Final Exam Review
Series and Sequences
1.
Consider the infinite geometric series 405 + 270 + 180 +....
(a)
For this series, find the common ratio, giving your answer as a fraction in its simplest form.
(b)
Find the fifteenth term of this series.
(c)
Find the exact value of the sum of the infinite series.
(Total 6 marks)
2.
A sum of $5000 is invested at a compound interest rate of 6.3% per annum.
(a)
Write down an expression for the value of the investment after n full years.
(b)
What will be the value of the investment at the end of five years?
(c)
The value of the investment will exceed $10000 after n full years,
(i)
Write down an inequality to represent this information.
(ii)
Calculate the minimum value of n.
(Total 6 marks)
3.
Let Sn be the sum of the first n terms of the arithmetic series 2 + 4 + 6 + ….
(a)
Find
(i)
S4;
(ii)
S100.
(4)
1 2
 .
Let M = 
0 1
(b)
(i)
Find M2.
(ii)
1 6
 .
Show that M3 = 
0 1
(5)
 1 2n 
n
 , for n  4. The sum Tn is defined by
It may now be assumed that M = 
0 1 
Tn = M1 + M2 + M3 + ... + Mn .
(c)
(i)
Write down M4.
(ii)
Find T4.
(4)
(d)
Using your results from part (a) (ii), find T100.
(3)
(Total 16 marks)
1
4.
(a)
Write down the first three terms of the sequence un = 3n, for n 1.
(1)
(b)
Find
20
(i)
 3n ;
n 1
100
(ii)
 3n .
n  21
(5)
(Total 6 marks)
5.
The first four terms of a sequence are 18, 54, 162, 486.
(a)
Use all four terms to show that this is a geometric sequence.
(2)
(b)
(i)
Find an expression for the nth term of this geometric sequence.
(ii)
If the nth term of the sequence is 1062 882, find the value of n.
(4)
(Total 6 marks)
Logarithms and Exponentials
6.
(a)
(b)
Let logc 3 = p and logc 5 = q. Find an expression in terms of p and q for
(i)
log c 15;
(ii)
log c 25.
Find the value of d if log d 6 =
1
.
2
(Total 6 marks)
7.
(a)
Given that log3 x – log3 (x – 5) = log3 A, express A in terms of x.
(b)
Hence or otherwise, solve the equation log3 x – log3 (x – 5) = 1.
(Total 6 marks)
8.
Let ln a = p, ln b = q. Write the following expressions in terms of p and q.
(a)
ln a3b
(b)
 a

ln 

b


(Total 6 marks)
2
2x
9.
Solve the equation 9x–1 =  1  .
3
(Total 4 marks)
Functions
10.
Part of the graph of the function y = d (x −m)2 + p is given in the diagram below.
The x-intercepts are (1, 0) and (5, 0). The vertex is V(m, 2).
(a)
(b)
Write down the value of
(i)
m;
(ii)
p.
Find d.
(Total 6 marks)
11.
The functions f (x) and g (x) are defined by f (x) = ex and g (x) = ln (1+ 2x).
(a)
Write down f −1(x).
(b)
(i)
Find ( f ◦ g) (x).
(ii)
Find ( f ◦ g)−1 (x).
(Total 6 marks)
12.
The population of a city at the end of 1972 was 250 000. The population increases by 1.3 per year.
(a)
Write down the population at the end of 1973.
(b)
Find the population at the end of 2002.
(Total 6 marks)
3
13.
The following diagram shows part of the graph of f (x) = 5 − x2 with vertex V (0, 5).
h
Its image y = g (x) after a translation with vector   has vertex T (3, 6).
k 
(a)
Write down the value of
(i)
h;
(ii)
k.
(2)
(b)
Write down an expression for g (x).
(2)
(c)
On the same diagram, sketch the graph of y = g (−x).
(2)
(Total 6 marks)
14.
A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of
the year 2000, there were 280 taxis in the city. After n years the number of taxis, T, in the city is given by
T = 280  1.12n.
(a)
(i)
Find the number of taxis in the city at the end of 2005.
(ii)
Find the year in which the number of taxis is double the number of taxis there were at the end
of 2000.
(6)
(b)
At the end of 2000 there were 25 600 people in the city who used taxis. After n years the number of
people, P, in the city who used taxis is given by
P=
2 560 000
.
10  90e – 0.1n
(i)
Find the value of P at the end of 2005, giving your answer to the nearest whole number.
(ii)
After seven complete years, will the value of P be double its value at the end of 2000? Justify
your answer.
(6)
(c)
Let R be the ratio of the number of people using taxis in the city to the number of taxis. The city will
reduce the number of taxis if R  70.
(i)
Find the value of R at the end of 2000.
(ii)
After how many complete years will the city first reduce the number of taxis?
(5)
(Total 17 marks)
4
15.
Let f (x) =
x  4 , x  − 4 and g (x) = x2, x 
(a)
Find (g ◦ f ) (3).
(b)
Find f −1(x).
(c)
Write down the domain of f −1.
.
(Total 6 marks)
Trigonometry and the Unit Circle
16.
Consider y = sin  x    .
9

