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Date May 2015 Marks available 4 Reference code 15M.2.sl.TZ2.2
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 2 Adapted from N/A

Question

The following diagram shows a perfume bottle made up of a cylinder and a cone.

The radius of both the cylinder and the base of the cone is 3 cm.

The height of the cylinder is 4.5 cm.

The slant height of the cone is 4 cm.

(i)     Show that the vertical height of the cone is \(2.65\) cm correct to three significant figures.

(ii)     Calculate the volume of the perfume bottle.

[6]
a.

The bottle contains \({\text{125 c}}{{\text{m}}^{\text{3}}}\) of perfume. The bottle is not full and all of the perfume is in the cylinder part.

Find the height of the perfume in the bottle.

[2]
b.

Temi makes some crafts with perfume bottles, like the one above, once they are empty. Temi wants to know the surface area of one perfume bottle.

Find the total surface area of the perfume bottle.

[4]
c.

Temi covers the perfume bottles with a paint that costs 3 South African rand (ZAR) per millilitre. One millilitre of this paint covers an area of \({\text{7 c}}{{\text{m}}^{\text{2}}}\).

Calculate the cost, in ZAR, of painting the perfume bottle. Give your answer correct to two decimal places.

[4]
d.

Temi sells her perfume bottles in a craft fair for 325 ZAR each. Dominique from France buys one and wants to know how much she has spent, in euros (EUR). The exchange rate is 1 EUR = 13.03 ZAR.

Find the price, in EUR, that Dominique paid for the perfume bottle. Give your answer correct to two decimal places.

[2]
e.

Markscheme

(i)     \({x^2} + {3^2} = {4^2}\)     (M1)

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

Accept correct alternative method using trigonometric ratios.

 

\(x = 2.64575 \ldots \)     (A1)

\(x = 2.65{\text{ }}({\text{cm}})\)     (AG)

Note: The unrounded and rounded answer must be seen for the (A1) to be awarded.

 

OR

\(\sqrt {{4^2} - {3^2}} \)     (M1)

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

 

\( = \sqrt 7 \)     (A1)

\( = 2.65{\text{ (cm)}}\)     (AG)

Note: The exact answer must be seen for the final (A1) to be awarded.

 

(ii)     \(\pi  \times {3^2} \times 4.5 + \frac{1}{3}\pi  \times {3^2} \times 2.65\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into the volume of a cylinder formula, (M1) for correct substitution into the volume of a cone formula, (M1) for adding both of their volumes.

 

\( = 152{\text{ c}}{{\text{m}}^3}\;\;\;(152.210 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}48.45\pi {\text{ c}}{{\text{m}}^3})\)     (A1)(G3)

a.

\(\pi {3^2}h = 125\)     (M1)

Note: Award (M1) for correct substitution into the volume of a cylinder formula.

Accept alternative methods. Accept \(4.43\) (\(4.42913 \ldots \)) from using rounded answers in \(h = \frac{{125 \times 4.5}}{{127}}\).

 

\(h = 4.42{\text{ (cm)}}\;\;\;\left( {4.42097 \ldots {\text{ (cm)}}} \right)\)     (A1)(G2)

b.

\(2\pi  \times 3 \times 4.5 + \pi  \times 3 \times 4 + \pi  \times {3^2}\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into curved surface area of a cylinder formula, (M1) for correct substitution into the curved surface area of a cone formula, (M1) for adding the area of the base of the cylinder to the other two areas.

 

\( = 151{\text{ c}}{{\text{m}}^2}\;\;\;(150.796 \ldots {\text{ c}}{{\text{m}}^2},{\text{ }}48\pi {\text{ c}}{{\text{m}}^2})\)     (A1)(G3)

c.

\(\frac{{150.796 \ldots }}{7} \times 3\)     (M1)(M1)

Notes: Award (M1) for dividing their answer to (c) by \(7\), (M1) for multiplying by \(3\). Accept equivalent methods.

 

\( = 64.63{\text{ (ZAR)}}\)     (A1)(ft)(G2)

Notes: The (A1) is awarded for their correct answer, correctly rounded to 2 decimal places. Follow through from their answer to part (c). If rounded answer to part (c) is used the answer is \(64.71\) (ZAR).

d.

\(\frac{{325}}{{13.03}}\)     (M1)

Note: Award (M1) for dividing \(325\) by \(13.03\).

 

\( = 24.94{\text{ (EUR)}}\)     (A1)(G2)

Note: The (A1) is awarded for the correct answer rounded to 2 decimal places, unless already penalized in part (d).

e.

Examiners report

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e.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.4 » Geometry of three-dimensional solids: cuboid; right prism; right pyramid; right cone; cylinder; sphere; hemisphere; and combinations of these solids.
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