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Mathematics Number & Algebra Revision Notes For
Higher Tier
Thomas Whitham Sixth Form
S J Cooper
Factors, Primes & Prime Factors
Approximations
Fractions: Equivalent, expressing as, fractions of a quantity.
Percentages: Expressing as, Percentage of a quantity, finding the
original amount, compound measures.
Proportion: Ratio, Direct proportion and Inverse proportion.
Standard from.
Surds
Index Notation
Algebra: Collection of like terms, Solving equations
Factorisation
Graphs: Linear & Quadratic.
Solving simultaneous equations using their graphs.
Thomas Whitham Sixth Form Page 1
Factors
A factor is any number which will divide into a given number an exact
number of times.
Example 3 is a factor of 12 since 3 divides into 12 exactly 4 times.
Example List all the factors of 18.
Factors of 18 = {1, 2, 3, 6, 9, 18}
Prime Factors
A prime number is a number who’s factors are itself and 1.
A prime factor is a factor which is a prime number. For example the prime
factors of 12 are found in the diagram below.
The prime factors of 12 are 2 x 2 x 3 = 22 x 3
Example
Express each of the following as products of their prime factors
(i) 135 (ii) 54
24
6
2
4
3 2 2
Thomas Whitham Sixth Form Page 2
(i) (ii)
135 = 5 x 3 x 3 x 3 54 = 3 x 3 x 3 x 2
= 5 x 33 = 33 x 2
NB Here the highest common factor (HCF) = 33 = 27
Approximation
Example
Find an approximate value of 02.0
8.43.54
02.0
8.43.54
135
5 27
3 9
3 3
54
9 6
3 2 3 3
Nearer 50 than 60 Nearer 5 than 4
Nearer 0 than 1. So
leave as 0.02
Thomas Whitham Sixth Form Page 3
02.0
250
02.0
550
02.0
8.43.54
2
25000
= 12500
Example
(a) Estimate 96.1
25.360.2 giving your answer to 2 significant figures
(b) Evaluate 96.1
25.360.2 giving your answer to 2 significant figures
(a) 5.42
9
2
33
96.1
25.360.2
(b) 3.496.1
45.8
96.1
25.360.2
Example Estimate the value of 02.0
83.7986.4 giving your answer to
one significant figure.
20002
4000
02.0
40
02.0
85
02.0
83.7986.4
Clearly here if we round 0.02 off we
have 0 which won’t do!
Simplify numerator
Get rid of decimal by multiplying
top and bottom by 100
Thomas Whitham Sixth Form Page 4
Example
Lewis uses his calculator to calculate 54.1 x 0.036 and gets
the answer 19.476.
Use estimation to work out whether his answer is reasonable.
54.1 x 0.036 50 x 0.04 = 2
Answer is unreasonable, as he is approximately a factor of 10 out.
Fractions
1. Equivalent fractions
Example
Express each of the following fractions in their simplest form.
(a) 6
10 (b)
24
42 (c)
60
84
(a) Here we notice that we have a common factor of 2. Since 2 will divide
into both 6 and 10.
6
10
2 3
2 5
3
5
(b) 24
42
6 4
6 7
4
7
Thomas Whitham Sixth Form Page 5
(c) 60
84
6 10
6 12
10
12
10
12
2 5
2 6
5
6
With practice you will not require the intermediate step but move from
the given fraction to the final answer.
Example
Eight pupils out of ninety two pupils failed to turn up for their Mathematics
examination.
What fraction of the group (a) failed to turn up? (b) did turn up for the
exam?
(a) Fraction who did not turn up = 8
92
2
23 {no marks are awarded for “8 out of 92”}
(b) fraction who did turn up = 21
23 {i.e. the rest}
Example
On Saturday 9000 people won ten pounds on the National Lottery draw.
However 360 people failed to claim their £10 prize. What fraction failed to
claim their prize?
Fraction failed to claim prize = 360
9000
36
900
4
100
1
25
Notice here we have not completely
cancelled down the fraction so we must
repeat the process.
Thomas Whitham Sixth Form Page 6
Example
What fraction has been shaded in for each of the following shapes?
(a)
Here there is a total of 15
squares, of which 3 are
shaded
fraction shaded = 3
15
1
5
(b)
Fraction shaded = 15
24
5
6
2. Fraction of a calculator (without calculator)
Example Find 1
448 of
1
448 of means
1
448 {i.e. 48 divided by 4}
Answer = 12
However we have a technique for showing our working as follows:
1
448 12
1
12
Example Find 3
570 of
Thomas Whitham Sixth Form Page 7
3
570 of means
3
570
{again we could find 1
5
of 70 by dividing 70 by 5}
{so we find 3
5 of 70 by dividing 70 by 5 and then multiplying the answer by 3}
3
570
3
570 42
1
14 of
Example Find 2
7238 of
2
7238
2
7238 68
1
34 of 7 23 8
3 42
Example
In a school with 720 pupils, 9
10 stay in school at lunch time, and
3
8 of
these pupils bring a packed lunch. How many pupils bring a packed lunch?
Number staying at school = 9
10720 648
1
72
7 2
9
2 4 3
Number with packed lunch = 3
8648 243
1
81 8 64881
More complicated divisions may require
some additional calculations at the side
of your page
Thomas Whitham Sixth Form Page 8
3. Fraction of a quantity (with calculator)
Example Find 2
37 41 of £ .
2
37 41 of £ . means
2
37 41 £ .
Using the calculator we have 2 7 41 3 .
Answer = £4.94 [Don’t forget the units]
Example Find 3
4 of 1488km
3
4
3
4 of 1488km = 1488 = 1116km
Example
Mr Ashworth earns £295.68 in one week. After tax is deducted, he receives
only five sevenths of this amount. How much does he receive?
We require 5
7
5
7 of £295.68 = 295.68 = 211.2
He receives £211.20 [remember currency has two decimal places]
4. Fractions to percentages/decimals
Example
Express each of the following as fractions in their simplest form.
