Business Mathematics - I

Step-by-Step Solution:

Given two points:

$$(x_1, y_1) = (3, 6) \quad \text{and} \quad (x_2, y_2) = (-5, 2)$$

Step 1: Compute the slope ($m$)

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 6}{-5 - 3} = \frac{-4}{-8} = \frac{1}{2}$$

Step 2: Use point-slope form

$$y - y_1 = m(x - x_1)$$

Substitute $(x_1, y_1) = (3, 6)$ and $m = \frac{1}{2}$:

$$y - 6 = \frac{1}{2}(x - 3)$$

$$2(y - 6) = x - 3$$

$$2y - 12 = x - 3$$

Rearrange into standard form $Ax + By + C = 0$:

$$x - 2y - 3 + 12 = 0$$

$$x - 2y + 9 = 0$$

Final Answer: The equation of the straight line is $x - 2y + 9 = 0$ (or $y = \frac{1}{2}x + \frac{9}{2}$).

Step-by-Step Solution:

Given:

• Total investment / budget ($M$) = $\text{Rs } 90,000$
• Unit price of material $X$ ($P_X$) = $\text{Rs } 1,500$
• Unit price of material $Y$ ($P_Y$) = $\text{Rs } 1,800$

Step 1: Formulate the Budget Constraint Equation

$$P_X \cdot X + P_Y \cdot Y = M$$

$$1,500X + 1,800Y = 90,000$$

Step 2: Determine the Slope

Expressing $Y$ as a function of $X$ ($Y$ on the vertical axis, $X$ on the horizontal axis):

$$1,800Y = 90,000 - 1,500X$$

$$Y = \frac{90,000}{1,800} - \frac{1,500}{1,800}X$$

$$Y = 50 - \frac{5}{6}X$$

The slope of the budget constraint line is:

$$\text{Slope} = -\frac{P_X}{P_Y} = -\frac{1,500}{1,800} = -\frac{5}{6} \approx -0.8333$$

Final Answer: The slope of the budget constraint is $-\frac{5}{6}$ (or approximately $-0.833$).

Step-by-Step Solution:

Given geometric series:

$$2 + 6 + 18 + \dots$$

Step 1: Identify components

• First term ($a$) = $2$
• Common ratio ($r$) = $\frac{6}{2} = 3$
• Term number ($n$) = $10$

Step 2: Compute the $n^{\text{th}}$ term

$$T_n = a \cdot r^{n-1}$$

$$T_{10} = 2 \cdot 3^{10 - 1} = 2 \cdot 3^9$$

Evaluating $3^9$:

$$3^9 = 19,683$$

$$T_{10} = 2 \times 19,683 = 39,366$$

Final Answer: The $10^{\text{th}}$ term of the geometric series is $39,366$.

Step-by-Step Solution:

Given system of linear equations:

$$x + 4y = 9 \quad \text{--- (1)}$$

$$2x + y = 8 \quad \text{--- (2)}$$

Method: Ratio of Coefficients / Determinant Test

For a system $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$:

• $a_1 = 1, b_1 = 4, c_1 = 9$
• $a_2 = 2, b_2 = 1, c_2 = 8$

1. Check the ratio of coefficients:

   $$\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{4}{1} = 4$$
   Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ ($\frac{1}{2} \neq 4$), the two straight lines are non-parallel and intersect at exactly one point.

2. Check via Coefficient Matrix Determinant:

   $$D = \begin{vmatrix} 1 & 4 \\ 2 & 1 \end{vmatrix} = (1)(1) - (4)(2) = 1 - 8 = -7 \neq 0$$
   Since $D \neq 0$, Cramer’s rule guarantees a unique solution.

(The unique solution is $x = \frac{23}{7}, y = \frac{10}{7}$).

Final Answer: The system of linear equations has a unique solution.

Step-by-Step Solution:

Given quadratic equation:

$$x^2 - 3x - 10 = 0$$

Step 1: Factorization

We look for two numbers whose product is $-10$ and whose sum is $-3$. These numbers are $-5$ and $+2$:

$$x^2 - 5x + 2x - 10 = 0$$

$$x(x - 5) + 2(x - 5) = 0$$

$$(x - 5)(x + 2) = 0$$

Step 2: Solve for $x$

$$x - 5 = 0 \implies x = 5$$

$$x + 2 = 0 \implies x = -2$$

Final Answer: The roots of the quadratic equation are $x = 5$ and $x = -2$.

