and sectors Paper EHard6 marks

Calculator

A sector of a circle has arc length 15.7 cm and area 62.8 cm². The sector is part of a circular game-board dial.

Diagram not to scale

(a)

Find the radius of the circle.

(3)

(b)

Find the angle of the sector, correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sector area ÷ arc length: (^x⁄₃₆₀πr²) ÷ (^x⁄₃₆₀2πr) = ^r⁄₂ (or equivalent method)	✓ 1
M1 r = 2 × 62.8 ÷ 15.7	✓ 1
A1 8.0 cm	✓ 1
Part (b)
M1 15.7 = ^x⁄₃₆₀ × 2 × π × 8 (use r = 8)	✓ 1
M1 x = 15.7 × 360 ÷ (2π × 8)	✓ 1
A1 112.4°	✓ 1

Exam tip: Compare the formulas for arc length and sector area: dividing one by the other cancels the angle and gives the radius directly.

+ Add to test✓ In your test200IGMC3-5-11r34-13Surface area and volume of solids Paper EHard8 marks

Calculator

Two water-storage tanks are mathematically similar. The small tank has height 50 cm, capacity 40 litres and an outside surface area of 0.45 m². The large tank has height 80 cm.

(a)

Calculate the capacity of the large tank, correct to 3 significant figures.

(3)

(b)

Calculate the outside surface area of the large tank, correct to 3 significant figures.

(2)

(c)

A third tank, similar to the small tank, has capacity 540 litres. Calculate its height, correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length scale factor 80 ÷ 50 = 1.6	✓ 1
M1 40 × 1.6³	✓ 1
A1 164 litres	✓ 1
Part (b)
M1 0.45 × 1.6²	✓ 1
A1 1.15 m²	✓ 1
Part (c)
M1 volume scale factor 540 ÷ 40 = 13.5	✓ 1
M1 length scale factor = ∛13.5 (= 2.381…)	✓ 1
A1 119 cm	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³; do not use the volume ratio directly as a length ratio.

+ Add to test✓ In your test201IGMC3-5-11r37-11Circles, arcs and sectors Paper EHard7 marks

Calculator

A sector of a circle has radius 12 cm and angle 120°. It is cut out of card and bent so that its two straight edges meet, forming the curved surface of a cone.
Volume of a cone = ⅓πr²h.

Diagram not to scale

(a)

Show that the radius of the base of the cone is 4 cm.

(3)

(b)

Calculate the perpendicular height of the cone.

(2)

(c)

Calculate the volume of the cone, correct to 3 significant figures. Use the unrounded height.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 arc length of sector = (120 ÷ 360) × 2 × π × 12 = 8π	✓ 1
M1 this equals the circumference of the base, so 2πr = 8π	✓ 1
A1 r = 4 cm	✓ 1
Part (b)
M1 h² = 12² − 4²	✓ 1
A1 h = √128 = 11.3 cm	✓ 1
Part (c)
M1 ⅓ × π × 4² × √128	✓ 1
A1 190 cm³	✓ 1

Exam tip: In a sector-to-cone problem, the radius of the sector becomes the slant height and the arc becomes the base circumference.

+ Add to test✓ In your test202IGMC3-5-11r37-12Surface area and volume of solids Paper EHard7 marks

Calculator

A water company has two similar cylindrical storage tanks. The small tank is 2 m high and holds 5.6 m³. The large tank is 3 m high. The outside surface area of the small tank is 14 m².

(a)

Calculate the volume of the large tank.

(2)

(b)

Calculate the outside surface area of the large tank.

(2)

(c)

A third tank, similar to the others, holds 44.8 m³. Calculate its height.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.6 × (3 ÷ 2)³	✓ 1
A1 18.9 m³	✓ 1
Part (b)
M1 14 × (3 ÷ 2)²	✓ 1
A1 31.5 m²	✓ 1
Part (c)
M1 volume scale factor = 44.8 ÷ 5.6 = 8	✓ 1
M1 length scale factor = ∛8 = 2	✓ 1
A1 2 × 2 = 4 m	✓ 1

Exam tip: For similar solids: lengths scale by k, areas by k² and volumes by k³.

