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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  c 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  c 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  c m 3 , correct to 3 significant figures.

[2]
a.

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

[3]
b.

Find the cost of 100  c m 3 of chocolate mousse.

[2]
c.

Show that there is 20  c m 3 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

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

( V = )   1 3 π ( 3.6 ) 2 × 11.8     (M1)

 

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

 

= 160.145   ( c m 3 )     (A1)

= 160   ( c m 3 )     (AG)

 

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

 

[2 marks]

a.

1 2 × 4 3 π r 3 = 225     (M1)(A1)

 

Notes:     Award (M1) for multiplying volume of sphere formula by 1 2 (or equivalent).

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

 

OR

4 3 π r 3 = 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 × 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]

c.

r 2 = 1.8     (A1)

V 2 = 1 3 π ( 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  c m 3     (AG)

OR

r 2 = 1 2 r     (A1)

V 2 = ( 1 2 ) 3 160     (M1)

 

Notes:     Award (M1) for multiplying 160 by ( 1 2 ) 3 . Award (A0)(M1) for 1 8 × 160 if 1 2 is not seen.

 

= 20   ( c m 3 )     (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.

20 100 × 7.42 + 140 100 × 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.

 

1.89 x + 2.66 y = 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

[N/A]
a.
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b.
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c.
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d.
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e.
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f.

Syllabus sections

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

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