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MATHS M.C.Q (1 Marks)

Linear Programming · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 157 questions.

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M.C.Q (1 Marks)

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157 questions · 14 auto-graded MCQ + 143 self-marked written.

Question 11 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:
  1. a = 2b
  2. 2a = b
  3. a = b
  4. 3a = b
Answer
  1. a = 2b
Solution:
4a + 0b = 2a + 4b
4a = 2a + 4b
4a - 2a = 4b
2a = 4b
a = 2b
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Question 21 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:
  1. a = 2b
  2. 2a = b
  3. a = b
  4. 3a = b
Answer
  1. a = 2b
Solution:
4a + 0b = 2a + 4b
4a = 2a + 4b
4a - 2a = 4b
2a = 4b
a = 2b
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Question 31 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:
  1. a = 2b
  2. 2a = b
  3. a = b
  4. 3a = b
Answer
  1. a = 2b
Solution:
4a + 0b = 2a + 4b
4a = 2a + 4b
4a - 2a = 4b
2a = 4b
a = 2b
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Question 41 Mark
Maximize Z = 7x + 11y, subject to $3\text{x}+5\text{y}\leq26,5\text{x}+3\text{y}\leq30,\text{x}\geq0,\text{y}\geq0.$
  1. 59 at$\Big(\frac{9}{2},\frac{5}{2}\Big)$
  2. 42 at (6, 0)
  3. 49 at (7, 0)
  4. 57.2 at (0, 5.2)
Answer
  1. 59 at$\Big(\frac{9}{2},\frac{5}{2}\Big)$
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Question 51 Mark
The maximum value of Z = 3x + 4y subjected to contraints $\text{x}+\text{y}\leq40,\text{x}+2\text{y}\leq60,\text{x}\geq0$ and $\text{y}\geq0$ is:
  1. 120
  2. 140
  3. 100
  4. 160
Answer
  1. 140
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Question 61 Mark
A linear programming of linear functions deals with:
  1. Minimizing
  2. Optimizing
  3. Maximizing
  4. None
Answer
  1. Optimizing
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Question 71 Mark
The objective function of a linear programming problem is:
  1. A constraint
  2. Function to be optimised
  3. A relation between the variables
  4. None of these
Answer
  1. Function to be optimised
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Question 81 Mark
The feasible region for an LPP is shown below:
Let Z = 3x - 4y be the objective function. Minimum of Z occurs at
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
Answer
  1. (0, 8)
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Question 91 Mark
Choose the correct answer from the given four options. The feasible solution for a LPP is shown in. Let Z = 3x - 4y be the objective function.
Maximum of Z occurs at:
  1. (5, 0)
  2. (6, 5)
  3. (6, 8)
  4. (4, 10)
Answer
  1. (5, 0)
Solution:
Corner points
Corresponding value of Z = 3x - 4y
(0, 0)
(5, 0)
(6, 5)
(6, 8)
(4, 10)
(0, 8)
0
15 (Maxmimum)
-2
-14
-28
-32
Hence, maximum of Z occurs at (5, 0) and its maximum value is 27.
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Question 101 Mark
The number of points in $(-\infty,\infty)$ for which $\text{x}^{2}-\text{x}\sin\text{x}-\cos\text{x}=0,$ is:
  1. 6
  2. 4
  3. 2
  4. None of the above
Answer
  1. 2
Solution:
Better approch is with graphs.Considering graphs in eqaution we get
 $\text{x}^{2}-\text{x}\sin\text{x}-\cos\text{x}=0$
$\text{x}^{2}=\text{x}\sin\text{x}+\cos\text{x}$
Let  $\text{f}(\text{x})=\text{x}^{2},\text{g}(\text{x})=\text{x}\sin\text{x}+\cos\text{x}$
Using graphical methods,we can do the graph of f(x) and g(x)
The graph f(x) and g(x) intersects at two points between $(-\infty,\infty)$
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Question 111 Mark
The feasible solution of an LP problem, is ________.
  1. Must satisfies all of the problems constraints simultaneously.
  2. Must be a corner point of the feasible region.
  3. Need not satisfy all of the constraints, only some of them.
  4. Must optimize the value of the objective function.
Answer
  1. 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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Question 121 Mark
Objective function of a L.P.P. is:
  1. A constant
  2. A function to be optimised
  3. A relation between the variables
  4. None of these 
Answer
  1. A function to be optimised
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Question 131 Mark
The corner point of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y. Compare the quantity in Column A and Column B.
Column A
Column B
Maximum of Z
325
  1. The quantity in column A is greater.
  2. The quantity in column B is greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined On the basis of the information supplied.
Answer
  1. The quantity in column B is greater.
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Question 141 Mark
An objective function in a linear program can be which of the following?
  1. A maximization function
  2. A nonlinear maximization function
  3. A quadratic maximization function
  4. An uncertain quantity
  5. A divisible additive function
Answer
  1. A maximization function
Solution:
Linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints.
The objective function in a linear program is a maximization function.
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Question 151 Mark
The given table shows the number of cars manufactured in four different colours on a particular day. Study it carefully and answer the question.
 
Number of cars manufactured
Colour
Vento
Creta
Wagonr
Red
65
88
93
White
54
42
80
Black
66
52
88
Sliver
37
49
74
Which car was twice the number of silver Vento?
  1. Silver WagonR
  2. Red WagonR
  3. Red Vento
  4. White Creta
Answer
  1. Silver WagonR
Solution:
The number of silver Vento car = 37 (from the table)
Twice the number of silver Vento cars = 2 × 37 = 74
Now from table we can see that silver WagonR is only car type having 74 cars
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Question 161 Mark
The __________ is the method available for solving an L.P.P
  1. Graphical method
  2. Least cost method
  3. MODI method
  4. Hungarian method
Answer
  1. 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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Question 171 Mark
The problem associated with LPP is:
  1. Single objective function
  2. Double objective function
  3. No any objective function
  4. None
Answer
  1. Single objective function
Solution:
The problem associated with LLP is single objective.
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Question 181 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:
  1. 13.
  2. 1.
  3. -13.
  4. -17.
Answer
  1. -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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Question 191 Mark
In linear programming, objective function and objective constraints are:
  1. Solved
  2. Linear
  3. Quadratic
  4. Adjacent
Answer
  1. Linear
Solution:
In linear programming, objective function and objective constraints are linear.
Any linear programming problem must have the following properties:-
The relationship between variables and constraints must be linear.
The constraints must be non-negative.
objective function must be linear.
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Question 201 Mark
Objective of LPP is:
  1. A constraint
  2. A function to be optimized
  3. A relation between the variables
  4. None of the above
Answer
  1. A function to be optimized
Solution:
The objective of Linear Programming Problems (LPP) is to minimize or maximize the function.
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Question 211 Mark
Which of the termis not used in a linear programming problem:
  1. Slack inequation
  2. Objective function
  3. Concave region
  4. Feasible Region
Answer
  1. Concave region
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Question 221 Mark
In linear programming, oil companies used to implement resources available is classified as:
  1. Implementation modeling
  2. Transportation models
  3. Oil model
  4. Resources modeling
Answer
  1. Transportation models
Solution:
In linear programming, transportation model are applied to problems related to the study of efficient transportation routes.
For oil companies, how effectively the available resources are transported to different destinations with minimum cost.
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Question 231 Mark
The corner points of the feasible region are A(0, 0), B(16, 0), C(8, 16) and D(0, 24). The minimum value of the objective function z = 300x + 190y is _______.
  1. 5440
  2. 4800
  3. 4560
  4. 0
Answer
  1. 0
Solution:
We know that, for a cartesian polygon, the maximum value occurs at the corner points or vertices of the polygon.
Given z = 300x + 190y
By substituting A(0, 0) in the equation we get z = 0
By substituting B(16, 0) in the equation we get z = 4800
By substituting C(8,16) in the equation we get z = 5440
By substituting D(0, 24) in the equation we get z = 4560
Hence the minimum value of Z occured at C(0, 0) with z = 0
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Question 241 Mark
If a = b then ax = ...........
  1. b + x
  2. bx
  3. b - x
  4. b ÷ x
Answer
  1. bx
Solution:
Given, a = b Multiplying both sides by x.
ax = bx.
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Question 251 Mark
What is the solution of $\text{x}\leq4,\text{y}\geq0$ and $\text{x}\leq-4,\text{y}\geq0$?
  1. $\text{x}\geq-4,\text{y}\leq0$
  2. $\text{x}\leq4,\text{y}\geq0$
  3. $\text{x}\leq-4,\text{y}=0$
  4. $\text{x}\geq-4,\text{y}=0$
Answer
  1. $\text{x}\leq-4,\text{y}=0$
Solution:
$\text{x}\leq4$ and $\text{x}\leq-4$
 $\Rightarrow\text{x}\leq-4$
Also, $\text{y}\geq0$ and $\text{y}\leq0$
$\Rightarrow\text{y}=0$
Hence the solutione is $\text{x}\leq-4,\text{y}=0.$
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Question 261 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:
  1. p = q
  2. p = 2q
  3. q = 2p
  4. q = 3p
Answer
  1. q = 3p
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Question 271 Mark
Which of the following statements is correct?
  1. Every LPP admits an optimal solution
  2. A LPP admits unique optimal solution
  3. If a LPP admits two optimal solution it has an infinite number of optimal solutions
  4. The set of all feasible solutions of a LPP is not a converse set
Answer
  1. If a LPP admits two optimal solution it has an infinite number of optimal solutions
Solution:
Optimal solution of LPP has three types.
  1. Unique
  2. Infinite
  3. Does not exist.
Hence, it has infinite solution if it admits two optimal solution.
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Question 281 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 __________:
  1. q = 2p
  2. p = 2p
  3. p = q
  4. q = 3p
Answer
  1. q = 3p
Solution:
Let z0​ be the maximum value of z in the feasible region.
Since maximum occurs at both (15, 15) and (0, 20)$, the value z0​ is attained at  both (15, 15) and (0, 20).
⟹ z0​ = p(15) + q(15) and z0​ = p(0) + q(20)
⟹ p(15) + q(15) = p(0) + q(20)
⟹ 15p = 5q
⟹ 3p = q
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MCQ 291 Mark
The ratio of the rate of flow of water in pipes varies inversely as the square of the radius of the pipes. What is the ratio of the rates of flow in two pipes diameters $2\ cm$ and $4\ cm?$
  • A
    $1 : 6$
  • ✓
    $1 : 4$
  • C
    $1 : 2$
  • D
    $3 : 1$
Answer
Correct option: B.
$1 : 4$
Given: $d_1​= 2\ cm$
$d_2​ = 4\ cm$
Since the diameter are $2\ cm$ and $4\ cm.$
The replacement ratio of the two pipes are $1\ cm$ and $2\ cm$
$r_1 ​= 1\ cm$
$r_2​ = 2\ cm$
Square of the ratio of the pipes are $1$ and $4$
$\therefore$ The ratio of rates of flow in two pipes $=1:\frac{1}{4}$
$\Rightarrow\frac{1}{4}$
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Question 301 Mark
Choose the correct answer from the given four options.
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Let F = 4x + 6y be the objective function.
The Minimum value of F occurs at.
  1. (0, 2) only.
  2. (3, 0) only.
  3. The mid point of the line sgment joining the points (0, 2) and (3, 0) only.
  4. Any point on the line segment joining the points (0, 2) and (3, 0).
Answer
  1. Any point on the line segment joining the points (0, 2) and (3, 0).
Solution:
Corner points
Corresponding value of F = 4x + 6y
(0, 2)
12 (Minimum)
(3, 0)
12 (Minimum)
(6, 0)
24
(6, 8)
72 (Maxmimum)
(0, 5)
30
 
