Business Mathematics - I 2024 Board Question Paper

Step-by-Step Solution:

Given two points:

$$(x_1, y_1) = (2, 5) \quad \text{and} \quad (x_2, y_2) = (3, 1)$$

Step 1: Calculate the slope ($m$)

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

Step 2: Write the equation using point-slope form

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

$$y - 5 = -4(x - 2)$$

$$y - 5 = -4x + 8$$

Rearranging into standard form:

$$4x + y - 5 - 8 = 0$$

$$4x + y - 13 = 0$$

Final Answer: The equation of the straight line is $4x + y - 13 = 0$ (or $y = -4x + 13$).

Step-by-Step Solution:

Given demand function:

$$P = 40 - 0.2Q$$

Step 1: Find quantity ($Q$) when $P = 20$$20 = 40 - 0.2Q$$

$$0.2Q = 40 - 20 = 20$$

$$Q = \frac{20}{0.2} = 100 \text{ units}$$

Step 2: Differentiate with respect to $P$

From $P = 40 - 0.2Q$:

$$\frac{dP}{dQ} = -0.2$$

Taking the reciprocal:

$$\frac{dQ}{dP} = \frac{1}{\frac{dP}{dQ}} = \frac{1}{-0.2} = -5$$

Step 3: Compute point price elasticity of demand ($\epsilon_d$)

$$\epsilon_d = -\frac{P}{Q} \cdot \frac{dQ}{dP}$$

$$\epsilon_d = -\frac{20}{100} \times (-5) = -0.2 \times (-5) = 1.0$$

In absolute magnitude:

$$|\epsilon_d| = 1.0$$

Final Answer: The price elasticity of demand is $|\epsilon_d| = 1.0$ (unitary elastic demand).

Step-by-Step Solution:

Let the given roots be:

$$\alpha = 5\sqrt{5} \quad \text{and} \quad \beta = -\sqrt{5}$$

Step 1: Calculate the sum of the roots ($S$)

$$S = \alpha + \beta = 5\sqrt{5} + (-\sqrt{5}) = 4\sqrt{5}$$

Step 2: Calculate the product of the roots ($P$)

$$P = \alpha \cdot \beta = (5\sqrt{5})(-\sqrt{5}) = -5 \times (\sqrt{5})^2 = -5 \times 5 = -25$$

Step 3: Form the quadratic equation

A quadratic equation with sum of roots $S$ and product of roots $P$ is given by:

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

Substituting $S = 4\sqrt{5}$ and $P = -25$:

$$x^2 - 4\sqrt{5}x - 25 = 0$$

Final Answer: The quadratic equation is $x^2 - 4\sqrt{5}x - 25 = 0$.

Step-by-Step Solution:

Given equation:

$$2^{x-2} \times 8^x = 16 + 5$$

Step 1: Simplify both sides

Left-hand side:

$$8^x = (2^3)^x = 2^{3x}$$

$$2^{x-2} \times 2^{3x} = 2^{(x-2) + 3x} = 2^{4x - 2}$$

Right-hand side:

$$16 + 5 = 21$$

Thus:

$$2^{4x - 2} = 21$$

Step 2: Solve for $x$ using logarithms

Take natural logarithms on both sides:

$$\ln(2^{4x - 2}) = \ln(21)$$

$$(4x - 2) \ln(2) = \ln(21)$$

$$4x - 2 = \frac{\ln(21)}{\ln(2)} = \log_2(21)$$

Evaluating the numerical values:

$$\ln(21) \approx 3.044522$$

$$\ln(2) \approx 0.693147$$

$$\frac{\ln(21)}{\ln(2)} \approx 4.392317$$

Now solve for $x$:

$$4x = 4.392317 + 2 = 6.392317$$

$$x = \frac{6.392317}{4} \approx 1.598$$

(Note: If the right side was a misprint for $16 = 2^4$, then $4x - 2 = 4 \implies x = 1.5$; if $16 \times 2 = 32 = 2^5$, $4x - 2 = 5 \implies x = 1.75$. As printed, the exact mathematical solution is $x = \frac{2 + \log_2(21)}{4}$).

