F1 Business Maths April

8/14/2019 F1 Business Maths April

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THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS &QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - APRIL 2006.

Time Allowed: 3 hours and 10 minutes to read the paper Answer 5 questionsOnly the first five questions answered will be marked.

All questions carry equal marks.

1. You are advising CPA Ltd. on two alternative proposals for implementing an IT system. The system willautomate a number of existing office routines and is projected to provide an acceptable return on theinvestment. Proposal 1 requires an expenditure of 6,500 now and a further 4,000, at the end of 2 years.It will have a cash inflow of 18,100 in 5 years time. Proposal 2 requires 4 annual investments of 2,200starting now, it will have a cash inflow of 16,250 in 5 years time.

(i) Calculate the Net Present Value (NPV) for each project using a cost of capital of 10%. (8 Marks)

(ii) Calculate the Internal Rate of Return (IRR) on both projects. (8 Marks)

(iii) Advise the company on the best option and the most profitable investment. (4 Marks)

[Total: 20 Marks]

2. The battery division of DIB Ltd. manufactures a range of commercial batteries for fork-lift trucks. It claimsthat the average length of life for its grade 1 battery is 60 months. The company offers a guarantee on itfor 36 months. If the standard deviation is known to be 10 months and the frequency distribution is known

to be a standard normal distribution, assuming the companys claim is true, determine:

(i) The percentage of the grade 1 batteries which will last more than 50 months. (6 Marks)

(ii) The percentage of the batteries that will last less than 40 months. (6 Marks)

(iii) Comment on the companys claim if your battery lasts 37 months. (8 Marks)

[Total: 20 Marks]

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3. NPA Remuneration Consultants have developed a data-base in which they compile labour costs for arange of categories of employees. The following data relates to the percentage change in costs of seniormanagers over the past 3 years.

% change in costs No. of Managers

-5 and less 0 40 and less 5 325 and less 10 25

10 and less 15 1015 and less 20 820 and less 25 325 and less 30 230 and less 35 535 and less 40 440 and less 100 3

100 and less 170 4

You are asked to calculate:

(i) The mean and median values of the annual percentage change in labour costs. (8 Marks)

(ii) The standard deviation. (8 Marks)

(iii) Comment on the differences in values derived in (i) above. (4 Marks)

[Total: 20 Marks]

4. At a management meeting, the marketing manager of Superior Products Ltd. contends that the level ofsales is directly related to marketing expenditure. The data is provided below.

Sales Advertising Expenditure

(000s) (000s)

25 8

35 12

29 11

24 5

28 14

12 3

18 6

27 8

17 430 9

Since expenditure appears to be large relative to the volume of sales, you are asked to prepare a reporton the managers claim. As part of your report:

(i) Derive a linear regression equation representing the data. (8 Marks)

(ii) Find the correlation coefficient between the data. (8 Marks)

(iii) Comment on the marketing managers claim. (4 Marks)

[Total: 20 Marks]

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5. The sales of kitchen units (in 000s) for Sno Products Ltd. for the years 2003 2005 are set out in the tablebelow:

Quarter

Year 1 2 3 4

2003 3 5 8 4

2004 4 6 10 52005 5 8 11 7

The financial manager asks you to prepare a report containing the following information.

(i) The trend in sales using a centred moving average. (10 Marks)

(ii) Calculate the seasonal variations for each quarter and adjust the data for 2005. (10 Marks)

[Total: 20 Marks]

6. As a CPA consultant, part of your brief is to assist clients to resolve a wide range of business problems.The following problems have been assigned to you for immediate delivery to a corporate client:

(i) Electricity company ELP charges customers for units of energy used on a kwhr basis. The companyhas a standing charge of 30.04 per two months and charges 2.39c per kwhr for electricity used.

Represent this model by means of a linear equation and calculate the annual cost for 2,400 kwhrs.(6 Marks)

(ii) According to Consumer Survey magazine, a random sample of 36 family-owned hotels taken fromthe Failte Ireland brochure showed that the average daily profit was 720 with a standard deviationof 110.