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Question 311 Mark
Corner points of the bounded feasible region for an LP problem are A(0, 5) B(0, 3) C(1, 0) D(6, 0). Let z = -50x + 20y be the objective function. Minimum value of z occurs at ______ center point.
  1. (0, 5)
  2. (1, 0)
  3. (6, 0)
  4. (0, 3)
Answer
  1. (6, 0)
Solution:
We check the value of the z at each of the corner points.
A (0, 5) -z = -50x + 20y = -50(0) + 20(5) = 100
At B (0, 3) -z = -50x + 20y = -50(0) + 20(3) = 60
At C (1, 0) -z = -50x + 20y = -50(1) + 20(0) = -50
At D (6, 0) -z = -50x + 20y = -50(6) + 20(0) = -300
Hence, we see that z is minimum at D(6, 0) and minimum value is -300.
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Question 321 Mark
Which of the following is an essential condition in a situation for linear programming to be useful?
  1. Linear constraints
  2. Bottlenecks in the objective function
  3. Non - homogeneity
  4. Uncertainty
  5. None of the above
Answer
  1. Linear constraints
Solution:
For linear programming, the constraints must be linear.
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Question 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:
  1. Unbounded solution
  2. Optimum solution
  3. Feasible solution
  4. None of these
Answer
  1. Feasible solution
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Question 341 Mark
If A = {1, 2, 3}; B = {3, 4, 5}; C = {4, 6}, then $\text{A}\times(\text{B}\cap\text{C})=?$
  1. {(2, 4)(1, 4)}
  2. {(2, 4)(3, 4)(5, 6)}
  3. {(1, 4)(2, 4)(3, 4)}
  4. None of these
Answer
  1. {(1, 4)(2, 4)(3, 4)}
Solution:
Given,
A = {1, 2, 3}
B = {3, 4, 5}
C = {4, 6}
Now, $\text{B}\cap\text{C}=\{{4\}}$
$\therefore\text{A}\times(\text{B}\cap\text{C})=\{(1,4),(2,4),(3,4)\}$
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Question 351 Mark
The optimal value of the objective function is attained at the points
  1. given by intersection of inequations with the axes only
  2. given by intersection of inequations with x-axis only
  3. given by corner points of the feasible region
  4. none of these
Answer
  1. given by corner points of the feasible region
Solution:
It is known that the optimal value of the objective function is attained at any of the corner point.
Thus, the potimal value of the objective function is attined at the points given by corner points of the feasible region.
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Question 361 Mark
The given table shows the number of cars manufactured in four different colours on a particular day. Study it carefully and answer the question.
 
Number of cars manufactured
Colour
Vento
Creta
WagonR
Red
65
88
93
White
54
42
80
Black
66
52
88
Sliver
37
49
74
What was the total number of black cars manufactured?
  1. 240
  2. 206
  3. 205
  4. 159
Answer
  1. 206
Solution:
The number of Black cars manufactured.
= no. of black V ento + no. of black Creta + no. of black W agon R.
= 66 + 52 + 88 = 206
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Question 371 Mark
While plotting constraints on a graph paper, terminal points on both the axes are connected by a straight line because:
  1. The resources are limited in supply.
  2. The objective function as a linear function.
  3. The constraints are linear equations or inequalities.
  4. All of the above.
Answer
  1. The constraints are linear equations or inequalities.
Solution:
The graph of the linear equation is a straight line.
If the terminal points are connected by a straight line then the given constraints are linear equations which may include inequalities.
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Question 381 Mark
Choose the correct answer from the given four options.
Let F = 3x - 4y be the objective function. Maximum value of F is:
  1. 0.
  2. 8.
  3. 12.
  4. -18.
Answer
  1. 12.
Solution:
The feasible region as shown in the figure, has objective function F = 3x - 4y
Corner points
Corresponding value of Z = 3x - 4y
(0, 0)
(12, 6)
(0, 4)
0
12 (maximum)
-16 (minimum)
Hence, the maximum value of F is 12.
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Question 391 Mark
The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (25, 20) and (0, 30). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (25, 20) and (0, 30) is _______.
  1. 5p = 2q
  2. 2p = 5q
  3. p = 2q
  4. q = 3p
Answer
  1. 5p = 2q
Solution:
Maximum of Z occurs at (25, 20) and at (0, 30).
Hence, equating the vales of Z at these points, we get 25p + 20q = 30q
$\therefore$ 5p = 2q
This is the required relation.
Also as p, q > 0, the value of Z is always positive and hence, is greater at (25, 20) and at (0, 30) than at (0,10) and (5, 5).
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MCQ 401 Mark
Conclude from the following: $n^2 > 10,$ and $n$ is a positive integer. $A: n^3 B: 50.$
  • ✓
    The quantity $A$ is may be greater or smaller than $B.$
  • B
    The quantity $B$ is greater than $A.$
  • C
    The two quantities are equal.
  • D
    The relationship cannot be determined from the information given.
Answer
Correct option: A.
The quantity $A$ is may be greater or smaller than $B.$
given, $n^2 > 10$ and $n > 0$ multiplying both equations we get $n^3 > 0$
so, it may be greater than or less than $50.$
Hence, quantity $A$ is may be greater or smaller than $B$
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Question 411 Mark
In linear programming context, sensitivity analysis is a technique to:
  1. Allocate resources optimally.
  2. Minimize cost of operations.
  3. Spell out relation between primal and dual.
  4. Determine how optimal solution to LPP changes in response to problem inputs.
Answer
  1. Determine how optimal solution to LPP changes in response to problem inputs.
Solution:
A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.
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Question 421 Mark
An article manufactured by a company consists of two parts X and Y. In the process of manufacture of the part X. 9 out of 100 parts may be defective. Similarly 5 out of 100 are likely to be defective in part Y. Calculate the probability that the assembled product will not be defective.
  1. 0.86
  2. 0.864
  3. 0.8456
  4. 0.8645
Answer
  1. 0.8645
Solution:
Let A = Part X is not defective
Probability of A is $\text{P}(\text{A})=\frac{91}{100}$
B = Part Y is not defective.
Probability of B is $\text{P}(\text{B})=\frac{95}{100}$
Required probability
 $=\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\text{P}(\text{B})=\frac{91}{100}\times\frac{95}{100}=\frac{8645}{10000}$
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Question 431 Mark
Vikas printing company takes fee of Rs. 28 to print a oversized poster and Rs. 7 for each colour of ink used. Raaj printing company does the same work and charges poster for Rs. 34 and Rs. 5.50 for each colour of ink used. If z is the colours of ink used, find the values of z such that Vikas printing company would charge more to print a poster than Raaj printing company.
  1. $\text{z}<4$
  2. $2\leq\text{z}\leq4$
  3. $4\leq\text{z}\leq7$
  4. $\text{z}>4$
Answer
  1. $\text{z}>4$
Solution:
$28+7\text{z}>34+5.50\text{z}$
$\rightarrow1.50\text{z}>6$
$\rightarrow\text{z}>\frac{6}{1.5}\text{z}>4$
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Question 441 Mark
The point at which the maximum value of x + y, subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x, y ≥ 0 isobtained, is:
  1. (30, 25)
  2. (20, 35)
  3. (35, 20)
  4. (40, 15)
Answer
  1. (40, 15)
Solution:
We need to maximize the function
Z = x + y
Converting the given inequations into equations, we obtain x + 2y = 70, 2x + y = 95, x = 0 and y = 0
Region represented by x + 2y ≤ 70:
The line x + 2y = 70 meets the coordinate axes at A(70, 0) and B(0, 35) respectively.
By joining these points we obtain the line x + 2y = 70.
Clearly (0, 0) satisfies the inequation x + 2y ≤ 70.
So, the region containing the origin represents the solution set of the inequation x + 2y ≤ 70.
Region represented by 2x + y ≤ 95:
The line 2x + y = 95 meets the coordinate axes at $\text{C}\Big(\frac{95}{2},0\Big)$ and D(0, 95) respectively.
By joining these points we obtain the line 2x + y = 95.
Clearly (0, 0) satisfies the inequation 2x + y ≤ 95.
So, the region containing the origin represents the solution set of the inequation 2x + y ≤  95.
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 x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), $\text{C}\Big(\frac{95}{2},0\Big)$, E(40, 15) and B(0, 35).
The values of Z at these corner points are as follows.
$\text{Corner point}$ $\text{Z} = \text{x} + \text{y}$
$\text{O}(0, 0)$ $0 + 0 = 0$
$\text{C}\Big(\frac{95}{2},0\Big)$ $\frac{95}{2}+0,2=\frac{95}{2}$
$\text{E}(40, 1)$ $40 + 15 = 55$
$\text{B}(0, 35)$ $0 + 35 = 35$
We see that the maximum value of the objective function Z is 55 which is at (40, 15).
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Question 451 Mark
The optimal value of the objective function is attained at the points.
  1. Given by intersection of inequation with y - axis only.
  2. Given by intersection of inequation with x - axis only.
  3. Given by corner points of the feasible region.
  4. None of these
Answer
  1. Given by corner points of the feasible region.
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Question 461 Mark
In linear programming, lack of points for a solution set is said to:
  1. Have no feasible solution
  2. Have a feasible solution
  3. Have single point method
  4. Have infinte point method
Answer
  1. Have no feasible solution
Solution:
If there is no point in the feasible set, there is no feasible solution of the linear programming model.
In linear programming, lack of points for a solution set is said to have no feasible solution.
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MCQ 471 Mark
If $\text{a},\text{b},\text{c}\in+\text{R}$ such that $\lambda\text{ abc}$ is the minimum value of $a(b^2 + c^2) + b(c^2 + a^2) + c(a^2 + b^2),$ then $\lambda=$
  • A
    $1$
  • B
    $3$
  • C
    $4$
  • ✓
    None of the above.
Answer
Correct option: D.
None of the above.

