Here, we are going to solve the Paper of Business Mathematics and Statistics, Solved Paper 2008, 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 has already published.
Table of Contents
Business Mathematics and Statistics, Solved Paper 2008, Punjab University, BCOM,ADCI
Solved by Iftikhar Ali, M.Sc Economics, M.Com Finance Lecture Statistics, Finance & Accounting
Q.1 Provide short answers for the following. Unnecessary details will be penalized.
(i) What is the difference between primary & secondary data?
Answer:
Primary Data: First hand, newly collected data or a data which is not collected by someone previously is called primary data. Ungrouped data is also called primary data.
Secondary Data: Second hand, previously collected data or a data that is already collected by someone previously is called secondary data. Grouped data is also called secondary data.
(ii) Define the term skewness.
Answer:
Skewness is a statistical measure that calculates lack of symmetry of the data. We can say that when mean, median and mode are not equal, there is a skewness.
It has two possible types’ positive skewness and negative skewness. In positive skewness, mean is greater than mode and it is right skewed whereas in negative skewness mean is less than mode and it is left skewed.
(iii) What is meant by mode?
Answer:
Mode is a type of average. In ungrouped data, most repeated value is mode whereas in grouped data, following formula is applied to calculate it:
\[ \widehat{\mathbf{X}}\mathbf{= L +}\frac{\mathbf{fm – f1}}{\left( \mathbf{fm – f}\mathbf{1} \right)\mathbf{+ (fm – f}\mathbf{2)}}\mathbf{\times h}\ \]
Normally it is used in forestry surveys.
(iv) Define correlation.
Answer:
The correlation coefficient is a statistical measure that calculates or quantifies the relationship or strength and direction of the linear relationship between two variables. In other words, it indicates how closely the values of two variables move together. Correlation Coefficient always denoted by the symbol “r” and its answer ranges between -1 and 1 both inclusive.
(v) What is meant by continuous variable?
Answer:
A variable in which a data has any possible value within a given range is called continuous variable or in continuous variable, data is measureable such as length, height, speed etc.
(vi) What is meant by population?
The whole group under-discussion is called population. For example all the students in a college, production of cars in a year etc.
(vii) Under what conditions, we cannot find Harmonic mean?
Answer:
We cannot calculate harmonic mean if a data contains any zero values, or negative values because we cannot calculate reciprocal values of zeros and negative values.
(viii) What is meant by quadratic equation?
Answer:
It is a second degree polynomial equation to calculate a variable x where a, b and c are constants where a ≠ 0. If a = 0 then it will become linear equation. It is usually written in the form:
\[ \mathbf{a}\mathbf{x}^{\mathbf{2}}\mathbf{+ bx + c = 0}\ \]
It can also be calculated through quadratic equation or through roots:
\[ \mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm}\sqrt{\mathbf{b}^{\mathbf{2}}\mathbf{- 4}\mathbf{ac}}}{\mathbf{2}\mathbf{a}}\ \]
(ix) Define symmetric matrix.
Answer:
Symmetric matrix is a matrix that is equals to its transpose. For example if matrix “A” is equals to 
then it is called symmetric matrix. Consider the following Example:
\[ \mathbf{A =}\begin{bmatrix}\mathbf{1} & \mathbf{2} & \mathbf{3} \\\mathbf{2} & \mathbf{4} & \mathbf{5} \\\mathbf{3} & \mathbf{5} & \mathbf{6} \\\end{bmatrix}\mathbf{,\ }\mathbf{A}^{\mathbf{t}}\mathbf{=}\begin{bmatrix}\mathbf{1} & \mathbf{2} & \mathbf{3} \\\mathbf{2} & \mathbf{4} & \mathbf{5} \\\mathbf{3} & \mathbf{5} & \mathbf{6} \\\end{bmatrix}\ \]
(x) Define the term Annuity.
Answer:
Annuity is a financial concept of time value of money in which multiple payment or receipts stream is used to calculate future or present values of money on certain percentage of interest & considering time. It has two types, one is called ordinary annuity and other one is called annuity due.
