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PART C:
Each correct answer is worth 6 marks
16. As a result of poor attendance at soccer matches it was decided to decrease
the ticket price by 20. At the next match the number of tickets sold increased by 20. Compared to the previous match, the income from the sale
of tickets A increased
by 20
B decreased by 20
C increased by 4
D decreased by 4
E remained the same
ANSWER: D Let ticket price original be x
And number attended be y
Income 1 = xy
i New ticket price = x 0,8
Number attended at new price = y 1,2 Income 2
= x 0,8 y 1,2 = xy
0,96 ii
Comparing i and ii Income decreased by from 1 to 0,96 thus by 4
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17. The mean average of n numbers is p . When the number q is removed
from the list of numbers of which the average was taken, the mean average increases by 2. The value of q is
A B
C D
E p
n p
n p
n p
n p
n −
− +
− − +
− +
2 2
2 2
2 2
2 ANSWER: B
Total = pn New total = pn – q
Number of numbers remaining = n – 1 New average = p + 2
Therefore the equation is: pn
q n
p pn
q pn
p n
q p
n −
− = +
∴ − =
− + −
∴ = − +
1 2
2 2
2 2
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18.
1
R
2
B
3
X
4
S
5
O
6
P
7
E
8
D
9
M
10
Z
11
L
12
K
13
A
14
G
15
C
16
T
17
N
18
J
19
F
20
U
21
H
22
V
23
W
24
Q
25
Y
26
I In the above table each letter of the alphabet is given a value.
The algebraic expression 4 3
x −
is used as a key to convert the letters
P S R X B O E into the word H A R M O N Y. Which one of the following keys is used to convert S R X B into G O L D ?
A B
C D
E x
x x
x x
+ −
+ +
+ 2
2 1
3 2
5 1
2 1
ANSWER: C First you need to work out how the key works.
Letters in code Letters in word
P = 6 H = 21
S = 4 A = 13
etc. Let the “letter in code’ be x
Then for P
H And for S
A =
− = × −
= −
= =
= −
= × − =
− =
= 6 4
3 4 6 3
24 3 21
4 4 3
4 4 3 16 3
13 ,
, x
x
Thus the key is a “formula” which converts the code-number of “Letter in code” into the code-number of “Letter in word”.
Now consider S R X B G O L D
S = 4 G = 14
R = 1 O = 5
X = 3 L = 11
B = 2 D = 8
It is not too difficult to see that the key here is 3 2
3 2
× + = +
x x
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19. ABC is an equilateral triangle
with sides of 2 units. Using A, B and C as centres of circles, arcs
BC, AC and AB are drawn.
The shaded area is
A 2 B
C D 2 3
E π
π π
π π
− −
− 3 3
3 2
ANSWER: A Area 1, area 2 and area 3 are equal.
Area of Area
of area of circle because
Area
Area Area of shaded regions
∆
∆ +
= = ×
= ×
=
= − =
∴ = =
− −
∴ =
− =
− 1
1 6
1 6
60 1
6 360
1 6
4 2
3 2
1 3
3 1
2 3
3 3
2 3
3 2
3 3
2
2 2
2
π π
π
π π
π r
i
h h
ii i
ii :
D D
A B
C
1
3 2
h
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20. Five children, Amelia, Bongani, Charles, Divine and Edwina, were in the
