RD Sharma Class 12 Exercise MCQ Linear Programing Solutions Maths - Download PDF Free Online

RD Sharma class 12th exercise MCQ is not your standard NCERT solution. This book is one of the top-rated self-study guides that has already won the trust and respect of innumerable students and teachers. The RD Sharma class 12 chapter 29 exercise MCQ comes with a ton of answers that are simple to read and easy to understand. It can be a game-changer for many students who struggle to study on their own.

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The RD Sharma class 12 chapter 29 exercise MCQ is a trusted NCERT solution that has already helped hundreds of students in their exams. The 29th chapter of the book is titled Linear Programming. Exercise MCQ has 28 questions that cover concepts: Objective function of LPP, Sets are convex, Maximum value and Minimum value subjected to constraints, Optimal Value, and Feasible region. The RD Sharma class 12th exercise MCQ will help you revise all concepts that you have learned in the chapter.

RD Sharma Class 12 Solutions Chapter29 MCQ Linear Programming - Other Exercise

• Chapter 29- Linear Programming - Ex-29.1

• Chapter 29- Linear Programming - Ex-29.2

• Chapter 29- Linear Programming - Ex-29.3

• Chapter 29- Linear Programming - Ex-29.4

• Chapter 29- Linear Programming - Ex-29.5

• Chapter 29- Linear Programming - Ex-FBQ

Linear Programming Exercise Multiple Choice Question 1

Answer: (b)Open half plane not containing the origin.
Hint: Put $x=0$ and $y=0$ in given equation
Given: The solution set in equation $\text{2x+y}>5$
Solution: Let $\text{x,y}$ plane

By the given equation $\text{2x+y}>5$
$x=0$ and $y=0$
$y>5$ and $x>\frac{5}{2}$
In the above drawn plane, option a, c are not satisfied.
So, the correct option is (b) which is open half plane but not containing the origin.

Linear Programming Exercise Multiple Choice Question 3

Answer: (d) $\{(x,y):y\ge 2,y\le 4\}$
Hint:
For convex set, all points should be inside the set.
Given:
$\begin{array}{rl}& A=x^2+y^2\ge 1,B=y^2\ge x,C=3x^2+4y^2\ge 5\\ & D=y\ge 2,y\le 4\end{array}$
Solution :
$\begin{array}{rl}& A=x^2+y^2\ge 1\\ & x^2+y^2=1\end{array}$
At $(0,0)$ we have$0\ge 1$ , which is wrong. So the region of set A will not contain the origin.

Here, its all points are outside the set. So option (a) is incorrect
$\begin{array}{rl}& B=y^2\ge x\\ & y^2=x\end{array}$

At $(0,0)$ we have $0\ge 0$, which is wrong. So the region of set B will not contain the origin.
We observe that option (b) is also incorrect because, it’s all points are outside the set
$\begin{array}{rl}& C=3x^2+4y^2\ge 5\\ & \Rightarrow \frac{x^2}{\frac{5}{3}}+\frac{y^2}{\frac{5}{4}}\ge 1\end{array}$

At $(0,0)$ we have $0\ge 1$, which is wrong. So, the region of set C will not contain the origin. We observe that option (c) also incorrect. Because it fails the condition of convex set
Now, $D=y\ge 2,y\le 4$
It is only possible when $y=3$

So, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 5

Answer: (d) None of the options
Hint: Having drawn the graph find ${\text{Z}}_{\text{max}}$
Given: ${\text{Z}}_{\text{max}}=\text{4x}+\text{2y}$ subject to the constraints $2x+3y\le 18,x+y\ge 10,x,y\ge 0$
Solution:
$2x+3y\le 18$ and $x+y\ge 10$
$\frac{x}{9}+\frac{y}{6}\le 1$

We observe having drawn the graph the inequalities of equation, $2x+3y\le 18$ is inward and of $x+y\ge 10$ is outward
So, we are not getting feasible region and ${\text{Z}}_{\text{max}}$ cannot be determined
Thus, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 6

Answer: (c) given by corner points of the feasible region.
Hint : Let $z=2x+3y$
Given :
The optimal value of objective function is attained at the point.
Solution:
Let we have a graph and plot some constraints

Now, we check the options

1. Given by intersection of inequalities with axes only which is incorrect.
2. Given by intersection of inequalities with x-axis only which is also incorrect
3. Given by corners points of feasible region which is correct. Because we get the optimal value in the feasible region.