+ Add to test✓ In your test203IGMC3-5-11r37-14Surface area and volume of solids Paper EHard7 marks

No calculator

A concrete foundation for a wind turbine is a solid cylinder of radius 6 m and height 2 m. Give all answers in terms of π.

(a)

Find the volume of the cylinder.

(2)

(b)

A cylindrical hole of radius 3 m is made through the full height of the foundation. Find the volume of concrete that remains.

(2)

(c)

A different hole of radius x m is made through the full height so that exactly half of the original concrete remains. Find x in the form a√b, where a and b are integers and b is as small as possible.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 π × 6² × 2	✓ 1
A1 72π m³	✓ 1
Part (b)
M1 hole volume = π × 3² × 2 = 18π	✓ 1
A1 72π − 18π = 54π m³	✓ 1
Part (c)
M1 hole volume must be 36π	✓ 1
M1 2πx² = 36π, so x² = 18	✓ 1
A1 x = 3√2	✓ 1

Exam tip: Leave π in your working and combine terms such as 72π − 18π before simplifying surds.

+ Add to test✓ In your test204IGMC3-5-09r38-10Surface area and volume of solids Paper EHard6 marks

Calculator

Two model wind-turbine nose cones are mathematically similar. The small model has height 12 cm and volume 216 cm³. The large model has height 30 cm.

(a)

Calculate the volume of the large model.

(3)

(b)

The outer surface area of the large model is 1150 cm². Calculate the outer surface area of the small model.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 linear scale factor 30 ÷ 12 = 2.5	✓ 1
M1 volume scale factor 2.5³ (= 15.625)	✓ 1
A1 3375 cm³	✓ 1
Part (b)
M1 area scale factor 2.5² (= 6.25)	✓ 1
M1 1150 ÷ 6.25	✓ 1
A1 184 cm²	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³; going from large to small means dividing, not multiplying.

+ Add to test✓ In your test205IGMC3-5-09r38-11Surface area and volume of solids Paper EHard7 marks

Calculator

The concrete foundation of a wind turbine is in the shape of a frustum. It is made by cutting a small cone from the top of a large cone, using a plane parallel to the base. The large cone has base radius 6 m and perpendicular height 9 m. The small cone that is removed has perpendicular height 3 m.

Diagram not to scale

(a)

Show that the radius of the top circular face of the frustum is 2 m.

(2)

(b)

Calculate the volume of the frustum.

(3)

(c)

The concrete has a density of 2400 kg/m³. Use the unrounded volume from part (b) to calculate the mass of the foundation, in tonnes, to 3 significant figures. (1 tonne = 1000 kg)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 scale factor 3 ÷ 9 = ⅓ (small cone is similar to large cone)	✓ 1
A1 radius = ⅓ × 6 = 2 m	✓ 1
Part (b)
M1 volume of large cone = ⅓ × π × 6² × 9 (= 108π)	✓ 1
M1 subtract volume of small cone = ⅓ × π × 2² × 3 (= 4π)	✓ 1
A1 104π = 327 m³ (326.73... before rounding)	✓ 1
Part (c)
M1 326.73... × 2400 (= 784142 kg to the nearest kg)	✓ 1
A1 784 tonnes	✓ 1

Exam tip: A frustum is a large cone minus a small similar cone; use the ratio of heights to find the missing radius.

+ Add to test✓ In your test206IGMC3-5-09r38-12Perimeter and area of plane shapes Paper EHard7 marks

Calculator

A circular swimming pool has radius r metres. A path of constant width 1.5 m surrounds the pool, so the outer edge of the path is also a circle. The area of the path alone is 36π m².

Diagram not to scale

(a)

Show that 3r + 2.25 = 36.

(3)

(b)

Find the value of r.

(2)

(c)

The path is tiled at a cost of £48 for each m². Calculate the total cost of tiling the path, to the nearest pound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 outer radius = r + 1.5	✓ 1
M1 π(r + 1.5)² − πr² = 36π	✓ 1
A1 r² + 3r + 2.25 − r² = 36, so 3r + 2.25 = 36, shown	✓ 1
Part (b)
M1 3r = 36 − 2.25 = 33.75	✓ 1
A1 r = 11.25	✓ 1
Part (c)
M1 36π × 48 (= 5428.67...)	✓ 1
A1 £5429	✓ 1

Common error: The area of the path is the difference of two circle areas; squaring (r + 1.5) must give three terms, not r² + 2.25.