Final Answer: $x = \frac{2 + \log_2(21)}{4} \approx 1.598$.

Step-by-Step Solution:

Given arithmetic series:

$$6 + 10 + 14 + \dots$$

Step 1: Identify terms

• First term ($a$) = $6$
• Common difference ($d$) = $10 - 6 = 4$
• $n^{\text{th}}$ term ($T_n$) = $62$

Step 2: Apply the general term formula

$$T_n = a + (n - 1)d$$

$$62 = 6 + (n - 1)4$$

$$62 - 6 = 4(n - 1)$$

$$56 = 4(n - 1)$$

$$n - 1 = \frac{56}{4} = 14$$

$$n = 14 + 1 = 15$$

Final Answer: The $15^{\text{th}}$ term of the arithmetic series is $62$.

Step-by-Step Solution:

Given geometric series:

$$1 + 3 + 9 + \dots$$

Step 1: Identify components

• First term ($a$) = $1$
• Common ratio ($r$) = $\frac{3}{1} = 3$
• Number of terms ($n$) = $8$

Step 2: Apply sum formula for a Geometric Series (since $r > 1$)

$$S_n = \frac{a(r^n - 1)}{r - 1}$$

Substitute $a = 1, r = 3, n = 8$:

$$S_8 = \frac{1(3^8 - 1)}{3 - 1}$$

Since $3^8 = 6,561$:

$$S_8 = \frac{6,561 - 1}{2} = \frac{6,560}{2} = 3,280$$

Final Answer: The sum of the geometric series to 8 terms is $3,280$.

Step-by-Step Solution:

Given:

• Principal ($P$) = $\text{Rs } 12,500$
• Time ($t$) = $3$ years
• Rate of interest ($r$) = $12\% = 0.12$ per annum

Step 1: Compute the accumulated compound amount ($A$)

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

$$A = 12,500(1 + 0.12)^3 = 12,500(1.12)^3$$

$$(1.12)^3 = 1.404928$$

$$A = 12,500 \times 1.404928 = \text{Rs } 17,561.60$$

Step 2: Compute Compound Interest ($CI$)

$$CI = A - P$$

$$CI = 17,561.60 - 12,500 = \text{Rs } 5,061.60$$

Final Answer: The compound interest is Rs 5,061.60.

Step-by-Step Solution:

Evaluate:

$$L = \lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$

Step 1: Check form

Direct substitution of $x = 2$ gives $\frac{2^2 - 4}{2 - 2} = \frac{0}{0}$ (indeterminate form).

Step 2: Factorize the numerator

$$x^2 - 4 = x^2 - 2^2 = (x - 2)(x + 2)$$

Since $x \to 2$, $x \neq 2$, so we can cancel $(x - 2)$:

$$L = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2)$$

Step 3: Evaluate the limit

$$L = 2 + 2 = 4$$

Final Answer: The value of the limit is $4$.

Step-by-Step Solution:

Given function:

$$f(x) = x^3 - 12x^2 + 2$$

Step 1: Compute first and second derivatives

$$f'(x) = \frac{d}{dx}(x^3 - 12x^2 + 2) = 3x^2 - 24x$$

$$f''(x) = \frac{d}{dx}(3x^2 - 24x) = 6x - 24$$

Step 2: Set $f''(x) = 0$ to find candidate inflection point

$$6x - 24 = 0 \implies 6x = 24 \implies x = 4$$

Step 3: Check third derivative

$$f'''(x) = 6 \neq 0$$

Since $f'''(4) \neq 0$, the concavity changes sign across $x = 4$, confirming an inflection point.

Step 4: Compute the $y$-value at $x = 4$$f(4) = 4^3 - 12(4^2) + 2 = 64 - 12(16) + 2 = 64 - 192 + 2 = -126$$

Final Answer: The point of inflection is $(4, -126)$.