Step-by-Step Solution:

Given roots:

$$\alpha = 1 + \sqrt{3} \quad \text{and} \quad \beta = 1 - \sqrt{3}$$

Step 1: Calculate Sum of the Roots ($S$)

$$S = \alpha + \beta = (1 + \sqrt{3}) + (1 - \sqrt{3}) = 1 + 1 = 2$$

Step 2: Calculate Product of the Roots ($P$)

Using the identity $(a + b)(a - b) = a^2 - b^2$:

$$P = \alpha \cdot \beta = (1 + \sqrt{3})(1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2$$

Step 3: Form the Equation

$$x^2 - Sx + P = 0$$

Substitute $S = 2$ and $P = -2$:

$$x^2 - 2x - 2 = 0$$

Final Answer: The required quadratic equation is $x^2 - 2x - 2 = 0$.

Step-by-Step Solution:

Given equation:

$$2^{9-x} \cdot 4^x = 2^{2x-1}$$

Step 1: Convert all bases to 2

Since $4 = 2^2$:

$$4^x = (2^2)^x = 2^{2x}$$

Substitute back into the equation:

$$2^{9-x} \cdot 2^{2x} = 2^{2x-1}$$

Step 2: Combine exponents on the left-hand side

$$2^{(9 - x) + 2x} = 2^{2x - 1}$$

$$2^{x + 9} = 2^{2x - 1}$$

Step 3: Equate the exponents

Since the bases are identical ($2 > 0, 2 \neq 1$):

$$x + 9 = 2x - 1$$

$$9 + 1 = 2x - x$$

$$x = 10$$

Verification:

• $\text{LHS} = 2^{9 - 10} \cdot 4^{10} = 2^{-1} \cdot 2^{20} = 2^{19}$
• $\text{RHS} = 2^{2(10) - 1} = 2^{19}$ (Matches)

Final Answer: $x = 10$.

Step-by-Step Solution:

Given arithmetic series:

$$3 + 5 + 7 + \dots$$

Step 1: Identify parameters

• First term ($a$) = $3$
• Common difference ($d$) = $5 - 3 = 2$
• Number of terms ($n$) = $20$

Step 2: Apply the sum formula for an Arithmetic Progression

$$S_n = \frac{n}{2} [2a + (n - 1)d]$$

$$S_{20} = \frac{20}{2} [2(3) + (20 - 1)2]$$

$$S_{20} = 10 [6 + 19 \times 2]$$

$$S_{20} = 10 [6 + 38] = 10 \times 44 = 440$$

Final Answer: The sum of the first twenty terms is $440$.

Step-by-Step Solution:

Given function:

$$y = 5x^3 - 3x^2 + 25$$

Step 1: Differentiate with respect to $x$

$$\frac{dy}{dx} = \frac{d}{dx}(5x^3 - 3x^2 + 25) = 15x^2 - 6x$$

Step 2: Set the derivative equal to zero

Stationary points occur where $\frac{dy}{dx} = 0$:

$$15x^2 - 6x = 0$$

$$3x(5x - 2) = 0$$

Thus:

$$3x = 0 \implies x_1 = 0$$

$$5x - 2 = 0 \implies x_2 = \frac{2}{5} = 0.4$$

Step 3: Find corresponding $y$-coordinates

• At $x = 0$:
  $$y(0) = 5(0)^3 - 3(0)^2 + 25 = 25 \implies (0, 25)$$
• At $x = 0.4$:
  $$y(0.4) = 5(0.4)^3 - 3(0.4)^2 + 25 = 5(0.064) - 3(0.16) + 25$$
  $$y(0.4) = 0.32 - 0.48 + 25 = 24.84 \implies (0.4, 24.84) \text{ or } \left(\frac{2}{5}, \frac{621}{25}\right)$$

Final Answer: The stationary points are $(0, 25)$ and $(0.4, 24.84)$ (or $\left(\frac{2}{5}, \frac{621}{25}\right)$).

