Business Statistics and Mathematics Solved Paper 2012, Punjab University, BCOM, ADC I

Business Statistics and Mathematics Solved Paper 2012, Punjab University, BCOM, ADC I
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In this Post, we are going to discuss the Paper of Business Statistics and Mathematics Solved Paper 2012, Punjab University, BCOM, ADCI in which Measures of Central Tendency, Measures of Dispersion, Correlation & Regression, Index Numbers, Matrix, Arithmetic Progression, Geometric Progression, Simultaneous Linear Equations, Annuity, Quadratic Equation is discussed and solved. Solved Paper 2007 Punjab University , Solved Paper 2008, Solved Paper 2009, Solved Paper 2010 & Solved Paper 2011 have already posted.

Business Statistics and Mathematics Solved Paper 2012, Punjab University, BCOM, ADC I

Solved by Iftikhar Ali, M.Sc Economics, MCOM Finance Lecturer Statistics, Finance and Accounting

Table of Contents

Section I Business Statistics

Q.1 The mid values of a frequency distribution are given as:

Mid-Values115125135145155165175185195
Frequency62548721166038222

Calculate: (i) A.M (ii) Mode (iii) Coefficient of Skewness

Solution

Method 1

Mid Values (X)ffxfx²
115669079350
125253125390625
135486480874800
14572104401513800
155116179802786900
1656099001633500
1753866501163750
185224070752950
195239076050
Sum389597259271725
 ∑f=∑fx=∑fx²=

    \[ \left( \mathbf{a} \right)\mathbf{\ A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{= \ }\frac{\mathbf{59725}}{\mathbf{389}}\mathbf{= 153.53\ }\ \]

    \[ \left( \mathbf{b} \right)\mathbf{Mode\ }\widehat{\mathbf{X}}\mathbf{= maximum\ frequency\ is\ 116\ so\ mode\ of\ this\ discrete\ data\ is\ 155}\ \]

    \[ \left( \mathbf{C} \right)\mathbf{\ Karl\ Pearson's\ Coefficient\ of\ Skewness =}\frac{\mathbf{mean - mode}}{\mathbf{S.D}}\ \]

    \[ \mathbf{S.D =}\sqrt{\left\lbrack \frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{- \ }\left( \frac{\mathbf{\sum fx}}{\mathbf{\sum f}} \right)^{\mathbf{2}} \right\rbrack}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{9271725}}{\mathbf{389}}\mathbf{- \ }\left( \frac{\mathbf{59725}}{\mathbf{389}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D =}\sqrt{\left\lbrack \mathbf{23834.76 - \ 23571.46} \right\rbrack}\ \]

    \[ \textbf{S.D = 16.226} \]

    \[ \mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ Coefficient\ of\ Skewness =}\frac{\mathbf{153.53 - 155}}{\mathbf{16.226}}\mathbf{= \ - 0.09}\ \]

Method 2

Mid Values (X)Classesffxfx²C.f
115110–1206690793506
125120–13025312539062531
135130–14048648087480079
145140–15072104401513800151
155150–160116179802786900267
165160–1706099001633500327
175170–1803866501163750365
185180–190224070752950387
195190–200239076050389
Sum 389597259271725 
  ∑f=∑fx=∑fx²= 

Note: Mean 153.53 & standard deviation 16.226 is same as above. To make classes subtract 115 from 125 & divide the result by 2 after that subtract the result from X which is lower class boundary and adding up the result will produce upper class boundary. So we will calculate mode using group data and skewness.

    \[ \mathbf{Mode = L +}\frac{\mathbf{fm - f}\mathbf{1}}{\left( \mathbf{fm - f}\mathbf{1} \right)\mathbf{+}\left( \mathbf{fm - f}\mathbf{2} \right)}\mathbf{\ \times \ h}\ \]

    \[ \mathbf{Mode = 150 +}\frac{\mathbf{116 - 72}}{\left( \mathbf{116 - 72} \right)\mathbf{+}\left( \mathbf{116 - 60} \right)}\mathbf{\ \times \ 10 = 154.4}\ \]

    \[ \mathbf{Karl\ Pearson's\ Coefficient\ of\ Skewness =}\frac{\mathbf{mean - mode}}{\mathbf{S.D}}\ \]

    \[ \mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ Coefficient\ of\ Skewness =}\frac{\mathbf{153.53 - 154.4}}{\mathbf{16.226}}\mathbf{= \ - 0.05}\ \]

