Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

In this Post, we are going to discuss the Paper of Business Statistics and Mathematics Solved Paper 2013, 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.

Other solved papers of Business Statistics & Mathematics

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

Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

Section I Business Statistics

Q.1 Weight of 175 male students at a university are given in the following frequency table…..Calculate Karl Pearson’s and Bowley’s Coefficient of Skewness.

Q.1 Weight of 175 male students at a university are given in the following frequency table:

Weight	Frequency
118-126	20
127-135	35
136-144	49
145-153	32
154-162	25
163-171	14

Calculate Karl Pearson’s and Bowley’s Coefficient of Skewness

Solution:

*Classes*	*Class Boundaries*	X	*Frequency (f)*	fx	*fx²*	*C.F*
118-126	117.5-126.5	122	20	2440	297680	20
127-135	126.5-135.5	131	35	4585	600635	55
136-144	135.5-144.5	140	49	6860	960400	104
145-153	144.5-153.5	149	32	4768	710432	136
154-162	153.5-162.5	158	25	3950	624100	161
163-171	162.5-171.5	167	14	2338	390446	175
			*∑f = n= 175*	*∑f = 24941*	*∑ fx² = 3583693*

$\mathbf{A.Mean\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{24941}}{\mathbf{175}}\mathbf{= 142.52}\$

Selection of Model Class for Median

$\frac{\mathbf{n}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{175}}{\mathbf{2}}\mathbf{= 87.5\ falls\ in\ C.F\ 104}\$

$\mathbf{Meadian = L +}\frac{\mathbf{h}}{\mathbf{f}}\mathbf{(}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{- \ C)}\$

$\mathbf{Meadian = 135.5 +}\frac{\mathbf{9}}{\mathbf{49}}\left( \mathbf{87.5 - \ 55} \right)\$

$\mathbf{Meadian = 135.5 +}\frac{\mathbf{9}}{\mathbf{49}}\left( \mathbf{32.5} \right)\$

$\mathbf{Meadian = 141.46}\$

Model Class for Mode

Maximum Frequency is 49 so L = 135.5, fm =49, f1 =35, f2 =32 & h=9

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

$\mathbf{Mode = 135.5 +}\frac{\mathbf{(49 - 35)}}{\left( \mathbf{49 - 35} \right)\mathbf{+ \ (49 - 32)}}\mathbf{\ \times \ 9}\$

$\mathbf{Mode = 135.5 +}\frac{\mathbf{14}}{\mathbf{31}}\mathbf{\times 9\ }\$

$\mathbf{Mode = 139.56\ })$

Selection of Model Class for Q1

$\frac{\mathbf{1}\mathbf{n}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{175}}{\mathbf{4}}\mathbf{= 43.75\ falls\ in\ C.F\ 55}\$

$\mathbf{Q}\mathbf{1 = L +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{4}}\mathbf{- \ C} \right)\$

$\mathbf{Q}\mathbf{1 = 126.5 +}\frac{\mathbf{9}}{\mathbf{35}}\left( \mathbf{43.75 - \ 20} \right)\$

Selection of Model Class for Q3

$\frac{\mathbf{3}\mathbf{n}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{3(175)}}{\mathbf{4}}\mathbf{= 131.25\ falls\ in\ C.F\ 136}\$

$\mathbf{Q}\mathbf{3 = L +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{3}\mathbf{n}}{\mathbf{4}}\mathbf{- \ C} \right)\$

$\mathbf{Q}\mathbf{3 = 144.5 +}\frac{\mathbf{9}}{\mathbf{32}}\left( \mathbf{131.25 - \ 104} \right)\$

$\mathbf{Q}\mathbf{3 = 152.16}\$

$\mathbf{S.D =}\sqrt{\frac{\mathbf{\sum fx²}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fx}}{\mathbf{\sum f}} \right)^{\mathbf{2}}}\$

$\mathbf{S.D =}\sqrt{\frac{\mathbf{3583693}}{\mathbf{175}}\mathbf{-}\left( \mathbf{142.52} \right)^{\mathbf{2}}}\$

