RD Sharma Solutions Class 12 Mathematics Chapter 30 MCQ

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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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RD Sharma Class 12 Solutions Chapter29 MCQ Linear Programming - Other Exercise
Linear Programming Excercise: MCQ
RD Sharma Chapter wise Solutions

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

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

Linear Programming Excercise: MCQ

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.

Linnear Programing Exercise Multiple Choice Question 2

Answer: (b) A function to be optimized
Hint: We know the condition $\text {Z=CX}$
Given: Objective function of LPP______
Solution:
Let, $Z_{\max / \min }=C X$
The condition may be
$\begin{aligned} &a_{1} x \geq b_{1} \\ &a_{1} x \leq b_{2} \\ &x \geq 0 \end{aligned}$
So, the objective function of LP is a function to be optimized.

Linear Programming Exercise Multiple Choice Question 3

Answer: (d) $\{(x, y): y \geq 2, y \leq 4\}$
Hint:
For convex set, all points should be inside the set.
Given:
$\begin{aligned} &A=x^{2}+y^{2} \geq 1, B=y^{2} \geq x, C=3 x^{2}+4 y^{2} \geq 5 \\ &D=y \geq 2, y \leq 4 \end{aligned}$
Solution :
$\begin{aligned} &A=x^{2}+y^{2} \geq 1 \\ &x^{2}+y^{2}=1 \end{aligned}$
At $(0,0)$ we have $0\geq 1$ , 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
$\begin{aligned} &B=y^{2} \geq x \\ &y^{2}=x \end{aligned}$

At $(0,0)$ we have $0\geq 0$ , 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
$\begin{aligned} &C=3 x^{2}+4 y^{2} \geq 5 \\ &\Rightarrow \frac{x^{2}}{\frac{5}{3}}+\frac{y^{2}}{\frac{5}{4}} \geq 1 \end{aligned}$

At $(0,0)$ we have $0\geq 1$ , 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, $D=y \geq 2, y \leq 4$
It is only possible when $y=3$

So, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 4

Answer: (b) $x=\lambda x_{1}+(1-\lambda) \times, 0 \leq \lambda \leq 1$ is an optimal solution.
Hint: As per known condition
Given: $\text {x}_{1}$ and $\text {x}_{2}$ are optimal solution.
Solution:
A set $A_{1}$ is convex if, for any two points $x_{1}, x_{2} \in A \text { and } \lambda \in[0,1]$ imply that $\lambda x_{1}+(1-\lambda) x_{2} \in A$
Since, here $\text {x}_{1}$ and $\text {x}_{2}$ are optimal solution.
Therefore, their convex combination will also be an optimal solution
Thus, option (b) $x=\lambda x_{1}+(1-\lambda) x, 0 \leq \lambda \leq 1$ gives an optimal solution.

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 $2 x+3 y \leq 18, x+y \geq 10, x, y \geq 0$
Solution:
$2 x+3 y \leq 18$ and $x+y \geq 10$
$\frac{x}{9}+\frac{y}{6} \leq 1$

We observe having drawn the graph the inequalities of equation, $2 x+3 y \leq 18$ is inward and of $x+y \geq 10$ 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

Given by intersection of inequalities with axes only which is incorrect.
Given by intersection of inequalities with x-axis only which is also incorrect
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 $3 x+2 y \geq 160,5 x+2 y \geq 200, x+2 y \geq 80, x, y \geq 0$
Solution :
By the given constraints,
$\begin{aligned} &3 x+2 y \geq 160,5 x+2 y \geq 200, x+2 y \geq 80 \\ &\left(\frac{x}{\frac{160}{3}}\right)+\left(\frac{y}{80}\right) \geq 1, \frac{x}{40}+\frac{y}{100} \geq 1 \\ &\frac{x}{80}+\frac{y}{40} \geq 1, x, y \geq 0 \end{aligned}$
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) $2 x-y \geq 10$
Hint : Draw the graph and puts points on it
Given : $z_{\min }=6 x+10 y$ subject to the constraints $x \geq 6 ; y \geq 2 ; 2 x+y \geq 10 ; x, y \geq 10$
Solution :
$\begin{aligned} &x \geq 6 ; y \geq 2 ; 2 x+y \geq 10 ; x, y \geq 10 \\ &x=6, y=2,2 x+y=10 \end{aligned}$
By the given condition,

