Download GCSE UNIT 2 Foundation

GCSE UNIT 2 Foundation
This is a list of all the topics that you need to know for Module 2 Foundation exam. An idea of
the grade level is given in brackets. All the really important topics have an example question
with workings to show you how to apply the knowledge in the exam. Answers are underlined.
General Number

Can square, cube a number and find the square root of a number
Example question:
a) Find 82
82 = 8 × 8
= 64
3
b) Find 2
23 = 2 × 2 × 2
=8
c) Find
64
64 = 8 (because 8 × 8 = 64)
 Can recognise factors and multiples of a number
Factors are numbers that go into another number
Multiples are numbers in the given numbers multiplication table
Example question:
a) Write down the factors of 24
1, 2, 3, 4, 6, 8, 12, 24
b) Write down a multiple of 7
14 (although any number in the 7 times table
would be okay. I could have had 21, 28,
70, 140, 280 etc.)

Can write a number as the product of prime factors – this is sometimes
called ‘prime factorising’
This means write a number as prime numbers multiplied together/ Use the
method below:
Example question:
a) Write 120 as a product of prime factors
120
60
30
15
3
2
2
2
5
Find two numbers
that multiply
together to give 120
Keep splitting the
numbers up until you
only have prime
numbers left – which
can’t be split up.
Answer = Prime Factorisation of 120 = 2 × 2 × 2 × 3 × 5 = 23× 5×3
b) Given that 80 can be prime factorised to be 24× 5 120 find the Highest Common Factor (HCF) of
120 and 80.
80 = 24× 5 = 2 × 2 × 2 × 2 × 5
120 = 23× 5×3 = 2 × 2 × 2 × 3 × 5
To find the HCF you need to make a new number using the numbers that appear in both lists. These
are underline. So HCF = 2 × 2 × 2 × 5 = 40

Key Words
Sum = Add,
Product = Times, Difference = Subtract

Fraction, Decimal and Percentage equivalences
Note 3  3  8  0.375  37.5%
Remember: 1% = 0.01 = 1/100
10% = 0.1 = 1/10
8
Means divide!2
30% = 0.2 = /10
Example question:
a) In a class of 20, 7 are
male. What percentage are
male?
7
35

 35%
20 100
20% = 0.2 = 2/10
30% = 0.3 = 3/10 etc
25% = 0.25 = 25/100 = 1/4
50% = 0.5 = 50/100 = 1/2
75% = 0.75 = 75/100 = 3/4
(Or 7  20  0.35  35%)

Finding a Percentage of an amount
1
Remember that to find 10% divide by 10.
To find 1% divide by 100. Use these to
,
build all the other amounts.
2
,
Example question:
a) Increase £60 by 36%
3
Find 10% of 60 = £6
(divide by 10) ,
1% of £60 = £0.60 (divide by 10 again)
4
Then 36% = 3 lots of 10% + 5% + 1% = 3×6+3+0.6
= £21.60
,
This needs to be added to the original amount = 60 + 21.60 = £81.60
6
, 20 to 23. What is the percentage increase?
b) The number of people at a party increases from
8 of the original amount.
Increase is 3. Need to write this as a percentage
,
So 3/20 = 15/100 = 15%

1
2
Can estimate a calculation by rounding
appropriately
,
Example question:
2.01  2.99 2  3 5 2
a) Estimate the value of
Answer =2.5


3.9  1.9
42 2 4
Remember this means
divide!

