Maths
4 Modules
1.
Number and Calculation
2.
Probability and statistics
3.
Algebra
4.
Number Systems
Module 1  NUMBER and CALCULATION
Different types of Numbers
1.
Natural Numbers
Natural
numbers are numbers above zero and are positive
integers.
Eg
1,2,3,51,99,etc
Numbers
that are not natural are, for example, 2, 1.65 etc
Integers
are whole numbers eg 3,2,1,1,2,3 etc
2.
Irrational Numbers
Irrational
numbers are numbers that cannot be written as a fraction as can ½ Ύ etc. An example of an irrational number could
be 3.943286452. This cannot be written as a fraction.
3.
Rational Numbers
Rational numbers can be
written as a fraction eg ½ Ύ 22/7 (3.142
repeats itself as 3.142857142857142857. For an irrational number look for no
recurring pattern. 0.33333 is rational whereas 0.33245673 is irrational.
It is easier to identify an irrational number than a rational one.
Something such as √7 =2.645751311 is irrational but √4 = 2 or
√9=3 are rational.
4.
Factors
If we split 6=2x3 then 2
and 3 become factors of 6. A factor is one that will divide into a number with
no remainder. Eg split 24 into all its factors
.1x24 2x12 3x8 4x6
. 1,2,3,4,6,8,12
&24 are the factors of 24.
To find factors
a/ Write factors in pairs
b/ Start at 1 and work up
through the natural numbers (1 2 3 4 etc) finding pairs.
c/ Stop when a pair is
repeated eg the factors of 10 are  1x10 and 2x5 = 1,2,5,10
Factors of 72 are 1x72 2x36
4x18 6x12 8x9 3x24
5. Prime Numbers
Prime numbers can only be
divided by themselves or 1 
2
3 5 7 11 12 17 19 23 29 etc nb I is
not a prime number
6. Square Numbers
A square number is a number
that is multiplied by itself. Eg 1x1=1 2x2=4 3x3=9 4x4=16. Therefore 1 4 9 16
25 etc are square numbers.
7.Cube Numbers
A cube is length x height x
width eg 2x2x2=8, 3x3x3=27, 4x4x4=64, 5x5x5=125
**An integer raised to the power of 3**
8. Multiples
Multiples of a number are
numbers which are in the same times table. Multiples of 3 are 3 6 9 12 15
18 21 etc
Multiples of 100 are 100
200 300 etc
9. Triangle Numbers
Square numbers are
generated as 1 4 9 16
Triangle numbers are
generated as 1 3
6 10 15
To find triangle numbers
a/ To find the third
triangle number we add all integers up to 3 1+2+3=6
To find the 10^{th}
triangle number add to 10 1+2+3+4+5+6+7+8+9+10=55
To find the Nth triangle
number add all positive integers up to Nth. N is any number you want.
b/ If we are on the third
triangle number and we want to find the next one we add 4. On the 15^{th}
and we want the next one we add 16. Always add the next integer in sequence
each time.
If we are at the Nth
triangle number and we want the next we add N+1
Sequences
Triangular numbers are an
example of a sequence. We will look at sequences and how we can find a general
rule to generate ant term in that sequence.
Example : 2,4,6,8,
.
Here we can see that the
next two terms are 10 and 12. What is the 50^{th} term? 2x50 = 100
n = term/number pattern n ¦ 1
2 3 4
.50
t = term in sequence t ¦
2 4 6 8
.100
common difference = d
The difference of 2 means
that as n increases by 1 our term increases by 2.
This means we know
that 
T=2n+?
Now we look at our first
term :
n=1 t=2 => substitute
into T=2n+?
2 = 2(1)+? => 2=2+? =>
2=2+? So ?=0
This gives a general formula for the Nth term as T=2n.
Example sequence
3,8,13,18
..
What are the next 2 terms?
Write an expression in
terms of n for the nth term of the sequence.
n¦1 2 3
4 5
t ¦3 8
13 18 23
d=5 t=3 T=5n
n=1
first term n=1 and t=3
3=5(1)+? => 3=5+? => 35=? => 2=?
T=5n2 therefore, 6^{th} term is 28.
28=5x62 where 28 equals T.
Substitute
into T=5n2 => 28=5(6)2 =>
28=302
Quick method If we have
sequence where a=1^{st} term in the sequence. D= common difference we
can give this formula T=dn+(ad)
A=3 d=5
T=5n+(35) =5n+(2) =5n2
n ¦ 1 2
3 4 5 6 7
t ¦ 2
5 8 11
14 17 20
T= 3n+? n=1
t=2 d=3 2=3(1)+?
2=3+? Therefore ?=1
T=3n1 Example 3x21=5 and 3x71=20
n= 1
2 3 4
5
t= 8
12 16 20 24 T=4n+?
n=1 t=8 d=4
8=4(1)+?
8=4+? So ?=4
T=4n+4 eg
4+4=8 and 10^{th} term is 44 **4x10+4
Dealing with
Negative Numbers
We
need to be competent at dealing with addition, subtraction, division and
multiplying with negative numbers
ve =negative
+ve
= positive
Multiplying and
dividing negative numbers
The
rule is If the signs of the two numbers are the same then the answer is
positive but, if the signs are different then the answer is negative.
Adding and
subtracting
The
rule is similar for addition and subtraction however this time we look at signs
in the middle of the numbers and decide whether to add or subtract. If the
signs are the same we add the numbers but if different then we subtract.
Examples
6 2=6+2=8, 6+9=3, 5 +2=7, 106=16
1.
For x or / decide using the rule if the answer is going to
be ve or +ve.
2.
For + or decide looking at only the signs between the numbers
whether to add or subtract.
Rounding and
Approximation
1.
Rounding to a given decimal place
2.
Rounding to a given significant figure
3.
Maximum and minimum values when rounding
4.
Approximation
We
may be asked to round off the following number
3.673459
So
if this was the answer to a problem we may need this level of accuracy or we
can round it off to a decimal place.
Eg
Rounded to one decimal place (DP) = 3.7
So
if we want a number with only one digit after the decimal place before cutting
off the other numbers we must take into account these numbers.
In this example we can see that 3.673459 lies
between 3.6 and 3.7 and is closer to 3.7 so to give a true picture of the
number we round it up.
**round
3.673459 to 2DP = 3.67**
When
rounding to a certain DP count the amount of places and put a cut off line eg
4.273¦6 to 3 DP = 4.274 because 6 rounds up to the next number (7 in this
case).
**Round
14.27569 to 2DP** = 14.28
(Nos to 10 are 0 1 2 3 4 5 6 7 8 9 and the
midway point is between 4 and 5
therefore if 4 or less we leave as is and 5 or
above we round up).
**round
the following to 1DP and 2DP **
4.9961 = 5.0 to 1DP and 5.00 to 2DP
You must put a number after the decimal point even if a zero.
Significant Figures ( Written as SF)
For
SF we round in exactly the same way however we change how to find our cut off
point. Instead of counting the numbers after the decimal point we count from
the first non zero figure.
Eg 2SF  3.216 = 3.2
4SF  32.2892 = 32.29
3SF  0.013462 = 0.0135
3SF  3.02161 = 3.02
2SF  1269.32 = 1300
When we cut off whole numbers i.e. numbers to the left of the
decimal point we must replace them with zeros.
Maximum and
Minimum Values when Rounding
Sometimes
we may have to answer a question like the following
**
A sunflower is measured as 67cms to the nearest cm.
a.
What is the minimum height of the sunflower?
b.
