Mathematics & Statistics (Commerce) - 2025 Board Paper Solution
Paper Code: J-316 | Max Marks: 80 | Time: 3 Hrs
SECTION - I
Q. 1. (A) Select and write the correct answer of the following multiple choice type of questions (1 mark each):
(i) If \(p\): He is intelligent, \(q\): He is strong. Then, symbolic form of statement "It is wrong that, he is intelligent or strong" is:
Explanation: "Intelligent or strong" is \(p \lor q\). "It is wrong that" implies negation. Thus, \(\sim (p \lor q)\).
(ii) \(\int (x + \frac{1}{x})^3 dx =\)
Solution: Expand \((x + x^{-1})^3 = x^3 + 3x + \frac{3}{x} + x^{-3}\).
Integrate term by term: \(\frac{x^4}{4} + \frac{3x^2}{2} + 3\log|x| + \frac{x^{-2}}{-2} + c\).
(iii) \(\int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9-x}} dx =\)
Solution: Using property \(\int_a^b f(x)dx = \int_a^b f(a+b-x)dx\). The integral \(I = \frac{b-a}{2} = \frac{7-2}{2} = \frac{5}{2}\).
(iv) The area of the region bounded by the curve \(y = x^2\) and the line \(y = 4\) is
Solution: Area \(= 2 \int_{0}^{2} (4 - x^2) dx = 2 [4x - \frac{x^3}{3}]_0^2 = 2(8 - \frac{8}{3}) = \frac{32}{3}\).
(v) The order and degree of the differential equation \((\frac{d^2y}{dx^2})^2 + (\frac{dy}{dx})^2 = a^x\) are _____ respectively.
(vi) The integrating factor of the differential equation \(\frac{dy}{dx} + \frac{y}{x} = x^3 - 3\) is
Solution: I.F. \(= e^{\int \frac{1}{x} dx} = e^{\log x} = x\).
Q. 1. (B) State whether the following statements are true or false (1 mark each):
(i) If \(A\) is a matrix and \(K\) is a constant, then \((KA)^T = K A^T\).
(ii) \(\int \log x dx = x \log x + x + c\).
(iii) The differential equation obtained by eliminating arbitrary constants from \(bx + ay = ab\) is \(\frac{d^2y}{dx^2} = 0\).
Q. 1. (C) Fill in the following blanks (1 mark each):
(i) The average revenue \(R_A\) is 50 and elasticity of demand \(\eta\) is 5, the marginal revenue \(R_M\) is _____.
(\(R_M = R_A(1 - \frac{1}{\eta}) = 50(1 - \frac{1}{5}) = 40\))
(ii) \(\int e^x (\frac{1}{x} - \frac{1}{x^2}) dx = \) _____ \(+ c\)
(iii) If \(f'(x) = x^2 + 5\) and \(f(0) = -1\) then \(f(x) = \) _____.
Q. 2. (A) Attempt any TWO of the following questions (3 marks each):
(i) Write the converse, inverse and contrapositive of the statement "If a triangle is equilateral then it is equiangular".
Statement: \(p \rightarrow q\)
Converse (\(q \rightarrow p\)): If a triangle is equiangular then it is equilateral.
Inverse (\(\sim p \rightarrow \sim q\)): If a triangle is not equilateral then it is not equiangular.
Contrapositive (\(\sim q \rightarrow \sim p\)): If a triangle is not equiangular then it is not equilateral.
