Mathematics & Statistics (Commerce)
March 2024 Board Paper Solution
Max. Marks: 80 | Time: 3 Hrs.
SECTION - I
Q. 1. (A) Select and write the correct answer (1 mark each)
(i) Which of the following is not a statement?
Answer: (d) Come here
Reason: It is an imperative sentence (command), hence not a logical statement.
(ii) If \(x+y+z=3\), \(x+2y+3z=4\), \(x+4y+9z=6\) then \((y,z) =\)
Answer: (b) (1, 0)
Solving the equations:
1) \(x+y+z=3\)
2) \(x+2y+3z=4\)
Subtract (1) from (2): \(y+2z=1\).
If we check option (b) where \(y=1, z=0\): \(1 + 2(0) = 1\). Correct.
1) \(x+y+z=3\)
2) \(x+2y+3z=4\)
Subtract (1) from (2): \(y+2z=1\).
If we check option (b) where \(y=1, z=0\): \(1 + 2(0) = 1\). Correct.
(iii) If \(y = \log(\frac{e^x}{x^2})\), then \(\frac{dy}{dx} = ?\)
Answer: (b) \(\frac{x-2}{x}\)
\(y = \log(e^x) - \log(x^2) = x\log e - 2\log x = x - 2\log x\)
\(\frac{dy}{dx} = 1 - \frac{2}{x} = \frac{x-2}{x}\)
\(\frac{dy}{dx} = 1 - \frac{2}{x} = \frac{x-2}{x}\)
(iv) The value of \(\int \frac{dx}{\sqrt{1-x}}\) is
Answer: (b) \(-2\sqrt{1-x} + c\)
(v) \(\int \frac{dx}{(x-8)(x+7)} = \_\).
Answer: (c) \(\frac{1}{15} \log |\frac{x-8}{x+7}| + c\)
(vi) The differential equation of \(y = k_1 e^x + k_2 e^{-x}\) is:
Answer: (a) \(\frac{d^2y}{dx^2} - y = 0\)
Q. 1. (B) State whether True or False (1 mark each)
- (i) \(\int_{a}^{b} f(x)dx = \int_{a}^{b} f(t)dt\)
Answer: True - (ii) For \(\int \frac{x-1}{(x+1)^3} e^x dx = e^x f(x) + c\), \(f(x) = (x+1)^2\)
Answer: False
(Correction: \(f(x)\) should be \(\frac{1}{(x+1)^2}\) or \((x+1)^{-2}\), not \((x+1)^2\)) - (iii) Order and degree of a differential equation are always positive integers.
Answer: True
Q. 1. (C) Fill in the blanks (1 mark each)
- (i) The slope of tangent at any point (a, b) is called as Gradient.
- (ii) If \(f'(x) = \frac{1}{x} + x\) and \(f(1) = \frac{5}{2}\) then \(f(x) = \log x + \frac{x^2}{2} + \) 2.
- (iii) A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called Particular solution.
Q. 2 & 3. Key Subjective Solutions & Activities
Q. 2. (A) Attempt any TWO of the following questions (3 marks each):
(i) Examine whether the following statement pattern is a tautology, a contradiction, or a contingency:
\( \sim p \rightarrow (p \rightarrow \sim q) \)
\( \sim p \rightarrow (p \rightarrow \sim q) \)
Solution:
Let's construct the truth table:
Since all entries in the last column are T, the given statement pattern is a Tautology.
Let's construct the truth table:
| \(p\) | \(q\) | \(\sim p\) | \(\sim q\) | \(p \rightarrow \sim q\) | \(\sim p \rightarrow (p \rightarrow \sim q)\) |
|---|---|---|---|---|---|
| T | T | F | F | F | T |
| T | F | F | T | T | T |
| F | T | T | F | T | T |
| F | F | T | T | T | T |
(ii) Find \( \frac{dy}{dx} \) if \( x = e^{3t} \), \( y = e^{(4t+5)} \)
Solution:
Given \( x = e^{3t} \)
Differentiating w.r.t \(t\):
\( \frac{dx}{dt} = e^{3t} \cdot \frac{d}{dt}(3t) = 3e^{3t} \)
Given \( y = e^{4t+5} \)
Differentiating w.r.t \(t\):
\( \frac{dy}{dt} = e^{4t+5} \cdot \frac{d}{dt}(4t+5) = 4e^{4t+5} \)
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
\( \frac{dy}{dx} = \frac{4e^{4t+5}}{3e^{3t}} \)
\( \frac{dy}{dx} = \frac{4}{3} e^{(4t+5) - 3t} \)
\( \frac{dy}{dx} = \frac{4}{3} e^{t+5} \)
Given \( x = e^{3t} \)
Differentiating w.r.t \(t\):
\( \frac{dx}{dt} = e^{3t} \cdot \frac{d}{dt}(3t) = 3e^{3t} \)
Given \( y = e^{4t+5} \)
Differentiating w.r.t \(t\):
\( \frac{dy}{dt} = e^{4t+5} \cdot \frac{d}{dt}(4t+5) = 4e^{4t+5} \)
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
\( \frac{dy}{dx} = \frac{4e^{4t+5}}{3e^{3t}} \)
\( \frac{dy}{dx} = \frac{4}{3} e^{(4t+5) - 3t} \)
\( \frac{dy}{dx} = \frac{4}{3} e^{t+5} \)
(iii) If \( A = \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix} \) then find \( A^T + 4B^T \)
Solution:
\( A^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} \)
\( B^T = \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} \)
\( 4B^T = 4 \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 8 \\ -8 & 4 \\ 12 & -16 \end{bmatrix} \)
\( A^T + 4B^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 0 & 8 \\ -8 & 4 \\ 12 & -16 \end{bmatrix} \)
\( = \begin{bmatrix} 7+0 & 0+8 \\ 3-8 & 4+4 \\ 0+12 & -2-16 \end{bmatrix} \)
\( = \begin{bmatrix} 7 & 8 \\ -5 & 8 \\ 12 & -18 \end{bmatrix} \)
\( A^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} \)
\( B^T = \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} \)
\( 4B^T = 4 \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 8 \\ -8 & 4 \\ 12 & -16 \end{bmatrix} \)
\( A^T + 4B^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 0 & 8 \\ -8 & 4 \\ 12 & -16 \end{bmatrix} \)
\( = \begin{bmatrix} 7+0 & 0+8 \\ 3-8 & 4+4 \\ 0+12 & -2-16 \end{bmatrix} \)
\( = \begin{bmatrix} 7 & 8 \\ -5 & 8 \\ 12 & -18 \end{bmatrix} \)
Q. 2. (B) Attempt any TWO of the following questions (4 marks each):
(i) Consider the following statements:
(a) If D is dog, then D is very good.
(b) If D is very good, then D is dog.
(c) If D is not very good, then D is not a dog.
(d) If D is not a dog, then D is not very good.
Identify the pairs of statements having the same meaning. Justify.
(a) If D is dog, then D is very good.
(b) If D is very good, then D is dog.
(c) If D is not very good, then D is not a dog.
(d) If D is not a dog, then D is not very good.
Identify the pairs of statements having the same meaning. Justify.
Solution:
Let \( p \): D is a dog.
Let \( q \): D is very good.
The symbolic forms of the given statements are:
(a) \( p \rightarrow q \) (Implication)
(b) \( q \rightarrow p \) (Converse)
(c) \( \sim q \rightarrow \sim p \) (Contrapositive)
(d) \( \sim p \rightarrow \sim q \) (Inverse)
Justification:
We know that a conditional statement is logically equivalent to its contrapositive.
Therefore, \( p \rightarrow q \equiv \sim q \rightarrow \sim p \).
Also, the converse is logically equivalent to the inverse.
Therefore, \( q \rightarrow p \equiv \sim p \rightarrow \sim q \).
Pairs having the same meaning:
1. (a) and (c) are equivalent.
2. (b) and (d) are equivalent.