What is the maximum height of the sunflower?
Firstly we need to remember that this has
been measured to the nearest cm and so its true height has been rounded off.
¦ ¦ ¦ ¦ ¦
66 66.5 67 67.5 68cms
Our
answer is 66.5cms because if the true measurement is any less it must round to
66cms. 66.5cms is the minimum that rounds up to 67cms.
The
maximum is 67.5 because
A
football match has 25000 people
attending it rounded to the nearest 1000.
Minimum
is 24500
Maximum
is 25499
Here
we have used discreet information as we must
count in whole numbers (people). Other data is continuous
data and is a measurement as in the sunflower example.
**Each
time we are looking for the midpoint between the measurement we are given and
the next possible measurement above and below it.**
Approximation and Estimation
Lets
look at an example of approximation/estimation
Eg
I buy 686 books at £2.75 each. It is useful to get an estimate as to the cost
and we can do this as follows
Round
to one significant figure: 686 =700 and £2.75 =£3.00=£1200
Estimate
an answer: 0.497 x 415.6 = .5 x 400 =1
4.32 x 51.6 4 x 50
In
real life we often check our calculations with an approximation to see if our
answer is roughly correct.
Eg (216 x 814.5)/ 0.54 = 320.75
(200 x 800)/ .5 = 320,000
The
answer is 100 times too small
Fractions
1.
Changing to equivalent
fractions
2.
Changing
fractions to their simplest form **
3.
Changing improper
fractions to mixed fractions and back
4.
Adding and
subtracting fractions **
5.
Multiplying and
dividing fractions **
6.
Converting
between percentages, decimals and fractions.
**can be worked out on a
calculator**
Equivalent
fractions
An equivalent fraction is a fraction which is
the same size as another but written in a different form eg ½ can be written as 2/4 or 4/8
To
change to an equivalent fraction you must multiply top and bottom by the same
number ½ x 2/2 = 2/4. 1/3 x 2/2 = 2/6.
3/8 = 9/24 (both x3). 7/20 = 35/100 (x5).
Question
In
a test eg 5/6 = ?/18. What is ?? 6x3 =
18. ?x3 = 15? Answer is 5 (5x3 = 15). Therefore the answer is 15/18.
Fractions to their Simplest Form
Whenever we deal with fractions we should give
our answer in its simplest form
Eg
2/4 = ½. So to change to the simplest form we are looking for a fraction with
the smallest numbers on the top and bottom.
To see if the fraction is in its simplest
form we must check to see if both top and bottom have a common factor. If there
is no common factor then it is in its simplest form. If there is a common
factor then we must divide top and bottom by that factor. Eg
15/30. 3,5 and 15 are common factors. Smallest
CF is ½. 6/8 = Ύ because 2 is a CF.
Changing Mixed Numbers to Common
Fractions
A mixed number is a number with a whole and a
fraction eg 1Ό. In an improper fraction the top is bigger than the bottom and
an improper fraction is more than a whole number eg 22/6 (sometimes called top
heavy).
To change from a mixed number to an improper
fraction we do the following
2
½ = ?/2 where 2 stays the same but the top changes. The whole number x bottom +
top gives the improper fraction (2x2)+1=5. = 5/2 .
6
3/8 = 51/8 ((6x8)+3)/8. 8 4/7 = 60/7
((8x7)+4)/7.
Changing from Improper Fractions to
Mixed Fractions
27/6 =
4 3/6 => 4 ½. **27 divided by 6 = 4
r3. 40/9 = 4 4/9.
Adding and Subtracting Fractions
When we add and subtract fractions we need the
bottoms to be the same and only add or subtract the top numbers eg ½ + ½ = 2/2
=> 1. Ό + 2/4 = Ύ.
To add or subtract fractions with different
bottoms do the following
½
+ 1/3 :
3x1. 3x2 = 3/6. 1x2. 3x2. = 2/6.
3/6 +2/6 = 5/6.
2/9
+ 1/6 : 2x6= 12 and 9x6= 54 so 12/54.
9x1=9 and 9x6=54 so 9/54. 12/54+9/54= 21/54
. 21 divided by 3 = 7 and 54
divided by 3 = 18. Simplest fraction is
7/18. So
. 2/9+1/6 = 7/18.
Ύ
 2/5
. 3x5 = 15. 4x5 = 20. 15/20. 2x4 = 8. 4x5 = 20. 8/20. 15/208/20 = 7/20.
When
you add mixed fractions you can add whole numbers then the fractions.
When
you subtract mixed numbers first change to improper numbers.
Eg
2 Ό  1 1/3 = 9/4 4/3. 27/12 16/12 = 11/12.
Multiplying and dividing fractions
Multiplying
½
x Ύ = 3/8. Top times top and bottom times bottom. You must change any mixed
number to an improper fraction before multiplying eg
1
½ x 2 Ό = 3/2 x 9/4 = 27/8 = 3 3/8.
Dividing Fractions
½
χ Ό = ½ x 4/1 = 4/2 = 2.
For
χ change to multiplication and invert the second fraction. If mixed numbers
change to an improper fraction first.
Multiplying
and dividing by a whole number.
Eg
5/6x2 = 5/6 x 2/1 = 10/6 = 1 4/6 = 1 2/3. Ύ χ 2 = Ύ χ 2/1 = Ύ x ½ = 3/8
Conversion Between Decimals, Percentages
and Fractions
Convert to Decimals
a/
To convert percentages to decimals we multiply by 100.
eg .59 x 100 = 59%
b/
To convert fractions to percentages we multiply by 100
eg 5/8 => 5/8x100/1 = 500/8 = 62.5% 1/25 x 100/1 = 100/25 = 4%
Convert to Fractions
To
convert decimals to fractions
0.7..
the number after the decimal place goes on top 7/? There is
always a multiple of 10 underneath i.e. 10, 100, 1000 etc. so count how many
numbers there are after the decimal place. eg one number equals one zero. So
7/? = 7/10
0.59
= 59/100 0.101 = 101/1000 1.37 = 137/100 => 1 37/100.
To convert s to fractions we divide
by 100 eg
79% = 79/100 137% = 137/100 => 1 37/100.
When converting to fractions you must
cancel down afterwards to simplest form eg 90% 90/100 = 9/10 to its simplest
form.
Ratio and Proportion
1/ Changing ratios to their simplest
form
2/ Splitting an amount into given
ratios
3/ Direct Proportion
A ratio describes one amount in
relation to another. Eg When mixing cement the ratio for sand to cement may be
described as 3:1 (sand:cement). This means that for every one part of cement we
have three parts of sand.
In real terms, if we had 1Kg of cement we need
3Kgs of sand eg 300:100Kg sand to cement.
Writing a Ratio in its Simplest Form
The method we use for this is exactly
the same as when cancelling down fractions.
Eg Write the following in their simplest
forms:
a/ 2:8 => common factor is 2 => 1:4
b/ 10:200 => common factor is 10 => 1:20
c/ 32:48 => common factor is 8 => 4:6 => 2:3
d/ 3:6:9 => common factor is 3 => 1:2:3
When we have more than 2 quantities
the rules we apply to ratios do not change. Slightly different examples :
We may be asked to simplify eg 2:1/3. In this example we are looking to eliminate
the fraction. To remove the fraction we must multiply both sides by the
denominator (bottom).
In this example we
multiply by 3: 2x3 and 1/3x3 =>6:1 as the answer.
4:2/5 => multiply by 5 = 4x5:2/5x5
=> 20:2 => 10:1
**
( 2/5x5/1 => 10/5 =>2).