Q.2: (a) The number of units produced by a process (x) and the cost of producing unit (y) were made as:

    \[ \mathbf{n = 4,\ \sum x = 20,\ \sum}\mathbf{x}^{\mathbf{2}}\mathbf{= 110,\ \sum y = 25,\ \sum}\mathbf{y}^{\mathbf{2}}\mathbf{= 162,\sum xy = 132}\ \]

Find (i) The coefficient correlation (ii) The Regression Equation of Y on X.

Solution: (i) The Coefficient Correlation

    \[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx \times Sy}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{20}}{\mathbf{4}}\mathbf{= 5\ ,\ }\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{25}}{\mathbf{4}}\mathbf{= 6.25\ }\ \]

    \[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{110}}{\mathbf{4}}\mathbf{- \ }\left( \frac{\mathbf{20}}{\mathbf{4}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{Sx =}\sqrt{\mathbf{27.5 - 25}}\ \]

    \[ \mathbf{Sx =}\sqrt{\mathbf{2.5}}\ \]

    \[ \mathbf{Sx = 1.58}\ \]

    \[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{y}^{\mathbf{2}}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{162}}{\mathbf{4}}\mathbf{- \ }\left( \frac{\mathbf{25}}{\mathbf{4}} \right)^{\mathbf{2}} \right\rbrack}\mathbf{\ }\ \]

    \[ \mathbf{Sy =}\sqrt{\mathbf{40.5 - 39.0625}}\ \]

    \[ \mathbf{Sy =}\sqrt{\mathbf{1.4375}}\ \]

    \[ \mathbf{Sy = 1.199}\ \]

    \[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx \times Sy}}\ \]

    \[ \mathbf{r =}\frac{\frac{\mathbf{132}}{\mathbf{4}}\mathbf{- \ }\left( \frac{\mathbf{20}}{\mathbf{4}} \right)\left( \frac{\mathbf{25}}{\mathbf{4}} \right)}{\left( \mathbf{1.58} \right)\mathbf{\times (1.199)}}\ \]

    \[ \mathbf{r =}\frac{\mathbf{33\ -\ 31.25}}{\mathbf{1.89442}}\ \]

    \[ \mathbf{r =}\frac{\mathbf{1.75}}{\mathbf{1.89442}}\ \]

    \[ \mathbf{r = 0.923} \textbf{(Positive Relation)} \]

Solution (ii) The Regression Equation Y on X

    \[ \widehat{\mathbf{Y}}\mathbf{= a + bx}\ \]

Where:

    \[ \mathbf{a =}\overline{\mathbf{Y}}\mathbf{- b}\overline{\mathbf{X}}\mathbf{\ \ \&\ }\mathbf{b}_{\mathbf{yx}}\mathbf{=}\mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{Sy}}{\mathbf{Sx}}\mathbf{\ \ \ }\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{= 0.923 =}\frac{\mathbf{1.199}}{\mathbf{1.58}}\mathbf{= 0.70}\ \]

    \[ \mathbf{a = 6.25 - (0.70)(5)}\ \]

    \[ \mathbf{a = 2.75}\ \]

    \[ \widehat{\mathbf{Y}}\mathbf{= a + bx}\ \]

    \[ \widehat{\mathbf{Y}}\mathbf{= 2.75 + 0.70x}\ \]

(b) Construct index number of prices with the help of following data by:

(a) Laspeyr’s (b) Paasche’s (c) Fisher’s (d) Marshall Edgeworth Formulae

CommodityBase YearCurrent Year
PriceQuantityPriceQuantity
A61408150
B1016012180
C168020110
D2010024120