Step-by-Step Solution:

Given:

$$y = \frac{1}{\sqrt{2x+1}} = (2x + 1)^{-1/2}$$

Step 1: Apply the Power Rule and Chain Rule

$$\frac{dy}{dx} = -\frac{1}{2}(2x + 1)^{-\frac{1}{2} - 1} \cdot \frac{d}{dx}(2x + 1)$$

$$\frac{dy}{dx} = -\frac{1}{2}(2x + 1)^{-3/2} \cdot (2)$$

Step 2: Simplify

$$\frac{dy}{dx} = -(2x + 1)^{-3/2} = -\frac{1}{(2x + 1)^{3/2}} = -\frac{1}{\sqrt{(2x + 1)^3}}$$

Final Answer: $\frac{dy}{dx} = -\frac{1}{(2x + 1)^{3/2}}$.

Step-by-Step Solution:

Given system of linear equations:

$$x + 2y + 3z = 13 \quad \text{--- (1)}$$

$$2x + 4y + z = 11 \quad \text{--- (2)}$$

$$3x + 2y + 2z = 14 \quad \text{--- (3)}$$

Step 1: Eliminate $x$ and $y$ using equations (1) and (2)

Multiply equation (1) by 2:

$$2(x + 2y + 3z) = 2(13)$$

$$2x + 4y + 6z = 26 \quad \text{--- (4)}$$

Now subtract equation (2) from equation (4):

$$(2x + 4y + 6z) - (2x + 4y + z) = 26 - 11$$

$$5z = 15$$

$$z = \frac{15}{5} = 3$$

Step 2: Substitute $z = 3$ into the system

From equation (2):

$$2x + 4y + 3 = 11 \implies 2x + 4y = 8 \implies x + 2y = 4 \implies x = 4 - 2y \quad \text{--- (5)}$$

From equation (3):

$$3x + 2y + 2(3) = 14$$

$$3x + 2y + 6 = 14 \implies 3x + 2y = 8 \quad \text{--- (6)}$$

Step 3: Solve for $y$ and $x$

Substitute equation (5) into equation (6):

$$3(4 - 2y) + 2y = 8$$

$$12 - 6y + 2y = 8$$

$$12 - 4y = 8$$

$$-4y = 8 - 12 = -4 \implies y = 1$$

Now find $x$:

$$x = 4 - 2(1) = 4 - 2 = 2$$

Step 4: Verification

• Eq (1): $2 + 2(1) + 3(3) = 2 + 2 + 9 = 13$ (Matches)
• Eq (2): $2(2) + 4(1) + 3 = 4 + 4 + 3 = 11$ (Matches)
• Eq (3): $3(2) + 2(1) + 2(3) = 6 + 2 + 6 = 14$ (Matches)

Final Answer: The solution is $x = 2$, $y = 1$, $z = 3$.

Step-by-Step Solution:

Given:

• Demand function: $P = 60 - 0.6Q$
• Supply function: $P = 20 + 0.2Q$

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a. Calculate Equilibrium Price ($P^*$) and Quantity ($Q^*$):

At equilibrium, quantity demanded equals quantity supplied ($P_d = P_s$):

$$60 - 0.6Q = 20 + 0.2Q$$

$$60 - 20 = 0.2Q + 0.6Q$$

$$40 = 0.8Q$$

$$Q^* = \frac{40}{0.8} = 50 \text{ units}$$

Substitute $Q^* = 50$ into either equation:

$$P^* = 60 - 0.6(50) = 60 - 30 = \text{Rs } 30$$

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b. Calculate Consumer Surplus ($CS$) and Producer Surplus ($PS$):

1. Consumer Surplus ($CS$): Demand price intercept (choke price) at $Q = 0$ is $P_{\text{max}} = 60$.

   $$CS = \int_0^{50} (60 - 0.6Q - 30) dQ = \int_0^{50} (30 - 0.6Q) dQ$$
   $$CS = \left[ 30Q - 0.3Q^2 \right]_0^{50} = 30(50) - 0.3(50^2) = 1,500 - 0.3(2,500) = 1,500 - 750 = 750$$
   (Or geometrically: $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 50 \times (60 - 30) = 750$).

