M.C.Q (1 Marks)

MCQ 11 Mark

If the minimum value of an objective function $Z=a x+$ by occurs at two points $(3,4)$ and $(4,3)$ then

A
$a+b=0$
B
$a=b$
C
$3 a=b$
D
$a=3 b$

Answer

Since, minimum value of $Z=a x+b y$ occurs at two points $(3,4)$ and $(4,3)$.
$\therefore \quad 3 a+4 b=4 a+3 b \Rightarrow a=b$

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MCQ 21 Mark

For the following LPP, maximise $Z=3 x+4 y$ subject to constraints $x-y \geq-1, x \leq 3, x \geq 0, y \geq 0$ the maximum value is

A
$0$
B
4
C
25
D
30

Answer

Given, $Z=3 x+4 y$
Subject to constraints, $x-y \geq-1, x \leq 3 ; x \geq 0, y \geq 0$

The shaded region $O A B C$ is the feasible region, where corner points are $O(0,0), A(0,1), B(3,4)$ and $C(3,0)$.
At $O(0,0), Z=3(0)+4(0)=0$
At $A(0,1), Z=3(0)+4(1)=4$
At $B(3,4), Z=3(3)+4(4)=25$
At $C(3,0), Z=3(3)+4(0)=9$
$\therefore \quad$ Maximum value of $Z$ is 25 , which occurs at $B(3,4)$.

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MCQ 31 Mark

The feasible region of an LPP is given in the following figure

Then, the constraints of the LPP are $x \geq 0, y \geq 0$ and

A
$2 x+y \leq 52$ and $x+2 y \leq 76$
B
$2 x+y \leq 104$ and $x+2 y \leq 76$
C
$x+2 y \leq 104$ and $2 x+y \leq 76$
D
$x+2 y \leq 104$ and $2 x+y \leq 38$

Answer

Clearly, the pair of points given in graph, and $(0,104) ;(52,0)$ and $(0,38) ;(76,0)$ satisfy the corresponding equations given in option(b) i.e., $2 x+y \leq 104$ and $x+2 y \leq 76$.

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MCQ 41 Mark

Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function $Z=3 x+9 y$ maximum?

A
Point $B$
B
Point $C$
C
Point $D$
D
every point on the line segment $C D$

Answer

We have,

Corner points	Value of Z=3x+9y
A(0,10)	3xx0+9xx10=90
B(5,5)	3xx5+9xx5=60
C(15,15)	3xx15+9xx15=180 (Maximum)
D(0,20)	3xx0+9xx20=180 (Maximum)

$\because \quad Z$ is maximum at $C(15,15)$ and $D(0,20)$.
$\therefore \quad Z$ is maximum at every point on the line joining $C D$.

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MCQ 51 Mark

In a linear programming problem, the constraints on the decision variables $x$ and $y$ are $x-3 y \geq 0, y \geq 0$, $0 \leq x \leq 3$. The feasible region

A
is not in the first quadrant
B
is bounded in the first quadrant
C
is unbounded in the first quadrant
D
does not exist

Answer

From the graph, we can say that the feasible region is bounded in the first quadrant.

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MCQ 61 Mark

In the given graph, the feasible region for a LPP is shaded. The objective function $Z=2 x-3 y$, will be minimum at

A
$(4,10)$
B
$(6,8)$
C
$(0,8)$
D
$(6,5)$

Answer

We have,

Corner points	Value of Z=2x-3y
(0,0)	2xx0-3xx0=0
(0,8)	2xx0-3xx8=-24 (Minimum)
(4,10)	2xx4-3xx10=-22
(6,8)	2xx6-3xx8=-12
(6,5)	2xx6-3xx5=-3
(5,0)	2xx5-3xx0=10

$\therefore \quad$ Value of $Z$ is minimum at $(0,8)$.