Solution:

CommodityBase YearCurrent Year
Price PoQuantity q0Price P1Quantity q1p0q0p1q1p1q0p0q1
A6140815084012001120900
B1016012180960216019201080
C16802011048022001600660
D201002412060028802400720
     2880844070403360
     ∑poqo=∑p1q1=∑p1qo=∑poq1=

    \[ \left( \mathbf{a} \right)\mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{\ }\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{0}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{0}}}\mathbf{\times \ 100}\ \]

    \[ \mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{\ }\mathbf{=}\frac{\mathbf{7040}}{\mathbf{2880}}\mathbf{\ \times \ 100 = \ 244.45}\ \]

    \[ \left( \mathbf{b} \right)\mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{1}}}\mathbf{\times \ 100}\ \]

    \[ \mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{=}\frac{\mathbf{8440}}{\mathbf{3360}}\mathbf{\times \ 100 = 251.19}\ \]

    \[ \left( \mathbf{c} \right)\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ =}\sqrt{\mathbf{L \times P}}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ =}\sqrt{\mathbf{244.45 \times 251.19}}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ = 247.79}\ \]

    \[ \left( \mathbf{d} \right)\mathbf{\ Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ =}\left( \frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{0}}\mathbf{+ \sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{0}}\mathbf{+ \sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{1}}} \right)\mathbf{\times 100}\ \]

    \[ \mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ =}\left( \frac{\mathbf{7040 + 8440}}{\mathbf{2880 + 3360}} \right)\mathbf{\times 100}\ \]

    \[ \mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ =}\left( \frac{\mathbf{15480}}{\mathbf{6240}} \right)\mathbf{\times 100}\ \]

    \[ \mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ = 248.07}\ \]

Q.3: Test independence of two classification in the following contingency table at 5% level of significance:

AttributesA1A2A3A4
B142837272
B233628264
B3371219390

(The Tabulated Value of Chi-Square is 12.59)

Solution:

AttributesA1A2A3A4Total
B142837272269
B233628264241
B3371219390341
Total112266247226851

(i) Testing the Hypothesis

Ho: There is no Association between Attributes A’s and B’s.

H1: There is Association between Attributes A’s and B’s.

(ii) Level of Significance = α=0.05

(iii) Test Statistics is:

    \[ \mathbf{\chi^2 = \ \sum}\frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\ \]

(iv) To find χ² first calculate expected frequencies fe:

 Calculation of expected frequencies (fe)
Attributes Attributes 
A1A2A3A4Total 
B1

    \[ \frac{269 \times 112}{851} = 35\ \]

    \[ \frac{269 \times 266}{851} = 84\ \]

    \[ \frac{269 \times 247}{851} = 78\ \]

    \[ \frac{269 \times 226}{851} = 72\ \]

269 
B2

    \[ \frac{241 \times 112}{851} = 32\ \]

    \[ \frac{241 \times 266}{851} = 75\ \]

    \[ \frac{241 \times 247}{851} = 70\ \]

    \[ \frac{241 \times 226}{851} = 64\ \]

241 
B3

    \[ \frac{341 \times 112}{851} = 45\ \]

    \[ \frac{341 \times 266}{851} = 107\ \]

    \[ \frac{341 \times 247}{851} = 99\ \]

    \[ \frac{341 \times 226}{851} = 90\ \]

341 
Total112266247226851 

(v) Calculation of χ²:

Table B. Computation of χ²
fofefo-fe(fo-fe)²

    \[ \frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\ \]

42357491.4
8384-110.011905
7278-6360.461538
7272000
3332110.03125
6275-131692.253333
8270121442.057143
6464000
3745-8641.422222
121107141961.831776
9399-6360.363636
9090000
    9.8328
    

    \[ \mathbf{\sum}\frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}}\mathbf{=}\ \]

(vi) Critical Region: Degree of Freedom d.f= (R-1)(C-1)

So d.f= (3-1)(4-1)=6

The Value of Tabulated χ²(0.05,6)=12.59

The Critical Region χ²cal>12.59

(vii) Conclusion: The calculated value of χ² is 9.8328 is less than the tabulated value of χ² 12.59 or 9.8328 falls in the acceptance region. We accept the Null Hypothesis Ho that there is no association between Attributes A’s and B’s.