$\mathbf{S.D =}\sqrt{\mathbf{20,478.24 - 20,311.95}\mathbf{\ }}\$

$\mathbf{S.D =}\sqrt{\mathbf{166.29}}\mathbf{= 12.89}\$

$\mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ Coefficient\ of\ \ Skewness = \ }\frac{\mathbf{Mean - Mode}}{\mathbf{S.D}}\$

$\mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ Coefficient\ of\ \ Skewness = \ }\frac{\mathbf{142.52 - 139.56}}{\mathbf{12.89}}\$

$\mathbf{Karl\ Pearso}\mathbf{n}^{\mathbf{'}}\mathbf{s\ Coefficient\ of\ \ Skewness = \ 0.2296}\$

$\mathbf{Bowly's\ Quartile}\mathbf{\ }\mathbf{\ Coefficient\ of\ \ Skewness = \ }\frac{\mathbf{Q}\mathbf{3 + Q}\mathbf{1 - 2}\mathbf{Median}}{\mathbf{Q}\mathbf{3 - Q}\mathbf{1}}\$

$\mathbf{Bowly's\ Quartile}\mathbf{\ }\mathbf{\ Coefficient\ of\ \ Skewness = \ }\frac{\mathbf{152.16 + 132.6 - 2(141.46)}}{\mathbf{152.16 - 132.6}}\$

$\mathbf{Bowly's\ Quartile}\mathbf{\ }\mathbf{\ Coefficient\ of\ \ Skewness = \ }\frac{\mathbf{284.76 - 282.92}}{\mathbf{19.56}}\$

$\mathbf{Bowly's\ Quartile}\mathbf{\ }\mathbf{\ Coefficient\ of\ \ Skewness = \ 0.0940}\$

Q.2 Calculate weighted index number of prices for the year 2012 from the following data taking 2008 as base and using formulae recommended by: Laspeyre, Fisher, Paasche’s and Marshall….

Q.2 Calculate weighted index number of prices for the year 2012 from the following data taking 2008 as base and using formulae recommended by: Laspeyre, Fisher, Paasche’s and Marshall

Year	A		B		C
Year	Price	Quantity	Price	Quantity	Price	Quantity
2008	5.0	80	3.6	90	3.1	20
2012	8.7	100	5.7	95	4.6	30

Solution:

Commodity	2008		2012
Commodity	*Price Po*	*Quantity q0*	*Price P1*	*Quantity q1*	*p0q0*	*p1q1*	*P1q0*	*P0q1*
A	5	80	8.7	100	400	870	696	500
B	3.6	90	5.7	95	324	541.5	513	342
C	3.1	20	4.6	30	62	138	92	93
Sum					786	1549.5	1301	935
					*∑p0q0=*	*∑p1q1=*	*∑P1q0=*	*∑P0q1=*

$\left( \mathbf{i} \right)\mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 2012}\mathbf{=}\frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{0}}{\mathbf{\sum p}\mathbf{0}\mathbf{q}\mathbf{0}}\mathbf{\times \ 100}\$

$\mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 2012}\mathbf{=}\frac{\mathbf{1301}}{\mathbf{786}}\mathbf{\ \times \ 100 = \ 165.52}\$

$\left( \mathbf{ii} \right)\mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 2012}\mathbf{=}\frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{1}}{\mathbf{\sum poq}\mathbf{1}}\mathbf{\times \ 100}\$

$\mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 2012}\mathbf{=}\frac{\mathbf{1549.5}}{\mathbf{935}}\mathbf{\times \ 100 = 165.72}\$

$\left( \mathbf{iii} \right)\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2012 =}\sqrt{\mathbf{L \times P}}\$

$\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2012 =}\sqrt{\mathbf{165.52 \times 165.72}}\$

$\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2012 = 165.61}\$

$\left( \mathbf{iv} \right)\mathbf{\ Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2012 =}\left( \frac{\mathbf{\sum p}\mathbf{1}\mathbf{q}\mathbf{0 + \sum p}\mathbf{1}\mathbf{q}\mathbf{1}}{\mathbf{\sum p}\mathbf{0}\mathbf{q}\mathbf{0 + \sum p}\mathbf{0}\mathbf{q}\mathbf{1}} \right)\mathbf{\times 100}\$

