f1 - business maths august 08

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NOTESYou 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.

BUSINESS MATHEMATICS &

QUANTITATIVE METHODSFORMATION 1 EXAMINATION - AUGUST 2008

The Institute of Certified Public Accountants in Ireland, 17 Harcourt Street, Dublin 2.


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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 automaticproduction machines at the year end. There are three machines available, each costing 20,000 but theprojected 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 thecompany carrying out an appropriate analysis. The projected cash flows are:

Year Machine A Machine B Machine C1 3,000 5,000 6,0002 4,000 6,000 6,0003 7,000 6,000 6,5004 8,000 4,000 4,0005 8,500 4,000 1,000

You are asked to:

(i) Compare the alternatives using Net Present Value. (8 Marks)

(ii) Derive the Internal Rate of Return on the most viable proposal. (8 Marks)

(iii) Describe the critical factors affecting the investment decision. (4 Marks)

[Total: 20 Marks]

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2. The Impart Union represents junior and middle management administrative staff. It compiles earnings datato 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 Unionclaims that there is a major variation in the weekly wages paid to employees in both companies.

Weekly Earnings No. Employees No EmployeesIn Company A In Company B

300 360 2 2360 420 6 3420 480 14 7480 540 18 3540 600 16 7600 650 10 9650 700 7 3700 800 10 16800 900 10 10900 1100 6 201100 1400 1 20

Analyse the data by:

(a) (i) Calculating the mean and standard deviation for both companies. (12 Marks)

(ii) Deriving the co-efficient of variation. (4 Marks)

(b) Comment on the Unions claim. (4 Marks)

[Total : 20 Marks]

3. The table below gives the number of disposable digital thermometers manufactured by Superior ProductsLtd. in thousands over the past 4 years.

Quarters

Year 1 2 3 4

2004 132005 14 16 9 142006 16 17 12 172007 18 20 13

By using the method of moving averages you are asked to:

(i) Smooth this data by means of a centred four quarterly moving average. (10 Marks)

(ii) Calculate the average seasonal variations. (10 Marks)

[Total: 20 Marks]

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4. DIY Ltd. produces a range of components for manufacturing printed circuit boards. The quantities producedover the past number of years are set out in the table below:

2004 2004 2005 2006 2007Component Average Price Quantity Quantity Quantity Quantity

(000) (000) (000) (000)

1 3.00 90 95 100 1352 4.50 180 200 150 90

3 1.00 1000 1080 1160 16004 7.00 20 20 20 20

The Management Accountant wishes to analyse this data. As a first step, you are required to:

(i) Construct a quantity index for the components listed for the period 2004 2007using 2004 prices as weights. (12 Marks)

(ii) Explain the purpose of preparing this index and comment on your results. (8 Marks)

[Total: 20 Marks]

5. As the Managing Partner of CPA consultants, you are asked to advise on a large range of business problems.

(i) The Cost Accountant of DIB Ltd. wishes to purchase office equipment for 37,500 on 1st October2008. The equipment will last for 5 years and will be replaced on 1st October 2013. The bank hasprovided a 5 year loan compounded annually at 12%. You are required to determine the size of theequal annual loan repayments.

(6 Marks)

(ii) DIB Ltd. produces a new product where its sales are expected to be 70 units per week when the priceis 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 ademand relationship between the price and quantity demanded and to find the cost of producing 100units.

(6 Marks)

(iii) DIB Ltd. also manufactures batteries. Its states that each battery will last for approximately 200 hourswith a standard deviation of 4 hours. You are asked to determine the probability that a random samplewill last not less than 198 hours and to support your calculation graphically, assuming that the data isnormally distributed.

(8 Marks)

[Total: 20 Marks]

6. You have attended a management conference on the subject Methods of Describing Sets of Data. Thepresentation covered:

G The data needed for management decision-making.G Methods of describing qualitative data such as charts and graphs.G Methods of describing quantitative data such as histograms, scattergrams, measures of central

tendency and dispersion and time series analysis.G 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

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THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS &QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - AUGUST 2008

SOLUTION 1

(i) Net Present Value.

