CIMA C03 Course Notes
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CIMA C03 Course Notes
Chapter 1
Introduction to Business
Mathematics
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1.
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Mathematical operations
Introduction
Welcome to the Astranti course notes for paper C03 business mathematics!
This is a very important paper as it ensures all students sitting CIMA have
the level of mathematical aptitude needed. The following set of notes will
prepare you for the exam whilst hopefully providing you with some
knowledge you can use in your everyday professional and personal life!
Mathematical operations
In simple terms, a mathematical operation is a symbol or operation that
creates a new value out of one or more starting values. The four core
operations of which I am sure you will already be aware are:
Addition (+) combining values together to give a total (2 + 2 = 4).
Subtraction (-) deducting one or more values from another (4 - 2 = 2).
Multiplication (x) taking a value and adding it to itself a set number of
times (2 x 3 = 2 + 2 + 2 = 6).
Division (÷) taking a value and splitting it into a number of equal parts:
6 ÷ 2 means to split 6 into 2 equal parts which would give us a value of 3!
Brackets
Often you may see values and operations contained within a pair of brackets
like so: (2 + 3) x 4.
In a case such as this we must work out the value of the equation contained
within the brackets before we multiply. This is because it is the value of 2 +
3 that is multiplied by 4:
(2 + 3) x 4 = 5 x 4 = 20
Order of operations
It is important to learn the hierarchy of operations (as in which one comes
first) as this can lead to incorrect outcomes:
9÷9x9+9-9=9
9 x 9 - 9 ÷ 9 + 9 = 17
9 ÷ 9 – 9 x 9 + 9 = -63
9 – 9 ÷ 9 + 9 x 9 = 81
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As you can see, simply be re-ordering the operations we were able to create
four completely different results from the same values. Therefore you must
learn and use the correct order.
The best way to do this is to use the acronym BEDMAS and think of it in the
sense that B is the most important and comes first, E the second and so on:
B = BRACKETS – work out the value of the sub equations contained within
brackets first. BEDMAS should be followed within a set of brackets and then
outside of it.
E = EXPONENTIAL – often referred to as powers (will be discussed later)
D = DIVISION
M = MULTIPLICATION
A = ADDITION
S = SUBTRACTION
For example:
(5 x 5) + (7 + 3) - 9
We do the brackets first:
25 + 10 - 9
Then we do the addition:
35 – 9
Finally the subtraction:
35 – 9 = 26
2.
Rounding
Rounding numbers essentially means replacing a value with a number that is
very close but perhaps shorter and simpler to use. For example, a company's
cost per unit may be £499.97 but they may round it up to £500 as this is very
much the same value but will be far easier to work with in cost equations
and accounts.
Rounding can be both up or down and sometimes will go both ways
dependent on which side of halfway it is. There is no set magnitude of
rounding - it could be to the nearest whole number, the nearest £1,000,000
or to one decimal place etc, for example:
£152,125,986.7875
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Rounded to the nearest whole number becomes:
£152,125,987
Rounded to the nearest million becomes:
£152,000,000
Rounded to six significant figures becomes:
£152,126,000
To one decimal place becomes:
£152,125,986.8
3.
Powers and roots
Powers
A power (or exponent) essentially means you multiply a value by itself a
certain number of times.
A power is represented by a smaller number to the top right-hand side of a
value (called a superscript), like so:
2
10
Which would be calculated as so:
10 x 10 = 100
Likewise:
4
10
Would be equal to:
10 x 10 x 10 x 10 = 10,000
Roots
Roots on the other hand represent going back the other way, as in what
value had to be multiplied by itself to give me the value we have? In other
words it is the reverse of a power. When 10 is squared we get 100, and so 10
must also be the square root of 100. A root is represented with the following
formula where n is the magnitude of root:
√n X
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To illustrate this lets take the two examples from the ‘powers’ section and
use them again here:
√100
So what number is multiplied by itself to give us 100? The answer is ten,
and;
√4 10000
What number is multiplied by itself four times and equals 10,000? The
answer again is 10.
4.
Fractions
A fraction is a number that represents a certain proportion of a whole. A
fraction has a denominator on the bottom which is the whole, and a
numerator on top which is the portion of the whole.
