BUSINESS MATHEMATICS &
QUANTITATIVE METHODS
FORMATION 1 EXAMINATION - AUGUST 2008
NOTES
You are required to answer any 5 questions.
(If you provide answers to all questions, you must draw a clearly distinguishable line through the answer not to
be marked. Otherwise, only the first 5 answers to hand will be marked).
All questions carry equal marks.
STATISTICAL FORMULAE TABLES ARE PROVIDED
DEPARTMENT OF EDUCATION MATHEMATICS TABLES ARE AVAILABLE UPON REQUEST
TIME ALLOWED:
3 hours, plus 10 minutes to read the paper.
INSTRUCTIONS:
During the reading time you may write notes on the examination paper but you may not commence
writing in your answer book.
Marks for each question are shown. The pass mark required is 50% in total over the whole paper.
Start your answer to each question on a new page.
You are reminded that candidates are expected to pay particular attention to their communication skills
and care must be taken regarding the format and literacy of the solutions. The marking system will take
into account the content of the candidates' answers and the extent to which answers are supported with
relevant legislation, case law or examples where appropriate.
List on the cover of each answer booklet, in the space provided, the number of each question(s)
attempted.
The Institute of Certified Public Accountants in Ireland, 17 Harcourt Street, Dublin 2.
THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND
BUSINESS MATHEMATICS &
QUANTITATIVE METHODS
FORMATION 1 EXAMINATION - AUGUST 2008
Time Allowed: 3 hours, plus 10 minutes to read the paper.
You are required to answer any 5 questions.
(If you provide answers to all questions, you must draw a clearly distinguishable line through the answer not to
be marked. Otherwise, only the first 5 answers to hand will be marked).
All questions carry equal marks.
1.
The accountant in DIY Products Ltd. estimates that the company will have to replace some of the automatic
production machines at the year end. There are three machines available, each costing €20,000 but the
projected annual cash flows for each machine are different. The machines have no residual value at the end
of year 5. The bank has agreed to fund the cost of the machine at an overdraft rate of 10% subject to the
company carrying out an appropriate analysis. The projected cash flows are:
You are asked to:
(i)
(ii)
(iii)
Year
1
2
3
4
5
Machine A
€3,000
€4,000
€7,000
€8,000
€8,500
Machine B
€5,000
€6,000
€6,000
€4,000
€4,000
Machine C
€6,000
€6,000
€6,500
€4,000
€1,000
Compare the alternatives using Net Present Value.
Derive the Internal Rate of Return on the most viable proposal.
Describe the critical factors affecting the investment decision.
1
(8 Marks)
(8 Marks)
(4 Marks)
[Total: 20 Marks]
2.
The Impart Union represents junior and middle management administrative staff. It compiles earnings data
to ensure that the terms and conditions of employment of its members over a range of companies are similar.
The data below has been provided by two companies relating to 100 employees. In negotiations the Union
claims that there is a major variation in the weekly wages paid to employees in both companies.
Weekly Earnings €
300
360
420
480
540
600
650
700
800
900
1100
Analyse the data by:
(a)
(b)
3.
(i)
(ii)
– 360
– 420
– 480
– 540
– 600
– 650
– 700
– 800
– 900
– 1100
– 1400
No. Employees
In Company A
2
6
14
18
16
10
7
10
10
6
1
No Employees
In Company B
2
3
7
3
7
9
3
16
10
20
20
Calculating the mean and standard deviation for both companies.
Deriving the co-efficient of variation.
Comment on the Unionʼs claim.
(12 Marks)
(4 Marks)
(4 Marks)
[Total : 20 Marks]
The table below gives the number of disposable digital thermometers manufactured by Superior Products
Ltd. in thousands over the past 4 years.
Year
2004
2005
2006
2007
1
14
16
18
Quarters
2
16
17
20
By using the method of moving averages you are asked to:
(i)
(ii)
3
9
12
13
4
13
14
17
Smooth this data by means of a centred four quarterly moving average.
Calculate the average seasonal variations.