Year Cash Cash Cash Discount PV PV PV

Flow Flow Flow Factor of of of

A B C @ 10% A B C

0 (20,000) (20,000) (20,000) 1.000 (20,000) (20,000) (20,000)1 3,000 5,000 6,000 0.909 2,727 4,545 5,454

2 4,000 6,000 6,000 0.826 3,304 4,956 4,9563 7,000 6,000 6,500 0.751 5,257 4,506 4,8814 8,000 4,000 4,000 0.683 5,464 2,732 2,7325 8,500 4,500 1,000 0.621 5,279 2,484 621NPV 2,031 (777) (1,356)

(2 Marks) (2 Marks) (2 Marks)

The only viable option for the company is purchase of machine A which provides a positive net present value.(2 Marks)

[8 Marks]

(ii) Since A has the highest NPV, this is the proposal for which the IRR is calculated. To derive the IRR byinterpolation or otherwise a discount factor of 14% is used.

Year Cash Flow A Discount Factor @14% PV of A

0 (20,000) 1.000 (20,000)

1 3,000 0.877 2,631

2 4,000 0.769 3,076

3 7,000 0.675 4,725

4 8,000 0.592 4,736

5 8,500 0.519 4,412NPV (420)(2 Marks)

Using the following formula for IRR

N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rateN1 - N2 I2 gives NPV N2.

Where N1 = 2,031, I1 = 10%; N2 = (420), I2 = 14% (2 Marks)

IRR = 2031 x 0.14 - 420 x 0.10 = 326.34 = 13.3%2031 - 420 2451

The IRR for the project (Machine A) is 13.3%. (4 Marks)[8 Marks]

SUGGESTED SOLUTIONS


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(iii) The process of investment for capital expenditure decisions is a vital part of policy making and topmanagement usually assumes direct responsibility for the authorisation of all large capital investmentdecisions because

- the sums involved are usually very large- the companys 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 investmentcould 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.

Company A

Weekly Freq (f) Mid point x fx (x x) (x x)2 f(x x)2

Earnings (000)

300-360 2 330 660 -288 82944 165888

360-420 6 390 2340 -228 51984 311904

420-480 14 450 6300 -168 28224 395136480-540 18 510 9180 -108 11664 209952

540-600 16 570 9120 -48 2304 36864

600-650 10 625 6250 7 49 490

650-700 7 675 4725 57 3249 22743

700-800 10 750 7500 132 17424 174240

800-900 10 850 8500 232 53824 538240

900-1100 6 1000 6000 382 145924 875544

1100-1400 1 1250 1250 632 399424 399424

100 61825 3130425

Mean = x = fx = 61825 = 618.25 (2 Marks)f 100

Standard deviation = = f (x - x)2 = 3130425f 100

= 31304.25 = 176.93 (4 Marks)


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Company B

Weekly Freq (f) Mid point x fx (x x) (x x)2 f(x x)2

Earnings (000)

300-360 2 330 660 -507 257049 514098

360-420 3 390 1170 -447 199809 599427

420-480 7 450 3150 -387 149769 1048383

480-540 3 510 1530 -327 106929 320787

540-600 7 570 3990 -267 71289 427734600-650 9 625 5625 -212 44944 449440

650-700 3 675 2025 -162 26244 78732

700-800 16 750 12000 -87 7569 121104

800-900 10 850 8500 13 169 1690

900-1100 20 1000 20000 163 26569 531380

1100-1400 20 1250 25000 413 170569 3411380

100 83650 7504155

Mean = x = fx = 83650 = 836.50 (2 Marks)f 100

Standard deviation = = f (x - x)2 = 7504155f 100

= 75041.55 = 273.93 (4 Marks)

(ii) Co-efficient of variation. /x

Company A: 176.93/618.09 = 28.62% (2 Marks)

Company B: 273.93/837.05 = 32.72% (2 Marks)

When considering different distributions with significantly different means, it is necessary to use other meansthan the variance or standard dev deviation to make a realistic comparison. The co-efficient of variationprovides a method of comparison of spread relative to the magnitude of data under consideration. In thepresent case the employees in company B appear to show substantial variability in both mean and standarddeviation. This is confirmed by calculating the co-efficient of variation which shows that employees incompany B show a greater variation in earnings. However, the co-efficient of variation is not of such a levelas 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. Thismay be due to age and seniority factors in this company while the sample from company A may contain asubstantial number of younger and less senior employees. Therefore there may be distortions in the sampledata which require further investigation. Using the mean and standard deviations in isolation may indicate thata problem exists with the earning data which may not be correct. It is obvious that the data in both companiesis positively skewed, that is, where the mean is greater than the median, therfore the Union claim is notnecessarily correct.