For example, we have a pie and we split it up into four equal parts, each
part can be represented as 1 in 4 or:
1
4
Here you can see the denominator is 4 and numerator is 1 representing what
proportion they make of the whole:
1 1 1 1 4
+ + + = =1
4 4 4 4 4
Note: if a fraction is has a larger numerator than denominator then this is a
‘top heavy’ fraction and is not wrong. It is simply because it may be easier
to write it out this way rather than write is as one and one half, e.g.
3
2
instead of
1
1
2
Adding and subtracting fractions
Believe it or not there is a very quick and simple way of adding and
subtracting fractions. As you will encounter in your studies, fractions are not
always nice and easy numbers like the one in the pie example, and you will
often be asked to add/subtract fractions where there is not a common
denominator.
The solution…find that common denominator!
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As we have discussed, fractions are nothing more than a proportion of a
whole, therefore it does not matter what the denominator is, for example:
1
1
equals one, just as:
95768
95768
equals one!
Just remember that when you change the denominator you must also
change the numerator in the same fashion, for example:
1
is the same as
4
2
but is NOT the same as
8
1
8
We doubled the bottom so we MUST double the top!
Knowing this we can then find a common denominator between two
fractions:
1 1
+
4 3
We cannot simply add them together as they have different denominators,
but we do know that both 4 and 3 go into 12. This is called the lowest
common denominator. We also know that 4 goes into 12 three times and 3
four times, so we times the respective numerators accordingly:
1 1 3 4
+ = +
4 3 12 12
Now we can simply add (or subtract if stipulated) the numerators and get
our answer:
3 4 7
+ =
12 12 12
Multiplying fractions
Multiplying fractions is quite simple really, all you need to do us multiply the
denominators and the numerators like so:
V Y V x Y VY
x =
=
W Z W x Z WZ
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So let’s see how this works with some actual values, using the same two
fractions as before:
1 1
x
4 3
So we take both our denominators and both our numerators and times them
together like so:
1 1 1 x1 1
x =
=
4 3 4 x 3 12
Simple!
Dividing fractions
Dividing fractions works in an almost identical fashion to multiplying with
one simple difference, we flip the denominator/numerator around on the
second fraction. We then multiply the two numerators and denominators
together as before:
1 ÷ 1
4
3
We start by switching the second fraction around:
1 becomes 3
3
1
Then we multiply the denominators and numerators:
1 3 1x 3 3
x =
=
4 1 4 x1 4
Again…simple!
5.
Ratios and percentages
Ratios and percentages are used in business for many types of calculation;
the difference between them is that a ratio is one value divided by
another where as a percentage is out of 100.
For example, if we wanted to express 8 as both a ratio and percentage of 50
it would look something like this:
Ratio
8
=0.16
50
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Or as a percentage:
8
x 100=16
50
= 16%
Ratios and percentages are an important part of measuring performance
within a business; here are a few key ratio/percentage equations:
Gross profit margin
Gross Profit is essentially the difference between revenue (or income)
and the costs of goods sold before additional costs such as overheads are
deducted. Generally it is calculated by deducting the cost of production
from the selling price.
For example, the pies from the fraction example cost £1.50 to make and sell
for £4.00, and we have sold 1000 therefore our gross profit will be:
£4,000 – £1,500 = £2,500
From this we can then work out the gross profit margin using the following
formula:
Gross profit X 100% = Gross profit margin,
Revenue
So, in our case:
£2,500 X 100% = 62.5%
£4,000
The larger the profit margin the more profitable the product is. When the
profit margin is small we need to manage costs very carefully if we are
going to continue making profits in the future.
Some businesses such as Ferrari tend to work on high profit margins; their
aim is to sell few cars at a high profit margins. Others tend to focus on high
volumes of sales at low margins – most supermarkets come into this bracket,
particularly the lowest priced ones.
Net profit margin
Net profit differs from gross profit in so far as that it is revenue minus all
costs. That is, it is the revenue minus total costs (including overheads and
interest payable etc.). It can be calculated in much the same way as gross
profit.
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For example, we know that our pie shop to has sales of £4,000 and the cost
of the production of goods sold comes to £1,500, but we know need to
deduct an additional £1,500 for overheads (costs such as rent):
£4,000 – £1,500 - £1,500 = £1,000
From this we can then work out the net profit margin using the following
formula:
Net profit X 100% = Net profit margin,
Revenue
So, in our case:
£1,000 X 100% = 25%
£4,000
Net profit margins are the key measure of profitability for shareholders, as
it considers all costs and not just those relating directly to products.
