A manufacturer makes two types of toys A and B. Three machines are needed for production with the following time constraints (in minutes): \[ \begin{array}{|c|c|c|} \hline \text{Machine} & \text{Toy A} & \text{Toy B} \\ \hline M1 & 12 & 6 \\ M2 & 18 & 0 \\ M3 & 6 & 9 \\ \hline \end{array} \] Each machine is available for 6 hours = 360 minutes. Profit on A = Rupee 20, on B = Rupee 30.
Formulate and solve the LPP graphically.
Four students of class XII are given a problem to solve independently. Their respective chances of solving the problem are: \[ \frac{1}{2},\quad \frac{1}{3},\quad \frac{2}{3},\quad \frac{1}{5} \] Find the probability that at most one of them will solve the problem.
If \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \] find \( A^{-1} \).
Using \( A^{-1} \), solve the following system of equations:
\[ \begin{aligned} 2x - 3y + 5z &= 11 \quad \text{(1)} \\ 3x + 2y - 4z &= -5 \quad \text{(2)} \\ x + y - 2z &= -3 \quad \text{(3)} \end{aligned} \]