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Jamie Smith

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 10th 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 15th 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 50th 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+? =>   3-5=? =>    -2=?

 

T=5n-2   therefore, 6th term is 28. 28=5x6-2 where 28 equals T.

Substitute into T=5n-2 => 28=5(6)-2 =>  28=30-2

 

Quick method – If we have sequence where a=1st term in the sequence. D= common difference we can give this formula T=dn+(a-d)

A=3   d=5   T=5n+(3-5)   =5n+(-2) =5n-2

 

 

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=3n-1   Example 3x2-1=5   and 3x7-1=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 10th 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, -10-6=-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

 

Let’s 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. 15/20-8/20 = 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

X4 => XxXxXxX   . 23 x 24 => (2x2x2)x(2x2x2x2)= 7

X4 x X2 = X6

 

Rule 1 –

X9 x X6 = X15… When we multiply 2 numbers (the same) in index form we add the indices together.

<<!--Important – the numbers must be the same!-->

Eg – 326 x 325 = 3211

What happens when we divide indices?

24 / 22 =     2x2x2x2 = 2x2 => 22

                      2x2

 

Y6 / Y3 =     YxYxYxYxYxY = Y3

                           YxYxY

 

Rule 2 –

X9 / X6 = X9-6     eg – 1711 / 173 = 178

Negative numbers –

113 / 116 =                  11x11x11        =        1          =  11 -3

                          11x11x11x11x11x11       113

 

When we have a negative number it means one over the indice to that power  eg    7 -4 =>  1/74    and      3 -3 =>  1/33

 

Rule 3 –

X-9 = 1/X9

What about 20? Any No to the power of zero = 1

Rule 4 –

X0 = 1

Why?  Lets look at 33 / 33  =  30

If we write this out in full for a value we get –

33 / 33 =       3x3x3 =  1  = 1 => 30

                   3x3x3      1

What happens when we have the following? –

(23)2 = 23 x 23 = 26        (54)3 = 512

Rule 5-

(Xa) b = X axb

 

Standard Form

 

Is a way of writing extremely big or small numbers.

Eg – 3.68 x 109 => 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     2.17x103 =  2170

 

Changing Ordinary Numbers into Standard Form

 

Eg – 3,100,000,000 to standard form = 3.1x109

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 =  3.624x103

 

Adding and Subtracting Standard Form

When we add or subtract standard forms we first change them to ordinary numbers then add/subtract them. Eg

(5.1x105) + (6.3x104) =>  510000 + 63000 =  573000 =  5.73x105

(5.1x105) – (6.3x104) =>  510000 – 63000 =  573000 =  5.73x105

 

Multiplying  and Dividing Standard Form

 

Multiplying

Eg

 (5x103) x (7x104) =   (7x5) x (103 +104) =  35x107 =  3.5x108

(6x108) x (7x10 -3) =  (6x7) x (108+10 -3) =  42 x 105 =  4.2x106

(4.36x105) x (1.4x10-3) =  (4.36x1.4) x(105+10 -3) =  6.104x102

 

Dividing

Eg

(8x105) / (4x102) =  (8/4) x (105 – 102) =  2x103

(2x103) / (4x105) =  (2/4) x (103 – 1-5) =  0.5x10 -2 = 5x10 -3

 

Some examples –

5100 = 5.1 x 103

700000 = 7 x 105

496000 = 4.96 x 105

63000000 = 6.3 x 107

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 doesn’t 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 people’s 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.

0-4, 5-9, 10-14, 15-19,…..

Another interval could be ‘10’

E.g.

0-9, 10-19, 20-29………

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

0-4     0m<5           0-9     0m<10

5-9     5m<10                  10-19 10m<20

10-15 10m<15                 20-29 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 shouldn’t be more than 1 degree from 360 degrees.

*                5’s 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 7th 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 = 3.5th 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 = 25-1 = 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 = 238-210 = 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     1-5               |                200 

            41             |     11    6-16              |                451

            42             |     24    17-40            |              1008

            43             |     16    41-56           |                 688

            44             |       4    57-60           |                 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 30.5th data item. Next find the 30.5th 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 (1-7)                     95                      7*95 =   665

100 ≤ x <110         16(8-23)                   105                  16*105 = 1680

110 ≤ x <120         15(24-38)                 115                  15*115 = 1725

120 ≤ x <130         10(39-48)                 125                  10*125 = 1250

130 ≤ x <140           8(49-56)                 135                    8*135 = 1080

140 ≤ x <150           4(57-60)                 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 don’t 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 = 75.5th 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 (75.5th).