We know that $ \text{A}.\text{M}.\geq\text{G}.\text{M}.$
Therefore, $ \frac{{\text{b}^2+\text{c}^2}​}{2}\geq\sqrt{\text{b}^2\text{c}^2}​$
$\Rightarrow\text{b}^{2}+\text{c}^{2}\geq2\text{bc}$
Multiplying a on both sides doesn’t change the inequality.
Since, given that a is positive.
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(1)$
Similarly, $\text{b}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(2)$
and $\text{c}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(3)$
adding $(1), (2)$ and $(3)$ we get
$\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{c}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq2\text{abc}+2\text{abc}+2\text{abc}$
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{b}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq6\text{abc}$
Therefore $\lambda$ is $6$

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Question 481 Mark
The point which does not lie in the half - plane 2x + 3y -12 < 0 is:
  1. (2, 1)
  2. (1, 2)
  3. (-2, 3)
  4. (2, 3)
Answer
  1. (2, 3)
Solution:
By putting the value of point (2, 3) in 2x + 3y - 12, we get;
2(2) + 3(3) = -12
= 4 + 9 - 12
= 13 - 12
= 1 which is greater than 0.
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Question 491 Mark
z = 10x + 25y subject to $0\leq\text{X}\leq3$ and $0\leq\text{X}\leq3,$ $\text{x}+\text{y}\leq5$ then the maximum value of z is:
  1. 80
  2. 95
  3. 30
  4. 75
Answer
  1. 95
Solution:
The end points of the figure which forms as per the given condition are (0, 0), (3, 0), (0, 3), (3, 2), (2, 3) We check the value of z at these points.
At (0, 3), z = 0 + 75 = 75 At (3, 0), z = 30 + 0 = 30 At (0, 0), z = 0 At (3, 2), z = 30 + 50 = 80 At (2, 3), z = 20 + 75 = 95
Therefore, the maximum value of z turns out to be 95.
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MCQ 501 Mark
By graphical method, the solution of linear programming problem
Maximize $Z = 3x_1 + 5x_2$
Subject to
$3x_1 + 2x_2 \leq 18$
$x_1 \leq 4$
$x_2 \leq 6$
$x1 \geq 0, x2 \geq 0,$ is:
  • A
    $x_1 = 2, x_2 = 0, Z = 6$
  • ✓
    $x_1 = 2, x_2 = 6, Z = 36$
  • C
    $x_1 = 4, x_2 = 3, Z = 27$
  • D
    $x_1 = 4, x_2 = 6, Z = 42$
Answer
Correct option: B.
$x_1 = 2, x_2 = 6, Z = 36$
We need to maximize the function $Z = 3x_4 + 5x_2$
First, we will convert the given inequations into equations, we obtain the following equations:
$3x_1 + 2x_2 = 18, x_1 = 4, x_2 = 6, x_1 = 0$ and $x_2 = 0$
Region represented by $3x_1 + 2x_2 \leq 18:$
The line $3x_1 + 2x_2 = 18$ meets the coordinate axes at $A(6, 0)$ and $B(0, 9)$ respectively.
By joining these points we obtain the line $3X1 + 2x2 = 18.$
Clearly $(0, 0)$ satisfies the inequation $3x_1 + 2x_2 = 18.$
So the region in the plane which contain the origin represents the solution set of the inequation$ 3x_1 + 2x_2 \leq 18.$
Region represented by $x_1 \leq 4:$
The line $x_1 = 4$ is the line that passes through $C(4, 0)$ and is parallel to the $Y$ axis.
The region to the left of the line $x_1 = 4$ will satisfy the inequation $x_1 \leq 4.$
Region represented by $x_2 \leq 6:$
The line $x_2 = 6$ is the line that passes through $D(0, 6)$ and is parallel to the $X$ axis.
The region below the line $x_2 = 6$ will satisfy the inequation $X_2 \leq 6.$
Region represented by $x_1 \geq 0$ and $x_2 \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x_1 \geq 0$ and $x_2 \geq 0.$
The feasible region determined by the system of constraints,$ 3x_1 + 2x_2 \leq 18, x_1 \leq 4, x_2 \leq 6, x_1 \geq 0$ and $x_2 \geq 0$ are as follows