$\mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2012 =}\left( \frac{\mathbf{1301 + 1549.5}}{\mathbf{786 + 935}} \right)\mathbf{\times 100}\$

$\mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2012 =}\left( \frac{\mathbf{2850.5}}{\mathbf{1721}} \right)\mathbf{\times 100}\$

$\mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2012 = 165.63})$

Q.3. A survey of 1600 families was conducted to observe that high and low income people send children to private and government school. The following results were obtained….Test whether income and type of school are independent at 5% level of significance (table Value is 3.841)

Q.3. A survey of 1600 families was conducted to observe that high and low income people send children to private and government school. The following results were obtained:

Income	School		Total
Income	Private	Government	Total
High	494	506	1000
Low	162	438	600
Total	656	944	1600

Test whether income and type of school are independent at 5% level of significance (table Value is 3.841)

Solution

(i) Testing the Hypothesis

Ho: There is no Association between Income & School Choice.

H1: There is Association between Income & School Choice.

(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)

\[ \frac{1000 \times 656}{1600} = 410\ \]

\[ \frac{1000 \times 944}{1600} = 590\ \]

(v) Calculation of χ²:

Computation of χ²

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

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

So d.f= (2-1)(2-1)=1

The Value of Tabulated χ²(0.05,1)=3.841

The Critical Region χ²cal>3.841

(vii) Conclusion: The calculated value of χ² is 77.784 is greater than the tabulated value of χ² 3.841 or 77.784 falls in the critical region. We reject the Null Hypothesis and accept alternative hypothesis. We can say that there is relationship between Income level and choice of school.

Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

Q.4. Given the six elements population 0, 3, 6, 12, 15 and 18. How many samples of size n = 3 can be drawn without replacement from this population. Form sampling distribution of sample means. Hence state and verify the relation between…..

Q.4. Given the six elements population 0, 3, 6, 12, 15 and 18. How many samples of size n = 3 can be drawn without replacement from this population. Form sampling distribution of sample means. Hence state and verify the relation between:

(i) Mean of the sampling distribution of the means and the population mean.

(ii) Variance of the sampling distribution of the mean and population variance.

Solution:

Population = 0,3, 6, 12, 15, 18

Population Size N = 6

Sample size n = 3

$Sample\ Space\ \eta(S) = \begin{pmatrix}N \\C \\n \\\end{pmatrix} = \begin{pmatrix}6 \\C \\3 \\\end{pmatrix} = 20\$

Sampling Distribution of Sample Means

S/No	Samples	Sum of Samples	Mean of Samples	S/No	Samples	Sum of Samples	Mean of Samples
1	0,3,6	9	3	11	3,6,12	21	7
2	0,3, 12	15	5	12	3,6,15	24	8
3	0,3, 15	18	6	13	3,6, 18	27	9
4	0,3, 18	21	7	14	3,12,15	30	10
5	0,6, 12	18	6	15	3, 12, 18	33	11
6	0,6, 15	21	7	16	3, 15, 18	36	12
7	0,6, 18	24	8	17	6, 12, 15	33	11
8	0, 12, 15	27	9	18	6, 12, 18	36	12
9	0, 12, 18	30	10	19	6, 15, 18	39	13
10	0, 15, 18	33	11	20	12, 15, 18	45	15

Sampling Distribution of Sample Means

X̅	f	*fX̅*	*fX̅²*
3	1	3	9
5	1	5	25
6	2	12	72
7	3	21	147
8	2	16	128
9	2	18	162
10	2	20	200
11	3	33	363
12	2	24	288
13	1	13	169
15	1	15	225
	20	180	1788
	*∑f =*	*∑fX̅ =*	*∑fX̅²=*

Mean & Variance of Sampling Distribution

$µ\overline{x\ } = \frac{\sum f\overline{x\ }}{\sum f} = \frac{180}{20} = 9\$

$\sigma{\overline{x\ }}^{2} = \frac{\sum f\overline{x\ }²}{\sum f} - \left( \frac{\sum f\overline{x}}{\sum f} \right)^{2}\$

