CUET UG Applied Mathematics Booster Test 1 - Graphical Method, Feasible Region, and Optimal Solution -

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CUET UG Applied Mathematics Booster Test 1 - Graphical Method, Feasible Region, and Optimal Solution

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 1 - Graphical Method, Feasible Region, and Optimal Solution

1 / 20

QUESTION 1 OF 20

📄 Passage

A cooperative society has \(50\) hectares of land to grow two crops \(X\) and \(Y\). The profits per hectare are ₹10,500 and ₹9,000 respectively. To control weeds, a herbicide is used at \(20\) liters per hectare for crop \(X\) and \(10\) liters per hectare for crop \(Y\), with a total availability limited to \(800\) liters. The objective is to formulate this situation as a Linear Programming Problem (LPP) to maximize total profit using graphical methods.

If x and y represent the hectares allocated to crops X and Y respectively, identify the correct mathematical formulation of the total land constraint.

Answer Review

📄 Passage

If x and y represent the hectares allocated to crops X and Y respectively, identify the correct mathematical formulation of the total land constraint.

A x + y ≥ 50

B 20x + 10y ≤ 50

C 10500x + 9000y ≤ 50

D x + y ≤ 50

💡 Short Explanation

The total land available is 50 hectares. The combined area under both crops cannot exceed 50 hectares. Therefore, the land constraint is x + y ≤ 50.

✅ Correct Answer Explanation

Let: x = hectares allocated to crop X y = hectares allocated to crop Y The cooperative society has a maximum of 50 hectares of land available. Therefore, the total cultivated area cannot exceed 50 hectares. Hence, the land constraint is: x + y ≤ 50 This inequality represents one of the linear constraints of the Linear Programming Problem. Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) x + y ≥ 50
Incorrect because the total cultivated area cannot exceed the available land. The inequality should be "≤", not "≥".
Option B) 20x + 10y ≤ 50
Incorrect because this represents neither the land constraint nor the available herbicide correctly.
Option C) 10500x + 9000y ≤ 50
Incorrect because this expression represents the objective function, not the land constraint.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Identify that the statement refers to the total available land and translate it directly into a mathematical constraint.

Final Logic:

Since the total land available is 50 hectares, the correct constraint is x + y ≤ 50.

🧠 Memory Trick

"Available resource ⇒ Use '≤'."

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

📄 Passage

Determine the correct objective function Z to be optimized.

A Maximize Z = 10500x + 9000y

B Minimize Z = 20x + 10y

C Maximize Z = 50x + 800y

D Maximize Z = 9000x + 10500y

💡 Short Explanation

The objective function represents the total profit. Multiply each crop area by its profit per hectare. Add the individual profits.

✅ Correct Answer Explanation

The profit earned is: ₹10,500 per hectare for crop X. ₹9,000 per hectare for crop Y. If: x = hectares of crop X y = hectares of crop Y Then, Total Profit Z = 10500x + 9000y Since the objective is to maximize profit, the objective function is: Maximize Z = 10500x + 9000y Therefore, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Minimize Z = 20x + 10y
Incorrect because it represents herbicide usage, not profit.
Option C) Maximize Z = 50x + 800y
Incorrect because these values correspond to resource limits rather than profits.
Option D) Maximize Z = 9000x + 10500y
Incorrect because the profit coefficients for crops X and Y are interchanged.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Assign the given profit per hectare as the coefficient of each decision variable.

Final Logic:

Profit equals 10500x + 9000y; therefore, Option A is correct.

🧠 Memory Trick

"Profit = Profit per unit × Quantity."

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

3 The boundary x = 5 is a vertical line. Evaluate the integral of a constant function:
\(\int_{0}^{5}\,10 dx\)

A 10

B 25

C 50

D 5

💡 Short Explanation

The integrand is a constant. Multiply the constant by the interval length. The value of the integral is 50.

✅ Correct Answer Explanation

The given definite integral is: ∫₀⁵ 10 dx Since the function is constant, Integral = 10 × (5 − 0) = 10 × 5 = 50 This also represents the area of a rectangle with: Height = 10 Width = 5 Therefore, the value of the integral is 50. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 10
Incorrect because it represents only the constant value.
Option B) 25
Incorrect because it is not the product of the height and width.
Option D) 5
Incorrect because it represents only the interval length.