Q.4: The population consists of five values 2, 4, 6, 8, 10. Take all possible samples of size n = 2 from this population without replacement. Find:

(i) Mean and Variance of Population

(ii) Mean and unbiased Variance of each sample.

(iii) Average of the means of all samples and average of the variances of all samples.

Solution:

Calculation of Sample Statistic

Population = 2, 4, 6, 8, 10

Population Size N = 5, Sample size n = 2

    \[ Sample\;Space\;\eta(s)=\begin{pmatrix}N\C\n\end{pmatrix}=\begin{pmatrix}5\C\2\end{pmatrix}=10 \]

S/NoSamplesSum of Samples(ii) Mean of Each Sample(ii) Variance of Each Sample
   

    \[ \mathbf{Sample\ }\mathbf{Mean =}\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\ \]

    \[ \mathbf{S}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum}\left( \mathbf{X -}\overline{\mathbf{X}} \right)^{\mathbf{2}}}{\mathbf{n}}\ \]

12,463

    \[ \frac{\left( \mathbf{2 - 3} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{4 - 3} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 1}\ \]

22,684

    \[ \frac{\left( \mathbf{2 - 4} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{6 - 4} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 4}\ \]

32,8105

    \[ \frac{\left( \mathbf{2 - 5} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{8 - 5} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 9}\ \]

42,10126

    \[ \frac{\left( \mathbf{2 - 6} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{10 - 6} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 16}\ \]

54,6105

    \[ \frac{\left( \mathbf{4 - 5} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{6 - 5} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 1}\ \]

64,8126

    \[ \frac{\left( \mathbf{4 - 6} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{8 - 6} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 4}\ \]

74,10147

    \[ \frac{\left( \mathbf{4 - 7} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{10 - 7} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 9}\ \]

86,8147

    \[ \frac{\left( \mathbf{6 - 7} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{8 - 7} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 1}\ \]

96,10168

    \[ \frac{\left( \mathbf{6 - 8} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{10 - 8} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 4}\ \]

108,10189

    \[ \frac{\left( \mathbf{8 - 9} \right)^{\mathbf{2}}\mathbf{+}\left( \mathbf{10 - 9} \right)^{\mathbf{2}}}{\mathbf{2}}\mathbf{= 1}\ \]

(a) Sampling Distribution of X̅

    \[ \overline X \]

f

    \[ f\overline X \]

    \[ f\overline X^2 \]

3139
41416
521050
621272
721498
81864
91981
    
1060390
 

    \[ \sum f \]

    \[ \sum f\overline X \]

    \[ \sum f\overline X^2 \]

(iii) Mean, Variance of Sampling Distribution

    \[ \mu\overline X=\frac{\sum f\overline X}{\sum f}=\frac{60}{10}=6 \]

    \[ \sigma\overline X^2=\frac{\sum f\overline X^2}{\sum f}-\left(\frac{\sum f\overline X}{\sum f}\right)^2 \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{390}}{\mathbf{10}}\mathbf{-}\left( \frac{\mathbf{60}}{\mathbf{10}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 39 - 36}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 3}\ \]

Calculation of Population Parameters

(i) Mean & Variance of Population

XX²
24
416
636
864
10100
  
∑X = 30∑X² = 220

    \[ \mu=\frac{\sum X}N=\frac{30}5=6 \]

    \[ \mathbf{\sigma^2 =}\frac{\mathbf{\sum X²}}{\mathbf{N}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{N}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma^2 =}\frac{\mathbf{220}}{\mathbf{5}}\mathbf{-}\left( \frac{\mathbf{30}}{\mathbf{5}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 44 - 36}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 8}\ \]

