Question 1
2 marks
Let the large number be x and the smaller number be y.
Then:
xy
xy
1  2 2x
x
y
8
 88
 96
 48
 40
1.
2.
Therefore the ratio x : y is 48 : 40 = 6 : 5
Question 2
2 marks
y
y 2   3 y   302
3y
2
30
y 2  9 y 2  900
10 y 2  900
y 2  90
3y 2
A
2
3  90 

2
 135 sq units
Question 3
4 marks
median  b  11
a  b  c  27 (as mean is 9)
a  c  11  27
a  c  16
c  a  10 (range is 10)
2c  26
c  13
a3
Question 4
2 marks
A
B
2
E
6
C
D
AE : ED
 2:6
 1: 3
Question 5
4 marks
The sequence is 3, x, x  3, 2x  6, 4 x  12, 8 x  24  80
8 x  24  80
8 x  56
x 7
The four missing numbers are 7, 10, 20, 40
The sum of these numbers is 77
Question 6
4 marks
4
3
Ratio of smaller triangle to larger triangle is 3 : 7
Area ratios  32 : 72  9 : 49
49
245
Area of larger triangle 
5 
9
9
Area of trapezium  Area of large triangle - area of smaller triangle

245
200
2
5 
or 22 sq units
9
9
9
Question 7
4 marks
A: Pay only three quarters of the normal price! (25% discount)
1
B: Buy two – get one free! (33 3 % discount)
C: Two for the price of four! (don’t shop here!)
D: One fifth off all prices! (20% discount)
E: 30% price cut (30% discount)
Answer: B
Question 8 6 marks
AShaded 
  2r 
4
2

r 2
4

3 2
r
4
4 r 2  r 2 3 r 2



4
4
4
6 r 2

4
3 r 2

2
2
3  2r 
3 r 2 1 2
A Unshaded 

 r
4
4
4
12 r 2 3 r 2  r 2



4
4
4
10 r 2

4
5 r 2

2
Ratio of shaded to unshaded area
3 r 2 5 r 2

:
2
2
 3:5
Question 9 6 marks
2
c  3t  8s 
 2c  6t  16s
3
4c  2t  6s 
12c  6t  18s
10c
c
 2s
 0.2s
4
c  3t  8s 
 4c  12t  32s
4c  2t  6s  
 4c  2t  6s
10t  26s
t  2.6s
For third set of scales we have
3c  4t  3  0.2s   4  2.6s 
 0.6s  10.4s
 11s
11 squares are needed
?
Question 10 4 marks
C
9 cm
15 cm
A
9 cm
BC  152  122
 225  144
 81
9
D
B
12 cm
A
ACD
99
2
 40.5cm2

Question 11 6 marks
Jill
Jack
The angle Jill moves
through before they
meet is given by
30
 360
80
 135

  135
225
Therefore Jack moves through
225° before they meet.
360
 30
225
72

 302
453
TJack 
30 s
144
3
 48 s

Question 12 6 marks
Splitting the regular hexagon into 6 congruent
triangles as shown below
A
B
Area of kite ABCE
4
 360
6
 4  60

C
F
 240 sq units
E
D
Question 13 8 marks
or y  x 2  6 x  13
  x  3   13  9
2
The coordinates of the
image become (-x, -y)
  x  3  4
2
 y    x   6   x   13
2
y   x  3  4
2
y
 x 2  6 x  13
y   x 2  6 x  13
13
If rotated 180° about
the origin this
becomes
(3, 4)
(-3, -4)
x
O
y    x  3  4
2


  x 2  6x  9  4
 x 2  6x  9  4
-13
  x 2  6 x  13
y    x  3  4
2
Question 14
8 marks
Let the fraction of the chocolate bar that
1
Pinkie has eaten be x then Perky has eaten x
3
The remaining fractions of the chocolate bar
uneaten are
Pinkie
Perky
1 x
1
1
x
3
1
x  2 1  x 
3
1
1  x  2  2x
3
3
x 1
2
2
x
3
When Perky has two times as much as has Pinkie we have 1 
8 marks
Question 15
Adding the following lines to the diagram so that we
have pairs of congruent triangles.
X
C
x
36
x
y
O
y
A
B
2 x  2y  180  36  360
Y
2 x  2y  36  180
2 x  2y  144
x  y  72
ˆ  72
XOY
Question 16
6 marks
For the cheetah
distance = speed × time
24
 80 
60  60
1
 80 
150
8

km.
15
For the snail
distance
speed
8
 15
1
30
8 230


1
1 15
time 
 16 hours
30 hr/km  1 km/30hr
1

km/h
30
C
Question 17
10 marks
The hypotenuse of the
triangle is
B
2A
8
r
22  2 2  8
2 2
2
Now consider triangle ABC and we can
deduce the missing lengths AC and BC and
apply Pythagoras theorem again.

 r 2  2  r 
2
8  4 8  4  r 2  4  4r  r 2
8  4 8  4r
C
8 2
2r
B
r
A

8 2
2
4 8 8
r
4
8 2
2 2 2
r 2


2 1
Question 18
12 marks
1 1 2
3
5
8
13
21
34
 







3 6 12 24 48 96 192 384 768
1 1  1
1  1
2   2
3   3
5   5
8 
  










3 6  12 12   24 24   48 48   96 96   192 192 
Let S 




1
S
4
S
1  1
1
2
3
5






3  12 24 48 96 192
1 1 1 1 2
3
5
   



3 4  3 6 12 24 48
1 1
1
 S S
3 4
2
1 3
 S
3 4
1
3
4
3
2
3
5
8
 1 1






  6 12 24 48 96 192
 
3
5
8
 1 1 1 2






 2  3 6 12 24 48 96 