2. Producer Surplus ($PS$): Supply price intercept at $Q = 0$ is $P_{\text{min}} = 20$.

   $$PS = \int_0^{50} (30 - (20 + 0.2Q)) dQ = \int_0^{50} (10 - 0.2Q) dQ$$
   $$PS = \left[ 10Q - 0.1Q^2 \right]_0^{50} = 10(50) - 0.1(50^2) = 500 - 0.1(2,500) = 500 - 250 = 250$$
   (Or geometrically: $\frac{1}{2} \times 50 \times (30 - 20) = 250$).

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c. Total Surplus ($TS$):

$$TS = CS + PS = 750 + 250 = 1,000$$

Final Answer:

• a. Equilibrium: $P^* = \text{Rs } 30$, $Q^* = 50 \text{ units}$
• b. Consumer Surplus: $750$, Producer Surplus: $250$
• c. Total Surplus: $1,000$

Step-by-Step Solution:

Given quadratic function:

$$y = x^2 - 5x + 6$$

Step 1: Intercepts

• $y$-intercept: When $x = 0$, $y = 6 \implies (0, 6)$.
• $x$-intercepts: Set $y = 0$:
  $$x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2, x = 3$$
  Intercept points are $(2, 0)$ and $(3, 0)$.

Step 2: Vertex and Minimum Value of $y$

Since the leading coefficient $a = 1 > 0$, the parabola opens upwards and has a global minimum at its vertex. The $x$-coordinate of the vertex:

$$x = -\frac{b}{2a} = -\frac{-5}{2(1)} = 2.5$$

The minimum value of $y$:

$$y_{\text{min}} = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 = -\frac{1}{4}$$

Step 3: Table of Values for Sketching

Sketch Description: Plot the points $(0, 6)$, $(1, 2)$, $(2, 0)$, vertex $(2.5, -0.25)$, $(3, 0)$, $(4, 2)$, and $(5, 6)$. Draw a smooth U-shaped parabolic curve symmetric about the vertical line $x = 2.5$.

Final Answer:

• Minimum value of $y$: $-0.25$ (or $-\frac{1}{4}$) at $x = 2.5$.
• Vertex: $(2.5, -0.25)$, intercepts at $(2, 0)$, $(3, 0)$, and $(0, 6)$.

Step-by-Step Solution:

Given resale value function:

$$V(t) = 250,000 e^{-0.6t}$$

where $t$ is the number of years since original purchase.

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1. Resale value after $t = 5$ years:

$$V(5) = 250,000 e^{-0.6(5)} = 250,000 e^{-3.0}$$

Using $e^{-3.0} \approx 0.04978707$:

$$V(5) = 250,000 \times 0.04978707 \approx \text{Rs } 12,446.77$$

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2. Resale value after $t = 10$ years:

$$V(10) = 250,000 e^{-0.6(10)} = 250,000 e^{-6.0}$$

Using $e^{-6.0} \approx 0.00247875$:

$$V(10) = 250,000 \times 0.00247875 \approx \text{Rs } 619.69$$

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3. Resale value after $t = 15$ years:

$$V(15) = 250,000 e^{-0.6(15)} = 250,000 e^{-9.0}$$

Using $e^{-9.0} \approx 0.00012341$:

$$V(15) = 250,000 \times 0.00012341 \approx \text{Rs } 30.85$$

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4. Resale value after $t = 20$ years:

$$V(20) = 250,000 e^{-0.6(20)} = 250,000 e^{-12.0}$$

Using $e^{-12.0} \approx 0.000006144$:

$$V(20) = 250,000 \times 0.000006144 \approx \text{Rs } 1.54$$

Final Answer:

• After 5 years: Rs 12,446.77
• After 10 years: Rs 619.69
• After 15 years: Rs 30.85
• After 20 years: Rs 1.54

Step-by-Step Solution:

Given:

• Principal ($P$) = $\text{Rs } 5,000$
• Accumulated Amount ($A$) = $\text{Rs } 20,000$
• Annual interest rate ($r$) = $5.5\% = 0.055$
• Let $n$ be the number of years.