Examples
Express the following in their lowest
terms
a.
2:4 b. 8:12
c. 60:150 d. 18:15 e. 4:1/2
Express the ratio of 5p to 75p in
lowest terms
Express ratio of 400m to 2Kms in
lowest terms
a.
1:2
b.
2:3
c.
6:15
=> 2:5
d.
6:5
e.
8:1
f.
1:15
g.
400:2000 => 4:20
=> 1:5
Splitting an Amount into a Given Ratio
Eg
Tom is going to split £100 between
his 2 sons, Dick and Harry, in the ratio of 3:7.
Tom splits the money into equal pots
of money to give to his sons. We can see that there are 10 pots of money (3 + 7
= 10). To find the amount in each we divide £100 into 10 pots so each pot has
£10. This means that Dick gets 3 pots of £10 and Harry gets 7 pots of £10.
Ratio is 30:70 => 3:7.
Method
1.
Find
total amount of parts in the ratio that we have.
2.
Divide
total amount we are splitting by the total parts.
3.
Multiply how much each part is worth by
amounts given in the ratio.
Eg
Jenny has 49 sweets that she is going
to split with her little sister in the ratio of 4:3 in her favour.
1.
4
+ 3 = 7 in total
2.
Splitting
49 into 7 parts => 49 / 7 = 7 sweets.
3.
Jenny
gets 4 x 7 = 28 and her sister gets 3 x 7 = 21 sweets. 28 + 21 = 49.
Split 90 into the ratio of 7:6:2
1.
Total
parts is 15
2.
90
/ 15 = 6
3.
ratio
is 42:36:12 42 + 36 + 12 = 90
Exercise
a. Divide 300 into ratio of 2:1
b. Divide 60 into ratio of 5:7
c. A line of 84 cms is split into
ratio of 2:3:7. Calculate length of each part.
d. £600 divided between 3 children in
ratio of their ages John is 5, Claire is 7 and Robert is 8. How much they
each get?
e. A sum of money is divided in the
ratio of 2:3. If the larger amount is £18, what is
1. Other amount? (Answer is £12). 2. Total sum of money? (Answer is £30).
Answers
a. 200 and 100. b. 25 and 35. c. 14, 21, 28. d. 150, 210, 240
Direct Proportion
Example
20 oranges cost £1.16. How much do 27
oranges cost?
This is an example of direct
proportion because as the amount of oranges increases so does the cost.
Likewise, as the amount of oranges decreases so does the cost.
First we must find the cost of one
orange  £1.60/20 = 8p. Once we know the cost of one orange we can find the
cost of any amount.
27 oranges cost 27 x 8p = 216p or
 27 x 0.08p = £2.16
1.
7
pears cost 84p. How much do 5 cost?
2.
5Kgs
of potatoes cost 40p. What is cost of 8Kgs?
3.
A
train travels 300Km in 5 hours. How long will it take to travel 450Kms?
4.
3m
of wood costs £2.25. How much does 7m cost?
5.
2
bottles of wine fills 8 glasses. How many glasses of wine can be poured from 8
bottles?
Answers
a.
84
/ 7 = 12. 12 x 5 = 60p.
b.
5 / .40 = 8. 8 x 8 = 64
c.
300
/ 2 = 150 x 3 = 450. 2 ½ x 3 = 7 ½ hours
d.
5.25
e.
32
Indices and Standard Form
·
Rules
of Indices
·
Standard
form converting to ordinary and back
·
Adding,
subtracting, dividing and multiplying with standard form
1.
Indices
Are the power of the index eg 2#3
where #3 is the power 1e 2x2x2
X^{4} => XxXxXxX . 2^{3} x 2^{4} =>
(2x2x2)x(2x2x2x2)= 7
X^{4} x X^{2} = X6
Rule 1
X^{9} x X^{6} = X^{15}
When we multiply 2 numbers (the same) in index form we add the indices
together.
<<!Important the
numbers must be the same!>
Eg 32^{6 }x 32^{5 }=
32^{11}
What happens when we divide indices?
2^{4} / 2^{2} = 2x2x2x2 = 2x2 => 2^{2}
2x2
Y^{6 }/ Y^{3} = YxYxYxYxYxY = Y^{3}
YxYxY
Rule 2
X^{9} / X^{6} = X^{9}6 eg 17^{11} / 17^{3} = 17^{8}
Negative numbers
11^{3} / 11^{6}
= 11x11x11 =
1 =
11 ^{3}
11x11x11x11x11x11 11^{3}
When we have a negative number it
means one over the indice to that power
eg 7 ^{4} => 1/7^{4} and
3 ^{3} => 1/3^{3}
Rule 3
X^{9} = 1/X^{9}
What about 2^{0}? Any No to
the power of zero = 1
Rule 4
X^{0} = 1
Why?
Lets look at 3^{3} / 3^{3} = 3^{0}
If we write this out in full for a
value we get
3^{3} / 3^{3 }= 3x3x3
= 1 = 1 => 3^{0}
3x3x3 1
What happens when we have the
following?
(2^{3})^{2} = 2^{3
}x 2^{3} = 2^{6}
(5^{4})^{3} = 5^{12}
Rule 5
(X^{a}) ^{b} = X ^{axb}
Standard Form
Is a way of writing extremely big or
small numbers.
Eg 3.68 x 10^{9} =>
368000000000
4.2
x
10 ^{9} => 0.00000000042
Standard form has to follow these
rules :
A x 10 ^{n}
Some number A x 10 to the
power of some number n (integer)
1 ≤A < 10
A is greater than or equal
to 1
A is less than 10
Dividing and Multiplying by Multiples of 10.
36.324 x 10 363.24
36.324 x 100 = 3632.4
36.324 x 1000 = 36324
The rule is when we multiply by 10
we move the decimal point n places to the right. Eg 4.78 x 10#^{6} =
4780000.
Dividing
36.324 / 10 = 3.6324. 36.324 / 100 =
.36324.
Because we are dividing we move the
point n places to the left.
36.324 / 10 => 36.324 x 10# ^{1}
=> 36.324 x 1/10 => 36.324
10
36.324 / 100 => 36.324 x 10# ^{2}
Rule 6
When we have a negative power of 10 we move the decimal point to the left. Eg
4.782 x 10 ^{6} = 0.000004782 7x10 ^{2} = 0.07
6.3x10 ^{4}= 0.00063
4.92x10 ^{3} =
0.00492 2.17x10^{3}
= 2170
Changing Ordinary Numbers into Standard Form
Eg 3,100,000,000 to standard form =
3.1x10^{9}
Move the decimal place so
only one whole number is to the left of the decimal place.
*if the original number is big it is
positive. If the original number is small then it is negative.