🎯 Strategy

used

Substitution

Application:

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

Final Logic:

10 × (5 − 0) = 50; therefore, Option C is correct.

🧠 Memory Trick

"Constant × Interval = Integral."

📚 Class XII – Mathematics – Chapter: Integrals – Topic: Definite Integral of a Constant Function

4 Why are graphical limits restricted to the first quadrant in an LPP?

A Objective functions cannot have negative coefficients

B All constraints intersect only in the first quadrant

C Decision variables represent physical quantities with x ≥ 0, y ≥ 0

D Y-axis simplifies complex variables

💡 Short Explanation

Decision variables represent real physical quantities. Physical quantities cannot be negative. Therefore, solutions are restricted to the first quadrant.

✅ Correct Answer Explanation

In Linear Programming Problems, the decision variables represent quantities such as production, transportation, labour, or land allocation. These quantities cannot be negative. Hence, the non-negativity constraints are: x ≥ 0 y ≥ 0 These constraints restrict the feasible region to the first quadrant of the Cartesian plane. Therefore, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Objective functions cannot have negative coefficients
Incorrect because objective functions may contain negative coefficients.
Option B) All constraints intersect only in the first quadrant
Incorrect because constraints may intersect in any quadrant.
Option D) Y-axis simplifies complex variables
Incorrect because this has no relation to graphical limits in LPP.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recognize that non-negativity constraints determine the allowable region for decision variables.

Final Logic:

Since x and y cannot be negative, the graphical solution is restricted to the first quadrant; therefore, Option C is correct.

🧠 Memory Trick

"Physical quantities ⇒ First quadrant."

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

5 Arrange the x-intercepts of the following lines in ascending order:
1. 2x + 5y = 20
2. x + y = 4
3. 3x + 2y = 18
4. x + 4y = 12

A 2, 1, 3, 4

B 2, 3, 1, 4

C 4, 1, 3, 2

D 3, 2, 4, 1

💡 Short Explanation

Find each x-intercept by putting y = 0. Arrange the obtained values in ascending order. Compare with the given options.

✅ Correct Answer Explanation

To find the x-intercept, substitute y = 0. 1. 2x + 5y = 20 2x = 20 x = 10 2. x + y = 4 x = 4 3. 3x + 2y = 18 3x = 18 x = 6 4. x + 4y = 12 x = 12 Thus, the x-intercepts are: Line 2 → 4 Line 3 → 6 Line 1 → 10 Line 4 → 12 Ascending order: 2, 3, 1, 4 Therefore, the mathematically correct answer is Option B. Note: The provided answer key is incorrect.

❓ Why Other Options Are Incorrect

Option A) 2, 1, 3, 4
Incorrect because Line 3 has a smaller x-intercept than Line 1.
Option C) 4, 1, 3, 2
Incorrect because the intercepts are not arranged in ascending order.
Option D) 3, 2, 4, 1
Incorrect because the sequence does not follow increasing x-intercepts.

🎯 Strategy

used

Substitution

Application:

Set y = 0 in each equation, calculate the x-intercept, and arrange the values from smallest to largest.

Final Logic:

The ascending order is 2, 3, 1, 4; therefore, Option B is correct, and the provided answer key is incorrect.

🧠 Memory Trick

"x-intercept ⇒ Put y = 0."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Boundary Lines and Intercepts

6 Identify the incorrect statement regarding half-planes.

A Substituting (0, 0) into 2x + y > 4 gives 0 > 4, so the origin lies inside the valid half-plane.

B A linear inequality divides the plane into two half-planes.

C The boundary is included for ≤ or ≥.

D A closed half-plane includes its boundary line.

💡 Short Explanation

Test the point (0, 0) in the inequality. Since 0 > 4 is false, the origin is not in the required half-plane. Therefore, Option A is incorrect. The feasible region is a right-angled triangle. Determine its base and height from the intercepts. Apply the triangle area formula.