Verification Formulas:

    \[ (i)\;\mu\overline X=\mu \]

    \[ \mathbf{6 = 6}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ }\mathbf{\sigma}_{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sigma^2}}{\mathbf{n}}\left\lbrack \frac{\mathbf{N - n}}{\mathbf{N - 1}} \right\rbrack\ \]

    \[ \mathbf{3 =}\frac{\mathbf{8}}{\mathbf{2}}\left\lbrack \frac{\mathbf{5 - 2}}{\mathbf{5 - 1}} \right\rbrack\ \]

    \[ \mathbf{3 = 3}\ \]

Section II Business Mathematics

Q.5 If A =

    \[ \begin{bmatrix}\mathbf{0} & \mathbf{1} & \mathbf{3} \\\mathbf{1} & \mathbf{2} & \mathbf{3} \\\mathbf{3} & \mathbf{1} & \mathbf{1} \\\end{bmatrix} \]

then obtain inverse of A.

Solution

Step 1 Minors

    \[ \left| \begin{matrix}\mathbf{2} & \mathbf{3} \\\mathbf{1} & \mathbf{1} \\\end{matrix} \right|\mathbf{= 2 \times 1 - 3 \times 1 = 2 - 3 = \ - 1}\ \]

    \[ \left| \begin{matrix}\mathbf{1} & \mathbf{3} \\\mathbf{3} & \mathbf{1} \\\end{matrix} \right|\mathbf{= 1 \times 1 - 3 \times 3 = 1 - 9 = \ - 8}\ \]

    \[ \left| \begin{matrix}\mathbf{1} & \mathbf{2} \\\mathbf{3} & \mathbf{1} \\\end{matrix} \right|\mathbf{= 1 \times 1 - 2 \times 3 = 1 - 6 = \ - 5}\ \]

    \[ \left| \begin{matrix}\mathbf{1} & \mathbf{3} \\\mathbf{1} & \mathbf{1} \\\end{matrix} \right|\mathbf{= 1 \times 1 - 3 \times 1 = 1 - 3 = \ - 2}\ \]

    \[ \left| \begin{matrix}\mathbf{0} & \mathbf{3} \\\mathbf{3} & \mathbf{1} \\\end{matrix} \right|\mathbf{= 0 \times 1 - 3 \times 3 = 0 - 9 = \ - 9}\ \]

    \[ \left| \begin{matrix}\mathbf{0} & \mathbf{1} \\\mathbf{3} & \mathbf{1} \\\end{matrix} \right|\mathbf{= 0 \times 1 - 1 \times 3 = 0 - 3 = \ - 3}\ \]

    \[ \left| \begin{matrix}\mathbf{1} & \mathbf{3} \\\mathbf{2} & \mathbf{3} \\\end{matrix} \right|\mathbf{= 1 \times 3 - 3 \times 2 = 3 - 6 = \ - 3\ }\ \]

    \[ \left| \begin{matrix}\mathbf{0} & \mathbf{3} \\\mathbf{1} & \mathbf{3} \\\end{matrix} \right|\mathbf{= 0 \times 3 - 3 \times 1 = 0 - 3 = \ - 3}\ \]

    \[ \left| \begin{matrix}\mathbf{0} & \mathbf{1} \\\mathbf{1} & \mathbf{2} \\\end{matrix} \right|\mathbf{= 0 \times 2 - 1 \times 1 = 0 - 1 = \ - 1}\ \]

    \[ \mathbf{Matrix\ of\ Minors =}\begin{bmatrix}\mathbf{- 1} & \mathbf{- 8} & \mathbf{- 5} \\\mathbf{- 2} & \mathbf{- 9} & \mathbf{- 3} \\\mathbf{- 3} & \mathbf{- 3} & \mathbf{- 1} \\\end{bmatrix}\ \]

