Here, we are going to solve the Paper of Business Statistics and Mathematics, Solved Paper 2009, 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 has already published.
Table of Contents
Business Statistics and Mathematics, Solved Paper 2009, Punjab University, BCOM,ADCI
Solved by Iftikhar Ali, M.Sc Economics, MCOM Finance Lecturer Statistics, Finance and Accounting
Part I Business Statistics
Q.1 For the following data obtain the: (a) Mode (b) Median (c) Coefficient of Variation
| Weekly Wages | 30–39 | 40–49 | 50–59 | 60–69 | 70–79 | 80–89 | 90–99 |
| No of Workers | 6 | 10 | 11 | 12 | 32 | 18 | 8 |
Solution
| Weekly Wages | Class Boundaries | f | X | fx | fx² | C.f |
| 30–39 | 29.5—39.5 | 6 | 34.5 | 207 | 7141.5 | 6 |
| 40–49 | 39.5—49.5 | 10 | 44.5 | 445 | 19802.5 | 16 |
| 50–59 | 49.5—59.5 | 11 | 54.5 | 599.5 | 32672.75 | 27 |
| 60–69 | 59.5—69.5 | 12 | 64.5 | 774 | 49923 | 39 |
| 70–79 | 69.5—79.5 | 32 | 74.5 | 2384 | 177608 | 71 |
| 80–89 | 79.5—89.5 | 18 | 84.5 | 1521 | 128524.5 | 89 |
| 90-99 | 89.5—99.5 | 8 | 94.5 | 756 | 71442 | 97 |
| 97 | 6686.5 | 487114.3 | ||||
| ∑f= | ∑fx= | ∑fx²= |
Maximum frequency is 32 so model class for mode is 69.5—79.5
\[ \left( \mathbf{a} \right)\mathbf{Mode = L +}\frac{\mathbf{fm – f}\mathbf{1}}{\left( \mathbf{fm – f}\mathbf{1} \right)\mathbf{+}\left( \mathbf{fm – f}\mathbf{2} \right)}\mathbf{\ x\ h = 69.5 +}\frac{\mathbf{32 – 12}}{\left( \mathbf{32 – 12} \right)\mathbf{+}\left( \mathbf{32 – 18} \right)}\mathbf{\ x\ 10 = 75.38}\ \]
Selection of class for median
\[ \left( \mathbf{b} \right)\mathbf{Median = L + \ }\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{2}}\mathbf{- \ C} \right)\mathbf{= 69.5 + \ }\frac{\mathbf{10}}{\mathbf{32}}\left( \mathbf{48.5 – \ 39} \right)\mathbf{= 72.468}\ \]
\[ \left( \mathbf{c} \right)\mathbf{Coefficient\ of\ Variation\ C.V = \ }\left( \frac{\mathbf{S.D}}{\mathbf{Mean}} \right)\mathbf{100}\ \]
\[ \mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{= \ }\frac{\mathbf{6686.5}}{\mathbf{97}}\mathbf{= 68.93\ }\ \]
\[ \mathbf{S.D =}\sqrt{\left\lbrack \frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{- \ }\left( \frac{\mathbf{\sum fx}}{\mathbf{\sum f}} \right)^{\mathbf{2}} \right\rbrack}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{487114.3}}{\mathbf{97}}\mathbf{- \ }\left( \frac{\mathbf{6686.5}}{\mathbf{97}} \right)^{\mathbf{2}} \right\rbrack}\ \]
\[ \mathbf{=}\sqrt{\left\lbrack \mathbf{5021.79 – \ 4751.34} \right\rbrack}\ \]
\[ \textbf{S.D = 16.44} \]
\[ \mathbf{Coefficient\ of\ Variation\ C.V = \ }\left( \frac{\mathbf{16.44}}{\mathbf{68.93}} \right)\mathbf{100 = 23.85\%}\ \]
Q.2 (a) Find the chain indices from the following price relatives of four commodities using the Geometric Mean as an average.
| Years | A | B | C | D |
| 1951 | 81 | 77 | 119 | 55 |
| 1952 | 62 | 54 | 128 | 52 |
| 1953 | 104 | 87 | 111 | 100 |
| 1954 | 93 | 75 | 154 | 96 |
| 1955 | 60 | 43 | 165 | 88 |
(b) A card is drawn from a well shuffled pack of 52 playing cards. What is the probability that it is?