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

Linear Programming Exercise Multiple Choice Question 9

Answer: (c) Infinite number of points
Hint: Firstly convert the inequality into equations
Given: $z_{\max }=4 x+3 y$ subject to the constraints $3 x+4 y \leq 24,8 x+6 y \leq 48, x \leq 5, y \leq 6, x, y \geq 0$
Solution:
Let’s convert the given inequalities into equations, we obtain the following equations,
$3 x+4 y=24,8 x+6 y=48, x=5, y=6, x=0, y=0$
The line $3 x+4 y=24$ meets the coordinate axis at $A(8,0)$ and $B(0,6)$ join these points to obtain the line $3 x+4 y=24$
Clearly, $(0,0)$ satisfies the inequality $3x+4y\leq 24$ . So, the region in $xy$ plane that contains the origin represents the solution set of given equation.
The line $8x+6y=48$ meets the coordinate axis at $C(6,0)$ and $D(0,8)$ join these points to obtain the line $8x+6y=48$
Clearly, $(0,0)$ satisfies the inequality $8x+6y\leq 48$ . So, the region in $xy$ plane that contains the origin represents the solution set of given equation.
$x=5$ is the line passing through $x=5$ in y-axis region represented by $x\geq 0$ and $y\geq 0$ . Since every point in the first quadrant satisfies these in inequations.
So, the first quadrant is the region represented by the inequations.
The corner points of the feasible region are $O(0,0), G(5,0), F\left(5, \frac{4}{3}\right), E\left(\frac{24}{7}, \frac{24}{7}\right)$ and $B(0,6)$
The values of $z$ at these corners points are as follows,

$\text {corner point}$	$\text{z}=\text {4x+3y}$
$O(0,0)$	$4(0)+3(0)=0$
$G(5,0)$	$4(5)+3(0)=20$
$F\left ( 5,\frac{4}{3} \right )$	$4(5)+3\left ( \frac{4}{3} \right )=24$
$E\left ( \frac{24}{7},\frac{24}{7} \right )$	$4(\frac{24}{7})+3(\frac{24}{7})=\frac{196}{7}=24$
$B(0,6)$	$4(0)+3(6)=18$

We see that maximum value of the objective function $z$ is 24 which is at $F\left(5, \frac{4}{3}\right)$ and $E\left(\frac{24}{7}, \frac{24}{7}\right)$ . Thus, the optimal value of $z$ is 24.
Therefore, the given objective function can be subjected at an infinite number of points.
The correct option is (c).

Linear Programming Exercise Multiple Choice Question 10

Answer:
(a) The problem is to be re-evaluated.
Hint:
As per LPP condition
Given:
If the constraints in a linear programming are changed
Solution:
The optimization of the objective function of LPP is governed by the constraints
Therefore, if the constraints in a linear programming problem are changed, then the problem needs to be revaluated.
So, the correct option is (a).

Linear Programming Exercise Multiple Choice Question 11

Answer:
(c) If a LPP admits two optimal solution. It has an infinite number of optimal solution
Hint:
LPP condition
Given:
a) Every LPP admits an optional solution
b) A LPP admits unique optimal solution
c) If a LPP admits two optimal solution. It has an infinite number of optimal solution
d) The set of all feasible solutions of a LPP is not a converse set
Solution:
It is known that the optimal solution of an LPP either exists uniquely, does not exist or exist infinitely.
So, if an LPP admits two optimal solution, it has an infinite number of optimal solution.
Thus, the correct option is (c).

Linear Programming Exercise Multiple Choice Question 12

Answer : (c)
$\{x:|x|=5\}$
Hint:
Convex set is a set in which all points joining the line segment lies inside the set
Given:
a) $\begin{aligned} &\{(x, y): 2 x+5 y<7\} \\ \end{aligned}$
b) $\begin{aligned} &\left\{(x, y): x^{2}+y^{2} \leq 4\right\} \\ \end{aligned}$
c) $\begin{aligned} &\{x:|x|=5\} \\ \end{aligned}$
d) $\begin{aligned} &\left\{(x, y): 3 x^{2}+2 y^{2} \leq 6\right\} \end{aligned}$
Solution :
$\left | x \right |=5$ is not a convex set as any two points from negative and positive x-axis, if are joined will not lie in the set.
Since, option (a), (b) and (d) is a convex set and option (c) is not a convex set. So, the correct option is (c).

