Macmillan Revision Guides for CSEC®Examinations
MATH EMATICS
5tephen Jackson
CSEC®is a registered trade mark of the
Caribbean Examinations Council (CXC).
MACMILLAN REVISION GUIDES FOR CSEC®
EXAMINATIONS: MATHEMATICS is an independent
publication and has not been authorized,
sponsored, or otherwise approved by Cxc.
~
MACM ILLAN
CARIBBEAN
Macmillan Education
Between Towns Road, Oxford OX4 3PP
A division of Macmillan Publishers Limited
Companies and representatives throughout the world
www.macmillan-caribbean.com
ISBN 978 0 333 77672 8
Text © 2002 Step hen ]ackson
Design and illustration © Macmillan Publishers Limited 2002
First published 2002
All rights reserved; no part of this publication may be
reproduced, stored in a retrieval system, transmitted in any
form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission
of the publishers.
DeSigned by]im Weaver
Illustrated by typesetter
Cover design by Gary Fielder at AC Design
Printed and bound in Malaysia
2016 2015 2014 2013 2012 2011 2010
10 9 8 7 6 5 4 3
( Contents
Introduction
A note to Teachers
Acknowledgements
vi
vi
Part 1 - Multiple Choice Paper - Paper 1
1
Topic 1
Topic 2
Topic 3
Topic 4
Topic 5
Topic 6
Topic 7
Topic 8
Topic 9
2
5
9
13
16
22
27
31
36
Sets
Relations, Functions and Graphs
Computation
Number Theory
Measurement
Consumer Arithmetic
Statistics
Algebra
Geometry
Part 2 - Essay and Short Answer
Paper - Paper 2
Section I
Topic 1
Topic 2
Topic 3
v
43
44
Sets
Relations, Functions and Graphs
Consumer Arithmetic and
Computation
Statistics and Probability
Algebra
Measurement
Geometry and Trigonometry
Vectors and Matrices
61
68
81
89
99
113
Section 11
Topic 1
Relations, Functions and Graphs
Topic 2
Geometry and Trigonometry
Topic 3
Vectors and Matrices
117
126
135
Topic 4
Topic 5
Topic 6
Topic 7
Topic 8
51
iii
iv
PART 3 - Practice Tests
143
Paper 1
Paper 2
Paper 1 - Solutions
Paper 2 - Solutions
144
149
154
157
Examination Procedure and Practice
164
Appendix I - Accuracy in Construction
Appendix 11 - Precision in Graph Work
Appendix III - Simplicity in Drawing
166
167
169
Contents
Introduction - An explanation of how to use this book effectively
IMPORTANT - READ THIS PAGE FIRST
In this text the focus is on candidates taking the
CSEC Mathematics examination. You may be in
school, or revising for another attempt out of
school. It will also be helpful to students entered
at the Basic Level or taking '0' levels.
There are already several revision texts on the
market that have been in print for some time
and yet the results each year continue to be
disappointing. As an Examiner for CXC in this
area I am faced each year with the perplexing
question 'Why do so many fail to obtain a
satisfactory grade?'
This book is written because it is my strong belief
that the answer to that question is firmly linked to
the teaching and learning of the higher thinking
skills. It is not enough to know the content and to
'swat' and memorize large chunks of material
from the textbook or from notes. Perhaps that
can work at a lower level in Primary school, for
the Grade Six Achievement Test (GSA1) or a
similar examination, but the CSEC Mathematics
examination requires more than simply the recall
of facts and formulae - it requires considerable
thinking in the examination itself
Because of this, any revision book that simply
supplies questions to be worked out, with
answers in the back, is insufficient except as a
supplement to strong teaching. Even a revision
book which works out the answers to past paper
questions in detail is insufficient, because those
questions will not be on this year's paper - and
this year's questions may well be sufficiently
different that you cannot just apply what has
been memorized blindly! It is the thought
processes that underlie the working out of the
problems that are critical to understand.
• How to decide which path to follow to the
solution.
• How to apply the content that you know
correctly and appropriately.
• How to take the shortest path to that
solution.
• How to recognize the hints and signposts
hidden in the questions that point you in the
right direction.
This book is, therefore, rather different
from others in that, as well as giving questions
to be worked and worked solutions, it
focuses on guiding the reader through the
thinking process that is required each time
you attempt to solve a problem at this
level.
This book will not replace your textbook - of
which there are many good ones on the market.
It is to help you to revise, so please:
1. Read the question - seek to understand what
it is asking.
2. Read and seek to answer any questions posed
under 'To think about'.
3. Close the book and try the problem for
yourself.
Only then, or if you are making no progress:
4. Look at the 'Framework solution'.
In order to make the best use of this book then,
do not run on ahead, rather take the time to
think through along with me the stages reqUired
in the planning and execution of each solution.
It is my fervent hope that in so dOing, my
thinking process will become yours and that by
thinking correctly you will indeed be successful
in your examination, but even more so you will
find Mathematics easier, rewarding and very
enjoyable.
Introduction v
A note to teachers
This book is written out of research done by the
author into 'metacognition'. Cognition is
thinking. Metacognition is 'thinking alongside
of thinking' or 'thinking about one's thinking'.
Some people (even some teachers) believe that
Mathematics is only for a gifted few. But of
what does that special gift consist? I submit it
consists of a developed sense of metacognition.
In some students this develops naturally
(specially gifted). In other students it does not
develop naturally but it can be taught.
The sad thing is that many teachers are not even
aware of what metacognition is, much less are
they actively modelling and teaching it! This
book attempts to teach metacognitive skills
whilst teaching the effective revision of CSEC
Mathematics. You can help your students by
doing the same kind of things as are done in
this book in your classroom at whatever level
you teach (Primary, Secondary or Tertiary).
Metacognitive students:
• continually monitor and check their own
thought processes
• are able to recognize when errors have been
made by doing automatic 'reasonableness'
checks alongSide the actual working of the
question
• are willing and able to backtrack to an
earlier point in the working and recognize
branches in a question's structure
• can explain their own thinking on request
• think on different levels at the same time
• are not distracted from the main plan they
are following towards a solution when they
have to perform some complex
computation using a calculator or pencil
and paper
• usually become the best students in the
Mathematics classroom.
Acknowledgements
I would like to thank:
• Macmillan for spurring me to write this text
• my students at Calabar, Cornwall College,
Priory, Camperdown and elsewhere, who
over the years led me to an understanding of
the priority of teaching metacognition
• my colleagues in the Ministry of Education,
Youth & Culture, Jamaica, who encourage
and inspire me daily
• my wife, Delrose, and children, Stacey and
Mark, who keep me young
• my teacher in grades 4 and 5, the long
departed Mrs Cooper, who taught me to be
vi
Introduction
)
metacognitive, long before the word even
existed
• My Lord and Saviour, Jesus Christ, who
being instrumental in creation gave us
wonderful minds, the ability to think and
reason; who instrumental in our redemption
can remove those burdens of care, confusion
and stress, which cause our thinking to blur,
deviate and stall at the most critical
moments of our lives.
Part
1
MULTIPLE CHOICE PAPER - Paper 1
This objective test requires the candidate to
answer 60 questions in 90 minutes. (90 seconds
per question!)
+
It is worth of the total mark for the
examination and is often overlooked in the
revision process.
It is possible with guided practice to score over
55 out of 60 even if you are not particularly
strong in the subject area.
Such a score would make Paper 2 a much
easier prospect.
For each section the same number of sample
questions as are on an actual paper will be
looked at, identifying the main variations and
critical ideas.
Finally, in Part 3, a complete practice test of 60
questions is given along with the solutions and
reasoning.
1
Topic
1
( Sets
4 questions out of 60)
TYPICAL QUESTION 1
{a}, {b}, {e}, {d} and { }
(the empty set is a subset of every set)
How many subsets does the set {a, b, e, d}
contain?
A.8
B. 14
C. 15
D. 16
Specific objective being tested
Calculate the number of subsets of a set with n
elements.
Count them up: 1 + 4+6 + 4+1 = 16
Answ erD.
Now I rem ember the formula! 'A set of n
elements always has 2n subsets.'
This set has 4 elements and, therefore, has
24 = 2 x 2 x 2 x 2 = 16 subsets.
(Doing it this way saves valuable time.)
To think about
• Do I know a formula for this? If I don't, or
am not sure what the formula is, I can still
do it by listing the subsets - but it will take
much longer than 90 seconds.
• Am I sure that I know what a subset is?
• Perhaps I need to come back to this question
later.
List the elements of {I, 2, 3, 5}n{3, 4,5, 6}.
A. {I, 2, 3,4, 5, 6}
C. {3}
B. {3,5}
D. { }
Specific objective being tested
Framework soLution
Determine and count the elements in the
intersection of not more than three sets.
By counting the subsets
{a, b, e, dJ,
(a set is always a subset of itself!)
To think about
{a, b, e}, {a, b, dJ, {a, e, dJ, {b, e, dJ,
(that's all the subsets with 3 elements in)
(miss out d, then miss out e, then miss out b,
then miss out a)
{a, b}, {a, e}, {a, dJ,
(no more with an 'a' in it)
{b, e}, {b, dJ,
(no more with a Ob' in it)
{e, dJ,
(that's all the subsets with 2 elements)
2
TYPICAL QUESTION 2
Revision Maths
• Do I know the sign on '?
• More important than its name
(intersection), is what it does or what it
stands for.
• It stands for the word 'AND'.
• What I want then is all the elements that are
in the first set AND in the second set.
Framework soLution
1 is in the first set but not the second - I don't
want 1
2 is in the first set but not the second - I don't
want 2
3 is in the first set AND the second - I need 3
5 is in the first set AND the second - I need 5
B is out because the values are not between I
and 6 at all.
A is out because I and 6 are included. (This
would be the right answer if the two signs were
~ instead of < )
Answer C.
4 is in the second set but not the first - I don't
want 4
The answer is {2, 3, 4, 5}.
6 is in the second set but not the first - I don't
want 6
TYPICAL QUESTION 4
So the elements needed are {3, 5}. Answer B.
Which of the following describes the shaded
area?
A. P' n Q
B. P U Q'
c. P' U Q' D. Q
TYPICAL QUESTION 3
.-------------------.u
If X = {x: I <x < 6 and x E Z} then X is equal to:
A. {l, 2, 3,4, 5, 6}
B. {... , -3, -2, -I,O}
C. {2, 3, 4, 5}
D. {lo5, 2.5, 3.5,4.5, 5.5}
Specific objective being tested
List the members of a set from a given
description.
To think about
•
•
•
•
•
•
•
How do I read the mathematical sentence?
If I can't read it, I can't do it!
No guesswork please!!!
What is < <?
What is:?
What is E?
What is Z?
Framework solution
Specific objective being tested
Construct and use Venn diagrams to show
subsets, complements, intersections and unions
of sets and solve problems involving not more
than three sets.
To think about
•
•
•
•
•
What does the symbol n mean?
What does the symb ol U mean?
What does the symbol' mean?
What is the shaded area really showing?
If an element is in the shaded area,
which circle is it inside of and which is it
outside of?
Translating the question into English:
'If the set X is made up of a lot of elements x
such that (:) x lies between ( < <) the values I
and 6 (but is not equal to either I or 6) and x is
an integer (Z) (Le. does not have any fractional
part) then what are the elements of X?'
If the elements have to be integers then D is out
(.5 is a fractional part).
Framework solution
Answer A means 'all things NOT (') in
P AND (n) (or better BUT) in Q. That's it!!
Put your finger on the shaded area. Your finger
is in(side) Q but not in (side) P. The answer
is A. What would the other answers look
like?
(
Topic 1
Sets
3
u
p
AnswerD. Q
Shade all of the inside of Q
Answer B. PUQ'
Everything in P (shade P) OR (U)
NOT (') in Q (shade outside of Q)
u
Answer C. P' U Q'
Everything NOT (') in P (shade everything
except P) OR (U) NOT (') in Q (shade everything
outside of Q)
4
Revision Maths
Q
Of course, in a test once you are sure of the
correct answer, you move on to the next
question. What we just did could be part of the
double checking that is done in the time
between finishing question 60 and the end of
the test.
2
Relations, Functions and Gra hs
Topic
6 questions out of 60
TYPICAL QUESTION 1
RHS = 3(3) - 4 = 9 - 4 = 5 different
e is NOT in the set
Which of the following ordered pairs is not
found in the set {(x,y): 2y = 3x - 4}?
Check D for yourself if you want but e is the
answer.
A. (0, - 2)
B. (2, 1)
C. (3, 3)
D. (4, 4)
Specific objective being tested
Describe a relation as a set of ordered pairs.
To think about
• Do I understand the terms spoken of in the
question?
• What are ordered pairs?
• What does {(x,y): 2y= 3x-4} mean?
• Is this question equivalent to asking which of
the four points (A, B, C, D) is not on the line
2y=3x-4?
• Yes it is! But instead of using graph-language
it is using relation-language!
Framework solution
Test each of the ordered pairs (points) A, B,
e, D in turn.
A (0, - 2) x=O andy=-2
LHS =2( -2) =-4
RHS = 3(0) -4=0-4= -4 same
A is in the set
B(2,1)x=2andy=1
LHS =2(1) = 2
RHS = 3(2) - 4 = 6 - 4 = 2 same
B is in the set
e (3, 3) x= 3 andy= 3
LHS =2(3) =6
Tip
In questions like this one, where trying each
answer in turn is clearly the best method, it is
often qUicker to start from D than A as the
correct answer will more often be placed at
CorD.
TYPICAL QUESTION 2
Ifj: X-7 3x-2, evaluateff(2).
A. 10
B. 6
C. 4
D. 3
Specific objective being tested
Use the functional notations e.g. f: x -7 x 2
or f(x) = x2 as well as y = f(x) for given
domains.
To think about
• Can I rewrite f: x -7 3x - 2 in another
(equation) form?
• Yes, it is the same as f(x) = 3x - 2 I think
I am more comfortable with that form.
• What does evaluate mean?
• It means find the value of, so my answer will
be just a number.
• But what isffi2)?
Topic 2 Relations, Functions and Graphs
5
Framework solution
f f(2) or p(2) which means the same thing, is
talking about the number you get when f
operates on f operating on the number 2.
A function f maps or moves or changes one
number into another. We are concerned with
what happens to the number 2 when fis
applied to it twice.
j(x) = 3x- 2 so thatf(2) = 3(2) - 2 = 6 - 2 = 4
so f(2) = 4 i.e. f changes 2 into 4
into 4 (these say the same thing)
or f maps 2
so f[f(2)] = f[4] = 3(4) - 2 = 12 - 2 = 10
Q 3. How many absences were recorded in
5T during the week shown?
A. 6
B. 12
C. 14
D. 15
Q4. On what day of the week were more
students present than any other day?
A. Monday
B. Tuesday
C. Wednesday D. Thursday
Specific objective being tested
so ff(2) = P(2) = f[f(2)] = 10 i.e.ff changes
2 into 10 or ff maps 2 into 10
Interpret data presented by bar chart, pie chart,
histogram, frequency polygon, as well as on the
rectangular Cartesian plane.
Answer A. In this question there is no way to
To think about
work through the four answers, A, B, C and D,
to identify the correct one without working
through all the steps above.
•
Note
f f does not mean f multiplied by f,
f f reads 'f operating on f operating on .. . '.
This is the most common misconception of
those who do not understand relations.
If the question had asked for fg(2) where f and g
were different functions, the first step would be
to calculate g(2) then find f of whatever you got
for g(2).
TYPICAL QUESTIONS 3 AND 4
6
5
~ 4
How does a bar graph or bar chart present
data?
• Where do I look to find the information
being asked about?
• What information is being presented? - One
particular week. ST. Number absent each day.
• I remember the teacher telling us to use a
ruler horizontally across the page so that we
can measure the heights of the bars
accurately against the vertical scale.
Framework solution
The bar chart gives us information in the
heights of the bars.
The chart is telling us that:
4 students were absent on Monday
2 students were absent on Tuesday
1 student was absent on Wednesday
6 students were absent on Thursday
2 students were absent on Friday
Q3. There are 4+ 2 + 1 + 6 + 2 = 15 absences for
the week. Answer D.
.l:l
~ 3
Q)
.l:l
E2
::>
z
0
M
6
The bar graph shows the number of students
absent in Form 5T of the Bridgetown High
School in a certain week.
Revision Maths
Tu
W
Th
Day of the week
F
Q4. Read it carefully. The answer is NOT D!
'More' and 'taller' bars seem to go together, but
not here. More present students means less
absent students.
The answer is C Wednesday, when the least
number were absent, therefore the most were
present.
Note
You need to review each type of chart listed in
the Specific objective. Remind yourself what it is
that indicates the information on that chart. Ask
yourself what sort of questions can be asked
about a chart of that type. Always read this type
of question very carefully. It may require some
rough working - especially for a pie chart.
y
D.
-3
-3
Specific objective being tested
Find by drawing and/or calculation, the
gradients and intercepts of graphs of linear
functions.
TYPICAL QUESTION 5
Which of the following graphs shows a line
with /gradient 1 and y-intercept 3?
To think about
•
Can I recognize a positive gradient and a
negative gradient?
• Where do I look for the y-intercept?
• Intercept means 'cross the path of like an
intercepting missile that shoots something
down.
• Gradient has to do with slope.
A.
-3
Framework solution
The positive direction of the y-axis is 'up the
page/.
The positive direction of the x-axis is 'to the
right'.
y
B.
x
Gradient is a measure of slope found by
3
vertical shift or Rise
horizontal shift or Run
c.
A line like / moves up as it moves to the right.
Rise and Run are both positive. A positive
divided by a positive gives a positive slope or
gradient.
y
x
3
-3
A line like \ moves down as it moves to the
right. Rise is negative but Run is positive. A
negative divided by a positive gives a negative
slope or gradient.
/ = positive, \ = negative. Remember now?
Topic 2 Relations, Functions and Graphs
7
In the question A and C can be eliminated
immediately as the lines on those graphs have a
negative slope and th e slope we want is 1
(that is + 1).
Specific objective being tested
Both Band D have gradient + 1 because each is
calculated from + 3/ + 3 = + 1 as the rise and
run are both + 3 in each case.
To think about
The y-intercept is where the line crosses
(intercepts) the y-axis. Which one crosses at 3
(that is + 3)? B does. D intercepts at - 3.
AnswerB.
TYPICAL QUESTION 6
W hich of the following mappings is not a
function?
A.
x
y
5
4
2
x
y
o -+--7"""-- '
x
c.
D.
8
Revision Maths
x
What makes a relation between two sets of
numbers into that specific type of relation
that we call a function?
• Not every relation is a function then.
• A relation is only a function in two specific
cases.
• Read the Specific objective again!
• 'Many' means more than one.
Framework solution
In Answer A one element in set x maps to one
and only one element in set y. This is what is
meant by a one-to-one mapping or relation. It is
a function.
x map to the element 1 in set y. Two elements
map to one element, so it is a many-to-one
mapping or relation. It is a function.
Similarly in Answer C, both 1 and 0 in set x map
to 0 in set y. This is again a many-to-one
mapping or relation. It is a function.
+----~
2
•
In Answer B both the elements 1 and -1 in set
3
B.
Define a function as a many-to-on e or a o n eto-one relation.
y
y
But in Answer D we have an example of a oneto-many mapping. 2 in set x maps to both 2 and
5 in set y. It is, therefore, not a function .
A nswer D .
Note
The most typical sort of mapping that describes
a relation that is not a function is one involving
a root, e.g. x ~ .y~ where 4 maps to both + 2
and - 2 at the same time. This is a one-to-many
relation, not a function.
Topic
3
.----~~~--
-
.......... ..........-.-.--------
utation
6 questions out of 60
TYPICAL QUESTION 1
No te
Evaluate 3 x 4 + 8 -:- 2
You can get each of the answers A, C and D by
A. 10
B. 16
C. 18
D. 24
Specific objective being tested
Perform any of the four basic operations with
rational numbers (whole numbers, fractions and
decimals).
To think about
• I see three different signs x, + and -:- .
I don't see any brackets telling me what to
do first. This question looks easy!
• But wait a minute! Do I just work from left
to right or is there any special order to work
these operations?
• BOMDAS. What does it mean again?
Hrackets first, Of, Multiply or Divide
next, Add and Subtract last.
• I'm glad I didn't rush this one, I nearly fell
into the trap of saying Answer A.
Framework solution
using a different (wrong) order of operations.
Always follow the BOMDAS order of
operations.
TYPICAL QUESTION 2
Express the decimal 0.625 as a common
fraction, in its lowest terms.
A 625
B _ 1_
C. 2.
D. 3
.
1000
.
625
8
5
Specific objective being tested
Convert fractions to decimals and vice versa.
To think about
• This is something to do with place value,
tenths hundredths, thousandths.
,
625
• 0.625 is 625 thousandths or 1000 ' Answer A,
but the question asks about 'lowest terms'.
• Can it be cancelled down?
•
_ 1_ is far too small to be the answer, 0.625 is
625
.
prob ably C
more
than 21 (0.5), the answer IS
or D. Let's see.
Framework solution
3 x 4 + 8-:-2
Because the multiplication and division must
be done first, it is as if there were brackets
around the numbers to be multiplied and
divided:
(3 x 4) + (8-:-2)
= 12+4
= 16
Answer B.
1000
625
0.625 x 1000 = 1000
625
5 125
1000 -:- "5 = 200
125 5 25
200 5 40
25 5 5
- -- - 40 5 8
Topic 3 (omputation 9
which cannot be cancelled down any further,
because 5 and 8 have no common factor except
1 (they are co-prime). Answer C.
To think about
•
Note
If you practise converting between decimals and
common fractions (and also percentages), you
will get to know some common equivalent
values such as:
1
10%=0.1=10
1
5
2
40%=0.4= 5
•
3
60%=0.6=-
Framework solution
20%=0.2=-
30%=0.3=~
10
1
50%=0.5=-
2
•
5
4
80%=0.8=-
70%=0.7=-.L
10
90%=0.9=~
10
1
1
12"2%=0.125=8
1
3
37"2%=0.375=8
1
5
1
100%= 1.0=1
1
25%=0.25=4
3
75%=0.75=4
5
62"2% = 0.625 =8
1
•
Estimate. Yes, this would be easy with a
calculator, but how are my estimation skills
without one - I can't take my calculator into
the exam for this paper.
Square root of 200. This is asking for a
number which when multiplied by itself
(squared) gives 200. I know 12 x 12 = 144.
C and 0 look too big. 40 x 40 is like 4 x 4
with two zeros = 1600 and the other values
are bigger still.
A and B look more likely.
7
8
87"2%=0.875=-
Let's try A.
13 x 13 = 169
14x14=196
Is 200 between
169 and 196?
No.
Working
14
x - 14
56
140
196
13
x - 13
39
130
169
Let's try B.
14x14=196
15 x 15=225
Is 200 between
196 and 225?
Yes (close to 196).
~200 is just a little more than 14.
Working
x
15
15
75
150
225
33+%=0.3333 recurring=l
3
66+% = 0.6666 recurring = 2.
3
TYPICAL QUESTION 3
~200 must lie between which of the following
whole numbers?
A. 13 and 14
C. 40 and 50
B. 14 and 15
D. 90 and 110
Specific objective being tested
Estimate the result of a computation and
determine a range in which the exact value
must lie.
10
Revision Maths_
)
AnswerB.
Note
Estimation is a very important mathematical
skill. It is much more than guessing. You have to
have a sound mathematical reason for your
answer.
This question required some long multiplication
as rough work. You must be willing to do rough
work calculations in many multiple choice
questions. The name 'rough work' does not
mean that the working must be untidy.
Nor should it be hurried. It should be neat
and readable. It is rough only in the sense that
it is not part of your presented answer.
TYPICAL QUESTION 4
Express 0.007092 correct to 2 significant
figures.
A. 0.0
B. 0.0070
C. 0.00709
O. 0.0071
Specific objective being tested
Approximate a value to a given number of
significant figures and express any decimal to a
given number of decimal places.
To think about
• What can I remember about significant
figures? [ think it is a way of rounding off a
number.
• Answer A looks strange - but it does have
two figures, the rest all have more.
• I don't like the look of all those zeros at the
front. Are those zeros significant?
• Do those zeros make a difference to the
number? No - it's still zero until you reach
the 7. I think my teacher would say that
the 7 is the first significant figure.
• Yes -leading zeros are not Significant, but if
a zero comes after the 7 then it will be
significant.
Framework soLution
o. 0 0 Z 0 9 2
first
Significant
figure
The first non-zero digit is the
first Significant figure.
So the second significant figure is the number
after the 7, that is the O. Chop the number off
right there.
0.0070[92]
Should the second Significant figure remain a 0
or should it be rounded up?
That depends on the next digit - a 9 - so round
up the 0 to 1, because 0.007092 is nearer to
0.0071 than it is to 0.0070.
0.0071
Answer D.
Note
If you round any number, the rounded value
must resemble the original in size.
If the 9 in the question above had been 0, 1, 2, 3
or 4 the answer would have been kept at 0.0070
(rounded down), but because it actually was 5, 6,
7, 8 or 9 it was rounded up to 0.007l.
0-4 round down.
5-9 round up.
TYPICAL QUESTION 5
Write 215 000 in scientific notation (standard
form).
A. 215 X 10 3
C. 2.15 X 105
B. 215 X 10 - 3
O. 2.15 X 10- 5
Specific objective being tested
Write any rational number in standard form
(scientific notation).
To think about
• Scientific notation is what scientists use to
express very large and very small numbers
easily.