Eg 0.0003624 = 3.624x10 ^{4} and
3624.00 = 3.624x10^{3}
Adding and Subtracting Standard Form
When we add or subtract standard
forms we first change them to ordinary numbers then add/subtract them. Eg
(5.1x10^{5}) + (6.3x10^{4})
=> 510000 + 63000 = 573000 =
5.73x10^{5}
(5.1x10^{5}) (6.3x10^{4})
=> 510000 63000 = 573000 =
5.73x10^{5}
Multiplying and Dividing Standard
Form
Multiplying
Eg
(5x10^{3}) x (7x10^{4}) = (7x5) x (10^{3} +10^{4)}
= 35x10^{7} = 3.5x10^{8}
(6x10^{8}) x (7x10 ^{3})
= (6x7) x (10^{8}+10 ^{3})
= 42 x 10^{5} = 4.2x10^{6}
(4.36x10^{5)} x (1.4x10^{3})
= (4.36x1.4) x(10^{5}+10 ^{3})
= 6.104x10^{2}
Dividing
Eg
(8x10^{5}) / (4x10^{2})
= (8/4) x (10^{5} 10^{2})
= 2x10^{3}
(2x10^{3}) / (4x10^{5})
= (2/4) x (10^{3} 1^{5})
= 0.5x10 ^{2} = 5x10^{ 3}
Some examples
5100 = 5.1 x 10^{3}
700000 = 7 x 10^{5}
496000 = 4.96 x 10^{5}
63000000 = 6.3 x 10^{7}
0.52 = 5.2 x 10 ^{1}
0.0076 = 7.6 x 10 ^{3}
0.0000127 = 1.27 x 10 ^{6}
0.000452 = 4.52 x 10 ^{4}
Module 2 PROBABILITY and STATISTICS
Statistics
Is
about collecting data and analysing it. What do we remember about statistics?
Bar charts/histograms, averages, pie charts, frequency polygons,
scatter charts, data (qualitative and quantative), standard deviation range
quartiles (spread).
Types
of Data
We
can split data into qualitative and quantative data.
Quantative
is any numerical data
E.g.
How
many people in a class, age, average earnings, shoe size, height etc.
Qualitative
is data that uses words.
E.g.
Favourite
colours, race, sex, holiday locations and data that contains both words and
figures such as address.
To
distinguish between these two types we ask the question does it contain words
or numbers?
We
can split quantative into two separate sub categories
Continuous and discrete data.
So
if you find you have quantative data you must say if it is continuous or
discrete.
Continuous
data is data that has almost an infinite range of values that it can take
E.g.
Someone
can weigh 75Kg, 75.2Kg, 75.23Kg, 75.234Kg, 75.2346Kg
Discrete
data is data obtained from via counting.
E.g.
Number
of chairs in a room, age, children etc
With
discrete data you usually have distinct values the data can take and is usually
whole numbers. Some discrete data doesnt always take whole numbers e.g. shoe
size or pints drunk.
To distinguish between continuous data and discrete data ask
yourself, do you measure it or do you count it?
How
do you use a Tally Chart?
Below
are 20 peoples marks out of 10 in a test. Collect this in a tally chart.
8,6,5,9,2,10,1,6,7,5,10,8,7,6,7,9,8,3,5,3.
Mark Tally Frequency
1 1 1
2 1 1
3 11 2
4 0
5 111 3
6 111 3
7 111 3
8 111 3
9 11 2
10 11 2 Total = 20
Grouping Data
If
you imagine a set of 25 exam marks as a %. With a & we have 101 different
possible marks (including 0) so to reduce the size of our table and tell us
more about the data we could have an interval size of 5
E.g.
04,
59, 1014, 1519,
..
Another
interval could be 10
E.g.
09,
1019, 2029
Intervals
of 5 and 10 are the most common as it makes it easier for data collection.
Interval Using Mathematical Notation
(M=mark)
E.g.
M>0
m is bigger than zero
M<5
m is less than 5
M m m
is greater or equal to 5
So
04 0m<5 09 0m<10
59 5m<10 1019 10m<20
1015 10m<15 2029 20m<30
These
intervals are used for continuous data.
Bar Charts and
Histograms
There
are two ways of representing our data graphically.
Bar Chart
Histograms
When
a histogram starts higher than zero we show this to show the axis
Pie Charts
A
pie chart represents the frequency of information in a graphical way. We often
use %s with pie charts. This means that we can compare samples of different
sizes.
E.g.
For
our work sheet we have a sample of 150 shoppers who travel into a town centre.
Using pie charts we can compare this to a sample of 60 shoppers.
We
need 60 equal parts 360/60 = 6. Therefore 6 = one person.
Transport Frequency Angle(o) %
Bus 12 12 x 6 = 72 20
Car 23 23 x 6 = 138 38.3
Park/ride 10 10 x 6 = 60 16.7
Walk 6 6 x 6 = 36 10
Cycle 5 5 x 6 = 30 8.3
Other 4 4 x 6 = 24 6.7
Angles should be
rounded to nearest whole degree
& should be
rounded to nearest 1 dp
Sometimes because
of rounding we have a small error but,
Both angles shouldnt
be more than 1 degree from 360 degrees.
5s should be no
more than 1% from 100%
Frequency Polygon
For
this we need a frequency table
Weight Males frequency Female frequency
30w<40 3 5
40w<50 7 9
50w<60 8 4
60w<70 2 2
Totals
 20 20
An
axis is exactly as we would for a histogram
The scale must include all male and female values to check min/max
values
Each
point is plotted midway between internal values.
Include
a key to show which is male and which is female
ό
Males are heavier
than females
ό
More females to
the lower end of the scale
ό
More males to the
higher end of the scale
Averages
Averages
tell us the typical value for a set of data and are useful for comparing sets
of data.
Three
types of averages we have are
Mean:
Total
of all data divided by the number of data.
Median:
The
middle value when the data is put in order.
Mode:
Most
commonly occurring piece of data.
Example
Calculate
the mean, mode and median for the following data
6,9,5,4,9,7,2,5,6,4,8,3,6
Mean
add all the numbers and divide by the number of pieces.
i.e. 74/13 = 5.7 1dp
Median
before doing the median we must put the data in order
2,3,4,4,5,5,6,6,6,7,8,9,9
Before
finding the median value (average) we
need to find the median place (where
in the list the data is). In this example the median place is the 7^{th}
digit and so the median value is 6.
Median Place = n+1 n=13
13+1 = 7_{}
2 2
Mode
the mode value is 6 because 6 is the most common number.
If
more than one common number then all can be the mode.
e.g 2,2,2,6,6,6,9,9, the mode is 2 and 6.
For
an even amount of data we have one extra step of median to complete
e.g 10,15,13,16,12,14
Find
the median 10,12,13,14,15,16
Find
the median place n+1 = 6+1 = 3.5^{th} place. Between 13
& 14
2
2
The
extra step for an even set of numbers is to add the middle two numbers together
and divide by 2
Median
value = 13+14 = 13.5
2
Note in this example we have NO mode
Range
To
find the range:
Range
= Max value Min value.
The
range gives us an idea about the spread of data. We use the spread of data to
determine how reliable the mean is.
e.g
3
boys and 3 girls take a test marked out of 25. We want to decide who succeeded
more, the boys or the girls.
Boys
marks 1,13,25
Girls
marks 11,13,15
Boys
mean = 1+13+25 = 39 = 13
3 3
Girls
mean = 11+13+15 = 39 = 13
3 3
Here
we can see that the girls marks are a better representation of the actual
marks. This is because the boys have extreme values.
The
range helps us identify this.
Boys
range = 251 = 24
Girls
range = 15 11 =4
Here
we can see that with a large range we have a less reliable mean.
1/
189,192,192,204,213,214,217
Mean
= 1421 / 7 = 203
Median
place = 7+1 = 8 = 4 therefore the Median value = 204
2 2
Range
= 217 189 = 28.
Mode
= 192 which occurs twice in the list.
2/
4,4,4,6,8,9,16,18,20,22,25
Mean
= 136 / 11 = 12.4
Median
place = 11+1 = 12 = 6
therefore the mean value = 9
2 2
Range
= 25 2 = 21
Mode = 4 which occurs 3 times.