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MCQ 71 Mark

The maximum value of $Z=3 x+4 y$ subject to the constraints $x \geq 0, y \geq 0$ and $x+y \leq 1$ is

A
7
B
4
C
3
D
10

Answer

We have to maximise $Z=3 x+4 y$
Subject to constraints, $x \geq 0, y \geq 0$ and $x+y \leq 1$

The shaded portion $O A B$ is the feasible region, where $O(0,0), A(1,0)$ and $B(0,1)$ are the corner points.
At $O(0,0), Z=3 \times 0+4 \times 0=0$
At $A(1,0), Z=3 \times 1+4 \times 0=3$
At $B(0,1), Z=3 \times 0+4 \times 1=4$
$\therefore \quad$ Maximum value of $Z$ is 4 , which occurs at $B(0,1)$.

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MCQ 81 Mark

If the corner points of the feasible region of an LPP are $(0,3),(3,2)$ and $(0,5)$, then the minimum value of $z=11 x+7 y$ is

A
21
B
33
C
14
D
35

Answer

Given, $Z=11 x+7 y$
At $(0,3), Z=11 \times 0+7 \times 3=21$
At $(3,2), Z=11 \times 3+7 \times 2=47$
At $(0,5), Z=11 \times 0+7 \times 5=35$
Thus, $Z$ is minimum at $(0,3)$ and minimum value of $Z$ is 21 .

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MCQ 91 Mark

The number of solutions of the system of inequations $x+2 y \leq 3,3 x+4 y \geq 12, x \geq 0, y \geq 1$ is

A
$0$
B
2
C
finite
D
infinite

Answer

Given,
\[x+2 y \leq 3,3 x+4 y \geq 12, x \geq 0, y \geq 1\]
The graph of given constraints is shown here.

Since, there is no common region, so, no solution exists.

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MCQ 101 Mark

For an objective function $Z=a x+b y$, where $a, b>0$; the corner points of the feasible region determined by a set of constraints (linear inequalities) are $(0,20),(10,10),(30,30)$ and $(0,40)$. The condition on $a$ and $b$ such that the maximum $Z$ occurs at both the points $(30,30)$ and $(0,40)$ is

A
$b-3 a=0$
B
$a=3 b$
C
$a+2 b=0$
D
$2 a-b=0$

Answer

As, $Z$ is maximum at $(30,30)$ and $(0,40)$.
$\Rightarrow \quad 30 a+30 b=40 b \Rightarrow b-3 a=0$

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MCQ 111 Mark

A linear programming problem is as follows:
Minimize $Z=30 x+50 y$
Subject to the constraints,
$\begin{array}{l}3 x+5 y \geq 15 \\ 2 x+3 y \leq 18 \\ x \geq 0, y \geq 0\end{array}$
In the feasible region, the minimum value of $Z$ occurs at

A
a unique point
B
no point
C
infinitely many points
D
two points only

Answer

Here, the feasible region is shaded.

Corner points	Value of Z=30 x+50 y
A(0,3)	30 xx0+50 xx3=150 (Minimum)
B(5,0)	30 xx5+50 xx0=150quad (Minimum)
C(9,0)	30 xx9+50 xx0=270
D(0,6)	30 xx0+50 xx6=300

Since, minimum value of $Z$ occurs at both $A$ and $B$. So, $Z$ is minimum at every point on the line joining $A B$. So, minimum value of Z occurs at infinitely many points.

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MCQ 121 Mark

The feasible region for an LPP is shown below: Let $z=3 x-4 y$ be the objective function. Minimum of $z$ occurs at

A
$(0,0)$
B
$(0,8)$
C
$(5,0)$
D
$(4,10)$

Answer

We know that minimum of objective function occurs at corner points.