(a)
The graph of y intersects the x-axis at point A. Find the x-coordinate of A, where 0  x  π.
(b)
Solve the equation sin  x    = – 1 , for 0  x  2.
9
2

(Total 6 marks)
17.
Let ƒ (x) = a sin b (x − c). Part of the graph of ƒ is given below.
Given that a, b and c are positive, find the value of a, of b and of c.
(Total 6 marks)
5
18.
The following diagram shows a sector of a circle of radius r cm, and angle  at the centre. The perimeter of
the sector is 20 cm.
20  2r
.
r
(a)
Show that  =
(b)
The area of the sector is 25 cm2. Find the value of r.
(Total 6 marks)
19.
The diagram below shows a circle of radius 5 cm with centre O. Points A and B are on the circle, and
AÔB is 0.8 radians. The point N is on [OB] such that [AN] is perpendicular to [OB].
A
5 cm
O
0.8
N
B
Find the area of the shaded region.
(Total 6 marks)
20.
If A is an obtuse angle in a triangle and sin A = 5 , calculate the exact value of sin 2A.
13
(Total 4 marks)
21.
(a)
Write the expression 3 sin2 x + 4 cos x in the form a cos2 x + b cos x + c.
(b)
Hence or otherwise, solve the equation
3 sin2 x + 4 cos x – 4 = 0,
0  x  90.
(Total 4 marks)
6
22.
The following diagram shows a triangle ABC, where AĈB is 90, AB = 3, AC = 2 and BÂC is .
(a)
Show that sin  =
5
.
3
(b)
Show that sin 2 =
4 5
.
9
(c)
Find the exact value of cos 2.
(Total 6 marks)
23.
The function f is defined by f : x  30 sin 3x cos 3x, 0  x 
π
.
3
(a)
Write down an expression for f (x) in the form a sin 6x, where a is an integer.
(b)
Solve f (x) = 0, giving your answers in terms of .
(Total 6 marks)
24.
(a)
Consider the equation 4x2 + kx + 1 = 0. For what values of k does this equation have two equal roots?
(3)
Let f be the function f ( ) = 2 cos 2 + 4 cos  + 3, for −360    360.
(b)
Show that this function may be written as f ( ) = 4 cos2  + 4 cos  + 1.
(1)
(c)
Consider the equation f ( ) = 0, for −360    360.
(i)
How many distinct values of cos  satisfy this equation?
(ii)
Find all values of  which satisfy this equation.
(5)
(d)
Given that f ( ) = c is satisfied by only three values of , find the value of c.
(2)
(Total 11 marks)
25.
The diagram below shows triangle PQR. The length of [PQ] is 7 cm, the length of [PR] is 10 cm, and PQ̂R
7
is 75.
(a)
Find PQ̂R.
(3)
(b)
Find the area of triangle PQR.
(3)
(Total 6 marks)
Matrices
26.
3
 
