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

99
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
84 8 4r
2
 4  4r  r
8 2

2
8  4 8  4r
C
8 2
2r
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 
11 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
 11 1









2
3
6
1
2
2
4
4
8
9
6