(10 Marks)
(10 Marks)
[Total: 20 Marks]
2
4.
DIY Ltd. produces a range of components for manufacturing printed circuit boards. The quantities produced
over the past number of years are set out in the table below:
Component
1
2
3
4
2004
Average Price
€
3.00
4.50
1.00
7.00
2004
Quantity
(000)
90
180
1000
20
2005
Quantity
(000)
95
200
1080
20
2006
Quantity
(000)
100
150
1160
20
2007
Quantity
(000)
135
90
1600
20
The Management Accountant wishes to analyse this data. As a first step, you are required to:
(i)
(ii)
5.
Explain the purpose of preparing this index and comment on your results.
(12 Marks)
(8 Marks)
[Total: 20 Marks]
As the Managing Partner of CPA consultants, you are asked to advise on a large range of business problems.
(i)
(ii)
(iii)
6.
Construct a quantity index for the components listed for the period 2004 – 2007
using 2004 prices as weights.
The Cost Accountant of DIB Ltd. wishes to purchase office equipment for €37,500 on 1st October
2008. The equipment will last for 5 years and will be replaced on 1st October 2013. The bank has
provided a 5 year loan compounded annually at 12%. You are required to determine the size of the
equal annual loan repayments.
(6 Marks)
DIB Ltd. produces a new product where its sales are expected to be 70 units per week when the price
is €50 per unit. However, the price was advertised at €70 per unit and only 30 units were sold. The
fixed costs are €1,500 per week and the variable costs per unit are €10. You are asked to develop a
demand relationship between the price and quantity demanded and to find the cost of producing 100
units.
(6 Marks)
DIB Ltd. also manufactures batteries. Its states that each battery will last for approximately 200 hours
with a standard deviation of 4 hours. You are asked to determine the probability that a random sample
will last not less than 198 hours and to support your calculation graphically, assuming that the data is
normally distributed.
(8 Marks)
[Total: 20 Marks]
You have attended a management conference on the subject “Methods of Describing Sets of Data”. The
presentation covered:
●
●
●
●
The data needed for management decision-making.
Methods of describing qualitative data such as charts and graphs.
Methods of describing quantitative data such as histograms, scattergrams, measures of central
tendency and dispersion and time series analysis.
How presentations may be used to distort the data.
Discuss the content of the presentation with appropriate descriptions of the material outlined.
[Total: 20 Marks]
END OF PAPER
3
SUGGESTED SOLUTIONS
THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND
BUSINESS MATHEMATICS &
QUANTITATIVE METHODS
SOLUTION 1
(i)
Net Present Value.
Year
0
1
2
3
4
5
NPV
Cash
Flow
A
(€20,000)
€3,000
€4,000
€7,000
€ 8,000
€ 8,500
FORMATION 1 EXAMINATION - AUGUST 2008
Cash
Flow
B
(€20,000)
€5,000
€6,000
€6,000
€4,000
€4,500
Cash
Flow
C
Discount
Factor
@ 10%
(€20,000)
€6,000
€6,000
€6,500
€4,000
€1,000
1.000
0.909
0.826
0.751
0.683
0.621
PV
of
A
(€20,000)
€2,727
€3,304
€5,257
€5,464
€5,279
€2,031
(2 Marks)
PV
of
B
(€20,000)
€4,545
€4,956
€4,506
€2,732
€2,484
(€777)
(2 Marks)
PV
of
C
(€20,000)
€5,454
€4,956
€4,881
€2,732
€ 621
(€1,356)
(2 Marks)
The only viable option for the company is purchase of machine A which provides a positive net present value.
(2 Marks)
(ii)
[8 Marks]
Since A has the highest NPV, this is the proposal for which the IRR is calculated. To derive the IRR by
interpolation or otherwise a discount factor of 14% is used.
Year
Cash Flow A
Discount Factor @14%
2
€4,000
0.769
0
1
3
4
5
NPV
(€20,000)
€3,000
1.000
(€20,000)
0.675
€4,725
0.877
€7,000
€ 8,000
0.592
€ 8,500
PV of A
0.519
Using the following formula for IRR
N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rate
I2 gives NPV N2.