(a) 45% (b) 0.34 (c) 2.6 (d) 12%
Thomas Whitham Sixth Form Page 9
(a) 45% = 45
100
9
20 (b) 0.34 =
34
100
17
50
(c) 2.6 = 26
102
3
5 (d) 12% =
25
3
100
12
5. Addition and subtraction of fractions
Example Work out 4
7
5
7
4
7
5
7
9
71
2
7
Which is easily done provided the denominators are the same.
Example Work out 3
4
5
12
First change the first fraction with its equivalent in twelfths.
3
4
3 3
3 4
9
12
3
4
5
12
9
12
5
12
14
121
2
121
1
6
Thomas Whitham Sixth Form Page 10
Example Work out 51
32
2
5
51
32
2
57
5
15
6
15
711
15
Example Work out 31
71
2
3
31
71
2
32
3
21
14
21
124
21
14
21
110
21
Example
A piece of wood is 72
5 metres long. If 1
7
15 metres is cut off, what length
of wood is left?
72
51
7
156
6
15
7
15
521
15
7
15
514
15
First add the whole numbers together.
Next the common denominator is 15.
hence replace each fraction with its
equivalent in terms of fifteenths.
Subtract the whole numbers and
place each fraction with common
denominator 21.
Since we cannot subtract 14 from 3 w
must use one of the whole numbers.
hence 321 becomes 24
21 and we now
have only 1 whole one.
Length of wood left =
Thomas Whitham Sixth Form Page 11
6. Multiplication and Division of fractions
Example Work out 2
3
5
7
2
3
5
7
2 5
3 7
10
21
Example Work out 4
9
3
7
4
9
3
7
4
213
1
Example Work out 11
22
2
5
11
22
2
9
3
2
20
9
10
3
31
3
1
1
10
3
To multiply any two fractions
together simply multiply
numerators together and then
multiply denominators together.
In the event that a number on the
numerator has a common factor to a
number on the denominator, cancel
to start with. i.e. 3 will divide exactly
into both 3 and 9.
Change fractions to top heavy fractions.
Next cancel where possible
Thomas Whitham Sixth Form Page 12
Example Work out 10
13
5
6
10
13
5
6
10
13
6
5
12
13
2
1
Example Work out 35
91
1
2
35
91
1
2
32
9
3
2
9
32
2
3
18
96
Example
Find the exact value of ba
11 when
3
2a and
5
4b
2
3
2
31
3
2
11
a That is the reciprocal of
3
2 is
2
3
Similarly the reciprocal of 5
4 is
4
5
1. Invert the second fraction and
change the ‘‘ to a ‘x’ sign.
2. Cancel where possible
3. multiply across.
change fractions to top heavy fractions
Invert second fraction and cancel where possible.
Finally multiply across
Thomas Whitham Sixth Form Page 13
4
32
4
11
4
5
4
6
4
5
2
311
ba
Example
Louise has 71
4m of ribbon.
She makes 9 skirts and uses 2
5 of a metre for each one. How much ribbon
does she have left?
5
33
5
18
5
29
Amount left = 5
33
4
17
= 5
3
4
14
m20
133
20
12
20
253
20
12
20
54
Place over common
denominators
Subtract whole quantities
Since we cannot subtract 12
from 5 here , use one of the
whole ones as 20
20
Thomas Whitham Sixth Form Page 14
Example
Express each of the following as fractions in their simplest form
a) 0.237
b) 0.2373737.....
c) 0.1373737....
a) 1000
237237.0
b) Let ...237237237.0x
...237237.2371000 x
Subtract from previous expression gives 237999 x
Hence 333
79
999
237x and since ...237237237.0x
333
79...237237237.0
c) Again Let ...1373737.0x
...73737.13100 x
Subtract from previous expression gives 6.1399 x
Hence 495
68
990
136
99
6.13x
Thomas Whitham Sixth Form Page 15
Percentages
1. Expressing as a percentage
Example Express each of the following percentages as
(a) Decimals (b) Fractions in their simplest form.
(i) 30% (ii) 46% (iii) 2% (iv) 154%
(a)(i) 30% = 3.0100
30 (ii) 46% = 46.0
100
46
(iii) 2% = 02.0100
2 (iv) 154% = 54.1
100
154
(b)(i) 30% = 10
3
100
30 (ii) 46% =
50
23
100
46
(iii) 2% = 50
1
100
2 (iv) 154% =
50
271
100
541
100
154
Example Mark scored 32 out of 40 in a recent mathematics test.
Express his score as a percentage.
Test result = %8010020
16
40
32 5
1
Thomas Whitham Sixth Form Page 16
Example The height of a tree increased from 2.73m to 2.98m in one
year. What percentage increase is this?
Increase = 2.98 – 2.73 = 0.25
Percentage increase = %16.910073.2
25.0
2. Percentage of a quantity
a) Without a calculator
Example What is 20% of 40 kg
Method 1: Using fractions
20% = 5
1
100
20 as a fraction
hence 20% of 40kg = 5
1 of 40 = kg840
5
1 8
1
Method 2 Using unity
10% of 40kg = 4 kg {i.e. divide by 10}
20% of 40 kg = 2 x 4 = 8kg
Example What is 35% of £60?
Method 1: Using fractions
35% = 20
7
100
35 as a fraction
Thomas Whitham Sixth Form Page 17
hence 35% of £60 = 20
7 of 60 = 21£60
20
7 3
1
Method 2 Using unity
10% of £60 = £6 {i.e. divide by 10}
5% of £60 = £3 {i.e. half of 10%}
30% of £60 = 3 x 6 = £18
35% of £60 = 18 + 3 = £21
Example What is 25% of 144cm?
Method 1: Using fractions
25% = 4
1
100
25 as a fraction
hence 25% of 144cm = 4
1 of 144 = cm36144
4
1 36
1
63
4144 2
Method 2 Using unity
10% of 144cm = 14.4cm {i.e. divide by 10}
5% of 144cm = 7.2cm {i.e. half of 10%}
20% of 144cm = 2 x 14.4 = 28.8cm
25% of 144cm = 28.8 + 7.2 = 36cm
63
4144 2
Division more complicated so
done at the side of our page
Thomas Whitham Sixth Form Page 18
Example
The price of a new television is £176 plus VAT at 17½%.