Step 1: Compound Interest Formula (Annual Compounding)

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

$$20,000 = 5,000(1 + 0.055)^n$$

$$\frac{20,000}{5,000} = (1.055)^n$$

$$4 = (1.055)^n$$

Step 2: Solve for $n$ using logarithms

Take natural logarithms on both sides:

$$\ln(4) = \ln((1.055)^n)$$

$$\ln(4) = n \cdot \ln(1.055)$$

$$n = \frac{\ln(4)}{\ln(1.055)}$$

Evaluate the numerical logs:

$$\ln(4) \approx 1.386294$$

$$\ln(1.055) \approx 0.053541$$

$$n = \frac{1.386294}{0.053541} \approx 25.892 \text{ years}$$

Converting the decimal portion to months:

$$0.892 \times 12 \approx 10.7 \text{ months} \approx 11 \text{ months}$$

(Note: If continuous compounding were assumed, $4 = e^{0.055n} \implies n = \frac{\ln 4}{0.055} \approx 25.21 \text{ years}$).

Final Answer: It will take approximately $25.89$ years (approx. 25 years and 11 months) for Rs 5,000 to grow to Rs 20,000.

Step-by-Step Solution:

Given:

• Purchase price of house = $\text{Rs } 5,000,000$
• Immediate down payment = $\text{Rs } 2,000,000$
• Loan balance to amortize ($PV$) = $5,000,000 - 2,000,000 = \text{Rs } 3,000,000$
• Number of equal annual installments ($n$) = $15$
• Interest rate ($i$) = $15\% = 0.15$ per annum
• Let $R$ be the equal annual installment amount.

Step 1: Ordinary Annuity Amortization Formula

The loan balance is the present value of an ordinary annuity of $n$ payments:

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

$$R = \frac{PV \times i}{1 - (1 + i)^{-n}}$$

Step 2: Compute $(1 + i)^{-n}$$(1 + 0.15)^{-15} = (1.15)^{-15} = \frac{1}{(1.15)^{15}} \approx \frac{1}{8.137062} \approx 0.122894$$

Step 3: Compute payment $R$$1 - (1.15)^{-15} \approx 1 - 0.122894 = 0.877106$$

$$\frac{i}{1 - (1 + i)^{-n}} = \frac{0.15}{0.877106} \approx 0.171017$$

$$R = 3,000,000 \times 0.171017 \approx \text{Rs } 513,051.05$$

Final Answer: Mr. Kumar has to pay Rs 513,051.05 annually.

Step-by-Step Solution:

(i) Find $\frac{dy}{dx}$ from $x^2 + y = 9$:

Rearrange explicitly for $y$:

$$y = 9 - x^2$$

Differentiating with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx}(9 - x^2) = -2x$$

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(ii) Find $\frac{dy}{dx}$ from parametric equations $y = t^4 + 2$ and $x = t^3 + 1$:

Differentiating each function with respect to parameter $t$:

$$\frac{dy}{dt} = \frac{d}{dt}(t^4 + 2) = 4t^3$$

$$\frac{dx}{dt} = \frac{d}{dt}(t^3 + 1) = 3t^2$$

Using the chain rule for parametric differentiation:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t^3}{3t^2} = \frac{4}{3}t$$

Final Answer:

• (i) $\frac{dy}{dx} = -2x$
• (ii) $\frac{dy}{dx} = \frac{4}{3}t$

Step-by-Step Solution:

This question contains two comprehensive analytical components:

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Part 1: Least Squares Method for Company Yearly Income

Given Data:

Number of years ($N$) = $7$ (odd). Let the middle year $2020$ be taken as the origin, defining transformed variable $u = \text{Year} - 2020$.