 1 x  1
 and B =  x  .
Let A = 
3 1 4 
2
 
(a)
Find AB.
(b)
 20 
The matrix C =   and 2AB = C. Find the value of x.
 28 
(Total 6 marks)
27.
 2p 3
 and det A = 14, find the possible values of p.
If A = 
 4p p
(Total 4 marks)
28.
 0 2
 .
Let A = 
 2 0
(a)
Find
(i)
A−1;
(ii)
A2.
(4)
 p 2
 .
 0 q
Let B = 
(b)
 2 6
 , find the value of p and of q.
Given that 2A + B = 
 4 3
(3)
(c)
Hence find A−1B.
(2)
(d) Let X be a 2 × 2 matrix such that AX = B. Find X.
(2)
8
29.
(a)
(b)
1  3 1 


Write down the inverse of the matrix A =  2 2  1
1  5 3 


Hence solve the simultaneous equations
x – 3y + z = 1
2x + 2y – z = 2
x – 5y + 3z = 3
(Total 6 marks)
Statistics
30.
The histogram below represents the ages of 270 people in a village.
(a)
Use the histogram to complete the table below.
Age range
Frequency
Mid-interval
value
0  age  20
40
10
20 ≤ age  40
40 ≤ age  60
60 ≤ age  80
80 ≤ age ≤100
(2)
(b)
Hence, calculate an estimate of the mean age.
(4)
(Total 6 marks)
9
31.
The 45 students in a class each recorded the number of whole minutes, x, spent doing experiments on
Monday. The results are x = 2230.
(a)
Find the mean number of minutes the students spent doing experiments on Monday.
Two new students joined the class and reported that they spent 37 minutes and 30 minutes respectively.
(b)
Calculate the new mean including these two students.
(Total 6 marks)
The cumulative frequency graph below shows the heights of 120 girls in a school.
130
120
110
100
Cumulative frequency
32.
90
80
70
60
50
40
30
20
10
0
150 155 160 165 170 175 180 185
Height in centimetres
(a)
(b)
Using the graph
(i)
write down the median;
(ii)
find the interquartile range.
Given that 60 of the girls are taller than a cm, find the value of a.
(Total 6 marks)
10
Probability
33.
A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after the other,
without replacement.
(a)
The first two apples are green. What is the probability that the third apple is red?
(b)
What is the probability that exactly two of the three apples are red?
(Total 6 marks)
34.
Dumisani is a student at IB World College.
The probability that he will be woken by his alarm clock is
7
.
8
If he is woken by his alarm clock the probability he will be late for school is
1
.
4
If he is not woken by his alarm clock the probability he will be late for school is
3
.
5
Let W be the event “Dumisani is woken by his alarm clock”.
Let L be the event “Dumisani is late for school”.
(a)
Copy and complete the tree diagram below.
L
W
L
L
W
L
(4)
(b)
Calculate the probability that Dumisani will be late for school.
(3)
(c)
Given that Dumisani is late for school what is the probability that he was woken by his alarm clock?
(4)
(Total 11 marks)
11
SOLUTIONS
35.
(a)
For taking an appropriate ratio of consecutive terms
r=
(b)
2
3
A1
For attempting to use the formula for the nth term of a GP
u15 = 1.39
(c)
(M1)
For attempting to use infinite sum formula for a GP
S = 1215
N2
(M1)
A1
N2
(M1)
A1
N2
[6]
36.
(a)
5000(1.063)n
A1
1
(b)
Value = $5000(1.063)5 (= $6786.3511...)
= $6790 to 3 sf (Accept $6786, or $6786.35)
A1
1
(i)
5000(1.063)n > 10000 or (1.063)n > 2
A1
1
(ii)
Attempting to solve the inequality «log (1.063) > log 2
n > 11.345...
12 years
Note: Candidates are likely to use TABLE or LIST
on a GDC to find n. A good way of communicating
this is suggested below.
(M1)
(A1)
A1
3
Let y = 1.063x
When x = 11, y = 1.9582, when x = 12, y = 2.0816
x = 12 ie 12 years
(M1)
(A1)
A1
3
(c)
[6]
37.
(a)
(i)
S4 = 20
(ii)
u1 = 2, d = 2
A1
N1
(A1)
Attempting to use formula for Sn
M1
S100 = 10100
A1
N2
12
(b)
(c)
(i)
 1 4

