Linear Programmig Problem MCQ Quiz in मल्याळम - Objective Question with Answer for Linear Programmig Problem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 15, 2025

നേടുക Linear Programmig Problem ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Linear Programmig Problem MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Linear Programmig Problem MCQ Objective Questions

Top Linear Programmig Problem MCQ Objective Questions

Linear Programmig Problem Question 1:

The feasible region for an LPP is shown in the Figure. Let F = 3x – 4y be the objective function.

F1 Savita CTET 10-5-22 D2

The minimum value of F is

  1. 0
  2. –16
  3. 12
  4. does not exist

Answer (Detailed Solution Below)

Option 2 : –16

Linear Programmig Problem Question 1 Detailed Solution

Explanation:

Corner points of the feasible region for the given figure are (0, 0), (0, 4), and (12, 6).
F = 3x - 4y

F1 Savita CTET 10-5-22 D2

 

Corner Points Corresponding Value of F = 3x - 4y
(0, 0) 3 × 0 - 4 × 0 = 0
(0, 4) 3 × 0 - 4 × 4 = - 16 (Minimum)
(12,6) 3 × 12 - 4 × 6 = 12 (Maximum)
 

So Minimum value of F = -16 at point (0,4).

The Correct Option is (2).

Linear Programmig Problem Question 2:

Consider the inequations 2x + y > 3, x + 2y > 6 and x, y > 0. Which one of the following points lies in the common region?

  1. (2, 1.5)
  2. (2, 2)
  3. (3, 1)
  4. (3, 5)

Answer (Detailed Solution Below)

Option 4 : (3, 5)

Linear Programmig Problem Question 2 Detailed Solution

Calculation:

Converting the inequations to equations, we obtain:

2x + y = 3, x + 2y = 6, x = 0, y = 0.

For 2x + y = 3 

  • Putting x = 0, 0 + y = 3 ⇒ y = 3
  • Putting y = 0, 2x + 0 = 3 ⇒ x = \(\frac{3}{2}\)
  • This line meets the x-axis at \(\left ( \frac{3}{2},0 \right )\) and the y-axis at (0, 3). Draw a line through these points.
  • We see that the origin (0, 0) does not satisfy the inequation 2x + y > 3
  • Therefore, the region that does not contain the origin is the solution of the inequality 2x + y > 3

For x + 2y = 6

  • Putting x = 0, 0 + 2y = 6 ⇒ y = 3
  • Putting y = 0, x + 0 = 6 ⇒ x = 6
  • This line meets the x-axis at (6, 0) and the y-axis at (0, 3). Draw a line through these points.
  • We see that the origin (0, 0) does not satisfy the inequation x + 2y > 6
  • Therefore, the region that does not contain the origin is the solution of the inequality x + 2y > 6

F3 Defence Savita 10-11-22 D2

The solution to the inequalities is the intersection of the 2x + y > 3, x + 2y > 6 and x, y > 0. Thus, the shaded region represents the solution set of the given set of inequalities.

Now solving 2x + y = 3 and x + 2y = 6, we get the point P(0, 3).

Now, Putting (3, 5) in 2x + y > 3,

2(3) + 5 > 3 ⇒ 11 > 3 which is true.

Again, putting (3, 5) in x + 2y > 6,

3 + 2(5) > 6 ⇒ 13 > 6 which is true.

∴ (3, 5) lies in the common region.

So, the correct option is (4)

Linear Programmig Problem Question 3:

Consider an objective function Z(x, y) = 11x + 7y

And the constraints:

2x + y ≤ 6

x ≤ 2

x ≥ 0, y ≥ 0

The maximum value of the objective function is.....

  1. 42
  2. 54
  3. 18
  4. 36

Answer (Detailed Solution Below)

Option 1 : 42

Linear Programmig Problem Question 3 Detailed Solution

Solution:

Given that, Z = 11x + 7y and the constraints 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0

Let 2x + y = 6

x 0 3
y 6 0
F3 Defence Savita 10-11-22 D5

Now, plotting all the constrain equations we see that the shaded area OABC is the feasible region determined by the constraints.

The feasible region is bounded. So, the maximum value will occur at a corner point of the feasible region.

Corner points are (0, 0), (2, 0), (2, 2) and (0, 6).

On evaluating the value of Z, we get

Corner Points

Value of Z

O(0, 0)

0

A(2, 0)

22

B(2, 2)

36

C(0, 6)

42

From the above table, it's seen that the maximum value of Z is 42.

Therefore, the maximum value of Z is 42 at (0, 6).

∴ The correct option is (1)

Linear Programmig Problem Question 4:

Corner points of the feasible region determined by the system of linear constrains 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. p = q/2
  3. p = 3q
  4. p = q

Answer (Detailed Solution Below)

Option 2 : p = q/2

Linear Programmig Problem Question 4 Detailed Solution

CONCEPT:

  • At the corner point of the feasible reason, we have to check the value of the objective function to get the maximum value.