(i) Black Card (ii) A Face Card
Solution (a)
| Years | A | L.R A | B | L.R B | C | L.R C | D | L.R D |
| 1951 | 81 | 81 | 77 | 77 | 119 | 119 | 55 | 55 |
| 1952 | 62 | \[ \frac{62}{81} \times 100 = 76.54\ \] | 54 | \[ \frac{54}{77} \times 100 = 70.13\ \] | 128 | \[ \frac{128}{119} \times 100 = 107.56\ \] | 52 | \[ \frac{52}{55} \times 100 = 94.54\ \] |
| 1953 | 104 | \[ \frac{104}{62} \times 100 = 167.74\ \] | 87 | \[ \frac{87}{54} \times 100 = 161.11\ \] | 111 | \[ \frac{111}{128} \times 100 = 86.72\ \] | 100 | \[ \frac{100}{52} \times 100 = 192.31\ \] |
| 1954 | 93 | \[ \frac{93}{104} \times 100 = 89.42\ \] | 75 | \[ \frac{75}{87} \times 100 = 86.21\ \] | 154 | \[ \frac{154}{111} \times 100 = 138.74\ \] | 96 | \[ \frac{96}{100} \times 100 = 96\ \] |
| 1955 | 60 | \[ \frac{60}{93} \times 100 = 64.52\ \] | 43 | \[ \frac{43}{75} \times 100 = 57.33\ \] | 165 | \[ \frac{165}{154} \times 100 = 107.14\ \] | 88 | \[ \frac{88}{96} \times 100 = 91.67\ \] |
| Years | Geometric Mean | \[ \mathbf{Chain\ Index =}\frac{\mathbf{Previous\ Chain\ \times Current\ Average}}{\mathbf{100}}\ \] |
| 1951 | \[ \mathbf{(81 \times 77 \times 119 \times 55)}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{= 79.93}\ \] | 79.93 |
| 1952 | \[ {\mathbf{(}76.54\mathbf{\times}70.13\mathbf{\times}107.56\mathbf{\times}94.54\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{= 85.95}\ \] | \[ \frac{\mathbf{79.93 \times 85.95}}{\mathbf{100}}\mathbf{= 68.70}\ \] |
| 1953 | \[ {\mathbf{(}167.74\mathbf{\times}161.11\mathbf{\times}86.72\mathbf{\times}192.31\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{= 145.70\ }\ \] | \[ \frac{\mathbf{68.70 \times 145.70}}{\mathbf{100}}\mathbf{= 100.06}\ \] |
| 1954 | \[ {\mathbf{(}89.42\mathbf{\times}86.21\mathbf{\times}138.74\mathbf{\times}96\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{= 100.66}\ \] | \[ \frac{\mathbf{100.06 \times 100.66}}{\mathbf{100}}\mathbf{= 100.72}\ \] |
| 1955 | \[ {\mathbf{(}64.52\mathbf{\times}57.33\mathbf{\times}107.14\mathbf{\times}91.67\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{= 77.64}\ \] | \[ \frac{\mathbf{100.72 \times 77.64}}{\mathbf{100}}\mathbf{= 78.20}\ \] |
Solution (b)
Total Cards (N) = 52
\[ \mathbf{Sample\ Space\ \eta}\left( \mathbf{s} \right)\mathbf{= 52}\ \]
Events:
\[ \left( \mathbf{i} \right)\mathbf{\ Black\ Card = \eta}\left( \mathbf{A} \right)\mathbf{= 26}\ \]
\[ \left( \mathbf{ii} \right)\mathbf{\ Face\ Card = \eta}\left( \mathbf{B} \right)\mathbf{= 12}\ \]
Probability
\[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta}\left( \mathbf{A} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{26}}{\mathbf{52}}\mathbf{= 0.5}\ \]
\[ \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta}\left( \mathbf{B} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{12}}{\mathbf{52}}\mathbf{= 0.230}\ \]
Q.3 A population consists of four values 2, 4, 6, 10. Take all possible sample of size n = 2 without replacement. Find the mean of each sample. From a frequency table of sample means, calculate mean and variance. Also verify that:
(i) µx̅ = µ \[ {\left( \mathbf{ii} \right)\sigma x\overline{}}^{2} = \frac{\sigma ²}{n}\left\lbrack \frac{N – n}{N – 1} \right\rbrack\ \]