(a) Work out the VAT to be added to the price of the television.
(b) What is the total cost of the television?
(a) 10% of £176 = £17.60
5% of £176 = £ 8.80 {i.e. half of 10%}
2½% of £176 = £ 4.40 {i.e. half of 5%}
17½% of £176 = £30.80 =VAT {i.e. add the previous answers together}
(b) Total cost = 176+30.80 = £206.80
Example The number of girls attending football matches is expected
to increased by 3% this year. If there were 12500 girls
attending matches last year, how many girls are expected
to attend this year?
1% of 12500girls = 125 girls {i.e. divide by 100}
3% of 12500girls = 3 x 125 = 375girls
Number of girls = 12500 + 375 = 12875
b) With a calculator
Example What is 23% of 40 kg
With a calculator there is no need to cancel down fractions or use the unity
method. Simply set out your sum as a fraction and use the calculator.
23% of 40 kg = kg2.940100
23
Thomas Whitham Sixth Form Page 19
Keys used:
(there are alternative combinations for inputs to the calculator)
Example What is 17% of £6.50?
17% of £6.50 = 11.1£105.150.6100
17
{here we must round up the answer since money involves 2 decimal places}
Example I pay 6% of my salary into a pension fund. How much do I
pay into my fund if my salary is £12450 per annum?
Amount paid = 6% of £12450 = 747£12450100
6
Example Every year a car loses 15% of its value at the beginning of
that year. If it was originally worth £5000, what will it be
worth after 2 years?
Method 1: Find the amount lost and then subtract from the original.
Amount lost = 15% of 5000 = 750£5000100
15
Value after 1st year = 5000 – 750 = £4250
Amount lost in 2nd year = 15% of 4250 = 50.637£4250100
15
Value after 2nd year = 4250 – 637.50 = £3612.50
X
4 0 ÷ 1 2 3 = 0 0
Thomas Whitham Sixth Form Page 20
Method 2: Using the percentage Loss we can determine the percentage
left!
If the value of the car is 15% less each year, then the new value will be
85% of the original.
(i.e. 85% + 15% = 100%)
value after 1st year = 85% of 5000 = 4250£5000100
85
Or better still 4250£500085.0
value after 2nd year = 85% of 4250 = 50.3612£4250100
85
{answer comes out quicker!}
Example
The population of Umbridge increased by 37% during the years 1960-80.
If the population in 1960 was 447, what was the population in 1980?
An increase of 37% means than the new population is 137% what it was!
{i.e. an increase implies we add on to 100%}
New population = 137% of 447 = 61239.612447100
137
Or better still 61239.61244737.1
Thomas Whitham Sixth Form Page 21
Example
This year nurses were given a 3.5% pay rise. If Susan earned £16 540 per
annum last year how much more will she get in her pay packet this year?
What will be her new salary?
Increase of 3.5% means that Nurses now earn 103.5% what they did the
year before!
New Wage = 103.5% of £16 540 = 75.17025£16450100
5.103
3. Finding the original percentage
Here we make use of the formula:
Example The population of Villanova has increased by 63% during
the last five years and is now 124 000. What was its
population five years ago?
Here we have been told the answer. That is the new value is 124 000.
after an increase of 63%
163% of original value = 124 000
{i.e. increase means 63+100%}
1% of original value = ...73.760163
124000
{i.e divide by 163 to find 1%}
New value = Percentage of Original value
Or
New value = Percentage x Original value
Thomas Whitham Sixth Form Page 22
100% of original value = original value = 76074...73.760100
NB original value is always 100% and we have either increased it or
decreased it to find the new value
Example The price of houses in Villanova has increased by 18%
during the last year. If the house costs $45 000 now, what
would it have cost a year ago?
Here we have been told the answer. That is the new value is $45 000.
after an increase of 18%
118% of original value = 45 000 {i.e. increase means 18+100%}
1% of original value = ...355.381118
45000
{i.e divide by 118 to find 1%}
100% of original value = original value = 59.38135$...355.381100
Example The attendance of Burnley football club fell by 7% in 2001.
If 2030 fewer people went to matches in 2001, how many
went in 2000?
Here we have been told the answer. That is the new value is 2030.
which represents the 7%!
7% of original value = 2030
1% of original value = 2907
2030 {i.e divide by 7 to find 1%}
100% of original value = original value = 29000.290100 people
Thomas Whitham Sixth Form Page 23
Example During a Grand Prix race, the tyres on a car are reduced in
weight by 3%. If they weigh 388 kg at the end of the race,
how much did they weigh at the start?
Here we have been told the answer. That is the new value is 388kg.
after an decrease of 3%
97% of original value = 388 {i.e. increase means 100 – 3%}
1% of original value = 497
388 {i.e divide by 97 to find 1%}
100% of original value = original value = kg4004100
Example A car, which failed its MOT test, was sold for £456, thereby
making a loss of 35% on the cost price. What was the cost
price?
Here we have been told the answer. That is the new value is £456. after
an decrease of 35%
65% of original value = 456 {i.e. increase means 100 – 35%}
1% of original value = ...015.765
456 {i.e divide by 65 to find 1%}
100% of original value = original value = 54.701£...015.7100
Thomas Whitham Sixth Form Page 24
Example
In 2002 the population of England was 49,561,800.
The population of England is increasing by an annual rate of 0.3 per cent.
a) Write down the single number that we must multiply 49,561,800 by
if we wish to calculate an estimate for what the population was in
2004.
b) Assuming that England’s population continues at this same rate,
calculate the population of England in 2012.
a) increase implies add on to 100%, therefore 100.3% = 1.003
This is the number which must be multiplied to 49,561,800 if we want the
population in 2003. For 2004 we must multiply by 1.003 again.