Let \( p \): D is a dog.
Let \( q \): D is very good.
The symbolic forms of the given statements are:
(a) \( p \rightarrow q \) (Implication)
(b) \( q \rightarrow p \) (Converse)
(c) \( \sim q \rightarrow \sim p \) (Contrapositive)
(d) \( \sim p \rightarrow \sim q \) (Inverse)
Justification:
We know that a conditional statement is logically equivalent to its contrapositive.
Therefore, \( p \rightarrow q \equiv \sim q \rightarrow \sim p \).
Also, the converse is logically equivalent to the inverse.
Therefore, \( q \rightarrow p \equiv \sim p \rightarrow \sim q \).
Pairs having the same meaning:
1. (a) and (c) are equivalent.
2. (b) and (d) are equivalent.
(ii) Determine the minimum value of the function:
\( f(x) = 2x^3 - 21x^2 + 36x - 20 \)
\( f(x) = 2x^3 - 21x^2 + 36x - 20 \)
Solution:
Given \( f(x) = 2x^3 - 21x^2 + 36x - 20 \)
Step 1: Find \( f'(x) \) and \( f''(x) \).
\( f'(x) = 6x^2 - 42x + 36 \)
\( f''(x) = 12x - 42 \)
Step 2: Find stationary points by setting \( f'(x) = 0 \).
\( 6(x^2 - 7x + 6) = 0 \)
\( x^2 - 6x - x + 6 = 0 \)
\( x(x-6) - 1(x-6) = 0 \)
\( (x-1)(x-6) = 0 \)
So, \( x = 1 \) or \( x = 6 \).
Step 3: Test the points in \( f''(x) \).
For \( x = 1 \):
\( f''(1) = 12(1) - 42 = -30 < 0 \) (Maxima)
For \( x = 6 \):
\( f''(6) = 12(6) - 42 = 72 - 42 = 30 > 0 \) (Minima)
Step 4: Calculate the minimum value at \( x = 6 \).
\( f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20 \)
\( f(6) = 2(216) - 21(36) + 216 - 20 \)
\( f(6) = 432 - 756 + 216 - 20 \)
\( f(6) = 648 - 776 \)
\( f(6) = -128 \)
The minimum value of the function is -128.
Given \( f(x) = 2x^3 - 21x^2 + 36x - 20 \)
Step 1: Find \( f'(x) \) and \( f''(x) \).
\( f'(x) = 6x^2 - 42x + 36 \)
\( f''(x) = 12x - 42 \)
Step 2: Find stationary points by setting \( f'(x) = 0 \).
\( 6(x^2 - 7x + 6) = 0 \)
\( x^2 - 6x - x + 6 = 0 \)
\( x(x-6) - 1(x-6) = 0 \)
\( (x-1)(x-6) = 0 \)
So, \( x = 1 \) or \( x = 6 \).
Step 3: Test the points in \( f''(x) \).
For \( x = 1 \):
\( f''(1) = 12(1) - 42 = -30 < 0 \) (Maxima)
For \( x = 6 \):
\( f''(6) = 12(6) - 42 = 72 - 42 = 30 > 0 \) (Minima)
Step 4: Calculate the minimum value at \( x = 6 \).
\( f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20 \)
\( f(6) = 2(216) - 21(36) + 216 - 20 \)
\( f(6) = 432 - 756 + 216 - 20 \)
\( f(6) = 648 - 776 \)
\( f(6) = -128 \)
The minimum value of the function is -128.
(iii) Find the area of the regions bounded by the line \( y = -2x \), the X-axis and the lines \( x = -1 \) and \( x = 2 \).
Solution:
The area \( A \) is given by \( \int_{-1}^{2} |y| dx \).
Since \( y = -2x \), the line passes through the origin.
- In the interval \([-1, 0]\), \( x \) is negative, so \( y = -2(\text{neg}) = \text{positive} \).
- In the interval \([0, 2]\), \( x \) is positive, so \( y = -2(\text{pos}) = \text{negative} \).
Therefore, we split the integral at \( x = 0 \):
\( A = |\int_{-1}^{0} (-2x) dx| + |\int_{0}^{2} (-2x) dx| \)
Part 1 (from -1 to 0):
\( A_1 = \int_{-1}^{0} -2x dx = -2 [\frac{x^2}{2}]_{-1}^{0} = - [x^2]_{-1}^{0} \)
\( A_1 = - [0^2 - (-1)^2] = - [0 - 1] = 1 \)
Part 2 (from 0 to 2):
\( A_2 = \int_{0}^{2} -2x dx = - [x^2]_{0}^{2} \)
\( A_2 = - [2^2 - 0^2] = -4 \)
Area cannot be negative, so we take absolute value: \( |A_2| = 4 \).
Total Area:
\( A = A_1 + |A_2| = 1 + 4 = 5 \) sq. units.
The area \( A \) is given by \( \int_{-1}^{2} |y| dx \).
Since \( y = -2x \), the line passes through the origin.
- In the interval \([-1, 0]\), \( x \) is negative, so \( y = -2(\text{neg}) = \text{positive} \).
- In the interval \([0, 2]\), \( x \) is positive, so \( y = -2(\text{pos}) = \text{negative} \).
Therefore, we split the integral at \( x = 0 \):
\( A = |\int_{-1}^{0} (-2x) dx| + |\int_{0}^{2} (-2x) dx| \)
Part 1 (from -1 to 0):
\( A_1 = \int_{-1}^{0} -2x dx = -2 [\frac{x^2}{2}]_{-1}^{0} = - [x^2]_{-1}^{0} \)
\( A_1 = - [0^2 - (-1)^2] = - [0 - 1] = 1 \)
Part 2 (from 0 to 2):
\( A_2 = \int_{0}^{2} -2x dx = - [x^2]_{0}^{2} \)
\( A_2 = - [2^2 - 0^2] = -4 \)
Area cannot be negative, so we take absolute value: \( |A_2| = 4 \).
Total Area:
\( A = A_1 + |A_2| = 1 + 4 = 5 \) sq. units.
Q. 3. (A) Attempt any TWO of the following questions (3 marks each):
(i) Find \( \frac{dy}{dx} \) if \( y = x^{e^x} \)
Solution:
Given \( y = x^{e^x} \)
Taking logarithm on both sides:
\( \log y = \log(x^{e^x}) \)
\( \log y = e^x \cdot \log x \)
Differentiating with respect to \( x \):
\( \frac{d}{dx}(\log y) = \frac{d}{dx}(e^x \cdot \log x) \)
\( \frac{1}{y} \frac{dy}{dx} = e^x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(e^x) \) (Product Rule)
\( \frac{1}{y} \frac{dy}{dx} = e^x (\frac{1}{x}) + \log x (e^x) \)
\( \frac{1}{y} \frac{dy}{dx} = e^x (\frac{1}{x} + \log x) \)
\( \frac{dy}{dx} = y \cdot e^x (\frac{1 + x\log x}{x}) \)
Substituting \( y = x^{e^x} \):
\( \frac{dy}{dx} = x^{e^x} e^x (\frac{1 + x\log x}{x}) \)
Given \( y = x^{e^x} \)
Taking logarithm on both sides:
\( \log y = \log(x^{e^x}) \)
\( \log y = e^x \cdot \log x \)
Differentiating with respect to \( x \):
\( \frac{d}{dx}(\log y) = \frac{d}{dx}(e^x \cdot \log x) \)
\( \frac{1}{y} \frac{dy}{dx} = e^x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(e^x) \) (Product Rule)
\( \frac{1}{y} \frac{dy}{dx} = e^x (\frac{1}{x}) + \log x (e^x) \)
\( \frac{1}{y} \frac{dy}{dx} = e^x (\frac{1}{x} + \log x) \)
\( \frac{dy}{dx} = y \cdot e^x (\frac{1 + x\log x}{x}) \)
Substituting \( y = x^{e^x} \):
\( \frac{dy}{dx} = x^{e^x} e^x (\frac{1 + x\log x}{x}) \)
(ii) If \( f'(x) = 4x^3 - 3x^2 + 2x + k \), \( f(0) = 1 \) and \( f(1) = 4 \), find \( f(x) \).
