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Date November 2017 Marks available 2 Reference code 17N.2.sl.TZ0.6
Level SL only Paper 2 Time zone TZ0
Command term Show that Question number 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 cm3.

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 cm3 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 cm3, correct to 3 significant figures.

[2]
a.

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

[3]
b.

Find the cost of 100 cm3 of chocolate mousse.

[2]
c.

Show that there is 20 cm3 of orange paste in each special dessert.

[2]
d.

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

[2]
e.

Find the value of x.

[3]
f.

Markscheme

(V=) 13π(3.6)2×11.8     (M1)

 

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

 

=160.145 (cm3)     (A1)

=160 (cm3)     (AG)

 

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

 

[2 marks]

a.

12×43πr3=225     (M1)(A1)

 

Notes:     Award (M1) for multiplying volume of sphere formula by 12 (or equivalent).

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

 

OR

43πr3=450     (A1)(M1)

 

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

 

(r=) 4.75 (cm) (4.75380)     (A1)(G2)

[3 marks]

b.

1.89×100225     (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]

c.

r2=1.8     (A1)

V2=13π(1.8)2×5.9     (M1)

 

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

 

=20 cm3     (AG)

OR

r2=12r     (A1)

V2=(12)3160     (M1)

 

Notes:     Award (M1) for multiplying 160 by (12)3. Award (A0)(M1) for 18×160 if 12 is not seen.

 

=20 (cm3)     (AG)

 

Notes:     Do not award any marks if the response substitutes in the known value (V=20) to find the radius of the cone.

 

[2 marks]

d.

20100×7.42+140100×0.84     (M1)

 

Note:     Award (M1) for the sum of two correct products.

 

$ 2.66     (A1)(ft)(G2)

 

Note:     Follow through from part (c).

 

[2 marks]

e.

x+y=50     (M1)

 

Note:     Award (M1) for correct equation.

 

1.89x+2.66y=111.44     (M1)

 

Note:     Award (M1) for setting up correct equation, including their 2.66 from part (e).

 

(x=) 28     (A1)(ft)(G3)

 

Note:     Follow through from part (e), but only if their answer for x is rounded to the nearest positive integer, where 0<x<50.

Award at most (M1)(M1)(A0) for a final answer of “28, 22”, where the x-value is not clearly defined.

 

[3 marks]

f.

Examiners report

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f.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.5 » Volume and surface areas of the three-dimensional solids defined in 5.4.
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