Step 1: Determine Trend Line Coefficients

Since $\sum u = 0$:

$$a = \frac{\sum y}{N} = \frac{413}{7} = 59$$

$$b = \frac{\sum uy}{\sum u^2} = \frac{52}{28} = \frac{13}{7} \approx 1.85714$$

The trend equation is:

$$y = 59 + 1.8571 u \quad (\text{where } u = \text{Year} - 2020)$$

Step 2: Estimate Income for 2024 and 2026

1. For Year 2024: $u = 2024 - 2020 = 4$$y_{2024} = 59 + \left(\frac{13}{7}\right)(4) = 59 + \frac{52}{7} = 59 + 7.4286 = \mathbf{66.43\text{ million Rs}}$$

2. For Year 2026: $u = 2026 - 2020 = 6$$y_{2026} = 59 + \left(\frac{13}{7}\right)(6) = 59 + \frac{78}{7} = 59 + 11.1429 = \mathbf{70.14\text{ million Rs}}$$

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Part 2: Open-Economy Keynesian Macroeconomic Model

Given:

• National Income Identity: $Y = C + I + G + X - M$
• Consumption function: $C = C_0 + 0.8Y_d = 90 + 0.8Y_d$
• Investment: $I = 500\text{m}$
• Government expenditure: $G = 700\text{m}$
• Exports: $X = 250\text{m}$
• Import function: $M = 0.3Y_d$
• Tax rate: $t = 0.4$, with $T = tY = 0.4Y$
• Disposable Income: $Y_d = Y - T = Y - 0.4Y = 0.6Y$

Step 1: Derive the Aggregate Expenditure Equation ($E$)

Substitute $Y_d = 0.6Y$ into $C$ and $M$:

$$C = 90 + 0.8(0.6Y) = 90 + 0.48Y$$

$$M = 0.3(0.6Y) = 0.18Y$$

Aggregate Expenditure:

$$E = C + I + G + X - M$$

$$E = (90 + 0.48Y) + 500 + 700 + 250 - 0.18Y$$

$$E = (90 + 500 + 700 + 250) + (0.48 - 0.18)Y$$

$$\mathbf{E = 1540 + 0.30Y}$$

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

Set $Y = E$:

$$Y = 1540 + 0.30Y$$

$$Y - 0.30Y = 1540$$

$$0.70Y = 1540$$

$$Y^* = \frac{1540}{0.70} = \mathbf{2,200\text{ million Rs}}$$

Step 3: Determine Equilibrium Consumption ($C^*$) and Total Tax ($T^*$)

• Consumption ($C^*$):
  $$C^* = 90 + 0.48(2,200) = 90 + 1,056 = \mathbf{1,146\text{ million Rs}}$$
• Total Tax ($T^*$):
  $$T^* = 0.4(2,200) = \mathbf{880\text{ million Rs}}$$

Final Answer:

• Least Squares Trend Line: $y = 59 + 1.8571(\text{Year} - 2020)$
  • 2024 Income: Rs 66.43 million
  • 2026 Income: Rs 70.14 million
• Expenditure Equation: $E = 1540 + 0.30Y$
• Equilibrium Income ($Y^*$): Rs 2,200 million
• Equilibrium Consumption ($C^*$): Rs 1,146 million
• Total Tax ($T^*$): Rs 880 million

Step-by-Step Solution:

Given market functions:

• Supply: $P_s = 3Q_s + 40$
• Demand: $P_d = -2Q_d + 80 = 80 - 2Q_d$
• Unit specific tax = $\text{Rs } t$.