Linear Programming Exercise Multiple Choice Question 13

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

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

So, the correct option is (b).

Linear Programming Exercise Multiple Choice Question 14

Answer: (c) bounded in first quadrant
Hint:
Convert the inequalities into equation
Given:
$x, y \geq 0, y \leq 6, x+y \leq 3$
Solution :
Let’s convert the given inequalities into equations, we obtain
$y=6, x+y=3, x=0, y=0$
$y=6$ is the line passing through $(0,6)$ and parallel to the x-axis. The region below the line $y=6$ will satisfy the given inequalities.
The line $x+y=3$ meets the coordinate axis at $A (3,0)$ and $B (0,3)$ . Join these point to obtain the line $x+y=3$ . Clearly $(0,0)$ satisfies the inequalities $x+y\leq 3$ . So, the region in $xy$ -plane that contains the origin represents the solution set of the given equation.
Region represented by $x\geq 0$ and $y\geq 0$
Since, every point in the first quadrant satisfies these inequalities. So, the first quadrant is the region represented by the inequalities.
So, the correct option is (c).

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 $x+2 y \leq 70,2 x+y \leq 95, x, y \geq 0$
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 $70$
$\begin{aligned} &\frac{x}{70}+\frac{2 y}{70}=\frac{70}{70} \\ &\frac{x}{70}+\frac{y}{35}=1 \end{aligned}$
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 $95$
$\begin{aligned} &\frac{2 x}{95}+\frac{y}{95}=\frac{95}{95} \\ &\frac{x}{\frac{95}{2}}+\frac{y}{95}=1 \end{aligned}$
The line $2x+y=95$ can be plot in the graph as a line passing through the points, $\left ( \frac{95}{2},0 \right )$ 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 $x\geq 0$ and $y\geq 0$ , 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 $z=x+y$
$\begin{aligned} &z_{\text {at }} O(0,0)=0+0=0 \\ \end{aligned}$
$z_{\text {at }} A(0,35)=0+35=35 \\$
$z_{\text {at }} B(40,15)=40+15=55 \\$
$z \text { at } C\left(\frac{95}{2}, 0\right)=\frac{95}{2}+0=\frac{95}{2}=47.5$
From the values, it is cler that $z$ maximized at $B(40,15)$ .

Linear Programming Exercise Multiple Choice Quetion 16

Answer:
(c) at any vertex of feasible region
Hint:
As per LPP conditions
Given:
The value of objective function is maximum under linear constraints_____
Solution:
In linear programming problem, we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value.
Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.
So, the correct option is (c).

Linear Programming Exercise Multiple Choice Question 17

Answer : (d) $q=3p$
Hint:
Find $z$ by the given corner points
Given:
Linear inequalities,
$\\2 x+y \leq 10, x+3 y \leq 15, x, y \geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5)$
Let $z=p x+q y$ where, $p, q>0$
Solution :
Let's find $z=p x+q y$ by the given corner points

Corner Points	Value of $z$
$(0,0)$	$0$
$(5,0)$	$5p$
$(3,4)$	$3p+4q$
$(0,5)$	$5q$

Now comparing corner points $(3,4)$ and $(0,5)$
$\text {Value on }(3,4)=\text {value on}(0,5)$
$3\text{p}+4\text{q}=5\text{q}$
$3\text{p}=\text{q}$
So, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 18

Answer : (d) $\text{q = 3p}$
Hint : Put $(15,15)$ and $(0,20)$ in $\text {z = px + qy}$
Given : Linear constraints are $(0,10),(5,5),(15,15),(0,20)$
Let $\text {z = px + qy}$ where $\text {p,q}>0$ . $\text {z}_{\text {max}}$ occurs at both points $(15,15)$ and $(0,20)$ ________
Solution :
$\begin{aligned} &z_{\max (15,15)}=z_{\max (0,20)} \\ \end{aligned}$
$\begin{aligned} &p(15)+q(15)=p(0)+q(20) \\ \end{aligned}$
$\begin{aligned} &p(15)=q(20)-q(15) \\ \end{aligned}$
$\begin{aligned} &15 p=5 q \\ \end{aligned}$
$\begin{aligned} &3 p=q \end{aligned}$
So, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 19