Solution:
Calculation of Sample Statistic
Population = 2, 4, 6, 10
Population Size N = 4
Sample size n = 2
Sample Space = \[ \begin{pmatrix}\mathbf{N} \\\mathbf{C} \\\mathbf{n} \\\end{pmatrix} \]
All possible Samples, their sums and Means of Samples
| S/No | Samples | Sum of Samples | Mean of Samples |
| 1 | 2,4 | 6 | 3 |
| 2 | 2,6 | 8 | 4 |
| 3 | 2,10 | 12 | 6 |
| 4 | 4,6 | 10 | 5 |
| 5 | 4,10 | 14 | 7 |
| 6 | 6,10 | 16 | 8 |
| (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 |
| 5 | 1 | 5 | 25 |
| 6 | 1 | 6 | 36 |
| 7 | 1 | 7 | 49 |
| 8 | 1 | 8 | 64 |
| 6 | 33 | 199 | |
| \[ \mathbf{\sum f\ =}\ \] | \[ \mathbf{\sum f}\overline{\mathbf{X}}\mathbf{\ =}\ \] | \[ \mathbf{\sum f}\overline{\mathbf{X}}\mathbf{² =}\ \] | |
(b) Mean, Variance of Sampling Distribution
\[ \mathbf{µ}\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{33}}{\mathbf{6}}\mathbf{= 5.5}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}²}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{199}}{\mathbf{6}}\mathbf{-}\left( \frac{\mathbf{33}}{\mathbf{6}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 33.167 – 30.25}\ \]
\[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 2.917}\ \]
Calculation of Population Parameters
| X | X² |
| 2 | 4 |
| 4 | 16 |
| 6 | 36 |
| 10 | 100 |
| ∑X = 22 | ∑X² = 156 |
\[ \mathbf{Mean\ }\overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{\sum X}}{\mathbf{N}}\mathbf{=}\frac{\mathbf{22}}{\mathbf{4}}\mathbf{= 5.5}\ \]
\[ \mathbf{\sigma ² =}\frac{\mathbf{\sum X²}}{\mathbf{N}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{N}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma ² =}\frac{\mathbf{156}}{\mathbf{4}}\mathbf{-}\left( \frac{\mathbf{22}}{\mathbf{4}} \right)^{\mathbf{2}}\ \]
\[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 39 – 30.25}\ \]
\[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 8.75}\ \]
Verification Formulas:
\[ \left( \mathbf{i} \right)\mathbf{\ µx}\mathbf{\overline{}}\mathbf{\ = \ }\mathbf{µ}\ \]
\[ \mathbf{5.5 = 5.5}\ \]
\[ \left( \mathbf{ii} \right)\mathbf{\ }\mathbf{\sigma}_{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sigma ²}}{\mathbf{n}}\left\lbrack \frac{\mathbf{N – n}}{\mathbf{N – 1}} \right\rbrack\ \]
\[ \mathbf{2.917 =}\frac{\mathbf{8.75}}{\mathbf{2}}\left\lbrack \frac{\mathbf{4 – 2}}{\mathbf{4 – 1}} \right\rbrack\ \]
\[ \mathbf{2.917 = 2.917}\ \]
Q.4 (a) Discuss the Association among 1000 school boys between the general ability and their mathematical ability from the following data. Using level of significance be 5%.
| Mathematical Ability | General Ability | |||
| Good | Fair | Poor | Total | |
| Good | 44 | 22 | 4 | 70 |
| Fair | 265 | 257 | 178 | 700 |
| Poor | 41 | 91 | 98 | 230 |
| Total | 350 | 270 | 280 | 1000 |
(b) Find Regression Coefficient of the following case: \[ \mathbf{\sum x = 17.6,\ \sum y = 32.8,\ \sum xy = 94.7,\ }\mathbf{\sum x}^{\mathbf{2}}\mathbf{= 49.64,\ }\mathbf{\sum y}^{\mathbf{2}}\mathbf{= 182,\ n = 8\ }\ \]
(i) Testing the Hypothesis
Ho: There is no Association between Mathematical & General Ability.
H1: There is Association between Mathematical & General Ability.
