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Date	May 2019	Marks available	2	Reference code	19M.2.SL.TZ2.T_3
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	T_3	Adapted from	N/A

Question

A factory packages coconut water in cone-shaped containers with a base radius of 5.2 cm and a height of 13 cm.

The factory designers are currently investigating whether a cone-shaped container can be replaced with a cylinder-shaped container with the same radius and the same total surface area.

Find the slant height of the cone-shaped container.

[2]

Find the slant height of the cone-shaped container.

[2]

Show that the total surface area of the cone-shaped container is 314 cm², correct to three significant figures.

[3]

Find the height, $h$ , of this cylinder-shaped container.

[4]

The factory director wants to increase the volume of coconut water sold per container.

State whether or not they should replace the cone-shaped containers with cylinder‑shaped containers. Justify your conclusion.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{\pi {{\left( {5.2} \right)}^2} \times 13}}{3}$ (M1)

Note: Award (M1) for correct substitution in the volume formula for cone.

368 (368.110…) cm³ (A1)(G2)

Note: Accept 117.173… $\pi$ cm³ or $\frac{{8788}}{{75}}\pi$ cm³.

[2 marks]

(slant height²) = (5.2)² + 13² (M1)

Note: Award (M1) for correct substitution into the formula.

14.0 (14.0014…) (cm) (A1)(G2)

[2 marks]

14.0014… × (5.2) × $\pi$ + (5.2)² × $\pi$ (M1)(M1)

Note: Award (M1) for their correct substitution in the curved surface area formula for cone; (M1) for adding the correct area of the base. The addition must be explicitly seen for the second (M1) to be awarded. Do not accept rounded values here as may come from working backwards.

313.679… (cm²) (A1)

Note: Use of 3 sf value 14.0 gives an unrounded answer of 313.656….

314 (cm²) (AG)

Note: Both the unrounded and rounded answers must be seen for the final (A1) to be awarded.

[3 marks]

2 × $\pi$ × (5.2) × $h$ + 2 × $\pi$ × (5.2)² = 314 (M1)(M1)(M1)

Note: Award (M1) for correct substitution in the curved surface area formula for cylinder; (M1) for adding two correct base areas of the cylinder; (M1) for equating their total cylinder surface area to 314 (313.679…). For this mark to be awarded the areas of the two bases must be added to the cylinder curved surface area and equated to 314. Award at most (M1)(M0)(M0) for cylinder curved surface area equated to 314.

( $h$ =) 4.41 (4.41051…) (cm) (A1)(G3)

[4 marks]

$\pi$ × (5.2)² × 4.41051… (M1)

Note: Award (M1) for correct substitution in the volume formula for cylinder.

375 (374.666…) (cm³) (A1)(ft)(G2)

Note: Follow through from part (d).

375 (cm³) > 368 (cm³) (R1)(ft)

“volume of cylinder is larger than volume of cone” or similar (R1)(ft)

Note: Follow through from their answer to part (a). The verbal statement should be consistent with their answers from parts (e) and (a) for the (R1) to be awarded.

replace with the cylinder containers (A1)(ft)

Note: Do not award (A1)(ft)(R0). Follow through from their incorrect volume for the cylinder in this question part but only if substitution in the volume formula shown.

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints

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Topic 3—Geometry and trigonometry

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