NEB Class 12 Management Business Mathematics Question Paper 2080 (Set B) Nepal

Since $A$ is an identity matrix ($I$), multiplying it by any matrix $B$ results in $B$. Therefore, $AB = B = \begin{bmatrix} 3 & 2 \\ 1 & 2 \end{bmatrix}$.

The equation $x^2 - 2x + 1 = 0$ simplifies to $(x-1)^2 = 0$, giving identical roots $x = 1, 1$. Since $A$ is a $2 \times 2$ matrix with leading diagonal elements as $1, 1$, and assuming it is diagonal or standard identity layout, $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$. Since $AB = I$, $B = A^{-1} = I^{-1} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

For a skew-symmetric matrix, $a_{12} = -a_{21}$. Thus, $x + 2 = -5 \implies x = -7$.

For $y = x^3$, $y' = 3x^2$ and $y'' = 6x$. Setting $y'' = 0$ gives $x = 0$. Since $y''$ changes sign across $x = 0$, $(0,0)$ is the point of inflection.

The equation $ydx + xdy = 0$ can be rewritten as $d(xy) = 0$. Integrating both sides gives $xy = c$.

Let the three terms in GP be $\frac{a}{r}, a, ar$. Their product is $\frac{a}{r} \cdot a \cdot ar = a^3 = 64 \implies a = 4$. The second term is $a = 4$.

$\text{IRR} = \frac{\text{Net Return}}{\text{Initial Investment}} \times 100\% = \frac{5000}{40000} \times 100\% = 12.5\%$.

The formula for the present value of an ordinary (immediate) annuity is $PV = \frac{A}{i}[1 - (1+i)^{-n}]$. Here $A = 1200, i = 0.1, n = 5$. Thus $PV = \frac{1200}{0.1}[1 - (1.1)^{-5}] = 12000[1 - (1.1)^{-5}]$.

Every linear programming problem (the primal) is fundamentally associated with another linear programming problem called its dual.

A negatively skewed distribution has a long tail to the left, which pulls the mean down the most. Hence, the relationship is $\text{mean} < \text{median} < \text{mode}$.

The variance of a binomial distribution with parameters $n$ and $p$ (where $q = 1-p$) is given by $npq$.

Let the cost of a chair, a table, and a desk be $x$, $y$, and $z$ respectively. From the given data, we can set up the system of linear equations:

1. $3x + 2y + z = 97000$
2. $2x + y + 3z = 86000$
3. $x + 3y + 2z = 99000$

Expressing this in matrix form $AX = B$: $\begin{bmatrix} 3 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 97000 \\ 86000 \\ 99000 \end{bmatrix}$

Find the determinant of $A$: $|A| = 3(2 - 9) - 2(4 - 3) + 1(6 - 1) = 3(-7) - 2(1) + 1(5) = -21 - 2 + 5 = -18$

Since $|A| \neq 0$, the inverse exists. Finding the adj(A) and solving $X = A^{-1}B$ gives: $x = 11000$ (Chair) $y = 22000$ (Table) $z = 10000$ (Desk)

Augmented matrix: $\begin{bmatrix} 3 & 1 & 1 & | & 5 \\ 1 & -4 & 1 & | & -2 \\ 1 & 1 & -3 & | & -1 \end{bmatrix}$ Swap $R_1$ and $R_2$: $\begin{bmatrix} 1 & -4 & 1 & | & -2 \\ 3 & 1 & 1 & | & 5 \\ 1 & 1 & -3 & | & -1 \end{bmatrix}$ Perform $R_2 \to R_2 - 3R_1$ and $R_3 \to R_3 - R_1$: $\begin{bmatrix} 1 & -4 & 1 & | & -2 \\ 0 & 13 & -2 & | & 11 \\ 0 & 5 & -4 & | & 1 \end{bmatrix}$ Solving via row operations or back substitution yields: $z = 1, y = 1, x = 1$.

Multiply $R_1$ by $a$, $R_2$ by $b$, and $R_3$ by $c$, and divide the determinant by $abc$: $\frac{1}{abc} \begin{vmatrix} abc & a^2 & a^3 \\ abc & b^2 & b^3 \\ abc & c^2 & c^3 \end{vmatrix}$ Take $abc$ common from $C_1$: $\frac{abc}{abc} \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix} = \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix}$. Hence Proved.

Given: Total Cost $(TC) = 2Q - Q^2$, Price/Demand Function $(P) = Q - 3$. a) Revenue Function $R = P \times Q = (Q - 3)Q = Q^2 - 3Q$. b) Profit Function $\pi = R - TC = (Q^2 - 3Q) - (2Q - Q^2) = 2Q^2 - 5Q$. c) Marginal Revenue function is the derivative of total revenue with respect to $Q$: $MR = \frac{dR}{dQ} = 2Q - 3$. d) Profit at $Q = 2$: $\pi(2) = 2(2)^2 - 5(2) = 8 - 10 = -2$.

To find points of intersection, set $x^2 + 2 = x \implies x^2 - x + 2 = 0$. This quadratic equation has non-real roots, which implies the curves do not intersect in the real plane. Note: There may be a typo in the question layout standard to NEB supplementary papers, such as a missing boundary or incorrect sign (e.g., $y = 2 - x^2$). Under literal interpretation, the area is undefined/infinite without vertical boundaries.

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = 1, Q = e^x$. Integrating factor $IF = e^{\int 1 dx} = e^x$. Solution is $y \cdot IF = \int (Q \cdot IF) dx + c \implies y e^x = \int e^{2x} dx + c \implies y e^x = \frac{e^{2x}}{2} + c \implies y = \frac{e^x}{2} + ce^{-x}$.