(ii) Level of Significance = α=0.05
(iii) Test Statistics is: \[ \mathbf{\chi}\mathbf{²}\mathbf{= \ }\mathbf{\sum}\frac{{\mathbf{(}\mathbf{fo}\mathbf{-}\mathbf{fe}\mathbf{)}}^{\mathbf{2}}}{\mathbf{fe}}\ \]
(iv) To find χ² first calculate expected frequencies fe:
| Calculation of expected frequencies (fe) | |||||
| Mathematical Ability | General Ability | ||||
| Good | Fair | Poor | Total | ||
| Good | \[ \frac{70 \times 350}{1000} = 24.5\ \] | \[ \frac{70 \times 270}{1000} = 18.9\ \] | \[ \frac{70 \times 280}{1000} = 19.6\ \] | 63 | |
| Fair | \[ \frac{700 \times 350}{1000} = 245\ \] | \[ \frac{700 \times 270}{1000} = 189\ \] | \[ \frac{700 \times 280}{1000} = 196\ \] | 630 | |
| Poor | [\[ \frac{230 \times 350}{1000} = 80.5\ \] | \[ \frac{230 \times 270}{1000} = 62.1\ \] | \[ \frac{230 \times 280}{1000} = 64.4\ \] | 207 | |
| Total | 350 | 270 | 280 | 900 | |
(v) Calculation of χ²:
| Table B. Computation of χ² | ||||
| fo | fe | fo-fe | (fo-fe)² | (fo-fe)²/fe |
| 44 | 24.5 | 19.5 | 380.25 | 15.52 |
| 22 | 18.9 | 3.1 | 9.61 | 0.5085 |
| 4 | 19.6 | -15.6 | 243.36 | 12.416 |
| 265 | 245 | 20 | 400 | 1.6327 |
| 257 | 189 | 68 | 4624 | 24.466 |
| 178 | 196 | -18 | 324 | 1.6531 |
| 41 | 80.5 | -39.5 | 1560.3 | 19.382 |
| 91 | 62.1 | 28.9 | 835.21 | 13.449 |
| 98 | 64.4 | 33.6 | 1129 | 17.53 |
| 106.56 | ||||
| \[ \mathbf{\sum}\frac{\left( \mathbf{fo – fe} \right)^{\mathbf{2}}}{\mathbf{fe}}\mathbf{=}\ \] | ||||
(vi) Critical Region: Degree of Freedom d.f= (R-1)(C-1)
So d.f= (3-1)(3-1)=4
The Value of Tabulated χ²(0.05,4)=9.488
The Critical Region χ²cal>9.488
(vii) Conclusion: The calculated value of χ² is 106.56 is greater than the tabulated value of χ² 9.488 or 106.56 falls in the rejection or critical region. We reject the Null Hypothesis Ho that there is no association between Mathematical & General Ability.
Solution (b)
\[ \mathbf{\sum x = 17.6,\ \sum y = 32.8,\ \sum xy = 94.7,\ }\mathbf{\sum x}^{\mathbf{2}}\mathbf{= 49.64,\ }\mathbf{\sum y}^{\mathbf{2}}\mathbf{= 182,\ n = 8\ }\ \]
\[ \mathbf{Mean}\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{17.6}}{\mathbf{8}}\mathbf{=}\mathbf{2.2}\mathbf{,\ }\mathbf{Mean}\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{32.8}}{\mathbf{8}}\mathbf{=}\mathbf{4.1}\ \]
Regression Coefficient of Y on X
\[ \mathbf{byx}\mathbf{= \ }\frac{\mathbf{\sum}\mathbf{xy – n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}\mathbf{- n}\mathbf{(}\overline{\mathbf{X}}\mathbf{)}\mathbf{²}}\mathbf{= \ }\frac{\mathbf{94.7 -}\left( \mathbf{8} \right)\left( \mathbf{2.2} \right)\mathbf{(}\mathbf{4.1}\mathbf{)}}{\mathbf{49.64 – 8}\mathbf{(}\mathbf{2.2}\mathbf{)}\mathbf{²}}\mathbf{= \ }\frac{\mathbf{22.54}}{\mathbf{10.92}}\mathbf{=}\mathbf{2.064}\ \]
Regression Coefficient of X on Y
\[ \mathbf{bxy}\mathbf{= \ }\frac{\mathbf{\sum}\mathbf{xy – n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{Y}^{\mathbf{2}}\mathbf{- n}\mathbf{(}\overline{\mathbf{Y}}\mathbf{)}\mathbf{²}}\mathbf{= \ }\frac{\mathbf{94.7 -}\left( \mathbf{8} \right)\left( \mathbf{2.2} \right)\mathbf{(}\mathbf{4.1}\mathbf{)}}{\mathbf{182 – 8}\mathbf{(}\mathbf{4.1}\mathbf{)}\mathbf{²}}\mathbf{= \ }\frac{\mathbf{22.54}}{\mathbf{47.52}}\mathbf{=}\mathbf{0.4743}\ \]