(ii) Find \(x, y, z\) if \(\left\{ 5 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{bmatrix} - 3 \begin{bmatrix} 2 & 1 \\ 3 & -2 \\ 1 & 3 \end{bmatrix} \right\} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} x-1 \\ y+1 \\ 2z \end{bmatrix}\)
\(5A - 3B = \begin{bmatrix} -6 & 2 \\ -4 & 6 \\ 2 & -4 \end{bmatrix}\)
Multiply by \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\):
\(\begin{bmatrix} -6(2) + 2(1) \\ -4(2) + 6(1) \\ 2(2) + (-4)(1) \end{bmatrix} = \begin{bmatrix} -10 \\ -2 \\ 0 \end{bmatrix}\)
Equating to RHS:
\(x - 1 = -10 \Rightarrow x = -9\)
\(y + 1 = -2 \Rightarrow y = -3\)
\(2z = 0 \Rightarrow z = 0\)
(iii) Evaluate: \(\int \frac{1}{x(x^6+1)} dx\)
Put \(x^6 = t \Rightarrow 6x^5 dx = dt\)
\(I = \frac{1}{6} \int \frac{dt}{t(t+1)} = \frac{1}{6} \int (\frac{1}{t} - \frac{1}{t+1}) dt\)
\(I = \frac{1}{6} (\log|t| - \log|t+1|) + c = \frac{1}{6} \log|\frac{x^6}{x^6+1}| + c\)
Q. 2. (B) Attempt any TWO of the following questions (4 marks each):
(i) Solve the following equations by the method of inversion:
\(2x - y + z = 1\)
\(x + 2y + 3z = 8\)
\(3x + y - 4z = 1\)
The given system of equations can be written in matrix form \(AX = B\), where
\(A = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 8 \\ 1 \end{bmatrix}\)
Step 1: Find determinant of A (\(|A|\))
\(|A| = 2(-8 - 3) - (-1)(-4 - 9) + 1(1 - 6)\)
\(|A| = 2(-11) + 1(-13) + 1(-5)\)
\(|A| = -22 - 13 - 5 = -40 \neq 0\)
Since \(|A| \neq 0\), \(A^{-1}\) exists.
Step 2: Find Matrix of Cofactors
\(A_{11} = -11, \quad A_{12} = 13, \quad A_{13} = -5\)
\(A_{21} = -3, \quad A_{22} = -11, \quad A_{23} = -5\)
\(A_{31} = -5, \quad A_{32} = -5, \quad A_{33} = 5\)
Cofactor Matrix \(C = \begin{bmatrix} -11 & 13 & -5 \\ -3 & -11 & -5 \\ -5 & -5 & 5 \end{bmatrix}\)
\(\text{adj } A = C^T = \begin{bmatrix} -11 & -3 & -5 \\ 13 & -11 & -5 \\ -5 & -5 & 5 \end{bmatrix}\)
Step 3: Find X using \(X = A^{-1}B\)
\(X = \frac{1}{|A|} (\text{adj } A) B\)
\(X = \frac{1}{-40} \begin{bmatrix} -11 & -3 & -5 \\ 13 & -11 & -5 \\ -5 & -5 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ 8 \\ 1 \end{bmatrix}\)
\(X = \frac{1}{-40} \begin{bmatrix} -11(1) -3(8) -5(1) \\ 13(1) -11(8) -5(1) \\ -5(1) -5(8) + 5(1) \end{bmatrix}\)
\(X = \frac{1}{-40} \begin{bmatrix} -11 - 24 - 5 \\ 13 - 88 - 5 \\ -5 - 40 + 5 \end{bmatrix}\)
\(X = \frac{1}{-40} \begin{bmatrix} -40 \\ -80 \\ -40 \end{bmatrix}\)
\(\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}\)
\(\therefore x = 1, y = 2, z = 1\)
(ii) Find MPC, MPS, APC and APS, if the expenditure \(E_c\) of a person with income \(I\) is given as: \(E_c = (0.0003)I^2 + (0.075)I\); when \(I = 1000\).
APC \(= \frac{E_c}{I} = 0.0003I + 0.075\)
At \(I=1000\): APC \(= 0.0003(1000) + 0.075 = 0.3 + 0.075 = 0.375\)
APS \(= 1 - APC = 1 - 0.375 = 0.625\)
MPC \(= \frac{dE_c}{dI} = 0.0006I + 0.075\)
At \(I=1000\): MPC \(= 0.0006(1000) + 0.075 = 0.6 + 0.075 = 0.675\)
MPS \(= 1 - MPC = 1 - 0.675 = 0.325\)
(iii) Evaluate: \(\int_1^2 \frac{dx}{x^2+6x+5}\)
Partial Fractions: \(\frac{1}{(x+1)(x+5)} = \frac{1}{4}(\frac{1}{x+1} - \frac{1}{x+5})\)
\(I = \frac{1}{4} [\log|x+1| - \log|x+5|]_1^2 = \frac{1}{4} [\log(\frac{x+1}{x+5})]_1^2\)
Upper limit: \(\log(\frac{3}{7})\), Lower limit: \(\log(\frac{2}{6}) = \log(\frac{1}{3})\)
\(I = \frac{1}{4} (\log \frac{3}{7} - \log \frac{1}{3}) = \frac{1}{4} \log(\frac{3}{7} \times 3) = \frac{1}{4} \log(\frac{9}{7})\).
