CUET UG Applied Mathematics Booster Test 2 - Feasible Regions, Optimal Solutions, and Graphical Methods -

🚀

CUET UG Applied Mathematics Booster Test 2 - Feasible Regions, Optimal Solutions, and Graphical Methods

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Feasible Regions, Optimal Solutions, and Graphical Methods

1 / 20

QUESTION 1 OF 20

A society has 50 hectares available for crops, leading to the constraint:
x + y ≤ 50
Herbicide usage imposes the constraint:
20x + 10y ≤ 800 ⇒ 2x + y ≤ 80
The point of intersection of:
x + y = 50 and 2x + y = 80
is:

Answer Review

1 A society has 50 hectares available for crops, leading to the constraint:
x + y ≤ 50
Herbicide usage imposes the constraint:
20x + 10y ≤ 800 ⇒ 2x + y ≤ 80
The point of intersection of:
x + y = 50 and 2x + y = 80
is:

A (20, 20)

B (10, 40)

C (30, 20)

D (40, 10)

💡 Short Explanation

Solve the two equations simultaneously. Eliminate one variable to obtain the other. The intersection point is (30, 20).

✅ Correct Answer Explanation

The boundary equations are: x + y = 50 2x + y = 80 Subtract the first equation from the second: (2x + y) − (x + y) = 80 − 50 x = 30 Substitute x = 30 into the first equation: 30 + y = 50 y = 20 Therefore, the intersection point is: (30, 20) This point satisfies both equations and represents the common point where the two boundary lines intersect. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) (20, 20)
Incorrect because 20 + 20 = 40, which does not satisfy x + y = 50.
Option B) (10, 40)
Incorrect because 2(10) + 40 = 60, not 80.
Option D) (40, 10)
Incorrect because 2(40) + 10 = 90, not 80.

🎯 Strategy

used

Substitution

Application:

Solve the simultaneous equations by eliminating one variable and then substitute the obtained value into the other equation.

Final Logic:

The simultaneous solution gives x = 30 and y = 20; therefore, Option C is correct.

🧠 Memory Trick

"Two equations → One common point."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Points by Solving Simultaneous Equations

2 At the optimal point (30, 20), evaluate:
Z = 10500x + 9000y

A 400000

B 500000

C 450000

D 495000

💡 Short Explanation

Substitute x = 30 and y = 20 into the objective function. Compute the total value. The result is 495000.

✅ Correct Answer Explanation

The objective function is: Z = 10500x + 9000y Substitute x = 30 and y = 20. Z = 10500(30) + 9000(20) = 315000 + 180000 = 495000 Therefore, the value of the objective function at the optimal point is 495000. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 400000
Incorrect because it is less than the calculated value.
Option B) 500000
Incorrect because it does not match the result obtained after substitution.
Option C) 450000
Incorrect because of incorrect multiplication or addition.

🎯 Strategy

used

Substitution

Application:

Insert the coordinates of the optimal point directly into the objective function and simplify.

Final Logic:

Substituting (30, 20) gives Z = 495000; therefore, Option D is correct.

🧠 Memory Trick

"Optimal point → Substitute into Z."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Evaluation of the Objective Function

3 Assertion (A): The non-negativity constraints x ≥ 0, y ≥ 0 can be ignored without affecting the feasible region.
Reason (R): Linear variables always take negative values in applied mathematics.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R is the correct explanation of A

D A is false, R is true

💡 Short Explanation

Non-negativity constraints define the feasible region. Decision variables usually represent physical quantities and cannot be negative. Therefore, both statements are false.

✅ Correct Answer Explanation

Assertion (A): The non-negativity constraints x ≥ 0 and y ≥ 0 can be ignored without affecting the feasible region. This statement is false. In linear programming, non-negativity constraints ensure that decision variables represent meaningful quantities such as production, labour, land, or transportation. Removing these constraints changes the feasible region and may introduce unrealistic solutions. Reason (R): Linear variables always take negative values in applied mathematics. This statement is also false. In most real-life applications of linear programming, variables represent physical quantities that cannot be negative. Therefore, both the Assertion and the Reason are false. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) A is true, R is false
Incorrect because the Assertion is false.
Option C) Both A and R are true, and R is the correct explanation of A
Incorrect because neither statement is true.
Option D) A is false, R is true
Incorrect because the Reason is also false.