3/
210,214,217,223,225,228,233,236,236,238
Mean
= 2260 / 10 = 226
Median
place = 10+1 =11 = 5.5 therefore the mean value = 226.5
2 2
Range = 238210 = 28.
Mode
= 236 which occurs twice.
4/
0.9,2,2,2.6,3.1,5.5,5.6,6.6,7.3,8.7,12.1
Mean
= 56.4 / 11 = 5.1
Median
place = 11+1 = 12 = 6
therefore the median value = 5.5
2 2
Range
= 12.1  .9 = 11.2
Mode
= 2 which occurs 3 times.
Averages from a Table
e.g  Find the mean, mode and median from the
following table
No of matches  Frequency  Frequency
total (fx)
40  5 15  200
41 
11 616  451
42  24
1740
 1008
43  16
4156  688
44  4
5760  176
Total 60 2523
Mode
= 42 which occurs 24 times.
Make sure you give the value and NOT the frequency.
The
median is very similar to before. First, find the median place.
60+1 = 30.5. this is the 30.5^{th} data item. Next find the
30.5^{th} value.
2
which is 42.
Mean
= total data = fx = 2523 = 42.1
matches
n n 60
Finding
averages with group data
Heights
of some children were measured and results were as follows
Height (x) Frequency Midpoint Estimate of totals
90 ≤x
<100 7 (17) 95 7*95 = 665
100
≤ x <110 16(823) 105 16*105 = 1680
110
≤ x <120 15(2438)
115 15*115 = 1725
120
≤ x <130 10(3948)
125 10*125 = 1250
130
≤ x <140 8(4956) 135 8*135 = 1080
140
≤ x <150 4(5760) 145 4*140 = 580
Totals n=60
fx = 6980
We use the midway point of intervals to estimate height. Using group
data the mode is called MODAL CLASS or
INTERVAL
Modal
interval = 100 ≤ x <110
Median
place = n+1 = 60+1 = 30.5th place
2 2
Median
interval = 110 ≤ x <120
With
group data we can only find an estimate of the mean because we dont have exact
values.
Estimate
of mean = Total of height / No of people = 6980 = 116.3
60
Cumulative Frequency
These
are the marks of 150 year 11 students as %s
MARK FREQUENCY
CUMULATIVE FREQUENCY
0 ≤
m <10 2 2
10
≤ m <20 4 6
20
≤ m <30 17 23
30
≤ m <40 22 45
40
≤ m <50 45 90
50
≤ m <60 26 116
60
≤ m <70 19 135
70
≤ m <80 9 144
80
≤ m <90 5 149
90
≤ m ≤100 1 150
TOTAL
 150
Find
the median from this.
Median
= n+1 = 150+1 = 75.5^{th} place
place
2 2
median
value = 40 ≤ m <50
Its
more usual that we use our cumulative table to draw a graph and take a reading
from that. To take a reading for our median from the graph we need the median
place (75.5^{th}).
Quartiles
Quartiles
are very similar to our median. We can look at our upper quartiles (uq), lower
quartiles (lq) and interquartile range (iqr).
Think
of a set of data as the following number line
Ό ½ Ύ
    
Lowest lq place median place uq
place highest
data n+1 n+1 n+1x3 data
value 4 2 2 value
Once
we have found our lq place we can use this to find our uq place.
Uq
= 3 x (n+1)/4 = 3 x lq place.
Here
we have shoe sizes of a family of 7 people
4,6,7,5,3,4,9
Find uq, lq, iqr.
First,
like the median, we must put the data into order
3,4,4,5,6,7,9
Lq
place = n+1 = 7+1
= 2^{nd} place
2 2
Uq
place = 3 x lq = 3 x 2
= 6^{th} place
Lq
value = 4 ( 2^{nd} place )
Uq
value = 7 ( 6^{th} place )
Iqr
= uq lq = 7 4 = 3 ( range of quartiles )
Finding Quartiles from a Graph
(referring
back to the example of the 150 students )
When
we find the quartiles from a cumulative frequency we follow the same method as
taking a reading for a median.
First
we find the lq place and uq place.
n
= 150 lq place = n+1 = 150+1
= 37.75^{th} place
2 2
Uq
place = 3 x lq place = 113.25^{th} place.
Standard Deviation
1.
What is standard deviation?
2.
How do you calculate sd?
3.
Alternative method of finding sd.
4.
Using a calculator to calculate sd.
1.
What is SD?
The
sd tells us the alternative amount each data item differs from the mean.
We
measure 100 sunflowers and calculate the mean = 70cms. Sd = 2cms (to nearest
cm).
This
means that the average of 100 sunflowers is within 2cms of the mean of 70cms.
We can say that the sunflowers are, on average, 70cms
+/
2cms. Another sample of 100 sunflowers found the mean was 73cms and the sd was
5cms. So, on average, the second group are, on average, 73cms +/ 5cms.
We
can use the sd to decide how reliable a mean is
the
higher the sd, the less reliable is the mean.
Consider
a school year of students who have a mean height of 1.75cms. Lets look at a
line graph showing frequency distribution of heights.
All
students are between 1.4m and 2.1m tall.
Any
statistical population which is said to be normal will have this shape
frequency distribution and should be bell shape.
x = mean
and r = sd
In
a normal distribution 68% of data values will be within 1 sd of the mean. So
the graph of our first sunflower example would look like
x = 70cms r= 2cms 68%
are between 68 and 72cms tall.
68cms
= x  r (70 2)
and 72 = x  r (70 + 2).
How to Calculate SD
Ex
Heights
of 5 men are 177.8, 175.3, 174.8, 179.1, 176.5cms
Find
the mean and sd of the data.
Mean
= x = ex = total
of data
n No of data
177.8+175.3+174.8+179.1+176.5 = 883.5
= 176.7cms (x = 176.7cms)
5 5
sdΕ e ( x  x)2 x
 d = x  x 
d2
n 174.8 176.7 174.8 1.9^{2} = 3.61
175.3 176.7 175.3 1.4^{2}= 1.96
176.5 176.7 176.5 .02^{2} = 0.04
177.8 176.7 177.8 1.1^{2 }= 1.21
179.1 176.7 179.1 2.4^{2} = 5.76
Totals  883.5
12.58
e(x  x)^{2}
Next
we find the variance.
r#^{2} = e ( x  x)^{2} = 12.58 =
2.516 variance
n 5
Finally
for the sd take the square root of the variance
r = Ε e ( x  x)^{2} = Ε 2.516 = 1.59 2dp
n
Method

(
we need the mean to calculate the sd )
1.
Draw a table and calculate d = x  x ( d = data value mean )
2.
Calculate d^{2} value
3.
Find total of d^{2} values ( e (x  x )^{2} )
4.
Find variance = e (x  x )^{2
}
n
5.
