Page 1 of 7

(1)

Page 1 of 7

THE INSTITUTE OF COST AND MANAGEMENT ACCOUNTANTS OF BANGLADESH CMA DECEMBER, 2020 EXAMINATION

BUSINESS LEVEL

SUBJECT: GE03. FUNDAMENTALS OF BUSINESS MATHEMATICS SOLUTION

Solution of Question No. 1 (a) Here profit = revenue – cost

= (22P²– 410P) – (27P²– 700P +4205)

= – 5P² + 290P - 4205; is the required expression.

In no loss, no profit situation, we have- – 5P² + 290P - 4205 = 0

 P² - 58P + 841 = 0

 (P – 29)² = 0

 P = 29, (b)

Selling price at end of year 2 = $30 x 1.05 x 1.06 = $33.39 Change in selling price in year 3 is $(35.73 – 33.39) = $2.34 Percentage change in year 3 was

= ($2.34/$33.39) x 100% = 7% (Answer).

Solution of Question No. 2

(a) The required equations are 5X + 7Y = 410 and 8X + 3Y = 490.

From the first equation we get 7Y = 410 – 5X, =>Y = (410 – 5X)/7

Putting this value of Y in the second equation, we get, 8X + 3 {(410 – 5X)/7} = 490

= > 56X + 1230 – 15X = 490 x 7

= > 41X + 1230 = 3430

= > 41X = 3430 – 1230

= > 41X = 2200

= > X = 2200/41=53.66 (rounded) So, Y = (410 – 5X)/7 = {410 – 5 (2200/41)} = 410 – 11000/41 = 5810/41 = 141.71 (rounded).

(b) Scroll Bars of spreadsheets: There are two scroll bars in excel, one is a vertical scroll bar which is used to view the data in excel from up and down and another scroll bar is horizontal scroll bar which is used to view the data from left to right.

i) Spreadsheets for a particulars budgeting applications will take time to develop he benefit of the spreadsheet must be greater than he cost of developing and maintaining it.

ii) Data can be accidentally changed (or deleted) without the user being aware of this occurring.

iii) Errors in design particularly in the use of formulae, can produces invalid output due to the complexity of the model these design error may be difficult to locate.

iv) A combination of errors of design together with flawed data may mean that decisions are made that are subsequently found out to be wrongs and cost the firm money.

v) Data used will be subject to a high degree of uncertainty, this may be forgotten and the data used to produce, what is considered to be an accurate report.

vi) Security issues such as the risk of unauthorized access (e.g. hacking or a loss of data due to fire or theft).

(a) If the inspector samples 100 tubes, then following the ratio 1:1:3, we would expected 20A, 20B and 60C to be selected.

10% of A tubes will be defective = 10% x 20 = 2. Similarly, 10% of B tubes = 10% x 20 = 2 will be defective and 20% of 60C tubes = 12 will be defective.

The following can be set up:

A B C Total OK 18 18 48 84 Defective 2 2 12 16 Total 20 20 60 100

(2)

Page 2 of 7

(i) P (A) = 20/100 = 0.2

(ii) P (defective) = 16/100 = 0.16 (iii) P (A|Defective) = 2/16 = 0.125.

(b)

(i) The probability of an undersized part is P(x ≤ 99) where x is the length (in mm) To transform this we have z = (x - μ)/ σ =(99 – 102)/2 = -1.5

Hence probability required = P(z ≤ -1.5)= 0.5 – 0.4332 = 0.0668, from normal table.

That is 6.68% of the parts will be undersized.

(ii) In the same way, probability of an oversize part is P(x ≥ 104.4) To transform this we have z = (x - μ)/ σ = (104.4 – 102)/2 = 1.2

Hence probability required = P(z ≥ 1.2) = 0.5 – 0.3849 = 0.1151, from normal table.

That is 11.51% of the parts will be oversized.

Solution of Question No. 4(a)(i)

} ) 1 ( 1

0

{

m r

m r A

PV

mn

A







} 2

15 . 0

2 ) 15 . 1 0 ( 1 { 2000

6 2

0







 PV

A

=15,475.00

(ii) Interest= Total Earning-Total Investment

=2,000x6x2-15,470

=8530.00

Investment Tk.15, 470.00 and Interest Tk.8,530.00 (b)(i) NPV=(CFATxPVIFA)-Initial Investment

=(35,000.00x3.352) - 95,000.00

=117,324.00 - 95,000.00

=T.22,324.00 (ii)

Year(t) Cash

inflows(CFAT)