Find the average profit of the hotels with a 95% confidence level. (6 Marks)

(iii) A Manufacturing Company claims that the wages of its employees have more than exceeded the rateof inflation which has increased at an average rate of 6.8% per annum. As an example it states thatthe wages of a typical employee 10 years ago were 12,000 and this has now risen to 25,200.

Verify its claim. (8 Marks)

[Total: 20 Marks]

END OF PAPER

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BUSINESS MATHEMATICS &QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - APRIL 2006.

Solution 1.

(i) Net Present Value for each proposal.

Cash flows for proposal 1.

Year Investment Inflows Net Cash PV Factor PV PV PV

Flow @ 10% Factor@ 14% PV

0 6,500 (6,500) 1.000 (6,500) 1.000 (6,500)

1 0.909 0.877

2 4,000 (4,000) 0.826 (3,304) 0.769 (3,076)

3 0.751 0.675

4 0.683 0.592

5 18,100 18,100 0.621 11,240 0.519 9,394

NPV 1,436 (182)

4 Marks

Cash flows for proposal 2.

Year Investment Inflows Net Cash PV Factor PV PV PVFlow @ 10% Factor

@ 14% PV

0 2,200 (2,200) 1.000 (2,200.0) 1.000 (2,200.0)

1 2,200 (2,200) 0.909 (1,999.8) 0.877 (1,929.4)

2 2,200 (2,200) 0.826 (1,817.2) 0.769 (1,691.8)

3 2,200 (2,200) 0.751 (1,652.2) 0.675 (1,485.0)

4 0.683 0.592

5 16,250 16,250 0.621 10,091.0 0.519 8,433.75

NPV 2,422 1,128

4 Marks

Due to the higher net present value, proposal 2 represents the best proposal for the company.

(ii) Internal rates of return.

The IRR can be derived by calculation or graphically. Using the formula

N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rate

N1 - N2 I2 gives NPV N2.

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SUGGESTED SOLUTIONS


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Proposal 1. N1 = 1436, I1 = 10%; N2 = (182), I2 = 14%

IRR = 1436 x 0.14 - (182) x 0.10 = 219.24 = 13.55%1436 - (182) 1618

4 Marks

Proposal 2. N1 = 2,422, I1 = 10%; N2 = 1,128, I2 = 14%

IRR = 2422 x 0.14 - 1128 x 0.10 = 226.28 = 17.5%2422 - 1128 1294

4 Marks

(iii) Initially the net present value for proposal 2 indicates that the company should accept this option. This isconfirmed by the internal rate of return which provides a increased return on the investment with respectto proposal 1. Proposal 2 is therefore the most profitable investment for the company.

4 Marks

[Total: 20 Marks]

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SOLUTION 2.

(i) If the distribution of length of life of the batteries is Normal with a mean of 60 months and a standarddeviation of 10 months, the diagram below represents the distribution.

This diagram represents the fact that 68% of battery lives fall within one standard deviation of the mean.i.e. x - , x + ; 95% fall within two standard deviations and 99% fall within three standard deviations.The figure shows (a) that 34% of the measurements fall between the mean and one standard deviation oneach side (b) 2.5% of the measurements lie beyond two standard deviations in each direction from themean.

The % of batteries lasting more than 50 months is 34% (between 50 and 60 months) plus 50% (greaterthan 60 months).

6 Marks

(ii) The % of batteries that lasts less than 40 months can thus be determined; 2.5% of the batteries should fail

prior to 40 months. 6 Marks

(iii) If the battery fails at 37 months a number of inferences can be made

- the battery was out of the 2.5% that fails prior to 40 months

- something about the companys claim is not true. The chances are so small that a battery failsbefore 40 months that there are substantial reasons to have doubts about the claim. A meansmaller than 60 months and/or a standard deviation greater than 10 months would bothincrease the likelihood of failure prior to 40 months.

8 Marks

The above probabilities can be determined by calculation using the population z-score wherez = x - and the Normal distribution tables.