Hence number must be 1.0032 = 1.006009
b) Population in 2012 = 1.00310 x 49561800 = 4,700,516.
Example
In 2008 the State Pension was increased by 2.5 per cent to £95.25 a
week.
What was the state pension before this increase?
Increase 102.5% of original value = 95.25
1.025 x Original pension = 95.25
Pension was 93.92£025.1
25.95
Thomas Whitham Sixth Form Page 25
Compound Measures
Example
A bank pays interest of 4% on money in deposit accounts. Mr Smith Puts
£2000 in the bank. How much has he after
a) One year b) three years?
a) After one year
Amount = 104% of 2000 = 1.04 x 2000 =£2080
b) After three years
Method 1 In 2nd year Amount = 104% of 2080 = 1.04 x 2080 = £2163.20
in 3rd year Amount = 104% of 2163.20 = 1.04 x 2163.20
= £2249.73
Method 2
After three years we have multiplied 1.04 a total of three times. i.e. 1.043
After three years Amount = 1.043 x 2000 = £2249.73
In general: For an initial amount of £P at an annual rate of interest r%, the
amount in the account after n years will be A where A is worked out using
the formula below.
However don’t learn the formula, just learn its meaning!
nr
PA
1001
Thomas Whitham Sixth Form Page 26
Example
The population of an island increases by 8% each year. If the population in
2009 was 10 million, what is the expected population of the island in 2015?
Here an increase by 8% means 108% or 1.08
2009 to 2015 means the increase over 6 years.
Expected population = 1.086 x 10 =15.87 million (rounded to 4 significant
figures)
Example
A new car is valued at £15 000. At the end of each year its value is reduced
by 15% of its value at the start of the year. What will it be worth after 6
years?
Here a decrease by 15% means it’s worth 85% or 0.85
Value after 6 years = 0.856 x 15000 = £5657.24
Ratio & Proportion
Example Simplify each of the following ratios
a) 4 : 28
b) 18 : 27
c) £3 : £1.80
d) 300m : 5.1km
e) 1250 cm2 : 3 litres
a) 4 : 28 = 1 : 7 {Divide both sides by 4}
b) 18 : 27 = 2 : 3 {Divide both sides by 9}
c) £3 : £1.80 = 300 : 180 {Change both into pence}
= 30 : 18 { Divide by 10}
= 5 : 3 { Divide by 6}
Thomas Whitham Sixth Form Page 27
d) 300m : 5.1km = 300 : 5100 { Change both into metres}
= 3 : 51 { Divide by 100}
= 1 : 17 { Divide by 3}
e) 1250 cm2 : 3 litres = 1250 : 3000 {Change both into cm2}
= 25 : 60 {divide both sides by 50}
= 5 : 12 {divide both sides by 5}
Example A school decides to give 20% of the proceeds of a jumble
sale to charity and the rest to the school fund. In what ratio
are the proceeds to be divided?
Charity to school fund = 20 : 80 = 1 : 4
Example One bottle of wine holds 750 cm2 whereas another holds
1.2 litres. Give the simplest ratio of their capacities.
Ratio = 750cm2 : 1.2 litres = 750 : 1200 {change units into cm2}
= 15 : 24
= 5 : 8
Example Jane took 45 minutes to do her homework, but her sister
Lucy took 1¼ hours. What is the simplest ratio of their
times taken?
Thomas Whitham Sixth Form Page 28
Jane to Lucy = 45 : 1¼ = 45 : 75 {change units into minutes}
= 9 : 15 = 3 : 5
Example The standard gauge of railway track is 1.43m. A model is to
be made with gauge 11mm. Calculate the scale in the form
1 : n, where n>1.
Model to Standard gauge = 11mm : 1.43m
= 11 : 1430 {change into mm}
= 1 : 130
Example £420 is divided between two people in the ratio 2 : 5. Work
out what each person will receive.
There are two ways of looking at this problem.
Method 1
There are a total of 7 parts 7 parts represents £420
1 part represents 60£7
420
2 parts = £60 x 2 = £120 and 5 parts = £60 x 5 = £300
Method 2
The fraction for the first person is 2 out of 7 parts, that is 7
2 and the
fraction for the second person will be 7
5
Thomas Whitham Sixth Form Page 29
First person = 120£4207
2 and
second person = 420 - 120 = £300
Both methods work for most cases.
For the following I will use method 1
Example A sum of money is divided in the ratio 3 : 4 and the smallest
share is equivalent to £27. What is
(a) the amount given to the largest share.
(b) the total amount of money shared?
The smallest share is worth 3 parts so here 3 parts was £27.
1 part = 9£3
27
(a) Largest share = 4 x 9 = £36
(b) Total amount = 36 + 27 = £63
Example Two lines have lengths in the ratio 5 : 2. If the longer line is
15 cm long, find the length of the other line.
Longer line = 5 parts so 5 parts = 15 cm
1 part = cm35
15
Other line = 2 parts = 2 x 3 = 6cm
Thomas Whitham Sixth Form Page 30
Example The ratio of my gas bill to my electricity bill was 13 : 5. If my
gas bill was £182, how much was my electricity bill?
Gas bill = 13 parts so 13 parts = £182
1 part = 14£13
182
Electricity bill = 5 parts = 5 x 14 = £70
Example Three people stake £10 on the national lottery and win
£850. Peter paid £2, John paid £4.50 and Claire paid the
rest towards the stake. The winnings are shared in the ratio
of the contributions. How much does Claire receive?
Ratio of share = Peter to John to Claire = £2 : £4.50 : £3.50
= 200 : 450 : 350
= 4 : 9 : 7
Hence out of 20 parts of the money Claire will receive 7 parts.
20 parts = £850
1 part = 50.42£20
850
Calire = 7 parts = 7 x 42.50 = £297.50
Thomas Whitham Sixth Form Page 31
Example
For every 9 teenagers who like pop music there are 2 that does not.
In a youth club of 187 members, how many do not like pop music?