Q. 3. (A) Attempt any TWO of the following questions (3 marks each):
(i) Find \(\frac{dy}{dx}\) if \(y = (x)^x + (a)^x\)
\(u = x^x \Rightarrow \log u = x \log x \Rightarrow \frac{1}{u}\frac{du}{dx} = 1 + \log x \Rightarrow \frac{du}{dx} = x^x(1+\log x)\)
\(v = a^x \Rightarrow \frac{dv}{dx} = a^x \log a\)
\(\frac{dy}{dx} = x^x(1+\log x) + a^x \log a\)
(ii) Find the area of the region bounded by the parabola \(y^2 = 25x\) and the line \(x = 5\).
Area \(= 2 \int_0^5 y dx = 2 \int_0^5 5\sqrt{x} dx = 10 \int_0^5 x^{1/2} dx\)
\(= 10 [\frac{x^{3/2}}{3/2}]_0^5 = \frac{20}{3} [5^{3/2}] = \frac{20}{3} (5\sqrt{5}) = \frac{100\sqrt{5}}{3}\) sq. units.
(iii) Find the differential equation by eliminating arbitrary constants from the relation \(y = Ae^{3x} + Be^{-3x}\).
Diff. again: \(y'' = 9Ae^{3x} + 9Be^{-3x} = 9(Ae^{3x} + Be^{-3x})\)
\(y'' = 9y \Rightarrow \frac{d^2y}{dx^2} - 9y = 0\)
Q. 3. (B) Attempt any ONE of the following questions (4 marks each):
(i) Using the truth table, verify \(p \lor (q \land r) = (p \lor q) \land (p \lor r)\)
We construct the truth table for the given logical statement.
| \(p\) | \(q\) | \(r\) | \(q \land r\) | \(p \lor (q \land r)\) (LHS) |
\(p \lor q\) | \(p \lor r\) | \((p \lor q) \land (p \lor r)\) (RHS) |
|---|---|---|---|---|---|---|---|
| T | T | T | T | T | T | T | T |
| T | T | F | F | T | T | T | T |
| T | F | T | F | T | T | T | T |
| T | F | F | F | T | T | T | T |
| F | T | T | T | T | T | T | T |
| F | T | F | F | F | T | F | F |
| F | F | T | F | F | F | T | F |
| F | F | F | F | F | F | F | F |
From the table, the entries in column 5 (LHS) and column 8 (RHS) are identical.
\(\therefore p \lor (q \land r) = (p \lor q) \land (p \lor r)\) is verified.
(ii) If \(x = \frac{4t}{1+t^2}, y = 3(\frac{1-t^2}{1+t^2})\), then show that \(\frac{dy}{dx} = \frac{-9x}{4y}\).
Thus \(\frac{x}{2} = \sin 2\theta\) and \(\frac{y}{3} = \cos 2\theta\).
Squaring and adding: \(\frac{x^2}{4} + \frac{y^2}{9} = 1\).
Differentiate w.r.t \(x\): \(\frac{2x}{4} + \frac{2y}{9}\frac{dy}{dx} = 0\).
\(\frac{x}{2} = -\frac{2y}{9}\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{-9x}{4y}\).
Q. 3. (C) Attempt any ONE of the following questions (Activity):
(i) Divide the number 84 into two parts such that the product of one part and square of the other is maximum.
\(f(x) = x^2(84-x) = 84x^2 - x^3\) (Note: Maximizing square of one times other).
\(f'(x) = 168x - 3x^2\)
For extreme values \(f'(x) = 0 \Rightarrow 3x(56-x) = 0\)
\(x = 0\) OR \(x = 56\)
\(f''(x) = 168 - 6x\)
If \(x=56, f''(56) = 168 - 336 = -168 < 0\)
Function attains maximum at \(x = 56\).