Step-by-Step Solution:

Given:

• Principal ($P$) = $\text{Rs } 1,000$
• Time ($t$) = $2$ years
• Annual interest rate ($r$) = $5\% = 0.05$

Step 1: Compute compound amount ($A$)

$$A = P(1 + r)^t$$

$$A = 1,000(1 + 0.05)^2 = 1,000(1.05)^2$$

$$A = 1,000 \times 1.1025 = \text{Rs } 1,102.50$$

Step 2: Compute Compound Interest ($CI$)

$$CI = A - P = 1,102.50 - 1,000 = \text{Rs } 102.50$$

Final Answer: The compound interest is Rs 102.50.

Step-by-Step Solution:

Given parabolic function:

$$y = x^2 - 2x - 8$$

Step 1: Find Intercepts

• $y$-intercept: Set $x = 0$:
  $$y = 0^2 - 2(0) - 8 = -8 \implies (0, -8)$$
• $x$-intercepts: Set $y = 0$:
  $$x^2 - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0$$
  $$x = 4, \quad x = -2$$
  Intercept points are $(4, 0)$ and $(-2, 0)$.

Step 2: Find Vertex and Minimum Value of $y$

The coefficient of $x^2$ is $a = 1 > 0$, confirming the parabola opens upwards and has a global minimum.

• Vertex $x$-coordinate:
  $$x_v = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$$
• Minimum value of $y$:
  $$y_{\text{min}} = 1^2 - 2(1) - 8 = 1 - 2 - 8 = -9$$
  Vertex: $(1, -9)$. Axis of symmetry: $x = 1$.

Step 3: Plotting Points

$x$	-3	-2	-1	0	1 (Vertex)	2	3	4	5
$y$	7	0	-5	-8	-9	-8	-5	0	7

Graph Sketch Instructions: Plot the symmetric U-shaped parabola passing through $(-2, 0)$, $(0, -8)$, with its lowest turning point at $(1, -9)$, and returning upwards through $(4, 0)$.

Final Answer: The minimum value of $y$ is $-9$ (at $x = 1$). Vertex is $(1, -9)$.

Step-by-Step Solution:

Given system of equations:

$$x + 2y + z = 9 \quad \text{--- (1)}$$

$$x + y - 2z = -9 \quad \text{--- (2)}$$

$$2x - y + z = -2 \quad \text{--- (3)}$$

Step 1: Eliminate $x$

Subtract equation (2) from equation (1):

$$(x + 2y + z) - (x + y - 2z) = 9 - (-9)$$

$$y + 3z = 18 \implies y = 18 - 3z \quad \text{--- (4)}$$

Multiply equation (1) by 2 and subtract equation (3):

$$2(x + 2y + z) - (2x - y + z) = 2(9) - (-2)$$

$$(2x + 4y + 2z) - (2x - y + z) = 18 + 2$$

$$5y + z = 20 \quad \text{--- (5)}$$

Step 2: Solve for $z$ and $y$

Substitute equation (4) into equation (5):

$$5(18 - 3z) + z = 20$$

$$90 - 15z + z = 20$$

$$90 - 14z = 20$$

$$-14z = 20 - 90 = -70$$

$$z = \frac{-70}{-14} = 5$$

Now find $y$:

$$y = 18 - 3(5) = 18 - 15 = 3$$

Step 3: Solve for $x$

Substitute $y = 3, z = 5$ into equation (1):

$$x + 2(3) + 5 = 9$$

$$x + 6 + 5 = 9$$

$$x + 11 = 9 \implies x = 9 - 11 = -2$$

Verification:

• Eq (1): $-2 + 2(3) + 5 = -2 + 6 + 5 = 9$ (Matches)
• Eq (2): $-2 + 3 - 2(5) = 1 - 10 = -9$ (Matches)
• Eq (3): $2(-2) - 3 + 5 = -4 - 3 + 5 = -2$ (Matches)

Final Answer: $x = -2$, $y = 3$, $z = 5$.