Ratio = Like pop to not like pop = 9 : 2
11 parts = 187
1 part = 1711
187
Not like pop = 2 parts = 2 x 174 = 34 people
Example A lorry is loaded up with fruit and vegetables for market.
The mass of fruit to vegetables is in the ratio of 7 : 8. If the
lorry’s load is 18.6 tonnes, find the mass of fruit and the
mass of vegetables it is carrying.
15 parts = 18.6 tonnes
1 part = 24.115
6.18 tonnes
Fruit = 7 parts = 7 x 1.24 = 8.68 tonnes
Vegetables = 8 parts = 8 x 1.24 = 9.92 tonnes
Example When £195 is divided in the ratio 2 : 4 : 7, what is the
difference between the largest share and the smallest?
Thomas Whitham Sixth Form Page 32
Difference between the largest and smallest share is 5 parts(7 -2)
13 parts = £195
1 part = 15£13
195
Difference = 5 parts = 5 x 15 = £75
Example
A man and a woman share a bingo prize of £1000 between them in the
ratio 1 : 4. The woman shares her part between herself, her mother and her
two daughters in the ratio 2 : 1 : 1
How much does the woman receive?
5 parts = £1000
1 part = 200£5
1000
Woman’s original share = 4 parts = 4 x 200 = £800
Hence 4 parts = £800
1 part = 200£4
800
Woman’s final share = 2 parts = 2 x 200 = £400
Thomas Whitham Sixth Form Page 33
Example £400 is divided between Ann, Brian and Carol so that Ann
has twice as much as Brian and Brian has three times as
much as Carol. How much does Brian receive?
If Carol received one part Brian would have to receive three parts hence
Ann would have to receive six parts. This gives the ratio
Ann : Brian : carol = 6 : 3 : 1
Hence 10 parts = £400
1 part = 40£10
400
Brian = 3 parts = 3 x 40 = £120
Example
Mrs Simms inherits £24 000.
She divides the money between her three children, aged 9, Alice, Brenda,
aged 7 and Charles, aged 8,in the ratio of their ages.
How much does Charles receive?
Ratio = 9 : 7 : 8
24 parts = 24 000
1 part = 1 000
Charles = 8 parts = £8 000
Thomas Whitham Sixth Form Page 34
Direct Proportion
If two quantities are directly proportional to one another then one can be
written as a constant (k) multiplied by the other.
In order to find the constant k, more information needs to be provided.
Example
W and P are both positive quantities.
W is directly proportional to the square of P.
When W = 12, P = 4.
(a) Express W in terms of P.
(b) What is the value of W when P = 6?
(c) What is the value of P when W = 75?
a) Using the definition above
W is directly proportional to the square of P means W = k x P2
Using the information W = 12, P = 4 12 = k x 42 or 12 = k x 16
k = 75.04
3
16
12
W = 4
3P2
b) P = 6 W = 27364
36
4
3 2
c) W = 75 2
4
375 P
= k x
Don’t forget once k is found to
write the equation down
Thomas Whitham Sixth Form Page 35
2
75.0
75P
2100 P
10P
Example
Y are X are both positive quantities.
Y is directly proportional to the square root of X.
When Y = 16, X = 16
a) Express Y in terms of X
b) What is the value of Y when X = 25?
c) What is the value of X when Y = 60?
a) XkY
Y = 16, X = 16 1616 k
4
416
k
k
XY 4
b) X = 25 2054254 Y
c) Y = 60 X460
X15
225
152
X
Thomas Whitham Sixth Form Page 36
Inverse Proportion
If two quantities are inversely proportional to one another then one can be
written as a constant (k) divided by the other.
In order to find the constant k, more information needs to be provided.
Example
Given that M varies inversely to P and that M = 12 when P = 4
a) Obtain an expression for M in terms of P
b) What is the value of M when P = 8?
c) What is the value of P when M = 0.5?
a) P
kM
M = 12, P = 4 4
12k
k412
48k
PM
48
b) P = 8 68
48M
k =
Thomas Whitham Sixth Form Page 37
c) M = 0.5 P
485.0
965.0
48P
Example
y is inversely proportional to the square root of x.
When y = 6, x = 9.
a) What is the value of y when x = 4.
b) What is the value of x when y = 10.
y is inversely proportional to the square root of x means x
ky
When y = 6, x = 9. 9
6k
or 3
6k
k36
18k x
y18
a) X = 4 92
18
4
18y
b) Y =10 x
1810
10
18x
8.1x
24.3x
Using the algebraic knowledge that
the P and 0.5 can be “swapped”
Thomas Whitham Sixth Form Page 38
Standard form
A number expressed in standard form is a number written between 1 and
10 multiplied by 10 to an appropriate power.
The use of standard form is to represent very small numbers or very large
numbers
For example
0.000 000 000 000 32 represented in standard form will be 13102.3
214000 000 000 represented in standard form will be 111014.2
Example Express each of the following in standard form
(i) 0.000 000 462
(ii) 0.004
(iii) 90 000
(iv) 5910 000 000
(i) 0.000 000 462 = 71062.4
(ii) 0.004 = 3104
(iii) 90 000 = 4109
(iv) 5910 000 000 = 91091.5
Example Express each of the following as ordinary numbers
(i) 8106.1
(ii) 61054.2
(iii) 12107
(iv) 210414.9
(i) 8106.1 = 160 000 000
(ii) 61054.2 = 0.000 00254
(iii) 12107 = 7000 000 000 000
(iv) 210414.9 = 0.09414
Thomas Whitham Sixth Form Page 39
Example The surface of the earth is about 509 970 000 km2. Express
this in standard form correct to two significant figures.