CALCULATION:

Given:

Corner points

Corresponding value of

Z = px + qy; p, q > 0

(0, 3) 3q
(1, 1) p + q
(3, 0) 3p


So, the condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) can be extracted by equating the minimum values of Z for (3, 0) and (1, 1).

⇒ p + q = 3p

⇒ 2p = q

∴ \(\rm p=\frac{q}{2}\)

  • So, the correct answer will be option 2.

Linear Programmig Problem Question 5:

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.

What is the condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1)?

  1. P = 2q
  2. P = 3q
  3. P = 1/2q
  4. p = q

Answer (Detailed Solution Below)

Option 3 : P = 1/2q

Linear Programmig Problem Question 5 Detailed Solution

CONCEPT:

At the corner point of the feasible reason, we have to check the value of the objective function to get the maximum value.

CALCULATION:

Given: Z = px + qy

Corner points

Corresponding value of

Z = px + qy; p, q > 0

(0, 3) 3q
(1, 1) p + q
(3, 0) 3p


So, the condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) can be extracted by equating the minimum values of Z for (3, 0) and (1, 1).

⇒ p + q = 3p

⇒ 2p = q

⇒ \(\rm p=\frac{q}{2}\)

∴  the correct answer will be option 3

Linear Programmig Problem Question 6:

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. p = \(\rm\frac{q}{2}\)
  3. p = 3q
  4. p = q

Answer (Detailed Solution Below)

Option 2 : p = \(\rm\frac{q}{2}\)

Linear Programmig Problem Question 6 Detailed Solution

Explanation:

 Z = px + qy, where p, q > 0

Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1), and (3, 0)

Let z0​ be the minimum value of z in the feasible region.

Since the minimum occurs at both (3, 0) and (1, 1).

The value z0​ is attained at both (3, 0) and (1, 1).

z0​ = p × 3 + q × 0     ---- (1)

z​= p × 1 + q × ​1     ---- (2)

From Equations (1) and (2) we get 

⟹ p × 3 + q × 0 = × 1 + q × ​1 

⟹ 3 p = p + q

⟹ 2 p = q

⟹ p = q/2

The Correct Option is (2).

Linear Programmig Problem Question 7:

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 (Detailed Solution Below)

Option 4 : any point on the line segment joining the points (0, 2) and (3, 0).

Linear Programmig Problem Question 7 Detailed Solution

Explanation:

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6,0), (6, 8) and (0,5).
F = 4x + 6y

Corner Points

Corresponding Value of

F = 4x + 6y

(0, 2) 4 × 0 + 6 × 2 = 12 (Minimum)
(3, 0) 4 × 3 + 6 × 0 = 12 (Minimum)
(6,0) 4 × 6 + 6 × 0 = 24
(6, 8) 4 × 6 + 6 × 8 = 72 (Maximum)  
(0,5)   4 × 0 + 6 × 5 = 30

 

So the Minimum value of F occurs at any point on the line segment joining the points (0, 2) and (3, 0).

The Correct Option is (4).

Linear Programmig Problem Question 8:

Consider an objective function Z(x, y) = 5x + 3y

And the constraints:

3x + 5y ≤ 15

5x + 2y ≤ 10

x ≥ 0, y ≥ 0

The maximum value of Z occurs at 

  1. \(\left ( \frac{13}{19},\frac{55}{19} \right )\)
  2. \(\left ( \frac{3}{19},\frac{5}{19} \right )\)
  3. \(\left ( \frac{20}{19},\frac{45}{19} \right )\)
  4. \(\left ( \frac{34}{19},\frac{56}{19} \right )\)

Answer (Detailed Solution Below)

Option 3 : \(\left ( \frac{20}{19},\frac{45}{19} \right )\)

Linear Programmig Problem Question 8 Detailed Solution

Solution:

Given that, Z = 5x + 3y and the constraints 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

Taking 3x + 5y = 15, we have

x 0 5
y 3 0

And, taking 5x + 2y = 10, we have

x 0 2
y 5 0

F3 Defence Savita 10-11-22 D3

Now, plotting all the constrain equations we see that the shaded area OABC is the feasible region determined by the constraints.

The feasible region is bounded. So, the maximum value will occur at a corner point of the feasible region.

Corner points are (0, 0), (2, 0), (0, 3) and \(\left ( \frac{20}{19},\frac{45}{19} \right )\)

On evaluating the value of Z, we get

Corner Points

Value of Z

O(0, 0)

0

A(2, 0)

10

B(0, 3)

9

C\(\left ( \frac{20}{19},\frac{45}{19} \right )\)

\(\frac{235}{19}\)

From the above table, it's seen that the maximum value of Z is \(\frac{235}{19}\).