Corner points are $O(0, 0), D(0, 6), F(2, 6), E(4, 3)$ and $C(4, 0).$
The values of the objective function at these points are given in the following table.
Points
Value of $Z$
$O(0, 0)$ $3(0) + 5(0) = 0$
$D(0, 6)$ $3(0) + 5(6) = 30$
$F(2, 6)$ $3(2) + 5(6) = 36$
$E(4, 3)$ $3(4) + 5(3) = 27$
$C(4, 0)$ $3(4) + 5(0) = 12$
We see that the maximum value of the objective function $Z$ is $36$ which is at $F(2, 6).$
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Question 521 Mark
Maximize Z = 3x + 5y, subject to constraints: $\text{x}+4\text{y}\leq24,3\text{x}+\text{y}\leq21,\text{x}+\text{y}\geq9,\text{x}\geq0,\text{y}\geq0.$
  1. 20 at (1, 0)
  2. 30 at (0, 6)
  3. 37 at (4, 5)
  4. 33 at (6, 3)
Answer
  1. 37 at (4, 5)
Solution:
Find the maximum value of Z = 3x + 5y referring to the explanation of Q.5.
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MCQ 531 Mark
Maximize $Z = 10 x_1 + 25 x_2,$ subject to $0\leq\text{x}_{1}\leq3,0\leq\text{x}_{2}\leq3,\text{x}_{1}+\text{x}_{2}\leq5.$
  • A
    $80$ at $(3, 2)$
  • B
    $75$ at $(0, 3)$
  • C
    $30$ at $(3, 0)$
  • ✓
    $95$ at $(2, 3)$
Answer
Correct option: D.
$95$ at $(2, 3)$
$95$ at $(2, 3)$
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Question 541 Mark
The optimal value of the objective function is attained at the points:
  1. On x - axis
  2. On y - axis
  3. Which are corner points of the feasible region
  4. None of these
Answer
  1. Which are corner points of the feasible region
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Question 551 Mark
The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 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:
  1. P = q
  2. p = 2q
  3. p = 3q
  4. q = 3q
Answer
  1. q = 3p
Solution:
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).
Value of Z at (3, 4) = Value of Z at (0,5)
= p(3) + q(4) = p(0) + 7(5)
= 3p + 4q = 5q
= q = 3p
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Question 561 Mark
The objective function Z = 4x + 3y can be maximised subjected to the constraints 3x + 4y ≤ 24, 8x + 6y ≤ 48, x ≤ 5, y ≤ 6, x, y ≥ 0
  1. at only one point
  2. at two points only
  3. at an infinite number of points
  4. none of these
Answer
  1. at an infinite number of points
Solution:
We need to maximize Z = 4x + 3y
First, we will convert the given inequations into equations, we obtain the following equations: 3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6, x = 0 and y = 0.
The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0, 6).
Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24.
So, the region in xy-plane that contains the origin represents the solution set of the 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 inequation 8x + 6y ≤ 48.
So, the region in xy plane that contains the origin represents the solution set of the given equation.
x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.
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.
These lines are drawn using a suitable scale.
and B (0,6).
The corner points of the feasible region are O(0, 0), G(5, 0), $\text{F}\Big(5,\frac{4}{3}\Big),\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$and B(0, 6).
The values of Z at these corner points are as follows.
$\text{Corner point}$ $\text{Z} = 4\text{x} + 3\text{y}$
$\text{O}(0, 0)$ $4 \times 0 + 3 \times 0= 0$
$\text{G}(5, 0)$ $4 \times 5 + 3 \times 0 = 20$
$\text{F}\Big(5,\frac{4}{3}\Big)$ $4\times5+3\times\frac{4}{3}=24$
$\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$ $4\times\frac{24}{7}+3\times\frac{24}{7}=\frac{196}{7}=24$
$\text{B}(0, 6)$ $4\times0+3\times6=18$
We see that the maximum value of the objective function Z is 24 which is at F(5, 4) and $\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big).$
Thus, the optimal value of Z is 24.
As, we know that if a LPP has two optimal solution, then there are an infinite number of optimal solutions.
Therefore, the given objective function can be subjected at an infinite number of points.
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Question 571 Mark
The value of objective function is maximum under linear constraints
  1. at the centre of feasible region
  2. at (0, 0)
  3. at any vertex of feasible region
  4. the vertex which is maximum distance from (0, 0)
Answer
  1. at any vertex of feasible region
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.
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MCQ 581 Mark
For the $\text{LPP};$ maximise $z = x + 4y$ subject to the constraints $\text{x}+2\text{y}\leq2,$ $\text{x}+2\text{y}\geq8,$ $\text{x},\text{y}\geq0.$
  • A
    $z_{max} ​= 4$
  • B
    $z_{max​} = 8$
  • C
    $z_{max}​ = 16$
  • ✓
    Has no feasible solution
Answer
Correct option: D.
Has no feasible solution
$\text{x}+2\text{y}\leq2$
$\text{x}+2\text{y}\geq8$
$\text{x},\text{y}\geq0.$
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Question 591 Mark
Which of the following sets are convex?
  1. $\{(\text{x},\text{y}):\text{x}^2+\text{y}^2\geq1\}$
  2. $\{(\text{x},\text{y}):\text{y}^2\geq\text{x}\}$
  3. $\{(\text{x},\text{y}):3\text{x}^2+4\text{y}^2\geq5\}$
  4. $\{(\text{x},\text{y}):\text{y}\geq2,\text{y}\leq4\}$
Answer
  1. $\{(\text{x},\text{y}):\text{y}\geq2,\text{y}\leq4\}$
Solution:
is the region between two parallel lines, so any line segment joining any two points in it lies in it.
Hence, it is a convex set.
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Question 601 Mark
The solution set of the inequation 3x + 2y > 3 is:
  1. Half plane not containing the origin
  2. Half plane containing the origin
  3. The point being on the line 3x + 2y = 3
  4. None of these
Answer
  1. Half plane not containing the origin
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Question 611 Mark
Maximize Z = 6x + 4y, subject to $\text{x}\leq2,\text{x}+\text{y}\leq3,-2\text{x}+\text{y}\leq1,\text{x}\geq0,\text{y}\geq0.$
  1. 12 at (2, 0)
  2. 16 at (2, 1)
  3. $\frac{140}{3}$ at $\Big(\frac{2}{3},\frac{1}{3}\Big)$
  4. 4 at (0, 1)
Answer
  1. $\frac{140}{3}$ at $\Big(\frac{2}{3},\frac{1}{3}\Big)$
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Question 621 Mark
Region represented by $\text{x}\geq0, \text{y}\geq0$ is:
  1. First quadrant
  2. Second quadrant
  3. Third quadrant
  4. Fourth quadrant
Answer
  1. First quadrant
Solution:
 All the positive values of x and y will lie in the first quadrant.
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Question 631 Mark
Which of the following is not true about feasibility?
  1. It cannot be determined in a graphical solution of an LPP.
  2. It is independent of the objective function.
  3. It implies that there must be a convex region satisfying all the constraints.
  4. Extreme points of the convex region gives the optimum solution.
Answer
  1. It cannot be determined in a graphical solution of an LPP. 
Solution:
There are various methods to solve the linear programming problems namely simplex method, ellipsoid method, graphical method, interior points method, etc.
Therefore a linear programming problem can be solved using the graphical method.
Hence, the feasibility of the linear programming problem can be determined by the graphical method.
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Question 641 Mark
The maximum value of Z = 4x + 3y subjected to the constraints $2\text{x}+3\text{y}\leq18,$ $\text{x}+\text{y}\geq10;\text{x},\text{y}\geq0$ is:
  1. 36
  2. 40
  3. 20
  4. None of these
Answer
  1. None of these
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Question 651 Mark
For a linear programming equations, convex set of equations is included in region of:
  1. Feasible solutions
  2. Disposed solutions
  3. Profit solutions
  4. Loss solutions
Answer
  1. Feasible solutions
Solution:
In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints.
Then the problem becomes convex and has a single optimum(maximum or minimum) solution.
Therefore the convex set of equations is included in the feasible region.
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Question 661 Mark
Which of the following is a property of all linear programming problems?
  1. Alternate courses of action to choose from.
  2. Minimization of some objective.
  3. A computer program.
  4. Usage of graphs in the solution.
  5. Usage of linear and nonlinear equations and inequalities.
Answer
  1. Alternate courses of action to choose from.
Solution:
According to Robbins, the resources(capital, land, labour, materials, ...) are always limited.
Every resource have multiple uses.
The problem before any organisation or manager is to choose the best alternatives which can maximize the profit or minimize the cost of production.
Linear programming is the method which is used to select the best possible alternatives from the all alternatives.
According to William M. Fox, "Linear programming is a planning technique that permits some objective function to be maximized or minimized within the framework of given situational restrictions"
Therefore, the linear programming is the process of selecting best courses of action to choose from various alternatives.
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Question 671 Mark
To write the dual; it should be ensured that
  1.  All the primal variables are non - negative.
  2. All the bi values are non - negative.
  3. All the constraints are $\leq$ type if it is maximization problem and $\geq$ type if it is a minimization problem.
  1. I and II
  2. II and III
  3. I and III
  4. I, II and IIl
Answer
  1. I and III
Solution:
To write the dual, then all the primal variables must be non-negative.
All the constraints are $\leq$ type if it ia maximization problem and $\geq$ type if it is a minimization problem.
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Question 681 Mark
If two constraints do not intersect in the positive quadrant of the graph, then.
  1. The problem is infeasible
  2. The solution is unbounded
  3. One of the constraints is redundant
  4. None of the above
Answer
  1. The problem is infeasible
Solution:
Any linear programming problem must have the following properties:-1.
The relationship between variables and constraints must be linear2.
The constraints must be non - negative.3.. objective function must be linear.
Non - negativity conditions are used because the variables cannot take negative values.
i.e., it is not possible to have negative resources (land, capital, labour cannot be negative).
Because of the non - negativity condition, the feasible region exists only in I quadrant.
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MCQ 691 Mark
Let $X_1$ and $X_2$ are optimal solutions of a $LPP,$ then:
  • A
    $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,\lambda\in R$ is also an optimal solution
  • ✓
    $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution
  • C
    $\text{X}=\lambda\ \text{X}_1+(1+\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution
  • D
    $\text{X}=\lambda\ \text{X}_1+(1+\lambda)\text{X}_2,\lambda\in R$ given an optimal solution
Answer
Correct option: B.
$\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution

A set $A$ is convex if, for any two points $X_1, X_{2 }$
$\in\text{A}$ and $\lambda\in0,1$ imply that $\lambda\times1+1-\lambda\times2\in\text{A}$.
Since, here $X_1$ and $X_2$ are optimal solution
Therefore, their convex combination will also be an optimal solution
Thus, $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ gives an optimal solution.