M2 = 
 0 1
A2
(ii)
  1 2  1 4
 
 
For writing M3 as M2  M or M  M2  or 
0
1
0
1


 
M1
 1 0
M3 = 
0 0
A2
4  2

0  1
 1 6

M3 = 
 0 1
AG
N0
(i)
 1 8

M4 = 
 0 1
A1
N1
(ii)
 1 2  1 4  1 6  1 8
  
  
  

T4 = 
 0 1  0 1  0 1  0 1
 4 20 

= 
 0 4
(d)
N2
 1 2  1 4
 1 200 
  
  ...  

T100 = 
1
 0 1  0 1
0
100 10100 

= 
100 
 0
(M1)
A1A1
N3
(M1)
A1A1
N3
[16]
38.
(a)
For taking three ratios of consecutive terms
54 162 486
 3


18 54 162
hence geometric
(M1)
A1
AG
N0
13
(b)
(i)
r=3
(A1)
un = 18  3n  1
(ii)
A1
For a valid attempt to solve 18  3 n  1 = 1062882
N2
(M1)
eg trial and error, logs
n = 11
A1
N2
[6]
39.
(a)
3, 6, 9
(b)
(i)
A1
Evidence of using the sum of an AP
eg
N1
M1
20
2  3  20 1 3
2
20
 3n  630
A1
N1
n 1
(ii)
METHOD 1
100
Correct calculation for
 3n
(A1)
n 1
eg
100
2  3  99  3,15150
2
Evidence of subtraction
(M1)
eg 15150  630
100
 3n 14520
A1
N2
n  21
METHOD 2
Recognising that first term is 63, the number of terms is 80 (A1)(A1)
eg
80
63  300 , 80 126  79  3
2
2
100
 3n 14520
A1
N2
n  21
[6]
14
40.
(a)
(i)
logc 15 = logc 3 + logc 5
=p+q
(ii)
logc 25 = 2 logc 5
= 2q
(b)
(A1)
A1
N2
(A1)
A1
N2
METHOD 1
1
2
d 6
M1
d = 36
A1
N1
METHOD 2
For changing base
eg
M1
log 10 6 1
 , 2 log 10 6  log 10 d
log 10 d 2
d = 36
A1
N1
[6]
41.
(a)
 x 
log3 x  log3 ( x  5)  log3 

 x 5
A
x
x 5
(A1)
(A1) (C2)
Note: If candidates have an incorrect or no answer to part (a) award
(A1)(A0)
 x 
if log 
 seen in part (b).
 x 5
15
(b)
EITHER
 x 
log3 
 1
 x 5
x
 31   3
x5
(M1)(A1)(A1)
x  3 x  15
 2 x  15
x
15
2
(A1) (C4)
OR
 x 
log10 