(4 Marks)

[Total: 20 Marks]

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SOLUTION 3

(i) The original data is presented below.

Quarters

Year 1 2 3 4

1984 131985 14 16 9 141986 16 17 12 171987 18 20 13

Using a four quartered centred moving average gives the trend in the table below in Col. 6.

Year Quarter Data Moving Moving Trend Deviation Seasonal

000s annual pair total variation

total

Col 1 Col 2 Col 3 Col 4 Col 5 Col 6 Col 7 Col 8

1984 4 13 0.375

1985 1 14 1.375

2 16 52 105 13.125 2.875 2.3753 9 53 108 13.500 - 4.500 - 4.125

4 14 55 111 13.875 0.125 0.375

1986 1 16 56 115 14.375 1.625 1.375

2 17 59 121 15.125 1.875 2.375

3 12 62 126 15.750 - 3.750 - 4.125

4 17 64 131 16.375 0.625 0.375

1987 1 18 67 135 16.875 1.125 1.375

2 20 68 2.375

3 13 - 4.125

2 Marks 2 Marks 2 Marks 4 Marks (10 Marks)

(ii) 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 adjustedto provide the seasonal variation as below. The total average deviation is zero. If there is an adjustment theoriginal data can be adjusted by the required amount to provide data incorporating the seasonal adjustments.

Quarter 1 2 3 4

1985 ------ 2.875 - 4.500 0.125

1986 1.625 1.875 - 3.750 0.6251987 1.125 ------- -------- -------Total 2.750 4.750 - 8.250 0.750Average 1.375 2.375 - 4.125 0.375 0Adjustment 0 0 0 0Seas Var 1.375 2.375 - 4.125 0.375

(10 Marks)

[Total: 20 Marks]

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SOLUTION 4

(i) Using base prices as weights gives the quantity index as the base weighted index.

2004 2004 2005 2006 2007

Component Average Price Quantity Quantity Quantity Quantity

(000) (000) (000) (000)

1 3.00 90 95 100 1352 4.50 180 200 150 90

3 1.00 1000 1080 1160 1600

4 7.00 20 20 20 20

Com. P0 Q0 Q1 Q2 Q3 P0Q0 P0Q1 P0Q2 P0Q3

1 3.00 90 95 100 135 270 285 300 4052 4.50 180 200 150 90 810 900 675 405

3 1.00 1000 1080 1160 1600 1000 1080 1160 16004 7.00 20 20 20 20 140 140 140 140Total 2220 2405 2275 2550

(2 Marks) (2 Marks) (2 Marks)

Deriving the Index of Production where 2004 is the base:

2005 = 2405/2220 = 108.3 (2 Marks)

2006 = 2275/2220 = 102.4 (2 Marks)

2007 = 2550/2220 = 114.8 (2 Marks)

(ii) The index for the three periods indicates that the index reduced for 2006. The change in production can bedue 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 asweights as in the present case. This implies that the quantities do not vary over time and in many cases thisis not correct, as in the present case. If the prices are kept the same, any variation in production must be dueto changes in quantities produced so a quantity or volume index can be calculated. This index tends tooverstate increases in P. To attempt to overcome this current year quantities may be used such as in thePaasche index. However, this tends to understate the affects of P. The Laspeyres index generally exceeds

the Paasche index.(8 Marks)

[Total: 20 Marks]

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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 whicheach 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 netpresent value of the annuity payments, A, over the period of the debt, that is, P = A/(1 + i) + .. A/(1+ i)

n, 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)

= A(3.605)

Therefore, A = 37,500/3.605 = 10,402. (3 Marks)

(ii) 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.

The demand equation is y = 85 - 0.5x. (3 Marks)

The cost of producing 100 units: C = Fixed costs + Variable costs per unit

C = 1500 + 10 x 100 = 2500 (3 Marks)

(iii) Any random variable which follows a normal distribution can be transformed to the standard normal variablez by using the transformation z = (x - )/.

Therefore when x = 198, z = (x - )/ (198 - 200)/4 = - 2/4 = 0.5.