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Question 701 Mark
Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is 100 and the profit in the manufacture of a unit of product E is100 and the profit in the manufacture of aunit of product E is 87. The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,000. Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?
  1. $5\text{D}+7\text{E}\leq5,000$
  2. $9\text{D}+3\text{E}\geq4,000$
  3. $5\text{D}+7\text{E}=4,000$
  4. $5\text{D}+9\text{E}\leq5,000$
  5. $9\text{D}+3\text{E}\leq5,000$
Answer
  1. $9\text{D}+3\text{E}\leq5,000$
Solution:
Given, product D takes 5 hours per unit of labour, and product E takes 7 hours per unit of labour.
Therefore, to produce D units of product D takes 5D hours andto produce E units of product E takes 7E hours Given, total labour hours per week are 4000 hours.
Hence, $5\text{D}+7\text{E}\leq4,000$
Given, product D takes 9 hours per unit of machine time, andproduct E takes 3 hours per unit of machine time.
Therefore, to produce D units of product D takes 9D hours andto produce E units of product E takes 3E hours Given, total machine hours per week are 5000 hours.
Hence, $9\text{D}+3\text{E}\leq5,000$
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Question 711 Mark
Z = 8x + 10y, subject to $2\text{x}+\text{y}\geq1,2\text{x}+3\text{y}\geq15,\text{y}\geq2,\text{x}\geq0,\text{y}\geq0.$ The minimum value of Z occurs at.
  1. (4.5, 2)
  2. (1.5, 4)
  3. (0, 7)
  4. (7, 0)
Answer
  1. (1.5, 4)
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Question 721 Mark
Unboundedness is usually a sign that the LP problem.
  1. Has finite multiple solutions.
  2. Is degenerate.
  3. Contains too many redundant constraints.
  4. Has been formulated improperly.
  5. None of the above.
Answer
  1. Has been formulated improperly.
Solution:
A linear programming problem is said to have unbounded solution if it has infinite number of solutions.
I.e., the problem has been formulated improperly.
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Question 731 Mark
The maximum value of the object function Z = 5x + 10y subject to the constraints $\text{x}+2\text{y}\leq120,\text{x}+\text{y}\geq60,\text{x}-2\text{y}\geq0,\text{x}\geq0,\text{y}\geq0$ is:
  1. 300
  2. 600
  3. 400
  4. 800
Answer
  1. 600
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MCQ 741 Mark
$Z = 4x_1 + 5x_2,$ subject to $2\text{x}_{1}+\text{x}_{2}\geq7,2\text{x}_{1}+3\text{x}2\leq15,\text{x}_{2}\leq3,\text{x}_{1},\text{x}_{2}\geq0.$ The minimum value of $Z$ occurs at$:$
  • ✓
    $(3.5, 0)$
  • B
    $(3, 3)$
  • C
    $(7.5, 0)$
  • D
    $(2, 3)$
Answer
Correct option: A.
$(3.5, 0)$
$(3.5, 0)$
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Question 751 Mark
If the feasible region for a solution of linear inequations is bounded, it is called as:
  1. Concave Polygon
  2. Finite Region
  3. Convex Polygon
  4. None of the above
Answer
  1. Convex Polygon
Solution:
A bounded feasible region will have both a maximum value and a minimum value for the objective function. It is bounded if it can be enclosed in any shape.
A convex polygon is a simple not self-intersecting closed shape in which no line segment between two points on the boundary ever goes outside the polygon.
Hence, the answer is convex polygon.
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Question 761 Mark
Choose the correct answer from the given four options.
Let F = 3x - 4y be the objective function.
Minimum value of F is:
  1. 0.
  2. -16.
  3. 12.
  4. Does not exist.
Answer
  1. -16
Solution:
the feasible region as show in the figure, has objective function F= 3x - 4y
Corner points
Corresponding value of z = 3x - 4y
(0, 0)
0
(12, 6)
12 (masimum)
(0, 4)
-16 (miminum)
We have minimum value of F is -16at (0, 4).
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Question 771 Mark
Z = 6x + 21y, subject to $ \text{x}+2\text{y}\geq3,\text{x}+4\text{y}\geq4,3\text{x}+\text{y}\geq3,\text{x}\geq0,\text{y}\geq0.$ The minimum value of Z occurs at.
  1. $(4,0)$
  2. $(28,8)$
  3. $\Big(2,\frac{7}{2}\Big)$
  4. $(0,3)$
Answer
  1. $\Big(2,\frac{7}{2}\Big)$
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Question 781 Mark
The taxi fare in a city is as follows. For the first km the fare is Rs.10 and subsequent distance is Rs.6/ km.Taking the distance covered as x km and fare as Rs y, write a linear equation.
  1. y = 4 + 6x
  2. y = 4 + 5x
  3. y = 3 + 6x
  4. y = 3 + 5x
Answer
  1. y = 4 + 6x
Solution:
First kmkm fare = Rs.10 Subsequent distance fare = Rs 6/ km
Then fare x km of distance y = (x - 1) × 6 + 10y = 6x - 6 + 10y = 6x + 4
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Question 791 Mark
In graphical solutions of linear inequalities, solution can be divided into.
  1. One subset
  2. Two subsets
  3. Three subsets
  4. Four subsets
Answer
  1. Two subsets
Solution:
In graphical solutions of linear inequalities, solution can be divided into two subsets.
for example, $2\text{x}+\text{y}\leq4$
One subset includes all values (x, y) that satisfy the equation 2x + y = 4 and the other subset includes all the values (x, y) that satisfy 2x + y < 4.
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Question 801 Mark
The feasible region for a LPP is shown shaded in the figure. Let Z = 3x - 4y be the objective function. Minimum of Z occurs at.
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
Answer
  1. (0, 8)
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Question 811 Mark
In Graphical solution the feasible solution is any solution to a LPP which satisfies.
  1. Only objective function.
  2. Non - negativity restriction.
  3. Only constraint.
  4. All the three
Answer
  1. Non - negativity restriction.
Solution:
The feasible region is the set of all the points that satisfy all the given constraints.
The variables of the linear programs must always take the non - negative values (i.e., $\text{x}\geq0$ and $\text{y}\geq0$).
These are used because x and y are usually the number of items produced and we cannot produce the negative number of items.
The least possible number of items could be zero.
Therefore, the feasible solution should satisfy the non - negativity restriction.
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Question 821 Mark
Which of the following is a type of Linear programming problem?
  1. Manufacturing problem
  2. Diet problem
  3. Transportation problems
  4. All of the above
Answer
  1. All of the above
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Question 831 Mark
If an iso-profit line yielding the optimal solution coincides with a constaint line, then:
  1. The solution is unbounded
  2. The solution is infeasible
  3. The constraint which coincides is redundant
  4. None of the above
Answer
  1. None of the above
Solution:
If an iso profit line which is yielding the optimal solution coincide with a constant line; then
→ the solution will b bounded, i.e there will be a definite bounded area where the solution would be optional.
→ Since the area is bounded,the solution is feasible
→ And the constant which coincides is not a redundant
Hence None of above is the answer.
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Question 841 Mark
The Convex Polygon Theorem states that the optimum (maximum or minimum) solution of a LPP is attained at atleastone of the ______ of the convex set over which the solution is feasible.
  1. Origin
  2. Corner points
  3. Centre
  4. Edge
Answer
  1. Corner points
Solution:
The fundamental theorem of programming (i.e., Convex Polygon Theorem) states that the optimum value(maximum or minimum) of a linear programming problem over a convex region occur at the corner points.
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Question 851 Mark
The solution of the set of constraints of a linear programming problem is a convex (open or closed) is called ______ region.
  1. Feasible
  2. Active
  3. Linear
  4. None of these
Answer
  1. Feasible
Solution:
Our experts are building a solution for this.
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Question 861 Mark
Given a system of inequation: $2\text{y}-\text{x}\leq4$ $-2\text{x}+\text{y}\geq-4$Find the value of s, which is the greatest possible sum of the x and y co - ordinates of the point which satisfies the following inequalities as graphed in the xy plane.
  1. 8
  2. 12
  3. 2
  4. 4
Answer
  1. 8
Solution:
First, rewrite each equation, so that it is in the slope - intercept form of a line, which is y = mx + b, where mm is the slope and b is the y - intercept of the line.
The first equation becomes 2y < x + 4 or $\text{y}\leq\frac{1}{2}\text{x}+2.$
The second equation becomes $\text{y}\geq2\text{x}-4.$
The greatest x + y is the point at which the two lines intersect.
Set the equations of the two lines, $\text{y}=\frac{1}{2}\text{x}+2$ and y = 2x - 4, equal to each other and solve for x.
The resulting equation is $\text{y}=\frac{1}{2}\text{x}+2$ 
and y = 2x - 4.
Solve for x to get $\text{y}=\frac{3}{-2}\text{x}+2=-4$ or $\frac{3}{-2}\text{x}=-6,$
$\Rightarrow\text{x}=4$
Next, plug 4 into one of the two equations to solve for y.
Therefore, y = 2(4) - 4 = 4 and x + y = 4 + 4 = 8.
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Question 871 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A): Assertion (A):For an objective function Z= 15x + 20y, corner points are (0, 0), (10, 0), (0, 15) and (5, 5). Then optimal values are 300 and 0 respectively.
Reason (R): The maximum or minimum value of an objective function is known as optimal value of LPP. These values are obtained at corner points.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false.
  4. A is false but R is true.
Answer
  1. Both A and R are true and R is the correct explanation of A
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Question 881 Mark
Maximize Z = 11 x + 8y subject to $\text{x}\leq4,\text{y}\leq6,\text{x}+\text{y}\leq6,\text{x}\geq0,\text{y}\geq0.$
  1. 44 at (4, 2)
  2. 60 at (4, 2)
  3. 62 at (4, 0)
  4. 48 at (4, 2)
Answer
  1. 60 at (4, 2)
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MCQ 891 Mark
Which of the following is not a convex set?
  • A
    ${(x, y) ; 2x + 5y \leq 7}$
  • B
    ${(x, y) : x^{_2} + y^2 \leq 4}$
  • ✓
    ${x : |x| = 5}$
  • D
    ${(x, y) : 3x^2 + 2y^2 \leq 6}$
Answer
Correct option: C.
${x : |x| = 5}$
$|x| = 5$ is not a convex set as any two points from negative and positive $x-$axis if are joined will not lie in set.
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Question 901 Mark
In an LPP, if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then the number of points of which Zmax occurs is:
  1. 0
  2. 2
  3. Finite
  4. Infinite
Answer
  1. Infinite
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Question 911 Mark
If the constraints in a linear programming problem are changed:
  1. Solution is not defined.
  2. The objective function has to be modified.
  3. The problems is to be re - evaluated.
  4. None of these.
Answer
  1. The problems is to be re - evaluated.
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Question 921 Mark
The maximum value of f = 4x + 3y subject to constraints $\text{x}\geq0,$ $\text{y}\geq0,$ $2\text{x}+3\text{y}\leq18;\text{x}+\text{y}\geq10$ is:
  1. 35
  2. 36
  3. 34
  4. None of these
Answer
  1. None of these
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Question 931 Mark
In Graphical solution the redundant constraint is:
  1. Which forms the boundary of feasible region.
  2. Which do not optimizes the objective function.
  3. Which does not form boundary of feasible region.
  4. Which optimizes the objective function.
Answer
  1. Which does not form boundary of feasible region.
Solution:
A constraint in an LP model becomes redundant when the feasible region doesnt change by the removing the constraint.
For example, $2\text{x}+\text{y}\geq10$ and $6\text{x}+3\text{y}\geq30$ are constraints.
$6\text{x}+3\text{y}\geq30$
$\Rightarrow3\times(2\text{x}+\text{y})\geq3\times10$
$\Rightarrow2\text{x}+\text{y}\geq10$
 which is same as the first constraint.
Therefore, $6\text{x}+3\text{y}\geq30$  can be removed.
By removing this constraint the feasible region doesnt change.
If the boundary of the feasible region is removed then feasible solution set changes.
Hence, redundant constraint cannot be the boundary of the feasible region.
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Question 951 Mark
Choose the correct answer from the given four options.
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is:
  1. p = 2q
  2. $\text{p}=\frac{\text{q}}{2}$
  3. p = 3q
  4. p = q
Answer
  1. $\text{p}=\frac{\text{q}}{2}$
Solution:
Corner point
Corresponding value of X = px + qy; p,q > 0
(0, 3)
3q
(1, 1)
p + q
(3, 0)
3p
So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is,
p + q = 3p
⇒ 2p = q
$\therefore\text{p}=\frac{\text{q}}{2}$
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Question 961 Mark
Choose the correct answer from the given four options.
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5)
Let F = 4x + 6y be the objective function Find the
Maximum of F - Minimum of F =
  1. 60.
  2. 48.
  3. 42.
  4. 18.
Answer
  1. 60.
Solution:
Corner points
Corresponding value of F = 4x + 6y
(0, 2)
12 (Minimum)
(3, 0)
12 (minimum)
(6, 0)
24
(6, 8)
72 (maxmimum)
(0, 5)
30
Maximum of F - Minimum of F = 72 - 12 = 60.
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Question 971 Mark
The region represented by the inequation system x, y ≥ 0, y ≤ 6, x + y ≤ 3 is:
  1. unbounded in first quadrant
  2. unbounded in first and second quadrants
  3. bounded in first quadrant
  4. none of these
Answer
  1. bounded in first quadrant
Solution:
Converting the given inequations into equations, we obtain
y = 6, x + y = 3, x = 0 and 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 inequation.
The line x + y = 3 meets the coordinate axis at A(3, 0) and B(0, 3).
Join these points to obtain the line x + y = 3
Clearly, (0, 0) satisfies the inequation x + y ≤ 3.
So, the region in xy-plane that contains the origin represents the solution set of the given equation.
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.
These lines are drawn using a suitable scale.