+ Add to test✓ In your test207IGMC3-5-09r39-12Surface area and volume of solids Paper EHard8 marks

Calculator

A marker buoy for ships is made from a solid cylinder with a solid hemisphere joined to one circular end. The cylinder and the hemisphere both have radius 0.6 m. The total volume of the buoy is 3.0 m³.

Diagram not to scale

(a)

Calculate the volume of the hemisphere, giving your answer to 3 significant figures.

(2)

(b)

Calculate the height of the cylinder, giving your answer to 3 significant figures. Use the unrounded volume of the hemisphere.

(4)

(c)

Calculate the total height of the buoy, giving your answer to 3 significant figures. Use the unrounded height of the cylinder.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ⅔ × π × 0.6³	✓ 1
A1 0.452 m³	✓ 1
Part (b)
M1 volume of cylinder = 3.0 − 0.4523… = 2.547…	✓ 1
M1 π × 0.6² × h = 2.547…	✓ 1
M1 h = 2.547… ÷ (π × 0.6²)	✓ 1
A1 2.25 m	✓ 1
Part (c)
M1 cylinder height + radius of hemisphere: 2.252… + 0.6	✓ 1
A1 2.85 m	✓ 1

Exam tip: Subtract the volume you already know from the total before using the cylinder formula, then solve for the height.

+ Add to test✓ In your test208IGMC3-5-09r39-13Surface area and volume of solids Paper EHard8 marks

Calculator

A theatre prop is made by cutting a sector from a flat sheet of card. The sector has radius 24 cm and angle 150°. The straight edges of the sector are joined together to make a hollow cone with no base.

Diagram not to scale

(a)

Show that the length of the arc of the sector is 20π cm.

(2)

(b)

Show that the radius of the base of the cone is 10 cm.

(2)

(c)

Calculate the perpendicular height of the cone, giving your answer to 3 significant figures.

(2)

(d)

Calculate the volume of the cone, giving your answer to 3 significant figures. Use the unrounded height from part (c).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 150/360 × 2 × π × 24	✓ 1
A1 20π cm (shown)	✓ 1
Part (b)
M1 the arc becomes the circumference: 2 × π × r = 20π	✓ 1
A1 r = 10 cm (shown)	✓ 1
Part (c)
M1 √(24² − 10²)	✓ 1
A1 21.8 cm	✓ 1
Part (d)
M1 ⅓ × π × 10² × 21.81…	✓ 1
A1 2280 cm³	✓ 1

Common error: The radius of the sector (24 cm) becomes the slant height of the cone, not its base radius or its height.

+ Add to test✓ In your test209IGMC3-5-09r39-14Surface area and volume of solids Paper EHard5 marks

Calculator

A model of a cargo ship is mathematically similar to the real ship. The deck of the model has area 0.18 m² and the deck of the real ship has area 1152 m². The volume of the hull of the model is 0.0096 m³.

(a)

Work out the length scale factor from the model to the real ship.

(2)

(b)

Calculate the volume of the hull of the real ship, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area scale factor 1152 ÷ 0.18 = 6400	✓ 1
A1 √6400 = 80	✓ 1
Part (b)
M1 volume scale factor 80³ = 512 000	✓ 1
M1 0.0096 × 512 000	✓ 1
A1 4920 m³ (accept 4915.2)	✓ 1

Common error: Lengths scale by k, areas by k² and volumes by k³. Take the square root of the area ratio to get k, then cube it for volume.

+ Add to test✓ In your test210IGMC3-5-10r48-09Perimeter and area of plane shapes Paper EHard10 marks

Calculator

A Formula 1 team makes a small fairing in the shape of a circular segment. The segment is cut from a circle of radius 15.0 cm by a straight chord. The chord subtends an angle of 100° at the centre of the circle.

Diagram not to scale

(a)

Calculate the area of the sector of the circle formed by the chord's two radii. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the area of the triangle formed by the two radii and the chord. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the area of the segment. Use unrounded values from (a) and (b) and give your answer correct to 3 significant figures.