N1 - N2
€2,631
€3,076
€4,736
€4,412
(€420)
(2 Marks)
Where N1 = €2,031, I1 = 10%; N2 = (€420), I2 = 14%
(2 Marks)
The IRR for the project (Machine A) is 13.3%.
(4 Marks)
[8 Marks]
IRR = 2031 x 0.14 - 420 x 0.10
2031 - 420
= 326.34 = 13.3%
2451
5
(iii)
The process of investment for capital expenditure decisions is a vital part of policy making and top
management usually assumes direct responsibility for the authorisation of all large capital investment
decisions because
- the sums involved are usually very large
- the companyʼs resources may be tied up for a considerable period of time
- investment decisions are long-term and have a major impact on the viability of the company.
The critical factors affecting the capital investment decision are
- the projected future increases in income or savings in costs
- the net amount of the investment. The investment should not be so large that the failure of the investment
could have an impact on the survival of the company
- a satisfactory rate of return. This return should be sufficient to finance the cost of borrowings
- the level of risk. It is necessary to consider the risk factor since, in all business areas, the future is uncertain.
(4 Marks)
[Total: 20 Marks]
SOLUTION 2
(i)
The mean and standard deviation for both companies.
Weekly
Freq (f)
Mid point x
360-420
6
390
Earnings €
300-360
420-480
480-540
540-600
600-650
650-700
700-800
2
14
18
16
10
7
10
800-900
10
∑
100
900-1100
1100-1400
Mean
6
1
= x = ∑fx
∑f
330
450
Company A
fx
(x – x)
(x – x)2
f(x – x)2
2340
-228
51984
311904
-48
2304
660
6300
510
570
-108
4725
57
6250
675
750
7500
850
8500
1000
6000
1250
1250
61825
= 61825 = €618.25
100
Standard deviation = σ = √ ∑ f (x - x)2 = √ 3130425
∑f
100
=
√ 31304.25
=
-168
9180
9120
625
-288
€176.93
6
7
132
232
382
632
(000)
82944
165888
28224
395136
49
490
11664
3249
17424
53824
145924
399424
209952
36864
22743
174240
538240
875544
399424
3130425
(2 Marks)
(4 Marks)
Weekly
Freq (f)
Mid point x
360-420
3
390
Earnings €
300-360
2
420-480
7
480-540
3
540-600
650-700
3
700-800
16
1100-1400
20
800-900
900-1100
∑
= x = ∑fx
∑f
625
5625
1250
(ii)
Co-efficient of variation.
Company A:
Company B:
σ/x
€176.93/€618.09
€273.93/€837.05
€273.93
=
-327
106929
-387
-267
-212
28.62%
32.72%
-87
13
413
514098
149769
1048383
44944
449440
169
1690
71289
26569
83650
=
257049
163
25000
=
599427
-507
20000
= 83650 = €836.50
100
√ 75041.55
199809
(000)
26244
Standard deviation = σ = √ ∑ f (x - x)2 = √ 7504155
∑f
100
=
f(x – x)2
-162
8500
1000
(x – x)2
2025
12000
850
100
-447
3990
750
20
1170
660
1530
675
10
(x – x)
3150
570
9
fx
450
510
7
600-650
Mean
330
Company B
7569
170569
320787
427734
78732
121104
531380
3411380
7504155
(2 Marks)
(4 Marks)
(2 Marks)
(2 Marks)
When considering different distributions with significantly different means, it is necessary to use other means
than the variance or standard dev deviation to make a realistic comparison. The co-efficient of variation
provides a method of comparison of spread relative to the magnitude of data under consideration. In the
present case the employees in company B appear to show substantial variability in both mean and standard
deviation. This is confirmed by calculating the co-efficient of variation which shows that employees in
company B show a greater variation in earnings. However, the co-efficient of variation is not of such a level
as to indicate that may be difficulties with the data or its interpretation. There may be other factors involved.