The shaded region represents the feasible region of the given LPP, which is bounded in the first quadrant.
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Question 981 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 __________.
  1. q = 2p
  2. p = 2p
  3. p = q
  4. q = 3p
Answer
  1. q = 3p
Solution:
Since Z occurs maximum at (15, 15) and (0, 20)
Therefore, 15p + 15q = 0.p + 20q
⇒ q = 3p.
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Question 991 Mark
The feasible solution for a LPP is shown in the following figure. Let Z = 3x - 4y be the objective function.
Minimum of Z occurs at:
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
Answer
  1. (0, 8)
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Question 1001 Mark
The objective function of LPP defined over the convex set attains its optimum value at.
  1. Atleast two of the corner points.
  2. All the corner points.
  3. Atleast one of the corner points.
  4. None of the corner points.
Answer
  1. Atleast one of the corner points.
Solution:
Let Z = ax + by be the objective function
When Z has optimum value(maximum or minimum), where the variables
x and y are subject to constraints described by linear inequalities, this optimum value must occur at a corner points of the feasible region.
Thus, the function attains its optimum value at one of the corner points.
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Question 1011 Mark
If the constraints in linear programming problem are changed.
  1. The problem is to be re - evaluated
  2. Solution is not defined
  3. The objective function has to be modified
  4. The change in constraints is ignored
Answer
  1. The problem is to be re - evaluated
Solution:
The above question asks for the impact of change in constraints on the Linear programming problem.
In this scenario, when there is a change in constraint, the solution will change definitely.
Whether the solution exists or not, we can only find once the problem is re - evaluated.
In an LPP, the objective function is related to the main objective of any problem, either we have to maximize or minimize the function based on the situation whereas the constraints is related to physical restrictions in achieving the defined objective function.
In real life problems, there might be situations when the constraints change, but objective function does not changes to accommodate the change in constraints.
Thus, if constraints in linear programming problem is changed, the problem has to be re - evaluated for the same objective function and after solving we can find whether the solution exists or not.
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Question 1021 Mark
Maximize Z = 11x + 8y, subject to $\text{x}\leq4,\text{y}\leq6,\text{x}\geq0,\text{y}\geq0.$
  1. 44 at (4, 2)
  2. 60 at (4, 2)
  3. 62 at (4, 0)
Answer
  1. 60 at (4, 2)
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MCQ 1031 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
    $0.52$
  • ✓
    $1$
  • C
    $0.01$
  • D
    $0.1$
Answer
Correct option: B.
$1$
Formula used$:$
$a^3 + b^3= (a + b)(a^2 - ab + b^2)$
$\frac{0.76\times0.76\times0.76+0.24\times0.24\times0.24}{0.76\times0.76-0.76\times0.24+ 0.24+0.24}$
$\frac{(0.76)^{3}+(0.24)^{3}}{0.76\times0.76-0.76\times0.24+0.24+0.24}$
$=\frac{(0.76+0.24)(0.76\times0.76-0.76\times0.24+0.24\times0.24)}{0.76\times0.76-0.76\times0.24+0.24\times0.24}$
$=(0.76+0.24)=1$
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Question 1041 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:
  1. 36
  2. 40
  3. 20
  4. none of these
Answer
  1. 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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Question 1051 Mark
In transportation models designed in linear programming, points of demand is classified as:
  1. Ordination
  2. Transportation
  3. Destinations
  4. Origins
Answer
  1. Destinations
Solution:
In linear programming, transportation modeltransportation model are applied to problems related to the study of efficient transportation routes.
i.e., how effectively the available resources are transported to different destinations with minimum cost.
Therefore, the points of demand is classified as destinations.
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Question 1061 Mark
Objective function of a LPP is:
  1. a constraint
  2. a function to be optimized
  3. a relation between the variables
  4. none of these
Answer
  1. a function to be optimized
Solution:
The objective function of a linear programming problem is either to be maximized or minimized i.e. objective function is to be optimized.
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Question 1071 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:
  1. 320
  2. 300
  3. 230
  4. none of these
Answer
  1. 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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Question 1081 Mark
Graphical method can be used only when the decision variables is:
  1. More than 3.
  2. More than 1.
  3. Two
  4. One
Answer
  1. Two
Solution:
Graphical method can be used only when the decision variables is two.
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Question 1091 Mark
A constraint in an LP model becomes redundant because:
  1. Two iso - profit line may be parallel to each other
  2. The solution is unbounded
  3. This constraint is not satisfied by the solution values
  4. None of the above
Answer
  1. None of the above
Solution:
A constraint in an LP model becomes redundant when the feasible region doesnt change by the removing the constraint.
For example, $\text{x}+2\text{y}\leq20$ and $2\text{x}+4\text{y}\leq40$ are the constraints.
$2\text{x}+4\text{y}\leq40$
$\Rightarrow2\times(\text{x}+2\text{y})\leq2\times20$
$\Rightarrow\text{x}+2\text{y}\leq20$
 which is same as the first constraint.
Therefore, $2\text{x}+4\text{y}\leq40$ can be removed.
By removing this constraint feasible region doesnt change.
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Question 1101 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:
  1. p = q
  2. p = 2q
  3. p = 3q
  4. q = 3p.
Answer
  1. 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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Question 1111 Mark
Linear programming model which involves funds allocation of limited investment is classified as:
  1. Ordination budgeting model
  2. Capital budgeting models
  3. Funds investment models
  4. Funds origin models
Answer
  1. Capital budgeting models
Solution:
In linear programming, Capital budgeting models to minimize the total capital cost. 
The solutions from the model are used to decide the best combination of capital resources and best times to start and finish projects and to determine the net capital cost.
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Question 1121 Mark
In order for a linear programming problem to have a unique solution, the solution must exist.
  1. At the intersection of the nonnegativity constraints.
  2. At the intersection of a nonnegativity constraint and a resource constraint.
  3. At the intersection of the objective function and a constraint.
  4. At the intersection of two or more constraints.
  5. None of the above.
Answer
  1. At the intersection of two or more constraints.
Solution:
In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints.
Then the problem becomes convex and has a single optimum (maximum or minimum).
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Question 1131 Mark
The feasible region for an LPP is shown shaded in the following figure. Minimum of Z = 4x + 3y occurs at the point.
  1. (0, 8)
  2. (2, 5)
  3. (4, 3)
  4. (9, 0) 
Answer
  1. (2, 5)
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Question 1141 Mark
The solution set of the inequation 2x + y > 5 is:
  1. half plane that contains the origin
  2. open half plane not containing the origin
  3. whole xy-plane except the points lying on the line 2x + y = 5
  4. none of these
Answer
  1. open half plane not containing the origin
Solution:
On putting x = 0, y = 0 in the given inequality, we get 0 > 5, which is absurd.
Therefore, the solution set of the given inequality does not include the origin.
Thus, the solution set of the given inequality consists of the open half plane not containing the origin.
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Question 1151 Mark
Which of the following statements about an LP problem and its dual is false?
  1. If the primal and the dual both have optimal solutions, the objective function values for both problems are equal at the optimum.
  2. If one of the variables in the primal has unrestricted sign, the corresponding constraint in the dual is satisfied with equality.
  3. If the primal has an optimal solution, so has the dual.
  4. The dual problem might have an optimal solution, even though the primal has no (bounded) optimum.
Answer
  1. The dual problem might have an optimal solution, even though the primal has no (bounded) optimum.
Solution:
If one of the problems (primal, dual) is infeasible then the other problem is infeasible.
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Question 1161 Mark
Objective of linear programming for an objective function is to:
  1. Maximize or minimize.
  2. Subset or proper set modeling.
  3. Row or column modeling.
  4. Adjacent modeling.
Answer
  1. Maximize or minimize.
Solution:
In linear programming, the objective function is the linear equation which is representing some quantity (such as profit gained, cost, ...) which is to be maximized or minimized subject to the given constraints.
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Question 1171 Mark
Maximize Z = 4x + 6y, subject to $3\text{x}+2\text{y}\leq12,\text{x}+\text{y}\geq4,\text{x},\text{y}\geq0.$
  1. 16 at (4, 0)
  2. 24 at (0, 4)
  3. 24 at (6, 0)
  4. 36 at (0, 6) 
Answer
  1. 36 at (0, 6)
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MCQ 1181 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
    $\text{m}\text{n}$
  • ✓
    $\text{m}+\text{n}$
  • D
    $\frac{\text{m}}{\text{n}}$
Answer
Correct option: C.
$\text{m}+\text{n}$
Constraints will be
$x_{11} ​+ x_{21} ​+ ..... + x_{m1} ​= b_{1​}$
$x_{12}​ + x_{22}​ + ..... + x_{m2}​ = b_{2​}$
$x_{1n}​ + x_{2n} ​+ ..... + x_{mn}​ = b_n​$
$x_{11}​ + x_{12}​ + ..... + x_{1n}​ = b_1​$
$x_{21​}+ x_{22} ​+ ..... + x_{2n} ​= b_{2​}$
$x_{m1} + x_{m2}​ + ..... + x_{mn}​ = b_{n​}$
So the total number of constraints $= m + n$
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Question 1191 Mark
The first step in formulating an LP problem is:
  1. Graph the problem.
  2. Perform a sensitivity analysis.
  3. Identify the objective and the constraints.
  4. Define the decision variables.
  5. Understand the managerial problem being faced.
Answer
  1. Understand the managerial problem being faced.
Solution:
The first step in formulating an linear programming problem is to understand the managerial problem being faced i.e., determine the quantities that are needed to solve the problem.
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Question 1201 Mark
If the constraints in a linear programming problem are changed
  1. the problem is to be re-evaluated
  2. solution is not defined
  3. the objective function has to be modified
  4. the change in constraints is ignored
Answer
  1. the problem is to be re-evaluated
Solution:
The optimisation of the objective function of a LPP is governed by the constraints.
Therefore, if the constraints in a linear programming problem are changed, then the problem needs to be re-evaluated.
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MCQ 1211 Mark
$Z = 20x_1 + 20x_2,$ subject to $\text{x}1\geq0,\text{x}_{2}\geq0,\text{x}_{1}+2\text{x}_{2}\geq8,3\text{x}_{1}+2\text{x}_{2}\geq15,5\text{x}_{1}+2\text{x}_{2}\geq20.$ The minimum value of $Z$ occurs at
  • A
    $(8, 0)$
  • B
    $\Big(\frac{5}{2},\frac{15}{4}\Big)$
  • ✓
    $\Big(\frac{7}{2},\frac{9}{4}\Big)$
  • D
    $(0, 10)$
Answer
Correct option: C.
$\Big(\frac{7}{2},\frac{9}{4}\Big)$
$\Big(\frac{7}{2},\frac{9}{4}\Big)$
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Question 1221 Mark
In order to maximize the profit of the company, the optimal solution of which of the following equations is required?
  1. P = x + y - 200
  2. P = 5y - 2x
  3. P = y - 80
  4. P = 200 - x
Answer
  1. P = 5y - 2x
Solution:
Let the number of normal calculators produced in a day be x andthe number of scientific calculators produced in a day be y the minimum of total calculators to be produced per day is 200
$\Rightarrow\text{x}+\text{y}\leq200$
Given, the minimum number of normal calculators to be produced per day is 100
$\Rightarrow\text{x}\geq100$
andthe minimum number of scientific calculators to be produced per day is 80
 $\Rightarrow\text{y}\geq80$
Also given, the maximum number of normal calculators can be produced per day is 200
$\Rightarrow\text{x}\leq200$
 andthe maximum number of scientific calculators can be produced per day is 170
$\Rightarrow\text{x}\leq170$
A normal calculator incurred a loss of Rs. 2
For x normal calculators, the loss is Rs. 2x
A scientific calculator gained a profit of Rs. 5
For xy scientific calculators, the gain is Rs. 5y
Therefore, profit of the manufacturer P = 5y - 2x.
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Question 1231 Mark
The maximum value of Z = 4x + 2y subject to the constraints $2\text{x}+3\text{y}\leq18,\text{x}+\text{y}\geq10,\text{x},\text{y}\leq0$ is:
  1. 36
  2. 40
  3. 30
  4. None of these
Answer
  1. None of these
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Question 1241 Mark
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called:
  1. A constraint
  2. Decision variables
  3. Objective function
  4. None of the above
Answer
  1. A constraint
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Question 1251 Mark
Feasible region (shaded) for a LPP is shown in the given figure. Minimum of z = 4x + 3y occurs at the point.
  1. (0, 8)
  2. (2, 5)
  3. (4, 3)
  4. (9, 0)
Answer
  1. (2, 5)
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Question 1261 Mark
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is:
  1. $\text{p}=2\text{q}$
  2. $\text{p}=\frac{\text{q}}{2}$
  3. $\text{p}=3\text{q}$
  4. $\text{p}=\text{q}$
Answer
  1. $\text{p}=\frac{\text{q}}{2}$
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Question 1271 Mark
Choose the most correct of the following statements relating to primal - dual linear programming problems:
  1. Shadow prices of resources in the primal are optimal values of the dual variables.
  2. The optimal values of the objective functions of primal and dual are the same.
  3. If the primal problem has unbounded solution, the dual problem would have infeasibility.
  4. All of the above.
Answer
  1. All of the above.
Solution:
From the primal - dual relationship, The shadow prices of resources in the primal are optimal values of the dual variables.
If one of the problems has an optimal feasible solution then the other problem also has an optimal feasible solution.
The optimal objective function value is same for both primal and dual problems.
If one problem has an unbounded solution then the other problem is infeasible.
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Question 1281 Mark
Linear programming used to optimize mathematical procedure and is:
  1. Subset of mathematical programming
  2. Dimension of mathematical programming
  3. Linear mathematical programming
  4. All of above
Answer
  1. Subset of mathematical programming
Solution:
Linear programming is an extremely powerful tool for addressing a wide range of applied optimization problems.
A short list of application areas is resource allocation, production scheduling, warehousing, layout, transportation scheduling, facility location, flight crew scheduling, portfolio optimization, parameter estimation.
So, linear programming is used to subset mathematical programming.
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Question 1291 Mark
The maximum value of Z = 3x + 2y, subjected to $\text{x}+2\text{y}\leq2,\text{x}+2\text{y}\geq8;\text{x},\text{y}\geq0 $ is:
  1. 32
  2. 24
  3. 40
  4. None of these
Answer
  1. None of these
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Question 1301 Mark
The feasible solution of a LPP belongs to:
  1. First and second quadrants
  2. First and third quadrants.
  3. Second quadrant
  4. Only firstquadrant.
Answer
  1. Only firstquadrant.
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Question 1311 Mark
If $\text{x}+\text{y}\leq2,$ $\text{x}\leq0,$ $\text{y}\leq0$ the point at which maximum value of 3x + 2y attained will be.
  1. $(0,0)$
  2. $\Big(\frac{1}{2},\frac{1}{2}\Big)$
  3. $(0,2)$
  4. $(2,0)$
Answer
  1. $(0,0)$
Solution:
$\text{x}\leq0$ and $\text{y}\leq0$ represents third Quadrant $\text{x}+\text{y}\leq2$ represents the region below the line $\text{x}+\text{y}\leq2$ (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 $\leq0$ in the third quadrant, the maximum value of 3x + 2y occurs at x = 0 and y = 0 and the maximum value is 0.
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Question 1321 Mark
The region represented by the inequalities $\text{x}\geq6,\text{y}\geq2,2\text{x}+\text{y}\leq0,\text{x}\geq0,\text{y}\geq{0}$ is:
  1. Unbounded
  2. A polygon
  3. Exterior of a triangle
  4. None of these
Answer
  1. None of these
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Question 1331 Mark
The objective function Z = 4x + 3y can be maximised subjected to the constraints $ 3\text{x}+4\text{y}\leq24,$ $8\text{x}+6\text{y}\leq48,$ $\text{x}\leq5,\text{y}\leq6;\text{x},\text{y}\leq0.$
  1. At only one point
  2. At two points only.
  3. At an infinite number of points.
  4. None of these
Answer
  1. At an infinite number of points.
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Question 1341 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:
  1. (5, 0)
  2. (6, 5)
  3. (6, 8)
  4. (4, 10)
Answer
  1. (5, 0)
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Question 1351 Mark
The corner points of the feasible region are A(0, 0), B(16, 0), C(8, 16) and D(0, 24). The minimum value of the objective function z = 300x + 190y is _______:
  1. 5440
  2. 4800
  3. 4560
  4. 0
Answer
  1. 0
Solution:
We know that, for a cartesian polygon , the maximum value occurs at the corner points or vertices of the polygon.
Given z = 300x + 190y
By substituting A(0, 0) in the equation we get z = 0
By substituting B(16, 0) in the equation we get z = 4800
By substituting C(8, 16) in the equation we get z = 5440
By substituting D(0, 24) in the equation we get z = 4560
Hence the minimum value of Z occured at C(0, 0) with z = 0
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Question 1361 Mark
Consider the objective function Z = 40x + 50y The minimum number of constraints that are required to maximize Z are:
  1. 4
  2. 2
  3. 3
  4. 1
Answer
  1. 3
Solution:
Since in the given function Z = 40x + 50y, two variables are used.
So, the two constraints will be $\text{x}\geq0,\text{y}\geq0$ and the third one will be of the type
$\text{ax}+\text{by}\geq\text{c}.$
Hence, at least 3 constraints are required.
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Question 1371 Mark
 Z = 7x + y, subject to $ 5\text{x}+\text{y}\geq5,\text{x}+\text{y}\geq3,\text{x}\geq0, y\geq0.$ The minimum value of Z occurs at:
  1. $(3, 0)$
  2. $\Big(\frac{1}{2},\frac{5}{2}\Big)$
  3. $(7, 0)$
  4. $(0,5)$
Answer
  1. $(0,5)$
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Question 1381 Mark
The optimal value of the objective function is attained at the points:
  1. On X - axis
  2. On Y - axis
  3. Corner points of the feasible region
  4. None of these
Answer
  1. Corner points of the feasible region
Solution:
Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.
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Question 1391 Mark
An iso-profit line represents:
  1. An infinite number of solutions all of which yield the same profit.
  2. An infinite number of solution all of which yield the same cost.
  3. An infinite number of optimal solutions.
  4. A boundary of the feasible region.
Answer
  1. An infinite number of solutions all of which yield the same profit.
Solution:
The graph of the profit function is called an iso profit line. It is called this because iso means same or equal and the profit anywhere on the line is the same.
So, an iso-profit lines represents an infinite number of solutions all of which yield the same profit.
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Question 1401 Mark
Choose the correct answer from the given four options.
The 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 function is Z = 4x + 3y.
Compare the quantity in Column A and Column B.
Column A
Column B
Maximum of Z
325
  1. The quantity in column A is greater .
  2. The quantity in column B is greater.
  3. The two quantities are equal.
  4. The relationship can not be determined on the basis of the information supplied.
Answer
  1. The quantity in column B is greater.
Solution:
Corner points
Corresponding value of Z = 4x + 3y
(0, 0)
0
(0, 40)
120
(20, 40)
200
(60, 20)
300 (Maximum)
(60, 0)
240
Hence, maxmimum value of Z = 300 < 325
So, the quantity in column B is greater.
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Question 1411 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:
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
Answer
  1. (0, 8)
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Question 1421 Mark
Consider a LPP given by
Minimum Z = 6x + 10y
Subjected to x ≥ 6, y ≥ 2, 2x + y ≥ 10, x ≥ 0, y ≥ 0
Redundant constraints in this LPP are
  1. x ≥ 0, y ≥ 0
  2. x ≥ 6
  3. 2x + y ≥ 10
  4. none of these
Answer
  1. 2x + y ≥ 10
Solution:
Consider, x = 6
and y = 2
Now 2x + y = 10
x
y
(x, y)
0
10
(0, 10)
5
0
(5, 0)
 