Part 2 Business Mathematics
Q.5 (a) Solve the following equation by any appropriate method:\[ \sqrt{\mathbf{5}\mathbf{x + 4}}\mathbf{- \ }\sqrt{\mathbf{3}\mathbf{x + 1}}\mathbf{= 1}\ \]
Solution
First arrange the equations
\[ \sqrt{\mathbf{5}\mathbf{x + 4}}\mathbf{= (1 + \ }\sqrt{\mathbf{3}\mathbf{x + 1}}\mathbf{)}\ \]
Taking Square root on both sides
\[ {\mathbf{(}\sqrt{\mathbf{5}\mathbf{x + 4}}\mathbf{)}}^{\mathbf{2}}\mathbf{= \ }{\mathbf{(}\mathbf{1 +}\mathbf{(}\sqrt{\mathbf{3}\mathbf{x + 1}}\mathbf{)}}^{\mathbf{2}}\ \]
Apply (a+b)² = a² + b² + 2ab on right hand side
\[ \mathbf{5}\mathbf{x + 4 = 1 + 2}\sqrt{\mathbf{3}\mathbf{x + 1}}\mathbf{+ 3}\mathbf{x + 1}\ \]
\[ \mathbf{5}\mathbf{x + 4}\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1 – 1}\mathbf{= 2}\sqrt{\mathbf{3}\mathbf{x + 1}}\ \]
\[ \mathbf{2}\mathbf{x + 2}\mathbf{= 2}\sqrt{\mathbf{3}\mathbf{x + 1}}\ \]
\[ \frac{\mathbf{2}\mathbf{x}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{2}}\mathbf{=}\sqrt{\mathbf{3}\mathbf{x + 1}}\mathbf{\ }\ \]
\[ \mathbf{x + 1}\mathbf{=}\sqrt{\mathbf{3}\mathbf{x + 1}}\ \]
Taking square root on both sides
\[ \mathbf{(x + 1)}\mathbf{²}\mathbf{= (}\sqrt{\mathbf{3x + 1}}\mathbf{)}\mathbf{²}\ \]
Apply (a+b)² = a² + b² + 2ab on left hand side
\[ \mathbf{x}\mathbf{²}\mathbf{+ 2x + 1 = 3x + 1}\ \]
\[ \mathbf{x}\mathbf{²}\mathbf{+ 2x – 3x + 1 – 1 = 0}\ \]
\[ \mathbf{x}\mathbf{²}\mathbf{- x = 0}\ \]
\[ \mathbf{x(x – 1) = 0}\ \]
\[ \mathbf{x = 1}\ \]
(b) Solve the equation for x: \[ \frac{\mathbf{x + 1}}{\mathbf{3}\mathbf{x}}\mathbf{= \ }\frac{\mathbf{1}}{\mathbf{x}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\ \]
Solution:
\[ \frac{\mathbf{x + 1}}{\mathbf{3}\mathbf{x}}\mathbf{= \ }\frac{\mathbf{1}}{\mathbf{x}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}}\ \]
\[ \frac{\mathbf{x + 1}}{\mathbf{3}\mathbf{x}}\mathbf{= \ }\frac{\mathbf{3}\left( \mathbf{1} \right)\mathbf{- x(1)}}{\mathbf{3}\mathbf{x}}\ \]
\[ \frac{\mathbf{x + 1}}{\mathbf{3}\mathbf{x}}\mathbf{= \ }\frac{\mathbf{3 – x}}{\mathbf{3}\mathbf{x}}\ \]
Cross Multiplication
(x+1)(3x) = 3x(3 – x)
3x² + 3x = 9x – 3x²
3x² + 3x² + 3x – 9x = 0
6x² – 6x = 0
6x(x – 1) = 0
x-1 = 0
x = 1
Q.6 (a) Solve the following system of equations:
2x + 6y + 4z = 320
6x + 6y + 4z = 480
3x + 2y + 4z = 192
Solution
\[\begin{bmatrix}\mathbf{2} & \mathbf{6} & \mathbf{4} \\\mathbf{6} & \mathbf{6} & \mathbf{4} \\\mathbf{3} & \mathbf{2} & \mathbf{4} \\\end{bmatrix}\mathbf{\ }\begin{bmatrix}\mathbf{x} \\\mathbf{y} \\\mathbf{z} \\\end{bmatrix}\mathbf{= \ }\begin{bmatrix}\mathbf{320} \\\mathbf{480} \\\mathbf{192} \\\end{bmatrix}\]
As we know that:
\[ \mathbf{x}\mathbf{= \ }\frac{\left| \mathbf{Dx} \right|}{\left| \mathbf{D} \right|}\mathbf{,\ }\mathbf{y}\mathbf{= \ }\frac{\left| \mathbf{Dy} \right|}{\left| \mathbf{D} \right|}\mathbf{\ }\&\ z = \ \frac{\left| \mathbf{Dz} \right|}{\left| \mathbf{D} \right|}\ \]
\[ \begin{bmatrix}\mathbf{2} & \mathbf{6} & \mathbf{4} \\\mathbf{6} & \mathbf{6} & \mathbf{4} \\\mathbf{3} & \mathbf{2} & \mathbf{4} \\\end{bmatrix} \]