Business Statistics
Section I
Q.2 From the following frequency distribution, find: (a) A.M (b) Mode (c) Semi Inter Quartile Range.
| Classes | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 | 170-179 | 180-189 | 190-199 |
| Frequency | 4 | 17 | 28 | 25 | 18 | 13 | 6 | 5 |
Solution
| Classes | f | Class Boundaries | X | fx | C.f |
| 120–129 | 4 | 119.5–129.5 | 124.5 | 498 | 4 |
| 130–139 | 17 | 129.5–139.5 | 134.5 | 2286.5 | 21 |
| 140–149 | 28 | 139.5–149.5 | 144.5 | 4046 | 49 |
| 150–159 | 25 | 149.5–159.5 | 154.5 | 3862.5 | 74 |
| 160–169 | 18 | 159.5–169.5 | 164.5 | 2961 | 92 |
| 170–179 | 13 | 169.5–179.5 | 174.5 | 2268.5 | 105 |
| 180–189 | 6 | 179.5–189.5 | 184.5 | 1107 | 111 |
| 190–199 | 5 | 189.5–199.5 | 194.5 | 972.5 | 116 |
| 116 | 18002 | ||||
| ∑f= | ∑fx= |
(a) \[ \mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{= \ }\frac{\mathbf{18002}}{\mathbf{116}}\mathbf{= 155.189\ }\ \]
(b) Mode
Maximum frequency is 28 so model class for mode is 139.5–149.5
\[ \left( \mathbf{b} \right)\mathbf{Mode = L +}\frac{\mathbf{fm – f}\mathbf{1}}{\left( \mathbf{fm – f}\mathbf{1} \right)\mathbf{+}\left( \mathbf{fm – f}\mathbf{2} \right)}\mathbf{\ x\ h = 139.5 +}\frac{\mathbf{28 – 17}}{\left( \mathbf{28 – 17} \right)\mathbf{+}\left( \mathbf{28 – 25} \right)}\mathbf{\ x\ 10 = 147.35}\ \]
\[ \left( \mathbf{c} \right)\mathbf{S.I.Q.R = \ }\frac{\mathbf{Q}\mathbf{3 – Q}\mathbf{1}}{\mathbf{2}}\ \]
Selection of class for Q3= \[ \frac{\mathbf{3}\mathbf{n}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{3(116)}}{\mathbf{4}}\mathbf{= 87\ falls\ in\ c.f\ of\ 92\ so\ model\ class\ is\ }\mathbf{159.5 – – 169.5}\mathbf{\ \ }\ \]
Selection of class for Q1= \[ \frac{\mathbf{n}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{116}}{\mathbf{4}}\mathbf{= 29\ falls\ in\ c.f\ of\ 49\ so\ model\ class\ is\ }\mathbf{139.5 – – 149.5}\mathbf{\ \ }\ \]
\[ \mathbf{Q}\mathbf{3 = L + \ }\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{3}\mathbf{n}}{\mathbf{4}}\mathbf{- \ C} \right)\mathbf{= 159.5 + \ }\frac{\mathbf{10}}{\mathbf{18}}\left( \mathbf{87 – \ 74} \right)\mathbf{= 166.73}\ \]
\[ \mathbf{Q}\mathbf{1 = L + \ }\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{4}}\mathbf{- \ C} \right)\mathbf{= 139.5 + \ }\frac{\mathbf{10}}{\mathbf{28}}\left( \mathbf{29 – \ 21} \right)\mathbf{= 142.35}\ \]
\[ \mathbf{S.I.Q.R = \ }\frac{\mathbf{166.73 – 142.35}}{\mathbf{2}}\mathbf{= 12.19}\ \]
Q.3 (a) From the following data find index no of 2005 on the bases of 2003 by using Fisher’s formula.