🎯 Strategy

used

Elimination

Application:

Evaluate the truth of the Assertion and the Reason independently before selecting the appropriate option.

Final Logic:

Since both statements are false, Option A is correct.

🧠 Memory Trick

"LPP variables cannot be negative."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Non-Negativity Constraints

4 The position vector of the point C(2, 0) from the origin is:

A \(2\hat{j}\)

B \(2\hat{i}\)

C \(\hat{i}+2\hat{j}\)

D \(2\hat{i}+2\hat{j}\)

💡 Short Explanation

The x-coordinate gives the coefficient of î. The y-coordinate gives the coefficient of ĵ. The position vector is 2î.

✅ Correct Answer Explanation

The position vector of a point (x, y) from the origin is written as: xî + yĵ For the point C(2, 0), x = 2 y = 0 Therefore, Position vector = 2î + 0ĵ = 2î Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 2ĵ
Incorrect because the y-coordinate is zero.
Option C) î + 2ĵ
Incorrect because the coordinates are not (1, 2).
Option D) 2î + 2ĵ
Incorrect because the y-coordinate is not 2.

🎯 Strategy

used

Substitution

Application:

Replace the coordinates directly into the position vector formula xî + yĵ.

Final Logic:

The point (2, 0) corresponds to the vector 2î; therefore, Option B is correct.

🧠 Memory Trick

"(x, y) → xî + yĵ."

📚 Class XI – Mathematics – Chapter: Introduction to Three-Dimensional Geometry – Topic: Position Vector

5 Match the linear constraints with their x-intercepts:

List I	List II
1. 2x + y = 4 → (2, 0)	a. (4, 0)
2. 3x + y = 9 → (3, 0)	b. (3, 0)
3. 5x + 5y = 20 → (4, 0)	c. (2, 0)
4. 4x + 2y = 6 → (1.5, 0)	d. (1.5, 0)

A 1–c, 2–b, 3–a, 4–d

B 1–b, 2–c, 3–a, 4–d

C 1–c, 2–a, 3–b, 4–d

D 1–d, 2–b, 3–a, 4–c

💡 Short Explanation

Put y = 0 in each equation. Calculate the x-intercept. Match each intercept with List II.

✅ Correct Answer Explanation

To find the x-intercept, substitute y = 0 into each equation. 1. 2x + y = 4 2x = 4 x = 2 → (2, 0) = c 1. 3x + y = 9 3x = 9 x = 3 → (3, 0) = b 1. 5x + 5y = 20 5x = 20 x = 4 → (4, 0) = a 1. 4x + 2y = 6 4x = 6 x = 1.5 → (1.5, 0) = d Thus, the correct matching is: 1–c, 2–b, 3–a, 4–d Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the first and second equations are mismatched.
Option C) Incorrect because the second and third equations are incorrectly paired.
Option D) Incorrect because the first and fourth equations are incorrectly matched.

🎯 Strategy

used

Substitution

Application:

Set y = 0 for every equation, determine the x-intercept, and compare it with the given matches.

Final Logic:

The calculated x-intercepts exactly match Option A.

🧠 Memory Trick

"x-axis ⇒ y = 0."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Graphical Representation of Linear Constraints

6 Find the area of the triangular region bounded by:
x + y ≤ 8,
x ≥ 4,
y ≥ 0

A 16

B 4

C 8

D 12

💡 Short Explanation

Identify the vertices of the triangular region. Use the formula for the area of a triangle. The area obtained is 8 square units.