Find square root of variance Ε e (x  x )^{2} = sd
n
338,354,341,351,353
= 1737 / 5 = 347.4 therefore x = 347.4
x
 d = x  x 
d^{2}
338 347.4 338 = 9.4 88.36
354 347.4 354 = (6.6) 43.56
341 347.4 341 = 6.4 40.96
351 347.4 351 = (3.6) 12.96
353 347.4 353 = (5.6) 31.63
Total
 e = 217.2 e = ( x  x)^{2}
Variance
= e / n =
217.2 / 5 = 43.44
Square
root of the variance = Ε43.44 = 6.59 therefore r (sd) = 6.59
0,5,10,15,20
= 50 / 5 = 10
X d = x  x
d^{2}
0 10 0 = 10 100
5 10 5 = 5 25
10 10 10 = 0 0
15 10 15 = (5) 25
20 10 20 = (10) 100 Total 250 / 5 =
50 (variance) r = 7.07
Quicker Method
We
can use an alternative formula
r = Ε e x^{2}
 x^{2}
n
Ex/
Find the mean and sd of 338,354,341,351,353
Mean
= x = ex = 347.4
where ex is the total of data / n: 1737 = 347.4
n
5
Simplified
method is as follows
x
 x^{2}
338 114,244
354
125,316
341
116,281
351 123,201
353 124,609
Total
= 603,651 ( sum of x^{2})
(ex#2)
Variance
= r^{2} = ex^{2}
 x^{2} = 603,651 347.4^{2 } =
n 5
120,730.2
120,686.76 = 43.44 sd (r) = Ε43.44 = 6.59 2dp
Example
26,28,31,38,29,24,23,35
= 234 / 8 = 29.25
x 
x^{2}^{}
26
676
28
784
31
961
38 1444
29
841
24
576
23
529
35 1225
Total
= 7036 / 8 = 879.5 29.25^{2 } (855.56) =
23.94 = 4.89 (r)
Probability
1/
Introduction to probability
2/
Cover Tree diagrams
If
we toss a coin the probability of getting a tail is ½, 50%. 0.5. It is more
usual to use the 0.5 or ½ but mostly we will use fractions in the problems we
deal with.
Rolling
a 5 on an 8 sided dice is 1/8^{th} (one chance in eight). Using
notation we can say:
a.
P(tail) = ½ b. P(5 on 8 sided die)
= 1/8
c.
P(event) = number of ways the event can occur
total number of possible outcomes
EG
I
have a bag with 5 red balls, 6 green balls, 8 blue balls and 2 white balls.
1.
Probability of a red ball  P(red) = 5
21
2.
Probability of a black ball P(black) = 0
21
3.
Probability of a blue or white ball P(blue)
= 10
21
The probability of a certain
event is 1
1
An 8 sided spinner has 2 red, 2 green, 3 blue, 1 yellow edges. What is the probability
of obtaining:
a.
Blue = 3 of 8. b. yellow = 1 of 8. Not
green = 6 of 8. Not blue = 5 of 8
2
an 8 sided spinner has numbers 1 to 8. What is the probability of getting:
a.
Odd number = 4/8 => ½. b. Square No =
2/8 => Ό
c.
No > 5 = 3/8. d. No < 3 = 2/8
=> Ό. e. Prime No = 3/8.
3
In the word Mississippi
what is the probability of getting:
a.
An I = 4/11. b. S = 4/11. c. P = 2/11. d. A consonant = 7/11.
e.
an M = 1/11
Tree Diagrams
There
are 2 important types of events in probability
Independent Events and Mutually Exclusive Events. Each of these has an
important rule used for Tree Diagrams associated with it.
Mutually Exclusive Events (MEE)
Are
two or more events that cannot happen at the same time. MEE have the OR rule
associated with them.
OR
rule Probability of A or B = P(A) + P(B)
Ex
Role
a six sided dice P(even) = P( 2 or 4 or 6 )
=
P(2) + P(4) + P(6) + 1/6 + 1/6 + 1/6 = 3/6 = ½
Independent Events
Are
events that when one happens it does not effect the probability of another
happening eg flipping a coin more than once.
An
example of events that are NOT independent is drawing the balls on the lottery.
So
P(13) first draw is 1/49
& 
P(7) first draw is 1/49. If a 13
comes out the P(7) changes to 1/48 so they are not independent.
AND
rule for independent events The probability of both events happening is found
by multiplying the probabilities together.
P(
A & B ) = P( A ) x P( B )
Ex
Flipping
a coin twice. What is the probability of getting 2 tails? If we flip a coin
twice we get
2
tails, 2 heads, head and tail, tail and head [ TT, HH, HT, TH ]
1.
P( HH ) 2. P( different outcomes ) 3. P(probability of at least one tail)
Now
we have the tree diagram we can answer the questions
1.
P( HH ) = Ό 2. P( D.O.) = P( HT ) or
P( TH ) = ½ (Ό & Ό)
3.
P( TT or TH or HT ) = P(TT ) + P( HT ) + P( TH ) = Ό + Ό + Ό = Ύ
Ex
Probability
that Tom gets up late is 0.2. If Tome gets up late the probability of missing
the bus is 0.5. If he gets up on time the probability of missing the bus is
0.1. What is the probability of
1.
P(miss the bus) 2. P(catches bus if up late)
First
consider the order in which these events are likely to happen.
L
= get up late NL = not late M = miss the bus NM = not miss bus
See Tree Diagram for
results
Answers
P(miss
bus) = P9kiss late or not late) = P( L & M ) + P( NL & M )
=
0.1 + 0.08 = 0.09
2.
P( L & NM ) = P( late and not miss bus)
= 0.1
Module 3  ALGEBRA
The
letters in algebra are just symbols that represent an unknown number.
For
example we could use a ? to represent an unknown number but we would not
have enough symbols if we had 2 or more unknowns.
Algebra
splits almost into 2 sections
Algebraic Expression
If
I buy x oranges and y lemons an expression would be x & y. Expressions only
deal with letters, not numbers. We never get a numerical answer. We simply use
expressions as symbolic representations of sums and to practice algebraic
manipulation.
Algebraic Equations
Actively
generate a numeric answer. Lets says a plumber costs £15 an hour and has a
callout charge of £20. An equation we use for the cost of the plumber is : cost
= 20 + 15. Equations always have an equals sign whereas expressions dont.
From
this equation knowing the cost we could find the amount of hours the plumber
has worked through algebraic manipulation.
Algebraic Expressions the Basics.
Ex
 x + x + x = 3x
Above
is an example of simplifying an expression.
Ex
7 x P = 7P Ex. 2x 2y = 2x 2y.
We cannot add or subtract different symbols.
Ex
6x 2x = 4x Ex 6x + 3y + x
2y = 7x + y
Ex
5p  2g + 3p + 5g = 8p + 3g
When collecting data put positive Nos first and negative Nos last.
Multiplying and Dividing Terms
Ex
5 x X x Y = 5xy Ex B x A x 7 = 7ab
It is important to write the numbers first followed by the letters in
alphabetical order.
Ex
A x A = A^{2} Ex B x B x B x
B^{2} = B^{5} 6A x 5B = 30AB
When
numbers are involved we can multiply and divide them as normal.
Eg
30a^{3} / 15a = (30/15) x ( a^{3} / a) = 2a^{2}
Simplifying Expressions with Indices
Think
of this numerically. Lets say a = 2  so
2 + 2^{a} = 2 + 4 = 6
The
mistake is to think that a + a^{2} = 2a^{2}. this is 2(0^{2}
= 2(4) =8
We
must only simplify terms that are indices which are exactly the same.
ab
+ a^{2b} + ab^{2} + 2ab + a^{2b2} = 3ab + a^{2b} +ab2 + a^{2b2}
These are the only terms that we can collect
 ab + 2ab
Removing Brackets
Ex
6( x + y ) = 6x+y WRONG!
The
correct way is to multiply all terms in the brackets by whats outside  6x + 6y.