 x 5 1
log10 3
(M1)(A1)
 x 
log10 
  log10 3
 x 5
(A1)
x  7.5
(A1) (C4)
[6]
42.
(a)
ln a3b = 3ln a + ln b
ln a3b = 3p + q
(b)
ln
a 1
 ln a  ln b
b 2
ln
a 1
 pq
b 2
(A1)(A1)
A1
N3
(A1)(A1)
A1
N3
[6]
16
2x
43.
1
9 =  
3
2x–2
3
= 3–2x
2x – 2 = –2x
1
x= 2
x–1
(M1) (A1)
(A1)
(A1) (C4)
[4]
44.
(a)
(b)
(i)
m=3
A2
N2
(ii)
p=2
A2
N2
Appropriate substitution
M1
eg 0 = d(1  3)2 + 2, 0 = d(5  3)2 + 2, 2 = d(3  1)(3  5)
d 
1
2
A1
N1
[6]
45.
1
(a)
f
(b)
(i)
x   ln x
A1
Attempt to form composite (f ◦ g) (x) = f (ln (1 + 2x))
(f ◦ g) (x) = eln (1 + 2x) = (= 1 + 2x)
(ii)
Simplifying y = eIn(1 + 2x) to y = 1 + 2x (may be seen in part
(i) or later)
Interchanging x and y (may happen any time)
eg
x = 1 + 2y
(f ◦ g)1 (x) =
x 1
2
N1
(M1)
A1
N2
(A1)
M1
x  1 = 2y
A1
N2
[6]
46.
(a)
253250
(accept 253000)
A1
N1
17
(b)
1972  2002 is 30 years, increase of 1.3%  1.013
(A1)(A1)
Evidence of any appropriate approach
Correct substitution 250000  1.01330
368000 (accept 368318)
(M1)
A1
A1
N3
[6]
47.
(a)
(b)
(i)
h=3
A1
N1
(ii)
k=1
A1
N1
A2
N2
g (x) = f (x  3) + 1, 5  (x  3)2 + 1, 6  (x  3)2,  x2 + 6x  3
(c)
y
T
V
–8
0
8
M1A1
Note:
x
N2
Award M1 for attempt to reflect through
y-axis, A1 for vertex at approximately ( 3, 6).
[6]
48.
(a)
(i)
n=5
(A1)
T = 280  1.12
5
T = 493
(ii)
(b)
(i)
evidence of doubling
eg 560
setting up equation
eg 280  1.12n = 560, 1.12n = 2
n = 6.116...
in the year 2007
A1
(A1)
A1
(A1)
A1
2560000
10  90 e 0.15
(A1)
P = 39 635.993...
(A1)
P
P = 39 636
N2
A1
N3
N3
18
(ii)
(c)
(i)
P
2560000
10  90 e 0.17 
P = 46 806.997...
not doubled
valid reason for their answer
eg P < 51200
A1
A1
R1
N0
correct value
A2
N2
e.g.
(ii)
25600
, 91.4 , 640 : 7
280
setting up an inequality (accept an equation, or reversed
inequality)
M1
P
2560000
e.g.  70 ,
 70
T
10  90e 0.1n 280 1.12 n


finding the value 9.31....
after 10 years
(A1)
A1
N2
[17]
49.
(a)
METHOD 1
f (3) =
7
(A1)
(g ◦ f) (3) = 7
A1
N2
METHOD 2
x4
(g ◦ f) (x) =
2
(= x + 4)
(g ◦ f) (3) = 7
(b)
For interchanging x and y (seen anywhere)
Evidence of correct manipulation
eg
x=
(A1)
A1
N2
(M1)
A1
y  4 , x2  y  4
f 1(x) = x2  4
A1
N2
19
(c)
x0
A1
N1
[6]
50.
(a)
when y  0 (may be implied by a sketch)
x
(b)
8π
or 2.79
9
(A1)
(A1) (C2)
METHOD 1
Sketch of appropriate graph(s)
Indicating correct points
x  3.32 or x  5.41
(M1)
(A1)
(A1)(A1)(C2)(C2)
METHOD 2
π
1

sin  x    
9
2

x
π 11π
π 7π
, x 

9 6
9
6
x
7π π
11π π
 , x

6 9
6 9
x
19π
31π
, x
18
18
( x  3.32, x  5.41)
(A1)(A1)
(A1)(A1)(C2)(C2)
[6]
51.
a = 4, b = 2, c =
  3 
 or etc 
2  2

A2A2A2
N6
[6]
52.
(a)
For using perimeter = r + r + arc length
20 = 2r + r

20  2r
r
(M1)
A1
AG
N0
20
(b)
Finding A =

1 2  20  2r 
2
r 
 10r  r
2  r 
For setting up equation in r
Correct simplified equation, or sketch
eg 10r – r2 = 25, r2 – 10r + 25 = 0
r = 5 cm

