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Date	November 2017	Marks available	2	Reference code	17N.2.SL.TZ0.T_6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	T_6	Adapted from	N/A

Question

A restaurant serves desserts in glasses in the shape of a cone and in the shape of a hemisphere. The diameter of a cone shaped glass is 7.2 cm and the height of the cone is 11.8 cm as shown.

N17/5/MATSD/SP2/ENG/TZ0/06

The volume of a hemisphere shaped glass is $225{\text{ c}}{{\text{m}}^3}$ .

The restaurant offers two types of dessert.

The regular dessert is a hemisphere shaped glass completely filled with chocolate mousse. The cost, to the restaurant, of the chocolate mousse for one regular dessert is $1.89.

The special dessert is a cone shaped glass filled with two ingredients. It is first filled with orange paste to half of its height and then with chocolate mousse for the remaining volume.

N17/5/MATSD/SP2/ENG/TZ0/06.d.e.f

The cost, to the restaurant, of $100{\text{ c}}{{\text{m}}^3}$ of orange paste is $7.42.

A chef at the restaurant prepares 50 desserts; $x$ regular desserts and $y$ special desserts. The cost of the ingredients for the 50 desserts is $111.44.

Show that the volume of a cone shaped glass is $160{\text{ c}}{{\text{m}}^3}$ , correct to 3 significant figures.

[2]

Calculate the radius, $r$ , of a hemisphere shaped glass.

[3]

Find the cost of $100{\text{ c}}{{\text{m}}^3}$ of chocolate mousse.

[2]

Show that there is $20{\text{ c}}{{\text{m}}^3}$ of orange paste in each special dessert.

[2]

Find the total cost of the ingredients of one special dessert.

[2]

Find the value of $x$ .

[3]

Markscheme

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

$(V = ){\text{ }}\frac{1}{3}\pi {(3.6)^2} \times 11.8$ (M1)

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

$= 160.145 \ldots {\text{ }}({\text{c}}{{\text{m}}^3})$ (A1)

$= 160{\text{ }}({\text{c}}{{\text{m}}^3})$ (AG)

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

[2 marks]

$\frac{1}{2} \times \frac{4}{3}\pi {r^3} = 225$ (M1)(A1)

Notes: Award (M1) for multiplying volume of sphere formula by $\frac{1}{2}$ (or equivalent).

Award (A1) for equating the volume of hemisphere formula to 225.

$\frac{4}{3}\pi {r^3} = 450$ (A1)(M1)

Notes: Award (A1) for 450 seen, (M1) for equating the volume of sphere formula to 450.

$(r = ){\text{ }}4.75{\text{ }}({\text{cm}}){\text{ }}(4.75380 \ldots )$ (A1)(G2)

[3 marks]

$\frac{{1.89 \times 100}}{{225}}$ (M1)

Note: Award (M1) for dividing 1.89 by 2.25, or equivalent.

$= 0.84$ (A1)(G2)

Note: Accept 84 cents if the units are explicit.

[2 marks]

${r_2} = 1.8$ (A1)

${V_2} = \frac{1}{3}\pi {(1.8)^2} \times 5.9$ (M1)

Note: Award (M1) for correct substitution into volume of a cone formula, but only if the result rounds to 20.

$= 20{\text{ c}}{{\text{m}}^3}$ (AG)

${r_2} = \frac{1}{2}r$ (A1)

${V_2} = {\left( {\frac{1}{2}} \right)^3}160$ (M1)

Notes: Award (M1) for multiplying 160 by ${\left( {\frac{1}{2}} \right)^3}$ . Award (A0)(M1) for $\frac{1}{8} \times 160$ if $\frac{1}{2}$ is not seen.