• 215000 is not very large, but it is large.
• It's also called standard form because
everybody agreed on a particular way to
write these numbers - as a standard way for
everyone. What is that form?
• What does that negative sign in Band 0
mean?
• What is 10 3 again? I know 10 x 10 x 10 and
10 - 3 means divide (not multiply) by
10 x 10 x 10.
• Standard form must always be a number
between 1 and 10 multiplied by a power
ofl0.
Framework soLution
A and B begin with a number (215) that is
definitely not between 1 and 10. They are
eliminated immediately.
C TOPic 3 : Computation
11
D is going to be tiny because you divide by
10 x 10 x 10 x 10 x 10.
C looks like the only candidate. Let's see.
2.15 x 105 = 2.15 x 10 x 10 x 10 x 10 x 10 =
(move the numbers past the decimal point 5
times)
ir~'5''5''o'.oO = 215 000
Answer C is correct.
Framework soLution
Note
Let's try sharing $8.
There are two things to remember in a question
like this. Standard form must begin with a
number greater than or equal to 1 but less than
10, and the number must be equal to that given.
Tip
It is usually easier to work from the standard
form back to the number.
104 means multiply by 10 four times.
10 - 4 means divide by 10 four times.
TYPICAL QUESTION 6
If Annique and Basdeo share some money
between them in the ratio 3 : 5, what fraction of
the whole does Annique get?
3
A. 5
5
B. 3
3
C. 8
8
D. 3
Specific objective being tested
Compare two quantities in a given ratio.
To think about
• What fraction of the whole? If the whole is 1,
then it sounds like the answer should be less
12
than 1 - because Basdeo got some of the
money too.
• Ratio? That's about sharing. If I was Annique
that would mean:
3 for me, 5 for you, 3 for me, 5 for you, .. .
• Can I try an actual amount of money? Say
$10. No, that wouldn't share easily into
3 and 5.
Revision Maths
Annique gets $3 and Basdeo gets $5 .
What fraction of the whole does Annique get?
Annique gets $3, the whole is $8.
. . $3
Th e fractlOn
IS $8
C
="83 A nswer.
Note
Try it with $16 or any other multiple of $8 and
it works quite easily to give the same answer
once the fraction is reduced to its lowest terms
(cancelled down).
Tip
Questions in the Computation category are
usually felt to be easy questions.
There is no such thing as an easy question.
There are questions that you can solve and
questions that you can't. There are questions
that you can manage and questions that
manage you.
Even an apparently easy question can be done
incorrectly. Time and thought should be spent
on every question, especially those that look
very easy at first glance.
Topic
4
C~~umber Th~_Q_!y___
4 questions out of 60)
TYPICAL QUESTION 1
In which of the following are the rational
numbers given in ascending order (smallest to
largest)?
A 12..1
3
0.75
-= 3 -;- 4=4 )3.00
4
2.8
.20
.20
.00
. 3' 10' 4
2..= 7-;-10= 0.70
10
c 2..11
. 10' 4' 3
The order is 0.66 then 0.70 then 0.75.
f, then [0' then i. Answer A.
Specific objective being tested
In other words
Order a set of rational numbers.
Note
To think about
A common fraction is a single number, and
must always be treated as such.
From a question like this, you can see why
converting from fractions to decimals is such an
important skill.
•
Smallest to largest - I remember not to look
just at the top number (numerator) or just at
the bottom number (denominator) but at the
entire fraction. If I hadn't remembered that, I
would probably have chosen Answer B,
because both the numerators and the
denominators are in ascending order - and
that would have been wrong.
• I have to do a little long division to find out
how big these numbers really are.
• It's so much easier to compare numbers in
decimal form than it is in fractional form.
Framework solution
2
0.66 ...
- = 2 -;- 3 = 3 )2.00
3
1.8
.20
.18
.02
TYPICAL QUESTION 2
Which of the following integers is composite?
A. 43
B. 53
C. 63
D. 73
Specific objective being tested
Describe positive integers as being prime or
composite.
To think about
• What does 'composite' mean? - composed,
or made up of, several factors .
• So if a number isn't composite, it's prime,
and if it isn't prime it's composite!
(
Topic 4 Number Theory 13
•
A prime number has only two factors: itself
and 1.
• A factor is a number that divides into
another number exactly, leaving no
remainder.
• For example: 6 is composite. It has factors 2
and 3 (as well as itself (6) and 1).
5 is prime. It has no factors
except itself (5) and 1.
Framework soLution
Maybe you can spot the answer to this one
quickly if you know your multiplication tables
well. If not:
Consider A: 43
will 2 divide into it? No. 21 remainder 1.
will 3 divide into it? No. 14 remainder 1.
Will 5 divide into it? No.
8 remainder 3.
Will 7 divide into it? No.
6 remainder 1.
2
Because 7 = 49 is bigger than 43 we can stop
there. Notice we only had to try dividing by
prime numbers 2, 3, 5, 7, .. . because if it won't
divide by 2 it won't divide by 4 or 6 either.
Consider B: 53
Will 2 divide into it? No. 26 remainder 1.
Will 3 divide into it? No. 17 remainder 2.
Will 5 divide into it? No. 10 remainder 3.
7 remainder 4.
Will 7 divide into it? No.
Will 11 divide into it? No. 4 remainder 9.
Because 112 = 121 is bigger than 53 we can stop
there.
Consider C: 63
Will 2 divide into it? No.
Will 3 divide into it? Yes.
31 remainder 1.
21 exactly.
63 = 3 x 21 = 3 x 3 x 7. This is a composite
number. Answer C.
(Check out D if you want - but not in an
exam.)
Specific objective being tested
Compute the HCF or LCM of two or more
positive integers.
To think about
•
HCF - Highest Common Factor - sounds
like something big (Answer D?), but is
usually quite small! It is the biggest factor
common to all the numbers given: the
biggest number that will divide into all of the
numbers in the question.
• LCM - Lowest Common Multiple - sounds
like something small (Answer A/B?), but is
usually quite big! It is the smallest multiple
common to all the numbers given: the
smallest number that all of the numbers
given will divide into.
• I remember bits of about four different ways
to calculate the HCF. I'm mixed up.
(There are several ways to calculate these
values, sometimes you may be confused by
methods taught by different teachers. If the
methods shown below are not the ones you
know best, and your methods work - use
yours, and stick to them.)
Framework soLution
What are the factors of 12?
12
Dividing
2 6
3 3
1
12=2x2 x 3
m
What are the factors of 18?
Dividing
TYPICAL QUESTION 3
3
What is the HCF of 12,18 and 42?
A. 3
14
B. 6
Revision Maths
C. 84
D. 252
t#18
3 9
3
1
18=2 x 3 x 3
What are the factors of 42?
•
Base 10 has 10 digits and when you get 10
you carry 1, so it seems reasonable that base 5
will have 5 digits (0, 1, 2, 3, 4 only) and every
time 5 is made you carry 1.
• I don't see anything in this question about
converting bases, so it looks like I should just
work in base 5 throughout.
2142
Dividing
3 21
7 7
1
42=2 x 3 x 7
Rearranging so that
the numbers in
each column are
the same:
12 = 2 x 2 x 3
18=2
x 3x 3
42 = 2
x3
x7
i
i
The HCF means looking for full columns
(indicated by the arrows).
HCF=2 x 3= 6. AnswerB.
12-;.-6 = 2, 18-;.-6=3, 42-;.-6=7
6 is the largest factor of the three given numbers
12, 18 and 42.
Framework soLution
Answers A and B are impossible. Those
numbers don't exist in base 5, because they
contain the digits 5 and 6 and in base 5 there are
only 5 digits (0, 1, 2, 3, and 4). So that leaves
C and D.
1
+
2 1 1
4
4
°
Note
To find the LCM, the same procedure could
be followed. In the final arrangement of the
factors , the LCM takes a number from every
column, i.e.
LCM = 2 x 2 x 3 x 3 x 7 = 252
(Notice that this was a distractor in the
question.)
252-;.-12=21 , 252-;.-18=14, 252-;.-42=6
252 is the smallest multiple of the three given
n umbers 12, 18 and 42.
4+1=5
carry 1
(1 lot of 5)
remainder
°
°
1
2 1 1
4
+ 4
°
2
+
°
2 1 1
4
4
°
1 + 1 =2
no carry
write down 2
2+4=6
carry 1
(1 lot of 5)
remainder 1
1 1 2 0
Answ erC.
TYPICAL QUESTION 4
Calculate, in base 5,
A. 615five
Note
211 five + 404five
B. I115five C. 1120five
D.1332five
Specific objective being tested
Perform simple operations in any base
(excluding division).
The nu mber of the base always tells you:
(i) how many digits the base uses and
(ii) at what value 1 is carried to the next place.
Base 2 (binary) uses 2 digits (0 and 1 only), if you
get a 2 then carry 1 to the next place.
Base 8 (octal) uses 8 digits (0, 1, 2, 3, 4, 5, 6, 7), if
you get an 8 you carry 1 to the next place.
To think about
•
In base 5 - how many digits are used and
when do you carry I?
(
Topic 4
Number Th eory 15
Topic
5
( Measurement
8 questions out of 60)
----_. _---_.*._. __.._._.
TYPICAL QUESTION 1
AGB
o
The perimeter, in centimetres, of the semicircle
shown, which has radius r cm, is equal to:
A. (2+n)r
C. (2 + 2n)r
B.
n~
2
D. 2nr
• This question is definitely to do with
distance around and not area.
Framework solution
Perimeter = A 0 + OB + arc BA
= r + r + half of circumference of
full circle
1
=r+r+- (2nr)
2
=2r+nr
=(2+n)r Answer A.
TYPICAL QUESTION 2
Specific objective being tested
Calculate the perimeter of a polygon and circle
(circumference), and their combinations.
To think about
• What does 'perimeter' mean?
• Where is the radius on the diagram?
• What formulae do I know that use n?
• '-meter' means distance and 'peri-' means
around. Perimeter is the distance around
the figure. So if I started walking at A,
walked to 0 , then to B, then around the
curve and back to A, I would have walked
the perimeter.
• AO and OB are both equal to the radius of
the semicircle.
• There are two formulae I know that use n.
The circumference (distance around) a whole
circle is 2nr.
The area of a whole circle is n~.
16
Revision Maths
Q
IOm
R
The area of the figure PQRS, in m 2, is
A. 16
B. 20
C. 30
D. 80
Specific objective being tested
Calculate the areas of regions enclosed by
rectangles, triangles, parallelograms,
trapeziums, circles and their combinations.
To think about
• What shape is the figure? Is it a shape that I
know? Is there a formula for its area or can I
split it up into shapes that I know?
• 'Area' is a measure of the surface inside
PQRS. In other words, if I drew the figure on
squared paper it would be the number of
squares inside PQRS.
• I notice that 3 sides are given but not the
fourth, and that two right angles of 90° are at
Q and R - which means that PQ and SR are
parallel.
• SO PQRS is 4-sided (a quadrilateral) with one
pair of sides parallel- it's a trapezium!
use the formula for the area of a triangle:
t x base x perpendicular height
TYPICAL QUESTION 3
Framework soLution
In the same way that many questions involving
the calculation of area can be worked out in
several ways, there is more than one way to
look at this question.
Method 1. I recognize that it is
a trapezium
Area of a trapezium = t(a +b) h
(where a and b are the parallel sides and h is the
perpendicular distance between them)
:. Area=t(4+2) 1O= t(6) 10=3 x 1O=30m2
AnswerC.
Method 2. I split the figure into two
IOm
The figure, not drawn to scale, is a special
wooden metre ruler. How many cubic
centimetres of wood were needed to make it?
A.9
B. 106
C. 300
0 . 450
Specific objective being tested
Calculate the surface area and volume of simple
right prisms and pyramids.
To think about
• What figure is shown? It has the same
(triangular) shape throughout its length.
That means it is a prism with a triangle as its
cross-section.
• 'cubic centimetres' means that the question
is asking about volume of wood in cm 3, in
the figure.
• 'volume of a prism' is the formula I must
recall. If I can't recall that formula, then I'm
in trouble with this one.
Framework soLution
Area = area of rectangle MQRS + area of
triangle MSP
Area = (10 x 2) + (t x 10 x 2)
Area = 20 + 10 = 30 m 2
Volume of a prism = area of cross-section
x perpendicular height
(or length)
=Ah
Answer C.
= area of triangle x length
of1m
Note
Once you split the figure as in Method 2, it is
easy to see that the area of the triangle is exactly
half of the rectangle, although you could just
Note
The length is in the wrong units. Other
dimensions are in cm and the answer is in cm 3
( ' Topic 5
Measurement
17
so this has to be changed to cm also.
1m=100cm.
•
:. Volume of a prism
=
(t x base x perp. height for triangle)
x 100
=
•
(t x 3 x 3) x 100
I estimate the answer to be a few centimetres
on a map. If I get a very small answer or a
very large answer I may have made an
arithmetic mistake (multiply instead of
divide or vice versa).
Answers Band C look more reasonable.
Framework soLution
=4.5 x 100=450cm
3
AnswerD.
Tip
Distance on ground = 12 km
= 12 x 1000 x 100cm
=1200000cm
Learn the formulae for the areas and volumes of
the following shapes today.
Remember
1km=1000m
1 m=lOOcm
Area of: square, rectangle, triangle,
parallelogram, trapezium, circle.
Volume and surface area of: prism (including
cylinder), pyramid (including cone) and sphere.
Distance on map
1
50000 x 1 200 000 cm
= 24 cm
(Check: Yes, the size of this answer is reasonable
for a distance on a map.)
TYPICAL QUESTION 4
The distance from Port of Spain to Tarisa Village
is actually 12 km. On a map drawn to a scale of
1 : 50 000, how many centimetres apart would
these two places be?
A. ~
25
B. 24
C. 60
D. 600000
Specific objectives being tested
Convert units oflength, area, volume, capacity,
time and speed within the SI system.
Make suitable measurements on maps or scale
drawings and use them to determine distances
and areas and vice versa.
18
Answer B.
Tip
Keep the two calculations separate in your
thinking: (i) convert km to cm, (ii) move to the
distance on the map. Problems usually occur
when an attempt is made to combine the two
steps into one.
TYPICAL QUESTION 5
A packing case contains 200 small boxes each
measuring 10 cm x 20 cm x 30 cm with no space
remaining. The volume of the packing case, in
cubic metres, is:
A. 1.2
B. 1.8
C. 3.6
D. 30
To think about
Specific objective being tested
• The actual distance is in kilometres (km), the
map distance is in centimetres (cm).
• How many centimetres are there in one
kilometre?
• The scale is a reduction in length of every
dimension. This question involves length, so
the reduction factor is simply 5~OO'
Convert units of length, area, volume, capacity,
time and speed within the SI system.
Revision Maths
To think about
•
To find the volume of the packing case,
I must first find the volume of each small
box.
•
•
Units are a problem here. The answer must
be in m 3, but the dimensions given are all in
cm. Should I convert first or at the end?
A cubic metre is quite a large measurement
(1 m x 1 m x 1 m). I would expect the answer
to be no more than 3 or 4, or it couldn't be
lifted easily.
Framework solution
Method 1. Converting units first
Volume of one small box
= length x width x height
= 10 x 20 x 30 (all in cm)
=0.1 x 0.2 x 0.3 (all in m)
= 0.006 m 3
(1 x 2 x 3 = 6 and three places of decimals)
Remember
at the end. If you do it at the end, remember
that
100cm=lm
but 1002 cm 2 = I m 2
and 100 3 cm 3 = I m 3
TYPICAL QUESTION 6
A boy on his bicycle can cycle 100 m downhill
in exactly 10 seconds. How fast is this in
kilometres per hour?
A. 10
B. 24
C. 36
D. 1000
Specific objective being tested
Convert units of length, area, volume, capacity,
time and speed within the SI system.
To think about
100cm=1 m
•
... Volume of packing case = 200 x 0.006
=2 xO.6
=1.2m 3
Answer A.
Method 2. Converting units at the end
Volume of one small box
= length x width x height
=10 x 20 x 30 = 6000 cm 3
.. . Volume of packing case = 200 x 6000
= 1200000cm 3
1200000
100 x 100 x 100
= 1.2 m 3 Answer A.
Remember
100cm=1 m
:.100 3 cm 3 = 1 m 3
i.e. 100 x 100 x 100 cm 3 = 1 m 3
Yes, this answer is a reasonable size for a
packing case.
Note
Some people are more comfortable
converting the units first, others prefer to do it
This is another question on converting units.
100 metres in 10 seconds, but answer in
kilometres per hour.
• 'how fast' - this is a question about speed metres/second, kilometres/hour - I need to
recall a formula to do with speed.
• 100 m - I know how far that is (100 m sprint,
the straight part of a running track). In
10 seconds is quite fast. I would estimate the
answer to be between 10 and 50km/h
(B and C look more likely) .
• 0 is incredibly fast, 1000 km/h is an
aeroplane speed, not a bicycle speed.
Framework solution
Should we convert the units at the start or
the end?
Method 1. At the end
distance travelled
time taken
Speed (strictly
average speed)
100 (m)
10 (s)
= 10m/s
Converting units
1000m=lkm
(
Topic 5
Measurement
19
TYPICAL QUESTION 7
Speed in km/s (no. of kilometres travelled
in a second must be much less than no. of
metres travelled in a second, so divide by
1000) = 10/1000 km/s
60 x 60s=lh
Speed in km/h (no. of kilometres travelled
in an hour must be much more than
no. of kilometres travelled in a second, so
multiply by 60 x 60) = (l~~O) X 60 x 60 km/h
= 36 km/h Answer C.
On the pair of scales shown, what additional
mass should be placed in which pan in order to
exactly balance the scales?
Method 2. At the start
Distance travelled = 100 m
A. 4 g in pan P
c. 400g in pan P
= 100 km
1000
Specific objective being tested
=_1 km
10
Use correctly the SI units of measure for area,
volume, mass, temperature and time (including
the 24-hour clock).
Time taken = 10 s
10
h
(60 x 60)
To think about
=~h
3600
1
=-h
360
Speed
distance travelled
time taken
360 360
1
1
1
=--;--=- x - = - 10 360 10
1
10
= 36km/h
(
Answer C.
Reasonable answer? Yes.
Tips
1. Always have an estimate in mind.
2. Always pause and think before multiplying
or dividing by a conversion factor.
• If the unit you are going into is a smaller
unit, you will need more of them than of
the larger unit in order to have the same
quantity.
• If the unit you are going into is a larger
unit, you will need less of them than of
the smaller unit in order to have the same
quantity.
20
Revision Maths
B. 4 g in pan Q
D. 400g in pan Q
• Which pan is heavier and which is lighter?
Which needs more mass to balance?
• How many grams make a kilogram?
• 2.4 kg. What is .4 of a kilogram?
• I think this is actually quite an easy question,
because I can see which pan is lighter and I
know .4 kg (approximate mass of a book) is
not equal to 4 g (approximate mass of a
small pencil).
Framework solution
Converting both sides to grams.
PanP
2.4 x 1000g
=2400g
PanQ
2000g
Remember
1000g=lkg
... Pan P is heavier.
Pan Q needs 2400 - 2000 = 400 g more in it to
balance. Answer D.
Tip
In a question like this one, go through the
formal calculation. Don't guess. Don't jump to
an answer too quickly. This is especially true for
temperature, time, mass and capacity questions.
TYPICAL QUESTION 8
New Town
Content Bridge
Bus Timetable no. 17
Boggy Marsh
New Town
West Terminal
Top Hill
Content Bridge
0800
0810
0825
0842
0905
0830
0840
0855
0912
0935
0900
0910
0925
0942
1005
An extract from the number 17 bus timetable is
shown. If you catch a later bus at New Town at
13:25, and traffic conditions are similar, at what
time would you expect to arrive at Content
Bridge?
A. 14:30
B. 14:20
C. 14:05
D. 13:40
Specific objective being tested
Solve simple problems involving time, distance
and speed (e.g. timetable extracts such as bus
and airline schedules).
0810
0905
0840
0935
0910
1005
• When would I expect to arrive? Well, if the
bus is taking the same route and the traffic is
no lighter or heavier, it should take the same
length of time for each journey.
Framework solution
The 0800 bus takes from 8:10 until 9:05 from
New Town to Content Bridge which is
55 minutes (1 hour more, S minutes less) .
The other buses also take 55 minutes.
A bus at New Town at 13:25 should also take
55 minutes
13
25
+1
- 5
14
20
= 14:20
Answer B.
To think about
• There are five places on the bus timetable
but the question mentions only New Town
and Content Bridge, so I must focus on these
two lines.
(
Topic 5
Measurement
21
Topic
6
((9 nsu mer Arith m~tif __________________~~
______ 8 questions out of 60)
TYPICAL QUESTION 1
Tip
Merlene bought a radio in Trinidad for the
equivalent of G$22 000 and sold it to her friend
in Guyana for G$27 500. What percentage profit
did she make?
A. 20
B. 25
C. 35
O. 5500
Do not try to work out a question like this
mentally! The numbers are far too awkward for
most students to work in their heads. Pencil and
paper are needed. Use the paper given to you for
such calculations. If none was given, use the
back of the question paper or its margins, but
working is needed to get it right.
Specific objective being tested
Calculate profit or loss as a percentage.
TYPICAL QUESTION 2
To think about
• This question is about profit, but percentage
profit, not actual profit in dollars.
• 100% profit would double the dollar
amount, so 5500% would be immense, that
looks like the profit in dollars not percentage
profit (0 is out).
• I know that there is a simple formula for
calculating percentage profit.
• This is another question where I need to
conSCiously slow down and work very
carefully - I shouldn't get it wrong!
Framework soLution
Actual profit (in G$) = 27 500 - 22 000
=5500
Percentage profit actual profit x 100%
original price
5500 x 100%
22000
550
22
= 25% Answer B.
22
Revision Maths
Item
-Coffee
Flour
Bread
Sugar
Cost
$6.80
$2.20
$1.90
$3.00
Tax status
+VAT at 10%
Zero rated
+ VAT at 10%
Zero rated
How much sales tax (VAT), in dollars, should be
added to the bill shown?
A. $0.77
C. $1.39
B. $0.87
O. $13.90
Specific objective being tested
Calculate discount, sales tax, profit or loss
when these are given as a percentage.
To think about
•
VAT is Value Added Tax - it has different
names in different countries.
• Some things are taxed and others are not.
'Zero rated' means that this item has no tax
on it - it is taxed at 0%.
• The tax on some of the things in this
question is 10%. 10% is an easy percentage, it
is just one tenth.
• I think I read this question carefully enough,
but let me just read it one more time to make
sure I know what I am being asked to find.
Framework solution
There is tax on Coffee and Bread only.
Tax on Coffee = 10% of $6.80
1
= - x 6.80 = $0.68
10
Tax on Bread = 10% of $1.90
1
= - x 1.90=$0.19
10
1
0.68
+ 0.19
$0.87
To think about
• This question looks long. It has a lot of
reading.
• Perhaps it would be good to list what is
given in a more mathematical way.
• Discount means a lowering of a price due to
a sale, bulk buying or for any other reason.
• 20% is twenty hundredths (l~O) or (one
fifth).
• One dozen is 12 apples in the bag.
+
Framework solution
(a) Given. 1 apple costs $6.50.
bag. 20% discount.
12 apples in
To find. The total cost of the bag of apples.
Answer B.
Hint
Always take care to the very end of the question.
Notice that poor addition could easily have
given Answer A, even after the question had
been properly understood throughout.
TYPICAL QUESTION 3
(a) Apples were being sold for $6.50 each.
Keisha bought a bag containing one dozen
apples. Because she bought so many, she
was given a 20% discount on each one. How
much did she pay for the bag of apples?
A. $5.20
C $62.40
B. $58.00
D. $93.60
(b) Keisha bought a melon, on sale at 20% off,
for $62.40. What was its original price?
A. $49.92
C $12.48
B. $78.00
D. $93.60
Specific objectives being tested
Should we calculate the discount at the start
or the end?
Method 1. At the start
1 apple (before discount) costs $6.50.
After discount it costs 100 - 20 = 80% of what it
cost before.
80
4
1 apple costs x 6.50 = - x 6.5
100
5
=4 x 1.3 =$5.20
:. Bag of apples costs 12 x 5.2 = $62.40
AnswerC
Method 2. At the end
1 apple (before discount) costs $6.50.
So 12 apples cost 12 x $6.50=$78.00
After discount they cost 100 - 20 = 80% of what
they cost before.
80
4
Bag costs- x 78=- x 78=4 x 15.6=$62.40
100
5
AnswerC.
Calculate marked price when cost price and
percentage profit, loss or discount are given.
(b) Many students would calculate 20% of
$62.40, but that is incorrect.
Calculate cost price when selling price and
percentage profit, loss or discount are given.
Let original price be $x.
New price = 100 - 20 = 80% of x.
(
Topic 6
Consumer Arithmetic
23
4
:. - X= 62.4
5
312
:. 4x=5 (62.4)=312 :. x =-=$78.00
4
80
:. -ofx=62.4
100
AnswerB.
Note
(a) and (b) are actually the reverse of each other;
look at the numbers carefully.
TYPICAL QUESTION 4
Mr Walcott's water bill each month consists of a
'flat rate' meter rental payment of $20 and a
consumption payment of $2.50 per unit used.
What would be his bill in a month when he uses
22 units?
A. $44.50
C. $75.00
B. $55.00
D. $495.00
Specific objective being tested
Solve problems involving (a) rates and taxes,
(b) utility bills, (c) invoices and shopping bills,
(d) insurance, (e) salaries and wages.