✅ Correct Answer Explanation

The given constraints are: x + y ≤ 8 x ≥ 4 y ≥ 0 The boundary lines are: x + y = 8 x = 4 y = 0 The vertices of the triangular region are: (4, 0) (8, 0) (4, 4) The base of the triangle is: 8 − 4 = 4 units The height is: 4 units Area of triangle = (1/2) × Base × Height = (1/2) × 4 × 4 = 8 square units Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 16
Incorrect because it represents the area of the enclosing rectangle, not the triangle.
Option B) 4
Incorrect because the triangle's area is twice this value.
Option D) 12
Incorrect because it is not obtained from the correct base and height.

🎯 Strategy

used

Substitution

Application:

Find the corner points by intersecting the boundary lines and then apply the area formula for a triangle.

Final Logic:

The triangle has base 4 units and height 4 units, giving an area of 8 square units; therefore, Option C is correct.

🧠 Memory Trick

"Triangle Area = ½ × Base × Height."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Feasible Region

7 Depot A holds 30,000 bricks and Depot B holds 20,000 bricks. The total available stock is:

A 50,000

B 40,000

C 30,000

D 60,000

💡 Short Explanation

Add the stock available at both depots. Total stock equals the sum of individual stocks. The total available stock is 50,000 bricks.

✅ Correct Answer Explanation

Depot A contains: 30,000 bricks Depot B contains: 20,000 bricks Therefore, Total stock = 30,000 + 20,000 = 50,000 bricks This follows the principle of additivity, where the total quantity is obtained by adding the contributions from each source. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 40,000
Incorrect because it underestimates the total stock.
Option C) 30,000
Incorrect because it considers only Depot A.
Option D) 60,000
Incorrect because it exceeds the total available stock.

🎯 Strategy

used

Substitution

Application:

Add the quantities given for both depots to obtain the total stock.

Final Logic:

30,000 + 20,000 = 50,000; therefore, Option A is correct.

🧠 Memory Trick

"Total = Sum of all stocks."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Formulation of Constraints

8 Minimize:
Z = –x + 2y
subject to:
–x + 3y ≤ 10
x + y ≤ 6
x – y ≤ 2
At the point (2, 0), the value of Z is:

A 0

B –2

C 2

D 4

💡 Short Explanation

Substitute the given coordinates into the objective function. Simplify the expression carefully. The value of Z is –2.

✅ Correct Answer Explanation

The objective function is: Z = –x + 2y Substitute the point (2, 0). Z = –2 + 2(0) = –2 + 0 = –2 The given point satisfies all the constraints: For –x + 3y ≤ 10: –2 + 0 = –2 ≤ 10 ✓ For x + y ≤ 6: 2 + 0 = 2 ≤ 6 ✓ For x – y ≤ 2: 2 – 0 = 2 ≤ 2 ✓ Thus, the point is feasible and the corresponding value of the objective function is –2. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 0
Incorrect because direct substitution gives –2.
Option C) 2
Incorrect because the negative sign before x has been ignored.
Option D) 4
Incorrect because it is not obtained from the objective function.

🎯 Strategy

used

Substitution

Application:

Replace x and y with the given coordinates and evaluate the objective function directly.

Final Logic:

Substituting (2, 0) gives Z = –2; therefore, Option B is correct.

🧠 Memory Trick

"Given point → Substitute into Z."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Objective Function

9 If 100 points are randomly selected from an infeasible region, the probability that any point satisfies all constraints is:

A 0

B 0.5

C 1

D 0.1

💡 Short Explanation

An infeasible region contains no feasible points. Therefore, no selected point can satisfy all constraints. The probability is zero.

✅ Correct Answer Explanation

An infeasible region is one in which no point satisfies all the given constraints simultaneously. Since there are no feasible points in the region, selecting any point from it will never satisfy all the constraints. Therefore, Probability = Number of favourable points / Total number of selected points = 0 / 100 = 0 Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 0.5
Incorrect because no feasible point exists.
Option C) 1
Incorrect because every selected point is infeasible.
Option D) 0.1
Incorrect because the probability is not positive.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recognize that an infeasible region has no valid solutions, making the probability of selecting a feasible point equal to zero.