Two parts of 84 are 56 and 28.
(ii) Solve the following differential equation
\((x^2 - yx^2)dy + (y^2 + xy^2)dx = 0\)
Separating the variables, the given equation can be written as:
\(x^2(1-y)dy + y^2(1+x)dx = 0\)
Dividing by \(x^2y^2\),
\(\left[ \frac{1-y}{y^2} \right] dy + \left[ \frac{1+x}{x^2} \right] dx = 0\)
\(\therefore (y^{-2} - \frac{1}{y})dy + (x^{-2} + \frac{1}{x})dx = 0\)
\(\left[ y^{-2} \right] dy - \frac{1}{y}dy + x^{-2}dx + \left[ \frac{1}{x} \right] dx = 0\)
Integrating we get,
\(\int y^{-2}dy - \int \frac{1}{y}dy + \int x^{-2}dx + \int \frac{1}{x}dx = 0\)
\(\therefore \frac{y^{-1}}{-1} - \left[ \log y \right] + \frac{x^{-1}}{-1} + \left[ \log x \right] = c\)
\(-\frac{1}{y} - \frac{1}{x} + \log x - \log y = c\)
\(\log x - \log y = \left[ \frac{1}{x} + \frac{1}{y} \right] + c\)
is the required solution.
SECTION - II
Q. 4. (A) Select and write the correct answer (1 mark each):
(i) An agent who gives guarantee to his principal that the party will pay the sale price of goods is called –
(ii) In an ordinary annuity, payments or receipts occur at
(iii) Moving averages are useful in identifying
(iv) If \(P_{01}(L)=90\) and \(P_{01}(P)=40\), then \(P_{01}(D-B)\) is
(v) The objective of an assignment problem is to assign
(vi) The expected value of the sum of two numbers obtained when two fair dice are rolled is
Q. 4. (B) State whether true or false (1 mark each):
(i) If \(b_{yx} + b_{xy} = 1.30\) and \(r = 0.75\), data is inconsistent. Answer: True (If sum is 1.3, max product is \(0.65^2 = 0.4225\), but \(r^2 = 0.75^2 = 0.5625\). Impossible).
(ii) Cyclic variation can occur several times in a year. Answer: False
(iii) Cost of living index number is used in calculating purchasing power of money. Answer: True
Q. 4. (C) Fill in the blanks (1 mark each):
(i) The amount paid to the holder of the bill after deducting banker's discount is known as Cash Value.
(ii) The simplest method of measuring trend of time series is Graphical Method.
(iii) Quantity index number by weighted aggregate method is given by \(\frac{\sum q_1 w}{\sum q_0 w} \times 100\).
Q. 5. (A) Attempt any TWO (3 marks each):
(i) Compute the appropriate regression equation for X: 1, 2, 3, 4, 5 and Y: 5, 7, 9, 11, 13.
\(b_{yx} = \frac{\sum(X-\bar{X})(Y-\bar{Y})}{\sum(X-\bar{X})^2} = \frac{20}{10} = 2\).
Equation of Y on X: \(Y - 9 = 2(X - 3) \Rightarrow Y = 2X + 3\).
(ii) L.P.P. Formulation: Min Cost.
Minimize \(Z = 20x + 6y\)
Subject to:
\(x + y \ge 5\) (Weight)
\(x \ge 4\) (Min cement)
\(y \le 2\) (Max sand)
\(x, y \ge 0\).
(iii) Find the mean of number of heads in three tosses of a fair coin.
Mean \(E(X) = np = 3 \times 0.5 = 1.5\).