Corner points	Value of z=3x-4y
(0,0)	0
(5,0)	15
(6,5)	-2
(6,8)	-14
(4,10)	-28
(0,8)	-32 larr Minimum

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MCQ 131 Mark

The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then:

A
a = 2b
B
2a = b
C
a = b
D
3a = b

Answer

a = 2b

Solution:

4a + 0b = 2a + 4b

4a = 2a + 4b

4a - 2a = 4b

2a = 4b

a = 2b

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MCQ 141 Mark

The maximum value of $Z=4 x+y$ for a L.P.P. whose feasible region is given below is:

A
50
B
110
C
120
D
170

Answer

We have, Max. $Z=4 x+y$
The corner points of feasible region are $O, A, B$ and $C$. Thus,
$
\begin{array}{l}
Z_{(0,0\rangle}=0 ; \\
Z_{\langle 0,50\rangle}=50 ; \\
Z_{\langle 20,30\rangle}=20 \times 4+30=110 ; \\
Z_{\{30,0\rangle}=4 \times 30=120 \\
\therefore \quad M a x Z=4 x+y \text { is } 120 .
\end{array}$

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MCQ 151 Mark

The common region determined by all the constraints of a linear programming problem is called:

A
an unbounded region
B
an optimal region
C
a bounded regio
✓
a feasible region

Answer

Correct option: D.

a feasible region

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MCQ 161 Mark

The corner points of the bounded feasible region determined by a system of linear constraints are $(0,3),(1,1)$ and $(3,0)$. Let $Z=p x+q y$, where $p, q>0$,. The condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3,0)$ and $(1,1)$ is

A
$p=2 q$
B
$p=\frac{q}{2}$
C
$p=3 q$
D
$p=q$

Answer

Corner point	Value of Z=px+qy;p,q > 0
(0,3)	p xx0+q xx3=3q
(1,1)	p xx1+q xx1=p+q
(3,0)	p xx3+q xx0=3p

The minimum of $Z$ occurs at $(3,0)$ and $(1,1)$
$
\therefore p+q=3 p \Rightarrow p=\frac{q}{2}
$

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MCQ 171 Mark

A linear programming problem (LPP) along with the graph of its constraints is shown below. The corresponding objective function is
Minimize: $Z=3 x+2 y$. The minimum value of the objective function is obtained at the corner point ( 2 , 0).
The optimal solution of the above linear programming problem $\qquad$

A
does not exist as the feasible region is unbounded.
B
does not exist as the inequality $3 x+2 y<6$ does not have any point in common with the feasible region.
C
exists as the inequality $3 x+2 y>6$ has infinitely many points in common with the feasible region.
✓
exists as the inequality $3 x+2 y<6$ does not have any point in common with the feasible region.

Answer

Correct option: D.

exists as the inequality $3 x+2 y<6$ does not have any point in common with the feasible region.

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MCQ 181 Mark

The feasible region of a linear programming problem is bounded. The corresponding objective function is $Z=6 x-7 y$.
The objective function attains $\qquad$ in the feasible region.

A
only minimum
B
only maximum
✓
both maximum and minimum
D
either maximum or minimum but not both

Answer

Correct option: C.

both maximum and minimum

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MCQ 191 Mark

The feasible region corresponding to the linear constraints of a Linear Programming Problem is given below

Which of the following is not a constraint to the given Linear Programming Problem?

A
$x+y \geq 2$
B
$x+2 y \leq 10$
C
$x-y \geq 1$
D
$x-y \leq 1$

Answer

We observe, $(0,0)$ does not satisfy the inequality
$x-y \geq 1$
So, the half plane represented by the above inequality will not contain origin therefore, it will not contain the shaded feasible region.

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MCQ 201 Mark

The solution set of the inequation $3 x+5 y<7$ is

A
whole $x y$-plane except the points lying on the line $3 x+5 y=7$.
B
whole $x y$-plane along with the points lying on the line $3 x+5 y=7$.
C
open half plane containing the origin except the points of line $3 x+5 y=7$.
D
open half plane not containing the origin.