Final Logic:

No feasible point exists in an infeasible region; therefore, Option A is correct.

🧠 Memory Trick

"Infeasible ⇒ Zero valid points."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Infeasible Region

10 The area bounded by:
y = 3,
x = 0,
x = 2
is given by:
\(\int_{0}^{2}\,3 dx\)
The value is:

A 3

B 2

C 5

D 6

💡 Short Explanation

The function is constant. Multiply the constant height by the width of the interval. The area obtained is 6 square units.

✅ Correct Answer Explanation

The definite integral is: ∫₀² 3 dx Since the function is constant, Area = 3 × (2 − 0) = 3 × 2 = 6 Graphically, this represents a rectangle with: Height = 3 units Width = 2 units Area = Height × Width = 3 × 2 = 6 square units Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 3
Incorrect because it represents only the height.
Option B) 2
Incorrect because it represents only the width.
Option C) 5
Incorrect because it is not obtained from the definite integral.

🎯 Strategy

used

Substitution

Application:

Evaluate the definite integral by multiplying the constant integrand by the length of the interval.

Final Logic:

The integral ∫₀² 3 dx equals 6; therefore, Option D is correct.

🧠 Memory Trick

"Constant × Interval = Area."

📚 Class XII – Mathematics – Chapter: Integrals – Topic: Definite Integral as Area Under a Curve

11 The intersection point of:
3x + 5y = 15
5x + 2y = 10
is:

A \(\left(1\ ,\ 2\right)\)

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

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

D \(\left(2\ ,\ 3\right)\)

💡 Short Explanation

Solve the two equations simultaneously. Find the common values of x and y. The intersection point is (20/19, 45/19).

✅ Correct Answer Explanation

The given equations are: 3x + 5y = 15 5x + 2y = 10 Using the elimination method: Multiply the first equation by 2: 6x + 10y = 30 Multiply the second equation by 5: 25x + 10y = 50 Subtract the first equation from the second: 19x = 20 x = 20/19 Substitute x = 20/19 into the second equation: 5(20/19) + 2y = 10 100/19 + 2y = 190/19 2y = 90/19 y = 45/19 Therefore, the point of intersection is: (20/19, 45/19) Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) (1, 2)
Incorrect because it does not satisfy both equations simultaneously.
Option C) (19/20, 19/45)
Incorrect because these values are not obtained by solving the given equations.
Option D) (2, 3)
Incorrect because substituting this point into the equations does not satisfy them.

🎯 Strategy

used

Substitution

Application:

Solve the simultaneous linear equations using the elimination method and substitute the obtained value into one equation.

Final Logic:

The simultaneous solution gives (20/19, 45/19); therefore, Option B is correct.

🧠 Memory Trick

"Intersection = Common solution of both equations."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Points by Solving Simultaneous Equations

12 For the system:
4x + 6y = 360
3x = 180
the corresponding value of y is:

A 10

B 30

C 20

D 40

💡 Short Explanation

Find x from the second equation. Substitute x into the first equation. The corresponding value of y is 20.

✅ Correct Answer Explanation

The given equations are: 4x + 6y = 360 3x = 180 From the second equation: x = 180/3 x = 60 Substitute x = 60 into the first equation: 4(60) + 6y = 360 240 + 6y = 360 6y = 120 y = 20 Therefore, the corresponding value of y is 20. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 10
Incorrect because substituting x = 60 gives y = 20.
Option B) 30
Incorrect because it does not satisfy the first equation.
Option D) 40
Incorrect because it gives 4(60) + 6(40) = 480, not 360.

🎯 Strategy

used

Substitution

Application:

Determine x from the simpler equation and substitute it into the remaining equation to calculate y.

Final Logic:

Substituting x = 60 gives y = 20; therefore, Option C is correct.