(A1)
M1
(A1)
A1
N2
[6]
53.
METHOD 1
1
Area sector OAB  (5)2 (0.8)
2
 10
(M1)
(A1)
ON  5cos0.8
  3.483...
(A1)
AN  5sin 0.8
  3.586.....
(A1)
1
Area of  AON  ON  AN
2
 6.249... (cm2 )
Shaded area
(A1)
 10  6.249..
 3.75 (cm2 )
(A1) (C6)
21
METHOD 2
A
O
N
B
F
1
Area sector ABF  (5)2 (1.6)
2
 20
(M1)
(A1)
1
Area OAF  (5)2 sin1.6
2
(M1)
 12.5
(A1)
Twice the shaded area  20  12.5 ( 7.5)
(M1)
1
Shaded area  (7.5)
2
 3.75 (cm2 )
(A1) (C6)
[6]
54.
sin A =
5
12
 cos A = 
13
13
But A is obtuse  cos A = –
(A1)
12
13
sin 2A = 2 sin A cos A
5  12 
  
=2×
13  13 
120
=–
169
(A1)
(M1)
(A1) (C4)
[4]
55.
(a)
3 sin2 x + 4 cos x = 3(1 – cos2 x) + 4cos x
= 3 – 3 cos2 + 4 cos x
(A1) (C1)
22
(b)
3 sin2 x + 4 cos x – 4 = 0 3 – 3 cos2 x + 4 cos x – 4 = 0
 3 cos2 x – 4 cos x + 1 = 0
(3 cos x – 1)(cos x – 1) = 0
1
cos x = 3 or cos x = 1
(A1)
x = 70.5° or x = 0°
(A1)(A1) (C3)
Note: Award (C1) for each correct radian answer, ie x = 1.23 or x = 0.
[4]
56.
Note:
(a)
Throughout this question, do not accept methods which involve
finding  .
Evidence of correct approach
eg sin  =
sin  =
(b)
BC
, BC  32  2 2  5
AB
5
3
AG
Evidence of using sin 2 = 2 sin  cos 
 5  2 
 
= 2
 3 
3


=
(c)
4 5
9
Evidence of using an appropriate formula for cos 2
eg
A1
N0
(M1)
A1
AG
N0
M1
4 5
4
5  80 
 , 2  1, 1  2  , 1  
9 9
9
9  81 
cos 2 = 
1
9
A2
N2
[6]
57.
(a)
Evidence of choosing the double angle formula
f (x) = 15 sin (6x)
(M1)
A1
N2
23
(b)
Evidence of substituting for f (x)
(M1)
eg 15 sin 6x = 0, sin 3x = 0 and cos 3x = 0
6x = 0, , 2
x = 0,
 
,
6 3
A1A1A1
N4
[6]
58.
(a)
METHOD 1
Using the discriminant  = 0
(M1)
k2 = 4  4  1
k = 4, k =  4
A1A1
N3
METHOD 2
Factorizing
(M1)
(2x  1)2
k = 4, k =  4
(b)
Evidence of using cos 2 = 2 cos2   1
A1A1
N3
M1
eg 2(2 cos   1) + 4 cos  + 3
2
f () = 4 cos2  + 4 cos  + 1
AG
N0
24
(c)
(i)
1
A1
(ii)
METHOD 1
Attempting to solve for cos 
cos  = 
N1
M1
1
2
(A1)
 = 240, 120,  240, 120 (correct four values only)
A2
N3
METHOD 2
Sketch of y = 4 cos2  + 4 cos  + 1
M1
y
9
–360
–180
Indicating 4 zeros
 = 240, 120, 240, 120 (correct four values only)
(d)
Using sketch
180
360
x
(A1)
A2
N3
(M1)
c=9
A1
N2
[11]
59.
(a)
choosing sine rule
correct substitution
(M1)
sin R sin 75

7
10
A1
sin R = 0.676148...
PR̂Q = 42.5
A1
N2
25
(b)
P = 180  75  R
P = 62.5
(A1)
substitution into any correct formula
eg area  PQR =
A1
1
 7 10  sin (their P)
2
= 31.0 (cm2)
A1
N2
[6]
60.
(a)
Attempting to multiply matrices
 3
 1 x  1    3  x 2  2    1  x 2  
 


  x   
 

 3 1 4   2   9  x  8   17  x  
 
(b)
Setting up equation
(M1)
A1A1
N3
M1
 1 x 2   20   2  2 x 2   20   1 x 2  10 
    , 
    , 
   
eg 2 





17  x   28   34  2 x   28  17  x  14 
 1  x 2 10 


17  x 14 


2  2 x 2  20
34  2 x  28
x = 3
(A1)
A1
N2
[6]
61.
2p2 + 12p = 14
p2 + 6p – 7 = 0
(p + 7)(p – 1) = 0
p = –7 or p = 1
(M1) (A1)
(A1)
(A1) (C4)
Note: Both answers are required for the final (A1).
[4]
62.
(a)
(b)
(i)