Return on capital employed
Return on capital employed or ROCE is a measure of how profitable the
business is compared with the investment made in that company. It’s a
measure of how efficiently they are using the capital at their disposal. If a
company has a very low ROCE it is likely that much of their capital is not
being used well at all.
ROCE is measured by dividing the net profit by the amount of capital
invested/employed:
Net profit
X 100 = Return on capital employed,
Capital employed
Our pie shop has net profits of £1000 and has £5,000 of capital invested in it
by the owner so:
£1,000 X 100% = 20%
£5,000
ROCE is a useful tool as it looks beyond the base numbers to assess the
efficiency and ability of a company to turn investment into profit.
For example, the shop next door has a net profit of £3,000 from only £6,000
of revenue, making them more profitable as a whole, and as a proportion of
sales. However, this shop has also had £20,000 invested into it, giving an
ROCE of only 15%. This illustrates that while the pie shop next door may
appear to be more profitable, investors are actually more likely to see a
better return when investing with us.
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Asset turnover
The asset turnover is the amount of sales/revenue generated per the
value of the assets (or investment). The Asset Turnover ratio is an indicator
of how efficient a company deploys and makes use of its assets, and so it is
similar to ROCE in this regard. It is calculated by dividing revenue by
capital employed:
Revenue
= Asset turnover
Capital employed
For our example this would be:
£4,000 = 0.8
£5,000
Generally, the higher the number, the better the asset turnover and the
better an investment opportunity it is.
For the shop next door to the pie shop the asset turnover is 0.15
(£3,000/£20,000) and since this is so much lower, it shows that it is
generating much lower revenues for each £ invested.
6.
Algebra
In maths and business alike letters are often used in place of numbers. This
is to give us a nice and simple way of seeing the relationship between
variables in equations.
For example:
Y = 2X
We do not know what the X or Y value is, but we do know that the X value is
always half the Y value. So, for example, if the X in the above equation is
replaced by a 9, then we know that Y must equal 18, since one Y is worth
two X's.
7.
Linear equations
Linear equations take the form of Y = a + bx and are simply an algebraic
representation of a straight line on a graph. Another feature is that they do
not contain powers, unlike quadratic equations (covered in the next
section).
In a linear equation:
Y is the dependent variable; this is because its value will be different
depending on the value of X
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X is the independent value; this value is used to help calculate the value of
the dependent variable (Y)
a represents the constant fixed amount, and on a graph this number
illustrates the intercept (where the line crosses) with the Y axis, and
finally;
b is the coefficient of the X value. In other words, this is the number the X
value is multiplied by. This is also a fixed value, and on graph this
represents the gradient of the line (how steep it is).
So, how would we use this in the real world?
Let’s return to our pie shop and look at pie seller Tim. Tim earns £200 a
week for working in the shop as a standard flat rate wage; however, he also
earns £1 commission for every pie he sells. We can input this information
into the form of Y = a + bx to give us an equation for calculating Tim’s
earnings for a given week:
Y = his total earnings
X = the number of pies he has sold on commission
a = his flat rate wage
b = the rate of commission
Last week Tim sold 120 pies, so, how much did Tim earn last week? We
know he gets £200 as standard, gets £1 for every pie he sells, and he sold
120. So when put into the equation above we get:
Y = a + bx
Y = 200 + (1 x 120)
Y = 200 + 120
Y = 320
Last week Tim made £320!!
Unknown variables
There may be times when you are not given the values of each variable and
will have to calculate them yourself. In a standard variable equation there
will be only one unknown, and if this is the case we can re-arrange the
formula so that the unknown is on its own on one side of the equation,
giving us its value. For example:
3 x−1
=2
10
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To start with we need to get rid of the division. We can do this by
multiplying the equation by the amount it is divided by. However, whatever
you do to one side of the equation you MUST do to the other. In this case
we want to get rid of the divide by ten, so we need to multiply both sides by
ten, which gives us:
3 x−1=20
Now we need to get rid of the minus one, and again, we do this by adding
one to both sides of the equation:
3 x=21
The final step is to divide both sides of the equation by the coefficient of
the X value to give us the actual value of X:
x=
21
=7
3
x=7
Simultaneous equations
Sometimes you may have to work out two unknowns (both the x and the y).