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Step 1: Equilibrium with Tax

With unit tax $t$ levied on supply:

$$P_{st} = P_s + t = 3Q + 40 + t$$

Equating demand and taxed supply ($P_d = P_{st}$):

$$80 - 2Q = 3Q + 40 + t$$

$$80 - 40 - t = 3Q + 2Q$$

$$40 - t = 5Q$$

$$Q = \frac{40 - t}{5} = 8 - 0.2t$$

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

$$T(t) = t \times Q = t(8 - 0.2t) = 8t - 0.2t^2$$

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Step 3: Maximize Tax Revenue

Differentiate with respect to $t$:

$$\frac{dT}{dt} = 8 - 0.4t = 0$$

$$0.4t = 8 \implies t = \frac{8}{0.4} = 20$$

Second derivative test:

$$\frac{d^2T}{dt^2} = -0.4 < 0 \quad (\text{Confirmed maximum})$$

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a. Equilibrium Price and Quantity at $t = 20$:

• Equilibrium Quantity ($Q^*$):
  $$Q^* = 8 - 0.2(20) = 8 - 4 = 4 \text{ units}$$
• Equilibrium Price paid by consumers ($P^*$):
  $$P^* = 80 - 2(4) = 80 - 8 = \text{Rs } 72$$
  (Price received by suppliers: $P_{\text{supplier}} = P^* - t = 72 - 20 = \text{Rs } 52 = 3(4) + 40$).

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b. Total Tax Raised:

$$T = t \times Q^* = 20 \times 4 = \text{Rs } 80$$

Final Answer:

• Optimal tax rate: $t = \text{Rs } 20$ per unit
• a. Equilibrium Price and Quantity: $P = \text{Rs } 72$, $Q = 4$ units
• b. Total Tax Raised: Rs 80

Step-by-Step Solution:

Given:

• Market discount rate ($k$) = $9\% = 0.09$
• Investment outlays and 1-year returns:

Formulas:

1. Net Present Value (NPV):
   $$NPV = -C_0 + \frac{C_1}{1 + k} = -C_0 + \frac{C_1}{1.09}$$
2. Internal Rate of Return (IRR):
   $$IRR = \frac{C_1 - C_0}{C_0} \times 100\%$$

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Calculations:

1. Project A:

   • $C_0 = 10,000, C_1 = 11,000$
   • $IRR_A = \frac{11,000 - 10,000}{10,000} = \frac{1,000}{10,000} = \mathbf{10.00\%}$
   • $NPV_A = -10,000 + \frac{11,000}{1.09} = -10,000 + 10,091.74 = \mathbf{+\text{Rs } 91.74}$
   • Viability: Viable ($NPV > 0$ and $IRR > 9\%$).

2. Project B:

   • $C_0 = 6,000, C_1 = 6,520$
   • $IRR_B = \frac{6,520 - 6,000}{6,000} = \frac{520}{6,000} \approx \mathbf{8.67\%}$
   • $NPV_B = -6,000 + \frac{6,520}{1.09} = -6,000 + 5,981.65 = \mathbf{-\text{Rs } 18.35}$
   • Viability: Not viable ($NPV < 0$ and $IRR < 9\%$).

3. Project C:

   • $C_0 = 9,000, C_1 = 9,800$
   • $IRR_C = \frac{9,800 - 9,000}{9,000} = \frac{800}{9,000} \approx \mathbf{8.89\%}$
   • $NPV_C = -9,000 + \frac{9,800}{1.09} = -9,000 + 8,990.83 = \mathbf{-\text{Rs } 9.17}$
   • Viability: Not viable ($NPV < 0$ and $IRR < 9\%$).

4. Project D:

   • Initial outlay $C_0 = 8,000$. The return $C_1$ was omitted in the source paper.
   • Break-even threshold condition: For Project D to be viable at $k = 9\%$, its return must satisfy $C_1 > 8,000 \times 1.09 = \text{Rs } 8,720$.
   • (If $C_1 = 8,500$ as in 2023: $IRR = 6.25\%, NPV = -\text{Rs } 201.83$, not viable).

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Profitability Ranking:

Final Answer:

• Viability: Only Project A is economically viable.
• Order of Profitability: Project A > Project C > Project B.