Q. 5. (B) Attempt any TWO of the following questions (4 marks each):
(i) Obtain the trend value for the following data using 4-yearly centered moving averages :
| Years | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 |
|---|---|---|---|---|---|---|---|---|---|---|
| Index | 0 | 2 | 3 | 3 | 2 | 4 | 5 | 6 | 7 | 10 |
Calculation of 4-Yearly Centered Moving Averages:
| Year | Index (Y) | 4-Year Moving Total | Centered Total (Sum of 2) | Trend Value (Centered Avg / 8) |
|---|---|---|---|---|
| 1976 | 0 | - | - | - |
| 1977 | 2 | - | - | - |
| 1978 | 3 | 8 | 18 | 2.25 |
| 1979 | 3 | 10 | 22 | 2.75 |
| 1980 | 2 | 12 | 26 | 3.25 |
| 1981 | 4 | 14 | 31 | 3.875 |
| 1982 | 5 | 17 | 39 | 4.875 |
| 1983 | 6 | 22 | 50 | 6.25 |
| 1984 | 7 | 28 | - | - |
| 1985 | 10 | - | - | - |
Note:
1. First 4-Year Total: \(0+2+3+3 = 8\)
2. Second 4-Year Total: \(2+3+3+2 = 10\) ... and so on.
3. Centered Total: \(8+10=18\), \(10+12=22\) ... and so on.
4. Trend Value = Centered Total \(\div 8\).
(ii) Find the sequence that minimizes the total elapsed time to complete the following jobs in the order AB. Find the total elapsed time and idle time for machine B.
Step 1: Determine the Optimal Sequence
Using Johnson's Algorithm:
- Min time is 5: Job VII on A (First) and Job VI on B (Last).
- Next min is 7: Job I on A (Second) and Job VII (Done).
- Next min is 10: Job IV on A and B. Place IV after I.
- Next min is 14: Job V on A (After IV), Job II on B (Before VI), Job III on B (Before II).
Optimal Sequence: VII \(\to\) I \(\to\) IV \(\to\) V \(\to\) III \(\to\) II \(\to\) VI
Step 2: Work Table
| Job Sequence | Machine A | Machine B | ||
|---|---|---|---|---|
| In | Out | In | Out | |
| VII | 0 | 5 | 5 | 12 |
| I | 5 | 12 | 12 | 24 |
| IV | 12 | 22 | 24 | 34 |
| V | 22 | 36 | 36 | 52 |
| III | 36 | 55 | 55 | 69 |
| II | 55 | 71 | 71 | 85 |
| VI | 71 | 86 | 86 | 91 |
Total Elapsed Time (T) = 91 hours/units
Idle Time for Machine B:
Total time - Sum of processing times on B
\(= 91 - (12+14+14+10+16+5+7)\)
\(= 91 - 78\)
\(= 13 \text{ hours/units}\)
(iii) Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. Find the probability that :
(a) all the five cards are spades
(b) only 3 cards are spades.
Let \(X\) denote the number of spades.
Total cards = 52, Spades = 13.
Probability of success (getting a spade) \(p = \frac{13}{52} = \frac{1}{4}\).
Probability of failure \(q = 1 - \frac{1}{4} = \frac{3}{4}\).
Number of trials \(n = 5\).
This is a Binomial Distribution: \(P(X=x) = {}^nC_x p^x q^{n-x}\).
(a) Probability that all five cards are spades (\(x=5\)):
\(P(X=5) = {}^5C_5 (\frac{1}{4})^5 (\frac{3}{4})^0\)
\(= 1 \cdot \frac{1}{1024} \cdot 1\)
\(= \frac{1}{1024}\)
(b) Probability that only 3 cards are spades (\(x=3\)):
\(P(X=3) = {}^5C_3 (\frac{1}{4})^3 (\frac{3}{4})^2\)
\(= \frac{5 \times 4}{2 \times 1} \cdot \frac{1}{64} \cdot \frac{9}{16}\)
\(= 10 \cdot \frac{9}{1024}\)
\(= \frac{90}{1024}\)
\(= \frac{45}{512}\)
Q. 6. (A) Attempt any TWO (3 marks each):
(i) Insurance calculation.
Premium = \(0.80\%\) of 6,00,000 = 4,800.
Agent Commission = \(9\%\) of 4,800 = 432.
(ii) Solve LPP graphically. Max \(z = 4x + 6y\).
Constraints: \(3x+2y \le 12\) (towards origin), \(x+y \ge 4\) (away from origin).
Feasible region is triangle with vertices A(4,0), B(0,4), C(0,6).
\(Z\) at A(4,0) = 16.
\(Z\) at B(0,4) = 24.
\(Z\) at C(0,6) = 36.
