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Date	May 2017	Marks available	4	Reference code	17M.2.SL.TZ1.T_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	T_4	Adapted from	N/A

Question

A pan, in which to cook a pizza, is in the shape of a cylinder. The pan has a diameter of 35 cm and a height of 0.5 cm.

M17/5/MATSD/SP2/ENG/TZ1/04

A chef had enough pizza dough to exactly fill the pan. The dough was in the shape of a sphere.

The pizza was cooked in a hot oven. Once taken out of the oven, the pizza was placed in a dining room.

The temperature, $P$ , of the pizza, in degrees Celsius, °C, can be modelled by

$P(t) = a{(2.06)^{ - t}} + 19,{\text{ }}t \geqslant 0$

where $a$ is a constant and $t$ is the time, in minutes, since the pizza was taken out of the oven.

When the pizza was taken out of the oven its temperature was 230 °C.

The pizza can be eaten once its temperature drops to 45 °C.

Calculate the volume of this pan.

[3]

a.

Find the radius of the sphere in cm, correct to one decimal place.

[4]

b.

Find the value of $a$ .

[2]

c.

Find the temperature that the pizza will be 5 minutes after it is taken out of the oven.

[2]

d.

Calculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.

[3]

e.

In the context of this model, state what the value of 19 represents.

[1]

f.

Markscheme

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

$(V = ){\text{ }}\pi \times {{\text{(17.5)}}^2} \times 0.5$ (A1)(M1)

Notes: Award (A1) for 17.5 (or equivalent) seen.

Award (M1) for correct substitutions into volume of a cylinder formula.

$= 481{\text{ c}}{{\text{m}}^3}{\text{ }}(481.056 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}153.125\pi {\text{ c}}{{\text{m}}^3})$ (A1)(G2)

[3 marks]

a.

$\frac{4}{3} \times \pi \times {r^3} = 481.056 \ldots$ (M1)

Note: Award (M1) for equating their answer to part (a) to the volume of sphere.

${r^3} = \frac{{3 \times 481.056 \ldots }}{{4\pi }}{\text{ }}( = 114.843 \ldots )$ (M1)

Note: Award (M1) for correctly rearranging so ${r^3}$ is the subject.

$r = 4.86074 \ldots {\text{ (cm)}}$ (A1)(ft)(G2)

Note: Award (A1) for correct unrounded answer seen. Follow through from part (a).

$= 4.9{\text{ (cm)}}$ (A1)(ft)(G3)

Note: The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.

[4 marks]

b.

$230 = a{(2.06)^0} + 19$ (M1)

Note: Award (M1) for correct substitution.

$a = 211$ (A1)(G2)

[2 marks]

c.

$(P = ){\text{ }}211 \times {(2.06)^{ - 5}} + 19$ (M1)

Note: Award (M1) for correct substitution into the function, $P(t)$ . Follow through from part (c). The negative sign in the exponent is required for correct substitution.

$= 24.7$ (°C) $(24.6878 \ldots$ (°C)) (A1)(ft)(G2)

[2 marks]

d.

$45 = 211 \times {(2.06)^{ - t}} + 19$ (M1)

Note: Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their $a$ in part (c)).

$(t = ){\text{ }}2.89711 \ldots$ (A1)(ft)(G1)

$174{\text{ (seconds) }}\left( {173.826 \ldots {\text{ (seconds)}}} \right)$ (A1)(ft)(G2)

Note: Award final (A1)(ft) for converting their ${\text{2.89711}} \ldots$ minutes into seconds.

[3 marks]

e.

the temperature of the (dining) room (A1)

OR

the lowest final temperature to which the pizza will cool (A1)

[1 mark]

f.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

[N/A]

f.

Syllabus sections

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

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