You can calculate these by working out two different equations at the same
time, and this is known as a simultaneous equation:
Take the following two equations:
4x + 2y = 12
6x – 2y = 18
We need to work out the values for both X and Y. The first step is to add the
two equations together, giving us:
(4x + 6x) + (2y – 2y) = (12 + 18)
= 10x = 30
This has got rid of the Y's, since 2y – 2y = 0, leaving us with a nice simple
equation to solve for X. Since we know the value of ten X's, all we need to
do is divide both sides by 10 to get the value of one X:
10 x=30
10
x=3
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Now that we have the X value, we can work out Y’s simply by picking one of
the original equations and replacing each instance of X with the answer we
have for X, which in this case is 3. So:
6x – 2y = 18
(6 x 3) – 2y = 18
18 – 2y = 18
y=0
In this instance Y is equal to 0, and we can now substitute our values for
both X and Y into the other equation just to be sure:
4x + 2y = 12
(4 x 3) + (2 x 0) = 12
12 + 0 = 12
8.
Quadratic equations
Quadratic equations contain a square and are representable by the
equation:
ax 2+ bx+ c=0
Often in quadratic equations there will be two values for X. This is
because a quadratic equation represents a curved line on a graph, and will
often intercept the X axis at two different points.
There are three main ways to calculate value of X in a quadratic equation:
•
The substitution/factorising method
•
The graphical method
•
The formula method
Let’s consider each of these in turn.
The substitution/factorising method
First things first: we need an equation. For this example we’ll use:
2
x +10 x +21=0
To factorise we need to identify two numbers that add to make the b value
(in this case 10) and multiply to make the c value (in this case 21). This
would give us 7 and 3:
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7 + 3 = 10
7 x 3 = 21
and we write this out in brackets in the following way:
(x + 3)(x + 7) = 0
We do this because when we multiply the brackets together, which we will
do next, we will get the original equation.
When it comes to multiplying brackets like the ones above, the easiest way
to remember the correct method is to remember the acronym FOIL.
FOIL tells us the order in which to multiply together each component in
within the brackets. So lets do that now to show that we havethe true
factorised form of the original equation:
F
=
first number in each bracket
=
X x X = X2
O
=
outside numbers
=
X x 7 = 7x
I
=
inside numbers
=
X x 3 = 3x
L
=
last number in each bracket
=
7 x 3 = 21
This, added together equals:
X² + 7x + 3x + 21 = 0,
which simplifies to:
2
x +10 x +21=0
So, now we know:
(x + 3)(x + 7) = 0
This means that either (x + 3) = 0, or (x + 7) = 0, since any number
multiplied by zero gives you zero.
At this stage, we take each bracket in turn, and assume that it equals zero.
So, if we assume that x + 3 = 0, we simply have to find x, which in this case
would be -3. And equally, if we assume that x + 7 = 0, then x = -7.
We now have our answer, since we have calculated that x must be either -3
or -7. The next step is to substitute each of these values for x into the
equation to check that it equals 0:
First we substitute every instance of X from our original formula with -3:
2
−3 +(10 x−3)+21=0
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9−30+21=0
−21+21=0
=0
And then with -7…
2
−7 +(10 x−7)+21=0
49−70+21=0
−21+21=0
=0
So there you have it! X is either – 7 or – 3 in this equation, meaning that
on a graph it will intercept the X axis at both -7 and -3!
The graphical method
The graphical method is more manual in nature and often used when
plotting a graph rather than simply working out the values of X. It can also
be used in instances where the graph/line in question does not have any
intercepts on the X axis.
Arguably the graphical method is the easiest as it only involves simple
mathematics but it is also the most time consuming. Essentially it is a
process of finding out the Y values for a given set of X values. You do not
need to ‘solve’ the equation, you just need to workout what Y would be
given a particular value of X.
Let’s say we have the equation:
2
3 x +2 x−8=0
Now, we want to work out its Y values for the X axis range (that's every
value of X) from -5 to 5, so how do we do this? We start by drawing a table
containing the values of X along the top, and the individual parts of the
equation along the side:
-5
-4
-3
-2
-1
0
1
2
3
4
5
3x2
+2x
-8
Y
Next step…we fill it in by replacing each instance of X on the side with the
value at the top, and working out all the values gives us:
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3x2
+2x
-8
Y
-5
75
-10
-8
57
-4
48
-8
-8
32
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-3
27
-6
-8
13
-2
12
-4
-8
0
-1
3
-2
-8
-7
0
0
0
-8
-8
1
3
+2
-8
-3
2
12
+4
-8
8
3
27
+6
-8
25
4
48
+8
-8
48
5
75
+10
-8
77
Where each value in for Y is given by the sum of the middle three rows (i.e.