Answer

Given inequation is $3 x+5 y<7$
Let us draw the graph of $3 x+5 y=7$

x	0	2.33
y	1.4	0

Substitute, $x=0$ and $y=0$ in the inequation, we get
$3(0)+5(0)<7$
i.e., $0<7$ which is true.
$\therefore \quad$ The solution set of the inequality is an open half plane containing the origin except the points on line $3 x+5 y=7$

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MCQ 211 Mark

The number of corner points of the feasible region determined by the constraints $x-y \geq 0,2 y \leq x+2$, $x \geq 0, y \geq 0$ is:

Answer

We have, $x-y \geq 0,2 y \leq x+2, x \geq 0$ and $y>0$. Let us draw the graph of given constraints, we get $x-y=0$

x	0	1
y	0	1

and 2y = x + 2

x	0	-2
y	1	0

The feasible region is unbounded.
$\therefore \quad$ There are two corner points as $(0,0)$ and $(2,2)$.

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MCQ 221 Mark

The corner points of the bounded feasible region of an LPP are $O(0,0), A(250,0), B(200,50)$ and $C(0,175)$. If the maximum value of the objective function $Z=2 a x+$ by occurs at the points $A(250,0)$ and $B(200,50)$, then the relation between $a$ and $b$ is:

A
$2 a=b$
B
$2 a=3 b$
C
$a=b$
D
$a=2 b$

Answer

Given, $Z=2 a x+b y$.........(i)
Putting $x=250$ and $y=0$ in (i), we get
$Z_{\max }=2 a(250)+b(0)=500 a$.........(ii)
Putting $x=200$ and $y=50$ in (i), we get
$Z_{\max }=2 a(200)+b(50)=400 a+50 b$..........(iii)
From (ii) and (iii), we get $500 a=400 a+50 b$
$\Rightarrow \quad 100 a=50 b \Rightarrow 2 a=b$

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MCQ 231 Mark

The point which lies in the half-plane $2 x+y-4 \leq 0$ is:

A
$(0,8)$
B
$(1,1)$
C
$(5,5)$
D
$(2,2)$

Answer

Substitute $x=1$ and $y=1$ in $2 x+y \leq 4$
$\Rightarrow 2(1)+1 \leq 4 \Rightarrow 3 \leq 4$ which is true.
So, $(1,1)$ lies in the half plane $2 x+y-4 \leq 0$

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MCQ 241 Mark

The corner points of the feasible region in the graphical representation of a linear programming problem are $(2,72),(15,20)$ and $(40,15)$. If $z=18 x+9 y$ be the objective function, then:

A
$z$ is maximum at $(2,72)$, minimum at $(15,20)$
B
$z$ is maximum at $(15,20)$, minimum at $(40,15)$
C
$z$ is maximum at $(40,15)$, minimum at $(15,20)$
D
$z$ is maximum at $(40,15)$, minimum at $(2,72)$

Answer

The objective function is given as $z=18 x+9 y$
The corner points are given as $(2,72),(15,20)$ and $(40,15)$
At $(2,72), z=18 \times 2+9 \times 72=36+648=684$
At $(15,20), z=18 \times 15+9 \times 20=270+180=450$
At $(40,15)=z=18 \times 40+9 \times 15=720+135=855$
$\therefore \quad z$ is maximum at $(40,15)$ and minimum at $(15,20)$.

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MCQ 251 Mark

Which of the following points satisfies both the inequations $2 x+y \leq 10$ and $x+2 y \geq 8$ ?

A
$(-2,4)$
B
$(3,2)$
C
$(3,2)$
D
$(4,2)$

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MCQ 261 Mark

The solution set of the inequality $3 x+5 y<4$ is

A
an open half-plane not containing the origin.
B
an open half-plane containing the origin.
C
the whole $X Y$-plane not containing the line $3 x+5 y=4$.
D
a closed half plane containing the origin.