It is obvious that in company B a large number of the sample are at the higher end of the earnings scale. This
may be due to age and seniority factors in this company while the sample from company A may contain a
substantial number of younger and less senior employees. Therefore there may be distortions in the sample
data which require further investigation. Using the mean and standard deviations in isolation may indicate that
a problem exists with the earning data which may not be correct. It is obvious that the data in both companies
is positively skewed, that is, where the mean is greater than the median, therfore the Union claim is not
necessarily correct.
(4 Marks)
[Total: 20 Marks]
7
SOLUTION 3
(i)
The original data is presented below.
Year
1984
1985
1986
1987
Quarters
2
1
14
16
18
16
17
20
3
4
13
14
17
9
12
13
Using a four quartered centred moving average gives the trend in the table below in Col. 6.
Year
Quarter
Col 1
Col 2
1984
1985
Col 3
Moving
annual
total
Col 4
Moving
pair total
2
16
1
16
4
13
1
14
3
9
4
1986
Trend
Deviation
Col 5
Col 6
Col 7
52
105
13.125
2.875
56
115
14.375
53
14
108
55
111
1
18
67
135
16.875
17
2
64
20
3
13
2 Marks
131
68
2 Marks
2 Marks
2.375
2.375
15.125
4
1.375
1.875
13.875
121
126
0.375
- 4.125
59
62
Col 8
- 4.500
17
12
Seasonal
variation
13.500
2
3
1987
(ii)
Data
000s
0.125
0.375
1.625
15.750
1.375
- 3.750
16.375
- 4.125
0.625
0.375
1.125
1.375
2.375
- 4.125
(10 Marks)
4 Marks
The deviation can be derived by using either the additive or multiplicative models. Using the additive model,
the deviation is found in Col. 7 by deducting the trend from the original data. This deviation is then adjusted
to provide the seasonal variation as below. The total average deviation is zero. If there is an adjustment the
original data can be adjusted by the required amount to provide data incorporating the seasonal adjustments.
Quarter
1985
1986
1987
Total
Average
Adjustment
Seas Var
1
-----1.625
1.125
2.750
1.375
0
1.375
2
2.875
1.875
------4.750
2.375
0
2.375
3
- 4.500
- 3.750
-------- 8.250
- 4.125
0
- 4.125
4
0.125
0.625
------0.750
0.375
0
0.375
0
(10 Marks)
[Total: 20 Marks]
8
SOLUTION 4
(i) Using base prices as weights gives the quantity index as the base weighted index.
Component
2004
Average Price €
2
4.50
1
3
7.00
P0
3.00
4.50
1.00
7.00
Q0
90
180
1000
20
2005
Quantity
(000)
2006
Quantity
(000)
2007
Quantity
(000)
1000
1080
1160
1600
90
95
180
1.00
4
Com.
1
2
3
4
Total
3.00
2004
Quantity
(000)
200
20
Q1
95
200
1080
20
20
Q2
100
150
1160
20
Q3
135
90
1600
20
Deriving the Index of Production where 2004 is the base:
2005 = 2405/2220 = 108.3
135
150
90
20
P0Q0
270
810
1000
140
2220
P0Q1
285
900
1080
140
2405
20
P0Q2
300
675
1160
140
2275
P0Q3
405
405
1600
140
2550
(2 Marks) (2 Marks) (2 Marks)
(2 Marks)
(2 Marks)
2006 = 2275/2220 = 102.4
(2 Marks)
2007 = 2550/2220 = 114.8
(ii)
100
The index for the three periods indicates that the index reduced for 2006. The change in production can be
due to changes in quantities produced and these changes can be measured by a volume or quantity index.
Using base prices as weights gives the Laspeyres quantity index. This index uses base year statistics as
weights as in the present case. This implies that the quantities do not vary over time and in many cases this
is not correct, as in the present case. If the prices are kept the same, any variation in production must be due
to changes in quantities produced so a quantity or volume index can be calculated. This index tends to
overstate increases in P. To attempt to overcome this current year quantities may be used such as in the
Paasche index. However, this tends to understate the affects of P. The Laspeyres index generally exceeds
the Paasche index.