Step 2 Matrix of Cofactors

    \[ \mathbf{Minors =}\begin{bmatrix}\mathbf{- 1} & \mathbf{- 8} & \mathbf{- 5} \\\mathbf{- 2} & \mathbf{- 9} & \mathbf{- 3} \\\mathbf{- 3} & \mathbf{- 3} & \mathbf{- 1} \\\end{bmatrix}\mathbf{\ \rightarrow}\begin{bmatrix}\mathbf{+} & \mathbf{-} & \mathbf{+} \\\mathbf{-} & \mathbf{+} & \mathbf{-} \\\mathbf{+} & \mathbf{-} & \mathbf{+} \\\end{bmatrix}\mathbf{\rightarrow Cofactors = \ }\begin{bmatrix}\mathbf{- 1} & \mathbf{8} & \mathbf{- 5} \\\mathbf{2} & \mathbf{- 9} & \mathbf{3} \\\mathbf{- 3} & \mathbf{3} & \mathbf{- 1} \\\end{bmatrix}\ \]

Step 3 Adjugate or Adjoint

    \[ \mathbf{Transpose\ of\ Cofactors = \ }\begin{bmatrix}\mathbf{- 1} & \mathbf{2} & \mathbf{- 3} \\\mathbf{8} & \mathbf{- 9} & \mathbf{3} \\\mathbf{- 5} & \mathbf{3} & \mathbf{- 1} \\\end{bmatrix}\mathbf{= Adjoint}\ \]

Step 4 Determinant

    \[ \textbf{Determinant = 0(-1) -- 1 (-8) + 3(-5)}\]

    \[ \textbf{Determinant = 0 + 8 -- 15 = -7} \]

    \[ \mathbf{A}^{\mathbf{- 1}}\mathbf{= \ }\frac{\mathbf{Adjoint\ of\ A}}{\left| \mathbf{A} \right|}\mathbf{= \ }\frac{\begin{bmatrix}\mathbf{- 1} &\mathbf{2} & \mathbf{- 3} \\\mathbf{8} & \mathbf{- 9} & \mathbf{3} \\\mathbf{- 5} & \mathbf{3} & \mathbf{- 1} \\\end{bmatrix}}{\mathbf{-7}}\mathbf{= \ -}\frac{\mathbf{1}}{\mathbf{7}}\begin{bmatrix}\mathbf{- 1} & \mathbf{2} & \mathbf{- 3} \\\mathbf{8} & \mathbf{- 9} &\mathbf{3} \\\mathbf{- 5} & \mathbf{3} & \mathbf{- 1} \\\end{bmatrix}\ \]

Q.6 (a) Solve for x the equation:

    \[ \mathbf{x = \ }\sqrt{\mathbf{x + 3}}\mathbf{- 3}\ \]

Solution

    \[ \mathbf{x = \ }\sqrt{\mathbf{x + 3}}\mathbf{- 3}\ \]

    \[ \mathbf{x + 3 = \ }\sqrt{\mathbf{x + 3}}\ \]

Take square root on both sides:

    \[ \mathbf{(x + 3)}^{\mathbf{2}}\mathbf{=}{\mathbf{(}\sqrt{\mathbf{x + 3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\ }\ \]

    \[ \mathbf{x}\mathbf{²}\mathbf{+ 6x + 9 = x + 3}\ \]

    \[ \mathbf{x}\mathbf{²}\mathbf{+ 6x - x + 9 - 3 = 0}\ \]

    \[ \mathbf{x}\mathbf{²}\mathbf{+ 5x + 6}\ \]

    \[ \mathbf{x}\mathbf{²}\mathbf{+ 3x + 2x + 6}\ \]

    \[ \mathbf{x}\left( \mathbf{x + 3} \right)\mathbf{+ 2\ (x + 3)}\ \]

    \[ \left( \mathbf{x + 2} \right)\mathbf{,}\left( \mathbf{x + 3} \right)\mathbf{= 0}\ \]

    \[ \left( \mathbf{x + 2} \right)\mathbf{= 0}\ \]

    \[ \mathbf{x = \ - 2}\ \]

‘

    \[ \left( \mathbf{x + 3} \right)\mathbf{= 0}\ \]

    \[ \mathbf{x = \ - 3}\ \]

S.S {-2, -3}

(b) Solve the following system of equations:

9x + 15y = 123

15x +93y = 201

Solution

9x + 15y = 123 …… (i)

15x +93y = 201…… (ii)

Multiply equation (i) by 15 & (ii) by 9, we get:

    \[ 135x\ + \ 225y\ = \ 1845\ \]

    \[ \pm 135x\ \pm \ 837y\ = \ \pm 1809\ \]

    \[ - 612y\ = \ 36\ \]

    \[ y\ = \ - 36/612\ = \ - \ 0.0588\ \]

    \[ \textbf{Put y = -0.0588 in equation (i)} \]

    \[ 9x\ + \ 15y\ = \ 123\ \]

    \[ 9x\ + \ 15( - 0.0588)\ = \ 123\ \]

    \[ 9x\ -\ 0.822\ = \ 123\ \]

    \[ 9x\ = \ 123\ + \ 0.822\ \]

    \[ 9x\ = \ 123.822\ \]

    \[ X\ = \frac{123.822}{9}\ = \ 13.758\ \]

Q.7 (a) Show that the sum of geometric series of 6 terms:

    \[ \frac{\mathbf{1}}{\mathbf{3}}\mathbf{,\ }\frac{\mathbf{- 1}}{\mathbf{9}}\mathbf{,\ }\frac{\mathbf{1}}{\mathbf{27}}\mathbf{,\ }\frac{\mathbf{- 1}}{\mathbf{81}}\mathbf{\ is\ }\frac{\mathbf{182}}{\mathbf{729}}\mathbf{\ }\ \]

Solution

It is the case of geometric series where:

    \[ \mathbf{a}\mathbf{1 =}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,\ r = \ -}\frac{\mathbf{1}}{\mathbf{9}}\mathbf{< 1}\ \]

First we will find 6th term with the formula:

    \[ \mathbf{an =}{\mathbf{a}\mathbf{1}\mathbf{r}}^{\mathbf{n - 1}}\ \]

    \[ \mathbf{a}\mathbf{6 =}{\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\mathbf{( -}\frac{\mathbf{1}}{\mathbf{9}}\mathbf{)}}^{\mathbf{6 - 1}}\ \]

    \[ \mathbf{a}\mathbf{6 =}{\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{9}} \right)}^{\mathbf{5}}\ \]

    \[ \mathbf{a}\mathbf{6 =}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{59049}} \right)\ \]

    \[ \mathbf{a}\mathbf{6 = -}\frac{\mathbf{1}}{\mathbf{177147}}\ \]

    \[ \mathbf{Sn = \ }\frac{\mathbf{a}\mathbf{1 - a}\mathbf{1}\mathbf{r}}{\mathbf{1 - r}}^{\mathbf{n}}\ \]

    \[ \mathbf{Sn = \ }\frac{\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{9}} \right)}{\mathbf{1 -}\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{9}} \right)}^{\mathbf{6}}\ \]

    \[ \mathbf{Sn = \ }\frac{\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\left( \frac{\mathbf{1}}{\mathbf{531441}} \right)}{\mathbf{1 -}\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{9}} \right)}\mathbf{\ }\frac{\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\left( \frac{\mathbf{1}}{\mathbf{1594323}} \right)}{\mathbf{1 -}\left( \mathbf{-}\frac{\mathbf{1}}{\mathbf{9}} \right)}\ \]

    \[ \mathbf{Sn = \ }\frac{\frac{\mathbf{531440}}{\mathbf{1594323}}}{\frac{\mathbf{10}}{\mathbf{9}}}\mathbf{\ }\ \]

    \[ \mathbf{Sn}\mathbf{=}\frac{\mathbf{531440}}{\mathbf{1594323}}\mathbf{\ \times}\frac{\mathbf{9}}{\mathbf{10}}\ \]

    \[ \mathbf{Sn}\mathbf{=}\frac{\mathbf{4782960}}{\mathbf{15943230}}\mathbf{\ \ }\mathbf{\ }\ \]

    \[ \mathbf{Sn = \ }\frac{\mathbf{4782960}}{\mathbf{15943230}}\ \]

    \[ \mathbf{Sn =}\frac{\mathbf{478296}}{\mathbf{1594323}}\ \]

    \[ \mathbf{Sn =}\frac{\mathbf{182}}{\mathbf{729}}\ \]

(b) The first term in A.P is 5, the last term 45 and the sum is 400. Find number of terms and common difference in the series.