Answer

The strict inequality represents an open half plane and it contains the origin, as $(0,0)$ satisfies it.

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MCQ 271 Mark

The corner points of the shaded unbounded feasible region of an LPP are $(0,4),(0.6,1.6)$ and $(3,0)$ as shown in the figure. The minimum value of the objective function $Z=4 x+6 y$ occurs at

A
$(0.6,1.6)$ only
B
$(3,0)$ only
C
$(0.6,1.6)$ and $(3,0)$ only
D
at every point of the line-segment joining the points $(0.6,1.6)$ and $(3,0)$

Answer

The minimum value of the objective function occurs at two adjacent corner points $(0.6,1.6)$ and $(3,0)$ and there is no point in the half plane $4 x+6 y<12$ in common with the feasible region.
So, the minimum value occurs at every point of the linesegment joining the two points.

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MCQ 281 Mark

In an LPP, if the objective function $z=a x+$ by has the same maximum value on two corner points of the feasible region, then the number of points at which $z_{\max }$ occurs is

A
$0$
B
2
C
finite
D
infinite

Answer

In an LPP, if the objective function $z=a x+b y$ has the same maximum value on two corner points of the feasible region, then the number of points at which $z_{\max }$ occurs is infinite.

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MCQ 291 Mark

The corner points of the feasible region determined by the system of linear inequalities are $(0,0),(4,0)$, $(2,4)$ and $(0,5)$. If the maximum value of $z=a x+b y$, where $a, b>0$ occurs at both $(2,4)$ and $(4,0)$, then

A
$a=2 b$
B
$2 a=b$
C
$a=b$
D
$3 a=b$

Answer

Since, maximum value of $z=a x+b y$ occurs at both
$(2,4)$ and $(4,0)$.
$
\therefore \quad 2 a+4 b=4 a+0 \Rightarrow 4 b=2 a \Rightarrow 2 b=a$

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MCQ 301 Mark

The objective function of an LPP is

A
a constant
B
a linear function to be optimised
C
an inequality
D
a quadratic expression

Answer

A linear function to be optimized is called an objective function

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MCQ 311 Mark

The graph of the inequality $2 x+3 y>6$ is

A
half plane that contains the origin
B
half plane that neither contains the origin nor the points of the line $2 x+3 y=6$.
C
whole $X O Y$-plane excluding the points on the line $2 x+3 y=6$.
D
entire XOY-plane.

Answer

From the graph of inequality $2 x+3 y>6$. It is clear that it does not contain the origin nor the points of the line $2 x+3 y=6$

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MCQ 321 Mark

The optimal value of the objective function is attained at the points:

A
On x - axis
B
On y - axis
C
Which are corner points of the feasible region
D
None of these

Answer

Which are corner points of the feasible region

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MCQ 331 Mark

A set of values of decision variables that satisfies the linear constraints and non - negativity conditions of an L.P.P. is called its:

A
Unbounded solution
B
Optimum solution
C
Feasible solution
D
None of these

Answer

Feasible solution

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MCQ 341 Mark

Which of the following is a type of Linear programming problem?

A
Manufacturing problem
B
Diet problem
C
Transportation problems
D
All of the above

Answer

All of the above

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MCQ 351 Mark

A linear programming of linear functions deals with:

A
Minimizing
B
Optimizing
C
Maximizing
D
None

Answer

Optimizing

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MCQ 361 Mark

The feasible, region for an LPP is shown shaded in the figure. Let Z = 3x - 4y be the objective function. A minimum of Z occurs at:

A
(0, 0)
B
(0, 8)
C
(5, 0)
D
(4, 10)

Answer

(0, 8)

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MCQ 371 Mark

The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is Maximum of Z occurs at:

A
(5, 0)
B
(6, 5)
C
(6, 8)
D
(4, 10)

Answer

(5, 0)

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MCQ 381 Mark

Which of the following statement is correct?