Certainty equivalent

factors

Certain CFAT PVIFat

10% PV

1 35,000 1.00 35,000 0.909 31,815

2 35,000 0.80 28,000 0.826 23,128

3 35,000 0.60 21,000 0.751 15,771

4 35,000 0.60 21,000 0.683 14,343

5 35,000 0.20 7,000 0.621 4,347

(3)

Page 3 of 7

Total 89,404

NPV=PV of NCB- PV of NCO

= 89,404.00 – 95,000.00= - 5596

(a)

Length of life in hours

m.p Company-A Company-B

m f1 d=(m-

1000)/200 f1d f1d² f2 f2d f2d²

700-900 800 10 -1 -10 10 3 -3 3

900-1100 1000 16 0 0 0 42 0 0

1100-1300 1200 26 1 26 26 12 12 12

1300-1500 1400 8 2 16 32 3 6 12

N=60 ∑f1d =32 Σf1d2 =68 Σf2 =60 Σf2d =15 Σf2d2=27

Company A: C.V=^𝜎

𝑋X100 X=A+^Σf1d

N Xi=1000+32/60X200=1000+106.67=1106.67 Standard deviation=√(⁶⁸₆₀)− (³²₆₀)²x200

= √(1.33 − .284)x200=.9214x200=184.28 C.V=^𝜎

𝑋X100=184.28/1106.67x100=16.65%

Company B: C.V=^𝜎

𝑋X100 X=A+^Σf2d

N Xi=1000+15/60X200=1050 Standard deviation=√(²⁷₆₀)− (¹⁵₆₀)²x200

= √(.45 − .625)x200=124.50 C.V=^𝜎

𝑋X100=124.5/1050x100=11.86%

Since coefficient of variance is less for the company B hence company B’s lamps are more uniform.

(b)

Year Chain Index Fixed base index nos.

2015 94 94

2016 104 (104x94)/100=97.76

2017 104 (104x97.76)/100=101.67

2018 93 (93x101.67)/100=94.553

2019 103 (103x94.553)/100=97.39

2020 102 (102x97.39)/100=99.34

(4)

Page 4 of 7

(a) Labeling vehicles as x and deaths as y, the following table shows the calculations for the result:

So, coefficient of Rank correlation-

𝑅 = 1 − 6 ∑ 𝑑²

𝑛(𝑛²− 1) = 1 −

6 × 127.00 12(12²− 1) =

6 × 127.00

12(143) = 𝟎.𝟓𝟔

Coefficient of Rank correlation, R = 0.56 indicates a moderate degree of positive correlation indicating that there is some correlation between vehicles owned and number of road deaths , but the relationship is not strong.

(b) To determine the average salary paid in the whole industry it is necessary to find the total number of workers in each salary group. This will be done by multiplying the average number of workers by the no. of firms and then in usual manner .we can find mean.

Calculate of Average Salary Income

Groups

No of Firms

Average no of employees

Total no.

of workers

1 20 3 f (4)=2x3 (5) m d=(m-275)/25 fd

100-150 20 5 100 125 -6 -600

150-200 15 8 120 175 -4 -480

200--250 11 6 66 225 -2 -132

250-300 8 4 32 275 0 0

300-400 12 3.5 42 350 3 126

400-500 16 2 32 450 7 224

N=392 Σfd= -862

X=A+^Σfd

Nxixi

A=375, Σfd = −862,N = 392,i = 25 Ie.X=275-862/392x25=275-55=220 Hence average salary =Tk.220.

x Rank of x y Rank of y d d²

30 3.5 30 9 -5.5 30.25

31 6 14 1 5 25

32 7 30 9 -2 4

30 3.5 23 4.5 -1 1

46 10.5 32 11 -.5 .25

30 3.5 26 6.5 -3 9

19 1 20 2 -1 1

35 8 21 3 5 25

40 9 23 4.5 4.5 20.25

46 10.5 30 9 1.5 2.25

57 12 35 12 0 0

30 3.5 26 6.5 -3 9

Total: 127.00

(5)

Page 5 of 7

(a)

Mean Temperature

(x)

2.5 0.8 4.6 8.9 14.6 15.1 18.3 21.2 20.3 14.2 10.6 6.5 137.6

Gas Consumption

(y)

26.7 33.1 28.6 22.4 17.3 10.1 5.1 4.2 4.4 11.6 23.5 24.5 211.5

x² 6.25 0.64 21.16 79.21 213.16 228.01 334.89 449.44 412.09 201.64 112.36 42.25 2101.1 y² 712.89 1095.61 817.96 501.76 299.29 102.01 26.01 17.64 19.36 134.56 552.25 600.25 4879.59 xy 66.75 26.48 131.56 199.36 252.58 152.51 93.33 89.04 89.32 164.72 249.1 159.25 1674

𝑟 = 𝑛 ∑ 𝑥𝑦 − (∑ 𝑥)(∑𝑦)

√[𝑛 ∑ 𝑥²− (∑ 𝑥)²][𝑛 ∑ 𝑦²− (∑ 𝑦)²)]

= 12 × 1674 − 137.6 × 211.5

√[12 × 2101.1 − (137.6)²][12 × 4879.59 − (211.5)²)]

= −0.968is the required correlation coefficient. It indicates a strong negatively correlation.

