1. 1. (a) Simplify x3
+ y + 5x2
− 2y − 2x2
+ 5 (2 marks)
x3
− y + 3x2
+ 5 (2 marks)
(b) Evaluate x2
y − y2
x when y = 3 and x = 2. (2 marks)
22
× 3 − 32
× 2 = 12 − 18 = −6 (2 marks)
2. (a) What is the equation of the line passing through the point (2, −6) with gradient
−3? (3 marks)
−6 = −3 × 2 + c (1 mark)
c = 0 (1 mark)
y = −3x (1 mark)
(b) Solve the simultaneous equations (4 marks)
3x + y = 0
x − y = −4
4x = −4 (1 mark)
x = −1 (1 mark)
3 × −1 + y = 0 (1 mark)
y = 3 (1 mark)
3. A company wants to know if there is a relationship between age and favourite drink. They get the
following results:
Coke Water Wine
16-20 7 0 3
21-40 5 9 6
41-60 0 3 7
(a) State the null hypothesis and the alternative hypothesis. (2 marks)
H0 : There is no correlation. (1 mark)
H1 : There is a correlation. (1 mark)
(b) What is the χ2
test statistic? (10 marks)
Totals:
7 0 3 10
5 9 6 20
0 3 7 10
12 12 16 40
Fit:
3 3 4
6 6 8
3 3 4
Residuals:
4 -3 -1
-1 3 -2
-3 0 3
χ2
table:
5.3333 3 0.25
0.1667 1.5 0.5
3 0 2.25
1

2. Test statistic χ2
= 16
2 marks per table (-0.5 per error), 2 marks for solution.
(c) If we test at a 5% level of significance what is the critical χ2
value? (2 marks)
df = (3 − 1)(3 − 1) = 4 (1 mark)
Critical value is 9.49. (1 mark)
(d) What can we deduce? (2 marks)
16 > 9.49 (1 mark)
There is a correlation. Reject H0. (1 mark)
4. A businessman keeps a note of who many hours overtime he works over 9 days.
1 4 2 2 5 3 2 2 1
(a) What is the mode? (1 marks)
Mode is 2. (1 mark)
(b) What is the median? (2 marks)
1 1 2 2 2 2 3 4 5
9 × 1
2 = 4.5
Median is 2.
1 mark working, 1 mark solution
(c) What is the interquartile range? (3 marks)
Q1 : 9 × 1
4 = 2.25
Q1 = 2. (1 mark)
Q3 : 9 × 3
4 = 6.75
Q3 = 3. (1 mark)
Interquartile range is 3 − 2 = 1 (1 mark)
5. Consider the following triangle.
43
A 60
(a) What is the angle A to 2 d.p.? (2 mark)
sin(A)
3 = sin(60)
4 (1 mark)
sin(A) = 3 sin(60)
4
A = sin−1
(
√
27
8 ) = 40.51 (1 mark)
(b) What is the area of the triangle to 1 d.p.? (3 marks)
180 − 60 − 40.51 = 79.49 (1 mark)
Area= 1
2 × 4 × 3 × sin(79.49) = 5.9 (2 marks)
2

3. 6. Solve the following quadratic equations using the method stated. No points will be awarded
if another method is used. Leave answers in surd form.
(a) 2x2
+ 10x + 9 = 0. Solve by using the quadratic formula. (4 marks)
a = 2, b = 10, c = 9. 1 mark for a,b,c
x = −10±
√
102−4×2×9
2×2
= −10±
√
28
4
= −5±
√
7
2
1 mark for working, 2 marks for soln
(b) x2
− 9x + 8 = 0. Solve by factorising. (2 marks)
(x − 8)(x − 1) = 0 (1 mark)
x = 8, 1 (1 mark)
(c) x2
− 6x + 2 = 0. Solve by completing the square. (5 marks)
(x − 3)2
+ 2 − 9 = 0 (2 marks)
(x − 3)2
= 7 (1 mark)
x − 3 = ±
√
7 (1 mark)
x = 3 ±
√
7 (1 mark)
7. The zookeepers at London zoo are asked to make sure the elephants are drinking enough water.
The elephants are supposed to drink a mean of 200 litres a day. The zookeepers keep track of the
drinking habits of 5 elephants to make sure the are not drinking too little.
220 180 182 193 160
(a) State the null hypothesis and the alternative hypothesis. (2 marks)
H0 : µ = 200 (1 mark)
H1 : µ < 200 (1 mark)
(b) What is the mean of the test data? (2 marks)
µ = 935
5 = 187
1 mark working, 1 mark soln
(c) What is the sample standard deviation of the test data? (4 marks)
Square the data: 48400 32400 33124 37249 25600
σ2
(X) = 176773−5×1872
5−1 = 482
σ(X) = 21.9545
1 mark for the squares, 1 mark for working, 1 mark for sample var, 1 mark for sample sd
(d) What is the T test statistic? (3 marks)
T = 187−200
21.9545/
√
5
= −1.324
2 marks identifying vars, 1 mark for soln
(e) The study is taken with a 1% level of significance. What is the critical T value? (2
marks)
df = 5 − 1 = 4 1 mark
1%, 1 tailed, C = −3.747 1 mark
0.5 mark if written correct sig. lev. and num. tails but incorrect C
(f) What can we deduce? (2 marks)
−3.747 < −1.324 1 mark
Accept H0. 1 mark
8. (a) Differentiate y = 2x3
+ 3x2
. (2 marks)
dy
dx = 6x2
+ 6x
1 mark per term
3