🧠 Memory Trick

"Find x first, then find y."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Solution of Simultaneous Linear Equations

📄 Passage

A diet problem involves nutrients \(X,Y,Z\). Product A costs ₹20 and Product B costs ₹40. The objective is: \(Minimize Z=20x+40y\) Subject to nutrient constraints: \(36x+6y\geq 108,3x+12y\geq 36,20x+10y\geq 100\)

Based on the nutrient constraints, the simplified inequality for nutrient Y is:

A x + 2y ≥ 12

B 3x + y ≥ 36

C x + 4y ≥ 12

D 4x + y ≥ 12

💡 Short Explanation

Identify the nutrient Y constraint. Divide all terms by the common factor. Obtain the simplified inequality.

✅ Correct Answer Explanation

The nutrient Y constraint is: 3x + 12y ≥ 36 Divide every term by 3. 3x/3 + 12y/3 ≥ 36/3 x + 4y ≥ 12 Thus, the simplified inequality is: x + 4y ≥ 12 This simplified form represents the same feasible region because every term is divided by the same positive constant. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) x + 2y ≥ 12
Incorrect because only the coefficient of x is simplified correctly.
Option B) 3x + y ≥ 36
Incorrect because the coefficients are not simplified.
Option D) 4x + y ≥ 12
Incorrect because the variables have been simplified incorrectly.

🎯 Strategy

used

Substitution

Application:

Divide each term of the inequality by the greatest common factor while keeping the inequality sign unchanged.

Final Logic:

Dividing 3x + 12y ≥ 36 by 3 gives x + 4y ≥ 12; therefore, Option C is correct.

🧠 Memory Trick

"Divide every term by the same number."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Formulation and Simplification of Constraints

📄 Passage

Based on the nutrient constraints, the simplified inequality for nutrient Z is:

A 2x + y ≥ 10

B x + 2y ≥ 10

C 2x + y ≥ 20

D x + y ≥ 100

💡 Short Explanation

Identify the nutrient Z constraint. Divide all terms by the common factor. The simplified inequality is 2x + y ≥ 10.

✅ Correct Answer Explanation

The nutrient Z constraint is: 20x + 10y ≥ 100 Divide every term by 10. 20x/10 + 10y/10 ≥ 100/10 2x + y ≥ 10 Since all terms are divided by the same positive number, the inequality remains unchanged. Therefore, the simplified inequality is: 2x + y ≥ 10 Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) x + 2y ≥ 10
Incorrect because the coefficients are interchanged.
Option C) 2x + y ≥ 20
Incorrect because the constant term is incorrectly simplified.
Option D) x + y ≥ 100
Incorrect because the entire inequality has not been simplified correctly.

🎯 Strategy

used

Substitution

Application:

Simplify the inequality by dividing every term by the greatest common factor.

Final Logic:

Dividing 20x + 10y ≥ 100 by 10 gives 2x + y ≥ 10; therefore, Option A is correct.

🧠 Memory Trick

"Simplify all terms together."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Formulation and Simplification of Constraints

15 Maximize:
Z = 3x + 2y
subject to:
–2x + y ≤ 1
x ≤ 2
x + y ≤ 3
x ≥ 0
y ≥ 0
Identify the correct optimal value and corresponding point:

A Maximum Z = 5 at (1, 1)

B Maximum Z = 6 at (0, 3)

C Maximum Z = 7 at (1.5, 1.5)

D Maximum Z = 8 at (2, 1)

💡 Short Explanation

Determine the feasible corner points. Evaluate the objective function at each corner point. The maximum value obtained is 8 at (2, 1).