Step-by-Step Solution:

Given macroeconomic model:

• Equilibrium condition: $Y = E$
• Aggregate expenditure: $E = C + I$
• Consumption function: $C = 600 + 0.24Y$
• Autonomous investment: $I = 280$

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Step 1: Substitute $C$ and $I$ into Aggregate Expenditure

$$E = (600 + 0.24Y) + 280$$

$$E = 880 + 0.24Y$$

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Step 2: Determine Equilibrium Level of National Income ($Y^*$)

Set $Y = E$:

$$Y = 880 + 0.24Y$$

$$Y - 0.24Y = 880$$

$$0.76Y = 880$$

$$Y^* = \frac{880}{0.76} = \frac{88,000}{76} = \frac{22,000}{19} \approx 1,157.89$$

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Step 3: Determine Equilibrium Level of Consumption ($C^*$)

Substitute $Y^* = \frac{22,000}{19}$ into the consumption function:

$$C^* = 600 + 0.24\left(\frac{22,000}{19}\right) = 600 + \frac{5,280}{19}$$

$$C^* = 600 + 277.8947 \approx 877.89$$

Verification:

$$C^* + I = 877.89 + 280 = 1,157.89 = Y^*$$

Final Answer:

• Equilibrium National Income ($Y^*$): Rs 1,157.89 (or $\frac{22,000}{19}$)
• Equilibrium Consumption ($C^*$): Rs 877.89 (or $\frac{16,680}{19}$)

Step-by-Step Solution:

Given consumption function:

$$C = 800 \left(1 - e^{-0.5y}\right)$$

where $y$ is the income level.

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Calculations for each income level:

1. At $y = 5$:

   $$C(5) = 800(1 - e^{-0.5 \times 5}) = 800(1 - e^{-2.5})$$
   Since $e^{-2.5} \approx 0.082085$:
   $$C(5) = 800(1 - 0.082085) = 800(0.917915) \approx 734.33$$

2. At $y = 10$:

   $$C(10) = 800(1 - e^{-0.5 \times 10}) = 800(1 - e^{-5.0})$$
   Since $e^{-5.0} \approx 0.006738$:
   $$C(10) = 800(1 - 0.006738) = 800(0.993262) \approx 794.61$$

3. At $y = 15$:

   $$C(15) = 800(1 - e^{-0.5 \times 15}) = 800(1 - e^{-7.5})$$
   Since $e^{-7.5} \approx 0.000553$:
   $$C(15) = 800(1 - 0.000553) = 800(0.999447) \approx 799.56$$

4. At $y = 20$:

   $$C(20) = 800(1 - e^{-0.5 \times 20}) = 800(1 - e^{-10.0})$$
   Since $e^{-10.0} \approx 0.0000454$:
   $$C(20) = 800(1 - 0.0000454) = 800(0.999955) \approx 799.96$$

5. At $y = 25$:

   $$C(25) = 800(1 - e^{-0.5 \times 25}) = 800(1 - e^{-12.5})$$
   Since $e^{-12.5} \approx 0.0000037$:
   $$C(25) = 800(1 - 0.0000037) \approx 800.00$$

(Note: The saturation level of consumption is $\lim_{y \to \infty} C(y) = 800$).

Final Answer:

• $C(5) \approx \mathbf{734.33}$
• $C(10) \approx \mathbf{794.61}$
• $C(15) \approx \mathbf{799.56}$
• $C(20) \approx \mathbf{799.96}$
• $C(25) \approx \mathbf{800.00}$

Step-by-Step Solution:

Given:

• Periodic payment ($R$) = $\text{Rs } 50,000$
• Duration ($n$) = $15$ years
• Annual interest rate ($i$) = $5\% = 0.05$
• “Amount of an immediate annuity” refers to the accumulated Future Value ($A$ or $FV$) of the annuity at the end of the period.

---

Step 1: Future Value Formula of an Ordinary (Immediate) Annuity

$$A = FV = R \left[ \frac{(1 + i)^n - 1}{i} \right]$$

Step 2: Compute $(1 + i)^n$

$$(1 + 0.05)^{15} = (1.05)^{15} \approx 2.07892818$$

Step 3: Compute $FV$

$$A = 50,000 \left[ \frac{2.07892818 - 1}{0.05} \right]$$

$$A = 50,000 \left[ \frac{1.07892818}{0.05} \right] = 50,000 \times 21.5785636$$

$$A \approx \text{Rs } 1,078,928.18$$

(Note: If the question intended the Present Value ($PV$) of the immediate annuity:

$$PV = R \left[ \frac{1 - (1 + i)^{-n}}{i} \right] = 50,000 \times 10.379658 = \text{Rs } 518,982.90$$

).