At market equilibrium, $P_d = P_s \implies 113 - x^2 = x^2 + 2x + 1 \implies 2x^2 + 2x - 112 = 0 \implies x^2 + x - 56 = 0 \implies (x+8)(x-7) = 0$. Since quantity must be positive, $x_0 = 7$. Equilibrium price $P_0 = 113 - (7)^2 = 113 - 49 = 64$. Consumer Surplus $(CS) = \int_{0}^{7} (113 - x^2) dx - (64 \times 7) = [113x - \frac{x^3}{3}]_0^7 - 448 = (791 - 114.33) - 448 = 228.67$. Producer Surplus $(PS) = (64 \times 7) - \int_{0}^{7} (x+1)^2 dx = 448 - [\frac{(x+1)^3}{3}]_0^7 = 448 - (\frac{512}{3} - \frac{1}{3}) = 448 - 170.33 = 277.67$.

Introduce slack variables $s_1, s_2$:

1. $4x + y + s_1 = 3$
2. $3x + y + s_2 = 1$ Objective function: $-12x - 5y + P = 0$ Constructing simplex tableaux and performing pivot operations yields the optimal values.

a) Formula 1: $S_k = \bar{x} - M_0$ Formula 2: $S_k = 3(\bar{x} - M_d)$ b) Empirical relationship: $\text{Mode} = 3\text{Median} - 2\text{Mean}$ (or $M_0 = 3M_d - 2\bar{x}$). c) $\text{Coefficient of Variation (C.V.)} = \frac{\sigma}{\bar{x}} \times 100\%$. d) Condition for positive skewness: $\bar{x} > M_d > M_0$.

Total gloves = 15. Ratio L:R = 2:1 $\implies$ Left = 10, Right = 5. a) $P(\text{Both Left}) = \frac{10}{15} \times \frac{9}{14} = \frac{2}{3} \times \frac{9}{14} = \frac{3}{7}$. b) $P(\text{One R, One L}) = P(RL) + P(LR) = (\frac{5}{15} \times \frac{10}{14}) + (\frac{10}{15} \times \frac{5}{14}) = \frac{10}{42} + \frac{10}{42} = \frac{20}{42} = \frac{10}{21}$. c) Assuming 'different process' implies with replacement: $P(\text{Both Right}) = \frac{5}{15} \times \frac{5}{15} = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.

Let's perform standard statistical calculations for $n=6$: $\sum x = 190$, $\sum y = 128$ $\sum x^2 = 6490$, $\sum y^2 = 2854$, $\sum xy = 3986$

a) Karl Pearson's Correlation Coefficient: $r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$ $r = \frac{6(3986) - (190)(128)}{\sqrt{[6(6490) - 190^2][6(2854) - 128^2]}} = \frac{23916 - 24320}{\sqrt{[38940 - 36100][17124 - 16384]}} = \frac{-404}{\sqrt{2840 \times 740}} = \frac{-404}{1449.69} \approx -0.279$

b) Regression line of $y$ on $x$: $y - \bar{y} = b_{yx}(x - \bar{x})$ $b_{yx} = \frac{6(3986) - 190 \times 128}{6(6490) - 190^2} = \frac{-404}{2840} \approx -0.1422$ $\bar{x} = 31.67, \bar{y} = 21.33$ $y - 21.33 = -0.1422(x - 31.67) \implies y = -0.1422x + 25.83$ For $x = 50$: $y = -0.1422(50) + 25.83 = -7.11 + 25.83 = 18.72 \approx 19$ appliances.

From simple interest for 3 years: $SI = 300 \implies$ SI for 1 year = Rs. 100. Thus, SI for 2 years = Rs. 200. Given Compound Interest for 2 years = Rs. 204. The difference between CI and SI for 2 years is $204 - 200 = \text{Rs. 4}$. This Rs. 4 is the interest generated by the first year's interest (Rs. 100). Therefore, Rate of interest $R = \frac{4}{100} \times 100\% = 4\%$. Using Simple Interest formula: $100 = \frac{P \times 4 \times 1}{100} \implies P = \text{Rs. 2500}$.

b) New Rate = $2 \times 4\% = 8\%$. $P = 2500$, Amount $A = 3149.28$. Using compound interest amount formula: $A = P(1 + R/100)^t \implies 3149.28 = 2500(1.08)^t$ $1.259712 = (1.08)^t \implies t = 3$ years.

Excess demand: $Q_d - Q_s = (500 - 5p) - (20p - 40) = 540 - 25p$. Given differential equation: $\frac{dp}{dt} = 0.02(540 - 25p) = 10.8 - 0.5p \implies \frac{dp}{dt} + 0.5p = 10.8$. This is a first-order linear differential equation. $IF = e^{0.5t}$. $P(t) e^{0.5t} = \int 10.8 e^{0.5t} dt + c = \frac{10.8}{0.5} e^{0.5t} + c = 21.6 e^{0.5t} + c \implies P(t) = 21.6 + ce^{-0.5t}$. Using initial condition $P(0) = 100 \implies 100 = 21.6 + c \implies c = 78.4$. Thus, the time path function is: $P(t) = 21.6 + 78.4e^{-0.5t}$.

1. For $t = 5$: $P(5) = 21.6 + 78.4e^{-2.5} \approx 21.6 + 78.4(0.08208) \approx 28.04$.
2. To find when price drops by Rs. 60, i.e., $P(t) = 100 - 60 = 40$: $40 = 21.6 + 78.4e^{-0.5t} \implies 18.4 = 78.4e^{-0.5t} \implies e^{-0.5t} = 0.2347 \implies -0.5t = -1.4495 \implies t \approx 2.9$ time periods.
3. Stability comment: Since $\lim_{t \to \infty} e^{-0.5t} = 0$, the price converges dynamically to the equilibrium price of 21.6. Hence, the market is dynamically stable.