| Year | A | B | C | D | E | |||||
| Price | Quantity | Price | Quantity | Price | Quantity | Price | Quantity | Price | Quantity | |
| 2003 | 9 | 10 | 6 | 80 | 3 | 17 | 9 | 20 | 6 | 30 |
| 2005 | 11 | 5 | 9 | 100 | 2 | 20 | 7 | 15 | 8 | 40 |
Solution:
| Commodity | 2003 | 2005 | ||||||
| Price Po | Quantity q0 | Price P1 | Quantity q1 | \[ \mathbf{p}_{\mathbf{o}}\mathbf{q}_{\mathbf{o}}\ \] | \[ \mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}\ \] | \[ \mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{o}}\ \] | \[ \mathbf{p}_{\mathbf{o}}\mathbf{q}_{\mathbf{1}}\ \] | |
| A | 9 | 10 | 11 | 5 | 90 | 55 | 110 | 45 |
| B | 6 | 80 | 9 | 100 | 480 | 900 | 720 | 600 |
| C | 3 | 17 | 2 | 20 | 51 | 40 | 34 | 60 |
| D | 9 | 20 | 7 | 15 | 180 | 105 | 140 | 135 |
| E | 6 | 30 | 8 | 40 | 180 | 320 | 240 | 240 |
| 981 | 1420 | 1244 | 1080 | |||||
| \[ \mathbf{\sum}\mathbf{p}_{\mathbf{o}}\mathbf{q}_{\mathbf{o}}\ \] | \[ \mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}\ \] | \[ \mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{o}}\ \] | \[ \mathbf{\sum}\mathbf{p}_{\mathbf{o}}\mathbf{q}_{\mathbf{1}}\ \] | |||||
\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2005 =}\sqrt{\left( \frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{o}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{o}}\mathbf{q}_{\mathbf{o}}}\mathbf{\ \times \ }\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{o}}\mathbf{q}_{\mathbf{1}}} \right)}\mathbf{\ \times \ 100}\ \]
\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2005 =}\sqrt{\left( \frac{\mathbf{1244}}{\mathbf{981}}\mathbf{\ \times \ }\frac{\mathbf{1420}}{\mathbf{1080}} \right)}\mathbf{\ \times \ 100}\ \]
\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2005 =}\sqrt{\left( \mathbf{1.2680\ \times \ 1.3148} \right)}\mathbf{\ \times \ 100}\ \]
\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2005 = 129.11}\ \]
(b) Two cards are drawn at random from a pack of 52 cards, what is the probability that: (i) Both are of same color (ii) Both are of different color
Solution:
Total Cards (N) = 52, Drawn Cards (r) = 2
\[ \mathbf{Sample\ Space\ \eta}\left( \mathbf{s} \right)\mathbf{=}\begin{pmatrix}\mathbf{52} \\\mathbf{C} \\\mathbf{2} \\\end{pmatrix}\mathbf{= 1326}\ \]
Events:
\[ \left( \mathbf{i} \right)\mathbf{\ Both\ are\ of\ same\ color\ red = \eta}\left( \mathbf{A} \right)\mathbf{=}\begin{pmatrix}\mathbf{26} \\\mathbf{C} \\\mathbf{2} \\\end{pmatrix}\begin{pmatrix}\mathbf{26} \\\mathbf{C} \\\mathbf{0} \\\end{pmatrix}\mathbf{= 325 \times 1 = 325}\ \]
\[ \left( \mathbf{ii} \right)\mathbf{\ Both\ are\ of\ different\ color = \eta}\left( \mathbf{B} \right)\mathbf{=}\begin{pmatrix}\mathbf{26} \\\mathbf{C} \\\mathbf{1} \\\end{pmatrix}\begin{pmatrix}\mathbf{26} \\\mathbf{C} \\\mathbf{1} \\\end{pmatrix}\mathbf{= 26 \times 26 = 676}\ \]
Probability
\[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta}\left( \mathbf{A} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{325}}{\mathbf{1326}}\mathbf{= 0.2450}\ \]
\[ \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta}\left( \mathbf{B} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{676}}{\mathbf{1326}}\mathbf{= 0.5098}\ \]