The only answer close to 80 is C. This could save
valuable time in an exam, especially where the
exact values are very awkward.
TYPICAL QUESTION 5
Mrs Brown invested $280 at 15% simple interest
3 years ago. What is the total value of her
investment today?
B. $398
A. $406
Solve problems involving simple interest,
compound interest and depreciation.
•
On a bill, what is a 'flat rate' payment? What
is consumption?
• Flat rate is for meter rental- it has nothing
to do with how much water is used. Indeed,
you would have to pay it if you used no
water at all. It is a single payment.
• 'Consumption' is the number of units used
or consumed. This payment is per unit, paid
once for every unit the customer uses.
•
Framework soLution
Framework soLution
In a question like this, an estimate may be
sufficient to identify the right answer, without
Revision Maths
')
D. $126
Use the simple interest formula (or otherwise)
to calculate simple interest, principal, time, rate
or amount.
To think about
Note
C. $298
Specific objectives being tested
To think about
Mr Walcott's bill = meter rental + (consumption
charge x no. of units used)
= 20 + (2.50 x 22)
=20+55
=$75.00 Answer C.
24
doing an exact calculation. $2.50 is about $3,
22 units is about 20 units (move one up and one
down for greater accuracy) . Bill is about
20+ 3 x 20=80.
'simple interest' are the key words in this
question.
• The formula is 1= ~~ . I may have to
transpose it, but that is the basic form.
• Let me read the question part again. It's not
just asking for interest gained is it?
• 'total value' - I think that's what my teacher
called 'Amount', the Principal with the
Interest added on.
I (Interest)
p (Principal) x R (Rate per annum) x T(Time in years)
100
280 x 15 x 3 14 x 15 x3
:. I
5
100
= 14 x 3 X 3 = $126 (Note: This is Answer D)
: . Amount ofInterest=P+I =280+ 126 =$406
Answer A.
Tip
Most people memorize 1= ~~~. Many do not
know exactly what each letter stands for. Many
also do not know A(Amount) = P + 1.
A = 10 000 (1 + (-10))3 (-10 because value is
100
being lost)
A=10000(~)3
=10000(~
X ~ X ~)
100
100 100 100
A = 10 x 9 x 9 x 9 = $7290
Answer B.
TYPICAL QUESTION 6
Note
A car was purchased for $10 000 in 1994. Its
value depreciates by 10% each year. What was it
worth in 1997 after three years of
(compounded) depreciation?
A. $7000
B. $7290
C. $7410
Compound interest or depreciation is limited to
three periods so there is really no need to learn
or apply this formula if you are happy with
Method 1.
D. $9970
TYPICAL QUESTION 7
Specific objectives being tested
Calculate compound interest, depreciation and
amount (for not more than three periods).
Solve problems involving simple interest,
compound interest and depreciation.
Casey McDonald went on holiday to Barbados.
He took with him £800 in travellers cheques.
If the exchange rate was £1 = B$3.2, how many
Barbadian dollars did he receive?
A. 800 x 3.2
To think about
• The word in brackets is helpful. I was
wondering whether this was simple or
compound interest. Yes, depreciation is one
of those things that is always compounded.
• I know there is a formula, but for only three
periods it should be easy to calculate it step
by step.
Framework solution
Method 1. No formula
1994 Car is worth $10000
1995 Car worth 100 - 10 = 90% of $10 000
=~ x 10000=$9000
100
1996 Car worth 100 -10 = 90% of $9000
=~x 9000=$8100
100
1997 Car worth 100 -10 = 90% of $8100
90
= - x 8100=$7290 AnswerB.
100
Method 2. Using formula
r
A=P(l+-)n
100
C. 800+ 3.2
B. 800
3.2
D. 800-3.2
Specific objective being tested
Solve problems involving measures and money
(including exchange rates).
To think about
• This question looks easy because there is
nothing to calculate.
• The four answers have the four main
operations x, -:--, + and -.
• English pounds and Barbadian dollars are
involved. Which is stronger?
• Whichever is stronger, less of those make
more of the weaker one.
• Answers C and 0 don't look sensible. When
exchanging between currencies, you always
have to multiply or divide by an exchange
rate. That is what the word rate means.
Framework solution
If £1 = B$3.2, the £ is stronger than the B$.
CTopic 6 ::Consumer Arithmetic 25
£1 can buy as much as would cost B$3.2.
To think about
Because the B$ is weaker, it will take more B$ to
equal any amount of £'s.
•
•
:. £800 = B$800 x 3.2 (B$3.2 for each and
every one of the £'s) Answer A.
•
Hint
When dealing with currencies with which you
are unfamiliar, take a moment to establish
which is stronger, and whether more or less of
the second currency will, therefore, be needed.
Framework soLution
Pounds
Barbadian dollars
The sizes of the bars are 50, 100 and 250 g.
3.2
6.4 ( = 2 x 3.2)
N x 3.2 ( = 3.2N)
800 x 3.2
All are multiples of 50 g.
1
2
N
800
---j
---j
---j
---j
•
I. 50 g bar costs $2.60 per 50 g
11. 100 g bar costs ($5;00) per 50 g
= $2.50 per 50 g
(100= 50 x 2)
TYPICAL QUESTION 8
Ill. 250 g bar costs ( $1~.80) per 50 g
= $2.56 per 50 g
(250=50 x 5)
'Milko' milk chocolate is sold in three sizes
1. 50 g for $2.60
11. 100 g for $5.00
Ill. 250 g for $12.80
Which of them is the 'best buy'? (Best value for
money?)
A.I
B. 11
CIIl
The best buy (lowest rate per 50 g) is the 100 g
bar (bar 11). Answer B.
Note
D. all the same
(
Specific objective being tested
Solve problems involving measures and money
(including exchange rates).
26
Each answer gives the mass of the chocolate
bar and its price.
The best buy is a sort of combination of
these.
I need to be able to compare like with like, to
reduce the pieces of chocolate to the same
size.
Answer 0 is probably unlikely, but you
never know!
Revision Maths
You could also find the cost per 1 g but the
divisions required would be a little harder, and it
is better not to deal with very small numbers.
Topic
7
( Statistics
6 questions out of 60)
TYPICAL QUESTION 1
• Let me make an estimate. Those going by
car is a little more than half of 48, that is a
little more than 24.
• Answer A is too small. Answer D is too big.
Framework solution
There are 360° in a full circle.
Bus is represented by 90°.
Walk is represented by 45°.
Car
:. Car is represented by 360° - 90° - 45° = 225°
~~;!:~ ~ing by car is ~~~ x 48 = 30 children.
In a class of 48 children, some go to school by
car, others by bus and others walk. From the pie
chart shown, the number who go to school by
car is:
A.3
B. 30
C. 35
D. 225
Specific objective being tested
Draw and use pie charts, bar charts,
line graphs, histograms and frequency
polygons.
To think about
• A pie chart is a circle (sometimes called a
circle graph).
• The number of degrees in a whole circle
is 360.
• I know the symbol in the 'bus' sector means
90 degrees.
• Read the question again. There are
48 children in the whole class
(in the whole circle).
[Alternatively, and probably quicker,
Bus: 90° is ~ of 360°, ~ of 48 = 12
Walk: 45° is ~ of360°, ~ of 48 = 6
So,
Car is 48 - 12 - 6 = 30.
Answer B.]
TYPICAL QUESTION 2
~~~:~ :~ -tl=-
:-:::=_:-)( =
_ :-:::=
_ :-)/=
'----:- =
7
_
:- ;:):_;~z='_
-::::=_-::::
¥=_-::::=
_ ::::
....
: =_-::::=_
~~I--~I~--~I--~~
--~~---1991
1992
1993
1994
1995
Year
The line graph represents sales figures for the
Betta Shoe Company. By how much did sales
increase between 1992 and 1994?
A. $150
C. $150000
B. $200
D. $200000
CTopic 7 Statistics 27
Specific objective being tested
Draw and use pie charts, bar charts, line
graphs, histograms and frequency polygons.
To think about
• A line graph is a simple graph. Each 'x' marks
a piece of information - in this case an
amount of sales (in thousands of dollars) in a
given year.
• Sales were steady between 1991 and 1992,
rose in 1993, rose again in 1994 and dropped
back a little in 1995. (Perhaps they sold a lot
of football boots in 1994 which was a World
Cup year!)
• Read the question again - what exactly is it
asking for? The sales increase 1992 to 1994.
Framework soLution
•
The mode is the most frequently occuring
number.
• The median (6 letters) is the middle
(6 letters) number when put in order.
• The mean is just the regular average - add
them up and divide by how many there are.
Just looking at the numbers, 3 looks low and
9 looks high. Its B or C.
Framework soLution
Mean
sum of values
number of values
3+5+8+3+7+12+4
7
= 42 =6 AnswerC.
7
[If the question had asked for mode instead:
Reading the graph up from the year to the 'x',
then across to the vertical scale.
Mode = most frequent number = 3 (Answer A)
which occurs twice - more than any other
number.
Sales in 1992 were 50.
If the question had asked for median instead:
Sales in 1994 were 200.
Rise in sales was 200 - 50 = 150.
First rearrange the numbers into ascending
order.
3, 3, 4, 5, 7, 8, 12
But, the numbers on this scale represent
thousands of dollars.
Median = middle number = 5 (Answer B)
:. Rise in sales was $150000. Answer C.
Answer D is the range! ]
TYPICAL QUESTION 4
TYPICAL QUESTION 3
Calculate the mean of the following numbers:
What range is represented by the following
values?
3, 5, 8, 3, 7, 12, 4
A. 3
B. 5
C. 6
1.5, 2.5, 2.5, 4, 1, 2.5, 1
D.9
Specific objective being tested
Determine mean, median and mode for a set
of data.
A. 0.5
B. 2
C. 2.5
D. 3
Specific objective being tested
Determine the range, interquartile and
semi-interquartile ranges for a set of data.
To think about
•
28
Can I remember which average is the
mean?
Revision Maths
To think about
•
What is range? It is a measure of spread.
•
I remember the teacher arranging the
numbers for range just like for median, then
saying it wasn't really necessary.
• The range of values is 'the biggest value
minus the smallest value'.
• I'm tempted by the 2.5 (Answer C) because
there are a lot of 2,S's in the question.
Framework solution
Arrange the numbers into ascending order if
you wish:
1, 1, 1.5, 2.5, 2.5, 2.5, 4
All you have to do is to pick out the biggest and
smallest values.
Specific objective being tested
Determine experimental and theoretical
probabilities of simple events.
To think about
•
What is a 'die' like? A cube with the numbers
1 to 6 on its faces.
• If you throw it once there are six possible
scores. For the second throw there are six
possible scores again. (That's where the 36's
in the answers come from I think
[36 = 6 x 6]).
• How can a sum of 3 be thrown? I think
that's the key - how to get 3 in two
throws.
Biggest value = 4
Framework solution
Smallest value = 1
Range=4-1 = 3
The first throw could only be a 1 or a 2.
Anything bigger and nothing would be left for
the second throw.
Answer D.
Note
x
For the same 7 values
1, 1, 1.5, 2.5, 2.5, 2.5, 4 with n = 7
The upper quartile (Q3) would be the _3-,-(n---,+_1..L)
3(7 + 1)
4
value =
value = 6th value = 2.5
4
The lower quartile (QI) would be the (n + 1)
(7 + 1)
4
value = - - value = 2nd value = 1
4
The interquartile range = (Q3 - QI)
=2.5-1 = 1.5
The semi-interquartile range =
(Q3 -QI) _ (2.5 - 1)
2
3
36
2
3
Throw 2
1
2
x
3
x
4
x
x
5
6
x
x
x
4
5
6
If the first throw was 1, the second would have
to be 2.
If the first throw was 2, the second would have
to be 1.
There are only two ways to get a total of 3: the
ordered pairs (1, 2) and (2, 1).
This out of 36 possible combinations (6 x 6).
Extension
What is the probability that if a die is thrown
twice, the sum of the numbers obtained is 3?
B.
1
2
36
AnswerA.
2
TYPICAL QUESTION 5
A. ~
36
Throw 1
C.
4
36
D. ~
36
3
Sum of 4 => (3, 1), (2, 2), (1, 3) => 36
4
Sum of 5 => (4, 1), (3, 2), (2, 3), (1,4) => 36
1
Sum of 12 => (6, 6) =>36
Topic 7 Statistics
29
Framework solution
TYPICAL QUESTION 6
There are 6 balls in a bag. 3 are black, 2 are red
and 1 is green. What is the probability of
choosing two balls (without replacement) and
finding that they are both red?
A. 2 x1.. B.2+1.. c. 2 +1..
666665
D.2 x1..
65
Specific objective being tested
Determine experimental and theoretical
probabilities of simple events.
To think about
®®®®®©
• Can I model it?
• Pick out one, then pick out another.
• 'No replacement' means that I don't put back
the first one before the second pick.
• These are separate events, not 2 outcomes of
a Single event, like 'a ball being red or black'.
The sign must be x and not + .You want a
Red AND a Red in 2 events ( x ) not a Red
OR a Black in 1 event ( + ).
30
Revision Maths
® ®
®
1
2
®
®
©
1
2
3
4
5
6
Pick one ball. Chance of it being Red is 2
(number of Reds) out of 6 (number of balls)
2
=6"'
One Red has now gone.
® @ ®
1
2
3
1
®
4
©
5
Pick second ball. Chance of it being Red
is 1 (number of Reds left) out of 5
1
(number of balls left) =-.
5
Probability both are Red
AnswerD.
=26 x ~5 (= 230 =~)
15
Note
If the word 'without' were changed to 'with'
(replacement), the answer would be
2 2
- x6 6
Topic
8
9 questions ou( of 60 )
[ab and ba are the same and cancel, one being
TYPICAL QUESTION 1
negative and one being positive]
= - ae + be
Expand and simplify the following:
.--R-e-m-e-m-b-e-r----,
a(b - c) - b(a -c)
A. be-ae
B. -2abe
- -gives a+
C. 2ab-ae-be
D. 0
Specific objectives being tested
Perform the four basic operations with algebraic
expressions.
Apply the distributive law to insert or remove
brackets in algebraic expressions, e.g.
ax± bx = (a ±b)x.
To think about
•
•
•
•
•
It's algebra - I can't do it! -let me just guess
one and move on. Help!
Alright, let me calm down - expand first,
then simplify. I know that the 'a' outside the
bracket means to multiply through the
bracket by 'a'.
There are a lot of minuses about. I need to
keep an eye on them.
I definitely need some paper to write down a
few lines of working for this one.
If all else fails I could substitute some 'easy'
numbers for a, band e in the given expression
and in each answer to see which agree.
Framework solution
Method 1. Main method
a(b - e) - b(a - e)
= a x (b - e) - b x (a - e)
=ab-ae-ba - -be
= be - ae Answer A.
Method 2. By substituting 'easy' values
Let a = 0, b = 1, e = 1 (or any other values)
Given expression =a(b - e) - b(a -e) =0(1-1)
- 1(0 - 1) =0(0) -1( -1) =0 + 1 = 1
Ans. A =be-ae= 1(1) -0(1) = 1-0= 1
Ans. B = - 2abe = - 2(0)(1)(1) = 0
Ans. c = 2ab-ae-be
= 2(0)(1) -0(1) -1 (1) = - 1
Ans.D=O
Only Answer A agrees with the given
expression. Answer A is correct.
Note
The second method could also be used as a
check. If more than one answer agrees with the
given expression, use different simple values to
find which one is correct.
TYPICAL QUESTION 2
.
What IS the value of
2b-4ac
3b
when a= - 3, b =2 and e=4?
A. - 61..
B. - 71..
c. 8 1.
333
D. 32
Topic 8 Algebra 31
Specific objective being tested
To think about
Substitute numerals for algebraic symbols in
simple algebraic expressions.
• These questions always give me trouble. It's
almost like gibberish. 'm years old' - what is
m? If it was an actual number I could deal
with it.
• I wonder if I replace m with an actual
number, ifI could see what is going on
better?
• 'Five years ago' - everyone was 5 years
younger than now.
• Three times as old' - her age multiplied by 3.
• 'How old is Kevin today?' - now.
To think about
• Algebra again - why does it look so
hard?
• I need to substitute carefully, calculate
carefully and again that rough work paper
(neatly done) is essential- otherwise I could
only guess, and anyone of those answers
could be right!
Framework solution
Framework solution
Method 1. Main method
2b -3b4ac
Nardia today is m years old
. h a= - 3, b = 2 an d c= 4'
WIt
gIves
Nardia 5 years ago was (m - 5) years old
2(2) - 4( - 3)(4)
3(2)
The brackets are put in to make sure that
I multiply and don't add (a negative x
a negative x a positive gives a positive).
= 4+48 =~=81.
6
6
3
AnswerC.
Kevin 5 years ago was 3(m- 5) years old
=3m-15
Kevin today is (3m - 15) + 5 years old
= 3m - lO AnswerB.
Method 2. Using a number for m as a check
Letm=lO
Note
There is really no way to check the correctness
of this answer except by going over the steps in
the calculation a second time and making sure
that no mistake has been made.
Nardia 5 years ago was 5 years old (10- 5)
Kevin 5 years ago was 15 years old (3 x 5)
Kevin today is 20 years old (15 + 5)
Ans. A= 30-5 =25
Ans. 0= 30+ 5=35
Ans. C= 30-15 =15
Ans. B=30-1O=20
TYPICAL QUESTION 3
Answer B is confirmed.
Nardia is m years old. Five years ago, Kevin was
exactly three times as old as Nardia was then.
How old is Kevin today?
TYPICAL QUESTION 4
A. 3m- 5
C. 3m-15
B. 3m-1O
D.3m+5
If P * q means p - q +pq,
A. -8
B. -2
evaluate 4* - 2.
C. 2
0.10
Specific objective being tested
Specific objective being tested
Translate verbal phrases into algebraic symbols
and vice versa.
32
Revision Maths
Use symbols to represent binary operations
(other than the four basic ones) and perform
simple computations with them.
To think about
• The * is a new operation. The question tells
me what it means.
• This * doesn't mean multiply, it means
p - q + pq where p is the number before the *
and q is the number after the *.
• Those minus signs are sure to give trouble
again. I must be very careful in each step of
the calculation.
negative value, or if the two sides of the
inequality are swapped around.
Framework soLution
2x+ 3~ 3x-4
Collect all the x's on one side so that there can
be a positive number of them.
2x+ 3-2x~ 3x-4-2x
3~x-4
Framework soLution
p * q = p - q+ pq
becomes:
and when p = 4 and q = - 2 it
Collect the numbers on the other side.
3+4~x-4+4
4*-2=4 - -2+4( -2) =4+2-8
=6-8= -2 AnswerB.
Note
The symbol used in a question like this does not
have to be *. It can be any symbol, e.g. @, #,
§, t, t. The question will define the symbol
for you.
TYPICAL QUESTION 5
7~x
Swap sides and change sign around.
:. x ~ 7
Answer A.
Note
Some people can do without so much working,
but better to be safe than sorry. Better to be
correct more slowly than quickly wrong.
TYPICAL QUESTION 6
If2x+ 3 ~ 3x-4 then:
A.x~7
B.x~ - l
C.x ~7
D.x ~-l
2X2
B. E..
12
c.~
D. llx
12
Specific objectives being tested
A.-
Find the solution set of linear equations and
inequalities in one unknown.
Specific objective being tested
Solve a simple linear inequality in one
unknown.
12
12
Simplify factors of the form aw + bdx where
cy
z
a, b, c, d are integers and w, x,y, z can be integers
or variables.
To think about
• A linear inequality. Again, the algebra feels
like it is going to overwhelm me. I have a
strong urge to simply guess and move on,
but I know I can do it ifI concentrate and
try. Take a deep breath!
• It's really just like an equation, except an eye
has to be kept on the sign.
• The sign reverses if the inequality is
multiplied through or divided through by a
To think about
• There are a lot of ways to simplify algebra.
This one looks like combining two fractions
to become one.
• I remember some work we did on common
denominators, and also changing to
equivalent fractions.
• All the denominators in the answers are 12,
so it's just the numerator that's the problem.
( ) opic 8
Algebra
33
'Remember don't simplify unless you obey
the rules.' I remember!
Framework solution
•
I think there are several ways to work this
question.
Framework solution
2x x
-+-
3 4
We need to put both fractions over the same
denominator.
Method 1. Using indices
24 x 2 5=24+5 =2 9
The LCM of 3 and 4 is 12 (the smallest number
that both 3 and 4 will divide into exactly) .
Putting both fractions over 12,
2x
2x
4
8x
-becomes- x
-3
3 x4
12
3x
12
~ becomes ~ x 3
4
4 x3
llx
8x 3x 8x+ 3x
12+12= 12
12
I
= 26 x2 = 23 =8
Answer C.
AnswerD.
Method 2. Without using indices
Note
This method of equivalent fractions may not be
exactly how you learned it, but as you can see it's
easy. If you do it differently, fine - so long as it
gives the right answer.
f24d = ) (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2 x 2)
..Jf
2x2x2
= "../2 x 2 x 2 x 2 x 2 x 2
= . .J (2 x 2 x 2) x (2 x 2 x 2)
TYPICAL QUESTION 7
=...J(2 x 2 x 2f=2 x 2 x 2=8 Answer C.
Tip
Simplify:
A. 2
B.4
C. 8
D. 16
Specific objective being tested
Follow through the steps in the working above
carefully. Revise the laws of indices and do
some practice questions from your textbook.
Did you remember that the index for square
. 1",
root IS 2 r
Use the laws of indices to manipulate
expressions with integral indices.
To think about
•
I can see my teacher writing The three laws
of indices' on the chalkboard.
• Can I remember what they are? Multiply
means add indices. Divide means subtract
indices. A power raised to a power means
multiply the powers.
34
Revision Maths ' : )
TYPICAL QUESTION 8
2
Make d the subject of the formula J=--x
d
A. d=J+x
B. d=_2_
2
J+x
c. d=2(j+x)
D. d
1
2(j+x)
Specific objective being tested
Change the subject of formulae including those
involving roots and powers.
TYPICAL QUESTION 9
If 6 men can paint a fence in 3 days, how long
will 4 men take if they work at the same rate?
To think about
A.
• This is the sort of question that I always
seem to get wrong because everything gets
mixed up.
• d is in an awkward place. There is only one d
so there is no collecting of d's together.
• On the same side as the d are the 2 above it,
the - x alongside it and the d is in the
denominator instead of the numerator.
• Three main steps need to be taken to
'unpack' the d and make it the subject.
C.
Framework soLution
J= ~ - x
Isolate the term containing d (add x to both
sides) .
2
2days
4t days
B.
2t days
D. 5 days
Specific objective being tested
Perform calculations involving direct and
inverse variations.
To think about
• If 6 men take a certain amount of time,
would 4 men take more time or less time?
• Less men would take longer. The answer
must be longer than 3 days. Answers C and
D look good.
• This is what my teacher called inverse
variation. One variable gets bigger, the other
gets smaller and vice versa.
J+x=d
Framework soLution
Swap two sides around to get d on the left.
2
6 men take 3 days.
The job takes 6 x 3 = 18 man-days.
Make right-hand side into a fraction by putting
over l.
(i.e. 1 man would take 18 days, 18 men could do
it in a day, 6 men take 3 days, 3 men take 6 days,
9 men take 2 days, 2 men take 9 days, etc.)
d=J+ x
2_ [+x
d-
4 men would take ~ = 4t days. Answer C.
1
Invert both sides to get d in numerator.
d
1
2= J+x
Multiply both sides by 2. (Remember if you
multiply a fraction by 2 it is only the numerator
that changes.)
d=_ 2_
J+x
Note
If x and y vary inversely, we write:
1
k
x ex: - or x = - or xy = k. k in this question is 18.
y
Y
The number of men (x) x the number of days
(y) = 18 (k)
Answ erB.
Topic 8 Algebra 35
Topic
9
Geomet
9 questions out of 60
TYPICAL QUESTION 1
Framework solution
The angle given (35°) is corresponding with
L.QPR which is also 35°.
In ~QPR:
L.R = 90° and L.P = 35°
p
R
so L. Q= x = 180 - 90 - 35 = 55
:. x= 55° Answ erC.
There are several other ways to argue it through,
using vertically opposite angles at Q, for
example.
In the figure x is equal to:
Specific objective being tested
Use the properties of rays, perpendiculars,
parallels and angles related to them to
(a) draw accurate geometrical figures,
(b) solve problems.
TYPICAL QUESTION 2
(
A. (1 , - 4)
To think about
• The two lines with arrows on them are
parallel. This makes me think about
corresponding angles, alternate angles and
allied (co-interior) angles.
• There is also a perpendicular line. 90° angles
are marked.
• There are also intersecting lines. I'm thinking
of vertically opposite angles.
• There is a triangle. Angles of a triangle add
up to 180°.
• Some of that stuff must be useful in this
question. Work from the 35° given to the XO
I want to find.
36
Revision Maths
A translation T maps point P (3, - 2) into
P' (- 3, 2). What point will T map Q (-1, 4)
into?
),
B. (2, 2)
C. (5, 0)
D. (- 7,8)
Specific objectives being tested
Specify translations in a plane as vectors,
written as column matrices, and recognize them
when so specified.
Locate the image of a set of points under the
transformations listed (including translation) .
To think about
• What does a translation do? I know
generally it's just a movement, so many
places to the right (x-direction), so many
places up (y-direction) .
•
•
Answer A looks likely because there is a
pattern to P and P' , Q and Q' , but I'm
always careful about obvious answers like
Answer A.