[Total: 20 Marks]

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Relativefrequency

Life length (months)

13.5%

3 2 + +2 +3

-3 -2 -1 0 +1 +2 +3

40 50 60 70 80

2.5%

34% 34%

13.5% 13.5%

2.5%


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Solution 3.

(i) The mean and median annual values of percentage on labour costs. Assuming that the data iscontinuous gives the mid points in the table.

Class f Cum freq Mid fx x - x (x x)2 f(x x)2

Interval point x

-5 0 4 4 - 2.5 -10.0 -19.575 383.18 1532.7

0 5 32 36 2.5 80.0 -14.575 212.43 6797.8

5 10 25 61 7.5 187.5 -9.575 91.68 2292.0

10 -15 10 71 12.5 125.0 -4.575 20.93 209.3

15 20 8 79 17.5 140.0 0.425 0.18 1.4

20 25 3 82 22.5 67.5 5.425 29.43 88.3

25 30 2 84 27.5 55.0 10.425 108.68 217.4

30 35 5 89 32.5 162.5 15.425 237.93 1189.6

35 40 4 93 37.5 150.0 20.425 417.18 1668.7

40 100 3 96 70.0 210.0 52.925 2801.05 8403.2

100 - 170 4 100 135.0 540.0 117.92 13906.30 55625.2

100 1707.5 78025.6

Mean = x = fx = 1707.5 = 17.075% 5 Marksf 100

Median: The median position lies in the class interval 5 - 10

Median = Lm + {N/2 - Fm-1} x Cm{ fm }

where

Lm = lower boundary of median class (5 - 10)

Fm-1 = cumulative frequency of class prior to median class (36)

Fm = frequency of median class (25)

Cm = median class width (5), N = f = 100

Therefore, median = 5 + {100.5/2 - 36}x 525

= 7.85%5 Marks

(ii) Standard Deviation.

= f (x - x)2 = 78025.6f 100

= 780.256 = 27.9%5 Marks

(iii) When the data set is distributed in a symmetrical way all three types of average the mean, median, mode have the same value. However, when a set of data is skewed in one direction the three types of averagewill not be identical; the mean will always be the most heavily influenced by the direction of skew, that is,it will be most affected by a few extreme values.

In this case the distribution is positively skewed and the mean is greater than the median.5 Marks

[Total: 20 Marks]

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Solution 4.

(i) y is the dependent variable and the data in the table can be represented by a linear regression equation.

Sales Advertising Expenditure(000s) (000s)

25 8

35 12

29 1124 5

28 14

12 3

18 6

27 8

17 4

30 9

The linear regression equation is: y = a + bx where

y = na + bxxy = a x + b x2 ; and

a = y - bxn n

b = nxy - xy,nx2 - (x)2

Sales Advert Exp(000s) (000s)

y x x2 y2 xy

25 8 64 625 200

35 12 144 1225 420

29 11 121 841 319

24 5 25 576 120

28 14 196 784 392

12 3 9 144 36

18 6 36 324 108

27 8 64 729 216

17 4 16 289 6830 9 81 900 270

245 80 756 6437 2149

Substituting the above values

b = 10 x 2149 - 80 x 24510 x 756 - 802

= 21,490 - 19,600 = 1890 = 1.637560 - 6400 1160

a = 245 - 1.63 x 80 = 11.4610 10

Therefore, y = 11.46 + 1.63x [10 Marks]

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(ii) The strength of the relationship is represented by the correlation co-efficient, r

where r = nxy - xy,[nx2 - (x)2][ny2 - (y)2]

= 10 x 2149 80 x 245

[10 x 756 802][10 x 6437 60,025]

= 1890 = 18905040200 2245

= 0.842

Since the value of r is close to +1, this suggests a strong positive relationship between the variables expenditure on advertising is directly related to sales, that is, 84.2% of the variation in sales is due to thevariation in the levels of advertising expenditure. This can be confirmed prior to the calculation of r bydrawing a scatter graph which will show the form and strength of the relationship.

[10 Marks]

[Total: 20 Marks]

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SOLUTION 5.