 

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  =  2nd place

                  2          2

Uq place = 3 x lq  =  3 x 2  =  6th place

Lq value = 4  ( 2nd place )

Uq value = 7 ( 6th 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 = 37.75th place

                               2          2

Uq place = 3 x lq place = 113.25th 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             1.92 = 3.61

                              175.3       176.7 – 175.3           1.42= 1.96

                              176.5       176.7 – 176.5           .022 = 0.04

                              177.8       176.7 – 177.8          -1.12 = 1.21

                              179.1       176.7 – 179.1          -2.42 = 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 d2 value

3. Find total of d2 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                 |            d2

  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              d2

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 x2 - x2 

          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      |        x2

338          114,244

354          125,316

341          116,281

351          123,201

353          124,609

Total =    603,651 ( sum of x2)

(ex#2)

 

Variance = r2 = ex2 - x2 = 603,651 – 347.42  =

                        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   |      x2

26        676

28        784

31        961

38      1444

29        841

24        576

23        529

35      1225

Total = 7036 / 8 = 879.5 – 29.252  (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 1/8th (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 don’t.

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 = A2   Ex B x B x B x B2 = B5 6A x 5B = 30AB

 

When numbers are involved we can multiply and divide them as normal.

Eg – 30a3 / 15a = (30/15) x ( a3 / a) = 2a2

 

Simplifying Expressions with Indices

 

Think of this numerically. Lets say a = 2 -  so 2 + 2a = 2 + 4 = 6

 

The mistake is to think that a + a2 = 2a2. this is 2(“02 = 2(4) =8

 

We must only simplify terms that are indices which are exactly the same.

ab + a2b + ab2 + 2ab + a2b2 = 3ab + a2b +ab2 + a2b2

 

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 what’s outside -   6x + 6y.   5(x – 6y) = 5x – 30y.    7p(q – 2p) = 7pq – 14p

 

Simplify –

3x(2y + 2x) – 3y(2x – 2y) =  6xy + 6x2 – 6yx – 6y2

 

6x2 + 6y2 = 12xy2

 

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.

 

Re-arranging 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    =>  3-3 ≤ 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 => -6-4 ≤4x ≤ 12-6 => -10 ≤ 4x ≤ 6 => -2 ½ ≤ x ≤ 1 ½

Sol: -2,-1,0,1

 

f/ -5 ≤ 5x + 8 ≤ 16 => -8-5 ≤ 5x ≤ 16-8 => -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.

 

84         83       82       81       80

4096   512     64       8         1

 

Convert 128 into decimal

 

(1 x 8) + (2 x 1) = 8 + 2 = 1010

 

358 = (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 810 into octal = 108  Convert 23 into octal = 27 Convert 213 = 325

 

84                83                82                81                80 

 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 8’s and 9’s.

27 + 33 = 62 because ---     27

                                          +33 =  7+3 = 1 r 2. 3+2+1 = 6 … 62

 

3758 + 4258 = 10228 (3758 = 27710) + (4258 = 25310) 27710+25310 = 53010.

53010 = 10228

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 –

 

163              162              161              160

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

 

163              162              161              160

4096            256              16                   1

                                        3                  C  3x16=48+12=6010

                     A                 2                  D  (10x256)+(2x16)+13=260510    

  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+16–A(10)=9. 7 becomes 8. B(11)-8=3. Ans = 39

 

AC – 8E = AC

                  8E

                  1E  ---  C(12)+16=28. 28-E(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 1st group of 3 digits covers 0-7

 1       0       1        1        0        1     Similar to 1st column of Octal system

 

In Decimal 4+1 = 5 in 1st column

In Octal 4+1 = 5 in 1st column

In Decimal 32+8 = 40 in columns 1 and 2

In Octal 32+8 = 5 in 2nd column.

So 101101 = 558    

 

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 = 6358       

 

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 = BD16  

 

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 9316 

 

 

 

Octal and Hexadecimal Conversion

 

Method –

 

1/ Convert number to binary

2/ Re-group binary

3/ Convert from binary into groups (octal or hd)

 

Ex.

 

Convert 6A16 to Octal:     =       1528

 

          |         |         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:            =       78F16

 

          |         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