509 970 000 = 8101.5 km2
Example
Given that 4106A and 7104B work out, without a calculator
(i) AB (ii) B
A
(i) (ii)
12
11
74
74
104.2
1024
10106
104106
AB
3
7
4
7
4
105.1
10
10
4
6
104
106
B
A
NB Here we used the laws of indices. That is nmnm aaa and
nm
n
mnm a
a
aaa
Example
Given that 5102.1 X and 9105 Y work out , in standard form
(i) XY (ii) Y
X
(i) (ii)
14
95
106
105102.1
XY
3
9
5
104.2
2400
105
102.1
Y
X
Here we use the button for the standard form and type for X EXP
Thomas Whitham Sixth Form Page 40
And for Y we type
Example
A rectangle has length AB = 5103.1 km and width BC = 4105.2 km.
Giving your answer in standard form, find
(i) the area,
(ii) the perimeter of this rectangle.
(i) Area = 3250000000105.2103.1 45
{this number is not in standard form}
Area = 91025.3 km2
(ii) Perimeter = 50000260000105.22103.12 45
= 310000 = 5101.3 km
Example
The weight of 1 grain of sand is given as 91043 grams.
a) (i) Write 91043 in standard form
(ii) What is the weight of 5 billion grains of sand? [1 billion is 1000 million]
b) A piece of sandstone weighs 1kg.
How many grains of sand is this equivalent to?
a) (i) 89 103.41043
(ii) Weight = 215103.45000000000 8 grams
b) Number of grains of sand = 10
81033.2
103.4
1000
{since 1kg = 1000g}
EXP 1 . 2 – 5
EXP 5 – 9
Thomas Whitham Sixth Form Page 41
Example 7104.5 p and 6105.3 q
Calculate the value of each of the following.
Give all your answers in standard form.
(a) qp
(b) qp 85
(c) qp
(a) 1467 1089.1105.3104.5 qp
(b)
867 1098.2298000000105.38104.5585 qp
Here the typing could be done as I have done above or you could break it
down. Simply type:
(c) 767 1005.550500000105.3104.5 qp
Example The space shuttle is covered with 4101.3 heat resistant
tiles. The total surface area of the shuttle is 15102 m2.
How many tiles per m2 does the shuttle have?
Number of tiles per m2 = 10
4
15
1045.6101.3
102
Example The mean weight of all men, women, children and babies in
the UK is 42.1kg. The population of the UK is 56 million.
Work out the total weight of the entire population giving
your answer in standard form.
Total weight = 9103576.223576000001.4256000000
EXP 5 . 4 7 + 5 X EXP 3 . 5 6 = 8 X
Thomas Whitham Sixth Form Page 42
Example
(a) At 300 000km/s, how long, to one significant figure, does it take
light to travel from the sun to Saturn, a distance of 9104.1
km?
(b) An electron carries a charge of 19106.1 coulomb. To one
significant figure, how many electrons are required for a total
charge of 1 coulomb?
(a) Using the formula for distance, speed time
Time taken = ssec..666.4666300000
104.1 9
(b) Number of electrons = 1818
191061025.6
106.1
1
(to 1 s.f.)
Example On an antique map (on which no scale is shown) the
distance between Oakford and Stanton is 3.6 cm whereas in
fact the distance as the crow flies is 19km. It is required to
express the scale in the form 1 : n to an accuracy of 2
significant figures in standard form.
Oakford to Stanton = 3.6 cm : 19km
= 3.6 : 1900000 (measurements in cm)
= 1 : 6.3
1900000
= 1 : ...7777.527777
= 1 : 5103.5
D
S T
Thomas Whitham Sixth Form Page 43
Example
The mass of one atom of oxygen is given as 231066.2 grams.
The mass of one atom of hydrogen is given as 241067.1 grams.
a) Find the difference in mass between one atom of oxygen and one
atom of hydrogen.
b) A molecule of water contains two atoms of hydrogen and one atom of
oxygen.
i) Calculate the mass of one molecule of water.
ii) Calculate the number of molecules in 1 gram of water.
a) difference = 232423 10493.21067.11066.2
b) i) Mass of 1 molecule of water =
232423 10994.21067.121066.2 grams
ii) Number of molecules = 22
231034.3
10994.2
1
Surds 5,3,2 are irrational numbers expressed in surd form.
A rational number is one which can be written in the form q
p where p and
q are integers. An irrational number cannot be written in this form.
2 for example is an irrational number, and so is the numbers .
Numbers written in the form a are called surds.
Laws
abba
b
a
b
a
Thomas Whitham Sixth Form Page 44
Special case
Example Simplify (i) 12 (ii) 27 (iii) 75
Hence simplify 752712
0
353332752712
= =
= =
= =
aaa
34 32
39 33
325 35
Thomas Whitham Sixth Form Page 45
Example
Express with rational denominator 5
20
Here we multiply by one! However we make one 5
5 as this will help us.
545
520
5
5
5
20
5
20
Example
Given that 3
91227 can be express in the form 𝑎 3. Find the
value a.
333927
323412
333
39
3
3
3
9
3
9
323332333
91227
From the fact that
555
Thomas Whitham Sixth Form Page 46
Example simplify (i) 1898 (ii) 4875
(i) 242327292491898
(ii) 4
5
34
35
316
3254875
Index notation
Laws
Special Cases 10 a
aa 2
1
, n aa n 1
,
n
n
aa
1 , n
na
a
1 ,
nn
a
b
b
a
nmnm aaa
nmnm aaa
mnnm aa
Thomas Whitham Sixth Form Page 47
n maa nm
mn a
Example 288 33
1
Example 3
1
9
1
9
19
2
1
2
1
Example
(a) Write down the value for each of the following
(i) 06 (ii) 32
27 (iii) 43
(i) 160 {anything to the power 0 is 1}
(ii) 93272722
332
(iii) 81
1
3
13
4
4
(b) Simplify 25.0 336
3
2
9
16
3
136336
2
25.0
Example
Simplify each of the following (i) 34 xx (ii) 75 yy (iii) 53t
(i) 734 xxx {Add the powers}
Power 0.5 means square root
Thomas Whitham Sixth Form Page 48
(ii) 275 yyy or 2
1
y {subtract the powers}
(iii) 1553 tt {multiply the powers}
Example
Simplify
24
323
6
23
ba
baa
ba
ba
ba
ba
baa
ba
baa
5
24
39
24
363
24
323
4
6
24
6
83
6
23
Calculator calculations
Example
a) Use your calculator to find 22 6.102.37
b) express this number correct to 3 significant figures.