4. (b) What are the x and y co-ordinates of the stationary points of the graph
y = 2x3
+ 3x2
? (4 marks)
dy
dx = 0
6x2
+ 6x = 0
6x(x + 1) = 0
x = −1, 0 (2 marks)
When x = 0 y = 2 × 03
+ 3 × 02
= 0 (1 mark)
When x = −1 y = 2 × −13
+ 3 × −12
= 1 (1 mark)
(c) What are the natures of these stationary points? (3 marks)
d2
y
dx2 = 12x + 6 (1 mark)
d2
y
dx2 (0) = 6 > 0 minimum (1 mark)
d2
y
dx2 (−1) = −6 < 0 maximum (1 mark)
(d) Sketch the graph y = 2x3
+ 3x2
making sure to label your sketch clearly. (5 marks)
−1.5 −1
1
x
y
1 mark per stat point, 1 mark per intercept, 1 mark for shape
(e) By integrating, find the area under the graph y = 2x3
+ 3x2
between the
values x = 1 and x = 2. (4 marks)
2
1
2x3
+ 3x2
dx = [1
2 x4
+ x3
]2
1
= (1
2 × 24
+ 23
) − (1
2 × 14
+ 13
)
= 16 − 3
2
= 14.5
1 mark per term in integral, 1 mark working, 1 mark soln
9. (a) Show that there is a solution to 100 cos(x) − 20 = 0 for some x between 78 and 79. (3
marks)
Set f(x) = 100 cos(x) − 20
f(78) = 0.79 (1 mark)
f(79) = −0.92 (1 mark)
As f(78) > 0 and f(79) < 0 there is a solution between 78 and 79. (1 mark)
(b) Use trial and improvement to find a solution to 1 d.p. (6 marks)
x f(x) Soln between
78.5 -0.06 78 and 78.5
78.3 0.28 78.3 and 78.5
78.4 0.11 78.4 and 78.5
78.45 0.002 78.45 and 78.5
4

5. Solution is 78.5 to 1 d.p.
3 marks working, 1 mark midpoint, 2 marks soln
10. A rocket is launched with speed v = 3t2
− 10t.
(a) What is the speed of the rocket when t = 10? (1 mark)
v = 3 × 102
− 10 × 10 = 200 (1 mark)
(b) What is the rockets acceleration as a function of time? (3 marks)
a = dv
dt = 6t − 10
1 mark for remembering to differentiate, 1 mark per term
(c) When t = 10 the distance x = 600. Write the distance the rocket travels as a
function of time. (5 marks)
x = vdt
= 3t2
− 10tdt
= t3
− 5t2
+ c
600 = 103
− 5 × 102
+ c
c = 100
x = t3
− 5t2
+ 100
1 mark remembering to integrate, 1 mark per term in integral, 1 mark subs, 1 mark soln
5

6. Formulae
Let X be a list of data of size n.
Mean:
µ(X) =
n
i=1 X[i]
n
Variance
σ2
(X) =
n
i=1(X[i])2
n
− µ2
(X)
Z-statistic
Z =
µ(X) − µ
σ/
√
n
Sample Variance
σ2
(X) =
n
i=1(X[i])2
− nµ2
(X)
n − 1
T-statistic
T =
µ(X) − µ
σ(X)/
√
n
Alternative notation
Mean
¯x =
x
n
Variance
V ar =
x2
n
− ¯x2
Z-statistic
Z =
¯x − A
σ/
√
n
Sample Variance
s2
=
x2
− n¯x2
n − 1
T-statistic
T =
¯x − A
s/
√
n
6

7. Pythagoras’ Theorem
a2
+ b2
= c2
tan(A) =
opp
adj
, cos(A) =
adj
hyp
, sin(A) =
opp
hyp
Sine rule
a
sin(A)
=
b
sin(B)
=
c
sin(C)
Cosine rule
a2
= b2
+ c2
− 2bc cos(A)
Area
Area = 1/2ab sin(C)
Quadratic formula
x =
−b ±
√
b2 − 4ac
2a
Equation of a straight line
y = mx + c
Gradient of a straight line
m =
y2 − y1
x2 − x1
7

8. Tables
Critical Z-values
Sig. Lev. 5% Sig. Lev. 1%
One-tail Two-tail One-tail Two-tail
Probability 0.05 0.025 0.01 0.005
Critical value 1.65 1.96 2.33 2.58
Critical T-values
Sig. Lev. 5% Sig. Lev. 1%
One-tail Two-tail One-tail Two-tail
d.f. 0.05 0.025 0.01 0.005
1 6.314 12.706 31.821 63.656
2 2.920 4.303 6.965 9.925
3 2.353 3.182 4.541 5.841
4 2.132 2.776 3.747 4.604
5 2.015 2.571 3.365 4.032
6 1.943 2.447 3.143 3.707
7 1.895 2.365 2.998 3.499
8 1.860 2.306 2.896 3.355
9 1.833 2.262 2.821 3.250
10 1.812 2.228 2.764 3.169
Critical χ2
value
5% significance 1% Significance
df Probability 0.05 Probability 0.01
1 3.84 6.63
2 5.99 9.21
3 7.81 11.3
4 9.49 13.3
5 11.1 15.1
6 12.6 16.8
7 14.1 18.5
8 15.5 20.1
9 16.9 21.7
10 18.3 23.2
11 19.7 24.7
12 21.0 262.
8