(i) Derivation of the trend by means of centred moving averages.

Quarter

Year 1 2 3 4

2003 3 5 8 42004 4 6 10 5

2005 5 8 11 7

The moving averages must coincide with the natural cycle of the series. In this case the moving averagemust have a period of 4. Since we are dealing with an even period, in order to calculate a satisfactory trendit is necessary to use centring where the calculated averages are averaged in successive overlapping pairs.This ensures that each trend value coincides with a time period.

Year Quarter Sales Moving Moving Centred Deviations Seasonal Seasonally(000s) totals of Average Moving Av Variation adjusted

4 data(y) (t) (y t) s

2003 1 3 -2.047 5.047

2 5 0.079 4.921

20 5

3 8 5.125 2.875 3.625 4.359

21 5.5

4 4 5.375 -1.375 -1.671 5.671

22 5.5

2004 1 4 5.75 -1.75 -2.047 6.047

24 6

2 6 6.125 -0.125 0.079 5.921

25 6.25

3 10 6.375 3.625 3.641 6.359

26 6.5

4 5 6.75 -1.75 -1.671 6.671

28 7

2005 1 5 7.125 -2.125 -2.047 7.047

29 7.25

2 8 7.5 0.5 0.079 7.991

31 7.75

3 11 3.641 7.359

4 7 -1.671 8.671

[5Marks] [5 Marks] [5 Marks]

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(ii) The last three columns in the table above outline the calculations for seasonally adjusted data. Seasonalvariation gives an average effect on the trend which is attributable to the season itself. They may beexpressed in terms of deviations or percentages of the trend. In this case the additative model is used, thatis,

y = t (trend) + s (seasonal) + r (residual variations)

The s values are found by calculating the mean of (y t) values and adjusting the averages so that their

total is zero, as follows.

Q1 Q2 Q3 Q4

2003 ------- ------- 2.875 -1.375

2004 -1.75 -0.125 3.625 -1.75

2005 -2.125 0.5 ------ -----

Totals -3.875 0.375 7.500 -3.125

Mean -1.938 0.188 3.750 -1.562 Total 0.438

Adjustment -0.109 -0.109 -0.109 -0.109Adj. average -2.047 0.079 3.641 -1.671 Total 0

[5 Marks]

Each of the adjusted average values are placed in column 8 of the table above and modify column 3 togive the adjusted figures in column 9.

[Total: 20 Marks]

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SOLUTION 6.

(i) The equation will have the general form

y = a + bx (y is the dependent variable, x is the independent variable)

where y = total charge, x = number of units (kwhrs) used

a = fixed costs, b = cost per unit of energy used (kwhr). [2 Marks]

Therefore the electricity model can be represented by

y = 30.04 + 0.0239x.[2 Marks]

The annual cost for 2400 kwhrs is

y = 30.04 x 6 + 0.0239 x 2400 = 180.24 + 57.36 = 237.60 [2 Marks]

(ii) 95% confidence interval is derived from the following and limits for _

x - 1.96 (s/ n) < < x + 1.96 (s/n)

where n = sample size = 36; s = sample standard deviation = 110;

x = average sample expenditure = 720 [3 Marks]

720 - 1.96 (110/36) < < 720 + 1.96 (110/36)

720 - 35.93 < < 720 + 35.93

684 < < 756 [3 Marks]

We can be 95% confident that the mean amount of profit is between 684 and 756.

(iii) The purpose is to determine the rate of interest when applied to 12.000 will achieve a level of 25,200,compounded over a 10 year period;

S = P(1 + r)10 where P = 12,000, S = 25,200, r = ? [4 Marks]

25,200 = 12,000(1+ r)10

(1+ r)10 = 25,200/12,000 = 2.1, that is, 1 + r = 102.1 = 1.077

Therefore r = 1.077 - 1 = 7.7%.[4 Marks]

Its claim is correct the increase in wages is greater than the rate of inflation at 6.8%.

[Total: 20 Marks]

P 12

F1 Business Maths April

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