Working out 2 cubed
Multiplying powers for a
Multiplying powers for b
Adding powers for a on
numerator Working out
3 x 8
Subtracting powers for b Dividing 24
by 6
Subtracting powers for a
Thomas Whitham Sixth Form Page 49
a) To ensure you get it right work out 37.22-10.62 first on the calculator.
Then square root.
37.22-10.62 = 1271.48
35.65781822
b) 3 significant figures means the first three numbers of value. Which is 35.6
However the number after the 6 is a 5 and therefore we must round up (as
we do with decimal places) Answer = 35.7
Example
Find the value of 65 107.4105.3 , giving your answer in standard
form.
Algebra
Collection of like terms
When collecting like terms remember we can collect together
equivalent letters i.e. aaa 1037 and we can collect together numerical terms i.e. 4 + 8 – 3 = 9
However we cannot collect together terms that are not alike i.e.
ba 54 cannot be simplified. Nor can 43 a Example Simplify each of the following
a) bababa 855439 {Remember to take note of the sign
in front of each letter}
b) yxyxyx 87626
c) fedfdefed 27647324
𝒙 Ans EXE
Thomas Whitham Sixth Form Page 50
d) qpqpqp 553782
Solving simple equations
Example Solve each of the following equations
a) 1743 x
b) 7325 xx
c) 9457 xx
d) 19231 xx
e) 1214 x
f) 7523 x
g) 53
52
x
h)
95
73
x
Golden rules The equation starts balanced and must remain balanced! So whatever you do to one side you must do to the other side.
i.e. (a) 1743 x
417443 x {Add 4 to both sides of the equation}
213 x {Simplify}
3
21
3
3
x {Divide both sides by 3}
7x {Simplify giving answer}
Thomas Whitham Sixth Form Page 51
An alternative way of thinking! When moving a number from one side of
the equal sign to the other we perform the opposite operation.
The opposite of Addition is subtraction and visa versa.
The opposite of Multiplication is Division and visa versa.
(b) 7325 xx
52
{Simplify} 52
subtract} andover 3 the{take 535
2}subtract andover 2 the{take 2735
7325
.x
x
xxx
xx
xx
(c) 9457 xx
33333.13
4
43
447
5947
9457
x
x
xx
xx
xx
(d) 19231 xx
Thomas Whitham Sixth Form Page 52
4
{Simplify} 520
add} andover 3 the{Move 3220
}add! andover 19 the{Move 23191
19231
x
x
xxx
xx
xx
(e) 1214 x
4
164
before} as rearrangecan we{now 4124
first} brackets {remove 1244
1214
x
x
x
x
x
(f) 7523 x
333333.13
4
6
8
86
1576
7156
7523
x
x
x
x
x
(g) 53
52
x
Thomas Whitham Sixth Form Page 53
5
102
5152
times}andover 3 the{take 1552
53
52
x
x
x
x
x
(h)
95
73
x
8
243
21453
bracket} the{Remove 45213
times}andover 5 the{Take 4573
95
73
x
x
x
x
x
x
Factorisation
Example Factorise each of the following
a) 342 xx
b) 652 xx
c) 1582 xx
d) 542 xx
e) 1522 xx
f) 62 xx
g) 2832 xx
Thomas Whitham Sixth Form Page 54
When factorising a quadratic such as 342 xx we express as two
brackets multiplied together. i.e. bxax where a and b are numbers
that
1. Multiply to give, in this case, 3
2. add to give 4.
This means they can be 1 and 3 or –1 and –3.
Since 1 and 3 add to give 4 then
31342 xxxx
(b) 652 xx
Here the numbers could be 1 and 6, –1 and –6, 2 and 3 or –2 and –3.
Since –2 and –3 add to give –5 the answer is found.
32652 xxxx
(c) 1582 xx
Here the numbers could be 1 and 15, -1 and –15, 3 and 5 or –3 and –5.
Since 3 and 5 add to give 8
531582 xxxx
(d) 542 xx
Here the numbers could be 1 and –5 or –1 and 5 (one positive and one
negative.)
Since 1 and –5 add to give –4.
51542 xxxx
Thomas Whitham Sixth Form Page 55
(e) 1522 xx
Here the numbers could be 1 and –15, –1 and 15, 3 and –5 or –3 and 5
Since –3 and 5 add to give 2
531522 xxxx
(f) 62 xx
Here the numbers could be 2 and –3, –2 and 3, 1 and –6 or –1 and 6
Since 2 and –3 add to give –1.
3262 xxxx
(g) 2832 xx
Here the numbers could be 1 and –28, –1and 28, 2 and –14, 14 and –2, 4
and –7 or –4 and 7.
Since –4 and 7 add to give 3.
742832 xxxx
Example Factorise each of the following:
a) xyx 32
b) qppq 22 4
c) mnmm 6104 2
d) abcbaba 223 37
Thomas Whitham Sixth Form Page 56
Here we cannot place into two brackets since it does not follow the
pattern of 2x followed by x, followed by a constant!
However we have a common factor. So place the common factor
outside a bracket.
a) 332 xyxxyx
b) pqpqqppq 44 22
c) nmmmnmm 35226104 2
d) cabaababcbaba 3737 2223
Difference of squares
Example factorise each of the following
(i) 𝑥2 − 36 (ii) 4𝑥2 − 49
(i) Here 𝑥2 − 36 represents the difference of two squares which
are 𝑥2 − 62. When we factorise 𝑥2 − 36 we are looking for
two numbers that have a sum of zero (for the middle term) this
is +6 − 6 = 0
Hence 𝑥2 − 36 = 𝑥 − 6 𝑥 + 6
(ii) 4𝑥2 − 49 = 2𝑥 − 7 2𝑥 + 7
Since
yxxyx 2
Since
xx 33
2𝑥 2 72
Thomas Whitham Sixth Form Page 57
Example (a) factorise the quadratic 1272 xx
(b) Hence solve the equation 01272 xx
(a) 431272 xxxx {since –3 and –4
add to give –7.