Final Answer: The accumulated amount (Future Value) of the annuity is Rs 1,078,928.18.

Step-by-Step Solution:

Given:

• Original cost of computer ($C$) = $\text{Rs } 50,000$
• Target depreciated value ($V$) = $\frac{C}{2} = \text{Rs } 25,000$
• Annual depreciation rate ($r$) = $5\% = 0.05$ on diminishing balance
• Let $n$ be the number of years.

Step 1: Diminishing Balance Formula

$$V = C(1 - r)^n$$

$$25,000 = 50,000(1 - 0.05)^n$$

$$\frac{25,000}{50,000} = (0.95)^n$$

$$0.5 = (0.95)^n$$

Step 2: Solve for $n$ using logarithms

Take natural logarithms on both sides:

$$\ln(0.5) = \ln((0.95)^n)$$

$$\ln(0.5) = n \cdot \ln(0.95)$$

$$n = \frac{\ln(0.5)}{\ln(0.95)}$$

Using natural logarithm values:

$$\ln(0.5) \approx -0.693147$$

$$\ln(0.95) \approx -0.051293$$

$$n = \frac{-0.693147}{-0.051293} \approx 13.513 \text{ years}$$

Converting decimal to months:

$$0.513 \times 12 \approx 6.16 \text{ months}$$

Final Answer: The computer will be reduced to half its original value in approximately $13.51$ years (or about 13 years and 6 months).

Step-by-Step Solution:

(i) Find $\frac{dy}{dx}$ from $x^2 + y^2 = 25$:

Differentiate both sides with respect to $x$ using implicit differentiation:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$

$$2x + 2y \frac{dy}{dx} = 0$$

Isolate $\frac{dy}{dx}$:

$$2y \frac{dy}{dx} = -2x$$

$$\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$$

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(ii) Find $\frac{dy}{dx}$ from parametric equations $x = 3t^2$ and $y = t^3$:

Differentiate each with respect to parameter $t$:

$$\frac{dx}{dt} = \frac{d}{dt}(3t^2) = 6t$$

$$\frac{dy}{dt} = \frac{d}{dt}(t^3) = 3t^2$$

Using the chain rule for parametric equations:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{6t} = \frac{t}{2}$$

Final Answer:

• (i) $\frac{dy}{dx} = -\frac{x}{y}$
• (ii) $\frac{dy}{dx} = \frac{t}{2}$

Step-by-Step Solution:

Given Data:

Number of years ($N$) = $7$ (odd). Let the origin be the middle year $2011$, defining transformed time variable $u = \text{Year} - 2011$.

Year	$u = \text{Year} - 2011$	Profit $y$ (million Rs)	$u^2$	$uy$
2008	-3	150	9	-450
2009	-2	180	4	-360
2010	-1	170	1	-170
2011	0	190	0	0
2012	1	160	1	160
2013	2	170	4	340
2014	3	210	9	630
Total	$\sum u = 0$	$\sum y = 1,230$	$\sum u^2 = 28$	$\sum uy = 150$

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Step 1: Compute Trend Line Coefficients

The trend equation is $y = a + bu$. Since $\sum u = 0$:

$$a = \frac{\sum y}{N} = \frac{1,230}{7} \approx 175.7143$$

$$b = \frac{\sum uy}{\sum u^2} = \frac{150}{28} = \frac{75}{14} \approx 5.3571$$

The least squares trend line is:

$$y = 175.7143 + 5.3571 u \quad (\text{where } u = \text{Year} - 2011)$$

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Step 2: Estimate Profit for the Year 2015

For Year 2015:

$$u = 2015 - 2011 = 4$$

$$y_{2015} = \frac{1,230}{7} + \left(\frac{75}{14}\right)(4) = \frac{1,230}{7} + \frac{150}{7} = \frac{1,380}{7} \approx 197.14$$

Final Answer:

• Equation of trend line: $y = 175.7143 + 5.3571(\text{Year} - 2011)$
• Estimated profit for 2015: Rs 197.14 million

Step-by-Step Solution:

Given market equations:

• Supply: $P_s = 6Q_s + 40$
• Demand: $P_d = -2Q_d + 120$
• Specific unit tax = $\text{Rs } t$.