Hence, P(x < 198) = P(z < - 0.5) = 0.5 - 0.1915 (from Normal Tables)

= 0.3085, that is 30.85%. (4 Marks)

The appropriate Normal Distribution is

(4 Marks)

[Total : 20 Marks]

10

x

y

x1

198

-0.5 0

= 200


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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 andultimately the decision made. Characteristics of a data set may contain the most frequent score, the variabilityin the score, the shape of the data, the highest and lowest scores, and whether or not the data set containsany unusual data. Interpreting or extracting data visually is, at a minimum, difficult since it may not be possibleto comprehend large volumes of information. Some formal methods for summarising and characterising theinformation in a data set are essential. Most populations are large data sets. Therefore, methods fordescribing such data sets are also essential for statistical inference. There are two key methods used fordescribing data one graphical and the other numerical. Both play an important role in statistics and bothmethods 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 acategorical or qualitative distribution. Therefore the value of a qualitative variable can be classified intocategories called classes. Such data can be summarized numerically in two ways by computing the classfrequency, that is, the number of observations in the data set that fall into each class or by computing theclass relative frequency, that is, the proportion of the total number of observations falling into each class. Twoof 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 bar is equal to the classfrequency. A pie chart shows the relative frequencies of the classes where the size of the slice apportionedto 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 acategorical or quantitative distribution. For describing, summarising and detecting patterns in such data themost 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 themeasurements falling into specified intervals. By looking at a histogram two important facts are apparent. Theproportion of the total area above the interval is equal to the relative frequency of measurements falling in theinterval. As the number of elements in the data set increase, a better description of the data set can beobtained by decreasing the width of the class intervals. When the class intervals become small enough arelative frequency histogram will appear as a smooth curve. While histograms provide good visualdescriptions 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 andthe successive points are connected. Applying a similar technique to a cumulative distribution (usually a lessthan distribution) an ogive can be obtained. This is where the cumulative frequencies are plotted at the classboundaries.

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 betweenthe variables by linear regression form of analysis. This can be visually and quantitatively presented and isan effective display.

A large number of numerical methods are available to describe quantitative data sets. Most of these methodsmeasure one of two data characteristics 1) the central tendency of the set, that is, the tendency of the datato cluster and 2) the variability of the set, that is, the spread of the data. The most popular measure of centraltendency is the arithmetic mean. This is the sum of the measurements divided by the number ofmeasurements contained in the data set. Another important measure of central tendency is the median. Thisis the middle number when the quantitative data set is arranged in ascending or descending order. This is ofmost 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 themean more than the median since these values are used explicitly to calculate the mean. The median is notaffected directly by extreme values since only the middle value is explicitly used to calculate the median.

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The modeis particularly useful for describing qualitative data. The modal category is the class that occursmost frequently. Because it emphasises data concentrations, the mode is also used with quantitative datasets 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 moremeaningful 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. Thedescription is incomplete without a measure of the variability of the data set. Knowledge of the datasvariability along with its centre can help us visualise the shape of the data as well as its extreme values. Thesimplest measure of the variability of a quantitative set is its range. This is equal to the largest less thesmallest measurement. The range is easy to compute and to understand but it is an insensitive measure ofdata variation when the data sets are large two data sets can have the same range but be vastly differentwith respect to data variation. The variation of such data can be obtained by measuring the distance betweeneach measurement and the mean. To cater for the + and signs of the deviations, the deviations are squaredto provide the sample variance, s2 = (xi x)2/(n 1). This is the preliminary step in calculating the standarddeviation of the data set, s2. Sample statistics like s2 are primarily used to estimate population parameterslike _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 overtime 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 aboutthe process that generated the data and to monitor the movement (trend) and changes (variations) in thevariable being examined.

(5 Marks)

Many of the presentations outlined can misrepresent or distort, or allow the data presented to bemisinterpreted. One common way to change the impression created by a pictorial or graphical presentationis to change the scale on either one or both axes. By stretching the vertical axis or by increasing the distancebetween vertical units can give a misleading visual impression of the data. In one case a histogram mayappear to be vertically elongated and horizontally compressed or vice versa and may lead to incorrectconclusions. A visual distortion can be achieved with bar graphs by making the width of the bars proportionalto the height. A similar effect can be achieved by using a scale break for the vertical axis. Further distortionscan also occur with numerical descriptive measures. If a measure of central tendency only is reported in asample, this can lead to a distortion of the information. Both a measure of central tendency and a measureof variability are needed to obtain an accurate mental image of a data set.

(5 Marks)

[Total : 20 Marks]

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f1 - business maths august 08

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