\[ \mathbf{\ }\left| \mathbf{D} \right|\mathbf{=}\mathbf{2}\left( \mathbf{6}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{2} \right)\mathbf{-}\mathbf{\ }\mathbf{6}\left( \mathbf{6}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{2} \right)\mathbf{+ \ }\mathbf{3}\mathbf{(}\mathbf{6}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{4}\mathbf{)}\ \]
\[ \left| \mathbf{D} \right|\mathbf{=}\mathbf{2}\left( \mathbf{24}\mathbf{-}\mathbf{8} \right)\mathbf{-}\mathbf{\ }\mathbf{6}\left( \mathbf{24}\mathbf{-}\mathbf{8} \right)\mathbf{+ \ }\mathbf{3}\mathbf{(}\mathbf{24}\mathbf{-}\mathbf{24}\mathbf{)}\ \]
\[ \left| \mathbf{D} \right|\mathbf{=}\mathbf{32}\mathbf{-}\mathbf{\ }\mathbf{96}\mathbf{+ \ }\mathbf{0}\ \]
\[ \left| \mathbf{D} \right|\mathbf{=}\mathbf{-}\mathbf{64}\ \]
\[ \begin{bmatrix}\mathbf{320} & \mathbf{6} & \mathbf{4} \\\mathbf{480} & \mathbf{6} & \mathbf{4} \\\mathbf{192} & \mathbf{2} & \mathbf{4} \\\end{bmatrix} \]
\[ \mathbf{\ }\left| \mathbf{Dx} \right|\mathbf{=}\mathbf{320}\left( \mathbf{6}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{2} \right)\mathbf{-}\mathbf{\ }\mathbf{480}\left( \mathbf{6}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{2} \right)\mathbf{+ \ }\mathbf{192}\mathbf{(}\mathbf{6}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{4}\mathbf{)}\ \]
\[ \left| \mathbf{Dx} \right|\mathbf{=}\mathbf{320}\left( \mathbf{24}\mathbf{-}\mathbf{8} \right)\mathbf{-}\mathbf{\ }\mathbf{480}\left( \mathbf{24}\mathbf{-}\mathbf{8} \right)\mathbf{+ \ }\mathbf{192}\mathbf{(}\mathbf{24}\mathbf{-}\mathbf{24}\mathbf{)}\ \]
\[ \left| \mathbf{Dx} \right|\mathbf{=}\mathbf{5120}\mathbf{-}\mathbf{\ }\mathbf{7680}\mathbf{+ \ }\mathbf{0}\ \]
\[ \left| \mathbf{Dx} \right|\mathbf{=}\mathbf{-}\mathbf{2560}\ \]
\[ \begin{bmatrix}\mathbf{2} & \mathbf{320} & \mathbf{4} \\\mathbf{6} & \mathbf{480} & \mathbf{4} \\\mathbf{3} & \mathbf{192} & \mathbf{4} \\\end{bmatrix} \]
\[ \mathbf{\ }\left| \mathbf{Dy} \right|\mathbf{=}\mathbf{2}\left( \mathbf{480}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{192} \right)\mathbf{-}\mathbf{\ }\mathbf{6}\left( \mathbf{320}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{192} \right)\mathbf{+ \ }\mathbf{3}\mathbf{(}\mathbf{320}\mathbf{x}\mathbf{4}\mathbf{-}\mathbf{4×480}\mathbf{)}\ \]
\[ \left| \mathbf{Dy} \right|\mathbf{=}\mathbf{2}\left( \mathbf{1920}\mathbf{-}\mathbf{768} \right)\mathbf{-}\mathbf{\ }\mathbf{6}\left( \mathbf{1280}\mathbf{-}\mathbf{768} \right)\mathbf{+ \ }\mathbf{3}\mathbf{(}\mathbf{1280}\mathbf{-}\mathbf{1920}\mathbf{)}\ \]
\[ \left| \mathbf{Dy} \right|\mathbf{=}\mathbf{2304}\mathbf{-}\mathbf{\ }\mathbf{3072}\mathbf{-}\mathbf{\ }\mathbf{1920}\ \]
\[ \left| \mathbf{Dy} \right|\mathbf{=}\mathbf{-}\mathbf{2688}\ \]
\[ \begin{bmatrix}\mathbf{2} & \mathbf{6} & \mathbf{320} \\\mathbf{6} & \mathbf{6} & \mathbf{480} \\\mathbf{3} & \mathbf{2} & \mathbf{192} \\\end{bmatrix} \]