Will a quick sketch help?
y co-ordinate
To think about
3 --7 - 3 means 'less by 6'
- 2 --7 2 means 'more by 4'
So, in the same way:
Q --7Q'
x co-ordinate - 1--7 ? must be 'less by 6'
y co-ordinate 4 --7 ? must be 'more by 4'
Q --7Q'
x co-ordinate
Y co-ordinate
State what are the relations between an object
and its image in a plane when reflected in a line
in that plane.
Locate the image of a set of points under the
transformations listed (including reflection).
Framework soLution
P--7P'
x co-ordinate
Specific objectives being tested
- 1 'less by 6' --7 - 7
4 'more by 4' --7 8
SO Q' is the point (-7,8)
•
•
•
Reflection - a sketch is definitely going to
help here.
'in the y-axis' means having the y-axis, that's
the vertical axis, as 'the mirror line.
No particular point is given. It is asking
about any point. So I could use any point to
test the truth of A, B, C and D.
Framework solution
Answer D.
y-axis
mirror
line
X P(2.1)
P' (-2 . I)X
2
-2
P' x
x
2
--~----~t-------~-----x
A sketch shows that Q' has a big negative
x co-ordinate and a big positive y co-ordinate, so
must be Answer D.
TYPICAL QUESTION 3
When a point is reflected in the y-axis which of
the following is true?
A. The x co-ordinate remains unchanged.
B. The x co-ordinate is multiplied by a factor
of - l.
e. The y co-ordinate and x co-ordinate are
interchanged.
D. The x co-ordinate is less than it was before.
Take the point P (2,1) as an example.
When reflected in the y-axis it becomes
P'(-2,1).
Answer A - No, the x co-ordinate is not the
same (the y co-ordinate is)
Answer B - Yes, x co-ordinate is multiplied by
- 1 (2 x -I = - 2)
Answer C - No, the co-ordinates are not
interchanged
Answer 0 - Yes, x co-ordinate is less than
before.
Because both Band 0 are possibilities, a second
example is required.
Use P' as the point, now P will be the image.
(- 2, 1) maps to (2, 1).
Answer B - Yes, x co-ordinate is multiplied by
- 1 ( - 2 x -I = 2)
Answer 0 - No, x co-ordinate is now more than
before. Answer B.
(
Topic 9
Geometry
37
e
Tip
J
A sketch is often vital to a proper understanding
of what is happening in a Geometry question.
TYPICAL QUESTION 4
Which of the following capital letters has no
line of symmetry?
A.H
B. I
eJ
No line of symmetry. The 'hook' on the bottom
spoils the symmetry.
D.
D. K
K
Specific objective being tested
Identify simple plane figures possessing
translational, bilateral and rotational symmetry.
1 line of symmetry - horizontal.
AnswerC.
To think about
• A line of symmetry is like a mirror line in a
reflection. It divides a figure into two halves,
one of which is an exact reflection of the
other.
• If you fold along the mirror line, one part
will fit exactly on the top of the other part.
• This is one of those questions that will save
me some time in the exam. I can see this
answer quite easily, because I can picture the
mirror lines on H, I and K in my mind.
Note
Think about folding along the line of symmetry,
or if you prefer, actually draw the shapes on
tracing paper and fold them to see that H, I and
K do have lines of symmetry.
TYPICAL QUESTION 5
What is the size of each interior angle of a
regular octagon?
Framework solution
A.
Specific objectives being tested
I I
Recognize (a) the properties of polygons, lines,
angles, (b) the properties of regular polygons.
Solve geometric problems using the properties
of polygons and circles.
2 lines of symmetry - horizontal and vertical.
To think about
B.
2 lines of symmetry - horizontal and vertical.
38
Revision Maths
)
• 'interior angle' is the angle inside the figure.
'regular' means equal sides and equal angles.
'octagon' (like octopus, octet) means 8 sides
and 8 angles.
• A sketch may help, but not much.
• What formulae do I know? There is one for
interior angles of a polygon, and one for
exterior angles, which is an easier formula to
remember. I can use either of them.
TYPICAL QUESTION 6
A bicycle wheel has radius 20 cm. Riding the
bicycle in a straight line, the wheel turns
12 times. How far does the bicycle travel, to the
nearest metre?
Framework soLution
B. 15
A.l3
C. 18
D. 32
Specific objectives being tested
Recognize the properties of circles.
Solve geometric problems using the properties
of polygons and circles.
i = interior angle
e = exterior angle
To think about
Method 1. Interior angle formula
Sum of interior angles = (n - 2) x 180°
[some books say (2n - 4) x 90°, it's the same
thing]
:. Sum of 8 interior angles
= (8-2) x 180
=6 x 180
= 1080°
•
•
•
•
:. Each interior angle
1080
--8
=135°
•
A wheel has the shape of a circle. When it
travels, the circumference moves along the
ground.
'12 times' means 12 times around the
circumference.
'nearest metre' is a hint that I should estimate
or approximate at some point (but not too
early).
What formula do I know for the length of
the circumference of a circle?
How many centimetres make a metre?
Framework soLution
AnswerD.
Method 2. Exterior angle formula
Sum of exterior angles = 360°
[this is always so for any polygon]
Each time the wheel turns, it travels the length
of the circumference.
. . Total distance travelled
=12 x C
:. Each of 8 exterior angles = 3:0
=45°
(C=2nr)
=12 x 2 x 3.14x20
(n:::::::3.14)
=480 x 3.14
:. Each interior angle = 180-45
: : : : 500 x 3 (move one number up a bit, the
other down a bit)
=135°
AnswerD.
(because interior angle + exterior angle = 180°)
= 1500cm
=
Tip
I personally prefer Method 2 here; even though
it has two steps, the calculation is almost always
easier.
= 12 x 2nr
I5m
(but 100 cm = 1 m)
AnswerB.
Tip
In a multiple choice question, when no
calculator is allowed, numbers will be set which
CTopic 9 Geometry 39
do not require complex long multiplication or
division. Est·imation, as done here, can be a lot
of help, is expected and will save time if done
with care.
In triangle XYZ (not drawn to scale), angle YXZ
is a right angle, XZ is 3 cm and YZ is 4 cm.
x
3cm
C. 5
B. 3
Xy2 = 16 - 9
Swap sides round
Xy2=7
Answer A.
Note
{7 is an answer in surd form. In other words the
square root sign can be left in, without having to
work it out. An answer in surd form often looks
strange to students, and they are reluctant to
choose it. It is just as valid as any other answer.
Its actual value is between 2 and 3 because -v7 is
between {ti and -V9.
z
4cm
What is the length, in centimetres, of side XY?
A. {7
Collect numbers on one side
XY = -17
TYPICAL QUESTION 7
y
16 - 9 =Xy2
D.
m
Specific objective being tested
TYPICAL QUESTION 8
In triangle PQR, what is the value of
(sin P + cosI{)?
Use Pythagoras' theorem to solve simple
problems (no formal proof required).
p
5
To think about
• What do I know about triangles?
• This triangle has a right angle at X. I know
two sides and need to find the other.
• Pythagoras' theorem says that a2 = b2 + c2,
where 'a' is the hypotenuse.
• I remember something about a 3,4,5
triangle. Could this be it?
• Which is the hypotenuse? YZ, that is 4 cm.
So, it's not the 3,4,5 triangle where the 5 is
the hypotenuse (the longest side).
• If 4 cm is the longest side then the answer
must be less than 4, so it's A or B.
A. 1.2
Q
4
B. 1.4
C. 1.6
R
D. 1.8
Specific objectives being tested
Determine the sine, cosine and tangent ratios of
acute angles in a right-angled triangle.
Use the sine, cosine and tangent ratios in the
solution of right-angled triangles.
Framework soLution
Using Pythagoras' theorem
a2=b 2+c2
42 = 32 + xy2
16=9+Xy2
40
Revision Maths
Substitute values
)
To think about
• It's a right-angled triangle. All three sides are
given. (This is the 3,4,5 triangle!)
• The question is asking about trig ratios sine
(sin) and cosine (cos) of angles P and R.
• I remember SOHCAHTOA:
opposite
.
adjacent
; cosme=
;
hypotenuse
hypotenuse
_ opposite,
tangent- d.
.
a pcent
I expected the answers to be fractions but
they are in decimal form.
sine
•
Framework soLution
.
SIn
P~
=
cos R =
opposite
hypotenuse
i (opposite to P)
5 (longest side)
adjacent =i (adjacent to R)
hypotenuse 5 (longest side)
Solve problems involving the gradients of
parallel and perpendicular lines.
To think about
•
Parallel lines have the same slope or gradient
as each other.
• The equation of a straight line is y = mx + c
(m is the gradient).
• If lines have the same m they are parallel.
• The problem with the equations Il and III is
that they are mixed up and need rearranging
as y = ... , so that I can see what the value of
m is in each of them.
Framework soLution
I.
(NB sin P and cos R are equal)
~
4 4 8
. ~
.·.(smP +cosR)=-+-=-=1.6
·
55
5
y=3x-4
6x=1O+2y
Il.
AnswerC.
1O+2y=6x
TYPICAL QUESTION 9
I. Y = 3x - 4
Il. 6x = 10 + 2y
A. I and Il
C. Il and III
B. I and III
D. I, Il and III
Isolate y term.
Divide through by 2.
y=3x-5
gradient is 3.
15x= 5y+ 7
5y+ 7= 15x
Ill. 15x - 5y = 7
Swap two sides.
q=6x-1O
Ill. 15x - 5y = 7
Which of the following lines are parallel to each
other?
gradient is 3.
Make term in y positive.
Swap two sides.
Isolate y term.
5y=15x-7
Divide through by 5.
y= 3x-l.4
gradient is 3.
Specific objectives being tested
: . All three lines are parallel.
Analyse the equation y = mx + c with respect to
its gradient (positive or negative) and its
y-intercept.
Note
Answer D.
Equations must be manipulated with care one step at a time.
(::To~ic 9
Geometry
41
Part
2
ESSAY AND SHORT ANSWER PAPER - Paper 2
This paper requires the candidate to answer all
the questions from Section I (compulsory
questions) and any two from six in Section 11 in
2 hours 40 minutes.
It is worth 1- of the total marks for the
examination.
Candidates often feel they have to rush this
paper, as time is short. It is better, however, to
work slowly and thoroughly and answer even
haltof the questions properly than it is to play
'tag' with the questions, touching each one and
running on to the next. Most of the marks in
these questions come nearer to the end of the
question than the beginning!
First we look at the topics which occur in
Section I, then those that occur in Section 11. It is
normal to go into the examination having a
preference for one or two of the areas covered in
Section 11.
Remember to try each question yourself
before looking at the Framework solution.
43
SECTION I
Topic
1
cSet~
10 marks out of 90 Sec!ion I )
The most common type of question in this
topic is a three-set Venn diagram question
where numbers must be inserted on the Venn
diagram in the various areas from information
given.
.-----------,u
A
c
.-----------, U
A
Diagram 1
.------------,u
B
Diagram 2
In a three-set situation the shaded area in
Diagram 1 represents AnB, in other words things in A AND B.
(Note: C could be removed from the picture
altogether and this will be seen clearly, just as in
the regular two-set Venn diagram.)
Diagram 3
c
Diagram 4
Similarly, Diagram 3 shows the whole of
set A shaded representing the things in A.
Diagram 4 shows the set AnB' nC'
representing the things in A but
not in B and not in C or, in short, the
things in A ONLY. There is a big
difference.
TYPICAL QUESTION 1
.------ - ---, u
In Diagram 2 the shaded area represents
AnBnC', in other words things inA AND B
but NOT C or putting this more concisely
things in A AND B ONLY.
The word 'only' is the key word in many of
these questions.
In a class of 50 students, each student does at
least one of three subjects: English, Geography
and History.
'Students doing Accounts and Biology' gives an
area as in Diagram l.
'Students doing Accounts and Biology only'
gives an area as in Diagram 2.
10 students do English and History only.
4 students do History only.
6 students do Geography and English only.
No student does Geography only.
Tip
44
Revision Maths
12 students do all thtee subjects.
2x students do Geography and History only.
x students do English only.
Which item of information can we use first?
Item 1 cannot help us as yet!
Think carefully and as the items are incorporated
in the Venn diagram, tick them off.
(a) Complete the Venn diagram to illustrate the
information in this question.
(b) Write an equation in x to represent the
composition of the class.
(c) Hence, calculate the number of students who
do English only.
u
,----------,u
Specific objectives being tested
Construct and use Venn diagrams to show
subsets, complements, intersection and union of
sets, and solve problems involving not more
than three sets.
Solve numerical problems arising from the
intersection of not more than three sets.
Identify and construct subsets of a given set.
Item 9 used
Tick it off
,@G~
10
Item 2 used
Tick it off
@ ,G~
10
o
4
L -_ _~H_~
To think about
•
•
•
How many items of information do I have?
I must list them and check them off as I use
them.
The order in which I use the items of
information may not be the same as the
order in which they are given in the
question.
Framework solution
(a) There are nine items of information:
1. 50 in the class.
2. Everyone does at least one of the three
subjects, in other words, no-one does
none of the three subjects.
3. 10 do E and H only.
4. 4 do H only.
5. 6 do G and E only.
6. 0 do G only.
7. 12 do all three subjects.
8. 2x do G and H only (don't worry about
the x here).
9. x do E only.
Item 4 used
Tick it off
Item 3 used
Tick it off
u
~
6
x
0
H
10
4
@G
6
x
---+
u
0
---+
10
0
4
0
H
H
Item 5 used
Tick it off
Item 6 used
Tick it off
u
u
~G
x
10
6
0
12
4
---+
0
0
H
Item 7 used
Tick it off
H
Item 8 used
Tick it off
Answer to part (i)
Note: Item 1 has still not been used!
(
Topic 1 Sets
45
(b) Add the number of students in all the areas
together - What should the total be? (Now is
the time for item 1.)
x+ 6+0+ 10+ 12+2x+4+0=50
(Tick off item 1)
(c) 2x+x=50-6-1O-12-4
3x=18
x=6
So the number of students who do English
only is x, which is 6.
[The Venn diagram actually looks like this.]
,----------,u
Solve numerical problems arising from the
intersection of not more than three sets.
Identify and construct subsets of a given set.
To think about
• List the items of information that I have.
Check them off as I use them.
• At first glance this looks like a four-set
question, but 'using the universal set as the set of
students who study English' tells me that it is
only the other three sets that are inside the
square that is the universal (English) set.
• One word that I do not see in the main
body of the question is only! That makes
this question rather different (and a little
more difficult) than the last question.
Framework solution
H
TYPICAL QUESTION 2
There are 40 students in Form V.
All students study English.
22 study Accounts.
13 study Economics.
22 study History.
7 study History, Economics and English.
13 study History, Accounts and English.
5 study Economics, Accounts and English.
5 study all four subjects.
(a) Draw a carefully labelled Venn diagram to
represent the data, using the universal set as
the set of students who study English.
(b) Determine the number of students who
study at least two subjects.
(c) Calculate the number of students who study
exactly two subjects.
(a) There are nine items of information (as there
usually are):
1. 40 in the class.
2. Everyone does E.
3. 22 do A.
4. 13 do Ec.
5. 22 do H.
6. 7 do Hand Ec (and E).
7. 13 do H and A (and E).
8. 5 do Ec and A (and E).
9. 5 do all four subjects (H and Ec and A
(and E)).
I keep putting E in brackets because in this
question it is not adding anything as E is the
universal set. It is the other information that is
important.
=
U {English}
Specific objectives being tested
Construct and use Venn diagrams to show
subsets, complements, intersection and union of
sets, and solve problems involving not more
than three sets.
46
Revision Maths
Item 2 used
Tick it off
t@9&
U= {English}
22 students do H (not 'H only'). This 22 is found
in the four regions inside H. That is the 5, the
8 and the 2 already placed on the diagram and
one other area which must be 22 - 5 - 8 - 2 = 7.
H
AC@JCU={EngliSh}
906
5
8
2
Item 9 used
Tick it off
7
-----.
3
H
U = {English}
Items 4, 3 and 1 used
Tick them off
Similarly, place the numbers 9, 6 and 3 so that
circle A contains 22, circle Ec contains 13 and
the whole universal set contains 40.
Item 8 used
Tick it off
The 5 who do both Ec and A are in two
categories: 5 of them also do Hand 5 - 5 = 0 do
not do H, i.e. none of them do Ec and A only.
EC
t@J
5
8
U = {English}
-----.
2
Note
H
There are many other types of Set question, and in
tackling them successfully one needs to be able to:
Items 7 and 6 used
Tick them off
Again the 13 who do H and A are in two
categories (5 and 8).
The 7 who do Hand Ec are split into 5 and 2.
At@J0EC
8
(b) The number doing at least 2 subjects means
all those who do English and are inside at
least one of the three circles.
That means everyone except those outside
all the three circles = 50 - 3 = 47
(c) n(Study exactly 2 subjects) =those who do
English and only one other subject
=9+6+7=22
2
U = {English}
1. Identify all the symbols and translate them
freely into English.
2. Keep a close track of all the information
given, used and required.
TYPICAL QUESTION 3
Given the following information:
U = {5, 6, 7, 8, 9, lO, ... , 16}
A = {even numbers}
B = {numbers less than lO}
A and B are subsets of U.
7
H
Item 5 used
Tick it off
(a) List the members of A and of B.
(b) Draw a Venn diagram to represent the
above data.
(c) State n(AUB').
Specific objectives being tested
List the members of a set from a given
description.
Construct and use Venn diagrams to show
subsets, complements, intersection and union of
sets, and solve problems involving not more
than three sets.
Find the complement of a given set, given the
universal set.
To think about
• Is A all of the even numbers? That would
be an infinite set and could not be shown
easily on a Venn diagram.
• Remember the line that said 'A and Bare
subsets ofU:
• What is U? Whole numbers from 5 to 16
only!
• What do those dots mean? Let's fill it out.
U={5, 6, 7,8,9, 10,11, 12,13,14,15, 16}
These are the only numbers under
consideration in this question (12 of them).
Framework soLution
(b) [Thinking: Do A and B intersect? Yes, 6 and 8
are common.
10, 12, 14 and 16 are only in A,
5, 7 and 9 are only in B and the other
numbers are in neither!]
.-------------------'u
B
10
8
15
[Check that all 12 numbers are on the
diagram.]
48
Revision Maths
.--------------------,u
I1
The same number, n, play football only and
swim only.
(a) Calculate the number who play football.
(b) State the information represented by the
shaded region of the Venn diagram.
(c) State the relationship between the members
of C and F, and between C and S.
Specific objectives being tested
6
13
TYPICAL QUESTION 4
F = {members who play football}
C= {members who play cricket}
S = {members who swim}
B = {5, 6,7,8, 9}
the numbers less than 10 in U
11
The' or complement sign means NOT.
Everything in A or (not B) means all
the numbers inside circle A or outside
circle B.
Namely 6,8,10, 12, 14 and 16 inside A
and also the numbers outside B, namely
10, 11, 12, 13, 14, 15, 16. Some ofthese
numbers appear twice, which is
unnecessary. In all we have: 6, 8, 10, 11, 12,
13, 14, 15, 16
How many numbers is that? Nine (Answer).
[There are nine numbers that are even or not
less than 10.]
The Venn diagram represents information
about the 60 members of a youth club.
(a) A={6,8, 10, 12, 14, 16}
the even numbers in U
A
(c) n(AUB') means the number of elements in
the set A or (not B).
Construct and use Venn diagrams to show
subsets, complements, intersection and union of
sets, and solve problems involving not more
than three sets.
Use set language.
To think about
• In this question only part (i) is of a numerical
nature requiring calculations. The
calculation cannot be very hard because it
can be worth only 1 or 2 marks, so must not
take much time.
• The other two parts of the question must not
be omitted just because they have no
numbers to work with.
• Think in terms of set language - union,
intersection, complement, subset - revise
these terms until you understand their
meaning in any problem setting.
(b) The shaded area is inside F and inside 5, it
therefore describes those who both play
football and swim.
(c) Cc F (C is a subset of F), in other words all
who play cricket also play football.
cns = 0 (C and 5 are disjoint); in other
words no-one both plays cricket and
swims.
In our next question there is no real-life story to
go with the question, the quantities given seem
more abstract but the information can still be
worked with.
TYPICAL QUESTION 5
Framework solution
(a) The information given:
(a) Draw and label a Venn diagram to show the
follOwing information:
1. The Venn diagram.
2. 60 in the club.
3. n play football only.
4. n swim only.
A, Band C are sets and U is the universal set
n(U) =40, CcA,
Place the two n's on the diagram, in the correct
places (items 1, 3 and 4).
,--------------------,u
n(A) = 26, n(B) = 16, n(C) = 21
n(AnB) = 9 and n(Bnq = 6
(b) Hence, determine
n(AUC)' and n(AnBnC)
Specific objectives being tested
11
Construct and use Venn diagrams to show
subsets, complements, intersection and union of
sets, and solve problems involving not more
than three sets.
n + 21 + 10+ n + 11 = 60 (item 2)
2n+42=60
2n=18
n=9
Now put this information back on the diagram.
Determine the number of elements in certain
subsets of two intersecting sets, given the
number of elements in some of the other
subsets.
u
To think about
11
The number who play football is 21 + 9 + 10
(all inside set F) = 40
• The symbols n , U, c and' all occur in this
question and I must be very clear what they
mean:
n means intersection, what is in one set
AND the other (things in both sets).
Topic 1 Sets
49
U means union, what is in anyone of the
two sets or both of them.
e means subset, the first set is contained
inside the other.
, means complement, or NOT whatever it
stands beside.
Item4(6+3+7=16}
Framework solution
(a) The information given:
1. Total number is 40.
2. C is inside A.
3. A contains 26.
4. B contains 16.
5. C contains 21.
6. A AND B contains 9.
7. BAND C contains 6.
Item 5 (15+6=21)
.--------,u
~B
What can we use first?
,--------:::--------, U
~'
~
----.
Item 3
(2 + 15 + 6 + 3 = 26)
Is this what we want?
No.
u
@J
u
B
(IS 63
2
7
7
Item 1
(All the numbers inside U add to 40.)
Item 2
~
u
----.
6 3
Item 6 (6 + 3 = 9)
50
Revision Maths
Look outside A and outside C.
Answer = 7 + 7 = 14
n(AnBnC) = the number of elements in
A and Band C = 6
Item 7
@D
(b) n(AUC)' = the number of elements in
(A or C) not, i.e. not in (A or C)
u
----.
That is the only number inside all three
circles.
Note
Now try lots of other questions like these from
your textbook.
Topic
2
Relations, Functions and Gra hs
10 marks out of 90 in Section I
This topic is wide ranging and is tested on the
multiple choice paper, and in Sections I and 11 of
Paper 2. In Section I, the questions tend to be
extensions of what may be found on the
multiple choice paper, though requiring a little
more thought and calculation.
This is one topic that requires thorough
revision, as most of the questions require special
knowledge and cannot be started from
background knowledge 'picked up' in Maths
classes along the way.
Before drawing the graphs, look at Appendix II
at the back of this book entitled 'Precision in
Graph Work' (p. 167).
TYPICAL QUESTION 1
A straight line is drawn through the points
To think about
• Am I going to draw a diagram or work out
the answer algebraically using the various
formulae?
• What are the relevant formulae?
vertical rise
• Gradient (m)
horizontal shift
Y2 -Yl
X2- XI
(where (XI ,YI) and (X2>Y2) are two points on
the line).
• Y = mx + c is the equation of a line
where m is the gradient, and c is the
y-intercept.
• y - Y1 = m(x - x1) is the equation of a line
where m is the gradient and (Xl' YI) is a point
on the line.
• Perhaps a quick sketch would be good.
M (2, - 3) and N ( -1, 3).
(a) Determine
(i) the gradient of MN,
(ii) the equation of the line MN.
(b) State the points which represent the
x- and y-intercepts of MN.
y
N
3
6 units down
3 units across
Specific objectives being tested
Recognize the gradient of a line as the ratio of
the vertical rise to the horizontal shift.
Write the equation of a line given the
co-ordinates of two points on the line.
Find by drawing and/or calculation, the gradients
and intercepts of graphs oflinear functions.
Yes, now I can see what is going on. I can see
the gradient is negative. I can see the vertical
rise and horizontal shift. I can see roughly
where the two intercepts are.
Topic 2 Relations, Functions and Graphs
51
Framework solution
(/ is a positive gradient, \ is a negative
gradient and - is a zero gradient.)
(a) (i) Method 1. From sketch graph
·
-6 (6 down)
Gra dlent=-3 (3 across to right)
-_ -2
TYPICAL QUESTION 2
Method 2. By algebra
Gradient=Y2-Yl = 3-(-3} =_6_=_2
X2-XI
-1-(2}
-3
(ii)
Y-Yl =m(x-xl}
:.y - (- 3}= -2(x-2}
y+3= -2x+4
Y= -2x+1
The y-intercept is clearly 1 on the y-axis
(c = 1 in the equation of the line)
i.e. the point (0, 1)
(Note: always give the intercept as a point.)
The x-intercept looks to be between 0 and 1.
Wheny=O,
0= -2x+1
2x=1
1
2
(a) Determine the equation of line 1.
(b) If line L passes through the point P (a, 6), find
the value of a.
(c) If line M is the perpendicular to line L
through P, and line M passes through the
point Q (2, b), find the value of b.
Specific objectives being tested
Write the equation of a line given the gradient
and one point on the line.