Maximum value is 36 at point (0,6).
(iii) Defects on plywood sheet occur at random with the average of one defect per 50 sq.ft. Find the probability that such a sheet has :
(a) no defect
(b) at least one defect
(use \(e^{-1} = 0.3678\))
Let \(X\) denote the number of defects on a plywood sheet.
Given average number of defects \(m = 1\) (per 50 sq.ft).
This follows a Poisson Distribution with parameter \(m=1\).
The probability mass function is \(P(X=x) = \frac{e^{-m} m^x}{x!} = \frac{e^{-1} (1)^x}{x!} = \frac{e^{-1}}{x!}\).
(a) Probability of no defect (\(x=0\)):
\(P(X=0) = \frac{e^{-1}}{0!} = \frac{0.3678}{1}\)
\(\therefore P(X=0) = 0.3678\)
(b) Probability of at least one defect (\(x \ge 1\)):
\(P(X \ge 1) = 1 - P(X = 0)\)
\(= 1 - 0.3678\)
\(\therefore P(X \ge 1) = 0.6322\)
Q. 6. (B) Attempt any ONE of the following questions (4 marks each):
(i) The equations of two regression lines are \(10x - 4y = 80\) and \(10y - 9x = -40\). Find
(a) \(\bar{x}\) and \(\bar{y}\)
(b) \(b_{yx}\) and \(b_{xy}\)
(c) \(r\)
(d) If \(var(Y) = 36\), obtain \(var(X)\).
(a) Finding \(\bar{x}\) and \(\bar{y}\):
Since the regression lines intersect at \((\bar{x}, \bar{y})\), we solve the equations simultaneously:
1) \(10x - 4y = 80\) (Multiply by 5) \(\Rightarrow 50x - 20y = 400\)
2) \(-9x + 10y = -40\) (Multiply by 2) \(\Rightarrow -18x + 20y = -80\)
Adding the two new equations:
\(32x = 320 \Rightarrow x = 10\)
Substitute \(x = 10\) in equation (1):
\(10(10) - 4y = 80 \Rightarrow 100 - 80 = 4y \Rightarrow 4y = 20 \Rightarrow y = 5\)
\(\therefore \bar{x} = 10\) and \(\bar{y} = 5\)
(b) Finding \(b_{yx}\) and \(b_{xy}\):
Let \(10x - 4y = 80\) be the regression line of X on Y.
\(10x = 4y + 80 \Rightarrow x = 0.4y + 8\)
\(\therefore b_{xy} = 0.4\)
Let \(10y - 9x = -40\) be the regression line of Y on X.
\(10y = 9x - 40 \Rightarrow y = 0.9x - 4\)
\(\therefore b_{yx} = 0.9\)
(Check: \(b_{xy} \times b_{yx} = 0.4 \times 0.9 = 0.36 < 1\), assumption is correct).
\(\therefore b_{yx} = 0.9\) and \(b_{xy} = 0.4\)
(c) Finding \(r\):
\(r = \pm \sqrt{b_{yx} \times b_{xy}}\)
Since both regression coefficients are positive, \(r\) is positive.
\(r = \sqrt{0.9 \times 0.4} = \sqrt{0.36} = 0.6\)
\(\therefore r = 0.6\)
(d) Finding \(var(X)\):
Given \(var(Y) = \sigma_y^2 = 36 \Rightarrow \sigma_y = 6\)
We know, \(b_{yx} = r \cdot \frac{\sigma_y}{\sigma_x}\)
\(0.9 = 0.6 \cdot \frac{6}{\sigma_x} \Rightarrow \sigma_x = \frac{3.6}{0.9} = 4\)
\(var(X) = \sigma_x^2 = 4^2 = 16\)
\(\therefore var(X) = 16\)
(ii) Find \(x\) if the cost of living index is 150 :
| Group | Food | Clothing | Fuel and electricity | House Rent | Miscellaneous |
|---|---|---|---|---|---|
| I | 180 | 120 | 300 | 100 | 160 |
| W | 4 | 5 | 6 | \(x\) | 3 |
We prepare the table for \(\sum W\) and \(\sum IW\):
| Group | I | W | IW |
|---|---|---|---|
| Food | 180 | 4 | 720 |
| Clothing | 120 | 5 | 600 |
| Fuel & Elec. | 300 | 6 | 1800 |
| House Rent | 100 | \(x\) | \(100x\) |
| Miscellaneous | 160 | 3 | 480 |
| Total | \(\sum W = 18 + x\) | \(\sum IW = 3600 + 100x\) |
Cost of Living Index (CLI) \(= \frac{\sum IW}{\sum W}\)
Given, CLI = 150.