75 – 10 – 8 = 57).
Finally, we plot the values for X and Y on a graph, like so:
The formula method
The formula method is perhaps the quickest and most versatile of the
three methods because it works for any quadratic equation. It is probably
the best method to use unless there is some specific reason to use the
factorising or graphical method.
Earlier on we saw that quadratic equations could be presented in this form:
ax 2 +bx+c=0
Well, by re-arranging any equation into this from, we can then, in turn, use
the following formula to work out the X values:
−b±√ b2−4 ac
2a
Note: This formula – called the Quadratic Formula - will be given to you in
the exam; do not worry about learning it off by heart.
So, we have the equation:
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x 2+40 x+100=0
Let’s put it into the formula and see what happens!
−40±√ 40 2−(4 x 1 x 100)
2x 1
−40±√ 1600−400
2
−40±√ 1200
2
−40±34.64
2
Now all we have to do is solve the equation twice, once assuming the
plus/minus symbol in the middle to be an addition and once as a
subtraction:
−40 +34.64 −5.36
=
=−2.68
2
2
−40−34.64 −74.64
=
=−37.32
2
2
So there you have it! X is equal to -2.68 and -37.32, which means that the
curve intercepts the X axis as these points!
9.
Manipulating inequalities
Inequalities are for the most part treated in exactly the same way as
standard equations except that one side of the equation may be larger or
smaller than the other. Inequalities are represented by one of the following
symbols:
Inequality
Symbol
Effect on the equation
The left hand side is less than the right hand
side e.g. 1 < 2
Less than
<
Greater than
>
The left hand side is greater than the right
hand side e.g. 2 > 1
Less than or
equal to
≤
The left hand side is less than or equal to the
right hand side
Greater than or
equal to
≥
The left hand side is greater than or equal to
the right hand side
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In business, inequalities are often used to demonstrate limitations in areas
such as production. For example, suppose each product that we make
utilises 5kgs of raw material, and we have 10,000kgs and not a single gram
more. Therefore we can construct an inequality equation for the maximum
number of units we can produce with 10,000kgs:
X = number of products:
5x ≤ 10,000
This is because it is impossible for us to use more than our limit of 10,000kgs
worth of material, and if each item requires 5kgs then the absolute limit on
products we can produce is:
5x ≤ 10,000
5
x ≤ 2,000
So the maximum number of units we can produce with 10,000kgs is 2000
units!!
Solving inequalities
You may be asked to calculate an inequality algebraically (just as we did in
linear/quadratic equations) and the principle is almost identical. Let's take:
6x + 16 > 10x + 4
Like with equations, the first step is to get all the x’s on one side and all the
numbers on the other. In this example we will deduct both 6x and 4 from
both sides, giving us:
= 16 > 4x + 4
= 12 > 4x
12 and 4 are both divisible by 4, which when simplified leaves us with:
3>x
This does not solve the equation in quite the same way as before, as there is
no definite answer for the value of x. But what it does show us is that the
equation will work for any value of x less than 3. Therefore, it will not
work if the value of x is greater than 3.
Double inequalities
A double inequality occurs when the value of the unknown variable is
restricted at both ends, i.e. X must be more than 2 but less than 10, which
looks a little something like this:
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2 < X < 10
Again, you may be asked to solve a double inequality, but no need to fear as
it is treated in exactly the same fashion as before, except you must
remember there are 3 separate sides, and so what is done to one must be
done to the other 2:
16 < 4x -4 < 32
When solving this equation, we would start by trying to get the X on its own.
We can do this by adding 4 to all sides:
= 16 + 4 < 4x – 4 + 4 < 32 + 4
= 20 < 4x < 36
The next step is to divide each side by 4 giving us the values for X:
20 < 4x < 36
4
=5<x<9
As you can see, the value of x MUST be less than 9 but simultaneously it
must be greater than 5!
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CIMA C03 Course Notes
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