Sharing in a given ratio
Example question:
Mr B and Mr Hancock win some money on a day out at the races. We win £45 and share it in the ratio
2:3. How much does each get?
Basically there are 5 parts (2+3) adding the ratios together.
Mr B gets 2/5 as he gets 2 out of the 5 parts
Mr Hancock gets 3/5 as he gets 3 out of the 5 parts.
So divide the 45 by 5 (the total of the ratios) = 9
Mr B gets 2 × 9 = 18 Mr Hancock gets 3 × 9 = 27 (Multiply this amount by the ratio numbers)
 Can find equivalent fractions and simplify fractions
To simplify/cancel a fraction divide both the top and bottom of the fraction by a
number that goes into them. Keep going until you cannot simplify any more.
Example question:
a) Cancel 24
Divide top and bottom by 2 12
30
15
Divide top and bottom by 3 4
Answer : 4
5
5
(Note you could divide top and bottom by 6
to get the answer in one go!)
To find an equivalent fraction you can just multiply the top and bottom by the same
number.
Example question:
?
a) 3 =
8
Answer: 15
40
40
(You can work this out because the bottom
number has been multiplied by 5 to get from 8
to 40. Therefore the top needs to be multiplied
by 5 as well)

Can add and subtract fractions
To add and subtract fractions you need the bottom numbers (called the
denominators) to be the same.
If the bottom numbers are the simply add/subtract the top numbers and
leave the bottom number alone
Example question:
a) Find 3 + 2
b) Find
8
3
7
Answer: 5
8
2
7
-
8
Answer: 1
7
If the bottom numbers are different you need to change the fractions into
equivalent fractions that have the same bottom number so that you can
add/subtract as before
Example question:
a) Find 2 + 3
5
Answer: 29
7
35
Bottom numbers different so use this method to get them the same:
27
5 7
Multiply both the
numbers in the first
fraction by the
bottom number in
the second fraction
+
3 5
75
=
Multiply both the
numbers in the
second fraction by
the bottom number
in the first fraction
14
35
+
15
35
=
29
35
Now the bottom
numbers are the
same so you can
add

Can find a fraction of an amount
To find a fraction of an amount divide by the bottom number and times
by the top.
Example question:
Find 2 of 63
7
Do 63 ÷ 7 = 9
9 × 2 = 18
Answer:18
2
Note: If you were asked to find  63 you would do the same thing, the answer
7
is still 18
Distance Speed and Time

Solve simple speed problems
(E)
Example question:
a) A car travels 5 kilometres in 15 minutes, what is its average speed?
We measure speeds in kilometres per hour so if it does 5 km in 15 minutes you would expect it to
travel 10 km in 30 minutes and 20 km in 60 mins.
The speed is therefore 20 km per hour
Answer 20 km per hour
b) A car travels 9 miles in 12 minutes, what is its average speed?
We measure speeds in miles per hour so if it does 9 miles in 12 mins would be the same as 45 miles in
60 minutes (by multiplying by 5). The speed is therefore 45 miles per hour
Answer 45 miles per hour

Solve more difficult speed problems (C)
Use the following triangle that shows that
Distance = Speed × Time
Speed = Distance ÷ Time
Time = Distance ÷ Speed
Distance
Speed Time
Example question:
a) A car travels 2500 m in 2 mins, what is its average speed?
We measure speeds in kilometres per hour
Convert 2500m into km (divide by 1000) = 2.5 km
Convert 2 mins into hours (divide by 60) = 0.0333333 hours
2.5÷ 0.0333333 = 75 km/h
Answer 75 miles per hour

Distance speed time graphs (C)
Remember that on a distance/time graph the gradient of the line represents the
speed. If the line is flat the item is not moving. The steeper the line the faster
the journey.
Use of symbols

Simplify expressions with one variable. (F)
Example question:
a) Simplify 3t + 5t – 2t
Answer: 6t
b) Simplify -4x + 2x – 3x
Answer: -5x
(Watch the negative numbers –4 + 2 = -2 then take 3 = -5)
c) Simplify 3x5 × 2x4
(Multiply the front numbers and add the powers)
Answer: 6x9
d) Simplify 8x7 ÷ 2x4
(Divide the front numbers and take the powers)
Answer: 4x3
Remember that: x1= x and x0= 1

Simplify expressions with more than one variable. (E)
Example question:
a) Simplify 3t + 5w– 2t + 3w
Answer: t + 8w
Answer:6t2 + 5t
b) Simplify 3t + 5t2 + 2t + t2
(Remember that t and t2 cannot be added together so keep them separate)