\[ \mathbf{\ }\left| \mathbf{Dz} \right|\mathbf{=}\mathbf{2}\left( \mathbf{6}\mathbf{x}\mathbf{192}\mathbf{-}\mathbf{480}\mathbf{x}\mathbf{2} \right)\mathbf{-}\mathbf{\ }\mathbf{6}\left( \mathbf{6}\mathbf{x}\mathbf{192}\mathbf{-}\mathbf{320}\mathbf{x}\mathbf{2} \right)\mathbf{+ \ }\mathbf{3}\mathbf{(}\mathbf{6}\mathbf{x}\mathbf{480}\mathbf{-}\mathbf{320}\mathbf{x}\mathbf{6}\mathbf{)}\ \]
\[ \left| \mathbf{Dz} \right|\mathbf{=}\mathbf{2}\left( \mathbf{1152}\mathbf{-}\mathbf{960} \right)\mathbf{-}\mathbf{\ }\mathbf{6}\left( \mathbf{1152}\mathbf{-}\mathbf{640} \right)\mathbf{+ \ }\mathbf{3}\mathbf{(}\mathbf{2880}\mathbf{-}\mathbf{1920}\mathbf{)}\ \]
\[ \left| \mathbf{Dz} \right|\mathbf{=}\mathbf{384}\mathbf{-}\mathbf{\ }\mathbf{3072}\mathbf{+ \ }\mathbf{2880}\ \]
\[ \left| \mathbf{Dz} \right|\mathbf{=}\mathbf{192}\ \]
\[ \mathbf{x}\mathbf{= \ }\frac{\left| \mathbf{Dx} \right|}{\left| \mathbf{D} \right|}\mathbf{\ = \ }\frac{\mathbf{-}\mathbf{2560}}{\mathbf{-}\mathbf{64}}\mathbf{=}\mathbf{40}\ \]
\[ \mathbf{y}\mathbf{= \ }\frac{\left| \mathbf{Dy} \right|}{\left| \mathbf{D} \right|}\mathbf{=}\frac{\mathbf{-}\mathbf{2688}}{\mathbf{-}\mathbf{64}}\mathbf{=}\mathbf{42}\ \]
\[ \mathbf{z}\mathbf{= \ }\frac{\left| \mathbf{Dz} \right|}{\left| \mathbf{D} \right|}\mathbf{= \ }\frac{\mathbf{192}}{\mathbf{-}\mathbf{64}}\mathbf{= \ }\mathbf{-}\mathbf{3}\ \]
\[ \]
(b) The 10th term of an Arithmetic Progression is 20 and 20th term is 40. Find the 7th term.
Solution:
α10 = 20, α20 = 40
αn = α1 + (n – 1)d
α10 = α1 + (10 – 1)d
20 = α1 + 9d………….(i)
αn = α1 + (n – 1)d
α20 = α1 + (20 – 1)d
40 = α1 + 19d………….(ii)
Subtract equation i & ii
20 = α1 + 9d ………….(i)
40 = α1 + 19d ………….(ii)
– = – –
-20 = -10d
-10d = -20
Now we will find the value of α1 by putting the value of d in any equation
20 = α1 + 9d ………….(i)
20 = α1 + 9(2)
20 = α1 + 18
α1 = 20 – 18
α1 = 2
Hence
α 7 = α1 + (7 – 1)d
α 7 = 2 + (6)(2)
α 7 = 2 + 12
α 7= 14
Q.7 (a) If the difference between the simple and compound interest for 3 year at 5% is Rs. 61. Find the Principal amount.
Solution
Assumed Principal Value = 100, n or N = 3, I or i = 5% = 0.05, Difference = 61
S.I = PIN
S.I = 100×0.05×3 = 15
\[ {\mathbf{C}\mathbf{.}\mathbf{I}\mathbf{=}\mathbf{P}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{)}}^{\mathbf{n}}\mathbf{-}\mathbf{p}\ \]
\[ {\mathbf{C}\mathbf{.}\mathbf{I}\mathbf{=}\mathbf{100}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{)}}^{\mathbf{3}}\mathbf{-}\mathbf{100}\ \]
\[ \mathbf{C}\mathbf{.}\mathbf{I}\mathbf{=}\mathbf{100}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{1576}\mathbf{)}\mathbf{-}\mathbf{100}\ \]
\[ \mathbf{C}\mathbf{.}\mathbf{I}\mathbf{=}\mathbf{115}\mathbf{.}\mathbf{76}\mathbf{-}\mathbf{100}\ \]
\[ \mathbf{C}\mathbf{.}\mathbf{I}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{76}\ \]
Difference between Simple & Compound Interest = 15.76 – 15 = 0.76
\[ \mathbf{Principal}\mathbf{= \ }\frac{\mathbf{100}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{61}}{\mathbf{0}\mathbf{.}\mathbf{76}}\mathbf{=}\mathbf{Rs}\mathbf{.}\mathbf{8026}\mathbf{.}\mathbf{32}\ \]
(b) Find the accumulated value of Rs. 5000 invested at the end of each quarter for 5 years at 8% compounded quarterly.
