Number
IB Studies Revision
Number types
Number type
Natural
numbers
Integers
Symbol Explanation
Positive whole numbers
Rational
numbers
Irrationals
Q
Real numbers
-
Whole numbers, positive, negative and the
number 0.
Any number that can be written as a
fraction. This includes all the integers.
A number that in it’s exact form can not be
written as a fraction.
All the numbers listed above.
Examples
1,2,3, ….
-3,-2,1,0,1,2,
……
1
1, -3, 0,
3
2, 3, Any of the
above
What is an exact number?
Sometimes you are required to write answers as exact numbers. This is when the
number is written as either a fraction or in square root from, not a rounded
number.
For example 3 in exact form is 3, not 1.73.
1
1
should be written as
not 0.333.
3
3
Significant figures and decimal places
These two are easily mixed up.
The first significant figure is the first digit in your list, excluding zero. Write out the
digits required, remembering to round up if the last significant number if the
number to the right is 5 or larger. Any digits not required change to zeros. Below
are some different examples of numbers rounded to 3 significant figures.
Number
6.5879
0.002562
10354
34.6582
To 3 significant
figures
6.59
0.00256
10400
34.7
Decimal places are easier to understand (and explain) simply write down the
number of digits required after the decimal point. Remember the last number
required may need to be rounded up. Below are some examples of numbers
rounded to 3 significant figures.
Number
6.5879
0.002562
0.20316
34.6582
To 3 decimal places
6.588
0.003
0.203
34.658
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Number
IB Studies Revision
Exam answers and working
On your final examination your answers must be to 3 significant figures (or higher)
or as an exact number. Students often loose marks on the examination on this
criteria. Here are some simple techniques:
• Do not round off numbers whilst working, try to develop techniques with
your TI83 to keep numbers in your calculator, or take the time to write out
the numbers in full.
• When you have your final answer if it is in decimal form write out to 3
significant figures. If you are unsure about this leave you answer in it’s full
format.
1
• Leave fractions as fractions, do not change
not 0.33.
3
Indices
23 – the number three is the index or power. Very simply 23 = 2 x 2 x 2 = 8
In the table below are some notes and facts about numbers in index form.
Index type
Power of 1
What does it do?
The number remains the same.
Power of 0
The number becomes 1.
Frational
powers
1
are equivalent to squares
2
1
are cube roots and
roots, powers of
3
so on.
The negative power simply turns the
number into
Negative
powers
Power of
Examples
1.
51 = 5
2.
451 = 45
3.
x1 = x
1.
50 = 1
2.
450 = 1
3.
x0 = 1
1.
161/2 = 16 = 4,-4
2.
271/3 = 3 27 = 3
1
5
1
=
16
1. 5-1 =
2. 4-2
3. 25-1/2 =
1
5
Simplifying in index form. For all these to work the larger non-index numbers must
be the same.
Operation
Multiply
Divide
Power of a
power
Explanation
Keep the base number the same
and add the indices
Keep the base number the same
and subtract the second from the
first
Multiply the powers together.
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Example
1.
2 4 x 25 = 29
2.
y3 x y-5 = y-2
1.
25 ÷ 22 = 23
2.
y3 ÷ y-7 = y10
1.
2.
(23)4 = 212
(y2)-6 = y-12
Number
IB Studies Revision
Standard form (a x 10n where 1 a 10)
Converting to standard form
Standard form is used to write large or small numbers in an easier format. The IB
does not use the term standard form but instead often phrases questions in the
form a x 10n where 1 a 10.
To convert numbers from standard form:
• Let’s choose a number 2,780,000
• First choose the a number, this is always a number between 1 and 10, round
this to 3 significant figures. So 2,780,000 becomes 2.78
• Secondly calculate your n number. You do this by counting how many times
you have moved the decimal place.
Moving the decimal point this way (left) n becomes negative
Moving the decimal point this way (left) n becomes positive
So in the number 2,780,000 we have moved the decimal point 5 places n
becomes 5.
The final answer is 2.78 x 105
Below are some more examples of numbers converted to standard form.
Number
Converted to standard form
3780124
3.78 x 106
0.002419
2.42 x 10-3
0.000075278
7.53 x 10-5
873.62
8.74 x 102
Calculations in standard form
When performing calculations using standard form apply the techniques shown
above with indices.
Multiplication: Multiply both the a numbers together. Add the n numbers together
and write again in the form a x 10n. Finally adjust the first a number so that it is
between 1 and 10, adjusting the n number accordingly.
Division: Divide the first a number by the second. Take the second n number form
the first and write again in the form a x 10n. Finally adjust the first a number so
that it is between 1 and 10, adjusting the n number accordingly.
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Number
IB Studies Revision
Guided example
x = 5.61 x 104 and y = 9.75 x 10-8.