Multiply out expressions with single brackets
(D)
Example question:
a) Expand 3(2x + 5)
Answer: 6x + 15
(because 3 × 2x = 6x
and 3 × 5 = 15)
b) Expand 4(2x - 3)
Answer: 8x - 12
(because 4 × 2x = 8x
and 4 × 3 = 12)

Multiply out expressions with double brackets
(C)
Example question:
a) Expand (x + 5)(x + 3)
Answer: x2 + 8x + 15
Use a method (either the ‘face method’, ‘the claw’ or ‘the grid method’ described below)
For the Grid Method, draw up a multiplication grid like this:
x
Second bracket
5
first bracket
x
5
x
x×x 5×x
x
x2
5x
3
3×x 3×5
3
3x
15
Complete each cell in the grid by multiplying the term at the top by the term at the side.
Then add up the parts of the grid x2 + 5x + 3x + 15 = x2 + 8x + 15

Factorise a linear expression (D)
Example question:
a) Factorise 6x + 8
Answer: 2(3x + 4)
(because 2 goes into both 6x and 8 write down 2(
)
then inside the bracket you need 3x because 2 × 3x = 6x and +4 because 2 × +4 = 8)
b) Factorise 12x - 30
Answer: 6(2x - 5)
(Notice that 2(6x – 15) would also be acceptable but worth less marks because you
want the biggest number you can in front of the bracket)

Expand and simplify harder expressions. (C)
Example question:
a) Expand 3(x + 5) + 2(4x – 5)
3x + 15 + 8x - 10
Answer: 11x -+ 5
(Expand each bracket separately and then simplify what you have left)
Formulae

Use a formula written in words (G)
Example question:
To convert Celsius to Fahrenheit double the temperature and add 32.
What is the temperature in Fahrenheit if it is 23 degrees Celsius
Answer: 68oF

Use a simple formula with positive and negative numbers (F/E/D)
Remember to use BIDMAS:
In particular to square before multiplying and to multiply before adding or
subtracting.
Remember:
- 4 × - 3 = 12
4 × - 3 = -12
- 4 × 3 = -12
4 ×
3 = 12
(-4)2 = 16
-ve × -ve = +ve
+ve × -ve = -ve
-ve × +ve = -ve
+ve × +ve = +ve
(-ve)2 = +ve
(That when you multiply or divide two negative numbers you get a positive
answer. When you square a negative number you also get a positive answer.)
Remember that when you add a negative number it is like taking away and that
when you take a negative number it is like adding.
8+-3=5
8 - -3 = 11
Example question:
If y = -3 and w = 2 find:
a) 4w + 2y
4w = 8
8 + -6 = 2
+- -
=
=
+
Answer: 2
2y = -6
b) 2y2 + w
y2 = 9 (not –9!)
2y2 = 18
2y2 + w = 20
Answer: 20

Write an expression from a problem
Example question:
Jim gets ‘x’ cards for his birthday. Ben gets three more than Jim, Amy gets twice as many as Jim. Sam
gets twice as many as Ben. Write an expression for:
a) How many cards Ben gets
Answer: x + 3
b) How many cards Amy gets
Answer: 2x
(Note 2x = x × 2 = 2 × x but only 2x will get a mark!)
c) How many cards Sam gets
Answer: 2(x + 3)
(The bracket is really important here. Note that (x + 3) × 2 would also get the marks!)

Solve a problem by forming & solving an equation (C/D)
Example question:
1) I am ‘x’ years old. Sally is two years younger than me, Joan is twice my age.
If the total of our ages is 118 write down this information as an equation.
My age x
Sally is two years younger thane me. Sally’s age is x – 2
Joan is twice as old as me. Joan’s age is 2x
The total of our ages is found by adding them up
So (x) + (x –2) + (2x) = 118
This simplifies to be 4x – 2 = 118
Answer: 4x – 2 = 118
2) Look at the rectangle below.
If the perimeter of the rectangle is 98 write down an equation for its perimeter.
2x + 3
x-2
The perimeter is found by adding up all 4 side lengths, so
(2x + 3) + (x – 2) + (2x + 3) + (x – 2) = 98
Simplifies to be 6x + 2 = 98

Answer: 6x + 2 = 98
Rearrange simple linear formulae (C)
This is a hard topic. Remember that when you switch a letter from one side of
the equation to the other it does the opposite thing on the other side. So ‘times
by w’ on one side becomes ‘divide by w’ on the other.
Example question:
a) Rearrange the equation c = u + at to make ‘t’ the subject.
This means rewrite the equation so that it is written as t = ….
Swap the u to the other side c – u = at
cu
Swap the a to the other side
t
a
b) Rearrange the equation w =
Answer:
cu
t
a
y
to make ‘y’ the subject.
x
This means rewrite the equation so that it is written as y = ….
Swap the x to the other side w × x = y
Answer: y= w × x
Equations

Solve one-step equations such as 3x = 12 or
x
 5 (E/F)
4
To solve 3x = 12 you need to understand that this means find a number for x so
that 3 × x = 12. In this case x is obviously 4 because 12 ÷ 4 = 3.
To solve
x
 5 you need to understand that this means find a number for x so
4
that x ÷ 4 = 5. In this case x is obviously 20 because 4× 5 = 20.

Solve two-step equations such as 3x – 1 = 9 (E/D)
Example questions:
a) Solve 3x – 1 = 9
Take the -1 over to the opposite side it becomes +1
Then
3x
=9+1
3x
= 10
Take the ×3 over to the opposite side it becomes ÷3
x
= 10 ÷ 3
Answer: x = 3.333…
b) Solve
Then
c) Solve
Then
2x + 5 = 12
2x
= 12 – 5
2x
=7
x
=7÷2
x
38
4
x
83
4
x
5
4
x=5×4
Take the +5 over to the opposite side it becomes -5
Take the ×2 over to the opposite side it becomes ÷2
Answer: x = 3.5
Take the +3 over to the opposite side it becomes -3
Take the ÷4 over to the opposite side it becomes ×4
Answer: x = 20

Solve equations with brackets such as 2 (5x + 1) = 28
(C)
The key here is to expand the brackets first then proceed as above
Example questions:
a) Solve 2(5x + 1) = 28
Then
10x + 5 = 28
10x
= 28 - 5
10x
= 23
x
= 23 ÷ 10
Answer: x = 2.3

Solve equations where the letter appear on both sides of the equation
(C)
The key here is to get rid of the unknowns from one side of the equation.
Example questions:
a) Solve 3x + 6
= x – 10
Move the +x over to the other side turn it into a -x
3x – x + 6 = -10
Then
2x
= -10 – 6
Move the +6 over to the other side turn into a -6
2x
= -16
x
= -16 ÷ 2
Move the ×2 over to the other side turn into a ÷2
x
= -8
Answer: x = -8
b) Solve
Then
5x - 4
-2x
3x - 4
3x
3x
x
= 2x + 7
-2x
=7
=7+4
= 11
= 11 ÷ 3
Move the ‘2x’ to the other side turn it into a ‘-2x’
Answer: x = 3.666..

Solve more complex equations combining several of the above such as
brackets and letter appearing twice (C)
Example questions:
a) Solve 4(x + 2) = 2x – 5
- expand brackets first
4x + 8 = 2x - 5
- subtract an 2x from both sides
Then
2x + 8 = -5
2x
= -13
x
= -13 ÷ 2
x
= -6.5
Answer: x = -6.5
Sequences (The word sequence means a number pattern)


Write the terms of a simple sequence. (G)
Describe the sequence using a term to term rule (E)
Example questions:
Look at this sequence of numbers: 7, 10, 13, 16, 19, ……
a) Write down the next two numbers in the sequence.
b) Describe the sequence
Answer: 22 and 25
Answer: Goes up by 3
We call this a term to term rule as it tells you how
to go from one number (term) in the sequence to
the next number (term)