✅ Correct Answer Explanation

The objective function is: Z = 3x + 2y The constraints are: –2x + y ≤ 1 x ≤ 2 x + y ≤ 3 x ≥ 0 y ≥ 0 The feasible corner points are: (0, 0) (0, 1) (1, 2) (2, 1) (2, 0) Evaluate the objective function at each point: At (0, 0): Z = 3(0) + 2(0) = 0 At (0, 1): Z = 2 At (1, 2): Z = 3 + 4 = 7 At (2, 1): Z = 6 + 2 = 8 At (2, 0): Z = 6 The largest value is 8, obtained at the point (2, 1). Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Maximum Z = 5 at (1, 1)
Incorrect because Z = 3(1) + 2(1) = 5, which is not the maximum.
Option B) Maximum Z = 6 at (0, 3)
Incorrect because (0, 3) does not satisfy the constraint x + y ≤ 3 and –2x + y ≤ 1 simultaneously.
Option C) Maximum Z = 7 at (1.5, 1.5)
Incorrect because although Z = 7.5, the point (1.5, 1.5) violates the constraint x + y ≤ 3 since 1.5 + 1.5 = 3 (it satisfies equality), but it violates –2x + y ≤ 1? Here, –3 + 1.5 = –1.5 ≤ 1, so it is feasible. However, Z = 3(1.5) + 2(1.5) = 4.5 + 3 = 7.5, not 7, making the option incorrect as stated.

🎯 Strategy

used

Substitution

Application:

Identify the feasible corner points and substitute each into the objective function to determine the maximum value.

Final Logic:

The maximum value of Z is 8 at (2, 1); therefore, Option D is correct.

🧠 Memory Trick

"Corner points decide the optimum."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Graphical Method of Maximization

16 Arrange the steps of the Iso-cost method:
1. Assign a convenient value to Z
2. Formulate the LPP
3. Draw lines parallel to the objective function
4. Identify the nearest point to the origin

A 1, 2, 3, 4

B 2, 1, 3, 4

C 2, 3, 1, 4

D 4, 3, 2, 1

💡 Short Explanation

Begin by formulating the LPP. Assign a convenient value to the objective function. Draw parallel iso-cost lines and identify the nearest feasible point.

✅ Correct Answer Explanation

The correct sequence for solving a minimization problem using the Iso-cost method is: Step 1: Formulate the Linear Programming Problem. Step 2: Assign a convenient value to the objective function Z and draw the corresponding iso-cost line. Step 3: Draw lines parallel to the objective function. Step 4: Move the iso-cost line towards the origin until it first touches the feasible region. This nearest point gives the minimum value. Thus, the correct order is: 2 → 1 → 3 → 4 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 1, 2, 3, 4
Incorrect because the LPP must be formulated before assigning a value to Z.
Option C) 2, 3, 1, 4
Incorrect because a value for Z must be assigned before drawing the iso-cost line.
Option D) 4, 3, 2, 1
Incorrect because it reverses the logical procedure.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recall the standard sequence of the graphical Iso-cost method used in linear programming.

Final Logic:

The correct order is Formulate → Assign Z → Draw parallel lines → Identify optimum point; therefore, Option B is correct.

🧠 Memory Trick

"Formulate → Draw → Slide → Select."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-cost Method

17 If evaluating:
ax + by > M
for maximizing Z shows that this inequality shares points with an unbounded feasible region, then:

A M is the minimum value

B M is the maximum value

C Z has no maximum value

D The region becomes bounded

💡 Short Explanation

The feasible region is unbounded. Objective values greater than M are still attainable. Therefore, no finite maximum value exists.

✅ Correct Answer Explanation

For a maximization problem, consider the inequality: ax + by > M This inequality represents objective function values greater than M. If this half-plane continues to intersect an unbounded feasible region, then larger and larger values of the objective function remain feasible. Since there is no upper limit to the objective function, a finite maximum value cannot be obtained. Therefore, the objective function has no maximum value. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) M is the minimum value
Incorrect because the problem concerns maximization.
Option B) M is the maximum value
Incorrect because values greater than M are still feasible.
Option D) The region becomes bounded
Incorrect because evaluating the objective function does not change the nature of the feasible region.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Interpret the graphical meaning of an unbounded feasible region together with the objective function.

Final Logic:

Since objective values can increase indefinitely, Option C is correct.