So, the correct option is (c).

Linear Programming Exercise Multiple Choice Question 7

Answer: (d) None of these
Hint:
Put the extracted intercepts on the graph
Given: ${\text{z}}_{\text{max}}=4x+3y$ subject to the constraints $3x+2y\ge 160,5x+2y\ge 200,x+2y\ge 80,x,y\ge 0$
Solution :
By the given constraints,
$\begin{array}{rl}& 3x+2y\ge 160,5x+2y\ge 200,x+2y\ge 80\\ & \left(\frac{x}{\frac{160}{3}}\right)+\left(\frac{y}{80}\right)\ge 1,\frac{x}{40}+\frac{y}{100}\ge 1\\ & \frac{x}{80}+\frac{y}{40}\ge 1,x,y\ge 0\end{array}$
Now let’s plot the points in the graph

Graph moves outwards as per given situation. Therefore, we are getting unbounded region. So $Z_{max}$ cannot be determined. Thus, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 8

Answer : (c) $2x-y\ge 10$
Hint : Draw the graph and puts points on it
Given : $z_{min}=6x+10y$ subject to the constraints $x\ge 6;y\ge 2;2x+y\ge 10;x,y\ge 10$
Solution :
$\begin{array}{rl}& x\ge 6;y\ge 2;2x+y\ge 10;x,y\ge 10\\ & x=6,y=2,2x+y=10\end{array}$
By the given condition,

The shaded region represent the feasible of the given LPP
We observe that the feasible region is due to constraint $x\ge 6;y\ge 2$
Therefore, the redundant constraint is $2x+y\ge 10$.

Linear Programming Exercise Multiple Choice Question 13

Answer : (b) $x_1=2,x_2=6,z=36$
Hint :
Convert the inequalities into equation
Given: $z_{max}=3x_1+5x_2$ Subject to the constraints are
$\begin{array}{rl}& 3x_1+2x_2\le 18\\ & x_1\le 4\\ & x_2\le 6\\ & x_1\ge 0,x_2\ge 0\end{array}$
Solution :
By the graphical method,
Now,
$\begin{array}{rl}& 3x_1+2x_2\le 18\\ & \frac{x_1}{6}+\frac{x_2}{9}\le 1\end{array}$

Checking coordinates,
$\begin{array}{rl}& \mathrm{At}(0,0)\to z=0\end{array}$
$\begin{array}{rl}& \mathrm{At}(4,0)\to z=12\end{array}$
$\begin{array}{rl}& \mathrm{At}(4,3)\to z=27\end{array}$
$\begin{array}{rl}& \mathrm{At}(2,6)\to z=36\end{array}$
$\begin{array}{rl}& \mathrm{At}(0,6)\to z=30\end{array}$
$\begin{array}{rl}& z_{max}={36}_{\mathrm{at}}(2,6)\end{array}$

So, the correct option is (b).

Linear Programming Exercise Multiple Choice Question 15

Answer : (d) $(40,15)$
Hint :
Convert the given inequalities into equation
Given:
$Z_{max}=x+y$ subject to the constraints $x+2y\le 70,2x+y\le 95,x,y\ge 0$
Solution:
Let us consider the mentioned constraints as equations for a while,
$x+2y=70$ ....(i)
$2x+y=95$ ....(ii)
Now, graph the equations by transforming the equations to intercept form of line.
Equation (i) dividing throughout by $70$
$\begin{array}{rl}& \frac{x}{70}+\frac{2y}{70}=\frac{70}{70}\\ & \frac{x}{70}+\frac{y}{35}=1\end{array}$
The line $x+2y=70$ can be plot in the graph as a line passing through the points, $(70,0)$ and $(0,35)$ as $70$ and $35$ are the intercepts of the line on the x-axis and y-axis respectively.
Similarly, Equation (ii) can be divided by $95$
$\begin{array}{rl}& \frac{2x}{95}+\frac{y}{95}=\frac{95}{95}\\ & \frac{x}{\frac{95}{2}}+\frac{y}{95}=1\end{array}$
The line $2x+y=95$can be plot in the graph as a line passing through the points, $(\frac{95}{2},0)$ and $(0,95)$ as $\frac{95}{2}$ and $95$ are the intercepts of the line on the x-axis and y-axis respectively.
By considering the constraints $x\ge 0$ and $y\ge 0$, thus clearly shows that the region can only be in the first quadrant. The graph of inequalities will look like,

