Download New tools and graphical aids for Business Mathematics 1

New tools for Business Mathematics and Statistics
Dr. Oded Tal, Conestoga College
1. Introduction
Teaching Business Mathematics and Statistics in college is a great challenge because of
three main reasons:

Students’ mathematical background is very varied, due to the multitude of Math
curricula in Canadian high schools.

Many students have struggled with Math in high school. In most cases they don’t
like Math and are even afraid of it.

Many students come to the college one year or more after finishing high
school, and they haven’t used Math at all during this period.
These facts contribute to the high failure rates in Business Mathematics and Statistics in
college.
One way of dealing with such high failure rates is simplifying the curriculum by one or
more of the following methods:

Using graphical aids, such as the Percentage Triangle or the Mark-up/Markdown chart, in order to make it easier for the students to understand and to
remember new concepts.

Using shortcut formulae in order to make certain calculations shorter, quicker,
and easier to memorize.

Using the calculator for more kinds of calculations in order to make them
simpler, quicker and less error-prone.

Generalizing two or more types of problems so that they can be solved using
one unified approach. Memorizing one approach instead of two or more
approaches is simpler. It also eliminates the chances of using the wrong approach.
The two following sections describe new tools which were successfully introduced to the
teaching of Business Mathematics 1 and Statistics I in Conestoga College during the Fall
2004 term.
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2. New tools for Business Mathematics 1
2.1
A graphical aid for solving Trade Discounts questions
This graphical aid is a combination of two triangles, based on the popular Percentage
Triangle (Base, Portion and Ratio):

The left triangle shows the relations among the List price (L), Net price (N) and 1
minus the rate of discount (1-d).

The right triangle shows the relations among the List price (L), Amount of
discount (D) and the rate of discount (d).
N
1-d
D
L
D
Within each triangle, any item can be calculated based on the two other items. The
students are told to use this together with L=N+D.
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2.2
Shortcut links between Rate of Markup and Gross Profit Margin
Each of the ratios RoM (Rate of Markup) and GPM (Gross Profit Margin) can be quickly
derived from the other ratio using the following formulae:
(1)
RoM=GPM/(1-GPM)
(2)
GPM=RoM/(1+RoM)
Derivation
Formula (1) can be derived as follows:
(3)
RoM=M/C=M/(S-M)
Where:
M:
Markup
C:
Cost
S:
Selling price
Dividing the numerator and the denominator by S:
(4)
RoM=(M/S)/[(S-M)/S]=(M/S)/[1-(M/S)]
Substituting GPM for M/S twice:
(1)
RoM=GPM/(1-GPM)
Formula (2) can be derived as follows:
(5)
GPM=M/S=M/(C+M)
Dividing the numerator and the denominator by C:
(6)
GPM=(M/C)/[(C+M)/C)]=(M/C)/[1+(M/C)]
Substituting RoM for M/C twice:
(2)
GPM=RoM/(1+RoM)
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2.3
A shortcut formula for calculating break-even dollar sales
Break-even dollar sales can be quickly calculated using the following formula:
(7)
BE dollar sales= FC/[1-(VC/S)]
Where:
FC:
Fixed cost
VC/S: Ratio of Variable Cost per unit to Selling price, or the ratio of Total Variable Cost
to Total Sales.
This formula works equally well in four different cases:

VC (variable cost per unit) and S (Selling price) are given

Total Variable Cost and Total Sales in dollars are given.

VC is given as a percentage of S

Total Variable Cost is given as a percentage of Total sales.
Derivation
Formula (7) can be derived as follows:
(8)
XBE=FC/(S-VC)
Where:
XBE:
The number of units required to break-even.
Multiplying both sides by S:
(9)
S*XBE=S*FC/(S-VC)
Or:
(10)
BE dollar sales= S*FC/(S-VC)
Dividing the numerator and the denominator by S:
(11)
BE dollar sales =FC/[(S-VC)/S]
Dividing each of the terms in brackets by S:
(7)
BE dollar sales= FC/[1-(VC/S)]
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2.4
A simpler Cost Volume Profit analysis when sales volume is given in dollars
instead of in units
Given TS (Total Sales), TVC (Total Variable Cost), FC (Fixed Cost) and Sales Capacity
in dollars, the following assumption enables transforming the situation into the simpler
and more common one, in which S (selling price per unit), VC (variable cost per unit),
FC (fixed cost) and Capacity in units are given.
The required assumption is: The company sold only one unit of product, whose sale
price per unit is the given TS (Total sales).
Based on this assumption, the given TVC becomes VC, and the Capacity in units is the
capacity in dollars divided by the Total Sales.
Hence:
S=TS, VC=TVC, FC=FC and Capacity in units=Capacity in $/(Total Sales).
Now:
Total Revenue=SX, TC=FC+(VC)X, XBE=FC/(S-VC) and BE $=XBE*S, just like when
S, VC and FC are given. Obviously, the number of units does not have to be an integer.
Example
The following information is available from the accounting records of Eva Corporation.
Sales (at 75% of capacity)
Fixed Costs
Variable Costs
$18,000
$4,000
$12,000
Total Costs
$16,000
Net Income
$ 2,000
Using an algebraic approach, calculate the break-even point in sales dollars and as a
percentage of capacity.
Solution
S=$18,000, FC=$4,000, VC=$12,000.
Capacity in dollars=18,000/0.75=$24,000
Capacity in units=Capacity in dollars/S=24,000/18,000=4/3 units
XBE=FC/(S-VC)=4,000(18,000-12,000)=2/3 units
XBE dollar amount=XBExS=2/3x18,000= $12,000
XBE/Capacity in units=(2/3)/(4/3)=0.5=50% of capacity.
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3. New tools for Statistics I
3.1
A shortcut method for calculating the Mean Deviation
3.1.1 The ordinary method
The ordinary method of calculating the mean deviation of a sample or a population is:
(1)
MD=  |Xi- μ|/n
Where:
Xi:
Individual values
μ:
The population mean or the sample mean
n:
Population size or sample size
Since calculators do not usually calculate the mean deviation, nor the absolute value, MD
has to be calculated step by step. The number of operations required is in the order of
magnitude of 2.5n (n subtractions, about 0.5n conversions of negative numbers to
positive numbers, n-1 additions and 1 division).
3.1.2 The new method
The mean deviation can be calculated using the following formula:
(2)
MD=[  Xl-  Xs-(nl-ns)μ]/n
Where:
 Xl: Sum of all values larger than the mean
 Xs: Sum of all values smaller than the mean
nl:
Number of values larger than the mean
ns:
Number of values smaller than the mean
μ:
The population mean or the sample mean
n:
Population size or sample size
Derivation
This formula can be derived as follows:
(1)
MD=  |Xi-μ|/n
The data can be divided into three categories: Values which are larger than the mean,
smaller than the mean and equal to the mean.
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
When Xi is larger than the mean, |Xi-μ| equals Xi-μ, and the sum of the deviations
equals the sum of values, minus the mean x the number of values in this category,
that is  Xl - nl*μ, where l stands for “larger than the mean”.