(2)

(d)

Calculate the perimeter of the segment, correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 100/360 × π × 15.0²	✓ 1
A1 196 cm²	✓ 1
Part (b)
M1 ½ × 15.0 × 15.0 × sin 100°	✓ 1
A1 111 cm²	✓ 1
Part (c)
M1 sector area − triangle area, 196.349... − 110.790...	✓ 1
A1 85.6 cm²	✓ 1
Part (d)
M1 arc length 100/360 × 2 × π × 15.0 = 26.18...	✓ 1
M1 chord length 2 × 15.0 × sin 50° or by the cosine rule	✓ 1
A1 chord = 22.98...	✓ 1
A1 49.2 cm	✓ 1

Exam tip: Segment area is sector area minus triangle area; the triangle's area uses ½ab sin C, with the angle in the correct units.

+ Add to test✓ In your test211IGMC3-5-10r48-11Surface area and volume of solids Paper EHard7 marks

Calculator

Two saline bags are mathematically similar in shape. The small bag has height 14 cm and holds 250 mL. The large bag has height 21 cm. The outside surface area of the small bag is 320 cm².

(a)

Write down the ratio of the height of the large bag to the height of the small bag as a decimal.

(1)

(b)

Calculate the volume of the large bag. Give your answer correct to 3 significant figures.

(2)

(c)

A label on the large bag has an area of 108 cm². Calculate the area of the corresponding label on the small bag.

(2)

(d)

Calculate the outside surface area of the large bag.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
1.5	✓ 1
Part (b)
M1 250 × 1.5³	✓ 1
A1 844 mL	✓ 1
Part (c)
M1 108 ÷ 1.5²	✓ 1
A1 48 cm²	✓ 1
Part (d)
M1 320 × 1.5²	✓ 1
A1 720 cm²	✓ 1

Common error: For similar shapes, areas scale by the square of the length ratio and volumes by the cube. Do not multiply by 1.5 for every quantity.

+ Add to test✓ In your test212IGMC3-5-10r48-12Surface area and volume of solids Paper EHard7 marks

Calculator

A Formula 1 team melts a solid aluminium sphere of radius 6.0 cm and casts all of the metal into a solid cone of height 36.0 cm. Volume of a sphere = 4/3 πr³; volume of a cone = 1/3 πr²h.

(a)

Calculate the volume of the sphere. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the radius of the base of the cone. Use the unrounded volume and give your answer correct to 3 significant figures.

(3)

(c)

The density of aluminium is 2.70 g/cm³. Calculate the mass of the sphere in kilograms. Use the unrounded volume and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4/3 × π × 6.0³	✓ 1
A1 905 cm³	✓ 1
Part (b)
M1 1/3 × π × r² × 36.0 = 904.77...	✓ 1
M1 r² = 24 (or 24.0)	✓ 1
A1 4.90 cm	✓ 1
Part (c)
M1 2.70 × 904.77... = 2442.9... g	✓ 1
A1 2.44 kg	✓ 1

Exam tip: When a solid is melted and recast, the volume stays the same: equate the two volume formulae.