Find by calculation, the gradients and intercepts
of graphs of linear functions.
To think about
X=-
(t,
i.e. the point
0)
(Note: always give the intercept as a point.)
Note
Because this question said 'determine' rather (
than 'calculate', an accurate graphical solution
would be quite acceptable.
If the question had said 'calculate', the graph
could only be used as your personal check, to
guard against arithmetical slip-ups in the
calculation.
Be careful, however, when reading values such
as from a graph. It may not be exactly as
graphs do not always give as accurate a value as
do calculations.
t
t
One of the most common errors regarding
gradients is the sign (+ or -). A sketch
will always ensure that you have the sign
.
right.
52
A straight line L, having a gradient of - 2 passes
through the point ( - 3,4).
Revision Maths
)
• A sketch may help again. So will all the
formulae.
vertical rise
Y2 -YI
• Gradient (m)
horizontal shift X2 - Xl
where (Xl,Yl) and (X2,Y2) are two points on
the line.
• y = mx + c where m is the gradient and c is the
y-intercept.
• y - Yl = m(x - Xl) where m is the gradient and
(Xl' Yl) is a point on the line.
• If two lines are perpendicular, the product of
their gradients is - 1. If you have one
gradient given, you invert it and change its
sign to get the other one.
Framework solution
(a) Method 1. Using y=mx+c
Y= - 2x+c
4= - 2( - 3)+c
To think about
4=6+c
6+c=4
c=4-6= -2
:.Y= -2x-2
Method 2. Using (Y-Yl) =m(x-xl)
y-4= -2(x- -3)
y-4= -2x-6
y= -2x-6+4
Framework soLution
:.Y= -2x-2
(a) g(x) = x 2 - 1 so g(3) = 32 - 1 = 9 - 1 = 8
(b) If (a, 6) is on line L, that is y = - 2x - 2
6= -2(a)-2
2a= -6-2=-8
:. a=-4
(c) If line M is perpendicular to line L, its
x -2= -1)
gradient must be
It also passes through P which is ( - 4, 6).
+t. (+t
1
2(y-6) = (x- -4)
2y-12=x+4
Line M is 2J = x + 16, and it passes through
(2, b).
:. 2b=2+ 16
2b=18
b=9
(c) Method 1. The 'easy' way
(Only works when there is a 'single' x in the
expression for f(x).)
For the inverse, invert the operations and
invert their order, in other words, subtract
1 from x first then divide by 2.
f - 1(x) = x-I
2
Method 2. The standard way
Note
There are other ways to work this last part, for
example using the gradient and the two points,
without having to find the equation of M.
Given that f: x ~ 2x+ 1
g : x ~ x2 - 1 calculate
(There are several variations to this
method - stick to what you know and are
happy with.)
f(x) =2x+ 1
Lety=j(x)
TYPICAL QUESTION 3
(b) fg(-2)
(b) fg(x) means f operating on g, whilst g is
operating on x. g operates on x first.
fg( - 2) = f[g( - 2)] =f[( - 2f-1]
=f[4-1] =f[3] =2(3) + 1 =6+ 1 =7
f(x) =2x+ 1
In other words, you multiply x by 2 then
you add 1.
:.y-6="2(x- -4)
(a) g(3)
• Let's translate these arrows into equation
form. f(x) = 2x + 1 and g(x) = x 2 - 1.
• Part (i) looks easy. Part (ii) is OK if! can
remember which function to do first. Part
(iii) is an inverse - but of an easy function I could use the 'easy' method.
:.y=2x+1
and
(c) f - 1(5)
Specific objectives being tested
Use the functional notations, e.g. f: x ~ x2 or
f(x) = x 2; as well as y = f(x) for given domains.
Interpret and make use of functional notations,
e.g. j(x), g(X),f- l(X), and their compositions.
y-1=2x+1- 1=2x
y-1
--=X
2
y-1
X= - 2
:·f- l(X) = x-I
2
(replacing x with f - 1(x) and y with x)
5-1
4
So, by either method,f- 1(5) = -2- ="2= 2
Topic 2 Relations, Functions and Graphs
53
TYPICAL QUESTION 4
(ii) gf (x) = g(f (x))
Given thatf:x~2-4x, g:x~ 2x and
3+x
3
h :x~ --wherexE 9\
6-x
(a) Calculate
(i)
fg(x)
(ii) gf(x)
(iii) j2(x)
=g(2 - 4x) = 2(2; 4X)
Specific objectives being tested
Use the functional notations, e.g. f: x ~ x2 or
fix) = x2, as well as y = f (x) for given domains.
Interpret and make use of functional notations,
e.g. f(x) , g(X),f- l(X), and their compositions.
8X
3
(iii) j2(x) = f f (x) = f (f(x)) = f (2 - 4x)
= 2 - 4(2 - 4x) =2 - 8+ 16x= 16x - 6
x
(iv) g2(x) = g g(x) = g (g(x)) = g (23 )
= ~ e;)= ~x
(v) f- 1 (x)
(b) (i) Determine the value of x for which
f - 1(x) = 3.
(ii) State the value of x for which h(x) is
undefined.
;
(v) f (x) = 2 - 4x in other words (multiply x by
- 4 then add 2)
For the inverse, invert the operations and
invert their order, in other words (subtract 2
from x first then divide by - 4)
f - 1(x) = x-2 = 2 -x
-4
4
(b) (i) Method 1. Using j - 1(x)
2- x
f - 1(x)=--=3 so 2-x=12
4
2-12 = x
The value of x is - 10.
To think about
• Translate the arrows into equations? No,
I can understand what they mean.
• What looks like the 'power' 2 in
parts (iii) and (iv) of (a) is not a power or
it would be [f(x)f Just the 'f' is raised,
i.e. the function operates twice like f f(x)
orf(f(x)).
• x E 9\ just m eans that all the x's are real
numbers. If a function is to be undefined, it
has to take a value that is not a real number
(the likeliest is 'infinity' which is not a real
number, but can sometimes occur in a
fraction) .
Framework solution
(a) (i) fg(x) = f (g(x))
=ft~)=2-4t~)= 2 - ~x
54
Revision Maths
Method 2. Using f(x)
If f - 1(x) = 3 then x=f(3)
The value of x is
f (3) = 2 - 4(3) = 2 - 12 = - 10
(ii) h(x) becomes infinite if the denominator
becomes zero. When 6 - x = 0 then x = 6,
the value for which h(x) is undefined.
TYPICAL QUESTION 5
(a) Copy and complete the table below for the
function y = x2 - 4x + 2.
,x
-2
, Y 14
-1
0
1
-1
2
3
4
2
5
'--
(b) Using a scale of 2 cm to one unit on the
x-axis and 1 cm to one unit on the y-axis,
draw the graph ofy=x2-4x+ 2 for
-2 ~ x ~ 5 .
To think about
(c) Using your graph, or otherwise, state the
range of values for which
(i) x2 - 4x + 2 is negative
(ii) x2 - 4x :::;; 2.
Specific objectives being tested
Draw and use graphs of the functions {(x, y) :
y = a+bx}, {(x,y) :y=a+bx+cx2} where a, band
c are integers.
Use graphs of a given function to determine
• the interval of a domain for which the
elements of the range may be positive or
negative.
• the interval of a domain for which the
elements of the range may be greater than or
less than a given value.
• The graph is a quadratic. Its shape is a
parabola. There should be some symmetry
in the shape and in the y-values.
• When plotting the graph, I must remember
all the things I know I should do. Correct
scales, label axes, use small, neat x's, join a
smooth curve, use a sharp pencil, etc.
• (c) (ii) is nearly, but not quite the same
equation. How can I make it the same?
• 'range of values' means using 'less than or
equal to' signs, like - 10 :::;; x:::;; 10.
• Before plotting the graph, I must consider
the values that I have for x and y so that I can
make a good decision as to where to place
the axes.
1\
'\
I,..,
\
\
'"
1\
\
0
\
1\
,
\\
/
11
h
r.:,
r\
1
\ ...
I1
1\
I~
-2
/
1
\
-I
...
"'"
~r--.
V
/
x
V
.,/
Topic 2 Relations, Functions and Graphs
55
Framework soLution
Specific objective being tested
(a) y = x2 - 4x+2.
Draw graphs to represent and solve linear
inequalities.
Whenx= -1
y = ( _1)2 - 4( - 1) + 2 = + 1 + 4 + 2 = 7
When x=O
When x = 2
Whenx= 3
When x= 5
y= (0)2 -4(0) + 2=0-0+ 2 = 2
y = (2)2_4(2) + 2=4-8 + 2 = -2
y =(3)2-4(3) + 2 =9-12+ 2=-1
y= (5)2-4(5) +2= 25 -20+ 2= 7
To think about
• All of these inequalities represent regions on
one side or another of the lines y = 0, x = - 4
and y = x + 1. So first I must identify these lines
and then decide which side of the line the
region is on (and if the line is included or not).
y~T ~TT -: 1~~gJn~l Framework soLution
(b) x-values go from - 2 to + 5 so the y-axis
should be to the left of centre.
y-values go from - 2 to + 14 so the x-axis
should be near the bottom.
(c) (i) Y = x2- 4x + 2. If x2- 4x + 2 is negative
then y is negative on the curve.
y is negative for values of x between
0.6 and 3.4, i.e. 0.6 ~ x ~ 3.4
(look where the curve crosses the x-axis).
(ii) y = x2 -4x+ 2. If x2-4x~ 2 then
x2-4x+ 2 ~4 andy~4.
y is less than 4 for values of x between
- 0.4 and 4.4, i.e. - 0.4 ~ x ~ 4.4
Oook at the circled points).
TYPICAL QUESTION 6
• y = 0 is the x-axis. y ~ 0 is the region below
the x-axis and on the x-axis.
• x = - 4 is a vertical line through - 4 on the
x-axis. x> - 4 is to the right of that line.
• Ify=x+ 1, when x=O, y=O+ 1 = 1;
whenx= -4,y= -4+ 1 = -3.
Y = x + 1 goes through (0, 1) and ( - 4, - 3).
Y > x + 1 is on which side of this line?
It is above and to the left of this line
because (0, 3) satisfies the inequality
whilst (0, 0) does not. [3 > 0 + 1 is true
but 0 > 0 + 1 is false.]
x=-4
y=x+1
y=_O__~-*~~~+-__~x
Copy the diagram and shade the region which
satisfies all of the following inequalities:
y~O
Two of the lines are broken because they are
strict inequalities without the 'equals' part.
The other line is solid because it represents
x>-4
y>x+1
y ~ O.
-6
-4
-2
2
-3
56
Revision Maths
)
x
TYPICAL QUESTION 7
Framework solution
Distance
(a) Part 1 of motion: speed = %= 1 ms - 1
Part 2 of motion: speed = Sf = 0 ms - 1
Part 3 of motion: speed = = 3 ms - 1
Part 4 of motion: speed = %= 0 ms - 1
Part 5 of motion: speed = %= 2.5 ms - 1
(m)6
t
/
\
I1
2
o
\
/
4
V
o
/
\
I
I
2
4
5
6
7
(b) The girl is stationary twice, for 1 second,
then for two seconds.
\
8 TImeW
This distance-time graph represents the
movement of a girl playing hopscotch. She
starts in the 'home square'. The motion has five
distinct parts.
Using the graph, determine
(a) The speed with which she moves during
each of the five parts of the motion.
(b) The number of times she remains stationary
and for how long.
(c) The furthest distance she moves away from
the 'home square'.
(d) After how long she returns to the 'home
square'.
Specific objectives being tested
Draw and use distance-time graphs and
speed-time graphs.
Use the gradient of the graph of a linear
function to determine the rate of change of one
variable with respect to the other.
To think about
• Distance is plotted against time.
• The gradient of the line represents
distance
speed
time
• She starts in the 'home square', so the
horizontal baseline (distance = 0) represents
that square.
• She moves away from the 'home square'
initially, then back towards it.
(c) She moves 5 metres away at the furthest
point. (Highest the graph reaches is 5.)
(d) She returns after 8 seconds. (Graph descends
to base line.)
TYPICAL QUESTION 8
v (m/s)
50
40
30
20
10
o
/
/
o
/
5
10
15
/
20
t(min)
The graph shows the motion of an ocean-going
liner.
(a) Calculate the acceleration of the liner in the
first part of the motion.
(b) Describe the motion betweel) t = 10 and
t=15 .
(c) Calculate the acceleration of the liner in the
last part of the motion.
(d) Calculate the total distance travelled, in
kilometres, in the 20 minutes shown.
Specific objectives being tested
Draw and use distance-time graphs and
speed-time graphs.
Use the gradient of the graph of a linear
function to determine the rate of change of one
variable with respect to the other.
Topic 2 Relations, Functions and Graphs
57
To think about
(d) Breaking the shape into 2 triangles and a
rectangle:
Distance = area under line
t is in minutes - danger, red flag, watch
out - an unusual unit to have against metres
per second.
• Its a speed-time (or better velocity-time)
graph, so the vertical axis is velocity (not
distance). The slope on this graph is
velocity
.
acceleratIOn
time
• A horizontal line ---7 zero gradient ---7 zero
acceleration ---7 constant velocity.
• The distance travelled is the area under the
line.
Note
'Velocity' implies a size (of speed) and a
direction (it is a vector quantity).
'Speed' implies only a size, not a direction (it is a
scalar quantity).
Framework solution
TYPICAL QUESTION 9
.
velocity
20
(a) AcceleratIOn.
600
tIme
1
=-ms - 2
30
(Time = 10 min = 600 seconds)
Using the graph shown ofj(x) = 6+ 5x-x 2
•
=
(~ x 600 x 20) + (~ x 300 x 20)
+ (600 x 20)
= 6000+ 3000+ 12 000
=21000m= 21km
_ _..L
(a) State the roots of the equation
6+5x-x 2 =0.
(b) State the maximum and minimum values of
j(x) over the domain - 2 ~ x ~ 6.
(b) The liner travels at a constant velocity of
20ms - 1 .
.
velocity
20
(c) AcceleratIOn.
300
tIme
1
- 2
=-ms
15
(Time = 5 min = 300 seconds)
(c) Estimate, by drawing a tangent to the curve,
its gradient at the point when x = 4.
((x)
I
12
8
/
V
/
V
........
V
V
/1
0
0
1
/
I1
58
Revision Maths
-A
~
"'"
4
I
2
""
12
~
'"
1\
\
x
6
Specific objectives being tested
• The slope of a tangent will be the same as the
slope of the curve at any point.
Draw and use graphs of the functions {(x, y):
y =a+ bx}, {(x,y):y=a+ bx+cx 2} where a, b
and c are integers. Read and interpret such
Framework soLution
graphs.
(a) The curve crosses the x-axis at x = - 1 and
x = 6 (the roots of the equation).
Use graphs to find the solution set of quadratic
equations.
Use graphs of a given function to determine
•
•
the roots of the given function.
the maximum and minimum values of the
function over a given interval of the domain.
Estimate the value of the gradient of a curve by
constructing a tangent to the curve at a given
point.
(b) Minimum value in this domain occurs at the
extreme left of the curve when x = - 2 and
fix) = -8.
Maximum value in this domain occurs at
the turning point of the curve when x = 2.5
(by symmetry) and fix) is a little more
than 12.
When x = 2.5, fix) = 6 + 5(2.5) - (2.5)2
= 6 + 12.5 - 6.25 = 12.25
Minimum value ofJ(x) =-8
Maximum value off(x) = 12.25
To think about
• The roots are found where the curve crosses
the x-axis because there fix) = 6 + 5x - x2 and
fix) = 0 therefore 6 + 5x - x2 = o.
• The maximum and minimum points are the
highest and lowest points the curve reaches.
Remember a parabola has symmetry.
(c) The tangent is drawn just to touch the curve.
It passes through (4, 10) and (6, 4).
10-4 =_6_=_3
4-6
-2
((x)
~
12
8
V
V
V
....-
~
-.....::::
"" '\
~~
/4
/
V
11
i'-. 4.10)
N~ .4)
1\
0
0
\
~
3
""
x
f
I
Topic 2 Relations, Functions and Graphs
59
Framework soLution
TYPICAL QUESTION 10
,
v-
.
\
.• i
,
1\
\
0
, 0-
,2
41\
\
~
8
6
\
j
-- -
--
-
L
~
~
I-
\
\
(a) Calculate the gradient.
(b) Determine the equation.
(c) Solve their equations simultaneously from
the graph.
Specific objectives being tested
Write the equation of a line given the graph of
the line.
Use graphs to find the solution set of
simultaneous linear equations.
To think about
vertical rise
"
• Gra dIent= - - - -- horizontal shift
• Positive gradient is I, negative gradient is \.
• (Y-YI) =m (X-Xl) should be useful here.
• To solve simultaneous linear equations
graphically, I only need to look for the point
where the lines cross each other; at that
point both equations are satisfied.
Revision Maths
)
2
10
S
L2 passes through the points (3, 3) and (4, 0)
0
For each of the lines Ll and L2
60
"
vertical rise
:. GradIent = h"
onzontal shift
1
=O.20T
X
-2
--4
i
i
\
(a) Ll passes through the points (0, -4) and
(10, -2)
"
vertical rise
:. Gra dIent=-----horizontal shift
- 3
-
1
=-3
(b) LI equation using point (10, - 2)
(Y-YI) =m (X-Xl)
1
(Y- - 2) =5 (x-10)
5(y+ 2) =x-lO
5y+ lO=x-lO
5y=x-20
L2 equation using point (4, 0)
(Y-YI)=m (X-Xl)
(y-O) = - 3 (x-4)
y= -3x+ 12
(c) Where lines cross x = 5, Y = - 3
Check
5y=x-20
LHS =5 (-3) = -15, RHS=5-20= -15
Matches.
Y = -3x+ 12
LHS = - 3, RHS = - 3(5) + 12
=-15+12=-3
Matches.
Topic
3
Consumer Arithmetic and Corn utation
10 ma rks out of 90 i~ctiO~
This question is usually found at the very
beginning of the paper as Question 1. Even
though it has been placed there because it is
considered to be the easiest question, many
students get confused because some of these
ideas have not been properly revised since they
were studied in first form and even at Primary
school!
Care must be taken when using a calculator in
this question as words and phrases such as
'exact value', may mean that the rounded value
given on a calculator may be insufficient.
A good rule of thumb is that 'a fraction is more
accurate than a decimal' so that 2+ is an accurate
value. 2.3 is not the same thing, nor is 2.33, nor
even 2.333, nor even 2.33333333!
• I must use common denominators in this
question, but first I must make the fractions
improper.
• A question like this needs careful calculation.
• I can check my work with a calculator,
especially if I have a fraction key. But I
must work by hand first and show my
w orking to get full marks for the
question.
Framework solution
1
1
1
2 - -3 - --;-14
8
2
=2
~ - (3 ~ --;-1 ~)
Working
(3~ - ;- 1~) = 2:--;-~
Now would be a good time to go back through
all those old textbooks that you used in first,
second and third forms, that contain examples
like these.
25 2 25
= - x -= 8
3 12
TYPICAL QUESTION 1
Calculate the exact value of 2l - 3 l --;- II
4
8
2
Specific objective being tested
Perform any of the four basic operations with
rational numbers: whole numbers, fractions and
decimals.
To think about
• The order of the operations (BOMDAS),
divide before subtract.
4
12
12
25
2
1
- =-= 12 12 6
TYPICAL QUESTION 2
Calculate the exact value of (0.42)2- 0.4 x 0.3
Specific objective being tested
Perform any of the four basic operations with
rational numbers: whole numbers, fractions and
decimals.
Topic 3 Consumer Arithmetic and Computation 61
To think about
•
•
Again the order of operations is important
BOMDAS, bracket, then multiply, then
subtract in this one.
If the question does not prohibit calculator
use, then I can use it so long as I write down
each single step.
•
When I multiply small numbers, the number
of decimal places in the answer is the sum of
the numbers of decimal places in the two
factors.
Framework solution
(0.42)2- 0.4 x 0.3
= (0.42)2- (0.4 x 0.3)
=0.1764-0.12
= 0.0564
NB Because an exact value is asked for, no
rounding should take place at all.
TYPICAL QUESTION 3
224 + (71..2 x 2)
5
Calculate the exact value of
51.. _ 31..
2
4
Write your answer (a) correct to 1 decimal
place, (b) to the nearest whole number.
Specific objectives being tested
Perform any of the four basic operations with
rational numbers: whole numbers, fractions and
decimals.
Approximate a value to a given number of
significant figures and express any decimal to a
given number of decimal places.
To think about
• This one has so many different operations.
It is easier to work the numerator and
denominator separately.
• Once I have an answer I must look back at
(a) and (b), but first I will work it exactly.
Framework solution
Remember
0.42
x 0.42
1680
84 +
0.1764
Answer has four places of decimals, because
there are two places in each factor.
0.4
x 0.3
0. 12
Two places of decimals (1 + 1 = 2).
0.1764
0.12
0.0564
Align decimal points carefully!
62
Revision Maths
3 (1 2)
Numerator = 2- + 7- x 4
2 5
=2i+(li
x2)
4
2
5
3
=2 - +3
4
3
= 54
Denominator = 5.1- 3.1
2
4
11
13
-- 2
4
13
22
4
4
9
4
3 9 23 9 23 4
23
Answer = 5 - -;- - = - -;- - = - x - = - 444449
9
(exact answer)
(a)
2:
= 2.5555 = 2.6 (correct to 1 decimal
place) (nearer to 2.6 than 2.5)
(b) 2.5555 = 3 (to the nearest whole number)
(nearer to 3 than 2)
0.0
11)0.429
-0
4
11 into 4 tenths goes 0 tenths, remainder 4.
Tip
In a question requiring several steps like this
one, the arrangement of the different parts on
the page and in your mind are critically
important. Make (brief) notes to yourself in the
working so that you remember which part of
the question it is that you are doing.
TYPICAL QUESTION 4
Calculate 0.429 --;- 11
(a) exactly
(b) correct to two decimal places.
Specific objectives being tested
Perform any of the four basic operations with
rational numbers: whole numbers, fractions and
decimals.
Approximate a value to a given number of
significant figures and express any decimal to a
given number of decimal places.
To think about
• This is a calculation question again. I know
that I must show all the working.
• I can use the calculator to check. I could even
use it first to make sure I don't make a
mistake in the working. But I know now that
to just use the calculator and write down the
answer will lose me some marks.
Framework solution
O.
11)0.429
11 into 0 goes 0 times.
0.03
11)0.429
-0
42
-33
9
11 into 42 hundredths goes 3 hundredths,
remainder 9.
0.039
11)0.429
-0
42
-33
99
-99
o
11 into 99 thousandths goes 9 thousandths
exactly.
(a) 0.039
(b) 0.03 19
Two decimal places are 0 and 3.
Because 9 is more than 5 the 3 is
rounded up to 4.
0.04 is the answer.
TYPICAL QUESTION 5
Calculate 9.64 x 6.2
(a) exactly
(b) correct to two decimal places
(c) correct to three significant figures
(d) in standard form.
Specific objectives being tested
Perform any of the four basic operations with
rational numbers: whole numbers, fractions and
decimals.
Topic 3 Consumer Arithmetic and Computation
63
Approximate a value to a given number of
significant figures and express any decimal to a
given number of decimal places.
Use calculators, logarithm tables, tables of
squares, square roots and reciprocals to
facilitate computation.
Write any number in standard form (scientific
notation).
To think about
To think about
• Another careful calculation required.
• I must take care with the decimal point.
9 x 6 = 54, so the answer should be more
than 54, but less than 10 x 7 = 70.
• Calculator can check once more.
Framework soLution
32
964
X
62
5784
1928
59768
2 dec. places
1 dec. place
but 3 dee. places, therefore:
• I never did like square roots.
• The only way I can calculate this is to use a
calculator.
• I think that the answer should be less than 2
because the square root of 4 is 2 and 3.009 is
less than 4.
• I am expecting 'one point something'.
Framework soLution
Press the" key. Enter 3.009 on the calculator.
This gives l.7346469 (or similar, some
calculators have a longer or shorter display). Is
this output reasonable? Yes.
l.73146469
= 1.73
Three sig. figs. are 1, 7 and 3.
Round down because 4 is less
than 5.
(a) 59.768
(b) 59.761 8 Two decimal places are 7 and 6.
=59.77 Round up because 8 is more
than 5.
(c) 59.7168
=59.8
Three sig. figures are 5, 9 and 7.
Round up because 6 is more
than 5.
(d) 59.768 = 5.9768 x 101
A number between 1 and 10, times a
power of ten.
TYPICAL QUESTION 6
Evaluate "3.009 giving your answer correct to 3
significant figures.
Specific objectives being tested
Approximate a value to a given number of
significant figures and express any decimal to a
given number of decimal places.
64
Revision Maths
Notes
In a question like this one I have often seen
'successive' rounding, where a student argues as
follows. 'Because of the 6 (in the fourth decimal
place) round up the 4 to a 5, then because it is
now a 5 round the 3 up to a 4, giving l.74.' This
is incorrect. 4 is less than 5,46 is less than 50.
Only one digit needs to be considered when
rounding to a required number of significant
figures or decimal places.
Although some students are still taught to use
logarithm tables, tables of squares, ete., it is
desirable and expected that students sitting an
examination like CXC (CSEC), should have a
suitable calculator. The calculator saves valuable
examination time.
TYPICAL QUESTION 7
A length of rope is divided into three pieces in
the ratio 2 : 5 : 6. The largest piece is 30 cm long.
How long is the rope altogether?
Specific objective being tested
Solve problems involving ratios and rates.
