MHT CET 2026 19th April Evening Shift | Linear Programming Question 3 | Mathematics

An airplane can carry a maximum of $250$ passengers. A profit of Rs $1500$ is made on each executive class ticket and a profit of Rs $900$ is made on each economy class ticket. The airline reserves at least $30$ seats for executive class. However at least $4$ times as many passengers prefer to travel by economy class than by executive class. Let $x_1$ be the number of passengers of executive class and $x_2$ be the number of passengers of economy class. Formulate the LPP in order to maximize the profit for the airline...

Maximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \leq 30$, $x_2 \leq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.

Minimize $z = 150x_1 + 90x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.

Minimize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.

Maximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

-0

The shaded region in the provided graph represents the solution set for which of the following systems of linear inequalities?
MHT CET 2026 19th April Morning Shift Mathematics - Linear Programming Question 1 English

MHT CET 2026 19th April Morning Shift Mathematics - Linear Programming Question 1 English

$2x + y \geq 2,\ x - y \geq 1,\ x + 2y \leq 8,\ x \geq 0,\ y \geq 0$

$x + 2y \geq 2,\ x - y \geq 1,\ x + 2y \leq 8,\ x \geq 0,\ y \geq 0$

$2x + y \geq 2,\ x - y \leq 1,\ x + 2y \leq 8,\ x \geq 0,\ y \geq 0$

$2x + y \geq 2,\ x - y \leq 1,\ 2x + y \leq 8,\ x \geq 0,\ y \geq 0$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

The feasible region for the constraints $x-y \geq 0, x-5 y \leq-5, x \geq 0, y \geq 0$ is shown by the figure:

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 5 English Option 1

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 5 English Option 2

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 5 English Option 3

MHT CET 2025 5th May Evening Shift Mathematics - Linear Programming Question 5 English Option 4

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

A manufacturing company produces two items, A and B. Each toy should be processed by two machines, I and II. Machine I can be operated for maximum 10 hours 40 minutes. It takes 20 minutes for an item of A and 15 minutes for B. Machine II can be operated for a total time at 8 hours 20 minutes. It takes 5 minutes for an item A and 8 minutes for B . The profit per item of $A$ is $Rs 25$ and per item of $B$ is ₹ 18 . The formulation of an L.P.P. to maximize the profit (where $x$ is number of items A and $y$ is the number of item $B$ ) is

$$ \begin{aligned} & \text { Maximize } \mathrm{z}=25 x+18 y \\ & \text { subject to } 20 x+15 y \leq 640 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 5 x+8 y \geq 500 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x, y \geq 0 \end{aligned} $$

Maximize $z=25 x+18 y$

$$ \begin{aligned} \text { subject to } 20 x+15 y & \leq 640 \\ 5 x+8 y & \leq 500 \\ x, y & \geq 0 \end{aligned} $$

$$ \begin{array}{r} \text { Maximize } z=25 x+18 y \\ \text { subject to } 20 x+5 y \leq 8 \\ 5 x+8 y \leq 10 \\ x, y \geq 0 \end{array} $$

$$ \begin{aligned} & \text { Maximize } \mathrm{z}=25 x+18 y \\ & \text { subject to } 4 x+3 y \leq 128 \\ & \qquad \begin{array}{r} 5 x+8 y \geq 500 \\ x, y \geq 0 \end{array} \end{aligned} $$

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