(b) 01272 xx is the same as saying 043 xx
Which means 03 x or 04 x
Since 0ba means 0a or 0b
3x or 4x
Example (a) factorise the quadratic 2762 xx
(b) Hence solve the equation 02762 xx
(a) 392762 xxxx
(b) 02762 xx
039 xx
09 x or 03x 9x or
3x
Difference of two squares
Example Factorise each of the following:
Thomas Whitham Sixth Form Page 58
Inequalities
Example Solve each of the following inequalities
a) 932 x
b) 1973 x
c) 8314 xx
d) 7335 xx
e) 9142 x
f) 73
5
x
g) 252 x
h) 812 x
Solving inequalities can be like solving equations. However don’t
forget to write the correct symbol; and not the equal sign!
a) 932 x
6
{Simplify} 122
side}other the to3 {Add 392
932
x
x
x
x
b) 1973 x
4
123
7193
1973
x
x
x
x
Thomas Whitham Sixth Form Page 59
c) 8314 xx
7
734
1834
8314
x
xx
xx
xx
d) 7335 xx
2
42
435
3735
7335
x
x
xx
xx
xx
e) 9142 x
8
11
118
298
928
9142
x
x
x
x
x
f) 73
5
x
Thomas Whitham Sixth Form Page 60
16
521
215
73
5
x
x
x
x
g) 252 x
5
252
x
x
However when dealing with a quadratic we have two solutions. NB
If 6x then 362 x which is also greater than 25. Hence
5x is a second solution.
h) 812 x
9 and 9
812
xx
x
Example
Given that n is an integer, find the values of n such that −7 ≤ 2𝑛 < 6
−3.5 ≤ 𝑛 < 3 {divide throughout by 2}
Hence n can equal −3, −2, −1 ,0 ,1 ,2
Whole number
Thomas Whitham Sixth Form Page 61
Example List the values of n such that n is an integer value and
a) 23 n
b) 51 n
c) 34 n
Here we are not asked to find the solution by solving an equation
but by listing the possible solutions.
An integer value means possible whole number answers whether
positive or negative.
a) 23 n means n can be –2 –1, 0 or 1 we write
1,0,1,2 n
b) 51 n Answer: 5,4,3,2,1,0n
c) 34 n Answer: 2,1,0,1,2,3,4 n
Graphs
1. The straight line
Example Draw the graph of 𝑦 = 3𝑥 − 4 for values of 𝑥 from −3 to 3
For any straight line we can get away with plotting three points and
then a line through these three points.
Three points are selected to make sure we have only one straight line.
X – 3 0 3
y – 13 – 4 5
13
49
433
y
4
403
y
11
415
453
y
Thomas Whitham Sixth Form Page 62
Example Draw the graph of 𝑥 + 𝑦 = 7 for values of 𝑥 from −1 to 7
X 4 0 3
y 3 7 4
3
74
y
y 70 y
4
73
y
y
2
4
6
-2
-4
-6
-8
-10
-12
-14
2 -2 0 x
y
Thomas Whitham Sixth Form Page 63
2. The quadratic curve
The quadratic Curve could appear on both papers but for non-
calculator papers it will be a very basic curve.
Example
(a) Complete the table of values of y given 𝑦 = 𝑥2 − 2𝑥 − 5
(b) Draw the graph of 𝑦 = 𝑥2 − 2𝑥 − 5
(c) Use the graph to state the values of x for which 𝑥2 − 2𝑥 − 5 = 0
(d) State the minimum value for y on the graph.
x –4 –3 –2 –1 0 1 2 3 4 y 19 10 –2 –5 –6 –2 3
2
4
6
8
10
-2
2 4 6 8 -2 0 x
y
Thomas Whitham Sixth Form Page 64
a)
X –4 –3 –2 –1 0 1 2 3 4
y 19 10 3 –2 –5 –6 –5 –2 3
b)
c) the curve equals zero at the points where the curve cuts the x-axis
𝑥 = −1.4, 𝑥 = 3.4
d) Minimum value of curve is where the curves gradient changes from
going down to going upwards. i.e. 𝑦 = −6 when 𝑥 = 1
3
544
52222
y
5
544
52222
y
5
10
15
20
-5
1 2 3 4 -1 -2 -3 -4 0 x
y
Thomas Whitham Sixth Form Page 65
Example
a) Complete the table below for the graph of 𝑦 = 2𝑥2 − 3𝑥
b) Draw the graph of 𝑦 = 2𝑥2 − 3𝑥
c) Draw on the same axis the graph of 𝑦 + 2𝑥 = 14
d) Use the graph to obtain a suitable approximation to solution of the
simultaneous equations given by 𝑦 = 2𝑥2 − 3𝑥 and 𝑦 + 2𝑥 = 14
a)
b)
x –3 –2 –1 0 1 2 3 4 y 27 14 0 –1 2 20
x –3 –2 –1 0 1 2 3 4 y 27 14 5 0 –1 2 9 20
Type in -1 and exe
Then type Ans2 – 3Ans
Type in 3 and exe
Then type Ans2 – 3Ans
5
10
15
20
25
30
-5
1 2 3 4 -1 -2 -3 0 x
y
𝑦 + 2𝑥 = 14
𝑦 = 2𝑥2 − 3𝑥
Thomas Whitham Sixth Form Page 66
c) 𝑦 + 2𝑥 = 14
X 0 1 3
y 14 12 8
Line drawn on the graph in part (b)
d) The curve 𝑦 = 2𝑥2 − 3𝑥 and the line 𝑦 + 2𝑥 = 14 are equal at the
points of intersection. Which occur at −2.4,19 and 2.9,8.1
approximately.
14
140
y
y
12
142
y
y
8
146
y
y
Thomas Whitham Sixth Form Page 67
Notes
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