To think about
•
•
•
A ratio question.
The largest piece is the '6' piece, the third
piece in order oflisting in the ratio.
The question asks for the total length of the
rope, so the 'shares' into which the ratio is
divided will have to be added together.
TYPICAL QUESTION 8
The simple interest on $850.00 invested for
2 years and 6 months is $255. Find the rate of
interest in effect.
Specific objective being tested
Use the simple interest formula (or otherwise)
to calculate simple interest, principal, time, rate
or amount.
Note
To think about
There are several ways to work this problem.
Use the one that you are most comfortable with.
•
A simple interest question - recall the
formula for simple interest.
• 1= PRT
Framework solution
100
(Interest, Principal, Rate, Time in years)
Method 1.
6 shares measure 30 cm.
30
1 share measures - = 5 cm
6
2 + 5 + 6 = 13 shares
•
Here we want to find R, so I need to change
the formula around, either at the start or at
the end.
Rope measures 13 x 5 = 65 cm
Framework solution
Method 1. Transpose formula at start
Method 2.
1= PRT
2:5 : 6
Swap sides.
100
If I want the 6 share to become 30 cm, I must
multiply by 5.
2: 5 : 6 becomes 10 : 25 : 30
PRT =1
100
PRT= 100 I
Multiply through by 100.
Divide through by PT.
Rope measures 10+ 25 + 30 = 65 cm
Method 3.
2 + 5 + 6 = 13 shares
6: 30 = 1: 5 [ -;- 6] = 13 : 65 [ x 13]
Rope measures 65 cm
R= 1001
PT
100 x 255
R
850 x 2.5
(6 months is 0.5 of a year)
R = 12% per annum
Method 4.
6 shares measure 30 cm.
2 shares measure
+of that 10 cm
=
Method 2. Wait till later on
I=PRT
100
5 shares measure 2} times that = 25 cm
Rope measures 30 + 10 + 25 = 65 cm
255 = 850 x R x 2.5
100
Topic 3 Consum er Arithmetic and Computation
65
255 = R x 2125
100
(850 x 2. 5 = 2125)
255 =R x 21.25
(2125-7100= 21.25)
R x 21.25 = 255
Swap sides.
255
R = - - = 12% per annum
21.25
Framework soluti on
(a) Hire purchase price = 5240
Deposit
= 800
4440
Total to be paid by instalment = $4440
Each instalment is $185.
.
4440
:. Number of mstalments =185
Tip
=24
In Method 1, using the calculator to evaluate the
.
100 x 255 h
1
expressiOn
850 x 2.5 ,t ere are severa ways to go
wrong. My advice - press 100, x, 255, =, -7-,
850, = , -7-, 2.5, = (in that order).
(b) (Read again - about three times.)
Hire purchase price = 5240
Cash price
= 4040 1200 ('the extra cost')
TYPICAL QUESTION 9
. d percentage 1S. 4040
1200 100
ReqUlre
x
The hire purchase price for a television set is
$5240. A deposit of $800 is made and the rest is
paid in equal instalments of $185.
('as a percentage' means x 100,
'of the cash price' means over 4040)
=29.7%
(use calculator)
(a) Calculate the number of monthly
instalments that must be paid.
TYPICAL QUESTION 10
(b) If the cash price is $4040, express, as a
percentage of the cash price, the extra cost of
buying on HP.
Mr Brown sells paper to make a profit of 20%.
Calculate
Specific objective being tested
(a) the price a customer will pay for a ream of
paper that cost Mr Brown $80
(b) the price Mr Brown paid for a ream of paper
that he sold for $198.
Solve problems involving payments by
instalments as in the case of hire purchase,
mortgages, ete. in simple cases.
Specific objectives being tested
To think about
Calculate marked price when cost price and
percentage profit, loss or discount are given.
• Hire purchase (HP) is a deposit, followed by
monthly payments (instalments) .
• I will read part (b) again after I have done
part (a).
• Usually instalments are paid for 6, 12, 18, 24
or 36 months.
• 'as a percentage', 'of the cash price', 'the extra
cost'. I think I've got it.
66
(use calculator)
Revision Maths
Calculate cost price when selling price and
percentage profit or loss are given.
To think about
• Buying and selling for a profit. I need to label
the prices carefully, and know for each
percentage, what dollar figure it is a
percentage of.
• These are the two types of question my
teacher told me about - one is easy, the
other harder. I think I have to use an X in
one of them.
• Let me re-read the question once more. First
(a), then (b).
(b) Cost price (to Mr Brown) = unknown = X
(say) Selling price = $198
Percentage profit is 20% of the cost price.
SP = CP + ( ~ x CP)
100
(a) Cost price (to Mr Brown) = $80
Selling price is to be found.
SP = CP + (~ x CP )
100
Term in bracket
represents pro f·it.
SP=80+(~
x 80 )=80+ 16=$96
100
formula as before.
198 = X + (~ x X) = X + O.2X = l.2X
100
:. l.2X=198
Framework solution
Percentage profit is 20% of the cost price.
Note: the same
X=198=$165
l.2
Tip
There are several variations to this method, but
my strong advice is to avoid methods which
require the memorization of several different,
difficult formulae.
Topic 3
Consumer Arithmetic and Computation
67
Topic
4
( Statistics and Probabi!!!~"
_
10 marks out of 90 in Section r)
The question on Statistics is usually found at
Question 7 or 8 of Section 1. It often involves a
graph, which should not be rushed, as drawing a
graph properly and accurately can obtain
several marks. There are a limited number
of types of question. It is easy to identify the
main types.
Students tend to reach this question just as they
are beginning to be concerned about the time
remaining in the exam. It can (if you are not
properly prepared) consume a lot of time. That
is why practice is needed in drawing graphs.
Deciding where to put axes, labelling axes,
plotting points correctly (using a small, neat 'x').
Drawing a curve, where required, smoothly, but
quickly. All of this should be almost automatic
by the time such a question is encountered in
the exam room.
Practise your graphs. Go and get a pack of 100
graph sheets and use them all!
Ages (x)
Frequency (f)
Cumulative
frequency
1-10
11-20
21-30
31-40
41-50
51-60
61-70
71-80
6
12
14
21
17
15
11
4
6
18
(c) Estimate from your graph (i) the lower
quartile, (ii) the upper quartile, (iii) the
median value.
(d) Hence, state the interquartile range.
(e) IfDr Vaughan's oldest patients were all
given a leaflet about 'influenza shots' and he
gave out 22 such leaflets in this particular
week, estimate the age of the youngest
patients among these who received a leaflet.
TYPICAL QUESTION 1
CumuLative Frequency Curve
The table gives the ages of 100 patients
Dr Vaughan saw in a particular week.
(a) Copy and complete the table.
(b) Plot a cumulative frequency curve (ogive) to
show this information using a scale of 2 cm
to represent 10 years on the horizontal axis
and 2 cm to represent 10 patients on the
vertical axis.
68
Revision
Maths
____
__
~JY
Specific objectives being tested
Construct a cumulative frequency table for a '
given set of data.
Determine from the cumulative frequency table
the proportion and/or percentage of the sample
above or below a given value.
Draw and use a cumulative frequency curve
(ogive) .
Determine the range, interquartile and
semi-interquartile ranges for a set of data.
• A cumulative frequency curve usually looks
like a slim'S'.
• Let me make sure the scales are correct
(even if the graph paper has to be turned
around with the shorter side vertical), the
points are clear, small x's are used, the
line is joined smoothly and the axes are
labelled.
• The quartiles are quarters, lower quartile of
the way up, median way up, upp er quartile
of the way up.
To think about
• 'Cumulative' frequency is like the word
'accumulate', meaning 'add up sll;ccessively'.
I can see the pattern from the first two
values 6 and 18 anyway.
• There are 100 patients, so the last number in
the cumulative frequency column must be
100.
t
i
100
To 22
pat en..,.
(01 est)
90
±
/
80
/
V
v ...-
/
/
/
70
c:
t!..,en 60
/
c:
Q)
'p
'"0.
'0
!....
Q)
I
/
/
SO
.!l
E
:J
/
Z
40
30
/
20
I
/
/
V
/
/
10
o /
0
V
Ag of y )ung st
p: t ien give
ale f1et
V
10
20
r
,/
30
Lower
quartile
/40
SO \
60
70
80
Age
Median
Upper
quartile
( Topic 4 Statistics and Probability 69
•
•
I will read parts (d) and (e) again later.
This is a long question. It will take time, but
should get me quite a few marks.
(a)
1-10
11-20
21-30
31-40
41-50
51-60
61-70
71- 80
Frequency (f)
6
12
14
21
17
15
11
4
Cumulative
frequency
6
18
32
53
70
85
96
70
Revision Maths
bi
(d) The interquartile range is 53 - 26 = 27.
(e) The age at which leaflets are first distributed
is 55 years.
Note
If your graph gave a value 1 more or 1 less than
those above, this would be considered correct
by an examiner, but more than lout would be
incorrect.
100
(check 100 total)
(b) See graph on page 69.
(c) (i) The lower quartile is 26.
(ii) The upper quartile is 53.
(iii) The median is 39.
(Values are read from the horizontal scale
looking at the points on the curve ~ and "2
way up the vertical scale.)
±,
Framework solution
Ages (x)
Answers to (c), (d) and (e) are all from the graph.
Please note that on the graph the points are
plotted at the upper boundary values. This is
critical. In other words, 6 is plotted at 10.5 (not
10), 18 at 20.5, 32 at 30.5, etc.
TYPICAL QUESTION 2
Line Graph
(c) Given that the policy was expected to gain
value increasingly year by year, in which
year might Mr Ali have been disappointed in
its performance? Why?
Mr Ali has an insurance policy. The value of the
policy is shown below over a five-year period
(as at December 31 of that year).
Year
Value (dollars)
1990
1991
1992
1993
1994
1995
280
300
340
390
460
520
(d) Predict the value of the policy at the end of
1996. Give a reason for your estimate.
Specific objective being tested
Draw and use pie charts, bar charts, line
graphs, histograms and frequency polygons.
To think about
.J
(a) On graph paper, using a scale of 2 cm to
represent 1 year on the horizontal axis, and
2 cm to represent $50 on the vertical axis,
draw a line graph to represent this data.
•
•
A line graph is not hard to draw.
Make sure the scale is right, use neat x's, join
the points with a ruler, in this case.
• When reading from the graph, I must be
certain what is being asked for, 'gain
most in value', 'disappointed in its
performance', 'predict', 'estimate',
'give a reason'.
From your graph:
(b) State the year during which the policy
gained most in value.
550
500
V
v
/
450
V
>..
.!:!
/
0Q.
(; 400
Q)
::J
c;;
>
/
350
V
/
//
300
V
V
~
250
f>
0
1990
1991
1992
1993
1994
1995
Year
Statistics and Probability
71
Specific objectives being tested
Framework soLution
Draw and use pie charts, bar charts, line graphs,
(a) See graph on page 71.
(b) The policy gained most in value in 1994
(steepest slope on graph).
(c) He may have been disappointed in
1995, because it gained less than in the
previous year, the first time this had
happened.
(d) [$570 to $610 very likely, the reason you
give depends on your answer.]
If you answer $570. Reason - increases
appear to have peaked and are dropping
off.
If you answer $590. Reason - expect an
increase a little more than the previous year
($60).
If you answer $610. Reason - 1995 an
unusual year, expect increases to resume
upward trend.
TYPICAL QUESTION 3
Histogram
Forty runners completed a cross country race.
The times taken are shown in the table.
Time (minutes)
10-14
15-19
20-24
25-29
30-34
Frequency
15
20
2
2
1
(a) State the modal class.
(b) Calculate the mean number of minutes the
athletes took for this race.
(c) Draw a histogram to represent the data.
(d) Give a reason why a histogram is more
suitable for this data than a bar chart.
(e) Calculate the probability that a runner,
selected at random, took more than 19
minutes to complete the course.
72
Revision Maths
histograms and frequency polygons.
Determine mean, median and mode for a set
of data.
Determine experimental and theoretical
probabilities of simple events.
To think about
• What can I remember about drawing a
histogram? Bars must touch, label the
cracks not the bars, edges of bars represent
boundary values, areas represent
frequencies, if widths are same then heights
represent frequencies.
• Mean value is (sum of values)/(number of
values).
• Mode is the most frequent value or class.
Framework soLution
(a) The modal class is 15-19. It has the highest
frequency (20).
(b) Using the mid-points of each class (12, 17,
22, 27, 32),
(12 x 15) + (17 x 20) + (22 x 2) + (27 x 2) + (32 x 1)
M ean= ~--~~--~~~~--~~~
40
180+ 340+44+ 54+ 32
40
= 16.25 minutes
650
40
(c)
I
20
'"a>
L-
c
c
:>
e 15
L-
L-
a>
.0
E
:>
..s 10
(Jc
a>
:>
0a>
L-
u..
5
o0
o
v 9.5
14.5
19.5
24.5
29.5
34.5
Time (minutes)
(d) The data is continuous (units of time), rather
than discrete, so a histogram is preferred.
(a) Calculate the mean score obtained by these
students.
(e) Probability that someone took more than 19
minutes to complete the course
(b) Estimate the probability that a student
selected at random would score 19 marks or
less on this test .
. 2+2+1 - 5 -1
IS
40
40
8
Tips
Please note how the horizontal axis is broken
with a special symbol between 0 and 9.5,
otherwise space would be wasted. Note also how
the horizontal axis is labelled, with the boundary
values on the bars, so that the plotting is precise.
(c) Using a scale of 1 cm to 1 unit on the
frequency axis and 2 cm to represent 5 units
on the score axis, use graph paper and draw
a frequency polygon to represent the
distribution shown.
Specific objectives being tested
Draw and use pie charts, bar charts, line graphs,
histograms and frequency polygons.
TYPICAL QUESTION 4
Frequency PoLygon
Determine mean, median and mode for a set of
data.
The table shows the distribution of the scores
of 50 students on a Geography test.
Determine experimental and theoretical
probabilities of simple events.
To think about
Score
Frequency
10-14
15-19
20-24
25-29
30-34
2
6
20
15
7
• The scores are grouped 10-14, that is 10, 11,
12, 13, 14, so the mid-pOint is 12.
• Mean score from a table like this is estimated
using the mid-points.
• Probability of 19 marks or less means the
probability of 10-19 marks from the table.
• Frequency polygon must be a closed polygon
otherwise it will be more like a line graph.
(
Topic 4 Statistics and Probability
73
(b) Probability (19 or less) = 2 ~
Framework solution
(a) Mean score = ~
Jd
6
8
50
=-=0.16
where fare the frequencies and d the
mid-points
_ ([2 x l2] + [6 x 17] + [20 x 22] + [15 x 27] + [7 x 32])
(2 + 6 + 20 + 15 + 7)
_ (24 + 102 + 440 + 405 + 224)
50
1195
50
= - - = 23.9 ~ 24 marks
(c)
20
11\
I\
18
1 \
I
I
16
~
1\
\"t
14
1\
cQ)
."
~
~
.D
E
I
I
:l
.s
?J 10
cQ)
:l
<T
~
LL.
\
\
1
1
12
Q)
1\
\
\
I1
8
J
6
/
\
\
1
4
\
/
!I
2
1\
\
\
/
V
o
o
5
10
15
20
Score
74
Revision Maths
)
25
30
35
40
TYPICAL QUESTION 5
Bar Chart
(a) Draw a bar chart to represent this
information. Use graph paper and a scale of
1 cm to 1 cm of rain on the horizontal axis
and 2 cm to 1 day on the vertical axis.
The table shows the rainfall in Port of Spain in a
month of 30 days.
Centimetres
of rain
Frequency
(days)
0-2
3-5
6-8
9-11
12-14
15-17
10
8
5
5
0
2
10
(b) (i) State the modal class.
(ii) Estimate the mean number of
centimetres of rain per day.
(iii) Estimate the probability that on the next
day following this month there are 6 cm
of rain or more.
I
9
8
r---r----
7
'"
;0-
r--
"'0
r--
..Q) 6
..0
E
:J
S
r--
~ 5
r--
Q)
:J
er
~
r----
I-
r----
r----
1-
r----
r--
f-----
r--
r--
-
r--
r----
-
r----
r----
-
r----
r--
-
r--
f-----
-
r----
I-
-
r----
u..
4
3
2
o
0-2
3-5
6-8
9-11
12- 14
15-17
Rain (cm)
Topic 4 Statistics and Probability
75
Specific objectives being tested
Draw and use pie charts, bar charts, line
graphs, histograms and frequency polygons.
Determine mean, median and mode for a set of
data.
Determine experimental and theoretical
probabilities of simple events.
To think about
•
•
•
•
•
•
(a) Draw an accurate pie chart to represent this
information.
(b) Explain why a pie chart is the preferred type
of graph to illustrate information of this
type.
(c) What percentage of the total expenditure
was made on Education?
Specific objective being tested
Rainfall over 30 days. Check
10+8+ 5+ 5+0+2=30. OK.
Groups of data: 0-2 is 0, 1, 2, so the
mid-point is 1.
A bar chart is to be drawn. The bars must be
labelled. Heights represent frequencies.
Mode = most frequently occuring class.
Mean and probability are as in the last
question.
6 cm or more means 6-17 cm here.
Framework solution
(a) See graph on page 75.
(b) (i) Largest frequency is 10.
:. Modal class is 0-2
(ii) Mean score = ~
Lf
= ([10 x 1] + [8 x 4]+ [5 x 7]+ [5 x 10] + [O x 13] +[2 x 16])
30
Draw and use pie charts, bar charts, line graphs,
histograms and frequency polygons.
To think about
0
• There are 360 in a full circle.
• The expenditures add up to
(8 + 5 + 4 + 10+ 13) = 40 million dollars.
• Each million dollars takes up
= 9 of the
whole.
• Use a protractor to measure the angles
accurately.
• 'Explain why a pie chart is preferred .. .' over
a bar chart, line graph, etc., presumably.
• The percentage is the fraction x 100.
3:
Framework solution
(a)
= (10+32+35+50+32) = 159 -53
30
30 - . cm
(iii) Probability (6 cm or more)
5+5+0+2 12
30
=30=0.4
TYPICAL QUESTION 6
Pie Chart
The Annual National Expenditure of Grenbados
is as follows (in millions of dollars):
Education
Health
Security
Works
Other
76
8
5
4
10
13
Revision Maths
Other
0
(b) Because a pie chart shows the whole
dearly, as well as the fraction taken up by
each sector. For a different whole
(expenditure) in a succeeding year, the
comparable fraction of the whole spent on
Education or anything else can be seen at a
glance, which is not so if using other chart
types.
(b) p(a pair passes)
(c) Percentage spent on Education
TYPICAL QUESTION 8
Mode, Median, Mean - Raw Data
=~ x 100=20%
= p(left hand passes) x p(right hand passes)
42 42
=- x 60 60
=0.7 x 0.7
=0.49
40
TYPICAL QUESTION 7
Probability
Given the following values:
2,3, 4, 10,4,5, 2,6, 2, 2
A firm making pairs of rubber gloves, tests each
one for flexibility and strength. On average
42 out of 60 pass the test.
(a) State the modal value.
Calculate
(d) Which one of these would be most useful, in
each case, if the values represented
(a) the probability that a glove selected at
random would fail the test
(b) the probability that a pair of gloves selected
at random would both pass the test.
Specific objective being tested
Determine experimental and theoretical
probabilities of simple events.
(b) State the median value.
(c) Calculate the mean value.
(i) the daily income of a company in
thousands of dollars
(ii) the sizes of paintbrushes, in inches, sold
in a hardware store
(iii) the handicap scores of golfers to be
divided equally into an 'A' team and a 'B'
team by ability level.
To think about
Specific objectives being tested
• A pair of gloves is two gloves, a left hand and
a right hand.
• Passing the test and failing the test are the
two possible outcomes of the test.
• p(passing) + p(failing) must = l.
• In part (ii) two separate independent events
are involved. The probabilities of the two
must, therefore, be multiplied.
Determine mean, median and mode for a set of
data.
(a) p(a glove passes) = :~ (given)
=0.3
To think about
• Mode = most frequently occuring.
• Median (6 letters) = middle (6 letters) number
when placed in order.
Framework solution
.
42
:. p(a glove falls) = 1- 60
Determine when it is most appropriate to use
mean, median or mode as the average for a set
of data.
60-42
60
• Mean=
18
60
sum of values
number of values
• 'most useful' - careful thinking here (but I
expect one of each as there are three).
(
Topic 4 Statistics and Probability 77
Framework solution
(a) State the modal class.
(a) The modal value = 2 (it occurs 4 times, more
than any other value)
(b) Estimate the median number of CXCs
obtained.
(b) The median value. First rearrange:
2,2,2,2,3,1 4,4,5,6,10
Half way between 3 and 4 is the middle.
Median = 3.5
(c) Calculate the mean number of CXCs
obtained.
(d) Explain why the mean or median is a better
measure of average for this distribution
rather than the mode.
(c) The mean value
~+2+2+2+3+4+4+5+6+1~
10
Specific objectives being tested
Determine mean, median and mode for a set of
data.
= 40 = 4
10
(d) (i) The daily income of a company. Some
days more, some days less. Key figure
would be mean (regular average) for
planning purposes.
(ii) The sizes of paintbrushes. Need to order
more of the size(s) sold more frequently.
Key figure would be the mode.
(iii) The handicap scores of golfers to be
divided into two teams. Key figure
would be the median. This would divide
the golfers into two groups by ability.
Determine when it is most appropriate to use
mean, median or mode as the average for a set
of data.
To think about
• To get mode, median, mean here, I need to
know exactly what the table is saying.
• 6 students got 0 or 1 CXCs, 9 got 2 or
3 - Yes, I see it now.
• 'Explain' - I must think carefully and give at
least one solid reason.
Framework solution
TYPICAL QUESTION 9
Mode, Median, Mean - Grouped Data
The table shows the number of CXC subjects
obtained by 28 students in form 53.
(a) Modal class = 6-7 (This class has the highest
frequency, that is 10.)
(b) If the 28 students were placed in order the
median would be between the 14th and
15th. I want to identify the 14th and 15th
students. Line them up.
1 2 3 4 5 6
~------~'
78
No. of (XC
subjects obtained
Frequency
0-1
2-3
4-5
6-7
8-9
6
9
1
10
2
Revision Maths
7 8 9 10 11 12 13 14 15
,
obtained 0 or 1
16
17 ...
4 or 5
6 or 7
/
obtained 2 or 3
The 14th and 15th students are right at the
top of the 2 or 3 subject category. Therefore,
it is likely that they both obtained 3 subjects.
[ (3~3) = 3] Therefore, median = 3 CXCs.
Specific objectives being tested
(c) Mean
(6 x Y2) + (9 x 2Y2) + (1 x 4Y2) + (10 x 6Y2) + (2 x 8Y2)
28
(3 + 22.5 + 4.5 + 65 + 17)
28
112
28
=4CXCs
TYPICAL QUESTION 10
CumuLative Frequency Curve
(A second example is given because this is the
most frequently asked type of question.)
The table shows the lengths of leather belts sold
in a shop.
40-49
50-59
60-69
70-79
80-89
90-99
Determine from the cumulative frequency table
the proportion and/or percentage of the sample
above or below a given value.
Draw and use a cumulative frequency curve
(ogive).
(d) The mode is misleading because this
distribution has a trough in the middle and
almost two modes (2- 3 and 6-7) which
counterbalance each other.
Length
(cm)
Construct a cumulative frequency table for a
given set of data.
f
"Lf
(Number of
belts)
(Cumulative
frequency)
3
6
3
10
14
5
2
(a) Copy and complete the table.
(b) Draw a cumulative frequency curve on
graph paper using a scale of 2 cm to
10 cm on the horizontal axis and 2 cm to
5 belts on the vertical axis.
Determine the range, interquartile and semiinterquartile ranges for a set of data.
To think about
• 'Cumulative' means 'add up succeSSively'.
The curve looks like a slim '5'.
• Let me make sure the scales are correct (even
if the graph paper has to be turned around
with the shorter side vertical), the points are
clear, small x's are used, the line is joined
smoothly and the axes are labelled.
• The quartiles are quarters, lower quartile
of the way up, upper quartile of the
way up.
i
+
Framework solution
(a) The required values are 9, 19, 33, 38 and 40.
(b) See graph on next page.
(c) (i) Lower quartile = 61
(ii) Upper quartile = 77
(d) Semi-interquartile range
(77-61) 16 8
2
2
(e) Line on graph meets vertical axis at 27.5, i.e.
27 below this and 13 above. Probability of a
belt being longer than 75 cm = ~ = 0.325.
(c) Using your graph, estimate (i) the lower
quartile, (ii) the upper quartile.
(d) Calculate the semi-interquartile range.
(e) If a belt is selected at random, what is the
probability of it being longer than 75 cm?
(
Topic 4
Stat~_tics !~~" Probability
79
(b)
40
->/
/
V
35
V
/'
'j
/
30
I1
/
25
1
....VI
Qj
.D
(;
....
C1I
.D
I1
1
20
E
:::J
.s
"-
/
H
15
V
/
10
I
5
~
80
Revision Maths
V
40
V
)v
50
V
60 \
7 0 ! 80
Lower
Upper
quartile
quartile
90
100
Length (cm)
Topic
5
15 marks out of 90 in Section I
------_.----------------
Algebra (from the Arabic al-jabr meaning
'breaking apart to heal', like setting a broken
limb), is the Mathematics that deals with
variables and unknowns. As such, it tends to be
feared and disliked by students, but it is one of
the most powerful ways of speaking
mathematically. A good grasp of Algebra
is critical for the scientist, social scientist
or information technologist, who wishes
to do well in future studies and
employment.