$\sigma{\overline{x\ }}^{2} = \frac{1788}{20} - (9)^{2}\$

Mean & Variance of Population

X	X²
0	0
3	9
6	36
12	144
15	225
18	324
54	738
*∑X =*	*∑X² =*

$Mean\ µ = \ \frac{\sum X}{N} = \frac{54}{6} = 9\$

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

$\sigma^2 = \frac{738}{6} - (9)^{2}\$

$\sigma^2 = 123 - 81\$

$\sigma^2 = 42\$

Verification Formulas:

$µ\overline{x}\ = \ µ\$

$9 = 9\$

${\sigma\overline{x}}^{2} = \frac{\sigma ²}{n}\left\lbrack \frac{N - n}{N - 1} \right\rbrack\$

$8.4 = \frac{42}{3}\left\lbrack \frac{6 - 3}{6 - 1} \right\rbrack\$

$8.4 = 8.4\$

Section II Business Mathematics

$Q.5:\;If\;A=\begin{bmatrix}1&3&5\\4&-2&7\\3&2&-4\end{bmatrix}then\;obtain\;A^{-1}\;inverse\;of\;A.$

Solution:

$A^{- 1} = \frac{Adj.\ A}{|A|}\$

Calculation of Determinant

$|A| = \left| \begin{matrix}1 & 3 & 5 \\4 & - 2 & 7 \\3 & 2 & - 4 \\\end{matrix} \right|\$

$|A| = 1\left| \begin{matrix}2 & 7 \\2 & - 4 \\\end{matrix} \right| - 4\left| \begin{matrix}3 & 5 \\2 & - 4 \\\end{matrix} \right| + 3\left| \begin{matrix}3 & 5 \\2 & 7 \\\end{matrix} \right|\$

$|A| = 1(8 - 14) - 4( - 12 - 10) + 3\lbrack 21 - ( - 10)\rbrack\$

$|A| = 1( - 6) - 4( - 22) + 3(31)\$

$|A| = - 6 + 88 + 93\$

$|A| = 175\$

Calculation of Adj. A

Step 1 Calculation of Minors

$\begin{bmatrix}\left| \begin{matrix}2 & 7 \\2 & - 4 \\\end{matrix} \right| & \left| \begin{matrix}4 & 7 \\3 & - 4 \\\end{matrix} \right| & \left| \begin{matrix}4 & - 2 \\3 & 2 \\\end{matrix} \right| \\\left| \begin{matrix}3 & 5 \\2 & - 4 \\\end{matrix} \right| & \left| \begin{matrix}1 & 5 \\3 & - 4 \\\end{matrix} \right| & \left| \begin{matrix}1 & 3 \\3 & 2 \\\end{matrix} \right| \\\left| \begin{matrix}3 & 5 \\2 & 7 \\\end{matrix} \right| & \left| \begin{matrix}1 & 5 \\4 & 7 \\\end{matrix} \right| & \left| \begin{matrix}1 & 3 \\4 & - 2 \\\end{matrix} \right| \\\end{bmatrix}$

$\begin{bmatrix}6 & - 37 & 14 \\22 & - 19 & - 7 \\31 & - 13 & - 14 \\\end{bmatrix}\$

Step 2: Calculation of Cofactors

$\begin{bmatrix}6 & - 37 & 14 \\22 & - 19 & - 7 \\31 & - 13 & - 14 \\\end{bmatrix} \times \begin{bmatrix}+ & - & + \\- & + & - \\+ & - & + \\\end{bmatrix}\$

$\begin{bmatrix}6 & 37 & 14 \\22 & - 19 & 7 \\31 & 13 & - 14 \\\end{bmatrix}\$

Step 3 Transpose of Cofactors = Adjoint A

$\begin{bmatrix}6 & 22 & 31 \\37 & - 19 & 13 \\14 & 7 & - 14 \\\end{bmatrix}\$

$A^{- 1}\mathbf{=}\frac{\mathbf{Adj.\ A}}{\left| \mathbf{A} \right|}\$

$A^{- 1}\mathbf{=}\frac{\begin{bmatrix}\mathbf{- 6} & \mathbf{22} & \mathbf{31} \\\mathbf{37} & \mathbf{- 19} & \mathbf{13} \\\mathbf{14} & \mathbf{7} & \mathbf{- 14} \\\end{bmatrix}}{\mathbf{175}}\$

$A^{- 1} = \begin{bmatrix}\mathbf{-}\frac{\mathbf{6}}{\mathbf{175}} & \frac{\mathbf{22}}{\mathbf{175}} & \frac{\mathbf{31}}{\mathbf{175}} \\\frac{\mathbf{37}}{\mathbf{175}} & \mathbf{-}\frac{\mathbf{19}}{\mathbf{175}} & \frac{\mathbf{13}}{\mathbf{175}} \\\frac{\mathbf{14}}{\mathbf{175}} & \frac{\mathbf{7}}{\mathbf{175}} & \mathbf{-}\frac{\mathbf{14}}{\mathbf{175}} \\\end{bmatrix}\$

Q.6 (a) Solve for x and y: 4x – 3y = 10, 5x – 7y = 6

Solution:

4x – 3y = 10 (i)

5x – 7y = 6 (ii)

Multiply equation (i) by 5 and (ii) by 4, subtract & get:

$20x\ -\ 15y\ = \ 50\$

$\pm 20x\ \mp \ 28y\ = \ \pm 24\$

$13y = 26\$

$y = \frac{26}{13} = 2\$

Put y =2 in equation (i) to get the value of x

$4x\ -\ 3y\ = \ 10\ (i)\$

$4x\ -\ 3(2)\ = \ 10\$

$4x\ -\ 6\ = 10\$

$4x\ = \ 10\ + \ 6\$

$4x\ = \ 16\$

$x = \frac{16}{4} = 4\$

$x = 4,\ y = 2\$

Prove:

4x – 3y = 10 (i)

4(4) – 3(2) = 10

10 = 10

(b) The area of a rectangular plot of land fenced all round is 2000 sq. yards and the total length of fencing is 180 sq. yards. Find the length and width of plot.

Solution:

Let the length of the plot be x

Let the width of the plot be y

Area = Length x Width

Area = 2000 sq. yards

Length of fencing = 180 sq. yards = Perimeter

Equation will be:

$Area\ = \ xy\$

$xy = \ 2000(i)\$

$2x\ + \ 2y\ = \ 180\$

$2(x + y)\ = \ 180\$

$x + y = \frac{180}{2}\$

$x + y = 90\ \ (ii)\$

$y = 90 - x\ \$

Put y = 90 – x into equation (i)

$xy = 2000\ \ (i)\$

$x(90 - x) = 2000\$

$90x - x^{2} = 2000\$

$90x - x^{2} - 2000 = 0\$

$- x^{2} + 90x - 2000 = 0\ or\$

$x^{2} - 90x + 2000 = 0\$

$a = 1,\ b = - 90,\ c = 2000\$

$\frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}\ \$

$\frac{- ( - 90) \pm \sqrt{{( - 90)}^{2} - 4(1)(2000)}}{2(1)}\ \$

$\frac{90 \pm \sqrt{8100 - 8000}}{2}\ \$

$\frac{90 \pm 10}{2}\ \$

$\frac{90 + 10}{2} = 50\ \&\ \frac{90 - 10}{2} = 40\$

$Put\ x = 50\$

$y = 90 - x\ \$

$y = 90 - 50\ \$

$y = 40\$

$Put\ y\ = \ 40\$

$y = 90 - x\ \$

$40 = 90 - x\ \$

$- x = 40 - 90\$

$- x = - 50\$

$x = 50\$

Hence

Length of the plot be x = 50

Width of the plot be y = 40

Q.7 (a) Show that the sum of geometric series of ten terms…..

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

$1.\ - \frac{1}{2},\frac{1}{4},\ - \frac{1}{8},,\frac{1}{16},\ - \frac{1}{32},\ldots\ldots is\ \frac{341}{512}\$

Solution:

$a = 1,\ n = 10\$

$Common\ Ratio\ r = \ \frac{- \frac{1}{2}}{1} = - \frac{1}{2}\$

$Since\ |r| = \left| - \frac{1}{2} \right| < 1\$

First we have to find 10^th term as:

$an = a1r^{n - 1}\$

$a10 = 1\left( - \frac{1}{2} \right)^{10 - 1}\$

$a10 = \left( - \frac{1}{2} \right)^{9} = - \frac{1}{512}\$

Applicable Formula

$S_{n} = \frac{a1 - a1r^{n}}{1 - r}\ where\ r < 1\$

$S_{10} = \frac{1 - 1{( - \frac{1}{2})}^{10}}{1 - ( - \frac{1}{2})}\$

$S_{10} = \frac{1 - \frac{1}{1024}}{1 + \frac{1}{2}}\$

$S_{10} = \frac{\frac{1023}{1024}}{\frac{3}{2}}\$

$S_{10} = \frac{1023}{1024} \times \frac{2}{3}\$

$S_{10} = \frac{2046}{3072}\$

$S_{10} = \frac{341}{512}\$

(b) A company offers two alternatives for the payment of salary for the post of a high executive. Either one may receive Rs. 240,000 per year or Rs. 100 in the first month, Rs. 200 in the second month, Rs. 400 in the third month and so on. Which of the two alternatives should be prefer…..

Solution:

First Alternative = 240,000 per year

Second alternative = 100, 200, 400…..12 terms

$a1 = 100,\ n = 12\$

$Common\ Ratio\ r = \ \frac{200}{100} = 2\$

$Since\ |r| = |2| > 1\$

Applicable Formula

$S_{n} = \frac{ra_{n} - a1}{r - 1}\ where\ r > 1\$

First we have to find 12^th term:

$a_{n} = a1r^{n - 1}\$

$a_{12} = 100 \times 2^{12 - 1}\$

$a_{12} = 100 \times 2^{11}\$

$a_{12} = 204800\$

Applicable Formula

$S_{n} = \frac{ra_{n} - a1}{r - 1}\ where\ r > 1\$

$S_{12} = \frac{ra_{12} - a1}{r - 1}\ \$

$S_{12} = \frac{2(204800) - 100}{2 - 1}\ \$

$S_{12} = 409500\$

Since the second option is greater than 1

409500>240,000 So Executive should prefer 2^nd option

Q.8 (a) Find the compound interest on Rs. 4500 in 3 years. If the rate of interest is 4% for the first year, 5% for the 2^nd year and 6% for the 3^rd year.

Solution:

$\emph{Principal Amount (P) = 4500}$

$\emph{Interest Rate (i) for 1\textsuperscript{st} year = 4\%}$

$\emph{Interest Rate (i) for 2\textsuperscript{nd} year = 5\%}$

$\emph{Interest Rate (i) for 3\textsuperscript{rd} year = 6\%}$

Calculation

$\mathbf{Amount = P(1 +}\mathbf{i)}^{\mathbf{n}}\$

$Amount\ 1st\ year = 4500(1 + {0.04)}^{1} = 4680\$

$Amount\ 2nd\ year = 4680(1 + {0.05)}^{1} = 4914\$

$Amount\ 3rd\ year = 4914(1 + {0.06)}^{1} = 5208.84\$

Compound Interest = 5208.84 – 4500 = 708.84

(b) Find the accumulated value of Rs. 5000 invested at the end of each quarter for 5 years at 8% compounded quarterly.

Solution:

$F.V\ = \ ?,\ R\ = \ 5000,\ n = 5 \times 4 = 20,i = \frac{8}{4} = 2\% = 0.02\$

Case = Ordinary Annuity, Quarterly Case

$F.V = R\left\lbrack \frac{(1 + i)^{n} - 1}{i} \right\rbrack\$

$F.V = 5000\left\lbrack \frac{(1 + 0.02)^{20} - 1}{0.02} \right\rbrack\$

$F.V = 5000(24.2973)\$

$F.V = 121487\$

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Introduction to Statistics Basic Important Concepts

Measures of Central Tendency, Arithmetic Mean, Median, Mode, Harmonic, Geometric Mean

Correlation Coefficient, Properties, Types, Important Formulas for Correlation Coefficient

fo	fe	*fo-fe*	*(fo-fe)²*	$\frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\$
494	410	84	7056	17.210
506	590	-84	7056	11.959
162	246	-84	7056	28.683
438	354	84	7056	19.932

				77.784
				$\mathbf{\sum}\frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}}\mathbf{=}\$

Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

Table of Contents

Section I Business Statistics

Q.1 Weight of 175 male students at a university are given in the following frequency table…..Calculate Karl Pearson’s and Bowley’s Coefficient of Skewness.