\(150 = \frac{3600 + 100x}{18 + x}\)
\(150(18 + x) = 3600 + 100x\)
\(2700 + 150x = 3600 + 100x\)
\(150x - 100x = 3600 - 2700\)
\(50x = 900\)
\(x = \frac{900}{50}\)
\(\therefore x = 18\)
Q. 6. (C) Activity (4 marks each):
(i) Bills of Exchange activity.
Given:
Sum Due (S.D.) = 18,000
Cash Value (C.V.) = 17,568
Rate (R) = 12% p.a.
Date of Discounting = 25th October 2017
Step 1: Calculate Banker's Discount (B.D.)
\(B.D. = \text{S.D.} - \text{C.V.}\)
\(B.D. = 18,000 - 17,568\)
\(\therefore B.D. = 432\)
Step 2: Calculate Period (n)
We know that, \(B.D. = \frac{\text{S.D.} \times n \times R}{100}\)
Substituting the values:
\(432 = \frac{18,000 \times n \times 12}{100}\)
\(n = \frac{432 \times 100}{18,000 \times 12}\)
\(n = \frac{1}{5} \text{ years}\)
Converting years to days: \(n = \frac{1}{5} \times 365 = 73 \text{ days}\)
Step 3: Calculate Legal Due Date
The period is 73 days from 25th October 2017.
| Month | Days Calculation | Days Taken |
|---|---|---|
| Oct 2017 | (31 - 25) | 6 |
| Nov 2017 | Full Month | 30 |
| Dec 2017 | Full Month | 31 |
| Jan 2018 | (Remaining to reach 73) | 6 |
| Total | 73 |
\(\therefore\) Legal Due Date = 6th January 2018
(ii) Solve the following assignment problem for minimization:
| I | II | III | IV | V | |
|---|---|---|---|---|---|
| 1 | 18 | 24 | 19 | 20 | 23 |
| 2 | 19 | 21 | 20 | 18 | 22 |
| 3 | 22 | 23 | 20 | 21 | 23 |
| 4 | 20 | 18 | 21 | 19 | 19 |
| 5 | 18 | 22 | 23 | 22 | 21 |
Step-I: Row Reduction
Subtract the smallest element of each row from every element of that row:
| 0 | 6 | 1 | 2 | 5 |
| 1 | 3 | 2 | 0 | 4 |
| 2 | 3 | 0 | 1 | 3 |
| 2 | 0 | 3 | 1 | 1 |
| 0 | 4 | 5 | 4 | 3 |
Step-II: Column Reduction
Subtract the smallest element of each column from every element of that column (Smallest elements for all columns are 0, except column V where it is 1):
| 0 | 6 | 1 | 2 | 4 |
| 1 | 3 | 2 | 0 | 3 |
| 2 | 3 | 0 | 1 | 2 |
| 2 | 0 | 3 | 1 | 0 |
| 0 | 4 | 5 | 4 | 2 |
Step-III & IV: Cover Zeros and Improve Matrix
Draw minimum lines to cover zeros. Here, minimum lines (4) < order of matrix (5). The smallest uncovered element is 1. Subtract it from uncovered elements and add to intersections.
Step-V: Final Assignment Matrix
| [0] | 5 | 0 | 1 | 3 |
| 2 | 3 | 2 | [0] | 3 |
| 3 | 3 | [0] | 0 | 2 |
| 3 | [0] | 3 | 1 | 0 |
| 0 | 3 | 4 | 3 | [0] |
Optimal Assignment:
1 → I (18)
2 → IV (18)
3 → III (20)
4 → II (18)
5 → V (21)
Minimum Value: 18 + 18 + 20 + 18 + 21 = 95