Calculate
(a)
xy
(b)
x÷y
Answer (a)
• 5.61 x 9.75 = 54.7
• 104 x 10 -8 = 10-4
• So we have 54.7 x 10-4 (note this is not in standard form)
• Convert to standard form 5.47 x 10-3
Answer (b)
• 5.61 ÷ 9.75 = 0.575
• 104 ÷ 10-8 = 1012
Sequences
and series
• So we have 0.575 x 1012 (note this is not in standard form)
At IB
level you need to know about two types of series: arithmetic sequence and
• Convert to standard form 5.75 x 1011
geometric progression. For each one you must be able to calculate the next term
and find the sum of terms.
What is the difference between the two?
An arithmetic sequence is one where the numbers go up (or down) by adding (or
subtracting) by the same number each time, known as the common difference.
A geometric sequence is one where the numbers go up (or down) by multiplying by
the same number each time, known of the common ratio.
Sequence
6, 9, 12, 15, 18, ……
1, 3, 9, 27, 81, ……
25, 20, 15, 10, 5,
……
1 1 1 1 1
, , , , , ……
2 4 8 16 32
Type
Arithmetic
Geometric
Arithmetic
Geometric
The first difficult problem with these questions is to establish if we have an AP or a
GP. Listing out the terms given should help you with this.
Using the formula sheet
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Number
IB Studies Revision
The notation
un = the next term of a sequence
Sn = the sum of a sequence
a = the first number of a sequence
d = the common difference for an arithmetic sequence
l = the last term of a sequence
r = the common ratio of a geometric sequence
n = the number of terms
Guided example
Devon has a savings scheme. He starts off by adding $10, the next month he
adds $12, the following month $14 and so on.
(a) Calculate how much Devon has adds on the 14th month.
(b) Calculate how much Devon has saved after 2 years.
Answer (a)
• Firstly, establish if it is an AP or a GP by listing out the sequence:
10, 12, 14, 16, 18, etc.
• As the same number is being added on each time the sequence is an AP.
• Find the numbers to go into the formula: a = 10, d = 2, n = 14 (as you are
finding the 14th month).
• Now put all these numbers into the formula un = 10 + (13 x 2).
• The answer is $36.
Answer (b)
• We have already established that it is an AP, and we know all the letters’
values.
• Use the formula for the sum of a formula: Sn = 24/2 (2x10 + (23x2) ), the
answer will be $792.
Guided example
A swimmer is training for a long distance race. She will start on the first day by
swimming 50m, and each day increase the number of metres she swims by
10% for the first 30 days.
(a) Calculate how far the swimmer must swim on the 30th day.
(b) Calculate how many metres has the swimmer swum in 30 days.
Answer (a)
• Like the question above start by listing the sequence:
50, 55, 60.5, 66.55, etc
• As the sequence is being multiplied each time it is a GP.
• Find the numbers to go into the formula: a = 50, r = 1.1, n = 30.
• Put these numbers into the formula un = 50 x 1.130.
• The answer is 872 metres.
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Number
IB Studies Revision
Answer (b)
• We have already established that this is a GP, and we have the letters’
values.
• Use the formula for Sn = [ 50 x (1.130 – 1) ] ÷ [1.1 – 1], the answer is 8225
metres.
Quadratic equations
A quadratic equation is an equation is one in the form ax2 + bx + c, where a can
not be 0.
Some examples of quadratic equations:
x2 – 5x + 6, x2 = 25, x2 + 16x, 2x2 – 3x + 16
You need to know how to factorise, solve, graph, and interpret graphs of quadratic
equations.
Factorisation
Factorising any equation simply means introducing brackets. When factorising
quadratic equations you will most probably need 2 brackets, but you may only need
1 bracket.
ax2 + bx or ax2 + c
ax2 + bx + c
1 bracket required
2 brackets required
Equations with 1 bracket can be factorised by looking for a common factor with in
coefficient (the number in front of the x), as shown in the examples below.
Quadratic equation
4x2 + 12
6x2 + 24x
10x2 – 15x
6 – 3x2
Factorised
4(x2 + 3)
6x(x + 4)
5x(2x – 3)
3(2 – x2)
Equations with 2 brackets can be factorised as follows:
• Remember that the general equation is given by x2 + bx + c
• Your bracket will factorise as follows: (x + p) (x + q), where p and q will all
be numbers. Do not get worried about all these letters, it is easy to follow
with an example.
• The two numbers p and q multiply to give c. They will add to give b. Choose
two numbers that multiply to give c. If they add to give b you have the two
numbers.
This works like so:
Factorise: x2 – 5x + 6
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Number
IB Studies Revision
•
•
All the pairs of numbers that multiply to give 6:
2 and 3
1 and 6
-2 and -3 -1 and -6
Only –2 and -3 add to give -5 so therefore the quadratic factorises to
give:
(x – 2) (x – 3)
Look at the table below for some more examples of quadratic equations that have
been factorised.