The points OABC is the feasible region of LPP
Now, form the points O,A,B and C the vertices of polygon formed by the constraints one of the points will provide the maximum solution $z=x+y$
Now, checking the points, O, A, B, and C by substituting in $z=x+y$
$\begin{array}{rl}& z_{\text{at }}O(0,0)=0+0=0\end{array}$
$z_{\text{at }}A(0,35)=0+35=35$
$z_{\text{at }}B(40,15)=40+15=55$
$z\text{ at }C(\frac{95}{2},0)=\frac{95}{2}+0=\frac{95}{2}=47.5$
From the values, it is cler that $z$ maximized at $B(40,15)$.

Linear Programming Exercise Multiple Choice Question 22

Answer : (b) 12
Hint : Putting the corner points in $z=3x-4y$
Given :
Let $z=3x-4y$

Solution :

Corner points	$z=3x-4y$
$(0,4)$	-16 (minimum)
$(0,0)$	0
$(12,6)$	$12(3)-4(6)=12\text{(maximum)}$

So, the correct option is (c) which is $z_{max}=12$

Linear Programming Exercise Multiple Choice Question 24

Answer : $(b)(0,8)$
Hint : Putting the corner points in $z=3x-4y$
Given :
Let $z=3x-4y$

Solution :

Corner points	$z=3x-4y$
$(5,0)$	$15$
$(6,5)$	$-2$
$(6,8)$	$-14$
$(4,10)$	$-28$
$(0,8)$	$-32\text{(minimum)}$

Minimum of $z=-32$ at $(0,8)$
So, the correct option is (b).

Linear Programming Exercise Multiple Choice Question 25

Answer : (a) $(5,0)$
Hint: Putting the corner points in $z=3x-4y$
Given :
Let $z=3x-4y$

Solution :

Corner points	$z=3x-4y$
$(5,0)$	$15\text{(maximum)}$
$(6,5)$	$-2$
$(6,8)$	$-14$
$(4,10)$	$-28$
$(0,8)$	$-32$

Maximum of $z=15$ at$(5,0)$
So, the correct option is (a).

Linear Programming Exercise Multiple Choice Question 26

Answer : (d)
Given let $z=3x-4y$
Hint :
Putting the corner points in $z=3x-4y$

Solution:

Corner points	$z=3x-4y$
$(5,0)$	$15\text{(maximum)}$
$(6,5)$	$-2$
$(6,8)$	$-14$
$(4,10)$	$-28$
$(0,8)$	$-32$

Maximum of $z=15$ aat $(5,0)$
Minimum of $z=-32(0,8)$
So maximum of z + minimum of $z=15-32=17$
Hence correct option is (d).

Linear Programming Exercise Multiple Choice Question 27

Answer : (b) half plane that neither contains the origin nor the points on the line $2x+3y=6$
Hint:
Convert the inequalities into equation
Given:
The graph of inequality,
$2x+3y>6$
Solution:
By the given inequality
$\begin{array}{rl}& 3y>6-2x\\ & y>\frac{6-2x}{3}\end{array}$
Now consider $y=\frac{6-2x}{3}$ and the plot the graph

x	0	3
y	2	0

We used dashed as
$y>\frac{6-2x}{3}$as there is no equal to sign
Therefore,$y$ is greater than , $\frac{6-2x}{3}$ we will shade the upper part because at (0,0) the inequality becomes 0 > 6, which is not correct.
Thus the graph of $2x+3y>6$drawn among the option (b)
So, the correct option (b), half-plane that neither contains origin nor the points of the line $2x+3y=6$

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RD Sharma Chapter wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable