1. Calculate the gradient of the line joining (1,5) to (7,2)
gradient =
y2  y1 2  5  3
1



x2  x1 7  1
6
2
11
2. Without a calculator find the value of 64 3
1 13
64
=
1
3
64  64  64  4  256
1
note that power 1/3 is cube root (and power ½ is square root)
3. The volumes of two similar jugs are 40cm3 and 135 cm3. If the surface area of the
smaller jug is 16cm2, find the surface area of the larger jug.
For volumes the scale factor is cubed, for areas it is squared.
135 27
SF 

40
8
3
So
so
SF  3
27 3

8
2
2
16

SF
 16 
Surface area larger jug =
9
 36cm 2
4
4. If s is inversely proportional to t and s = 4 when t = 5, find s when t = 2.
Proportionality question so s = k x function of (t)
Inversely proportional so
k
s
t
20
s

When inversely proportional k = s x t = 20 so
t
When t = 2
s
20
 10
2
5. Find the nth term of the sequence 7, 11, 15, 19, …
Constant first difference so un = an + b
Difference is 4 so a = 4
zeroth term is 3 so b = 3
un = 4n + 3
6. Find x, giving your
answer to 1 dp.
Height h2 = 142 - 122 = 196 - 144 = 52 so h = 7.21110
sin x 
h
 0.55470
13
x  sin 1 (0.55470)  33.69007o  33.7o
7. A Shopkeeper buys a case of 24 cans of cola for £4.90 and then sells each can for 35p.
What is his % profit?
Total takings = 24 x 35p = £8.40
%profit 
profit
8.40  4.90
3.50
 100% 
 100% 
 100%  71.4%
cost
4.90
4.90
8. The probability of Tim serving an ace is 0.15. Find the probability that Tim will not serve
any aces in 3 consecutive serves?
p(not an ace) = 1 – 0.15 = 0.85
p( not ace 3 times) = 0.853 = 0.614 (3dp)
9. Simplify (4b3 )2  5b4
(4b3 )2  5b4  (16b6 )  5b4  80b10
10. The bearing of B from A is 030 and the bearing of C from B is 105. Calculate ABC.
ABC = 30 + 73 = 105
11. Solve
LCD = 12
3x  5 2 x  1

3
4
6
3(3 x  5) 2(2 x  1)

3
12
12
9 x  15  4 x  2
3
12
5 x  17  36
5 x  19
19
4
x
3
5
5
12
5
1km  miles
8
5
so 3km  3 
8
15
 miles
8
15
 1760 miles
8
 15  220 yards
So 3 km  3300 yards
13
£18 000 after 1 year is
£18 000 x 0.95 = £17 100
after 2 years is £17 100 x 0.95 = £16 245
£18000
In general, after n years is  £18000  (0.95) n
4 8  2
3x
14
4x
2 x 1
First choose a lowest common base : 2
4 = 22; 8 = 23
2   2 
2 3x
3 4x
 2 2 x1
26 x  212 x  2 2 x1
218 x  2 2 x1
18 x  2 x  1
16 x  1
1
x
16
15 x3 – 7x2 + 10x = x(x2 – 7x + 10) = x(x - 5)(x - 2)
16
Beats/min
Frequency
61 to 70
71 to 80
81 to 90
Total
6
8
4
18
typical
value
65.5
75.5
85.5
Estimate of mean = 1339/18 = 74.3888
= 74.4 (3sig fig)
17
x 2  3x x( x  3) x


cx  3c c( x  3) c
fx
393
604
342
1339
18
19 18  400 = 0.045 = 4.5 x 10-2
20 cos x = 0.75
x = 41.4º and 360 - 41.4º = 318.6º
21 104% = £20 540 so 100% = £20 540  1.04 = £19 750
22 (a) out of 6 x 4 = 24 outcomes,
(6,1), (5,2), (4,3) and (3,4) 4 have total 7, so p= 4/24=1/6.
(b) 14/24 = 7/12
1
1
2
3
4
23
2
X
3
X
X
26
26
l
 2r 
 2 15  6.81
360
360
4
X
X
X
5
X
X
X
X
6
X
X
X
X
24 (a) (2,3+4) = (2,7) (b) (2+5,3) = (7,3)
(c) (2,2x3) = (2,6)
(d) (-2,3)
25 removed
26 Total girls mass = 6 x 40 = 240
Total boys mass = 4 x 45 = 180
Total mass = 420 Total of 10 children, so average = 420/10 = 42.
27 1 ¼ hours = 5/4 hours for 80 km
Speed = distance / time = 80  5/4 = 80 x 4/5 = 16 x 4 = 64 km / hr
28
y2  x2  z 2
y2  z 2  x2
x   y2  z2
29
4.25 x 1017  3.68 x 10-12 = 1.15x1029
30
52.05 (2dp) or 52.1 (3 sig fig)
31
32 e.g. 2r = 10
so r = 5/
33 Common denominator = (x+3)(x-3)
x
2
x( x  3)
2( x  3)



x  3 x  3 ( x  3)( x  3) ( x  3)( x  3)
x 2  3x  2 x  6

( x  3)( x  3)
x2  x  6

( x  3)( x  3)
34 Total mark = 6 x 74 = 444
Total new mark = 7 x 75 = 525
Mark required = 525 – 444 = 81
35
Exterior angle = 360/15 = 24
interior angle = 180 – 24 = 156
36
y = x3 + 2x + 3
0 = x3 – x + 5
subtract y = 3x - 2
37. Proportionality question so E = k x function of (x)
proportional to x squared so
E  kx2
126  k  32
k  126 / 9  14
E  14 x 2
When x = 4, E = 14 x 42 = 224J
R(Q x)  P  x
RQ Rx  P  x
Expand brackets
RQ P  x  Rx
Collect x terms on one side
RQ P  x  Rx
Collect non-x terms on other side
RQ P  x(1  R)
Factorise
x(1  R)  RQ  P
Reorder
RQ  P
x

Divide
1 R
38 First get rid of fractions
39 H = 60 tan 13 = 13.85m; tan x = H/40 x = tan-1 (H/40) = 19.1
40 Sin x = -0.8 x = sin-1 (0.8) = -53.1
To find value between 0 and 360, add 360 to get 306.9
Then by symmetry find 180 + 53.1 = 233.1
41 Vol = r2h so 350 = r2 x 8 rearranging
r2 
350
 13.93 so r = 3.73 cm
8
42 First note different units. Use cm for length as can convert to ml.
5 m/s = 500 cm/s so volume in 1 sec is 500 x 4 = 2000ml = 2l
Volume in 1 hour = 2l x 3600 = 7200l.
43 Total interior and exterior angles are 9 x 180 = 1620
but 360 external so total interior is 1620 - 360 = 1260
44 (x - 6)(x + 2) = 0 x = 6 or x = -2
A (0,-12)
B (0,6) and C (0, -2)
45 £18 + 3 x £ 6.50 + £7.45 = £18 + £19.50 + £7.45 = £44.95
46 298 000 x 60 x 60 x24 x 365 = 9.40 x 1012
47
48 ax – ay + bx – by
49
= a(x – y) + b(x – y)
= (a + b)(x – y)
r1 = 1.03r
h1 = 0.97h
V1 = πr12h1
= π(1.03r)2(0.97h)
= π(1.03r)2(0.97h)
=1.029 πr2h
So volume has gone up by 2.9%
50 x2 + 6x +2 = (x+3)2 – 9 + 2 = (x+3)2 – 7
p = 3; q = -7
Minimum value occurs when (x+3)2 = 0 x = -3, and then value is -7
51 gradient = (8 – (-1))/(4 – 1) = 9/3 =3
y = 3x + c
goes through (1,-1) so
y = 3x – 4
52
5x + 7y = 3
3x – 4y = 10
Equate y terms
(1) x 4
20x – 28y = 12
(2) x 7
21x – 28y = 70
Add
41x
= 82
(1)
(2)
=> x = 2
-1 = 3 + c c = -4
sub in (1) 10 + 7y = 3 => y = -1
check in (2) 3 x 2 – 4 x (-1) = 10
53
½ l = 500ml = 500 cm3 so total volume = 512 cm3
x3 = 512 so x = 8cm
54
(5t + 6)(4t – 3)
55
total of 2+3+4 = 9 parts represents 180 deg
So 1 part = 180/9 = 20 deg
Angles are 40°, 60°, 80°
56
346 x 72 = 346 x 70 + 346 x 2
= 21000 + 2800 + 420 + 692
= 23800 + 1112
= 24912
57
8500/24 multiple of 24: 24, 48, 72, 96 choose 7200 (300 boxes)
Take away from 8500 leave 1300 next choose 1200 (50 boxes)
Take away from 1300 leaves 100 – 4 boxes + a bit so need 5 boxes
Total = 300 + 50 + 5 = 355 boxes
58
6x – 8 < 2x + 3
4x < 11
x<2¾
but x is integer, so greatest is 2.
= 20t2 – 15t + 24t - 18
= 20t2 + 9t – 18
Check: x = 2 satisfies inequality but x = 3 does not
60
a) circle radius 3cm centre B
b) perpendicular bisector of AB
c) Striaght lines 2cm above/below line, with semi-circles at each end