Q.2 Calculate weighted index number of prices for the year 2012 from the following data taking 2008 as base and using formulae recommended by: Laspeyre, Fisher, Paasche’s and Marshall….

Q.3. A survey of 1600 families was conducted to observe that high and low income people send children to private and government school. The following results were obtained….Test whether income and type of school are independent at 5% level of significance (table Value is 3.841)

Q.4. Given the six elements population 0, 3, 6, 12, 15 and 18. How many samples of size n = 3 can be drawn without replacement from this population. Form sampling distribution of sample means. Hence state and verify the relation between…..

Section II Business Mathematics

$Q.5:\;If\;A=\begin{bmatrix}1&3&5\\4&-2&7\\3&2&-4\end{bmatrix}then\;obtain\;A^{-1}\;inverse\;of\;A.$

Q.6 (a) Solve for x and y: 4x – 3y = 10, 5x – 7y = 6

(b) The area of a rectangular plot of land fenced all round is 2000 sq. yards and the total length of fencing is 180 sq. yards. Find the length and width of plot.

Q.7 (a) Show that the sum of geometric series of ten terms…..

(b) A company offers two alternatives for the payment of salary for the post of a high executive. Either one may receive Rs. 240,000 per year or Rs. 100 in the first month, Rs. 200 in the second month, Rs. 400 in the third month and so on. Which of the two alternatives should be prefer…..

Q.8 (a) Find the compound interest on Rs. 4500 in 3 years. If the rate of interest is 4% for the first year, 5% for the 2^nd year and 6% for the 3^rd year.

(b) Find the accumulated value of Rs. 5000 invested at the end of each quarter for 5 years at 8% compounded quarterly.

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Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

Table of Contents

Section I Business Statistics

Q.1 Weight of 175 male students at a university are given in the following frequency table…..Calculate Karl Pearson’s and Bowley’s Coefficient of Skewness.

Q.2 Calculate weighted index number of prices for the year 2012 from the following data taking 2008 as base and using formulae recommended by: Laspeyre, Fisher, Paasche’s and Marshall….

Q.3. A survey of 1600 families was conducted to observe that high and low income people send children to private and government school. The following results were obtained….Test whether income and type of school are independent at 5% level of significance (table Value is 3.841)

Q.4. Given the six elements population 0, 3, 6, 12, 15 and 18. How many samples of size n = 3 can be drawn without replacement from this population. Form sampling distribution of sample means. Hence state and verify the relation between…..

Section II Business Mathematics

Q.6 (a) Solve for x and y: 4x – 3y = 10, 5x – 7y = 6

(b) The area of a rectangular plot of land fenced all round is 2000 sq. yards and the total length of fencing is 180 sq. yards. Find the length and width of plot.

Q.7 (a) Show that the sum of geometric series of ten terms…..

(b) A company offers two alternatives for the payment of salary for the post of a high executive. Either one may receive Rs. 240,000 per year or Rs. 100 in the first month, Rs. 200 in the second month, Rs. 400 in the third month and so on. Which of the two alternatives should be prefer…..

Q.8 (a) Find the compound interest on Rs. 4500 in 3 years. If the rate of interest is 4% for the first year, 5% for the 2nd year and 6% for the 3rd year.

(b) Find the accumulated value of Rs. 5000 invested at the end of each quarter for 5 years at 8% compounded quarterly.

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Q.8 (a) Find the compound interest on Rs. 4500 in 3 years. If the rate of interest is 4% for the first year, 5% for the 2^nd year and 6% for the 3^rd year.