Algebra is the big brother of Arithmetic. If, at
times, Algebra looks daunting, turn to the little
brother, Arithmetic, put in some actual
numbers and seek to judge what the Algebra is
doing.
The same rules apply to Algebra as apply in
Arithmetic. In simplifying expressions or
solving equations, great care must be taken with
negative signs, keeping equations balanced and
using brackets correctly. The layout of work in
Algebra is even m.ore critical than in most
other areas of the syllabus. Neat figure-work,
ensuring that fraction bars sit under all that
makes up the numerator, and generally working
tidily, is vital.
Usually an Algebra question appears as
Question 2 on the paper, but CXC have voiced
their intention to put more weight on this
topic as it is generally the worst done topic
on the paper. Students who fare badly
on Algebra questions are almost always
those who can barely read their own
writing!
TYPICAL QUESTION 1
Given that a = 3, b = - 1 and c = - 2, calculate
the value of
ab - c2
a+c
Specific objective being tested
Substitute numerals for algebraic symbols in
simple algebraic expressions.
To think about
• My teacher is always telling me to bracket
the negative values to make sure to stop 'sign
errors'.
Framework solution
ab - c2
a+c
--=
3( -1) - (- 2)2
(- 3) - (+ 4)
=
=- 7
3 + (-2)
1
Remember
3 x -I = - 3, - 2 x - 2 = + 4
3 + (- 2) = 3 - 2 = 1
(-3)-( + 4) = (-3)+( - 4)= -7
Tip
Try to make sure you know when you are
adding and w hen you are multiplying these
numbers.
Topic 5 Algebra
81
TYPICAL QUESTION 2
TYPICAL QUESTION 3
Factorize
Factorize
(a) 9p2- 3p
(a) 4a 2 - b2
(b) 9p2- 3p - 2
(b) 2m-9n+6mn-3
(c) 9p2- 9s 2
(c) 4h 2-8h
Specific objective being tested
Specific objective being tested
Factorize expressions of the forms: a2 - b2,
a2+ 2ab+b 2, ax+bx+ay+by, ax 2+ bx+c
where a, band c are integers.
To think about
• I need to identify what sort of expression
each one is so that I know how to
handle it.
• (i) is the 'difference of two squares' and
I know a2- b2 becomes (a + b) (a - b).
• (ii) has four terms, so it must be factorized 'in
pairs'. Some rearranging is needed as the first
two terms have no common factor. I know
that ax + bx + ay + by factorizes to become
Factorize expressions of the forms: a2 - b2,
a2+ 2ab+ b2, ax+bx+ay+by, ax 2+bx+c
where a, band c are integers.
To think about
• Again, I need to identify what sort of
expression each one is so that I know how to
handle it.
• (i) is the easiest type, 'take out a common
factor'.
• (ii) has three terms, it's a quadratic. It's like
ax 2 + bx + c and I must split the x term as a
first step.
• (iii) is the 'difference of two squares' and I
know a2 - b2 becomes (a + b) (a - b), but a
common factor has to be taken out first.
(a+b) (x+y).
•
(iii) is the easiest type, 'take out a common
factor'.
Framework solution
(a) 4a 2-b 2=(2a)2-(b)2
=(2a+b)(2a-b)
(b) 2m - 9n + 6mn - 3 = 6mn + 2m - 9n - 3
= 2m(3n + 1) - 3(3n + 1)
=(3n+ 1)(2m-3)
(c) 4h2-8h=4h(h-2)
Framework solution
(a) 9p2- 3p = 3p(3p-l)
(b) 9p2- 3p - 2
I need two numbers that add to - 3 and
multiply to give - 18 (9 x - 2)
Those numbers are - 6 and + 3
=9p2-6p+ 3p-2 = 3p(3p-2) + 1(3p- 2)
= (3p - 2)(3p+ 1)
(c) 9p2- 9s2=9(p2-S2)
= 9(p - s)(P+ s)
Tip
In each case, multiply out the answer and see if
you get the expression in the question. This
check is essential and should not take very long.
Check also that no further simplification of the
answer is possible.
82
Revision Maths
Tips
• Remember, multiply out the answer and see
if you get the expression in the question.
• Factorizing a quadratic expression is not
easy for many students. There are several
variations as to how the middle term is to be
split. Most rely on a student having done a
lot of practice questions. Now is the time to
practice factorizing quadratic expressions.
Do at least so.
TYPICAL QUESTION 5
Simplify as far as possible
1
2 1
-+--3y Sy 3
Specific objective being tested
TYPICAL QUESTION 4
Factorize completely
Simplify fractions of the form aw + bdx where a,
0'
z
b, c, d are integers and w, x,y, z can be integers or
variables.
(a) q3-4q
To think about
(b) 2xyz - 4yz+xz - 2z
Specific objective being tested
Factorize expressions of the forms: a2 - b2,
a2+ 2ab+ b2, ax+ bx + ay+ by, ax 2+ bx+cwhere
• We need a common denominator if we are
to make these three fractions into one.
• It looks a bit unusual with the y's in the
denominator, but they just represent another
number.
a, band c are integers.
Framework soLution
1
To think about
2
Sy
1
3
-+---
• Again, I need to identify what sort of
expression each one is so that I know how to
handle it. The 'completely' means there may
be more than one step in these qu estions.
• (i) is a cubic equation. I don't know any way
to factorize a cubic, but the first question (as
always) that I ask is whether there is a
common factor. Yes there is and then it
looks like 'difference of two squares' again.
• (ii) has four terms, it's factorising by pairs,
but wait - there's another common factor
first.
3y
Lowest common multiple of 3y, Sy and 3 is 15y .
..l.+ 1.-_1 = _ S_+_6_ _ .JL
3y Sy 3 lSy l Sy lSy
S+6-Sy
lSy
11 - 5y
15y
This cannot be Simplified any more. For
example, the y's cannot cancel because y is not a
factor of the entire top line.
Framework soLution
TYPICAL QUESTION 6
(a) q3_ 4q = q(q2 _ 4) = q(q2_ 22)
= q(q - 2)(q + 2)
Solve the equatlOn
(b) 2xyz-4yz +xz-2z=z(2xy - 4y+x - 2)
Specific objective being tested
=z(2y[x-2] + 1[x - 2])
=z(2y+ l)(x- 2)
Note
Do not overlook the necessity of beginning by
taking out a common factor in many of these
questions.
.
2x+l
2+x
-----=0
3
4
Solve linear equations in one unknown.
To think about
•
Get the lowest common multiple in the
denominator, or perhaps not if the righthand side is zero.
Topic 5 Algebra 83
• Yes, the easier way is to move one fraction
across, or multiply by 12 throughout.
Framework soLution
Method 1. Moving one fraction across
2x+ 1 _ 2+x =0
3
4
. 2x+ 1 _ 2+x
..
3 - 4
4(2x + 1) = 3(2+x)
8x+4=6+ 3x
8x- 3x=6-4
5x=2
x=2=o.4
5
Method 2. Multiplying through by 12
2x+1 _ 2+x=0
3
4
:.4(2x+ 1) - 3(2+x) =0
8x+4-6- 3x=0
5x-2=0
5x=2
x=2=O.4
5
TYPICAL QUESTION 7
Solve simultaneously
2x - 4y=16
3x+ y=3
• Because the - 4y in equation (1) gives such a
big positive answer (16), I suspect that y is
negative.
• I must choose as the method of solution
either elimination or substitution. In this
question either is easy. Elimination because
it is easy to get the y coefficients the same.
Substitution because equation (2) has a
single y, so that substitution will not bring in .
any fractions.
Framework soLution
Method 1. Elimination
2x - 4y= 16
3x + y = 3
equation (1)
equation (2)
Multiply equation (2) by 4
12x + 4y = 12 equation (3)
+ 2x-4y=16 equation (1)
14x+ 0=28
Method 2. Substitution
2x - 4y = 16
3x + Y = 3
equation (1)
equation (2)
Rearrange equation (2)
y=3-3x
Substitute this for y in equation (1)
2x-4(3 - 3x) = 16
2x - 12+12x=16
2x
+12x=16+12
14x=28
So, by either method
x=~=2
14
Specific objective being tested
Solve simultaneous linear equations
algebraically.
Substituting back in equation (2)
6+y=3
y=3-6= - 3
Solution is (2, - 3)
To think about
• These are two simultaneous equations. They
have just one solution in x and y. On a graph
they would be two straight lines crossing at
one point.
84
Revision Maths
TYPICAL QUESTION 8
If
ba = bC + 2 express b in terms of a an d c.
Specific objective being tested
Framework soLution
Change the subject of formulae including those
involving roots and powers.
E._1E..= 1
3
4
• There are two different b's which have to be
collected together somehow.
To clear (get rid of) the fractions, we need to
multiply through by the lowest common
multiple of the denominators 3 and 4,
i.e. multiply by 3 x 4 = 12. (Don't forget to
multiply the right-hand side as well.)
Framework soLution
..!lE. _l§.. = 12
To think about
• b is in the denominator - a little awkward.
a
c
b
b
- = -
+ 2 ** Collect terms in b on same
3
*Put 9q on the side
where it can be positive.
side.
Put fractions over common
denominator.
a-c
:. -b-=2
* Multiply both sides by b.
:. a-c= 2b
Swap sides.
:.2b=a-c
:.b= a-c
4p = 12 + 9q
12 + 9q = 4p
Note
This was the way I did it first, but there are at
least two improvements if you can spot them:
(i) At line * you could invert both sides, then
multiply through by (a - c).
(ii) At line ** you could multiply through by b
getting a = c + 2b and being near the answer
already.
3q
If - - - = 1
3
4
Divide through by 9.
q= 4p-12
9
Note
Some teachers would criticize the method
above for using too many steps, but several
small steps often lead to greater accuracy than
fewer larger steps. I did not want to multiply the
equation through by :- 1 at * as this leads to all
sorts of errors.
TYPICAL QUESTION 10
If r =
TYPICAL QUESTION 9
Swap sides.
Subtract 12 from each
side.
9q = 4p - 12
Divide both sides by 2.
2
p
Cancel.
4
4p-9q=12
J~ +
1
express s in terms of r.
Specific objective being tested
express q in terms of p.
Specific objective being tested
Change the subject of formulae including those
involving roots and powers.
To think about
• There is only one 'q' in this equation.
• It has to be 'divorced' from the - 3 and the
4 and made the subject of the formula.
Change the subject of formulae including those
involving roots and powers.
To think about
• There is only one's' in this equation.
• It has to be 'divorced' from the 4, the 3,
the + 1 and the root and isolated as the
subject of the formula.
• The order in which I do the 'divorcing' is
very important.
(
Topic 5 Algebra
85
Framework soLution
r=J~S+1
:.r2= 4s + 1
3
:.r2-1 = 4s
3
3 (r2 - 1) = 4s
4s = 3 (r2 - 1)
Square both sides.
TYPICAL QUESTION 12
IfY 'varies directly as' x and y = 4 when x = 9,
calculate the value of x when y = x-IS.
Subtract 1 from each side.
Specific objectives being tested
Multiply through by 3.
Swap sides.
Divide through by 4.
3{r2 -1)
4
s=-----'-----'-
Represent direct and inverse variations
symbolically.
Perform calculations involving direct and
inverse variations.
To think about
If a * b= ,j(a + 2b) where the positive root is
taken, calculate (1 * 4) * 11.
'varies directly' means that 'x goes up when y
goes up' and 'x goes down when y goes
down'.
• When x doubles, y doubles, when x is halved,
y is halved.
• The equation for direct variation is y = kx
where k is a constant.
Specific objective being tested
Framework soLution
Use symbols to represent binary operations
(other than the four basic ones) and perform
simple computations with them.
y oc x
•
TYPICAL QUESTION 11
Soy=kx
Wheny=4, x=9
To think about
• * is a newly defined operation.
• 'the positive root' - yes, because usually
when you take a square root the answer can
be positive or negative.
• I know I must start by working the one in
the bracket first.
Framework soLution
a*b= ,j(a+2b)
So, work out the bracket first,
1*4= ,j(1 + 2(4)) =~ ={(9) = 3
So, (1 *4)*11 = 3*11 = ,j(3 + 2(11)) = ,j(3 + 22)
= ,j(25) = 5
:.4=k x 9
k=±
9
4x
y=9
4x
Wheny=x-15 then -=x-15
9
Multiply by 9 throughout.
4x=9x-l35
Add 135 to each side.
l35+4x=9x
Subtract 4x from each side.
l35=9x-4x
l35 = 5x
So x=27
86
Revision Maths
TYPICAL QUESTION 13
TYPICAL QUESTION 14
Calculate the exact value of
On a quiz show, each contestant begins with
$170. They are asked 10 questions. For each
correct answer they receive $10, for each
incorrect answer they lose $35. They only
win if they end up with a positive score.
Assuming that a contestant gets x questions
correct, show that any cash winnings must be
divisible by 45, and state how many questions
the contestant must answer correctly in order to
win any money.
1
(a) 8 3
(c)
(:)1
Specific objective being tested
Manipulate algebraic expressions involving
rational indices.
To think about
• I think there are three basic things to
remember about these indices.
One, is that a number in the numerator of
the index means raise to that power
(square, cube, etc.).
Two, is that a number in the denominator of
the index means take that root.
Three, is that a negative sign in the index
means take the reciprocal.
• A reciprocal is one over a number,
e.g. the reciprocal of 3 is
t.
Solve linear equations in one unknown.
Use linear equations to solve word problems.
To think about
• Reading the question, I feel lost.
• Perhaps I should list what is given.
• Because it's an algebra question, I know it
will lead to some sort of equation to solve.
Framework soLution
Framework soLution
I
(a) 83 means the cube root of 8 to the power
1 = the cube root of 8 = 2
(b) (t) - 2 means the reciprocal of(t) squared
= the reciprocal of
Specific objectives being tested
(±) = 4
)
(c) (~) "2 means the cube of the square root of
~4 = the cube of 1.2 ='Q
8
Note
The order in which the root, reciprocal and/or
power are taken does not matter.
Start with $170. lO questions.
Correct answer + $lO. Wrong answer - $35.
x correct means (10 - x) incorrect. (10 altogether)
:. Winnings = 170+ (x x lO) - (lO-x) x 35
= 170+ lOx- 350+ 35x
=45x-180
=45(x-4)
which has a factor of 45 and so must always
be divisible by 45.
For the winnings to be positive x must be at
least 5. (When x=4, winnings =0)
Topic 5 Algebra
87
TYPICAL QUESTION 15
A teacher shared a tin of 320 peanuts between
three students Michael, Nkala and Opal. Opal
received 20 more than Michael. Nkala received
twice as many as Michael. Calculate how many
peanuts Michael received.
Specific objectives being tested
Solve linear equations in one unknown.
Use linear equations to solve word problems.
To think about
• Let's make x the number that Michael
received, so the question becomes - what is
the value of x?
• I expect to get a linear equation in x to solve.
88
Revision Maths ~ )
Framework solution
Michael
x peanuts
Opal
(20 + x) peanuts
2x peanuts
Nkala
x+ (20+x) + 2x= 320
4x+20=320
4x=300
300
x=-=75
4
Michael received 75 peanuts.
Note
In questions like these last two, the first step is
to recognize that the question is an algebra
question. Secondly introduce a variable for the
main unknown quantity (like x). Finally, do not
'force' the algebra, just allow the algebra to take
you where it wishes to go.
• Perhaps you have changed your mind by
now about how difficult algebra really is?
I hope so.
Topic
6
Measurement
10 marks out of 90 in Section I
Measurement questions come in three basic
categories.
Volume of a cylinder = nr2h (a prism with a
circular base)
1. Solids, their surface areas and volumes
2. Plane shapes, including sectors of circles,
perimeters and areas
3. Measurement conversions, average speeds
and scale drawings.
Volume of a pyramid = +Ah where A is area of
In these questions there are a lot of formulae
that need to be learned and recalled and I list the
main ones here.
Circumference of a circle = 2nr
Area of a circle = nr2
e
Arc length of a sector=-- x 2nr
e 360
Area of a sector = 360 x nr2
Perimeter of a square = 4 x side
Area of a square = side 2
Perimeter of a rectangle = 2 (length + width)
Area of a rectangle = length x width
1
Area of a triangle = x base x perpendicular
height
Area of a parallelogram = base x perpendicular
height
Area oCa trapezium
(a + b)h
2 ' where a and b
are the parallel sides and h is the perpendicular
height
Volume of a prism = Ah where A is
cross-sectional area and h is perpendicular
height
base and h is perpendicular height
Volume of a cone =
circular base)
+n ~h (a pyramid with a
Volume of a sphere = 1nr3
Curved surface area of a cylinder = 2nrh
Curved surface area of a cone = nrl where I is the
slant height
Surface area of a sphere = 4nr2
(r in all of the above refers to radius)
TYPICAL QUESTION 1
(a) A cylindrical oil drum of height 1.5 m can
hold 500 litres when full. Calculate the
diameter of the drum.
(b) If the full drum is emptied at a rate of 4 litres
per minute, calculate the height of oil
remaining after 45 minutes.
(c) If this remaining oil was used to fill some
small containers, cuboid in shape, measuring
15 cm by 15 cm by 25 cm, how many such
containers could be completely filled?
(Take n to be 272 in this question.)
Specific objectives being tested
Calculate the surface area and volume of simple
right prisms and pyramids.
Solve problems involving measurements.
(
Topic 6 Measurement
89
To think about
•
320000
A cylinder is a prism with a circular base, like
the figure shown.
h= 22
7 x1060.6
=96cm
(iii) Remaining oil 320 L.
Capacity of cuboid
=15 x 15 x 25cm 3 =5625cm 3 =5.625 L
• I need to recall the formulae for the volume
of a cylinder and the volume of a cuboid
(type of prism, with a rectangular base).
• 1000 cm 3 = 1 litre and 100 cm = 1 m.
• Diameter (d) of a circle = 2 x radius (r) of
circle.
• Remember to use the value of n that has
been given.
Number that can be filled = 320 = 56.88
5.625
So, 56 small containers can be completely
filled.
TYPICAL QUESTION 2
A cone as shown is made from thin cardboard.
The base diameter is 14 cm and the slant height
is 25 cm.
p
Framework solution
(i) Volume of a cylinder = nr2h
:. (500 x 1000) =1l x r2 x 150
7
(Putting everything in centimetre units)
. (500 x 1000) = r2
1l x 150
..
7
Take n to be
r2=1060.6
r= 32.566
d=2r= 65 .133
Diameter~65.1 cm
(ii) 4 L per min. for 45 minutes will lose
4 x 45 = 180 litres
y
22
7'
Calculate
(a) the height of the cone OP
(b) the area of the base
(c) the total surface area of the cone
Oil remaining = 500 - 180 = 320 L
(d) the volume of the cone.
Method 1. By ratios
(e) How tall would a cylinder have to be, of the
same base diameter as this cone, to be of
equal volume?
320
500
h
1.5
h= 320 x 1.5
500
90
x
0.96 m=96cm
Specific objectives being tested
Method 2. By formula
Calculate the surface area and volume of simple
right prisms and pyramids.
320000=1l x r2 x h
7
Solve problems involving measurements.
Revision Maths
To think about
• I need a number of formulae here again, and
need to know which one is which in my
mind.
• Otherwise this question looks fairly
straightforward.
• Slant height is 1, perpendicular height is h.
• Take your time substituting the values into
the formula.
• Take your time cancelling and/or calculating
the final value. Always have an idea of the
value you should get from your calculator,
1 22
1
e.g. 3" x ---:;- x 7 x 7 x 24~3" x 3 x 50 x 24
=50 x 24
=1200~1232
Framework soLution
TYPICAL QUESTION 3
(a)
p
The diagram, not drawn to scale, shows the
frustum of a cone, formed by cutting a small
cone from the top of a larger cone.
Sem
x
24em
In triangle OPX, I have two measurements and
need the third in a right-angled triangle.
-----ISe
XO =radius=..!±= 7 cm
---
2
XP = slant height = 25 cm
The height of the frustum is 24 cm, its top and
bottom radii are 5 cm and 15 cm respectively.
By Pythagoras' theorem
25 2=72+0P2 625=49+0P2
576=OP2
: .OP=24cm
(b) Area of base (circle) = nr2
=~ X7X 7=154cm2
7
(c) Total surface area = nrl + 154
=~ x 7 x 25+ 154=550+ 154=704cm2
Take n= 3.14
Calculate
(a) the height of the small cone which was cut
off to form the frustum
(b) the curved surface area of the frustum
(c) the volume of the frustum
(d) the mass of the frustum, to the nearest
kilogram, ifits density is 2g/cm 3.
7
(cl) Volume=lnr2h=l x ~ x 7 x 7 x 24
3
3
7
=22 x 56= 1232cm3
(e) Volume of cylinder = nr2h = 1232
:.~ x 7 x 7 x h= 1232 and h=8cm
7
Tips
• Take your time in writing d9wn the correct
formula - if the formula is wrong,
everything is wrong.
Specific objectives being tested
Calculate the surface area and volume of simple
right prisms and pyramids.
Use correctly the SI units of measure for area,
volume, mass, etc.
Recognize and solve problems using the
concept of Similarity.
Use Pythagoras' theorem to solve simple
problems (no formal proof required).
(
Topic 6
Measurement
91
To think about
•
•
•
•
•
I am going to need several sketches to see
what is happening in this question.
I know it has something to do with similar
triangles.
The curved surface area is the part on the
outside, excluding the top and the base.
The curved surface area of the frustum is the
difference between the curved surface areas
of the original and missing cones.
The volume of the frustum is the difference
between the volumes of the original and
missing cones.
Framework solution
= (3.14 x 15 x 39) - (3.14 x 5 x 13)
= 1836.9 - 204.1 = 1632.8 ~ 1630 cm 2
(c) Volume = volume oflarge cone (n~h/3)
-volume of small cone (n~h/3)
= [(3.14 x 15 2 x 36) - (3.14 x 52
x 12)]/3 = [25 434- 942]/3 = 8164
~8160cm3
(d) Mass = density x volume
Mass = (2 x 8160) g
8160
=2 x --kg~ 16kg
1000
TYPICAL QUESTION 4
c
24 cm
The diagram, not drawn to
scale, shows a square
inscribed in a circle.
If the length of a side of
the square is 4 cm, and
2
taking n = 27 , calculate
- - - - - - - - - -15 cm- -
(a) The large and small triangles in the diagram
are similar. All corresponding angles are the
same.
h h+24
5
15
15h= 5(h+ 24)
15h= 5h+ 120
lOh=120
h=12cm
(b) By Pythagoras' theorem
(Slant height of small conef = 52+ 122
= 25 + 144 = 169
:. Slant height of small cone = "169 = 13 cm
(Slant height oflarge cone) 2= 15 2+ 36 2
=225 + 1296
=1521
:. Slant height of small cone = "1521 = 39 cm
So, Curved Surface Area of frustum
= CSA oflarge cone (nrl)
-CSA of small cone (nrl)
92
Revision Maths
(a) the radius of the circle
(b) the circumference of the circle
(c) the perimeter of the segment ABC
(d) the area of the segment ABC.
Specific objectives being tested
Calculate the perimeter of a polygon and circle
(circumference), and their combinations.
Calculate the length of an arc of a circle using
angles at the centre whose measures are factors
of 360° (e.g. 15°,45°, 60l
Calculate the areas of regions enclosed by
rectangles, triangles, parallelograms,
trapeziums, circles and their combinations.
Calculate areas of sectors of circles using angles
.
at the centre whose measures are factors of
360° (e.g. 15°,45°, 60l
Use Pythagoras' theorem to solve simple
problems (no formal proof required).
e
1
x area of square
360
4
90
22
1
= - x - X 2.8285 2 - - x (4 x 4)
360
7
4
= 6.286 - 4 = 2.286 ~ 2.29 cm 2
To think about
= - - x ny2 - -
• What a lot of objectives! Let me recall
everything I know about a square - sides
equal, opposite sides parallel, 90 angles.
• I need formulae for area and circumference
of a circle, arc length and sector area, and
also area of a triangle, although some of
them may not really be needed here.
• This is another question to proceed
slowly on.
0
Note
In.an~ question where an a~fro~mate v~lue for
n IS gIven, whether 3.14 or 7 ' thIs value IS only
accurate to 3 figures. Nevertheless, in order not
to round off too soon, values are used having
(at least) 4 figures in all working (5.65 7, 2.8285,
4.445,6.286), rounding to 3 figures only at
the final step in each case. No answer more
accurate than 3 figures should be given (2.83,
17.8, 8.45, 2.29).
Framework solution
(a) Draw in a diagonal of the
square and label point D.
Because angle A is a right
angle, BD is a diameter
(circle theorem - the angle
in a semicircle is 90°).
c
TYPICAL QUESTION 5
By Pythagoras' theorem BD2= 4 + 4
=16+16=32
The trapezium
ACDE shown has
area 100 cm 2•
: . BD=m=5.657 ~ 5.66 cm = diameter
diameter 5.657
:. Radius
If the area of P is
the area of Q,
calculate
2
2
2
2
±
A
= 2.8285 ~ 2.83 cm
(b) Circumference of the circle
= 2nr= 2 x ~ x 2.8285 = 17.78 ~ 17.8 cm
7
(c) Perimeter of segment ABC
= curved part + straight part
=
!
Specific objectives being tested
of circumference + side of the square
Calculate the perimeter of a polygon and circle
(circumference), and their combinations.