Solution

    \[ \mathbf{Sn = \ }\frac{\mathbf{n}}{\mathbf{2}}\left( \mathbf{a + L} \right)\mathbf{\ Where\ Sn = 400,\ a = 5,\ L = 45}\ \]

    \[ \mathbf{400 = \ }\frac{\mathbf{n}}{\mathbf{2}}\mathbf{(5 + 45)}\ \]

    \[ \mathbf{400 = \ }\frac{\mathbf{n}}{\mathbf{2}}\mathbf{(50)}\ \]

    \[ \frac{\mathbf{n}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{400}}{\mathbf{50}}\mathbf{= 8}\ \]

    \[ \mathbf{n = 8\ \times \ 2 = 16}\ \]

    \[ \mathbf{L = a +}\left( \mathbf{n - 1} \right)\mathbf{d}\ \]

    \[ \mathbf{45 = 5 +}\left( \mathbf{16 - 1} \right)\mathbf{d}\ \]

    \[ \left( \mathbf{15} \right)\mathbf{d = 45 - 5}\ \]

    \[ \left( \mathbf{15} \right)\mathbf{d = 40}\ \]

    \[ \mathbf{d =}\frac{\mathbf{40}}{\mathbf{15}}\mathbf{= 2.67}\ \]

Q.8 Mohsin had a note for Rs. 15000 with an interest rate of 6%. The Note was dated January 12, 2003 and maturity date was 90 days after date. On January 27, 2003 he took the note to his bank, which discounted it at a discount rate of 7%. How much did he receive? (Take 360 days in the year)

Solution

Total Days = 90

Days of January = 19

Days of February = 30

Days of March = 30

Days of April = 11

Maturity Date 11th April

    \[ \mathbf{P\ = \ 15000,\ N\ = \ 3\ months,\ I\ = \ }\left( \frac{\mathbf{6}}{\mathbf{12}} \right)\mathbf{= 0.5\ = \ }\left( \frac{\mathbf{0.5}}{\mathbf{100}} \right)\mathbf{\ = \ 0.005}\ \]

    \[ \mathbf{S.I\ = \ PIN}\ \]

    \[ \mathbf{S.I\ = \ 15000\ \times \ 3\ \times \ 0.005\ = \ 225}\ \]

    \[ \mathbf{Amount\ or\ future\ value\ of\ Note\ = \ 15000\ + \ 225\ = \ 15225}\ \]

Discounted Date 27th January

Days of January = 4

Days of February = 30

Days of March = 30

Days of April = 11Total Days = 75

    \[ \mathbf{Discount\ rate\ = \ 7\%\ = \ 0.08}\ \]

    \[ \mathbf{Principal\ Amount\ = \ 15225}\ \]

    \[ \mathbf{Interest\ = \ 15225\ \times \ 0.07\ = \ 1065.75}\ \]

    \[ \mathbf{Interest = \ }\frac{\mathbf{1065.75}}{\mathbf{360}}\mathbf{\ \times \ 75 = 222\ }\ \]

    \[ \mathbf{Discounted\ Value\ of\ interest\ bearing\ Note\ = \ 15225\ -\ 222\ = \ 15003}\ \]

You may also interested in the following:

Business Statistics & Mathematics, Solved Paper 2011, Punjab University, BCOM,ADCI

Business Statistics & Mathematics, Solved Paper 2010, Punjab University, BCOM, ADCI

Business Statistics & Mathematics, Solved Paper 2009, Punjab University, BCOM, ADCI

Business Mathematics and Statistics, Solved Paper 2008, Punjab University, BCOM,ADCI

Business Mathematics and Statistics, Solved Paper 2007, Punjab University, BCOM,ADCI

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