Transcript Slide 1
Question 1
2 marks
Let the large number be x and the smaller number be y.
Then:
x y 8
1.
x y 88
2.
1 2 2x
96
x 48
y 40
Therefore the ratio x : y is 48 : 40 = 6 : 5
Question 2
2 marks
y
3y
y
30
2
3 y 30
y
2
2
2
9y
2
900
10 y
2
900
y
2
90
A
3y
2
2
3 90
2
1 3 5 sq u n i t s
Question 3
4 marks
m e d ia n b 1 1
a b c 2 7 ( a s m e a n is 9 )
a c 11 27
a c 16
c a 1 0 ( ra n g e is 1 0 )
2c 2 6
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, 2 x 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
A rea ratios 3 : 7 9 : 49
2
A re a o f la rg e r tria n g le
49
2
5
245
9
9
A re a o f tra p e ziu m A re a o f la rg e tria n g le - a re a o f sm a lle r tria n g le
245
9
5
200
9
or 22
2
9
sq u n its
Question 7
4 marks
A: Pay only three quarters of the normal price! (25% discount)
1
B: Buy two – get one free! ( 3 3 3 % d i sco u n t )
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
AS h a d e d
4 r
2
4
A
6 r
2
r
4
r
2
4
2
3 r
4
3
4
r
2
2
4
2
4
3 r
2
2
U n sh a d e d
1 2 r
2
4
2r
3 r
3 r
4
2
r
2
2
1
4
r
2
Ratio of shaded to unshaded area
4
3 r
2
4
2
2
2
4
4
1 0 r
5 r
3 2 r
2
3:5
2
:
5 r
2
2
Question 9 6 marks
2
c 3 t 8 s 2 c 6 t 1 6 s
3
4 c 2 t 6 s 1 2 c 6 t 1 8 s
2s
10c
0 .2 s
c
4
c 3 t 8 s 4 c 1 2 t 3 2 s
4 c 2 t 6 s 4 c 2 t
6s
10t 26s
t 2 .6 s
For third set of scales we have
3 c 4 t 3 0 .2 s 4 2 .6 s
0 .6 s 1 0 .4 s
1 1s
11 squares are needed
?
Question 10 4 marks
C
9 cm
15 cm
A
D
9 cm
BC
15 12
2
2
225 144
81
9
B
12 cm
A
ACD
99
2
4 0 .5 c m
2
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.
30 s
T Ja ck
360
72
225
453
30
144
3
48 s
302
Question 12 6 marks
Splitting the regular hexagon into 6 congruent
triangles as shown below
A
B
Area of kite ABCE
C
F
4
6
360
4 60
2 4 0 s q u n its
E
D
Question 13 8 marks
or y x 2 6 x 1 3
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 6 x 13
2
y x 6 x 13
2
13
If rotated 180° about
the origin this
becomes
(3, 4)
(-3, -4)
x
O
y x 3 4
2
x 6x 9 4
2
x 6x 9 4
2
-13
x 6 x 13
2
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
When Perky has two times as much as has Pinkie we have 1
1
3
1
1
x 2 1 x
x 2 2x
3
3
x 1
2
x
2
3
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
X Oˆ Y 7 2
6 marks
Question 16
For the cheetah
For the snail
distance = speed × time
tim e
80
80
8
15
24
60 60
1
150
d is ta n c e
speed
8
3 0 h r/k m 1 k m /3 0 h r
15
1
30
km .
8
1
30
23 0
1
1 15
1 6 h o u rs
k m /h
C
Question 17
10 marks
The hypotenuse of the
triangle is
2 2
2
2
B
2A
r
8
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
84 8 4r
2
4 4r r
8 2
2
8 4 8 4r
C
8 2
2r
B
r
A
2
4 8 8
4
r
8 2
2 2 2
r 2
2 1
2
Question 18
Let S
1
3
1
1
4
S
S
12
3
24
5
48
8
96
13
192
1
1
1
3
1
3
4
3
21
384
34
768
1 1
2 2
3 3
5 5
8
1
6 12 12 24 24 48 48 96 96 192 192
11 1
2
3
5
4 3 6 12 24 48
1
S
4
1
1
2
3
5
1
3 12 24 48 96 192
3
2
1
3
6
3
1
12 marks
3
4
1
2
S
S
1
2
3
5
8
1
6 12 24 48 96 192
2
3
5
8
11 1
2
3
6
1
2
2
4
4
8
9
6