Solution
\[ \mathbf{Data:\ R\ = \ 5000,\ n\ = \ 5 \times 4 = \ 20,\ i =}\frac{\mathbf{8}}{\mathbf{4}}\mathbf{= 2 = 0.02,\ F.V\ = \ ?\ }\ \]
Case: Ordinary Annuity
\[ \mathbf{F.V = R}\left\lbrack \frac{\mathbf{(1 + i)}^{\mathbf{n}}\mathbf{- 1}}{\mathbf{i}} \right\rbrack\ \]
\[ \mathbf{F.V = 5000}\left\lbrack \frac{\mathbf{(1 + 0.02)}^{\mathbf{20}}\mathbf{- 1}}{\mathbf{0.02}} \right\rbrack\ \]
\[ \mathbf{F.V = 5000}\left( \mathbf{24.2973} \right)\ \]
\[ \mathbf{F.V = 121486.5}\ \]
Q.8 Give short answers of the following and unnecessary details will be penalized.
(i) Define Matrix
Answer: Matrix is a mathematical concept of linear algebra that is used in linear equations to solve the variables. Matrix rules perform various function such as addition, subtraction, multiplication, division, transposition, determinant, ad-joint and inverse. It consists on rows and columns. Numbers used in matrix are called elements. Dimension of matrix depends on number of rows and columns.
(ii) Define a Common Ratio.
Answer: Common ratio is a concept of geometric progression or sequence in mathematics. In geometric progression each term is found by multiplying a number with non-zero constant term, except first number is called common ratio.
(iii) Define Compound Interest.
Answer: Compound interest is a concept of time value of money in which interest is calculated on balance or compound amount for each period of time. Its formula is given below:
\[ \mathbf{C.I = P}\left( \mathbf{1 + i} \right)^{\mathbf{n}}\mathbf{- P}\ \]
\[ \mathbf{Where\ }\mathbf{“}\mathbf{P}\mathbf{“}\mathbf{\ is\ Principal\ Amount,\ }\mathbf{“}\mathbf{i}\mathbf{“}\mathbf{\ is\ interest\ rate\ and\ }\mathbf{“}\mathbf{n}\mathbf{“}\mathbf{\ is\ time\ period}\ \]
(iv) Define Annuity Due.
Answer: Annuity due is a financial concept of Time Value of Money in which payment streams or cash flows are made at the start of time period. Formulas of Future and Present Values of Annuity Due are given below:
\[ \mathbf{F.V = R}\left( \frac{\left( \mathbf{1 + i} \right)^{\mathbf{n}}\mathbf{- 1}}{\mathbf{i}} \right)\mathbf{(1 + i)}\ \]
\[ \mathbf{P.V = R}\left( \frac{\mathbf{1 -}\left( \mathbf{1 + i} \right)^{\mathbf{- n}}}{\mathbf{i}} \right)\mathbf{(1 + i)}\ \]
\[ \mathbf{Where}\mathbf{\ }\text{R}\mathbf{\ is\ cash\ flows\ or\ payment\ stream,\ i\ is\ interest\ rate,\ n\ is\ time\ period}\ \]
(v) Define the population.
Answer: The whole group under discussion for research purpose is called population. For example all the students of the college, total production of cars in a particular year is called population.
(vi) What is the difference between sample and sampling?
Answer: Sample is a part of population whereas sampling is a statistical process to make inference about population without studying whole population because to study whole population is too much costly and time consuming.
(vii) Define the term 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.
(viii) Define Standard Deviation.
Answer: Standard deviation is an absolute measure of dispersion that calculates the spread or dispersion of the data around its mean value. It is an important measure of risk in finance also. There are various ways to calculate it, however formula for ungrouped and grouped data are given below:
Formula for ungrouped data
\[ \mathbf{S.D\ or\ S =}\sqrt{\frac{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{n}} \right)^{\mathbf{2}}}\ \]
Formula for grouped data
\[ \mathbf{S.D\ or\ S =}\sqrt{\frac{\mathbf{\sum f}\mathbf{X}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fX}}{\mathbf{\sum f}} \right)^{\mathbf{2}}}\ \]
(ix) What do you understand by Measures of Central Tendency?
Answer: Measures of central tendency or average is a statistical concept of descriptive statistics that calculates the most central point of the data. There are five different ways to calculate it. Arithmetic mean, Median, Mode, Harmonic Mean and Geometric Mean.
(x) Define Weighted Mean.
Answer: Weighted mean is a statistical concept of average in which weights are assigned to each value according to its importance and preference and after getting sum of product of value and weights is divided by sum of weights. Its formula is given below:
\[ \overline{\mathbf{X}}\mathbf{w =}\frac{\mathbf{\sum WX}}{\mathbf{\sum W}}\ \]
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