0
A1 = 
1

2
1

2
0 

(ii)
 4 0

A2 = 
 0 4
 0 4  p 2  2 6

  
  

 4 0  0 q   4 3
p = 2, q = 3
(c)
Evidence of attempt to multiply
A2
N2
A2
N2
(M1)
A1A1
N3
(M1)
26

0
1
eg A B = 
1

2
1

2
0 

 2 2


 0 3
3

0

A1B = 
2

1 1 
(d)



 0
 accept

1 p


2

1 
q
2 
1  

A1
Evidence of correct approach
N2
(M1)
eg X = A1B, setting up a system of equations
3

0

X= 
2

1 1 



 0
 accept

1 p


2

1 
q
2 
1  

A1
N2
[11]
63.
(a)
 0.1 0.4 0.1


  0.7 0.2 0.3 
  1.2 0.2 0.8 


(b)
 x  1
   
For recognizing that the equations may be written as A  y    2 
 z   3
   
 x
1
 
 
1
for attempting to calculate  y   A  2 
z
 3
 
 
A2
  1.2  
  
   0.6  
  1.6  
  
x = 1.2, y = 0.6, z = 1.6 (Accept row or column vectors)
3
(M1)
M1
A2
3
[6]
27
64.
(a)
Age range
Frequency
Mid - interval value
0  age < 20
40
10
20  age < 40
70
30
40  age < 60
100
50
60  age < 80
50
70
80  age  100
10
90
A1A1
(b)
For attempting to find
f x
Correct substitution
N2
(M1)
(A1)
eg 40  10 + ... + 10  90 = 11900
For dividing by 270
eg
(M1)
11900
270
Mean = 44.1
A1
N4
[6]
65.
(a)
(b)
(i)
m = 165
(ii)
Lower quartile (1st quarter) = 160
Upper quartile (3rd quarter) = 170
IQR = 10
Recognize the need to use the 40th percentile, or 48th student
eg a horizontal line through (0, 48)
a = 163
A1
N1
(A1)
(A1)
A1
N3
(M1)
A1
N2
[6]
66.
(a)
mean  
x  2230 


n  45 
x  49.6 (Accept 50)
(M1)
(A1) (C2)
28
(b)
y
 y (may be implied)
(M1)
n2
 y  2230  37  30
y
(A1)
2297
47
(A1)
 48.9 (Accept 49)
(A1) (C4)
[6]
67.
(a)
P=
22
(= 0.957 (3 sf))
23
(A2) (C2)
(b)
R
3
23
21
24
R
G
22
25
etc
3
25
G
(M1)
OR
P = P (RRG) + P (RGR) + P (GRR)
22 21 3 22 3 21 3 22 21
 

 

 
25 24 23 25 24 23 25 24 23
693
=
(= 0.301 (3 sf))
2300
(M1)
(M1)(A1)
(A1) (C4)
[6]
29
68.
(a)
7
8
1
4
L
3
4
L'
3
5
L
2
5
L'
W
1
8
W'
(A1)(A1)(A1)(A1)
4
 7 1 3
Note: Award (A1) for the given probabilities  , ,  in the correct
8 4 5
positions, and (A1) for each bold value.
(b)
Probability that Dumisani will be late is
47
(0.294)
160
(A1)
7 1 1 3
  
8 4 8 5
(A1)(A1)
=
(c)
P(W  L)
P(L)
7 1
P(W  L) = 
8 4
47
P(L) =
160
7
P(WL) = 32
47
160
35
=
(= 0.745)
47
(N2)
3
P(WL) =
(A1)
(A1)
(M1)
(A1) (N3)
4
[11]
30