Solution:
Given \( f'(x) = 4x^3 - 3x^2 + 2x + k \)
Integrate both sides to find \( f(x) \):
\( f(x) = \int (4x^3 - 3x^2 + 2x + k) dx \)
\( f(x) = 4\frac{x^4}{4} - 3\frac{x^3}{3} + 2\frac{x^2}{2} + kx + c \)
\( f(x) = x^4 - x^3 + x^2 + kx + c \) ... (Equation 1)
Step 1: Find c
Given \( f(0) = 1 \). Substitute \( x=0 \) in Eq 1:
\( 1 = 0^4 - 0^3 + 0^2 + k(0) + c \)
\( c = 1 \)
Step 2: Find k
Update Eq 1: \( f(x) = x^4 - x^3 + x^2 + kx + 1 \)
Given \( f(1) = 4 \). Substitute \( x=1 \):
\( 4 = 1^4 - 1^3 + 1^2 + k(1) + 1 \)
\( 4 = 1 - 1 + 1 + k + 1 \)
\( 4 = 2 + k \)
\( k = 2 \)
Final Answer:
\( f(x) = x^4 - x^3 + x^2 + 2x + 1 \)
Given \( f'(x) = 4x^3 - 3x^2 + 2x + k \)
Integrate both sides to find \( f(x) \):
\( f(x) = \int (4x^3 - 3x^2 + 2x + k) dx \)
\( f(x) = 4\frac{x^4}{4} - 3\frac{x^3}{3} + 2\frac{x^2}{2} + kx + c \)
\( f(x) = x^4 - x^3 + x^2 + kx + c \) ... (Equation 1)
Step 1: Find c
Given \( f(0) = 1 \). Substitute \( x=0 \) in Eq 1:
\( 1 = 0^4 - 0^3 + 0^2 + k(0) + c \)
\( c = 1 \)
Step 2: Find k
Update Eq 1: \( f(x) = x^4 - x^3 + x^2 + kx + 1 \)
Given \( f(1) = 4 \). Substitute \( x=1 \):
\( 4 = 1^4 - 1^3 + 1^2 + k(1) + 1 \)
\( 4 = 1 - 1 + 1 + k + 1 \)
\( 4 = 2 + k \)
\( k = 2 \)
Final Answer:
\( f(x) = x^4 - x^3 + x^2 + 2x + 1 \)
(iii) Obtain the differential equation whose general solution is \( x^3 + y^3 = 35ax \)
Solution:
Given equation: \( x^3 + y^3 = 35ax \) ... (1)
Here, \( a \) is the arbitrary constant.
Rewrite (1) to isolate \( a \):
\( \frac{x^3 + y^3}{x} = 35a \)
Differentiate w.r.t \( x \) to eliminate \( 35a \):
\( \frac{d}{dx} \left( \frac{x^3 + y^3}{x} \right) = 0 \)
Using Quotient Rule \( \frac{v u' - u v'}{v^2} \):
\( \frac{x \frac{d}{dx}(x^3 + y^3) - (x^3 + y^3) \frac{d}{dx}(x)}{x^2} = 0 \)
\( x(3x^2 + 3y^2 \frac{dy}{dx}) - (x^3 + y^3)(1) = 0 \) (Since \( x^2 \neq 0 \))
\( 3x^3 + 3xy^2 \frac{dy}{dx} - x^3 - y^3 = 0 \)
\( 2x^3 - y^3 + 3xy^2 \frac{dy}{dx} = 0 \)
Required Differential Equation:
\( 3xy^2 \frac{dy}{dx} = y^3 - 2x^3 \)
Given equation: \( x^3 + y^3 = 35ax \) ... (1)
Here, \( a \) is the arbitrary constant.
Rewrite (1) to isolate \( a \):
\( \frac{x^3 + y^3}{x} = 35a \)
Differentiate w.r.t \( x \) to eliminate \( 35a \):
\( \frac{d}{dx} \left( \frac{x^3 + y^3}{x} \right) = 0 \)
Using Quotient Rule \( \frac{v u' - u v'}{v^2} \):
\( \frac{x \frac{d}{dx}(x^3 + y^3) - (x^3 + y^3) \frac{d}{dx}(x)}{x^2} = 0 \)
\( x(3x^2 + 3y^2 \frac{dy}{dx}) - (x^3 + y^3)(1) = 0 \) (Since \( x^2 \neq 0 \))
\( 3x^3 + 3xy^2 \frac{dy}{dx} - x^3 - y^3 = 0 \)
\( 2x^3 - y^3 + 3xy^2 \frac{dy}{dx} = 0 \)
Required Differential Equation:
\( 3xy^2 \frac{dy}{dx} = y^3 - 2x^3 \)
Q. 3. (B) Attempt any ONE of the following questions (4 marks each):
(i) Find the inverse of \( A = \begin{bmatrix} 3 & 1 & 5 \\ 2 & 7 & 8 \\ 1 & 2 & 5 \end{bmatrix} \) by adjoint method.
Solution:
Let \( A = \begin{bmatrix} 3 & 1 & 5 \\ 2 & 7 & 8 \\ 1 & 2 & 5 \end{bmatrix} \)
Step 1: Find the determinant \(|A|\)
\( |A| = 3(35 - 16) - 1(10 - 8) + 5(4 - 7) \)
\( |A| = 3(19) - 1(2) + 5(-3) \)
\( |A| = 57 - 2 - 15 = 40 \)
Since \( |A| \neq 0 \), \( A^{-1} \) exists.
Step 2: Find the Cofactors
\( A_{11} = (-1)^{1+1}(35-16) = 19 \)
\( A_{12} = (-1)^{1+2}(10-8) = -2 \)
\( A_{13} = (-1)^{1+3}(4-7) = -3 \)
\( A_{21} = (-1)^{2+1}(5-10) = -(-5) = 5 \)
\( A_{22} = (-1)^{2+2}(15-5) = 10 \)
\( A_{23} = (-1)^{2+3}(6-1) = -5 \)
\( A_{31} = (-1)^{3+1}(8-35) = -27 \)
\( A_{32} = (-1)^{3+2}(24-10) = -14 \)
\( A_{33} = (-1)^{3+3}(21-2) = 19 \)
Step 3: Matrix of Cofactors and Adjoint
Cofactor Matrix \( C = \begin{bmatrix} 19 & -2 & -3 \\ 5 & 10 & -5 \\ -27 & -14 & 19 \end{bmatrix} \)
\( \text{adj } A = C^T = \begin{bmatrix} 19 & 5 & -27 \\ -2 & 10 & -14 \\ -3 & -5 & 19 \end{bmatrix} \)
Step 4: Find Inverse
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( A^{-1} = \frac{1}{40} \begin{bmatrix} 19 & 5 & -27 \\ -2 & 10 & -14 \\ -3 & -5 & 19 \end{bmatrix} \)
Let \( A = \begin{bmatrix} 3 & 1 & 5 \\ 2 & 7 & 8 \\ 1 & 2 & 5 \end{bmatrix} \)
Step 1: Find the determinant \(|A|\)
\( |A| = 3(35 - 16) - 1(10 - 8) + 5(4 - 7) \)
\( |A| = 3(19) - 1(2) + 5(-3) \)
\( |A| = 57 - 2 - 15 = 40 \)
Since \( |A| \neq 0 \), \( A^{-1} \) exists.