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Step 1: Equilibrium with Tax

With unit tax $t$, the supply price paid by buyers is:

$$P_{st} = 6Q + 40 + t$$

At equilibrium, $P_d = P_{st}$:

$$120 - 2Q = 6Q + 40 + t$$

$$120 - 40 - t = 6Q + 2Q$$

$$80 - t = 8Q$$

$$Q = \frac{80 - t}{8} = 10 - \frac{t}{8} = 10 - 0.125t$$

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Step 2: Total Tax Revenue Function ($T$)

$$T(t) = t \times Q = t\left(10 - \frac{t}{8}\right) = 10t - \frac{t^2}{8}$$

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Step 3: Maximize Total Tax Revenue

Differentiate with respect to $t$:

$$\frac{dT}{dt} = 10 - \frac{2t}{8} = 10 - \frac{t}{4}$$

Set to zero for maximum:

$$10 - \frac{t}{4} = 0 \implies \frac{t}{4} = 10 \implies t = 40$$

Second derivative test:

$$\frac{d^2T}{dt^2} = -\frac{1}{4} < 0 \quad (\text{Confirmed strict maximum})$$

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Step 4: Calculate Equilibrium Values and Maximum Tax

• Equilibrium Quantity ($Q^*$):
  $$Q^* = 10 - \frac{40}{8} = 10 - 5 = 5 \text{ units}$$
• Equilibrium Price ($P^*$):
  $$P^* = 120 - 2(5) = 120 - 10 = \text{Rs } 110$$
  (Price received by suppliers: $110 - 40 = \text{Rs } 70 = 6(5) + 40$).
• Total Tax Raised ($T_{\text{max}}$):
  $$T_{\text{max}} = t \times Q^* = 40 \times 5 = \text{Rs } 200$$

Final Answer:

• Optimal tax rate: $t = \text{Rs } 40$ per unit
• Equilibrium price: Rs 110
• Equilibrium quantity: 5 units
• Total tax raised: Rs 200

Step-by-Step Solution:

Given:

• Discount rate ($r$) = $15\% = 0.15$
• Discount factors:
  • Year 1: $\frac{1}{1.15} \approx 0.869565$
  • Year 2: $\frac{1}{(1.15)^2} \approx 0.756144$
  • Year 3: $\frac{1}{(1.15)^3} \approx 0.657516$

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Step 1: Calculate Net Present Value for Project A

• Initial Cost ($C_0$) = $\text{Rs } 24,000$
• Cash flows: $C_1 = 6,000$, $C_2 = 12,000$, $C_3 = 16,000$

$$PV_A = \frac{6,000}{1.15} + \frac{12,000}{(1.15)^2} + \frac{16,000}{(1.15)^3}$$

$$PV_A = 6,000(0.869565) + 12,000(0.756144) + 16,000(0.657516)$$

$$PV_A = 5,217.39 + 9,073.72 + 10,520.26 = 24,811.37$$

$$NPV_A = PV_A - C_0 = 24,811.37 - 24,000 = \mathbf{+\text{Rs } 811.37}$$

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Step 2: Calculate Net Present Value for Project B

• Initial Cost ($C_0$) = $\text{Rs } 24,000$
• Cash flows: $C_1 = 4,000$, $C_2 = 10,000$, $C_3 = 20,000$

$$PV_B = \frac{4,000}{1.15} + \frac{10,000}{(1.15)^2} + \frac{20,000}{(1.15)^3}$$

$$PV_B = 4,000(0.869565) + 10,000(0.756144) + 20,000(0.657516)$$

$$PV_B = 3,478.26 + 7,561.44 + 13,150.32 = 24,190.02$$

$$NPV_B = PV_B - C_0 = 24,190.02 - 24,000 = \mathbf{+\text{Rs } 190.02}$$

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Step 3: Comparative Decision Analysis

• Both projects have $NPV > 0$, indicating that both yield returns higher than the 15% required rate of return.
• Comparing the profitability:
  $$NPV_A (\text{Rs } 811.37) > NPV_B (\text{Rs } 190.02)$$
• Therefore, Project A adds significantly more net economic value and is the more profitable investment.

Final Answer:

• $NPV_A$ = +Rs 811.37
• $NPV_B$ = +Rs 190.02
• Conclusion: Project A is the most profitable project and should be chosen.