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × 9 × 6.4	✓ 1
A1 28.8 cm²	✓ 1
Part (b)
M1 \(\frac{1}{2}\) × 12 × h = 28.8 or h = 28.8 × 2 ÷ 12	✓ 1
A1 4.8 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × (7 + 13) × 5	✓ 1
A1 50 cm²	✓ 1
Part (b)
B1 1 m² = 10 000 cm²	✓ 1
A1 200 tiles	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{40}{360}\) × 2 × π × 9	✓ 1
A1 6.28 cm	✓ 1
Part (b)
B1 24.3 cm, follow through their arc length + 18	✓ 1
Part (c)
M1 for \(\frac{40}{360}\) × π × 9²	✓ 1
A1 28.3 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{4}{3}\) × π × 6³	✓ 1
A1: 905 cm³ (904.77…)	✓ 1
Part (b)
M1: 4 × π × 6²	✓ 1
A1: 452 cm² (452.38…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(2\pi\times3.2\)	✓ 1
A1 20.1 m	✓ 1
Part (b)
M1 \(\pi\times3.2^2\)	✓ 1
A1 32.2 m²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(240\times150\)	✓ 1
A1 36 000 m²	✓ 1
Part (b)
B1 3.6 hectares	✓ 1
Part (c)
M1 \(2\times(240+150)\)	✓ 1
A1 780 m	✓ 1
Part (d)
M1 \(780\times4.50\) (ECF from the perimeter)	✓ 1
A1 €3510	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times1.1\)	✓ 1
A1 3.46 m	✓ 1
Part (b)
M1 \(800\times\pi\times1.1\)	✓ 1
A1 2760 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(5\times3\times2\)	✓ 1
A1 30 m³	✓ 1
Part (b)
B1 30 000 litres (ECF)	✓ 1
Part (c)
M1 \(30\,000\div150\) (ECF)	✓ 1
A1 200 minutes	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{60^2 + 80^2}\)	✓ 1
A1 100 m	✓ 1
Part (b)
B1 240 m (ECF)	✓ 1
Part (c)
M1 \(\frac{1}{2}\times60\times80\)	✓ 1
A1 2400 m²	✓ 1
Part (d)
M1 \(2400\times25 = 60\,000\) g (ECF from the area)	✓ 1
A1 60 kg	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × 24 × 10 or 4 × \(\frac{1}{2}\) × 12 × 5	✓ 1
A1 120 cm²	✓ 1
Part (b)
M1 side = √(12² + 5²) = 13	✓ 1
A1 52 cm	✓ 1
Part (c)
M1 13 × h = 120	✓ 1
A1 9.23 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{\theta}{360}\) × 2 × π × 12 = 15	✓ 1
M1 for θ = 15 × 360 ÷ (24π)	✓ 1
A1 71.6°	✓ 1
Part (b)
M1 for \(\frac{\text{their }\theta}{360}\) × π × 12² or \(\frac{1}{2}\) × 12 × 15	✓ 1
A1 90.0 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{75}{360}\) × π × r² = 50	✓ 1
M1 for r² = 50 × 360 ÷ (75π) [= 76.39…]	✓ 1
A1 8.74 cm	✓ 1
Part (b)
M1 for \(\frac{75}{360}\) × 2 × π × their r + 2 × their r	✓ 1
A1 28.9 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{120}{360}\) × 2 × π × 11	✓ 1
A1 23.0 cm	✓ 1
Part (b)
B1 angle 210° seen or implied	✓ 1
M1 for \(\frac{210}{360}\) × π × 11²	✓ 1
A1 222 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{4}{3}\) × π × 1.5³	✓ 1
A1: 14.1 cm³ (14.137…)	✓ 1
Part (b)
M1: 8³ ÷ their (a), i.e. 512 ÷ 14.137… (= 36.2…)	✓ 1
A1: 36	✓ 1
Part (c)
M1: 512 − 36 × 14.137…	✓ 1
A1: 3.06 cm³ (3.06…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{150}{360}\times2\pi\times9\)	✓ 1
A1 23.6 m	✓ 1
Part (b)
M1 \(\frac{150}{360}\times\pi\times9^2\)	✓ 1
A1 106 m²	✓ 1
Part (c)
M1 arc length + 9 + 9	✓ 1
A1 41.6 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi r^2=2500\)	✓ 1
A1 28.2 m	✓ 1
Part (b)
M1 \(2\pi\times\sqrt{2500/\pi}\)	✓ 1
A1 177 m	✓ 1
Part (c)
M1 5 × 177.2…	✓ 1
A1 886 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}(14.2+9.8)\times6.5\)	✓ 1
A1 78.0 m²	✓ 1
Part (b)
M1 \(49=\frac{1}{2}(8.4+5.6)h\)	✓ 1