A
Every LPP admits an optimal solution.
B
Every LPP admits unique optimal solution.
C
If a LPP gives two optimal solutions it has infinite number of solutions.
D
None of these

Answer

Every LPP admits unique optimal solution.

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MCQ 391 Mark

The linear inequalities or equations or restrictions on the variables of a linear programming problem are called:

A
A constraint
B
Decision variables
C
Objective function
D
None of the above

Answer

A constraint

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MCQ 401 Mark

The feasible region for an LPP is shown shaded in the following figure. Minimum of Z = 4x + 3y occurs at the point.

A
(0, 8)
B
(2, 5)
C
(4, 3)
D
(9, 0)

Answer

(2, 5)

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MCQ 411 Mark

The feasible solution of an LP problem, is ________.

A
Must satisfies all of the problems constraints simultaneously.
B
Must be a corner point of the feasible region.
C
Need not satisfy all of the constraints, only some of them.
D
Must optimize the value of the objective function.

Answer

Must satisfies all of the problems constraints simultaneously.

Solution:

The feasibe solution of a inear programming probem (LP) is a solution that must satisfy all of the problems constraints simultaniously.

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MCQ 421 Mark

Refer to Question 18 maximum of Z occurs at:

A
(5, 0)
B
(6, 5)
C
(6, 8)
D
(4, 10)

Answer

(5, 0)

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MCQ 431 Mark

Maximize Z = 11x + 8y, subject to $\text{x}\leq4,\text{y}\leq6,\text{x}\geq0,\text{y}\geq0.$

A
44 at (4, 2)
B
60 at (4, 2)
C
62 at (4, 0)

Answer

60 at (4, 2)

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MCQ 441 Mark

The value of $\frac{0.76\times0.76\times0.76+0.24\times0.24\times0.24}{0.76\times0.76-0.76\times0.24+ 0.24+0.24}$is:

A
0.52
B
1
C
0.01
D
0.1

Answer

Solution:

Formula used:

a³ + b³= (a + b)(a² - ab + b²)

$\frac{0.76\times0.76\times0.76+0.24\times0.24\times0.24}{0.76\times0.76-0.76\times0.24+ 0.24+0.24}$

$\frac{(0.76)^{3}+(0.24)^{3}}{0.76\times0.76-0.76\times0.24+0.24+0.24}$

$=\frac{(0.76+0.24)(0.76\times0.76-0.76\times0.24+0.24\times0.24)}{0.76\times0.76-0.76\times0.24+0.24\times0.24}$

$=(0.76+0.24)=1$

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MCQ 451 Mark

The maximum value of Z = 4x + 2y Subjected to the constraints $2\text{x}+3\text{y}\leq18,\text{x}+\text{y}\geq10,\text{x},\text{y}\geq0$ is:

A
36
B
40
C
20
D
none of these

Answer

none of these

Solution:

Consider, 2x + 3y = 18

x	y	(x, y)
0	6	(0, 6)
9	0	(9, 0)

Consider, x + y = 10

x	y	(x, y)
0	10	(0, 10)
10	0	(10, 0)

From the graph we conclude that no feasible region exist.

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MCQ 461 Mark

The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80, x, y ≥ 0 is:

A
320
B
300
C
230
D
none of these

Answer

none of these

Solution:

We need to maximize the function Z = 4x + 3y

Converting the given inequations into equations, we obtain

3x + 2y = 160, 5x + 2y = 200, x + 2y = 80, x = 0 and y = 0

Region represented by 3x + 2y ≥ 160:

The line 3x + 2y = 160 meets the coordinate axes at A1603,0 and B(0, 80) respectively.

By joining these points we obtain the line 3x + 2y = 160.

Clearly (0, 0) does not satisfies the inequation 3x + 2y ≥ 160.

So, the region in xy plane which does not contain the origin represents the solution set of the inequation 3x + 2y ≥ 160.