Quadratic equation
x2 + 7x + 12
x2 - 6x + 5
x2 – 3x - 10
x2 + 2x - 10
Factorised
(x + 4) (x +
3)
(x - 5) (x – 1)
(x + 2) (x – 5)
(x + 5) (x - 3
)
Many aspects of maths leading into examinations are all about practice and this is
one area where lots of practice is required. Do 20 questions from a text book and
this subject becomes much easier.
Solving quadratic equations
Factorised quadratic equations can be solved if there is an equal sign. There will
always be 2 answers to the equation, even if one is a repeated (this is explained
below).
Each bracket must equal 0.
(x – 3) (x + 4) = 0
Make this bracket = 0
x+4=0
x = -4
Make this bracket = 0
x–3=0
x=3
So the two solutions are x = 3 and -4.
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Number
IB Studies Revision
Look at the table below for some more examples of solved quadratic equations.
Quadratic
equation
2
x + 7x + 12 = 0
x2 - 6x + 5 = 0
x2 – 3x – 10 = 0
x2 + 2x – 10 = 0
x2 - 4x + 4
x2 + 10x - 25
Factorised
Solved
(x + 4) (x + 3) =
0
(x - 5) (x – 1) =
0
(x + 2) (x – 5) = 0
(x + 5) (x - 3 ) =
0
(x – 2) (x – 2) = 0
(x + 5) (x + 5) =
0
x = -4 and x = 3
x = 5 and x = 1
x = -2 and x = 5
x = -5 and x = 3
x = 2 and x = 2
x = -5 and x = 5
The final two solutions where x has the same number are called a repeated
solution.
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Number
IB Studies Revision
Guided example
A classroom is be built in the shape of a rectangle. The width is 3 metres less
than the length and the area is 28m2
(a)
(b)
(c)
If the length is x metres, write down an expression for the width.
Write a quadratic equation using the area of the rectangle.
Find the length and width of the classroom.
Answer (a)
• As the width is 3 m less than the length we will have the answer x – 3
Answer (b)
• To get the area = length x width, so in this case x(x – 3) = 28
• Multiplying this out we have x2 – 3x = 28. This is now a quadratic equation as
it has an x2 in it. However by simplifying we get x2 – 3x – 28 = 0.
Answer (c)
• Now factorise the quadratic equation to get (x – 7) (x + 4) = 0.
• We have the solutions x = 7 and x = -4. As we are looking for a length, it can
not be negative so we must choose x = 7.
• The length is 7 metres and the width is 4 metres.
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Number
IB Studies Revision
Graphing and interpreting quadratic equations
Remember a quadratic equation is in the form ax2 + bx + c.
Look at the two graphs below.
Diagram 1 y = x2 - 2x - 3
Diagram 2 y = x2 – 7x + 6
Some features of the graph can clearly be seen by looking at where the graph
intercepts the x and y axis.
The y intercept will always be the letter c in the equation ax2 + bx + c.
The solutions to the graph will be where it intercepts the x-axis.
Diagram 1
The solutions are x = 3 and x = -1 and the graph will factorise into (x – 3) (x + 1).
Diagram 2
The solutions are x = 1 and x = 6 and the graph will factorise into (x – 1) (x – 6).
One last feature to note is the line of symmetry of the graph. This shows the point
at which the graph has it’s minimum or maximum value. The x-coordinate will be
half-way between the two x-axis intercepts. You can calculate the value by adding
up the two intercepts and dividing by 2. By substituting this value into the equation
you can calculate the y coordinate.
For diagram 1 this can be as follows:
• Add up – 1 and 3 and divide by 2, giving the answer 1 – the x-coorinate.
• Substitute 1 into the equation y = x2 - 2x – 3 and y = -4.
• The coordinate of the minimum point of the graph is (1, -4).
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Number
IB Studies Revision
What about when we have –x2?
The graph below shows the graph of y = -x2 + 3x + 10
The first thing to note is that the graph has changed from a into an shape.
The c number at the end is still the intercept on the y axis.
The solutions to this equation are still the intercepts on the x-axis, so x = -5 and x
= 2.
The graph will factorise into (x – 2) and (-x – 5). Note the change in the second
bracket where a negative x has been used and the value of the 5 has had to change
accordingly.
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Number
IB Studies Revision
Guided example
The diagram shows the graph of y = x2 – 2x – 10
(a)
(b)
(c)
Factorise x2 – 2x – 10
Find the missing letters in each of the in each of the coordinate (0,a)
which is a point on the graph.
Find the coordinate of the minimum point of the graph.
Answer (a)
• As the solutions to the graph are x = -2 and x = 5, we have can factorise the
equation (x – 5) (x + 2)
Answer (b)
• When x = 0 we are looking for the y-axis intercept. a = -10
Answer (c)
• By adding -2 and 5 (the x-intercepts together) and dividing by 2 we get 1.5
• Substitute 1.5 into the equation x2 – 2x – 10 and we get -12.25.
• The coordinates will be (1.5, -12.25)
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