Minimum Z will be at 2x + y ≥ 10.
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Question 1431 Mark
In solving the LPP: “minimize f = 6x + 10y subect to constraints $\text{x}\geq6,\text{y}\geq2,2\text{x}+\text{y}\geq10,\text{x}\geq0,\text{y}\geq0$” redundant constraints are:
  1. $\text{x}\geq6,\text{y}\geq2$
  2. $2\text{x}+\text{y}\geq10,\text{x}\geq0,\text{y}\geq0$
  3. $\text{x}\geq6$
  4. $\text{None of these}$
Answer
  1. $2\text{x}+\text{y}\geq10,\text{x}\geq0,\text{y}\geq0$
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MCQ 1441 Mark
Minimize $Z = 20x_1 + 9x_2,$ subject to $\text{x}_{1}\geq0,\text{x}_{2}\geq0,2\text{x}_{1}+2\text{x}_{2}\geq36,6\text{x}_{1}+\text{x}_{2}\geq60.$
  • A
    $360$ at $(18, 0)$
  • ✓
    $336$ at $(6, 4)$
  • C
    $540$ at $(0, 60)$
  • D
    $0$ at $(0, 0)$
Answer
Correct option: B.
$336$ at $(6, 4)$
$336$ at $(6, 4)$
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MCQ 1451 Mark
If $x + y = 3$ and $xy = 2,$ then the value of $x^3 - y^3$ is equal to.
  • A
    $6$
  • ✓
    $7$
  • C
    $8$
  • D
    $0$
Answer
Correct option: B.
$7$
Formula used:
$\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)$
$=(\sqrt{(\text{x}+\text{y})^{2}-4\text{xy}})[(\text{x}+\text{y})^{2}-\text{xy}]$
$=(\sqrt{(3)^{2}-4(2})[(3)^{2}-2]$
$=(\sqrt{1})(7)$
$=7$
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Question 1461 Mark
Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem (using simplex), we find that.
  1. The values of decision variables obtained by rounding off are always very close to the optimal values.
  2. The value of the objective function for a maximization problem will likely be less than that for the simplex solution.
  3. The value of the objective function for a minimization problem will likely be less than that for the simplex solution.
  4. All constraints are satisfied exactly.
  5. None of the above.
Answer
  1. The value of the objective function for a maximization problem will likely be less than that for the simplex solution.
Solution:
Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.
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Question 1471 Mark
Mark the wrong statement:
  1. The primal and dual have equal number of variables.
  2. The shadow price indicates the change in the value of the objective function, per unit increase in the value of the RHS.
  3. The shadow price of a non - binding constraint is always equal to zero.
  4. The information about shadow price of a constraint is important since it may be possible to purchase or, otherwise, acquire additional units of the concerned resource.
Answer
  1. The primal and dual have equal number of variables.
Solution:
The number of variables in dual is equal to the number of constraints in the primal and the number of variables in primal is equal to the number of constraints in the dual.
Therefore, the primal and dual doesnt have equal number of variables.
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Question 1481 Mark
Choose the correct answer from the given four options.
The feasible solution for a LPP is shown in. Let Z = 3x - 4y be the objective function.