A1 7.0 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 area of rectangle \(8\times5 = 40\)	✓ 1
M1 area of semicircle \(\frac{1}{2}\times\pi\times2.5^2\) \((=9.82)\)	✓ 1
A1 49.8 m²	✓ 1
Part (b)
M1 curved edge \(\frac{1}{2}\times\pi\times5\) \((=7.85)\)	✓ 1
M1 straight edges \(8 + 8 + 5 = 21\) (the shared side is not included)	✓ 1
A1 28.9 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\times(400+650)\times220\)	✓ 1
A1 115 500 m²	✓ 1
Part (b)
M1 \(115\,500\times0.15\) (ECF)	✓ 1
A1 17 325 m³	✓ 1
Part (c)
M1 \(17\,325\div12\) \((=1443.75)\) (ECF)	✓ 1
A1 1444 (rounded up to a whole number)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{120}{360}\times2\times\pi\times45\)	✓ 1
A1 94.2 cm	✓ 1
Part (b)
M1 \(\frac{120}{360}\times\pi\times45^2\)	✓ 1
A1 2120 cm²	✓ 1
Part (c)
M1 arc length + two radii \(= 94.25 + 2\times45\)	✓ 1
A1 184 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times44\times30\) (or \(2\pi\times22\times30\))	✓ 1
A1 4150 cm²	✓ 1
Part (b)
M1 area of one end \(\pi\times22^2\) \((=1520)\)	✓ 1
M1 curved area + 2 × one end	✓ 1
A1 7190 cm²	✓ 1
Part (c)
M1 \(\pi\times22^2\times30\div1000\)	✓ 1
A1 45.6 litres	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{140}{360}\) × 2 × π × 9	✓ 1
A1 7π cm	✓ 1
Part (b)
B1 18 + 7π cm	✓ 1
Part (c)
M1 \(\frac{140}{360}\) × π × 9²	✓ 1
A1 \(\frac{63}{2}\)π or 31.5π cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{2}\) × 8.5 × 11.2 × sin 47°	✓ 1
A1 34.8 cm²	✓ 1
Part (b)
M1 QR² = 8.5² + 11.2² − 2 × 8.5 × 11.2 × cos 47°	✓ 1
A1 QR² = 67.8(4)…	✓ 1
A1 QR = 8.24 cm	✓ 1
Part (c)
M1 8.5 + 11.2 + their QR	✓ 1
A1 27.9 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{120}{360}\) × π × 6²	✓ 1
A1 12π cm²	✓ 1
Part (b)
M1 for \(\frac{120}{360}\) × 2 × π × 6 [= 4π]	✓ 1
A1 4π + 12 cm	✓ 1
Part (c)
M1 for 2πr = 4π	✓ 1
A1 2 cm	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{1}{2}\) × \(\frac{4}{3}\) × π × 9³	✓ 1
A1: 1530 cm³ (1526.8…)	✓ 1
Part (b)
M1: their (a) × 2.7, giving 4122.4… g (or 4131 g from 1530)	✓ 1
A1: ÷ 1000 to give 4.12 kg (4.122…); accept 4.13 kg from using 1530	✓ 1
Part (c)
M1: curved surface \(\frac{1}{2}\) × 4 × π × 9²	✓ 1
M1: adding flat circle π × 9²	✓ 1
A1: 763 cm² (763.4…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\frac{1}{3}\) × 10² × 12	✓ 1
A1: 400 cm³	✓ 1
Part (b)
M1: √(12² + 5²)	✓ 1
A1: 13 cm	✓ 1
Part (c)
M1: 10² + 4 × \(\frac{1}{2}\) × 10 × 13	✓ 1
A1: 360 cm²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{2}{3}\pi\times6.5^3\)	✓ 1
A1 575 m³	✓ 1
Part (b)
M1 \(2\pi\times6.5^2\)	✓ 1
A1 265 m²	✓ 1
Part (c)
M1 265.46… ÷ 4.5	✓ 1
A1 59.0 litres	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{2.4^2+1.8^2}\)	✓ 1
A1 \(\sqrt{9}=3.0\) m	✓ 1
Part (b)
M1 \(\frac{1}{3}\pi\times2.4^2\times1.8\)	✓ 1
A1 10.9 m³	✓ 1
Part (c)
M1 \(\pi\times2.4\times3.0\)	✓ 1
A1 22.6 m²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{1}{3}\times\pi\times3.2^2\times2.1\)	✓ 1
A1 22.5 m³	✓ 1
Part (b)
M1 \(22.52\times750\div1000\)	✓ 1
A1 16.9 tonnes	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\pi\times2.4^2\times6.5\)	✓ 1
A1 118 m³	✓ 1
Part (b)
M1 \(\frac{1}{2}\times\frac{4}{3}\times\pi\times2.4^3\)	✓ 1
A1 29.0 m³	✓ 1
Part (c)
M1 \((117.6 + 28.9)\times0.78\) using unrounded volumes	✓ 1
A1 114 tonnes	✓ 1