Step 2: Find the Cofactors
\( A_{11} = (-1)^{1+1}(35-16) = 19 \)
\( A_{12} = (-1)^{1+2}(10-8) = -2 \)
\( A_{13} = (-1)^{1+3}(4-7) = -3 \)
\( A_{21} = (-1)^{2+1}(5-10) = -(-5) = 5 \)
\( A_{22} = (-1)^{2+2}(15-5) = 10 \)
\( A_{23} = (-1)^{2+3}(6-1) = -5 \)
\( A_{31} = (-1)^{3+1}(8-35) = -27 \)
\( A_{32} = (-1)^{3+2}(24-10) = -14 \)
\( A_{33} = (-1)^{3+3}(21-2) = 19 \)
Step 3: Matrix of Cofactors and Adjoint
Cofactor Matrix \( C = \begin{bmatrix} 19 & -2 & -3 \\ 5 & 10 & -5 \\ -27 & -14 & 19 \end{bmatrix} \)
\( \text{adj } A = C^T = \begin{bmatrix} 19 & 5 & -27 \\ -2 & 10 & -14 \\ -3 & -5 & 19 \end{bmatrix} \)
Step 4: Find Inverse
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( A^{-1} = \frac{1}{40} \begin{bmatrix} 19 & 5 & -27 \\ -2 & 10 & -14 \\ -3 & -5 & 19 \end{bmatrix} \)
(ii) The consumption expenditure \( E_c \) of a person with income \( x \), is given by \( E_c = 0.0006x^2 + 0.003x \).
Find average propensity to consume (APC), marginal propensity to consume (MPC) when his income is 200. Also find his marginal propensity to save (MPS).
Find average propensity to consume (APC), marginal propensity to consume (MPC) when his income is 200. Also find his marginal propensity to save (MPS).
Solution:
Given \( E_c = 0.0006x^2 + 0.003x \) and \( x = 200 \).
1. Average Propensity to Consume (APC)
\( \text{APC} = \frac{E_c}{x} = \frac{0.0006x^2 + 0.003x}{x} \)
\( \text{APC} = 0.0006x + 0.003 \)
At \( x = 200 \):
\( \text{APC} = 0.0006(200) + 0.003 \)
\( \text{APC} = 0.12 + 0.003 = \mathbf{0.123} \)
2. Marginal Propensity to Consume (MPC)
\( \text{MPC} = \frac{d(E_c)}{dx} = \frac{d}{dx}(0.0006x^2 + 0.003x) \)
\( \text{MPC} = 0.0006(2x) + 0.003 \)
\( \text{MPC} = 0.0012x + 0.003 \)
At \( x = 200 \):
\( \text{MPC} = 0.0012(200) + 0.003 \)
\( \text{MPC} = 0.24 + 0.003 = \mathbf{0.243} \)
3. Marginal Propensity to Save (MPS)
We know that \( \text{MPC} + \text{MPS} = 1 \)
\( \text{MPS} = 1 - \text{MPC} \)
\( \text{MPS} = 1 - 0.243 \)
\( \text{MPS} = \mathbf{0.757} \)
Given \( E_c = 0.0006x^2 + 0.003x \) and \( x = 200 \).
1. Average Propensity to Consume (APC)
\( \text{APC} = \frac{E_c}{x} = \frac{0.0006x^2 + 0.003x}{x} \)
\( \text{APC} = 0.0006x + 0.003 \)
At \( x = 200 \):
\( \text{APC} = 0.0006(200) + 0.003 \)
\( \text{APC} = 0.12 + 0.003 = \mathbf{0.123} \)
2. Marginal Propensity to Consume (MPC)
\( \text{MPC} = \frac{d(E_c)}{dx} = \frac{d}{dx}(0.0006x^2 + 0.003x) \)
\( \text{MPC} = 0.0006(2x) + 0.003 \)
\( \text{MPC} = 0.0012x + 0.003 \)
At \( x = 200 \):
\( \text{MPC} = 0.0012(200) + 0.003 \)
\( \text{MPC} = 0.24 + 0.003 = \mathbf{0.243} \)
3. Marginal Propensity to Save (MPS)
We know that \( \text{MPC} + \text{MPS} = 1 \)
\( \text{MPS} = 1 - \text{MPC} \)
\( \text{MPS} = 1 - 0.243 \)
\( \text{MPS} = \mathbf{0.757} \)
Q. 3. (C) Attempt any ONE of the following questions (Activity) (4 marks each):
(i) Complete the following activity to evaluate \( \int_{0}^{2} \frac{dx}{4+x-x^2} \)
Solution:
\( I = \int_{0}^{2} \frac{dx}{4+x-x^2} = \int_{0}^{2} \frac{dx}{-x^2 + \mathbf{\fbox{ x }} + \mathbf{\fbox{ 4 }}} \)
\( = \int_{0}^{2} \frac{dx}{-x^2 + x + \frac{1}{4} - \mathbf{\fbox{ \(\frac{1}{4}\) }} + 4} \)
\( = -\int_{0}^{2} \frac{dx}{(x - \frac{1}{2})^2 - (\mathbf{\fbox{ \(\frac{\sqrt{17}}{2}\) }})^2} \)
Using formula \( \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \), but with negative sign outside:
\( = \frac{1}{2(\frac{\sqrt{17}}{2})} \log \left( \frac{\frac{\sqrt{17}}{2} + (x - \frac{1}{2})}{\frac{\sqrt{17}}{2} - (x - \frac{1}{2})} \right) \Bigg|_{0}^{2} \)
\( = \frac{1}{\sqrt{17}} \log \left( \frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right) \)
\( I = \int_{0}^{2} \frac{dx}{4+x-x^2} = \int_{0}^{2} \frac{dx}{-x^2 + \mathbf{\fbox{ x }} + \mathbf{\fbox{ 4 }}} \)
\( = \int_{0}^{2} \frac{dx}{-x^2 + x + \frac{1}{4} - \mathbf{\fbox{ \(\frac{1}{4}\) }} + 4} \)
\( = -\int_{0}^{2} \frac{dx}{(x - \frac{1}{2})^2 - (\mathbf{\fbox{ \(\frac{\sqrt{17}}{2}\) }})^2} \)
Using formula \( \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \), but with negative sign outside:
\( = \frac{1}{2(\frac{\sqrt{17}}{2})} \log \left( \frac{\frac{\sqrt{17}}{2} + (x - \frac{1}{2})}{\frac{\sqrt{17}}{2} - (x - \frac{1}{2})} \right) \Bigg|_{0}^{2} \)
\( = \frac{1}{\sqrt{17}} \log \left( \frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right) \)
(ii) The rate of growth of population is proportional to the number of inhabitants. If the population doubles in 25 years and the present population is 1,00,000, when will the city have population 4,00,000?
Solution:
Let \( P \) be the population at time \( t \).
\( \frac{dP}{dt} \propto P \Rightarrow \frac{dP}{dt} = kP \Rightarrow \frac{dP}{P} = k \cdot dt \)
On integrating, \( \log P = kt + c \) ... (i)
(a) When \( t = 0 \), \( P = 1,00,000 \)
From (i), \( \log(1,00,000) = k(0) + c \)
\( \therefore c = \) \( \log(1,00,000) \)
Equation becomes: \( \log \left( \frac{P}{1,00,000} \right) = kt \) ... (ii)
(b) When \( t = 25 \), \( P = 2,00,000 \) (doubles)
From (ii), \( \log \left( \frac{2,00,000}{1,00,000} \right) = 25k \)
\( \log 2 = 25k \)
\( \therefore k = \) \( \frac{1}{25} \log 2 \)
(c) When \( P = 4,00,000 \), find \( t \).
\( \log \left( \frac{4,00,000}{1,00,000} \right) = \left( \frac{1}{25} \log 2 \right) \cdot t \)
\( \log 4 = \frac{t}{25} \log 2 \)
\( 2 \log 2 = \frac{t}{25} \log 2 \)
\( \therefore t = \) 50 years.
Let \( P \) be the population at time \( t \).