1
=- x 17.78+4
4
=4.445+4=8.445~8.45 cm
(d) Angle AOB = 90° because
the diagonals of a square
(of which OA and OB are
parts) intersect at right
angles.
(a) the length BE
(b) the length AB
(c) the perimeter of the trapezium to the nearest
centimetre.
Use Pythagoras' theorem to solve simple
problems (no formal proof required).
D
c
o
B
Area of segment ABC
= area sector OACB-area triangle OAB
To think about
• This problem is to do with areas, but the
areas are given, not to be calculated. I still
need to know the formula for the area of a
trapezium, though.
Topic 6
Measurement
93
• This trapezium is a rectangle (square
possibly?) and a triangle stuck together. It
may be possible to treat them that way.
• Let me mark the sides I am being asked to
find: BE and AB.
• The perimeter is the distance right around all
four sides of the trapezium.
Framework soLution
(a)
1
Area P=- x Area Q
4
But
Total area = 100 cm 2
.. 1
Area P + Area Q = 100
(- x Area Q) + Area Q = 100
4
5
- x Area Q = 100
4 4
Area Q=- x 100 =80 cm2
5
Area P= 100 -80=20cm 2
If Area Q = 80, we know that Q is a rectangle.
.". length x width = area
.". BE x 10=80
.". BE=8cm
(b) If Area P= 20, we know that P is a rightangled triangle.
.". 1 x base x perpendicular height = area
2
1
.". - x AB x 8 = 20
2
.". AB=5cm
(c) Perimeter of trapezium
=AC+CD+DE+EA
We know all of these except EA.
(CD =BE and CB =DE in the rectangle.)
By Pythagoras' theorem
EA 2 = 82 + 5 2
EA 2 =64+25=89
EA = -V89 = 9.434 cm
Perimeter = (5 + 10) + 8 + 10+ 9.434
=42.434 ~ 42 cm (to the nearest cm)
Note
Do not forget to check at the end of a
question like this one and make sure
94
Revision Maths
)
that you have
1. Answered all parts of the question
2. Used the correct units in your answer
3. Used the correct degree of accuracy in
your answer.
It would be unfortunate to throwaway the final
mark(s) by not rounding to the nearest
centimetre at the last step in the question
above.
TYPICAL QUESTION 6
The diagram represents the
sweep of a submarine radar
generator from point A. BC is
an arc of a circle, centred at A.
(a) If the distance AB is 10 km
and the area of ABC is
39.25km 2,
(i) what is the size of L BAC?
(ii) how long is arc BC?
(b) If an enemy ship is 5 km away from the
submarine (direction unknown), what is the
probability that this sweep would reveal its
presence?
Take n= 3.14
Specific objectives being tested
Calculate the length of an arc of a circle using
angles at the centre whose measures are factors
of 360° (e.g. 15°, 45°, 60°).
Calculate areas of sectors of circles using angles
at the centre whose measures are factors of 360°
(e.g. 15°, 45°, 60°).
To think about
• I like this question being about a submarine
and the Navy. Before I get too carried away
by that, however, let's try to identify the
things this question needs.
• It's a sector of a circle, so sector area and arc
length formulae.
Make suitable measurements on maps or scale
drawings and use them to determine distances
and areas and vice versa.
• It's strange that probability comes into this
question, but I'll see what happens when I
reach it. If the enemy ship was to the bottom
left (South-West?) of A it couldn't be seen
from this sweep.
To think about
Framework solution
• How many cm in a km? 100 cm = 1 m ;
1000m=1 km; 100000cm= 1 km.
(a) (i) Area of sector =_0_ x nr2 = 39.25 (given)
o 360 2
:. 360 x 3.14 x 10 = 39.25
• Areas scale up and down by ( 25
:.0
39.25 x 360
314
45° L BAC = 45°
(ii) Length of arc BC = _0_ x 2nr
360
= £ x 2 x 3.14xlO
360
= 7.85 km
(b) If enemy ship was in this sector it would be
seen.
of the whole
This sector represents £
360
circle.
45
1
:. 'Pro b ab'l'
Ilty=-=360 8
TYPICAL QUESTION 7
A map of St. Thomas has a scale of 1 : 25 000.
(a) On the map, the distance from the railway
station to the school is 14 cm. How far apart
are they on the ground in kilometres?
(b) On the ground, the distance from the railway
station to the police station is 2 km. How far
apart are they on the map in centimetres?
(c) If the area of Pondside Wood is 2.4 cm 2 on
the map, what area does it cover on the
ground (i) in km 2, (ii) in hectares?
Specific objectives being tested
Convert units of length, area, volume, capacity,
time and speed within the SI system.
• Distances scale up and down by 25 ~OO
~OO Y-
• How many cm 2 in a km 2?
(100000) 2 cm 2= 1 km 2. Plenty!
• What is this hectare thing again? It's a
struggle to remember. 1 hectare = 10 000 m 2;
100 hectares = 1 km 2.
Framework solution
(a) Actual distance = (14 x 25 000) cm
= (14 x 25 000) km
100000
= 3.5 km
Converting to a bigger unit means you
will have far fewer of them, hence divide by
100000.
(b) Distance on map = 25 ~OO km
2 x 100 000 cm
25000
=8cm
Converting to a smaller unit means you
will have far more of them, hence multiply by
100000.
(c) (i) Actual area = (2.4 x 25 000 2) cm 2
= (2.4 x 25 000 2) km 2
1000002
=O.15km 2
Converting to a bigger unit means you
will have far fewer of them, hence divide by
1000002.
(ii) 0.15km 2=(0.15 x 100) hectares
= 15 hectares
(
Topic 6
Measurement
95
TYPICAL QUESTION 8
A train T} leaves Station Pat 12:25 h and arrives
at Station Q at 15:05 h the same day.
(a) Calculate the time taken for the journey.
(b) Calculate the average speed of the train in
km/h if P and Q are 120 km apart.
(c) At the same time that train T} left P, train T2
left Q travelling to P. If train T2 travelled at an
average speed of 50 km/h, how many
minutes more or less would it take than train
T} to complete its journey?
Specific objective being tested
Solve simple problems involving time, distance
and speed (e.g. timetable extracts such as bus
and airline schedules).
Time = 120 =2.4h=2h 24min
50
(0.1 of an hour is 6 min)
:. Train T2 takes 16 minutes less
(2h 40min -2h 24min)
Note
The average speed formula has three versions.
In an exam you should be able to generate the
unusual ones:
Time taken
d
distance travelled
an
average speed
Distance travelled = average speed
x time taken
from the more familiar:
distance travelled
Average speed
time taken
TYPICAL QUESTION 9
To think about
• The 24 h clock is being used here. Remember
60 minutes make an hour.
distance travelled
• Average speed
time taken
• I must be careful in adding and subtracting
times because they are not decimal 'borrow l ' means borrowing 60 not 10.
A piece of wire, 40 cm long, is bent to form a
circle.
(a) Calculate the radius of the circle.
(b) Calculate the area it encloses.
If the wire is now bent to form a rectangle, twice
as long as it is wide,
(c) calculate the area of this rectangle.
Framework solution
(a)
hours
15
12
min
05
25 means
hours
14
12
2
min
65
25
40
(d) Explain why your answer to (c) is less than
your answer to (b) when the length of wire
remains the same.
(Take 11: = 3.14)
Specific objectives being tested
Time for journey = 2 h 40 min
distance travelled
(b ) Average speed
time taken
120
120 = 45 km/h
2-'J2h 40min
distance travelled
Time taken
average speed
(same formula changed round)
96
Revisi on Maths
Calculate the perimeter of a polygon and circle
(circumference), and their combinations.
Calculate the areas of regions enclosed by
rectangles, triangles, parallelograms,
trapeziums, circles and their combinations.
To think about
• Three 'calculate's and an 'explain'. I don't like
to explain in Maths, although my teacher
said I should only write one or two specific
points in such a situation.
• Formulae I need include area of circle and
rectangle and circumference of circle.
(b) If the map has a scale of 1 : 50000, what area
in square kilometres does the island actually
cover?
Specific objectives being tested
Framework solution
Calculate or estimate the areas of irregularly
shaped figures.
(a) Circumference of a circle = 2nr
Convert units oflength, area, volume, capacity,
time and speed within the SI system.
:.40=2x3.14xr
:.40=6.28 x r
40
:. r=--=6.369~6.37 cm
6.28
(b) Area of a circle = nr2 = 3.14 x 6.369 2
= 127.37 ~ 127 cm 2
(c) L = 2W and Perimeter = 40
:.2L+2W=40
2(2W) + 2W = 40
•
Make suitable measurements on maps or scale
drawings and use them to determine distances
and areas and vice versa.
To think about
4W+2W=40
6W=40
W=61.cm
3
1
L=2W=133"cm
:.Area=L x W=6f x 13t =88%~ 88.9cm2
(d) 88.9 is less than 127, because the closer an
area is to a regular polygon and the more
sides that regular polygon has, the greater its
area, once the perimeter is constant.
The circle can be considered as a regular
polygon of an uncountable number of sides;
the rectangle has only 4 sides and it is far
from regular.
(a) If each square is
1 cm by 1 cm, how
many square
centimetres does
the island cover on
the map?
Framework solution
~
1
'h
1
1 (
-
Y2
~ 1 ~
A 1 Y2l
TYPICAL QUESTION 10
This is a map of the
island ofpardos.
• Counting the squares and estimating the
pieces is the most appropriate way to
estimate the area of this island.
• When counting squares, it is important to
mark the ones that have been counted so
that no square is counted twice. Pieces can
be marked as fractions.
• Again I have to be careful of the conversion
factor when converting areas (everything is
squared!). 100000 cm = 1 km.
~ 1
~
1
~
(
I~
l/
I~
')
l
Number of square centimetres covered
= seven wholes + seven halves
=10.5cm 2
10.5 X (50000)2
Area actually covered
1000002
= 2.625 ~ 2.6 km 2
CTopi~
6
Measurement
97
Note
It is normal to count squares to the nearest half
square. Greater accuracy is not possible.
Similarly the final answer could not be left as
98
Revision Maths
2.625 as this would imply great accuracy when
the value is derived from an estimate. Two
figures suffice.
Topic
7
Geomet
and Tri onomet
20 marks out of 90 in Section I
The questions in Geometry do overlap to some
extent the questions already covered in
Measurement in the areas of co-ordinate
geometry and arc length and sector area. For
that reason only 18 questions are listed which
cover all the relevant areas. A sizeable number
of marks are available for fully correct responses
to questions in this category.
Questions are wide ranging. A construction
question occurs on almost every paper, as well
as a question involving some type of
transformation geometry.
TYPICAL QUESTION 1
Using ruler and compasses only, construct
triangle ABC having LA = 45°, LB = 60° and
AB= 10 cm.
To think about
• This is a construction question. In such a
question I know I should draw a sketch first
so that I can anticipate where the drawing is
going to go - otherwise it might drop off the
edge of the page!
• I can construct 45° by first constructing 90°
and bisecting it.
• I can construct 60° by constructing a small
equilateral triangle.
• The perpendicular bisector divides the line
into two equal parts - that's OK.
• I must remember that it's only in TO that I
must rub out the construction lines - in
Maths I must leave them in, and make them
visible.
Framework solution
Measure and state the length of AC.
Draw a sketch first please. See next page.
Construct the perpendicular bisector of AB to
meetACatM.
AC=8.9 cm (±0.1 cm)
Measure and state the length of MC and the size
of LAMB.
Specific objectives being tested
Use instruments to draw and measure angles
and line segments.
Use instruments (not necessarily restricted to
ruler and compasses) to construct triangles.
Use the properties of rays, perpendiculars,
parallels and angles related to them to draw
accurate geometrical figures.
MC= 2.0 cm (±0.1 cm)
L AMB = 90
0
(
± 1°)
Note
Using a protractor is not allowed in the
construction of the angles in Question 1
(except to check their accuracy). Many
students use a protractor and then put 'fake'
construction lines in afterwards by hand.
Examiners of Mathematics can spot these very
easily and all marks are thereby lost. Please
don't do it.
(
Topic 7 Geometry and Trigonometry 99
c
c
0)
3 arcs establish
60°
+-----0)-2-A..-L---------j-----J......--'-------+---------"I B
CD
I arc establishes
length AB
Total of 6
arcs establish 45°
Y
Wrcs for perpendicular
l i\, bisector
TYPICAL QUESTION 2
Construct a quadrilateral PQRS such that
PQ=5 cm, QR=6 cm,RS=8 cm,
LPQR = 70 and LQRS = 100°.
0
Measure and state
(a) the length of SP
(b) the size of angle PQS.
Specific objectives being tested
Use instruments to draw and measure angles
and line segments.
100
Revision Maths
Use instruments (not necessarily restricted to
ruler and compasses) to construct quadrilaterals.
To think about
• Remember to draw a sketch first.
• 'Construct' is not qualified by any phrase
such as 'using ruler and compasses only' so
any instrument may be used, specifically a
protractor - although the drawing must still
be done as accurately as possible.
• P, Q, Rand S must follow in order (either
clockwise or anticlockwise) around the figure.
• 100° is an obtuse angle, so I must be careful
when using the protractor here.
Specific objective being tested
TYPICAL QUESTION 4
Use the properties of the faces, edges and
vertices of simple solids to draw twodimensional representations of those solids.
A quadrilateral ABeD has angles xo, xo, 2xoand
2xo, in no particular order.
To think about
• A 'net' is a solid shape that has been peeled
apart and flattened out.
• I need to imagine it being folded up again.
The square looks like it might be the base
and then I could fold the triangles upwards
to a point.
• An edge is a line between and joining two
flat surfaces called faces.
• Vertices are (sharp) points where edges (and
faces) meet.
Framework solution
(a)
(a) Calculate the size of each angle.
(b) State the possible types of quadrilateral it
may be and sketch each possibility.
Specific objectives being tested
Recognize the properties of polygons, lines,
angles.
Solve geometric problems using the properties
of polygons and circles.
To think about
• A quadrilateral is a four-sided figure.
• The four angles of a quadrilateral add up
to 360°.
• 'Possible types' means there must be at least
two possibilities.
Framework solution
(a) x+x+ 2x+ 2x= 360
6x= 360
x=60°
(b) Number of edges = 4 on base + 4 from base
to apex=8
Number of faces = 1 square base
+ 4 triangles = 5
Number of vertices = 4 corners of
base + apex = 5
The angles are 60°, 60°, 120° and 120°
(b) Possibility 1: Parallelogram
(c) The figure is a square-based pyramid.
Note
Always try to be very precise when describing
or naming anything. Pyramid here would be
insufficient as there are many varieties of
pyramid, each one having a different shaped
base. The word 'exact' in the question is warning
you of this.
102
Revision Maths
Possibility 2: Trapezium
(it is also a cyclic quadrilateral)
To think about
TYPICAL QUESTION 7
• This is a transformation question - about a
translation and a reflection.
• In a translation, all the points move as if
locked together in a certain direction.
• In a reflection, a shape takes on its mirror
image.
• 'Similarities' and 'differences' must be in
terms of possible changes that can occur
when things are transformed.
Point P is (2. 1) and point Q is (4, 2), pi is (-1,2)
and Q' is (0,0).
(a) Sketch these four points and join the line
segments PQ and pi Q I •
(b) State precisely two possible
transformations which would together map
PQ onto P'Q'.
Specific objectives being tested
Framework soLution
Identify and describe a transformation, given an
object and its image.
(a) (i)
'le _, _
I
Perform successive transformations combining
any two of enlargements, translations, rotations
and reflections.
IV " I
t-
;--!-+-+
To think about
L.
'.
--4 1T
: -2
'
',.': •
~',
I ~
K ~\ ' 2
I' "
j': '
" Z"
+-.
L
f
• The line segments are the same length so
there's no enlargement, its hard to see where
a mirror line would be, so it looks like
translation + rotation.
Framework soLution
(a)
1
(ii) axyz and ax'y' Z ' are both the same
size and are both the same shape (and
both are the same way around [same
orientation]). Their positions are
different.
(b) (i) and (ii) see graph.
(iii)Z" is (5, 1).
(iv) axyz and ax"y" Z" are both the same
size and are both the same shape. Their
orientations are different (they are
different ways around).
y
p,
~
Q
IL
P~
\
1
V
\
\
\
¥'
-I
x
\
\
\
(b) There are several possibilities, including:
a rotation of - 90° about P (dotted line)
followed by a translation of ( - ~ )
or
a translation of (=~)
followed by a rotation of-90° about Q.
104
Revision Maths
)
TYPICAL QUESTION 10
To think about
• Finding the centre of rotation means joining
A to A' and B to B' and drawing the
perpendicular bisectors. Where they meet is
the centre of rotation.
• Even if you're not sure why it works you
can do it, but it seems to be because when a
point is rotated it's always the same distance
from the centre of rotation, no nearer, no
further away.
B
ISm
A~--L------------4C
o
Framework soLution
ABCD represents the plan view of a children's
(a) The centre of rotation is where the
perpendicular bisectors of AA' and BB'
cross, that is (1, - 3)
,
12 "-
-4
-6
V
B'
YYI\
\ / 2 \
\\
~
playground.
Calculate
V
~
-2
\
A
x
~ '\ /'"V2
4
0
>~\
Z f.-:- ~c.:tr. ofroU:io~
--~
,
-4
i
the vector
\~
(= ~ )
14
~ ~ priiir
Q
4
16
106
- ---~
Revision Maths
Use the sine and cosine formulae in the solution
of triangles.
Calculate the area of a triangle given two sides
and the included angle by means of the formula
Area of ~ABC = ab sin C.
1\
\
2
\\
0
To think about
x
A
2
4
,
l--
Use the sine, cosine and tangent ratios in the
solution of right-angled triangles.
t
B
. ("
l1:,ef ' ~ l--
Specific objectives being tested
Use Pythagoras' theorem to solve simple
problems.
~
(b) If AB is rotated about the origin by 90°, the
images are (0, 2) and (-4,1) respectively.
To map (0, 2) into A' (- 2, - 2) and ( - 4,1)
into B' (- 6, - 3) it needs a translation by
,
(a) the distance AC
(b) the distance BC
(c) the angle ABC and hence angle BCD.
(d) If children need an average of 2 m 2 each in
which to play, how many children could this
playground accommodate?
, \ 11
I~ r~
4.6m
IOm
-L
i
-
• There are two different types of triangle
'stuck together' here. One has a right angle,
the other doesn't.
• For the right-angled triangle, I can use trig
ratios and ~ x base x height for the area.
• For the other triangle, I need the sine or
cosine rule and ~ ab sin C for the area.
Framework solution
Specific objectives being tested
(a) vector j
moves from 2 to 6 horizontally = + 4
moves from 2 to 4 vertically = + 2
Perform pre- and post-multiplication of
matrices.
j=
(~)
given point P (a,b) where
(b) vector k
moves from to 3 horizontally = + 3
moves from 6 to 3 vertically = - 3
°
k=(_~)
(c) j -2k=
Associate a position vector oP=
(~) -2(_~) = (~) -(_~) = (-;)
(~) with a
°is the origin (0,0).
To think about
• This is another question that requires some
manipulation of vector algebra. Remember
to multiply rows by columns (Roman
Catholic!).
• The vector which is a position vector must
start from the origin.
• 'why' means a sentence must be written with
a reason.
y
Framework solution
(a)
(:)= (_ ~ -~)(!)
a) _ ( 3(8) + -4(b))
( 6 - -1(8) + 2(b)
The angle that j - 2k makes with the
y-axis = tan - 1 ( ~ ) :::::; 14
a=24-4b and 6= -8+2b
2b=6+8=14
b=7
a=24-4(7)=24 - 28 a=-4
0
~
(b) OA is the position vector, because it begins
at 0, the origin.
TYPICAL QUESTION 4
(a)If(:)=(_~ -~) (~)
TYPICAL QUESTION 5
find the values of a and b.
(b) Which of the vectors shown is a position
vector and why?
If M = (~ _ ~) find the 2 by 2 matrix N such
thatMN=(_!
y
~~)
A
Specific objectives being tested
c
x
Perform pre- and post-multiplication of
matrices.
(
Topic 8
Vectors and Matrices
115
SECTION II
In this section candidates are required to do any two questions from six.
If more than two are attempted, the best two will be graded.
This section is designed to clearly identify the grade 1 students.
It is normal to prefer one or two of the areas in this section over the other(s), even to having a good
idea before going into the examination which category(ies) you will probably attempt to answer
questions in.
Topic
1
Relations, Functions and Gra hs
2 optional questions worth 15 marks each in Section II
Having been tested on both the Multiple Choice
Paper and in Section I, candidates now meet the
hardest questions in this category. The main
topics include:
• Graphs of trigononometric functions and
other, more complex, functions
Completing the square to solve a quadratic
•
Solving simultaneously one linear and one
non-linear equation
12
(c) Using your graph,
(i) Estimate the gradient of the curve when
n
X=-
4
(ii) Solve the equation sin x + tan x = 4 - 6x
• More complex examples of topics to be
found in Section 1.
Before drawing the graphs, look at Appendix II
at the back of this book entitled 'Precision in
Graph Work' (p. 167).
(a) Complete the table below for
j(x) = sin x + tan x, correct to 3 decimal places.
,
f(x)
(0°)
n
12
(15°)
0
0.527
0
'---..~
-
-n
(d) Sketch (do not plot) for the same domain,
the graphs of
(i) j(x)= -sinx-tanx
(ii) j(x) = 2(sin x + tan x)
Specific objectives being tested
Draw the graphs of the circular functions sine,
cosine and tangent of () where () is in radian
measure for - 2n ~ () ~ 2n.
TYPICAL QUESTION 1
x
unit on the j(x) axis, draw the graph ofj(x) for
5n
o ~x~-.
• Inequalities and linear programming
•
(b) Using a scale of 2 cm to represent"!!:"" radians
12
(15°) on the x-axis and 2 cm to represent 1
n
n
5n
12
6
4
3
(3 0°) (45°) (60°) (75°)
-
1.707
.- .. _-_._ ..
_--_.._. ._._.__......._.._......
'\
Estimate the value of the gradient of a curve by
constructing a tangent to the curve at a given
point.
To think about
• This question has angles in radians and
degrees, although I know that if either were
Topic 1 Relations, Functions and Graphs
117
omitted I could calculate them because n
radians is the same as 180°.
I must make sure my calculator is in the
right mode (degrees or radians).
The curves in part (d) are closely related to
the original curve.
To estimate the gradient of the curve, I will
have to draw a tangent to the curve.
To solve the equation, I can draw fix) = 4 - 6x
and see where it crosses the curve.
•
•
•
•
(ii) All fix) values are now twice what they
were before, i.e.
fi'llOl
)
al~x
o
5rr / 12
TYPICAL QUESTION 2
Framework solution
(a)
x
n
12
(0°) (15°)
n
6
(30°)
n
5n
n
4
3
12
(45°) (60°) (75°)
f(x)
0
1.077
1.707 2.598 4.698
-
-
0
0.527
'\
-
"
f(x)5
....
'1
~
<fV
\\
17
\;r.-r\
2
V
\
./
~/
~
V
V
V
IT
""6
rr
rr
T
4
Srr
IT
x
27
(c) (1.) Gra d'lent at -n = 3.2 -0.4
n
~.
4
30° 60°
1.5
90° 120° 150°
0
180° '\
-2.6
"
2
Speci ic objectives being tested
, \
rr
f(x)
2
l,..-:;: j;?", 1
\
rr
0°
(d) So ve the equation cos x = "3 for
(i) 0° ~ x ~ 180° (ii) 180° ~ x ~ 360°
\V
o,/
o
x
(b) u~ng a scale of 2 cm to represent 30° on the
x- is and 2 cm to represent 1 unit on the fix)
a . ,draw the graph offix) for 0° ~ x ~ 180°.
(c) Us ng your graph, estimate the area between
th curve and the x-axis for x between 0 and
-n a d'lans.
(b)
4
(a) Complete the table below for fix) = 3 cos x,
correct to 1 decimal place.
_
3
Draw he graphs of the circular functions sine,
cosin and tangent of () where () is in radian
meas re for - 2n ~ () ~ 2n.
Estim te the area under graphs by 'counting
squar s' or by the trapezium rule.
(ii) Where lines cross x = 1. 9n ~ 0.5
12
(d) (i) All fix) values are now negative what they
were before, i.e.
o
5" 112
o~x
-f(x) 51
118
Revision Maths
\
•
I
ust make sure my calculator is in degree
m de.
• To estimate the area under the curve, I can
co nt squares although now the x-axis
val es have to be in radians.
• To solve the equation, I will have to draw the
lin · fix) = 2 and see where the line crosses the
MacmjJJan Revision Guides for CSE(JP Examinations are the
perfect companions for exam preparation and success.
•
Total coverage of theCSEC®syllabus
•
Numerous practice exercises and worked examples
•
Exam tips and practical advice from teachers and examiners
•
Vital last-minute revision and practice to boost grades and
improve chances of success
•
Ideal for self-study
•
Revision checklists
Rev. Stephen George Jackson B.A. (Hons) M.A. Dip.Ed has lived
in Jamaica for almost 25 years, working at the Calabar High
School, Cornwall College and the Priory International School (as
Principal). He is a pioneer of the use of 'Metacognition' in the
classroom - a successful and proven method of teaching CXC
Mathematics, which he has applied to great effect in this
comprehensive Macmillan Revision Guide. He is currently Senior
Education Officer for Mathematics at the Ministry of Education,
Jamaica.
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