Note: There are 26 Red and 26 Black cards so same color probability for black cards will also be 0.2450
Q.4 From the following data calculate: (i) Correlation Coefficient (ii) Both Regression Coefficients
| Supply (X) | 80 | 82 | 86 | 91 | 83 | 85 | 89 | 96 | 93 |
| Price (Y) | 145 | 140 | 130 | 124 | 133 | 127 | 120 | 110 | 116 |
Solution: (i) Correlation Coefficient
Formula
\[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx \times Sy}}\ \]
| S/No | X | Y | XY | X² | Y² |
| 1 | 80 | 145 | 11600 | 6400 | 21025 |
| 2 | 82 | 140 | 11480 | 6724 | 19600 |
| 3 | 86 | 130 | 11180 | 7396 | 16900 |
| 4 | 91 | 124 | 11284 | 8281 | 15376 |
| 5 | 83 | 133 | 11039 | 6889 | 17689 |
| 6 | 85 | 127 | 10795 | 7225 | 16129 |
| 7 | 89 | 120 | 10680 | 7921 | 14400 |
| 8 | 96 | 110 | 10560 | 9216 | 12100 |
| 9 | 93 | 116 | 10788 | 8649 | 13456 |
| SUM | 785 | 1145 | 99406 | 68701 | 146675 |
| ∑X = | ∑Y = | ∑XY = | ∑X² = | ∑Y² = |
\[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{785}}{\mathbf{9}}\mathbf{= 87.23\ ,\ }\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{1145}}{\mathbf{9}}\mathbf{= 127.23\ }\ \]
\[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]
\[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{68701}}{\mathbf{9}}\mathbf{- \ }\left( \frac{\mathbf{785}}{\mathbf{9}} \right)^{\mathbf{2}} \right\rbrack}\ \]
\[ \mathbf{Sx =}\sqrt{\mathbf{7633.45 – 7607.71}}\ \]
\[ \mathbf{Sx =}\sqrt{\mathbf{25.74}}\ \]
\[ \mathbf{Sx = 5.073}\ \]
\[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{y}^{\mathbf{2}}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]
\[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{146675}}{\mathbf{9}}\mathbf{- \ }\left( \frac{\mathbf{1145}}{\mathbf{9}} \right)^{\mathbf{2}} \right\rbrack}\mathbf{\ }\ \]
\[ \mathbf{Sy =}\sqrt{\mathbf{16297.23 – 16175.49}}\ \]
\[ \mathbf{Sy =}\sqrt{\mathbf{121.74}}\ \]
\[ \mathbf{Sy = 11.03}\ \]
\[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx \times Sy}}\ \]
\[ \mathbf{r =}\frac{\frac{\mathbf{99406}}{\mathbf{9}}\mathbf{- \ }\left( \mathbf{87.23} \right)\left( \mathbf{127.23} \right)}{\left( \mathbf{5.073} \right)\mathbf{\times (11.03)}}\ \]
\[ \mathbf{r =}\frac{\mathbf{11045.11\ –\ 11098.27}}{\mathbf{55.95}}\ \]
\[ \mathbf{r =}\frac{\mathbf{- 53.16}}{\mathbf{55.95}}\ \]
\[ \mathbf{r = – 0.95}\ \] (Negative Correlation)
Solution (ii) Both Regression Coefficients
Regression Coefficient Y on X
\[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{S}_{\mathbf{y}}}{\mathbf{S}_{\mathbf{x}}}\ \]
\[ \mathbf{b}_{\mathbf{yx}}\mathbf{= – 0.95 =}\frac{\mathbf{11.03}}{\mathbf{5.073}}\mathbf{= – 2.065}\ \]
Regression Coefficient X on Y
\[ \mathbf{b}_{\mathbf{xy}}\mathbf{=}\mathbf{r}_{\mathbf{xy}}\mathbf{=}\frac{\mathbf{S}_{\mathbf{x}}}{\mathbf{S}_{\mathbf{y}}}\ \]
\[ \mathbf{b}_{\mathbf{xy}}\mathbf{= – 0.95 =}\frac{\mathbf{5.073}}{\mathbf{11.03}}\mathbf{= – 0.436}\ \]
Q.5 Give that population consists of six values 1, 5, 7, 11, 15, 17. How many samples of size n = 2 can be drawn without replacement from this population? By forming a sampling distribution of the sample means, state and verify the relation between:
(i) Mean of the sampling distribution of the mean and the population mean.
(ii) Variance of the sampling distribution of the means and the population variance.
Solution:
Calculation of Sample Statistic
Population = 1, 5, 7, 11, 15, 17
Population Size N = 6Sample size n = 2
\[ \textbf{Possible Samples =} \begin{pmatrix}\mathbf{N} \\\mathbf{C} \\\mathbf{n} \\\end{pmatrix}\mathbf{= \ }\begin{pmatrix}\mathbf{6} \\\mathbf{C} \\\mathbf{2} \\\end{pmatrix}\mathbf{= 15}\ \]
| S/No | Samples | Sum of Samples | Mean of Samples |
| 1 | 1,5 | 6 | 3 |
| 2 | 1,7 | 8 | 4 |
| 3 | 1,11 | 12 | 6 |
| 4 | 1,15 | 16 | 8 |
| 5 | 1,17 | 18 | 9 |
| 6 | 5,7 | 12 | 6 |
| 7 | 5,11 | 16 | 8 |
| 8 | 5,15 | 20 | 10 |
| 9 | 5,17 | 22 | 11 |
| 10 | 7,11 | 18 | 9 |
| 11 | 7,15 | 22 | 11 |
| 12 | 7,17 | 24 | 12 |
| 13 | 11,15 | 26 | 13 |
| 14 | 11,17 | 28 | 14 |
| 15 | 15,17 | 32 | 16 |
| (a) Sampling Distribution of X̅ | |||
| f | \[ \mathbf{f}\overline{\mathbf{X}}\ \] | \[ \mathbf{f}\overline{\mathbf{X}}\mathbf{²}\ \] | |
| 3 | 1 | 3 | 9 |
| 4 | 1 | 4 | 16 |
| 6 | 2 | 12 | 72 |
| 8 | 2 | 16 | 128 |
| 9 | 2 | 18 | 162 |
| 10 | 1 | 10 | 100 |
| 11 | 2 | 22 | 242 |
| 12 | 1 | 12 | 144 |
| 13 | 1 | 13 | 169 |
| 14 | 1 | 14 | 196 |
| 16 | 1 | 16 | 256 |
| 15 | 140 | 1494 | |
| \[ \mathbf{\sum f\ =}\ \] | \[ \mathbf{\sum f}\overline{\mathbf{X}}\mathbf{\ =}\ \] | \[ \mathbf{\sum f}\overline{\mathbf{X}}\mathbf{² =}\ \] | |
Mean, Variance of Sampling Distribution
\[ \mathbf{µ}\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{140}}{\mathbf{15}}\mathbf{= 9.33}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}²}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1494}}{\mathbf{15}}\mathbf{-}\left( \frac{\mathbf{140}}{\mathbf{15}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 99.6 – 87.12}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 12.48}\ \]
Calculation of Population Parameters
| X | X² |
| 1 | 1 |
| 5 | 25 |
| 7 | 49 |
| 11 | 121 |
| 15 | 225 |
| 17 | 289 |
| ∑X = 56 | ∑X² = 710 |
\[ \mathbf{Mean\ }\overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{\sum X}}{\mathbf{N}}\mathbf{=}\frac{\mathbf{56}}{\mathbf{6}}\mathbf{= 9.33}\ \]
\[ \mathbf{\sigma ² =}\frac{\mathbf{\sum X²}}{\mathbf{N}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{N}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma ² =}\frac{\mathbf{710}}{\mathbf{6}}\mathbf{-}\left( \frac{\mathbf{56}}{\mathbf{6}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 118.33 – 87.11}\ \]
\[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 31.22}\ \]
Verification Formulas:
\[ \left( \mathbf{i} \right)\mathbf{\ µx\overline{}\ = \ µ}\ \]
\[ \mathbf{9.33\ = \ 9.33}\ \]
\[ \left( \mathbf{ii} \right)\mathbf{\ }\mathbf{\sigma}_{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sigma ²}}{\mathbf{n}}\left\lbrack \frac{\mathbf{N – n}}{\mathbf{N – 1}} \right\rbrack\ \]
\[ \mathbf{12.48 =}\frac{\mathbf{31.22}}{\mathbf{2}}\left\lbrack \frac{\mathbf{6 – 2}}{\mathbf{6 – 1}} \right\rbrack\ \]
\[ \mathbf{12.48 = 12.48}\ \]
Business Mathematics
Section II