\( \frac{dP}{dt} \propto P \Rightarrow \frac{dP}{dt} = kP \Rightarrow \frac{dP}{P} = k \cdot dt \)
On integrating, \( \log P = kt + c \) ... (i)
(a) When \( t = 0 \), \( P = 1,00,000 \)
From (i), \( \log(1,00,000) = k(0) + c \)
\( \therefore c = \) \( \log(1,00,000) \)
Equation becomes: \( \log \left( \frac{P}{1,00,000} \right) = kt \) ... (ii)
(b) When \( t = 25 \), \( P = 2,00,000 \) (doubles)
From (ii), \( \log \left( \frac{2,00,000}{1,00,000} \right) = 25k \)
\( \log 2 = 25k \)
\( \therefore k = \) \( \frac{1}{25} \log 2 \)
(c) When \( P = 4,00,000 \), find \( t \).
\( \log \left( \frac{4,00,000}{1,00,000} \right) = \left( \frac{1}{25} \log 2 \right) \cdot t \)
\( \log 4 = \frac{t}{25} \log 2 \)
\( 2 \log 2 = \frac{t}{25} \log 2 \)
\( \therefore t = \) 50 years.
SECTION - II
Q. 4. (A) Select and write the correct answer (1 mark each)
(i) The difference between face value and present worth is called
Answer: (b) True discount
(ii) In an ordinary annuity, payments or receipts occur at
Answer: (b) end of each period
(iii) \(b_{xy}\) and \(b_{yx}\) are
Answer: (b) Independent of change of origin but not of scale
(iv) Dorbish-Bowley's Price Index Number is given by
Answer: (a) \(\frac{\frac{\Sigma p_{1}q_{0}}{\Sigma p_{0}q_{0}} + \frac{\Sigma p_{1}q_{1}}{\Sigma p_{0}q_{1}}}{2} \times 100\)
It is the Arithmetic Mean of Laspeyres and Paasche Index Numbers.
(v) Objective function of L.P.P. is
Answer: (b) a function to be maximised or minimised
(vi) To use the Hungarian method, a profit maximization assignment problem requires
Answer: (a) Converting all profits to opportunity losses
Hungarian method is a minimization algorithm. Maximization problems must first be converted to minimization (opportunity loss) by subtracting all elements from the largest element.
Q. 4. (B) State whether True or False (1 mark each)
- (i) Broker is an agent who gives a guarantee to seller that the buyer will pay the selling price of goods.
Answer: False (This describes a Del Credere Agent). - (ii) \(\sum \frac{p_0 q_0}{p_1 q_1} \times 100\) is the Value Index Number by simple aggregate method.
Answer: False - (iii) The optimum value of the objective function of L.P.P. occurs at the center of the feasible region.
Answer: False (It occurs at a vertex/corner point).
Q. 4. (C) Fill in the blanks (1 mark each)
- (i) The banker's discount is always greater than the true discount.
- (ii) The cost of living index number using Weighted Relative Method is given by \(\frac{\sum IW}{\sum W}\).
- (iii) The time interval between starting the first job and completing the last job including the idle time is called Total Elapsed Time.
Q. 5 & 6. Key Subjective Solutions & Activities
Q. 5. (A) Attempt any TWO of the following questions (3 marks each):
(i) Deepak's salary was increased from ₹ 4,000 to ₹ 5,000. The sales being the same, due to reduction in the rate of commission from 3% to 2%, his income remains unchanged. Find his sales.
Solution:
Let the Sales be \( x \).
Condition 1 (Old):
Salary = ₹ 4,000
Commission Rate = 3%
Total Income = \( 4000 + \frac{3}{100}x = 4000 + 0.03x \)
Condition 2 (New):
Salary = ₹ 5,000
Commission Rate = 2%
Total Income = \( 5000 + \frac{2}{100}x = 5000 + 0.02x \)
Since the income remains unchanged:
\( 4000 + 0.03x = 5000 + 0.02x \)
\( 0.03x - 0.02x = 5000 - 4000 \)
\( 0.01x = 1000 \)
\( x = \frac{1000}{0.01} \)
\( x = 1,00,000 \)
Ans: Deepak's sales are ₹ 1,00,000.
Let the Sales be \( x \).
Condition 1 (Old):
Salary = ₹ 4,000
Commission Rate = 3%
Total Income = \( 4000 + \frac{3}{100}x = 4000 + 0.03x \)
Condition 2 (New):
Salary = ₹ 5,000
Commission Rate = 2%
Total Income = \( 5000 + \frac{2}{100}x = 5000 + 0.02x \)
Since the income remains unchanged:
\( 4000 + 0.03x = 5000 + 0.02x \)
\( 0.03x - 0.02x = 5000 - 4000 \)
\( 0.01x = 1000 \)
\( x = \frac{1000}{0.01} \)
\( x = 1,00,000 \)
Ans: Deepak's sales are ₹ 1,00,000.
(ii) For a bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of variance of Y if variance of X is 9.
Solution:
Given:
\( b_{yx} = 0.4 \) (Regression coefficient of Y on X)
\( b_{xy} = 0.9 \) (Regression coefficient of X on Y)
Variance of X \( V(X) = \sigma_x^2 = 9 \Rightarrow \sigma_x = 3 \)
We know the relationship:
\( b_{yx} = r \frac{\sigma_y}{\sigma_x} \) and \( b_{xy} = r \frac{\sigma_x}{\sigma_y} \)
Multiplying both:
\( b_{yx} \times b_{xy} = r^2 \)
\( 0.4 \times 0.9 = r^2 \)
\( r^2 = 0.36 \Rightarrow r = 0.6 \) (Taking positive root as both \( b_{yx}, b_{xy} > 0 \))
Now, using \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \):
\( 0.4 = 0.6 \times \frac{\sigma_y}{3} \)
\( 0.4 = 0.2 \sigma_y \)
\( \sigma_y = \frac{0.4}{0.2} = 2 \)
Variance of Y \( V(Y) = \sigma_y^2 = 2^2 = 4 \).
Ans: Variance of Y is 4.
Given:
\( b_{yx} = 0.4 \) (Regression coefficient of Y on X)
\( b_{xy} = 0.9 \) (Regression coefficient of X on Y)
Variance of X \( V(X) = \sigma_x^2 = 9 \Rightarrow \sigma_x = 3 \)
We know the relationship:
\( b_{yx} = r \frac{\sigma_y}{\sigma_x} \) and \( b_{xy} = r \frac{\sigma_x}{\sigma_y} \)
Multiplying both:
\( b_{yx} \times b_{xy} = r^2 \)
\( 0.4 \times 0.9 = r^2 \)
\( r^2 = 0.36 \Rightarrow r = 0.6 \) (Taking positive root as both \( b_{yx}, b_{xy} > 0 \))
Now, using \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \):
\( 0.4 = 0.6 \times \frac{\sigma_y}{3} \)
\( 0.4 = 0.2 \sigma_y \)
\( \sigma_y = \frac{0.4}{0.2} = 2 \)
Variance of Y \( V(Y) = \sigma_y^2 = 2^2 = 4 \).
Ans: Variance of Y is 4.
(iii) The following table shows the index of industrial production for the period from 1976 to 1985... Obtain the trend values using 4 yearly centered moving averages.
Solution:
Trend Values: 2.25, 2.75, 3.25, 3.875, 4.875, 6.25.
| Year | Index (Y) | 4-Yearly Moving Total |
Centered Total (Sum of 2) |
Trend Value (Centered Total / 8) |
|---|---|---|---|---|
| 1976 | 0 | - | - | - |
| 1977 | 2 | - | - | - |
| 8 | ||||
| 1978 | 3 | 18 | 2.25 | |
| 10 | ||||
| 1979 | 3 | 22 | 2.75 | |
| 12 | ||||
| 1980 | 2 | 26 | 3.25 | |
| 14 | ||||
| 1981 | 4 | 31 | 3.875 | |
| 17 | ||||
| 1982 | 5 | 39 | 4.875 | |
| 22 | ||||
| 1983 | 6 | 50 | 6.25 | |
| 28 | - | - | ||
| 1984 | 7 | - | - | - |
| 1985 | 10 | - | - | - |
Q. 5. (B) Attempt any TWO of the following questions (4 marks each):
(i) If for the following data, Walsh's Price Index Number is 150, find \( x \):
Solution:
Walsh's Price Index Formula: \( P_{01}(w) = \frac{\sum p_1 \sqrt{q_0 q_1}}{\sum p_0 \sqrt{q_0 q_1}} \times 100 \)
Let \( W = \sqrt{q_0 q_1} \).