🧠 Memory Trick

"Unbounded + Maximization = No maximum."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Unbounded Feasible Region and Optimal Solution

18 Minimize:
Z = 20x + 40y
At the point (2, 6), the value of Z is:

A 280

B 300

C 320

D 260

💡 Short Explanation

Substitute the given coordinates into the objective function. Perform the multiplication and addition carefully. The value of Z is 280.

✅ Correct Answer Explanation

The objective function is: Z = 20x + 40y Substitute x = 2 and y = 6. Z = 20(2) + 40(6) = 40 + 240 = 280 Thus, the value of the objective function at the point (2, 6) is 280. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 300
Incorrect because it is not obtained by correctly substituting x = 2 and y = 6.
Option C) 320
Incorrect because it results from incorrect multiplication or addition.
Option D) 260
Incorrect because the correct value of the objective function is 280.

🎯 Strategy

used

Substitution

Application:

Substitute the coordinates of the given point directly into the objective function and simplify.

Final Logic:

Substituting (2, 6) gives Z = 280; therefore, Option A is correct.

🧠 Memory Trick

"Given point → Substitute into Z."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Evaluation of the Objective Function

19 Identify the incorrect statement regarding alternate optimal solutions:

A Occurs when an iso-profit line is parallel to a constraint

B Gives the same optimal Z at multiple points

C Exists along the line segment joining optimal corner points

D Implies that the LPP has no solution

💡 Short Explanation

Alternate optimal solutions mean more than one optimal solution exists. They occur when the objective function is parallel to a boundary of the feasible region. The LPP still has valid optimal solutions.

✅ Correct Answer Explanation

Alternate optimal solutions occur when the objective function is parallel to one of the constraint boundaries that forms an edge of the feasible region. In such situations: Every point on the relevant boundary segment gives the same optimal value of the objective function. The LPP has more than one optimal solution. The feasible region is not empty. Therefore, the statement "Implies that the LPP has no solution" is incorrect. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Occurs when an iso-profit line is parallel to a constraint
This is a correct statement because parallelism between the objective function and a constraint boundary leads to alternate optimal solutions.
Option B) Gives the same optimal Z at multiple points
This is correct because every point on the optimal boundary segment produces the same objective value.
Option C) Exists along the line segment joining optimal corner points
This is correct because alternate optimal solutions occur along the boundary segment connecting optimal corner points.

🎯 Strategy

used

Elimination

Application:

Evaluate each statement using the properties of alternate optimal solutions and eliminate the correct statements.

Final Logic:

Alternate optimal solutions mean multiple optimal solutions exist, not the absence of a solution; therefore, Option D is correct.

🧠 Memory Trick

"Alternate optimum = Many solutions, not no solution."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Multiple Optimal Solutions

20 For:
Z = x + 2y
if both (6, 0) and (0, 3) give Z = 6, then at the point (3, 1.5), the value of Z is:

A 3

B 6

C 9

D 12

💡 Short Explanation

Substitute the coordinates into the objective function. Compute the value of Z. The value obtained is 6.

✅ Correct Answer Explanation

The objective function is: Z = x + 2y Substitute x = 3 and y = 1.5. Z = 3 + 2(1.5) = 3 + 3 = 6 Since both endpoints (6, 0) and (0, 3) produce the same objective value, every point on the line segment joining them also gives the same value because the objective function is parallel to that boundary. Therefore, the point (3, 1.5) also gives: Z = 6 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 3
Incorrect because substituting the coordinates gives Z = 6.
Option C) 9
Incorrect because it is not obtained from the objective function.
Option D) 12
Incorrect because the correct value is 6.

🎯 Strategy

used

Substitution

Application:

Substitute the coordinates directly into the objective function and verify the result.

Final Logic:

Substituting (3, 1.5) gives Z = 6; therefore, Option B is correct.

🧠 Memory Trick

"Same boundary → Same Z."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Multiple Optimal Solutions

CUET UG Applied Mathematics Booster Test 2 - Feasible Regions, Optimal Solutions, and Graphical Methods

Test Complete!

Answer Review