Answer : (b) $\text {q = 2p}$
Hint : Put the given corner points in $\text {z = px + qy}$
Given : Linear constraints $(0,3),(1,1),(3,0)$
Let $\text {z = px + qy}$ where $\text {p,q}>0$
Solution:
$\text {z}_{\text {min}}=\text {px + qy}$
By the given corners,
$\begin{aligned} &\mathrm{At}(1,1), z_{1}=p+q \\ \end{aligned}$
$\begin{aligned} &\mathrm{At}(3,0), z_{2}=3 p \end{aligned}$
Now,
$\begin{aligned} &z_{1}=z_{2} \\ \end{aligned}$
$\begin{aligned} &p+q=3 p \\ \end{aligned}$
$\begin{aligned} &q=2 p \end{aligned}$

Linear Programming Exercise Multiple Choice Question 20

Answer: (d) any point on the line segment joining the points $(0,2)$ and $(3,0)$
Hint: Put the corners on $\text {z = 4x + 6y}$
Given: Corner points of the feasible region for an LPP are $(0,2),(3,0),(6,0),(6,8),(0,5)$
Let $\text {z = 4x + 6y}$
Solution:
Let’s put the corners on $\text {z = 4x + 6y}$

Corner points	Corresponding value of $\text {z = 4x + 6y}$
$(0,2)$	12 (minimum)
$(3,0)$	12 (minimum)
$(6,0)$	24
$(6,8)$	72 (maximum)
$(0,5)$	30

The maximum value of $z$ occurs at any point on the line segment joining the points $(0,2)$ and $(3,0)$
So, the correct option is (d).

Linear Programming Exercise Multiple Choice Question 21

Answer : (a) $60$
Hint : Put the given corners point on $\text {z = 4x+6y}$
Given : Corner points of the feasible region for an LPP are $(0,2),(3,0),(6,0),(6,8),(0,5)$
Let $\text {z = 4x+6y}$
Solution :
Let's put the corners on $\text {z = 4x+6y}$

Corner points	Corresponding value of $\text {z = 4x+6y}$
$(0,2)$	12 (minimum)
$(3,0)$	12 (minimum)
$(6,0)$	24
$(6,8)$	72 (maximum)
$(0,5)$	30

According to the question $Z_{\max }-z_{\min }$
$\begin{aligned} &z_{\max }=72 \text { and } z_{\min }=12 \\ \end{aligned}$
$\begin{aligned} &=z_{\max }-z_{\min } \\ \end{aligned}$
$\begin{aligned} &=72-12 \\ \end{aligned}$
$\begin{aligned} &=60 \end{aligned}$
So, the correct option is (a).

Linear Programming Exercise Multiple Choice Question 22

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

Solution :

Corner points	$z=3 x-4 y$
$(0,4)$	-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 23

Answer : (b) quantity in column B is greater
Hint : Putting the corner point in $z=4 x+3 y$
Given :
Corner points of the feasible region determined by the system of linear constraints are
$(0,0),(0,40),(20,40),(60,20),(60,0)$
The objective finction is $z=4 x+3 y$
Solution :

Corner points	$z=4x+3y$
$(0,0)$	$0$
$(0,40)$	$120$
$(20,40)$	$200$
$(60,20)$	$300\; \text {(maximum)}$
$(60,0)$	$240$

$z_{max}=300<325$
So, the correct option is (b) which is quantity in column B is greater.

Linear Programming Exercise Multiple Choice Question 24

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

Solution :

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

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 $z=3 x-4 y$
Given :
Let $z=3 x-4 y$

Solution :

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

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

Linear Programming Exercise Multiple Choice Question 26

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

Solution:

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

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
$\begin{aligned} &3 y>6-2 x \\ &y>\frac{6-2 x}{3} \end{aligned}$
Now consider $y=\frac{6-2 x}{3}$ and the plot the graph

x	0	3
y	2	0

We used dashed as
$y>\frac{6-2 x}{3}$ as there is no equal to sign
Therefore, $y$ is greater than , $\frac{6-2 x}{3}$ 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 $2x+3y=6$

Linear Programming Exercise Multiple Choice Question 28

Answer:
(b) A linear function to be optimized
Hint:
We know the LPP condition
Given:
The objective function of an LPP is ____
Solution:
The objective of linear programming problem (LPP) is to minimise or maximise the function
So, the correct option is (b) which is a function to be optimized

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