Given \( P_{01}(w) = 150 \):
\( 150 = \frac{345}{130 + 6x} \times 100 \)
\( \frac{150}{100} = \frac{345}{130 + 6x} \)
\( \frac{3}{2} = \frac{345}{130 + 6x} \)
\( 130 + 6x = \frac{345 \times 2}{3} \)
\( 130 + 6x = 115 \times 2 = 230 \)
\( 6x = 230 - 130 \)
\( 6x = 100 \)
\( x = \frac{100}{6} = \mathbf{16.67} \)
Walsh's Price Index Formula: \( P_{01}(w) = \frac{\sum p_1 \sqrt{q_0 q_1}}{\sum p_0 \sqrt{q_0 q_1}} \times 100 \)
Let \( W = \sqrt{q_0 q_1} \).
| Comm. | \(p_0\) | \(q_0\) | \(p_1\) | \(q_1\) | \(W = \sqrt{q_0 q_1}\) | \(p_0 W\) | \(p_1 W\) |
|---|---|---|---|---|---|---|---|
| A | 5 | 3 | 10 | 3 | \(\sqrt{9}=3\) | 15 | 30 |
| B | \(x\) | 4 | 16 | 9 | \(\sqrt{36}=6\) | \(6x\) | 96 |
| C | 15 | 5 | 23 | 5 | \(\sqrt{25}=5\) | 75 | 115 |
| D | 10 | 2 | 26 | 8 | \(\sqrt{16}=4\) | 40 | 104 |
| Total | - | - | - | - | - | \(130 + 6x\) | 345 |
\( 150 = \frac{345}{130 + 6x} \times 100 \)
\( \frac{150}{100} = \frac{345}{130 + 6x} \)
\( \frac{3}{2} = \frac{345}{130 + 6x} \)
\( 130 + 6x = \frac{345 \times 2}{3} \)
\( 130 + 6x = 115 \times 2 = 230 \)
\( 6x = 230 - 130 \)
\( 6x = 100 \)
\( x = \frac{100}{6} = \mathbf{16.67} \)
(ii) A toy manufacturing company produces five types of toys. Each toy has to go through three machines A, B and C in the order ABC. Find the total elapsed time and also find idle time for machine B.
Solution:
Step 1: Check condition for 3 machines
Min \(A = 12\) (Type 3), Min \(C = 8\) (Type 1), Max \(B = 12\) (Type 2).
Since Min \(A \ge\) Max \(B\) (\(12 \ge 12\)), the condition is satisfied.
Step 2: Convert to 2 fictitious machines G and H
\(G_i = A_i + B_i\) and \(H_i = B_i + C_i\)
Step 3: Determine Sequence (Johnson's Algorithm)
Min value is 16 (Type 3 on G) &to; Place 3 first.
Next Min is 18 (Type 1, 4, 5 on H) &to; Place 1, 4, 5 last (order can vary, e.g., 5-4-1).
Remaining is Type 2.
Optimal Sequence: 3 - 2 - 5 - 4 - 1
Step 4: Total Elapsed Time Table
Total Elapsed Time (T) = 102 hours.
Step 5: Idle Time for Machine B
Idle time = Total Time - Sum of processing times of B
Sum of B = \(10+12+4+6+8 = 40\) hours.
Looking at the table, B is idle from:
0 to 12 (12 hrs), 16 to 32 (16 hrs), 44 to 54 (10 hrs), 62 to 68 (6 hrs), 74 to 84 (10 hrs), 94 to 102 (8 hrs).
Total Idle Time for B = \(102 - 40 = \mathbf{62 \text{ hours}}\).
Step 1: Check condition for 3 machines
Min \(A = 12\) (Type 3), Min \(C = 8\) (Type 1), Max \(B = 12\) (Type 2).
Since Min \(A \ge\) Max \(B\) (\(12 \ge 12\)), the condition is satisfied.
Step 2: Convert to 2 fictitious machines G and H
\(G_i = A_i + B_i\) and \(H_i = B_i + C_i\)
| Type | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| G (A+B) | 26 | 32 | 16 | 20 | 30 |
| H (B+C) | 18 | 30 | 20 | 18 | 18 |
Min value is 16 (Type 3 on G) &to; Place 3 first.
Next Min is 18 (Type 1, 4, 5 on H) &to; Place 1, 4, 5 last (order can vary, e.g., 5-4-1).
Remaining is Type 2.
Optimal Sequence: 3 - 2 - 5 - 4 - 1
Step 4: Total Elapsed Time Table
| Seq | Machine A | Machine B | Machine C | |||
|---|---|---|---|---|---|---|
| In | Out | In | Out | In | Out | |
| 3 | 0 | 12 | 12 | 16 | 16 | 32 |
| 2 | 12 | 32 | 32 | 44 | 44 | 62 |
| 5 | 32 | 54 | 54 | 62 | 62 | 72 |
| 4 | 54 | 68 | 68 | 74 | 74 | 86 |
| 1 | 68 | 84 | 84 | 94 | 94 | 102 |
Total Elapsed Time (T) = 102 hours.
Step 5: Idle Time for Machine B
Idle time = Total Time - Sum of processing times of B
Sum of B = \(10+12+4+6+8 = 40\) hours.
Looking at the table, B is idle from:
0 to 12 (12 hrs), 16 to 32 (16 hrs), 44 to 54 (10 hrs), 62 to 68 (6 hrs), 74 to 84 (10 hrs), 94 to 102 (8 hrs).
Total Idle Time for B = \(102 - 40 = \mathbf{62 \text{ hours}}\).
(iii) A random variable X has the following probability distribution:
Find (a) k (b) \(P(X < 3)\) (c) \(P(X > 6)\)
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | k | 2k | 2k | 3k | \(k^2\) | \(2k^2\) | \(7k^2+k\) |
Solution:
(a) Since \(\sum P(x) = 1\):
\( k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1 \)
\( 10k^2 + 9k = 1 \)
\( 10k^2 + 9k - 1 = 0 \)
\( 10k^2 + 10k - k - 1 = 0 \)
\( 10k(k+1) - 1(k+1) = 0 \)
\( (10k - 1)(k + 1) = 0 \)
\( k = \frac{1}{10} \) or \( k = -1 \).
Since probability cannot be negative, \( k = 0.1 \).
(b) \( P(X < 3) = P(X=1) + P(X=2) \)
\( = k + 2k = 3k \)
\( = 3(0.1) = \mathbf{0.3} \)
(c) \( P(X > 6) = P(X=7) \)
\( = 7k^2 + k \)
\( = 7(0.1)^2 + 0.1 \)
\( = 7(0.01) + 0.1 \)
\( = 0.07 + 0.1 = \mathbf{0.17} \)
(a) Since \(\sum P(x) = 1\):
\( k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1 \)
\( 10k^2 + 9k = 1 \)
\( 10k^2 + 9k - 1 = 0 \)
\( 10k^2 + 10k - k - 1 = 0 \)
\( 10k(k+1) - 1(k+1) = 0 \)
\( (10k - 1)(k + 1) = 0 \)
\( k = \frac{1}{10} \) or \( k = -1 \).
Since probability cannot be negative, \( k = 0.1 \).
(b) \( P(X < 3) = P(X=1) + P(X=2) \)
\( = k + 2k = 3k \)
\( = 3(0.1) = \mathbf{0.3} \)
(c) \( P(X > 6) = P(X=7) \)
\( = 7k^2 + k \)
\( = 7(0.1)^2 + 0.1 \)
\( = 7(0.01) + 0.1 \)
\( = 0.07 + 0.1 = \mathbf{0.17} \)
Q. 6. (A) Attempt any TWO of the following questions (3 marks each):
(i) The building is insured for 75% of its value. The annual premium at 0.70 percent amounts to ₹ 2,625. If the building is damaged to the extent of 60% due to fire, how much can be claimed under the policy?
Solution:
Let the Property Value be \( V \).
Policy Value = 75% of \( V = 0.75V \).
Rate of Premium = 0.70% = \( 0.007 \).
Premium = ₹ 2,625.
Step 1: Find Property Value
\( \text{Premium} = \text{Policy Value} \times \text{Rate} \)
\( 2625 = 0.75V \times \frac{0.70}{100} \)
\( 2625 = V \times 0.00525 \)
\( V = \frac{2625}{0.00525} = 5,00,000 \)
Property Value = ₹ 5,00,000.
Policy Value = \( 0.75 \times 5,00,000 = \text{₹ } 3,75,000 \).
Step 2: Calculate Claim
Loss = 60% of Value = \( 0.60 \times 5,00,000 = \text{₹ } 3,00,000 \).
Since the property is under-insured (Policy < Value), the Average Clause applies.
\( \text{Claim} = \frac{\text{Policy Value}}{\text{Property Value}} \times \text{Loss} \)
\( \text{Claim} = \frac{3,75,000}{5,00,000} \times 3,00,000 \)
\( \text{Claim} = 0.75 \times 3,00,000 \)
\( \text{Claim} = \mathbf{2,25,000} \)
Ans: The amount that can be claimed is ₹ 2,25,000.
Let the Property Value be \( V \).
Policy Value = 75% of \( V = 0.75V \).
Rate of Premium = 0.70% = \( 0.007 \).
Premium = ₹ 2,625.
Step 1: Find Property Value
\( \text{Premium} = \text{Policy Value} \times \text{Rate} \)
\( 2625 = 0.75V \times \frac{0.70}{100} \)
\( 2625 = V \times 0.00525 \)
\( V = \frac{2625}{0.00525} = 5,00,000 \)
Property Value = ₹ 5,00,000.
Policy Value = \( 0.75 \times 5,00,000 = \text{₹ } 3,75,000 \).
Step 2: Calculate Claim
Loss = 60% of Value = \( 0.60 \times 5,00,000 = \text{₹ } 3,00,000 \).
Since the property is under-insured (Policy < Value), the Average Clause applies.
\( \text{Claim} = \frac{\text{Policy Value}}{\text{Property Value}} \times \text{Loss} \)
\( \text{Claim} = \frac{3,75,000}{5,00,000} \times 3,00,000 \)
\( \text{Claim} = 0.75 \times 3,00,000 \)
\( \text{Claim} = \mathbf{2,25,000} \)
Ans: The amount that can be claimed is ₹ 2,25,000.
(ii) Three new machines \(M_1, M_2, M_3\) are to be installed in a machine shop. There are four vacant places A, B, C, D. Due to limited space, machine \(M_2\) cannot be placed at B. The cost matrix (in hundred) is as follows. Determine the optimum assignment schedule and find the minimum cost.
(Note: Assuming standard A, B, C, D column order based on typical problem structures).
| A | B | C | D | |
|---|---|---|---|---|
| \(M_1\) | 13 | 10 | 12 | 11 |
| \(M_2\) | 15 | - | 13 | 20 |
| \(M_3\) | 5 | 7 | 10 | 6 |
Solution:
Since there are 3 Machines and 4 Places, this is an unbalanced problem. We add a dummy machine \(M_4\) with 0 cost. Also, assign infinite cost to \(M_2\) at B.
Step 1: Matrix Setup
Step 2: Row Reduction (Subtract min from each row)
\(R_1\) (min 10): 3, 0, 2, 1
\(R_2\) (min 13): 2, ∞, 0, 7
\(R_3\) (min 5): 0, 2, 5, 1
\(R_4\) (min 0): 0, 0, 0, 0
Step 3: Make Assignments (Zero selection)
- \(R_2\) has single zero at C. Assign \(M_2 \to C\) (Cost 13).
- \(R_1\) has single zero at B. Assign \(M_1 \to B\) (Cost 10).
- \(R_3\) has single zero at A. Assign \(M_3 \to A\) (Cost 5).
- \(R_4\) has zero at D (others taken). Assign \(M_4 \to D\) (Cost 0).
Optimal Schedule:
\(M_1 \to B\) (10)
\(M_2 \to C\) (13)
\(M_3 \to A\) (5)
(Place D is left vacant)
Minimum Cost = 10 + 13 + 5 = 28 (hundreds) = ₹ 2,800.
Since there are 3 Machines and 4 Places, this is an unbalanced problem. We add a dummy machine \(M_4\) with 0 cost. Also, assign infinite cost to \(M_2\) at B.
Step 1: Matrix Setup
| 13 | 10 | 12 | 11 |
| 15 | ∞ | 13 | 20 |
| 5 | 7 | 10 | 6 |
| 0 | 0 | 0 | 0 |
Step 2: Row Reduction (Subtract min from each row)
\(R_1\) (min 10): 3, 0, 2, 1
\(R_2\) (min 13): 2, ∞, 0, 7
\(R_3\) (min 5): 0, 2, 5, 1
\(R_4\) (min 0): 0, 0, 0, 0
Step 3: Make Assignments (Zero selection)
- \(R_2\) has single zero at C. Assign \(M_2 \to C\) (Cost 13).
- \(R_1\) has single zero at B. Assign \(M_1 \to B\) (Cost 10).
- \(R_3\) has single zero at A. Assign \(M_3 \to A\) (Cost 5).
- \(R_4\) has zero at D (others taken). Assign \(M_4 \to D\) (Cost 0).
Optimal Schedule:
\(M_1 \to B\) (10)
\(M_2 \to C\) (13)
\(M_3 \to A\) (5)
(Place D is left vacant)
Minimum Cost = 10 + 13 + 5 = 28 (hundreds) = ₹ 2,800.
(iii) The eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg in the lot of 10 eggs.
Solution:
This is a Binomial Distribution problem \( X \sim B(n, p) \).
Probability of defective egg \( p = 10\% = 0.1 \).
Probability of non-defective egg \( q = 1 - 0.1 = 0.9 \).
Number of trials \( n = 10 \).
We need to find probability of at least one defective egg: \( P(X \ge 1) \).
\( P(X \ge 1) = 1 - P(X = 0) \)
\( P(X = 0) = {^{10}C_0} (0.1)^0 (0.9)^{10} = 1 \cdot 1 \cdot (0.9)^{10} \)
\( P(X \ge 1) = 1 - (0.9)^{10} \)
Ans: \( 1 - (0.9)^{10} \)
This is a Binomial Distribution problem \( X \sim B(n, p) \).
Probability of defective egg \( p = 10\% = 0.1 \).
Probability of non-defective egg \( q = 1 - 0.1 = 0.9 \).
Number of trials \( n = 10 \).
We need to find probability of at least one defective egg: \( P(X \ge 1) \).
\( P(X \ge 1) = 1 - P(X = 0) \)
\( P(X = 0) = {^{10}C_0} (0.1)^0 (0.9)^{10} = 1 \cdot 1 \cdot (0.9)^{10} \)
\( P(X \ge 1) = 1 - (0.9)^{10} \)
Ans: \( 1 - (0.9)^{10} \)
Q. 6. (B) Attempt any ONE of the following questions (4 marks each):
(i) Following table shows the all India infant mortality rates (per '000) for years 1980 to 2010. Fit the trend line to the above data by the method of least squares.
Solution:
Let \( n = 7 \). Middle year is 1995. Let \( u = \frac{Year - 1995}{5} \).
Equation of trend line: \( y = a + bu \)
\( a = \frac{\sum y}{n} = \frac{30}{7} \approx 4.286 \)
\( b = \frac{\sum uy}{\sum u^2} = \frac{-44}{28} = -\frac{11}{7} \approx -1.571 \)
Trend Line Equation:
\( y = 4.286 - 1.571 \left( \frac{t - 1995}{5} \right) \)
Let \( n = 7 \). Middle year is 1995. Let \( u = \frac{Year - 1995}{5} \).
| Year (t) | IMR (y) | \(u\) | \(u^2\) | \(uy\) |
|---|---|---|---|---|
| 1980 | 10 | -3 | 9 | -30 |
| 1985 | 7 | -2 | 4 | -14 |
| 1990 | 5 | -1 | 1 | -5 |
| 1995 | 4 | 0 | 0 | 0 |
| 2000 | 3 | 1 | 1 | 3 |
| 2005 | 1 | 2 | 4 | 2 |
| 2010 | 0 | 3 | 9 | 0 |
| Total | \(\sum y = 30\) | \(\sum u = 0\) | \(\sum u^2 = 28\) | \(\sum uy = -44\) |
\( a = \frac{\sum y}{n} = \frac{30}{7} \approx 4.286 \)
\( b = \frac{\sum uy}{\sum u^2} = \frac{-44}{28} = -\frac{11}{7} \approx -1.571 \)
Trend Line Equation:
\( y = 4.286 - 1.571 \left( \frac{t - 1995}{5} \right) \)
(ii) Minimize: \( z = 6x + 2y \)
Subject to: \( x+2y \ge 3 \), \( x+4y \ge 4 \), \( 3x+y \ge 3 \), \( x \ge 0, y \ge 0 \).
Subject to: \( x+2y \ge 3 \), \( x+4y \ge 4 \), \( 3x+y \ge 3 \), \( x \ge 0, y \ge 0 \).
Solution:
We plot the lines and find the feasible region (unbounded above).
1. \( x + 2y = 3 \) passes through (3, 0) and (0, 1.5).
2. \( x + 4y = 4 \) passes through (4, 0) and (0, 1).
3. \( 3x + y = 3 \) passes through (1, 0) and (0, 3).
Corner Points of the Feasible Region:
A(0, 3) [Y-intercept of line 3]
B: Intersection of \( 3x+y=3 \) and \( x+2y=3 \).
Solving: Multiply (ii) by 2: \( 6x+2y=6 \). Subtract \( x+2y=3 \).
\( 5x = 3 \Rightarrow x = 0.6 \). Then \( y = 3 - 3(0.6) = 1.2 \). B(0.6, 1.2).
C: Intersection of \( x+2y=3 \) and \( x+4y=4 \).
Solving: \( 2y = 1 \Rightarrow y = 0.5 \). Then \( x = 3 - 1 = 2 \). C(2, 0.5).
D(4, 0) [X-intercept of line 2].
Value of Z = 6x + 2y:
At A(0, 3): \( 6(0) + 2(3) = 6 \)
At B(0.6, 1.2): \( 6(0.6) + 2(1.2) = 3.6 + 2.4 = \mathbf{6} \)
At C(2, 0.5): \( 6(2) + 2(0.5) = 12 + 1 = 13 \)
At D(4, 0): \( 6(4) + 2(0) = 24 \)
Ans: The minimum value is 6. It occurs at point A(0, 3) and B(0.6, 1.2) and all points on the segment AB.
We plot the lines and find the feasible region (unbounded above).
1. \( x + 2y = 3 \) passes through (3, 0) and (0, 1.5).
2. \( x + 4y = 4 \) passes through (4, 0) and (0, 1).
3. \( 3x + y = 3 \) passes through (1, 0) and (0, 3).
Corner Points of the Feasible Region:
A(0, 3) [Y-intercept of line 3]
B: Intersection of \( 3x+y=3 \) and \( x+2y=3 \).
Solving: Multiply (ii) by 2: \( 6x+2y=6 \). Subtract \( x+2y=3 \).
\( 5x = 3 \Rightarrow x = 0.6 \). Then \( y = 3 - 3(0.6) = 1.2 \). B(0.6, 1.2).
C: Intersection of \( x+2y=3 \) and \( x+4y=4 \).
Solving: \( 2y = 1 \Rightarrow y = 0.5 \). Then \( x = 3 - 1 = 2 \). C(2, 0.5).
D(4, 0) [X-intercept of line 2].
Value of Z = 6x + 2y:
At A(0, 3): \( 6(0) + 2(3) = 6 \)
At B(0.6, 1.2): \( 6(0.6) + 2(1.2) = 3.6 + 2.4 = \mathbf{6} \)
At C(2, 0.5): \( 6(2) + 2(0.5) = 12 + 1 = 13 \)
At D(4, 0): \( 6(4) + 2(0) = 24 \)
Ans: The minimum value is 6. It occurs at point A(0, 3) and B(0.6, 1.2) and all points on the segment AB.
Q. 6. (C) Attempt any ONE of the following questions (Activity) (4 marks each):
(i) For a bivariate data \(\bar{x}=10\), \(\bar{y}=12\), \(V(X)=9\), \(\sigma_y=4\) and \(r=0.6\). Estimate \(y\) when \(x=5\).
Solution:
Line of regression of Y on X is:
\( Y - \bar{y} = b_{yx}(X - \bar{x}) \)
\( Y - 12 = r \cdot \frac{\sigma_y}{\sigma_x}(X - 10) \)
(Here \( V(X)=9 \implies \sigma_x = 3 \))
\( Y - 12 = 0.6 \times \frac{4}{\mathbf{\fbox{ \(\mathbf{3}\) }}} (X - 10) \)
When \( x = 5 \):
\( Y - 12 = \) 0.8 \( (5 - 10) \)
(Because \( 0.6 \times \frac{4}{3} = 0.8 \))
\( Y - 12 = -4 \)
\( Y = \) 8
Line of regression of Y on X is:
\( Y - \bar{y} = b_{yx}(X - \bar{x}) \)
\( Y - 12 = r \cdot \frac{\sigma_y}{\sigma_x}(X - 10) \)
(Here \( V(X)=9 \implies \sigma_x = 3 \))
\( Y - 12 = 0.6 \times \frac{4}{\mathbf{\fbox{ \(\mathbf{3}\) }}} (X - 10) \)
When \( x = 5 \):
\( Y - 12 = \) 0.8 \( (5 - 10) \)
(Because \( 0.6 \times \frac{4}{3} = 0.8 \))
\( Y - 12 = -4 \)
\( Y = \) 8
(ii) If \(X \sim P(m)\) with \(P(X=1) = P(X=2)\) then find the mean and \(P(X=2)\).
Given \(e^{-2} = 0.1353\).
Given \(e^{-2} = 0.1353\).
Solution:
Since \( P(X=1) = P(X=2) \)
Using Poisson formula \( P(X=x) = \frac{e^{-m}m^x}{x!} \):
\( \frac{e^{-m}m^1}{1!} = \frac{e^{-m}m^2}{2!} \)
Dividing both sides by \( e^{-m} \cdot m \):
\( 1 = \frac{m}{2} \)
\( m = \) 2
Now, find \( P(X=2) \):
\( P(X=2) = \frac{e^{-2} \cdot 2^2}{2!} \)
\( P(X=2) = \frac{0.1353 \times 4}{2} \)
\( P(X=2) = \) 0.2706
Since \( P(X=1) = P(X=2) \)
Using Poisson formula \( P(X=x) = \frac{e^{-m}m^x}{x!} \):
\( \frac{e^{-m}m^1}{1!} = \frac{e^{-m}m^2}{2!} \)
Dividing both sides by \( e^{-m} \cdot m \):
\( 1 = \frac{m}{2} \)
\( m = \) 2
Now, find \( P(X=2) \):
\( P(X=2) = \frac{e^{-2} \cdot 2^2}{2!} \)
\( P(X=2) = \frac{0.1353 \times 4}{2} \)
\( P(X=2) = \) 0.2706