When Xi is smaller than the mean, |Xi-μ| equals μ-Xi, and the sum of the
deviations equals the mean x the number of values in this category, minus the sum
of values, that is ns*μ –  Xs, where s stands for “smaller than the mean”.

When Xi equals the mean, the value does not contribute to MD.
Therefore we have:
(3)
MD= (  Xl - nl*μ + ns*μ –  Xs)/n =[  Xl-  Xs-(nl-ns)μ]/n
The maximum number of operations required is n+3 (n-2 additions, 3 subtractions, 1
multiplication and 1 division). This number decreases if any of the values are identical to
the mean. In addition, this method also requires counting the number of values which are
larger than the mean and the number of the values which are smaller than the mean. This
can be considered as easier and less error-prone than arithmetic operations.
3.1.3 Comparing the two methods
The difference between the numbers of operations required for the ordinary and the new
method is in the order of magnitude of 2.5n-(n+2)=1.5n-2, which is positive for n>3.
Obviously, less arithmetic operations means less chances of making mistakes.
3.1.4 Example
Find the mean deviation of the following data set: 95,103,105,110,104,105,112, 90. The
sample mean is 103 and the sample size is 8.
The required calculations using the ordinary method are:
MD=[(95-103)x(-1)+(103-103)+(105-103)+(110-103)+(104-103)+(105-103)+(112103)+(90-103)x(-1)]/8=5.25.
The number of operations required is 18 (8 subtractions, 7 additions, 2 multiplications, 1
division).
The required calculations using the new method are:
MD=[(105+110+104+105+112)-(90+95)-(5-2)x103]/8=5.25
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The number of operations required is only 10 (3 subtractions, 5 additions, 1
multiplication and 1 division).
3.2
Using a calculator to find the mean and variance of a discrete probability
distribution
3.2.1 The ordinary method
The ordinary method of calculating the mean (μ) and variance (  2) of a discrete
probability distribution is by using the following formulae:
(4)
μ=  [xiP(xi)]
(5)
 2=  [(xi- μ)2P(xi)]
Where xi are the discrete values and P(xi) are their corresponding probabilities.
3.2.2 The new method
Given a table of values, xi, and their corresponding probabilities, P(xi), multiply all the
probabilities by a number M, such that all the products MP(xi) would become integer
numbers. Enter the xi values and the MP(xi) products into the calculator as if you were
dealing with grouped data. Once you finish, select the Statistics mode on your calculator
and read the mean and standard deviation.
3.2.3
Comparing the two methods
The ordinary method requires a large number of operations. For n discrete values, the
required number of operations is as follows:
For finding the mean: 2n-1 operations (n multiplications, n-1 additions).
For finding the variance: 4n-1 operations (n subtractions, n times raising to the 2nd power,
n multiplications, n-1 additions)
The total number of required operations is, therefore, 6n-2.
The new method requires n multiplications of the probabilities by a certain number, M.
All the rest is done by the calculator. One more operation may be required if the
calculator calculates the standard deviation, but not the variance. The difference in the
numbers of calculations is, therefore, 5n-3, which is positive for any n>0.
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3.2.4 Example
Find the mean and variance of the following discrete probability distribution:
x
P(x)
0
0.1
1
0.2
2
0.3
3
0.3
4
0.1
Total
1.0
The ordinary method
μ=0x0.1+1x0.2+2x0.3+3x0.3+4x0.1=2.1
Number of operations: 9
 2=[(0-2.1) 2x0.1+(1-2.1) 2x0.2+(2-2.1)2x0.3+(3-2.1)2x0.3+(4-2.1)2x0.1]=1.29
Number of operations: 19
Total number of operations: 28
The new method
Multiplying the probabilities by 10 we get:
x
P(x)
10P(x)
0
0.1
1
1
0.2
2
2
0.3
3
3
0.3
3
4
0.1
1
Total
1.0
10
Once the x and 10P(x) values are entered into the calculator as grouped data, we get
n=10, μ=2.1, and  =1.135781669, which after squaring becomes  2=1.29.
Total number of operations: 6 (5 multiplications, 1 squaring).
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