Region represented by 5x +2y ≥ 200:

The line 5x + 2y = 200 meets the coordinate axes at C(40, 0) and D(0, 100) respectively.

By joining these points we obtain the line 5x + 2y = 200.

Clearly (0, 0) does not satisfies the inequation 5x +2y ≥ 200.

So, the region which does not contain the origin represents the solution set of the inequation 5x +2y ≥ 200.

Region represented by x +2y ≥ 80:

The line x + 2y = 80 meets the coordinate axes at E(80, 0) and F(0, 40) respectively.

By joining these points we obtain the line x + 2y = 80.

Clearly (0, 0) does not satisfies the inequation x + 2y ≥ 80.

So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 80.

Region represented by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 3x + 2y ≥ 160,5x+2y ≥ 200, x +2y ≥ 80, x ≥ 0, and y ≥ 0 are as follows.

Here, we see that the feasible region is unbounded.

Therefore,maximum value is infinity.

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MCQ 471 Mark

The __________ is the method available for solving an L.P.P

A
Graphical method
B
Least cost method
C
MODI method
D
Hungarian method

Answer

Graphical method

Solution:

There are different methods to solve an linear programming problem.

Such as Graphical method, Simplex method, Ellipsoid method, Interior point methods.

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MCQ 481 Mark

The corner points of the feasible region determined by the following system of linear inequalities:
$2\text{x}+\text{y}\le10,\ \text{x}+3\text{y}\le15,\ \text{x},\ \text{y}\ge0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is:

A
p = q
B
p = 2q
C
p = 3q
D
q = 3p.

Answer

q = 3p.

Explanation:

The maximum value of Z is unique.

It is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5).

$\therefore$ Value of z at (3, 4) = Value of z at (0, 5)

⇒ p(3) + q(4) = p(0) + q(5)

⇒ 3p + 4q = 5q

⇒ q = 3p

Hence, the correct answer is D.

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MCQ 491 Mark

Choose the correct answer from the given four options.

The feasible solution for a LPP shown in Fig. 12.12. Let z = 3x - 4y be objective functio. (Maximum value of Z + Minimum value of Z) is equal to:

A
13.
B
1.
C
-13.
D
-17.

Answer

-17.

Solution:

Corner points	Corresponding value of Z = 3x - 4y
(0, 0) (5, 0) (6, 5) (6, 8) (4, 10) (0, 8)	0 15 (Maximum) -2 -14 -28 -32 (Minimum)

Here, maximum value of Z + minimum value of Z = 15 - 32 = -17.

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MCQ 501 Mark

Minimise $\text{Z}=\sum\limits^{\text{n}}_{\text{j}=1}\sum\limits^{\text{m}}_{\text{i}=1}\text{c}_{\text{ij}}\cdot\text{x}_{\text{ij}}$ Subject to $\sum\limits^{\text{m}}_{\text{i}=1}\text{x}_{\text{ji}}=\text{b}_{\text{j}},\text{j}=1,2,....\text{n}$ $\sum\limits^{\text{n}}_{\text{j}=1}\text{x}_{\text{ji}}=\text{b}_{\text{j}},\text{j}=1,2,.....,\text{m}$ is a LPP with number of constraints.

A
$\text{m}-\text{n}$
B
$\text{m}\text{n}$
C
$\text{m}+\text{n}$
D
$\frac{\text{m}}{\text{n}}$

Answer

$\text{m}+\text{n}$

Solution:

Constraints will be

x₁₁ + x₂₁ + ..... + x_m1 = b₁

x₁₂ + x₂₂ + ..... + x_m2 = b₂

x_1n + x_2n + ..... + xmn = b_n

x₁₁ + x₁₂ + ..... + x_1n = b₁

x₂₁+ x₂₂ + ..... + x_2n = b₂

x_m1 + x_m2 + ..... + x_mn = b_n

So the total number of constraints = m + n

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Maths M.C.Q (1 Marks)

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