Q.6 (a) Solve the equation for x (a) \[ \frac{\mathbf{2}\mathbf{x – 10}}{\mathbf{x + 4}}\mathbf{+ 3\ \ \ \ \ \ \ \ \ = \ }\frac{\mathbf{x – 2}}{\mathbf{x – 3}}\mathbf{+ 4}\ \], (b) \[ \left( \mathbf{y – 3} \right)\left( \mathbf{y + 2} \right)\mathbf{= \ – 4}\ \]
Solution
\[ \frac{\mathbf{2}\mathbf{x – 10}}{\mathbf{x + 4}}\mathbf{+ 3\ \ \ \ \ \ \ \ \ = \ }\frac{\mathbf{x – 2}}{\mathbf{x – 3}}\mathbf{+ 4}\ \]
\[ \frac{\left( \mathbf{2}\mathbf{x – 10} \right)\mathbf{+ 3(x + 4)}}{\mathbf{x + 4}}\mathbf{= \ }\frac{\left( \mathbf{x – 2} \right)\mathbf{+ 4(x – 3)}}{\mathbf{x – 3}}\ \]
\[ \frac{\mathbf{2}\mathbf{x – 10\ + 3}\mathbf{x + 12}}{\mathbf{x + 4}}\mathbf{= \ }\frac{\mathbf{x – 2\ + 4}\mathbf{x – 12}}{\mathbf{x – 3}}\ \]
\[ \frac{\mathbf{5}\mathbf{x + 2}}{\mathbf{x + 4}}\mathbf{= \ }\frac{\mathbf{5}\mathbf{x – 14\ }}{\mathbf{x – 3}}\ \]
Cross Multiplication
\[ \left( \mathbf{5}\mathbf{x + 2} \right)\left( \mathbf{x – 3} \right)\mathbf{= (5}\mathbf{x – 14)(x + 4)}\ \]
\[\mathbf{5}\mathbf{x² – 15}\mathbf{x + 2}\mathbf{x – 6 = 5}\mathbf{x}^{\mathbf{2}}\mathbf{+ 20}\mathbf{x – 14}\mathbf{x – 56}\ \]
\[ \mathbf{5}\mathbf{x² – 13}\mathbf{x – 6 = \ 5}\mathbf{x}^{\mathbf{2}}\mathbf{+ 6}\mathbf{x – 56}\ \]
\[ \mathbf{5}\mathbf{x² – 5}\mathbf{x}^{\mathbf{2}}\mathbf{- 13}\mathbf{x – 6}\mathbf{x = \ – 56 + 6}\ \]
\[ \mathbf{- 19}\mathbf{x = \ – 50}\ \]
\[ \mathbf{x =}\frac{\mathbf{50}}{\mathbf{19}} \textbf{= 2.63} \]
(b) \[ \left( \mathbf{y – 3} \right)\left( \mathbf{y + 2} \right)\mathbf{= \ – 4}\ \]
Solution:
\[ \left( \mathbf{y – 3} \right)\left( \mathbf{y + 2} \right)\mathbf{= \ – 4}\ \]
\[ \mathbf{y² + 2}\mathbf{y – 3}\mathbf{y – 6 = \ – 4}\ \]
\[ \mathbf{y² – y – 6 + 4 = \ 0}\ \]
\[ \mathbf{y² – y – 2 = \ 0}\ \]
\[ \mathbf{y² – 2}\mathbf{y + y – 2 = \ 0}\ \]
\[ \mathbf{y}\left( \mathbf{y – 2} \right)\mathbf{+ 1(y – 2) = \ 0}\ \]
\[ \mathbf{(y + 1)(y – 2) = \ 0}\ \]
\[ \mathbf{y = \ – 1,\ y = 2}\ \]
\[ \textbf{S.S{ -1, 2}} \]
Q.7 (a) If \[ \mathbf{A = \ }\begin{bmatrix}\mathbf{2} & \mathbf{- 3} & \mathbf{5} \\\mathbf{k} & \mathbf{4} & \mathbf{6} \\\mathbf{2} & \mathbf{0} & \mathbf{8} \\\end{bmatrix}\mathbf{\ is\ singular\ matrix\ then\ find\ K.}\ \]
Solution:
If A is a singular matrix then \[ \left| \mathbf{A} \right|\mathbf{= 0}\ \]
\[ \textbf{So} \left| \begin{matrix}\mathbf{2} & \mathbf{- 3} & \mathbf{5} \\\mathbf{k} & \mathbf{4} & \mathbf{6} \\\mathbf{2} & \mathbf{0} & \mathbf{8} \\\end{matrix} \right|\mathbf{= 0}\ \]
\[ \mathbf{2}\left| \begin{matrix}\mathbf{4} & \mathbf{6} \\\mathbf{0} & \mathbf{8} \\\end{matrix} \right|\mathbf{-}\left( \mathbf{- 3} \right)\left| \begin{matrix}\mathbf{k} & \mathbf{6} \\\mathbf{2} & \mathbf{8} \\\end{matrix} \right|\mathbf{+ 5\ }\left| \begin{matrix}\mathbf{k} & \mathbf{4} \\\mathbf{2} & \mathbf{0} \\\end{matrix} \right|\mathbf{= 0}\ \]
\[ \mathbf{2}\left( \mathbf{32 – 0} \right)\mathbf{+ 3\ }\left( \mathbf{k}\mathbf{8 – 12} \right)\mathbf{+ 5}\left( \mathbf{0 – 8} \right)\mathbf{= 0}\ \]
\[ \mathbf{64 + 24}\mathbf{k – 36 – 40 = 0}\ \]
\[ \mathbf{24}\mathbf{k – 12 = 0}\ \]
\[ \mathbf{24}\mathbf{k = 12}\ \]
\[ \mathbf{k =}\frac{\mathbf{12}}{\mathbf{24}}\mathbf{= 0.5}\ \]