(8 Marks)
[Total: 20 Marks]
9
SOLUTION 5
(i)
The amortization method of debt repayment is required. If the amount of money is borrowed over a period,
the money may be repaid by means of an amortization annuity. This consists of a regular payment in which
each payment consists of both capital and interest. The debt is said to be amortized if this method is used.
Many of the mortgages for home purchases are of this type. The amount borrowed, P, is equal to the net
present value of the annuity payments, A, over the period of the debt, that is, P = A/(1 + i) + ……….. A/(1
n
+ i) , where i = interest rate and n = period of loan.
(3 Marks)
Since P
= €37,500, i = 12% = 0.12, n = 5
€37,500
= A/1.12 + A/1.122 + A/1.123 + A/1.124 + A/1.125
= A(0.893 + 0.797 + 0.712 + 0.636 + 0.567)
Therefore, A
(ii)
= A(3.605)
(3 Marks)
= 37,500/3.605 = €10,402.
Since the relationship is linear, it can be represented as y = a + bx where y = price, x = quantity.
Therefore, when x = 70, y = 50; 50 = a + 70b.
when x = 30, y = 70; 70 = a + 30b.
Solving the equations for values of a and b gives a = 85 and b = - 0.5.
(3 Marks)
The demand equation is y = 85 - 0.5x.
The cost of producing 100 units: C = Fixed costs + Variable costs per unit
(3 Marks)
C = 1500 + 10 x 100 = €2500
(iii)
Any random variable which follows a normal distribution can be transformed to the standard normal variable
z by using the transformation z = (x - µ)/σ.
Therefore when x = 198, z = (x - µ)/σ (198 - 200)/4 = - 2/4 = 0.5.
Hence, P(x < 198) = P(z < - 0.5) = 0.5 - 0.1915 (from Normal Tables)
(4 Marks)
= 0.3085, that is 30.85%.
The appropriate Normal Distribution is
x1
198
-0.5
x
µ= 200
0
y
10
(4 Marks)
[Total : 20 Marks]
SOLUTION 6
Methods for Describing Sets of Data
In all accountancy, management and consultancy areas, a large volume of data is subject to analysis,
interpretation and description. The methods and techniques used can distort the description of the data and
ultimately the decision made. Characteristics of a data set may contain the most frequent score, the variability
in the score, the ʻshapeʼ of the data, the highest and lowest scores, and whether or not the data set contains
any unusual data. Interpreting or extracting data visually is, at a minimum, difficult since it may not be possible
to comprehend large volumes of information. Some formal methods for summarising and characterising the
information in a data set are essential. Most populations are large data sets. Therefore, methods for
describing such data sets are also essential for statistical inference. There are two key methods used for
describing data – one graphical and the other numerical. Both play an important role in statistics and both
methods can be used for describing both qualitative and quantitative data.
(5 Marks)
Describing qualitative data. When data is grouped into non-numerical categories, the resulting table is a
categorical or qualitative distribution. Therefore the value of a qualitative variable can be classified into
categories called classes. Such data can be summarized numerically in two ways – by computing the class
frequency, that is, the number of observations in the data set that fall into each class or by computing the
class relative frequency, that is, the proportion of the total number of observations falling into each class. Two
of the most widely used graphical methods for describing qualitative data are bar graphs and pie charts. The
bar graph plots the class frequency against the class where the height of the ʻbarʼ is equal to the class
frequency. A pie chart shows the relative frequencies of the classes where the size of the ʻsliceʼ apportioned
to each class is proportional to the class relative frequency.
(5 Marks)
Describing quantitative data. When data are grouped according to numerical size, the resulting table is a
categorical or quantitative distribution. For describing, summarising and detecting patterns in such data the
most common graphical methods for describing frequency distributions are histograms, frequency curves
(such as polygons or ogives) and scattergrams.
Histograms. Histograms can be used to display either the frequency or relative frequencies of the
measurements falling into specified intervals. By looking at a histogram two important facts are apparent. The
proportion of the total area above the interval is equal to the relative frequency of measurements falling in the
interval. As the number of elements in the data set increase, a better description of the data set can be
obtained by decreasing the width of the class intervals. When the class intervals become small enough a
relative frequency histogram will appear as a smooth curve. While histograms provide good visual
descriptions of data sets they do not allow the identification of individual measurements.
Another form of presentation is the frequency polygon. The frequencies are plotted at the class marks and
the successive points are connected. Applying a similar technique to a cumulative distribution (usually a ʻless
thanʼ distribution) an ogive can be obtained. This is where the cumulative frequencies are plotted at the class
boundaries.
Scattergram. A common method to describe the relationship between two quantitative variables is to plot the
data on a scattergram. Both a relationship, and the strength of that relationship, can be determined between
the variables by linear regression form of analysis. This can be visually and quantitatively presented and is
an effective display.
A large number of numerical methods are available to describe quantitative data sets. Most of these methods
measure one of two data characteristics 1) the central tendency of the set, that is, the tendency of the data
to cluster and 2) the variability of the set, that is, the spread of the data. The most popular measure of central
tendency is the arithmetic mean. This is the sum of the measurements divided by the number of
measurements contained in the data set. Another important measure of central tendency is the median. This
is the middle number when the quantitative data set is arranged in ascending or descending order. This is of
most value in describing large data sets and is the point where 50% of the data lies above the mid point and
50% below it. In certain situations the median may be a better measure of central tendency than the mean.
It is less sensitive to extremely large or small values. In general, extreme values (large or small) affect the
mean more than the median since these values are used explicitly to calculate the mean. The median is not
affected directly by extreme values since only the middle value is explicitly used to calculate the median.
11
The mode is particularly useful for describing qualitative data. The modal category is the class that occurs
most frequently. Because it emphasises data concentrations, the mode is also used with quantitative data
sets to locate the region in which much of the data is concentrated. However, for some quantitative data sets,
the mode may not be very meaningful since there may be more than one mode in the sample. A more
meaningful measure is the modal class. This can be obtained from a relative frequency histogram.
These measures of central tendency provide only a partial description of a quantitative data set. The
description is incomplete without a measure of the variability of the data set. Knowledge of the dataʼs
variability along with its centre can help us visualise the shape of the data as well as its extreme values. The
simplest measure of the variability of a quantitative set is its range. This is equal to the largest less the
smallest measurement. The range is easy to compute and to understand but it is an insensitive measure of
data variation when the data sets are large – two data sets can have the same range but be vastly different
with respect to data variation. The variation of such data can be obtained by measuring the distance between
each measurement and the mean. To cater for the + and – signs of the deviations, the deviations are squared
to provide the sample variance, s2 = ∑(xi – x)2/(n – 1). This is the preliminary step in calculating the standard
deviation of the data set, √s2. Sample statistics like s2 are primarily used to estimate population parameters
like _2; (n -1) is preferred to n when defining the sample variance.
However, data of interest to managers are often produced over a time period. When data is produced over
time it is important to record both the measurements and the time period associated with each measurement.
With this information a time series plot can be constructed to describe the time series data and to learn about
the process that generated the data and to monitor the movement (trend) and changes (variations) in the
variable being examined.
(5 Marks)
Many of the presentations outlined can misrepresent or distort, or allow the data presented to be
misinterpreted. One common way to change the impression created by a pictorial or graphical presentation
is to change the scale on either one or both axes. By stretching the vertical axis or by increasing the distance
between vertical units can give a misleading visual impression of the data. In one case a histogram may
appear to be vertically elongated and horizontally compressed or vice versa and may lead to incorrect
conclusions. A visual distortion can be achieved with bar graphs by making the width of the bars proportional
to the height. A similar effect can be achieved by using a scale break for the vertical axis. Further distortions
can also occur with numerical descriptive measures. If a measure of central tendency only is reported in a
sample, this can lead to a distortion of the information. Both a measure of central tendency and a measure
of variability are needed to obtain an accurate mental image of a data set.
(5 Marks)
[Total : 20 Marks]
12