5(x 6y) = 5x 30y. 7p(q
2p) = 7pq 14p
Simplify
3x(2y
+ 2x) 3y(2x 2y) = 6xy + 6x^{2}
6yx 6y^{2}
6x^{2}
+ 6y^{2 }= 12xy^{2}
a/
3(2x + 1) + x => 6x + 3 + x = > 7x + 3
b/
5(2a + 1 ) 3a => 10a + 5 3a => 7a + 5
c/
2(x + 3y ) + x + y => 2x + 6y + x + y = 3x + 7y
d/
3( 3p q ) + 4( p + 2q) => 9p 3q + 4p + 8q => 13p 5q
e/
4( 3a 2b) + 2(a + b) => 12a + 8b + 2a + 2b =. 14a 6b
Solving Equations
An
example of solving an equation is
4x
= 32. Find x . x = 32/4 => x = 8
1/
5x = 45 => x = 9 2/ 7x = 63 => x
= 9 3/ 10y = 120 => y = 12
4/
8z = 48. => z = 6 5/ 13m = 169. m =
13
Solve
x + 6 = 10. x = 10 6 = 4.
1/ y 7 = 1 => y = 7 + 1 = 8. 2/ z + 3 =5 => 5 3 = z
3/
m 10 = 3 => m = 13 10 = 3. 4/ a
+13 = 25 => a = 12 +13 =25
5/
s 10 = 23 => s = 33 10 = 23
We
can think of solving equations as a game with a set of rules. The idea is to
get x, y, z
on its own by following a set of rules.
Rule
1
First
rule we have learned is
When
we take a number across an = sign it does the opposite thing eg a + becomes a
and a becomes a +. A x becomes a / and / becomes a x.
Solve
2x
+ 1 = 9 => 2x = 9 1 => 2x = 8 => x = 8/2 => x = 4
3x
1 = 14 => 3x = 14 + 1 => 3x = 15 => x = 15/3 => x = 5
Rule
2
Always
move the numbers first and leave any numbers attached to a letter last.
a/ 47 = 5m + 2 => 47 2 = 5m => 45 = 5m
=> 45/5 = m => m = 9
b/
48 = 4n + 20 => 48 20 = 4n => 28 = 4n => 28/4 = n => n = 7
c/
37 = 5p 3 => 37 + 3 = 5p => 40 = 5p => 40/5 = p => p = 8
Solve
3( 2x + 1) = 15
Because
of the bracket figures we cannot move it so we must get rid of the bracket
6x
+ 3 = 15 => 15 3 = 6x => 12 = 6x => 12/6 = x => x = 2
Solve
3y + 1 = 5
5
3y
+ 1 = 5 x 5 => 3y + 1 = 25 => 3y = 25 1 => 3y = 24 => y = 24/3
=> y = 8
Rule
3
If
one side is being divided then move the number first.
Multiply
any brackets out
Move
all numbers to one side leaving x on its own.
Rearranging Equations
We
are now going to use the rules we learned in a slightly different way.
In
these questions we will not get a numerical answer.
Ex
Make x the subject ( get x on its own )
3x
+ y = z => 3x = z y => x = z
 y
3
Ex
A plumber costs £x an hour and £z is his callout charge. He works for y hours.
C = cost.
C
= X x Y x 3
Find
an expression for the number of hours worked. Ie make Y the subject.
C
= xy + 3 => c 3 = xy => c 3 = y => hours = cost &
callout
x hourly rate
1/
y = 3x + 5 => y 5 = 3x => y 5 /3 = x
2/
p = 2q 3 => p + 3 2q => p + 3/2 = q
3
v = 4u + 7 => v 7 = 4u => v 7/4 = u
4
a = 1/2b + 1 => a 1 = 1/2b => a 1/1/2 = b => 2(a 1) =b
SURDS
A
surd is, for example  square root of 17 = 4.1231056 irrational number. An irrational number is one that
cannot be represented as a fraction.
We
use surds to give exact answers.
Simplifying Surds
Like
when we would simplify 31/62 to ½. Because we identify it more we will do
something similar to surds.
Eg/
Ε24 => Ε4x6 => Ε4 x Ε6 => 2Ε6
Ε72 => Ε9x8 => Ε9 Ε8 => 3Ε8 => 3Ε4 Ε2 => 3x2Ε2 = 6Ε2
Method
1.
Look for factors of the number that are square.
2.Squareroot
the square factor and make outside the squareroot leaving the non square
factor.
3.
Check for more square factors of the number.
Ex/
Ε98 => Ε49Ε2 => 7Ε2 Ε99 => Ε9x11 => 3Ε11
Ε40 => Ε4Ε10 => 2Ε10 Ε300 => Ε100x3 => 10Ε3
Ε192 => Ε64x3 => 8Ε3 Ε108 => Ε36x3 => 6Ε3
Ε28 => Ε4x7 => 2Ε7 Ε243 => Ε81x3 => 9Ε3
Inequalities
x
> 5 x is greater than 5
x
< 4 x is less than 4
x m 7 x
is greater or equal to 7
x
≤ 3 x is less or equal to
3
Write
down all integer solutions to the following
2
< x < 10 x is greater than 2 but
less than 10. x = 3,4,5,6,7,8,9
1
≤ x < 5 x is greater or equal
to 1 and less than 5. x = 1, 0, 1,2,3,4
6
< x ≤ 1 x is greater than 6 and less or equal to 1. x =
5,4,3,2,1, 0, 1
1
≤ x ≤ 7 x is greater or equal to 1 and less or equal to 7. x=
1,2,3,4,5,6,7
Solving Equations for Inequalities
Ex/
8
≤ 2x ≤ 4. Find all integer solutions of x
Take
2 to both sides : 4 ≤ x ≤ 2. Sol: 4,3,2,1,0,1,2
Nb
on the left, 8 divided by 2 = 4 and on the right, 4 divided by 2 = 2
Ex/
a/
8 ≤ 3x ≤ 10 => 8/3
≤ x => x m 2 2/3 . x
≤ 10/3 => x ≤ 3 1/3
Sol:
2,1,0,1,2,3
b/
3≤ 4x 3 ≤5 => 33 ≤ 4x ≤ 5+3 => 0 ≤ 4x ≤ 8 =
0,1,2
c/
5 ≤ 2x + 5 ≤ 9 => 5 5 ≤2x ≤ 9 5 => 10
≤ 2x ≤ 4 => 5 ≤ x ≤ 2 =
5,4,3,2,1,0,1,2
d/
6 ≤ 3x 2 ≤ 8 => 6+2 ≤ 3x ≤ 8+2 => 4 ≤
3x ≤ 10 => (4/3 = 1 1/3)
(10/3
= 3 1/3). Sol: 1,0,1,2,3
e/
4 ≤ 4x +6 ≤ 12 => 64 ≤4x ≤ 126 => 10 ≤
4x ≤ 6 => 2 ½ ≤ x ≤ 1 ½
Sol:
2,1,0,1
f/
5 ≤ 5x + 8 ≤ 16 => 85 ≤ 5x ≤ 168 => 13
≤5x ≤ 8 => 2.6 ≤ x ≤ 1.6
Sol:
2,1,0,1
Modulus Functions
When
looking at the modulus of a number we are purely looking at the size of a
number (we can use a technical term of magnitude).
Ex/
3
=3 3 = 3 We ignore whether a number is negative or
positive.
If
x = 5 what is x? x = 5 OR 5
2x+1
= 11 Find x.