(b)

C4 = B4*$B$9 and D4 = B4 + C4 are the required formula.

C10 = SUM(C4:C6) or =C4+C5+C6

D10= SUM(D4:D6) or =D4+D5+D6 or can be copied the formula in C10.

(a) Here, present value or principal PV=200.000, Interest rate r=6% =0.06, n=180 months, Installment A =? From the formula,

𝑃𝑉 = 𝐴

𝑟 12

× {1 − 1 (1 +₁₂^𝑟)^𝑚×𝑛}

=> 200,000 = 𝐴

0.06 12

× {1 − 1 (1 +^0.06₁₂)¹⁸⁰}

=> 𝐴 = $1687.71 (Answer)

(6)

Page 6 of 7

(b) Effective annual interest rate is

= (1+0.0221)¹²– 1

= 1.30 – 1

= 0.30

= 30% (Answer).

(c) The value of X would be-

𝑋 =𝐴

𝑟 𝑚

× {(1 + 𝑟 𝑚)

𝑚𝑛}

=14568.92

0.02 ×{(1 + 0.02)¹⁶} = 𝑇𝑘.1000,000 (Ans.) Solution of Question No. 9

(a)

(i) After 5 years the depreciated value is D=$12,000; The original book value B=$220,000; number of time period n=5. If the depreciation rate be i, then

D=B(1-i)ⁿ

 12000 = 220,000 (1-i)⁵

 (1-i) = (12000/220000)^1/5 = 0.558923

 i = 1 – 0.558923

 i = 0.4411 = 44.11%, is the required depreciation rate.

(ii) After 3^rd year the book value is = 220000(1-0.4411)³=$38,412.96. (Ans.)

(iii) If the straight line method of depreciation is used, then the annual depreciation is

=(220000-12000)/5 = $41,600

So, after three years the book value will be =$222,000-(3x$41,600)=$95,200.

So that the book value, using this method would be (95,200-38,412.96) =$56,787.04 (Ans.) (b) Computation of straight line trend

Year Sales( in

Thousand)of Tk. X XY X2 Trend values

Yc

2016 25 -2 -50 4 26

2017 30 -1 -30 1 32

2018 40 0 0 0 38

2019 50 1 50 1 44

2020 45 2 90 4 50

N=5 𝚺𝐘=190 0 𝚺𝐗𝐘=60 𝚺𝐗²⁼10 𝐘𝐜=190

The equation of the straight line trend is Y=a+bX, A=^ΣY

N=190/5=38 and b=^ΣXY

∑X

2

^=60/10=6 i.e, Y=38+6X

When X=-2 Y=-2

Y=38+6(-2)=26, Thus for 1996 Yc=26

The other values can be obtained by adding b to the preceding values. Thus for 2017 Yc =26+6=32 for 2018, Yc=32+6=38 etc.

(7)

Page 7 of 7

(a)

Particulars Delhi Kolkata Chennai Mumbai Total

Yes 45 55 60 50 210

No 35 45 35 45 160

No Opinion 5 5 5 5 20

Total 85 105 100 100 390 Let the events of A denote that a consumer selected at random preferred brand A

i) P(A)=210/390=7/13=0.5385 ii) (A∩C)=60/390=2/13=0.1538

iii) P(A/C)=P(A∩M) /P(M) = [60/390]/[100/390]=3/5=0.6 iv) P(M/A)=P(B∩A) /P(A)=[50/390]/[210/390]=5/21=0.238 (b)

Price relative Base-year quantity Base-year value (Rel, P

1

/P

0

) (Q

0

) (V

0

) Rel×Q

0

Rel×V

0

A 1.157 65 57.85 75.22 66.95 B 1.182 23 32.89 27.19 38.88 C 1.155 37 47.73 42.74 55.13 D 1.816 153 74.97 277.85 136.15 Total 278 213.44 423.00 297.11 Base-weighted relative price indices are:

*END OF QUESTION-SOLUTION PAPER*