Therefore, the maximum value of the function Z is \(\frac{235}{19}\) at \(\left ( \frac{20}{19},\frac{45}{19} \right )\).

∴ The correct option is (3)

Linear Programmig Problem Question 9:

Consider a product manufacturing company that makes products A and B. The company uses electricity and diesel as energy sources. For the production of A, 1 unit (not kWh) of electricity and 3 units of diesel are required and for B, 1 unit of electricity and 2 units of diesel are required. The company's captive power plant supplies 5 units of electricity and 12 units of diesel.

Product

Electricity required (units)

Diesel required (units)

A

1

3

B

1

2

On the sale of A and B, the company makes a profit of Rs 60 and 50 respectively. Which of the following is representing the objective function subject to the constraints? Take x units of A and y be the units of B being sold. 

  1. Max Z = 60x + 50y, x + y ≤ 5, 3x + 2y ≤ 12

  2. Max Z = 50x + 60y, x + y ≤ 5, 3x + 2y ≤ 12

  3. Min Z = 60x + 50y, x + 2y ≤ 5, 3x + y ≤ 12

  4. Min Z = 60x + 50y, x + y ≤ 5, 3x + 2y ≤ 12

Answer (Detailed Solution Below)

Option 1 :

Max Z = 60x + 50y, x + y ≤ 5, 3x + 2y ≤ 12

Linear Programmig Problem Question 9 Detailed Solution

Concept: 

There are three main components of linear programming:

  • Decision Variables: These variables are the activities that share the available resources while also competing with one another. In other words, these are interrelated in terms of resource utilization and need to be solved simultaneously. They are assumed to be non-negative and continuous.  
  • The Objective Function: Each linear programming problem is aimed to have an objective to be measured in quantitative terms such as profits, sales, cost, time, or some other parameter, etc which needs to be maximized or minimized. The objective function will vary depending upon the business requirements or other factors.
  • Constraints: These represent real-life limitations such as money, time, labor or other restrictions and the objective function needs to be maximized or minimized while satisfying the constraints. They are represented as linear equations or inequations in terms of the decision variables.

For example, suppose x and y are the decision variables. The objective function will be given by:

Z = ax + by ….(1)

Where a and b are constants and Z is the function to be maximized or minimized. There will be conditions x ≥ 0, and y ≥ 0, which indicates the non-negative constraints on the decision variable.

The equation looks very simple since there are various assumptions involved while forming a linear programming example. These are mentioned below:

  • Parameters such as resources available, profit contribution of unit decision variable and resource used by unit decision variable needs to be known.
  • Decision variables are continuous. Hence the outputs can be an integer or a fraction.
  • The contribution of each decision variable in the objective function is directly proportional to the objective function.

Calculation:

Given:

x, y: Number of units of A and B being sold respectively.

It is very beneficial if the information is converted to a tabular form to make the data visualization much easier. As per the above info, the table is formed below:

Product

Electricity required (units)

Diesel required (units)

Profit made by selling

A

1

3

Rs 60

B

1

2

Rs 50

Total resource available

5

12

 

 

  • Since x and y cannot be negative, x ≥ 0      …(2) and y ≥ 0      ….(3)
  • One unit of A takes 1 unit of electricity and 3 units of diesel.   
  • Similarly, one unit of B takes 1 unit of electricity and 2 units of diesel.   

x + y ≤ 5       ….(4) and 3x + 2y ≤ 12       ….(5)

  • The objective function will be the profit obtained by selling x units of A and y units of B.

⇒ Z = 60x + 50y ….(6)

  • Hence, the problem is represented mathematically by (2), (3), (4), (5), and (6).
  • So, the correct answer is option 1.

Linear Programmig Problem Question 10:

Which of the following represents the general form of an objective function of a Linear Programming Problem?

  1. Z = ax + y
  2. Z = ax + by + c
  3. ax + by + c = 1
  4. ax + by + c = 0 

Answer (Detailed Solution Below)

Option 1 : Z = ax + y

Linear Programmig Problem Question 10 Detailed Solution

Concept

Objective Function: In a Linear Programming Problem (LPP), the objective function is a linear equation involving decision variables that we aim to either maximize or minimize. The general form of an objective function is:

\( Z = ax + by \)

Linear Programming: Linear programming is a method for finding the maximum or minimum value of a linear function, subject to certain constraints. The objective function and the constraints are linear functions of the decision variables.

Calculation

Given the objective function:

\( Z = ax + y \)

This is a valid and simplified form of a linear objective function in Linear Programming.​

Important Points\(Z=ax+by+c\) 

This is also a valid form for the objective function, but it is more general than Option 1. It involves two variables, x and y, with coefficients a and b, and an additional constant term  c" id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0"> c .

Options 3 and 4 are constraint equations, not an objective function.

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