MCQ Questions for Class 12 Commerce Maths Linear Programming Quiz 6

$x+y \leq 2, x\leq 0, y\leq 0$ the point at which maximum value of

$3x+2y$ attained will be.

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$(0, 0)$
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$\left ( \dfrac{1}{2}, \dfrac{1}{2} \right )$
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$(0, 2)$
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$(2, 0)$

Explanation

$x \le 0$ and

$y \le 0$ represents third Quadrant

$x+y \le 2$ represents the region below the line

$x+y \le 2$

$($ the region which contains origin

$)$

The common region of given set of equations is third quadrant

$($ including negative

$x$ axis and negative

$y$ axis

$)$

Since

$x$ and

$y$ values are

$\le 0$ in the third quadrant , the maximum value of

$3x+2y$ occurs at

$x=0$ and

$y=0$ and the maximum value is

$0$

Therefore the correct option is

$A$

In figure 32, the shaded region within the triangle is the intersection of the sets of ordered pairs described by which of the following inequalities?

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y < x, x < 2
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y < 2x, x < 2
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y < 2x, x < 2, x > 0
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y < 2x, y < 2, x > 0
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y < 2x, x < 2, y > 0

The linear programming problem:
Maximize

$z={ x }_{ 1 }+{ x }_{ 2 }$
Subject to constraints

$\quad { x }_{ 1 }+2{ x }_{ 2 }\le 2000,{ x }_{ 1 }+{ x }_{ 2 }\le 1500,\quad { x }_{ 2 }\le 600,\quad { x }_{ 1 }\ge 0$

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No feasible solution
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Unique optimal solution
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A finite number of optimal solutions
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Infinite number of optimal solutions

Solve the following LPP graphically. Maximize or minimize

$Z = 3x + 5y$ subject to

$3x - 4y \geq -12$

$2x - y + 2\geq 0$

$2x + 3y - 12\geq 0$

$0 \leq x \leq 4$

$y \geq 2$ .

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Min. value $19$ at $(5, 2)$ and Max. value $42$ at $(4, 6)$ .
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Min. value $30$ at $(3, 2)$ and Max. value $42$ at $(4, 6)$ .
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Min. value $19$ at $(3, 2)$ and Max. value $42$ at $(4, 6)$ .
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Min. value $8$ at $(3, 2)$ and Max. value $42$ at $(4, 6)$ .

Explanation

To maximise :

$3x+5y$

Constraints:

$3x-4y\ge -12$

$2x-y\ge -2$

$2x+3y\ge 12$

$0\le x\le 4$

$y\ge 2$

The shaded region satisfies the given constraints.

From critical points;

Minimum at

$(3,2)$

$Z=3(3)+5(2)=9+10=19$

Maximum at

$(4,6)$

$Z=4(3)+5(6)=42$

To sum up,

min.

$19$ at

$(3,2)$ and

max.

$42$ at

$(4,6)$

2108754_846574_ans_0f10387e682742478216fbd6172ab134.png

Let

$P(-1, 0), Q(0, 0)$ and

$R(3, 3\sqrt{3})$ be three points. The equation of the bisector of the angle PQR is?

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$x+\sqrt{3}y=0$
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$\sqrt{3}x+y=0$
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$x+\dfrac{\sqrt{3}}{2}y=0$
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$\dfrac{\sqrt{3}}{2}x+y=0$

Explanation

Given points are

$P(-1,0),Q(0,0)$ and

$R(3, 3\sqrt{3})$

Slope of

$l\, ne \,QR=\sqrt 3$

Slope of lime

$QM=\tan \dfrac{2\pi}{3}$

$m=(\sqrt 3)$

Hence, eqn of line

$QM$ is

$y=mx+c$

$\therefore y=0\, -\sqrt {3x} +0$ [taking

$Q(0,0)$ ]

$\boxed {\sqrt{3x}+y=0}$

2113936_1058931_ans_04ad7c884a66429fa50ab6dc251c4939.png

Use graph paper for this question:

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Plot the points $A(-4, 2)$ and $B(2, 4)$
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${A}^{1}$ is the image of $A$ when reflected in the line $x=0$ . Write the co-ordinates of ${A}^{1}$ .
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${B}^{1}$ is the image of $B$ when reflected in the line $A{A}^{1}$ . Write the co-ordinates of ${B}^{1}$ .
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Write the geometrical name of the figure $AB{A}^{1}{B}^{1}$

Explanation

1) plot points

$A(-4,2) and\, B(2,4)$

2) A' image of A in

$x=0$

$\therefore A'\equiv (4,2)$

3) B' image of B in AA'

$\therefore B'=(2,0)$

4) shape of ABA'B' is shaped quadrilateral.

2110446_1037556_ans_33b4ca87eadd4fcba379bb3840f34cf6.png

Minimise and Maximise

$Z = x + 2y$ subject to the following constraints

$x + 2y \ge 100, ~2x - y \le 0, ~y \le 200$ and

$x, y \ge 0$

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Minimum $200$ , Maximum $400$
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Minimum $100$ , Maximum $500$
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Minimum $400$ , Maximum $500$
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Minimum $100$ , Maximum $400$

Let

$X_1 are X_2$ are optimal solution of a LPP, then

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$x = \lambda x_1 + (1 - \lambda) x_2, \lambda \epsilon R$ is also an optimum solution
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$X = \lambda x_1 + (1 - \lambda) X_2, 0 \le \lambda \le I$ gives an option
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$X = \lambda x_1 + (1 + \lambda). X_2, 0 \le \lambda \le 1$ gives an optimal solution
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$X = \lambda X_1 + (1 + \lambda) X_2, \lambda \epsilon R$ gives an optimal

Which of the following statement id correct?

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Every $LLP$ admits an optimal solution
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An $LLP$ admits unique optimal solution
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If an $LPP$ admits two optimal solutions it has infinite number of optimal solution
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None of these

The point which provides the solution to the linear programming problem : Max P= 2x+3y subject to constraints :

$x\geq 0, y\geq 0,2x+2y\leq 9,2x+y\leq 7,x+2y\leq 8,$ is

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(3,2.5)
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(2,3.5)
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(2,2.5)
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(1,3.5)

Feasible region is the set of points which satisfy

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the objective function
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all the given constraints
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some of the given constraints
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only one constraint

If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0), then the point of minimum z = 3x + 2y is

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(2, 2)
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(0, 10)
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(4, 0)
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(3, 4)

The maximum value of z=10x + 6y subject to the constraints

$3x + y \leq 12, 2x + 5y \leq 34, x \geq 0, y \geq 0$ is

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56
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65
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55
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66

The point of which the maximum value of x + y subject to the constraints

$x + 2y \leq 70, 2x + y \leq 95, x \geq 0, y \geq 0.$

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(30, 25)
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(20,25)
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(35, 20)
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(40, 15)

The maximum value of z = 5x + 3y subject tot the constraints

$3x + 5y \leq 15, 5x + 2y \leq 10, x, y \geq 0$ is

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235
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$\dfrac{235}{9}$
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$\dfrac{235}{19}$
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$\dfrac{235}{3}$

If the corner points of the feasible solution are (0, 0), (3, 0), (2, 1),

$\left ( 0, \dfrac{7}{3} \right )$ the maximum value of z = 4x + 5y is

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12
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13
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$\dfrac{35}{3}$
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0

CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Linear Programming Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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