Minimum of Z occurs at:
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
Answer
  1. (0, 8)
Solution:
Corner points
Corresponding value of Z = 3x - 4y
(0, 0)
(5, 0)
(6, 5)
(6, 8)
(4, 10)
(0, 8)
0
15 - 2
-14
-28
-32 (Minimum)
Hence, the minimum of Z occurs at (0, 8) and its minimum value is (-32).
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Question 1491 Mark
Which of the following statement is correct?
  1. Every LPP admits an optimal solution.
  2. Every LPP admits unique optimal solution.
  3. If a LPP gives two optimal solutions it has infinite number of solutions.
  4. None of these
Answer
  1. Every LPP admits unique optimal solution.
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Question 1501 Mark
The corner points of the feasible region determined by the following system of linear inequalities: $2\text{x}+\text{y}\leq10,\text{x}+3\text{y}\leq15, \text{x},\text{y}\geq0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Conditions on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is:
  1. p = 3q
  2. p = 2q
  3. p = q
  4. q = 3p
Answer
  1. q = 3p
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Question 1511 Mark
The minimum value of Z = 3x + 5y subjected to constraints $\text{x}+3\text{y}\geq3,\text{x}+\text{y}\geq2,\text{x},\text{y}\geq0$ is:
  1. 5
  2. 7
  3. 10
  4. 11
Answer
  1. 7
Solution:
The feasible region determined by the system of constraints, $\text{x}+3\text{y}\geq3,\text{x}+\text{y}\geq2,$ and $\text{x},\text{y}\geq0$ is given below

It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A(3, 0), $ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big)$ and C(0, 2)
The values of Z at these corner points are given below
Corner point
z = 3x + 5y
 
A (3, 0)
9
 
$ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big)$
7
Smallest
C (0, 2)
10
  
7 may or may not be the minimum value of Z because the feasible region is unbounded
For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check the resulting half - plane have common points with the feasible region or not.
Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7.
Thus, the minimum value of Z is 7 at point $ \text{B}\Big(\frac{3}{2},\frac{1}{2}\Big).$
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Question 1521 Mark
In linear programming, objective function and objective constraints are:
  1. Solved
  2. Linear
  3. Quadratic
  4. Adjacent
Answer
  1. Linear
Solution:
In linear programming, objective function and objective constraints are linear.
Any linear programming problem must have the following properties:-1.
The relationship between variables and constraints must be linear 2.
The constraints must be non - negative.3.. objective function must be linear.
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Question 1531 Mark
Linear programming model which involves funds allocation of limited investment is classified as:
  1. Ordination budgeting model
  2. Capital budgeting models
  3. Funds investment models
  4. Funds origin models.
Answer
  1. Capital budgeting models
Solution:
In linear programming, Capital budgeting models to minimize the total capital cost.
The solutions from the model are used to decide the best combination of capital resources and best times to start and finish projects and to determine the net capital cost.
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Question 1541 Mark
The minimum value of Z = 4x + 3y subjected to the constraints $3\text{x}+2\text{y}\geq160,$ $5+2\text{y}\geq200,$$ 2\text{y}\geq80;\text{x},\text{y}\geq0$ is:
  1. 220
  2. 300
  3. 230
  4. None of these
Answer
  1. 220
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Question 1551 Mark
The number of constraints allowed in a linear program is which of the following?
  1. Less than 5
  2. Less than 72
  3. Less than 512
  4. Less than 1,024
  5. Unlimited
Answer
  1. Unlimited
Solution:
There is no limit on constraints allowed in linear programming.
so the number of constraints is unlimited.
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Question 1561 Mark
The maximum value of Z = 3x + 4y subjected to constraints $\text{x}+\text{y}\leq4,\text{x}\geq0$ and $\text{y}\geq0$ is:
  1. 12
  2. 14
  3. 16
  4. None of the above
Answer
  1. 16
Solution:
 The feasible region determined by the constraints, $\text{x}+\text{y}\leq4,\text{x}\geq0,$ $\text{y}\geq0,$ is given below

O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given below:
Corner Point
z = 3x + 4y
O(0, 0)
0
A(4, 0)
12
B(0, 4)
16
Hence, the maximum value of Z is 16 at point B (0, 4)
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Question 1571 Mark
In equation $3\text{x}-\text{y}\geq3$ and 4x - 4y > 4.
  1. Have solution for positive x and y.
  2. Have no solution for positive x and y.
  3. Have solution for all x.
  4. Have solution for all y.
Answer
  1. Have solution for positive x and y.
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