2x
+ 1 = 11 or 2x +1 = 11
2x
= 11 1 2x
= 11 1
2x
= 12 2x
= 10
x
= 6 x
= 5
What
we have found is that x = 6 or x = 5
Find
the modulus of 3x + 3 when x = 2
Substitute
x = 2 into 3x + 3
3(2) + 3 => 6
+ 3 => 3 = 3 = 3
Answer
 3x + 3 = 3
2w
= 5 2w = 5
w
= 5/2 w = 5/2
w
= 2.5 w = 2.5
3r
+ 4 = 7 3r +4 = 7
3r
= 7 4 3r = 7 4
3r
= 3 3r = 11
r
= 1 r = 3 2/3
2q
1 = 5 2q 1 = 5
2q
= 5 + 1 2q = 5 + 1
2q
= 6 2q = 4
q
= 3 q = 2
Module 4  NUMBER SYSTEMS
Converting octal numbers to decimal
The
Octal number system works in base 8.
8^{4} 8^{3} 8^{2} 8^{1} 8^{0}
4096 512 64
8 1
Convert
12_{8} into decimal
(1
x 8) + (2 x 1) = 8 + 2 = 10_{10}
_{ }
35_{8
}= (3 x 8) + (5 x 1) = 24 + 5 = 29 (base 10)
213
= (2 x 64) + (1 x 8) + (3 x 1) = 128 + 8 + 3 = 139 (base 10)
3216
= (3 x 512) + (2 x 64) + (1 x 8) + (6 x 1) = 1536+128+8+6 = 1678
Converting Decimal to Octal
Convert
8_{10} into octal = 10_{8 }Convert
23 into octal = 27 Convert 213 = 325
8_{4} 8_{3} 8_{2} 8_{1} 8_{0 }
4096
512 64 8 1
1 0
2 7
3 2 5
3 7 4 7
8
divided by 8 = 1 r 0
23
divided by 8 = 2 r 7
213
divided by 64 = 3 r 21. 21 divided by 8 = 2 r 5
2023
divided by 512 = 3 r 487. 487 divided by 64 = 7 r 39. 39 divided by 8 = 4 r 7.
Therefore answer is 3747
Adding and Subtracting Octal Numbers
Most
important thing to remember is that we cannot have 8s and 9s.
27
+ 33 = 62 because  27
+33
= 7+3 = 1 r 2. 3+2+1 = 6
62
375_{8}
+ 425_{8} = 1022_{8} (375_{8} = 277_{10}) + (425_{8
}= 253_{10}) 277_{10}+253_{10} = 530_{10}.
530_{10}
= 1022_{8 }
34
+ 26 = 62 321 + 266 = 607
103
+ 77 = 202 6352 + 1335 = 7707
Octal Subtraction
423
121 = 423 323 145 = 323
 121 145
302 156
Remember base 8 so 3 5 = 11 5 = 6. 10 5 = 5. 3 2 = 1. Answer
156
Hexadecimal
Is
a number system that works in base 16. For HD we set up a table as follows
16^{3} 16^{2} 16^{1} 16^{0}
4096 256 16
1 numbers 0 15 are
represented in this row ΰ#
The
HD system can transmit more information in shorter strings than can decimal.
For the numbers 10 15 we use letters to represent them to avoid confusion.
The numbers are as follows
A
10
B
11
C
12
D
13
E
14
F
15
Converting Decimal to HD
We
use the same method as employed for converting to binary and octal.
Convert
2710 into HD = 1B16
. 27/16 = 1 r 11 REM: 11 is B
Convert
93 into HD = 5D
93/16 = 9 r 3
Convert
216 into HD = D8
. 216/16 = 13 (D) r 8
Convert
257 into HD = 101
257/256 = 1 r 1
Converting HD into Decimal
This
again will use the same method for conversion as binary and octal.
Ex/
Convert 3C into decimal. Ex/ Convert A2D into decimal
16^{3} 16^{2} 16^{1} 16^{0}
4096 256 16 1
3 C
3x16=48+12=60_{10}
A 2 D
(10x256)+(2x16)+13=2605_{10
}
_{ }4
A 9 F
ΰ
(4x4096)+(10x256)+(9x16)+15= 2605
Adding of HD
The
+/ of HD follows the same principles as +/ in decimal, binary and octal. The
only new element we need to learn to adapt to is the use of letters.
Ex/ 13+28 = 13
+ 28
3B  8+3=11. 11 in HD is B.
59+7B
= 59
+ 7B
D4 
9+B(11) = 20. 20 = 16 r 4. 5+7+1= 13(D). Ans = D4
AD+43
= AD
+ 43
F0 
D(13)+3 = 16 r 0. A(10)+4+1=F(15). Ans = F0
Subtraction of HD
Methods
are the same as in decimal, binary and octal but remember that letters are
involved and need to be converted into numbers.
Ex/ 9A 48 = 9A
48
52 
A(10)8 = 2. 4 9 = 5. Ans = 52
B3
7A = B3
7A
39 
3+16A(10)=9. 7 becomes 8. B(11)8=3. Ans = 39
AC
8E = AC
8E
1E 
C(12)+16=28. 28E(14)=E(14). C(10)9=1 Ans = 1E
Binary to Octal Conversion
Consider
the binary number 101101
32 16 8 4 2 1 The
1^{st} group of 3 digits covers 07
1
0 1 1 0 1 Similar to 1^{st} column of Octal
system
In
Decimal 4+1 = 5 in 1^{st} column
In
Octal 4+1 = 5 in 1^{st} column
In
Decimal 32+8 = 40 in columns 1 and 2
In
Octal 32+8 = 5 in 2^{nd} column.
So
101101 = 55_{8 }
In
order to convert binary to octal we simply split the binary number into blocks
of 3 then convert each block of 3 into decimal. This is because each number
from 0 to 7 can be represented by a block of 3 binary numbers
Ex/
Convert 110011101 into octal = 635_{8 }
256 128 64
 32 16
8  4 2 1
4
2 1 
4 2 1  4
2 1
1
1 0 
0 1 1 
1 0 1
 
 
4+2 =6
 2+1 = 3  4+1
= 5
Always
start groupings from the right.
To
convert from octal to binary we simply represent each octal column as a 3 digit
block of binary.
Ex/
Convert 21 in octal into binary = 10001
REM: The 4 2 1 sequence.
2  1 Convert 726 = 7 2 6
= 111010110
010 
001
111 010 110
(421) (421) (4+2+1) (0+2+0)
(4+2+0)
Converting Between HD and Binary
Consider
the number 10111101
In
HD we split the number into blocks of 4
8
4 2 1
 8 4 2 1
1
0 1 1
 1 1
0 1
11
= B in base 16 13 = D in base
16 Answer = BD_{16 }
Convert
into decimal then into HD.
Convert
10010011 into HD
8 4 2 1 8 4 2 1
1 0 0 1 0 0 1 1
=
8+1 = 9 and 2+1 = 3 so answer is 93_{16 }
Octal and Hexadecimal Conversion
Method
1/
Convert number to binary
2/
Regroup binary
3/
Convert from binary into groups (octal or hd)
Ex.
Convert
6A_{16} to Octal: = 152_{8}
  6 
A  6 =
0110 & A = 1010
1.   0110  1010  HD is
groups of 4: 8,4,2,1
2. 
001  101  010  Octal is
groups of 3: 4,2,1
3. 
1  5  2 
Convert
36178 into HD: = 78F_{16}
 3  6  1  7 
1. 
011  110  001  111 
2/   0111  1000  1111 
3/ 
 7  8  F 
Blocks
of 4 into 3 for HD to Octal
Blocks
of 3 into 4 for Octal to HD
My thanks to Jamie Smith for allowing me to recreate his class notes for this page