(b) \[ \mathbf{A = \ }\begin{bmatrix}\mathbf{1} & \mathbf{3} & \mathbf{2} \\\mathbf{3} & \mathbf{2} & \mathbf{1} \\\mathbf{4} & \mathbf{5} & \mathbf{6} \\\end{bmatrix}\mathbf{,\ B = \ }\begin{bmatrix}\mathbf{- 2} & \mathbf{5} & \mathbf{4} \\\mathbf{0} & \mathbf{3} & \mathbf{- 5} \\\mathbf{- 1} & \mathbf{4} & \mathbf{2} \\\end{bmatrix}\mathbf{then\ find\ AB\ \&\ (2}\mathbf{A – 3}\mathbf{B)\ }\ \]
Solution
\[ \mathbf{AB = \ }\begin{bmatrix}\mathbf{1}\left( \mathbf{- 2} \right)\mathbf{+ 3}\mathbf{x}\mathbf{0 + 2( – 1)} & \mathbf{1}\mathbf{x}\mathbf{5 +3}\mathbf{x}\mathbf{3 + 2}\mathbf{x}\mathbf{4} & \mathbf{1}\mathbf{x4 + 3( – 5) + 2}\mathbf{x}\mathbf{2} \\\mathbf{3}\left( \mathbf{- 2} \right)\mathbf{+ 2}\mathbf{x}\mathbf{0 + 1( – 1)} & \mathbf{3}\mathbf{x}\mathbf{5 + 2}\mathbf{x}\mathbf{3 + 1}\mathbf{x}\mathbf{4} & \mathbf{3}\mathbf{x}\mathbf{4 + 2( – 5) + 1}\mathbf{x}\mathbf{2} \\\mathbf{4}\left( \mathbf{- 2} \right)\mathbf{+ 5}\mathbf{x}\mathbf{0 + 6( – 1)} & \mathbf{4}\mathbf{x}\mathbf{5 +5}\mathbf{x}\mathbf{3 + 6}\mathbf{x}\mathbf{4} & \mathbf{4}\mathbf{x}\mathbf{4 + 5( – 5) + 6}\mathbf{x}\mathbf{2} \\\end{bmatrix}\ \]
\[ \mathbf{AB = \ }\begin{bmatrix}\mathbf{- 2 + 0 – 2} & \mathbf{5 + 9 + 8} & \mathbf{4 – 15 + 4} \\\mathbf{- 6 + 0 – 1} & \mathbf{15 + 6 + 4} & \mathbf{12 – 10 + 2} \\\mathbf{- 8 + 0 – 6} & \mathbf{20 + 15 + 24} & \mathbf{16 – 25 + 12} \\\end{bmatrix}\ \]
\[ \mathbf{AB = \ }\begin{bmatrix}\mathbf{- 4} & \mathbf{22} & \mathbf{- 7} \\\mathbf{- 7} & \mathbf{25} & \mathbf{4} \\\mathbf{- 14} & \mathbf{59} & \mathbf{3} \\\end{bmatrix}\ \]
\[ \mathbf{2}\mathbf{A = \ }\begin{bmatrix}\mathbf{2} & \mathbf{6} & \mathbf{4} \\\mathbf{6} & \mathbf{4} & \mathbf{2} \\\mathbf{8} & \mathbf{10} & \mathbf{12} \\\end{bmatrix}\mathbf{3}\mathbf{B = \ }\begin{bmatrix}\mathbf{- 6} & \mathbf{15} & \mathbf{12} \\\mathbf{0} & \mathbf{9} & \mathbf{- 15} \\\mathbf{- 3} & \mathbf{12} & \mathbf{6} \\\end{bmatrix}\ \]
\[ \mathbf{2}\mathbf{A – 3}\mathbf{B = \ }\begin{bmatrix}\mathbf{2 + 6} & \mathbf{6 – 15} & \mathbf{4 – 12} \\\mathbf{6 – 0} & \mathbf{4 – 9} & \mathbf{2 + 15} \\\mathbf{8 + 3} & \mathbf{10 – 12} & \mathbf{12 – 6} \\\end{bmatrix}\ \]
\[ \mathbf{2}\mathbf{A – 3}\mathbf{B = \ }\begin{bmatrix}\mathbf{8} & \mathbf{- 9} & \mathbf{- 8} \\\mathbf{6} & \mathbf{- 5} & \mathbf{17} \\\mathbf{11} & \mathbf{- 2} & \mathbf{6} \\\end{bmatrix}\ \]
Q.8 (a) Find the sum of all odd integers from 7 to 7291 inclusive.
Solution
Arithmetic Progression of Odd integer is 7, 9, 11, 13, ……7291
Here a = 7, an= 7291, d = 2, n = ?
an= a +(n – 1)d
7291 = 7 + (n – 1)2
7291 – 7 = (n – 1)2
(n – 1)2 = 7284
n – 1 = 7284/2
n =3642 +1 = 3643
\[ \mathbf{Sn =}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{\lbrack a}\mathbf{1 + an\rbrack}\ \]
\[ \mathbf{S3643 =}\frac{\mathbf{3643}}{\mathbf{2}}\mathbf{\lbrack 7 + 7291\rbrack}\ \]
\[ \mathbf{S}\mathbf{3643 = 13293307}\ \]
Alternative Method
\[ \mathbf{Number\ of\ Odd\ Integers = \ }\frac{\mathbf{(7291 – 7)}}{\mathbf{2}}\mathbf{+ 1 = 3643}\ \]
\[ \mathbf{Sum\ of\ Odd\ Integers = \ }\frac{\mathbf{(7 + 7291)}}{\mathbf{2}}\mathbf{\times \ 3643 = 13293307}\ \]
(b) A house worth Rs. 160, 000 twenty years ago has increased in value by 10% each year because of inflation. What is its worth today?