Write the terms of a sequence or a series of diagrams given the nth
term. (D)
Example questions:
The ‘n’ th term of a sequence is given by the rule 4n – 5
What is the value of the 8th term?
When you put 8 into the formula for n the number you get will be the 8th number in the sequence

Answer: 27
Write the ‘n’th term of a sequence or a series of diagrams. (C)
Example questions:
Look at this sequence of numbers: 7, 10, 13, 16, 19, ……
Write down the rule for the ‘n’th term
This means find a formula so that when you put 1 into it you get 7, when you put 2 in you get 10 etc.
Since the formula goes up in threes the formula will be based on the 3 times table = 3 × n
But if the formula were 3 × n then the sequence would be 3, 6, 9, 12 and this sequence is 4 less than
the one we want, so the formula for the one we want must be 3 × n + 4
Answer: 3× n + 4
Co-ordinates



Use coordinates in all four quadrants. (F)
Find the midpoint of a line segment
Use and understand coordinates in three dimensions (C)
Graphs of linear functions

Plot the graphs of straight lines such as x = 3 and y = 4
x = 3, x = -2, x = 0.7, x= -1.5 etc are all vertical straight lines
y = 2, y = -5, y = ½ y = 5.2 etc are all horizontal lines
7
6
5
4
3
2
1
y=2
-7 -6 -5 -4 -3 -2 -1
-2
-3
-4
-5
-6
-7
1 2 3 4 5 6 7
x = -4
x=3
Y = x and y = -x are the diagonal lines shown:
y = -x
y=x
6
5
4
3
2
1
-7 -6 -5 -4 -3 -2 -1
-2
-3
-4
-5
-6

1 2 3 4 5 6 7
Draw the graph of an equation like y = 2x + 3 (E) or y = x2 +2x - 1 (C)
To draw a graph for an equation find coordinates (x,y) that fit.
The easiest way to do this is to draw up a table of values
Example question:
1) Draw the graph of y = 2x - 3
X
Y
2×-3 - 3 = -9
-3
-9
-2
-7
-1
-5
0
-3
1
-1
2
1
2×-2 - 3 = -7
The plot the points on the coordinate axes:
6
5
4
3
2
1
-7 -6 -5 -4 -3 -2 -1
-2
-3
-4
-5
-6
1 2 3 4 5 6 7
3
3



Solve problems involving graphs, such as finding where the line
y = x + 2 crosses the line y = 1 (D)
Recognise the equations of straight line graphs
Notice that all equations that look like this (y = ax + b) will make a straight line:
y = 3x + 1 y= 2x + 4
y =6 – 3x
y = 2x + 1 y = -0.5x – 3
The gradient of the line is the number in front of the ‘x’
The place that the line cuts the vertical (y axis) is the number on its own.
Find the gradients of straight line graphs (C)
To find the gradient of a line find how many squares it goes up as it
travels a distance of one square to the right. If the line goes down the
gradient is negative.
Example question:
1) Find the gradient of these lines:
Gradient = 2
Gradient = -1
Inequalities


Solve linear inequalities such as 4x – 3 < 17 (C)
Just like solving equations (4x –3 =17) but keep the sign in the middle.
Eg 4x – 3 < 17
4x
< 20
x
<5
Represent sets of solutions on a number line (C)
Eg  2  x  3
(This means the numbers between –2 and 3,
including –2 and not including 3. We use a filled
dot for  or  )
-4 -3 -2 -1 0
1
2
3
4
Substitution

Use BIDMAS
- BRACKETS, INDICES, MULTILPLICATION/DIVDE, ADD/SUBTRACT
Example Question:
If a = 3, b = 4 and c = 6 find:
1) ab2
Answer = 3 × 42 = 3 × 16 = 48 (square before times!!)
2) ab + ½c
Answer = 3 × 4 + ½ of 6 = `12 + 3 = 15 (add before times!!)
REMEMBER TO HAVE A GO AT ALL QUESTIONS. EVEN WHEN A
QUESTION LOOKS HARD WRITING SOMETHING DOWN CAN BE
USEFUL AND MAY WELL GET YOU MARKS!!!!