Solution
\[ \mathbf{F.V =}{\mathbf{P}\left( \mathbf{1 + i} \right)}^{\mathbf{n}}\ \]
\[ \mathbf{F.V =}{\mathbf{160,000}\left( \mathbf{1 + 0.10} \right)}^{\mathbf{20}}\ \]
\[ \mathbf{F.V = 160,000(6.72749)}\ \]
\[ \mathbf{F.V = 1076400}\ \]
Q.9 (a) A machine in shoe factory purchased for Rs. 8500. After 8 years useful life its scrap value will be Rs. 3,000. Find the depreciation rate.
Solution
Depreciation Rate Case of Fixed Installment Method
\[ \mathbf{Annual\ Installment\ D =}\frac{\mathbf{Cost – Scrap\ Value}}{\mathbf{Useful\ Life}}\mathbf{=}\frac{\mathbf{8500 – 3000}}{\mathbf{8}}\mathbf{= 687.5}\ \]
\[ \mathbf{Rate\ of\ Depreciation =}\frac{\mathbf{D}}{\mathbf{(C – SV)}}\mathbf{=}\frac{\mathbf{687.5}}{\mathbf{(8500 – 3000)}}\mathbf{= 0.125\ or\ 12.5\%}\ \]
Depreciation Rate Case of Reducing Balance Method
\[ \mathbf{Sn =}\mathbf{P(1 – r)}^{\mathbf{n}}\ \]
Where Sn = Scrap Value, P = Original Cost, r = percentage rate of depreciation
& n = useful life of the tangible asset
So Sn = 3000, P = 8500, n = 8, r = ?
\[ \mathbf{3000 =}\mathbf{8500(1 – r)}^{\mathbf{8}}\ \]
\[ \mathbf{(1 – r)}^{\mathbf{8}}\mathbf{=}\frac{\mathbf{3000}}{\mathbf{8500}}\mathbf{= 0.35294}\ \]
\[ \mathbf{1 – r = \ }\mathbf{(0.35294)}^{\mathbf{1/8}}\ \]
\[ \mathbf{- r = 0.8779 – 1}\ \]
\[ \mathbf{r = 0.1221 = 12.21\%} \textbf{= Depreciation Rate} \]
(b) A house is rented for Rs. 9,000 per month, with each month’s rent payable in advance. If the interest rate is 12% compounded monthly and the rent is deposited in an account. What will be the amount of rent for two years?
Solution:
\[ \mathbf{Here\ R\ = \ 9000,\ i = 12\%\ Monthly =}\frac{\mathbf{12}}{\mathbf{12}}\mathbf{\ = 1\% = 0.01,\ n\ = \ 12\ x\ 2\ = \ 24,\ F.V\ = \ ?}\ \]
Case: Annuity Due
\[ \mathbf{F.V = R}\left\lbrack \frac{\mathbf{(1 + i)}^{\mathbf{n + 1}}\mathbf{- 1}}{\mathbf{i}} \right\rbrack\mathbf{- R}\ \]
\[ \mathbf{F.V = 9000}\left\lbrack \frac{\mathbf{(1 + 0.01)}^{\mathbf{24 + 1}}\mathbf{- 1}}{\mathbf{0.01}} \right\rbrack\mathbf{- 9000}\ \]
\[ \mathbf{F.V = 9000}\left( \mathbf{28.2431995} \right)\mathbf{- 9000}\ \]
\[ \mathbf{F.V = 245188.80}\ \]
Alternative Formula
\[ \mathbf{F.V = R}\left\lbrack \frac{\left( \mathbf{1 + i} \right)^{\mathbf{n}}\mathbf{- 1}}{\mathbf{i}} \right\rbrack\mathbf{(1 + i)}\ \]
\[ \mathbf{F.V = 9000}\left\lbrack \frac{\left( \mathbf{1 + 0.01} \right)^{\mathbf{24}}\mathbf{- 1}}{\mathbf{0.01}} \right\rbrack\mathbf{(1 + 0.01)}\ \]
\[ \mathbf{F.V = 9000}\left( \mathbf{26.973464} \right)\mathbf{(1 + 0.01)}\ \]
\[ \mathbf{F.V = 245188.80}\ \]
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