✅ Correct Answer Explanation

Consider the inequality: 2x + y > 4 Substitute the origin (0, 0): 2(0) + 0 > 4 0 > 4 This statement is false. Hence, the origin does not lie in the required half-plane. The remaining statements are standard properties of linear inequalities: Every linear inequality divides the plane into two half-planes. The symbols ≤ and ≥ include the boundary line. Such regions are called closed half-planes. Therefore, Option A is the incorrect statement. The boundary line is: x + y = 6 Intercepts: When y = 0: x = 6 Point = (6, 0) When x = 0: y = 6 Point = (0, 6) The feasible region is a triangle with vertices: (0, 0), (6, 0), and (0, 6) Area = (1/2) × Base × Height = (1/2) × 6 × 6 = 18 square units Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) A linear inequality divides the plane into two half-planes.
Correct because every linear inequality separates the plane into two regions.
Option C) The boundary is included for ≤ or ≥.
Correct because equality includes the boundary line.
Option D) A closed half-plane includes its boundary line.
Correct because closed half-planes contain all boundary points.
Option B) 36 sq units
Incorrect because it represents the area of the enclosing square.
Option C) 12 sq units
Incorrect because it is not obtained from the triangle area formula.
Option D) 24 sq units
Incorrect because the correct area is 18 square units.

🎯 Strategy

used

Substitution

Application:

Determine the intercepts of the boundary line and use them to calculate the area of the triangle.

Final Logic:

The triangle has base 6 and height 6, giving an area of 18 square units.

🧠 Memory Trick

"Triangle Area = ½ × Base × Height."

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

7 Find the area of the region defined by:
x + y ≤ 6,
x ≥ 0,
y ≥ 0

A 18 sq units

B 36 sq units

C 12 sq units

D 24 sq units

✅ Correct Answer Explanation

The feasible region is a right-angled triangle.
Determine its base and height from the intercepts.
Apply the triangle area formula.

❓ Why Other Options Are Incorrect

The boundary line is:

x + y = 6

Intercepts:

When y = 0:

x = 6

Point = (6, 0)

When x = 0:

y = 6

Point = (0, 6)

The feasible region is a triangle with vertices:

(0, 0), (6, 0), and (0, 6)

Area

= (1/2) × Base × Height

= (1/2) × 6 × 6

= 18 square units

Hence, Option A is the correct answer.

🎯 Strategy

Option B) 36 sq units

Incorrect because it represents the area of the enclosing square.

Option C) 12 sq units

Incorrect because it is not obtained from the triangle area formula.

Option D) 24 sq units

Incorrect because the correct area is 18 square units.

🧠 Memory Trick

"Triangle Area = ½ × Base × Height."

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

8 Assertion (A): The constraints x ≤ 5, y ≤ 5, x + y ≥ 12 form a bounded feasible region.
Reason (R): The maximum value of x + y under x ≤ 5, y ≤ 5 is 10, making x + y ≥ 12 impossible.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R explains A

D A is false, R is true

💡 Short Explanation

The largest possible value of x + y is 10. Therefore, x + y ≥ 12 cannot be satisfied. No feasible region exists.

✅ Correct Answer Explanation

Given: x ≤ 5 y ≤ 5 The largest possible value is: x + y = 5 + 5 = 10 However, another constraint requires: x + y ≥ 12 Since 12 is greater than the maximum attainable value of 10, no point satisfies all three constraints simultaneously. Therefore: Assertion (A) is false. Reason (R) is true because it correctly explains why no feasible region exists. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the Reason is true.
Option B) Incorrect because the Assertion is false.
Option C) Incorrect because the Assertion itself is false.

🎯 Strategy

used

Elimination

Application:

Evaluate the maximum possible value of x + y using the given constraints before checking the assertion.

Final Logic:

Since x + y can never reach 12, Option D is correct.

🧠 Memory Trick

"Maximum possible first, then compare."

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

9 Given: \(\vec{a}=4\hat{i},\vec{b}=4\hat{j}\)
Find: \(∣\vec{a}-\vec{b}∣\)

A 4

B 4√2

C 8

D 16

💡 Short Explanation

Express both vectors in component form. Subtract the vectors. Find the magnitude using the distance formula.

✅ Correct Answer Explanation

The vectors are: a⃗ = (4, 0) b⃗ = (0, 4) Therefore, a⃗ − b⃗ = (4, −4) Magnitude = √[(4)² + (−4)²] = √(16 + 16) = √32 = 4√2 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 4
Incorrect because both vector components contribute to the magnitude.
Option C) 8
Incorrect because vector magnitudes are not added directly.
Option D) 16
Incorrect because it is the square of one component, not the magnitude.

🎯 Strategy

used

Substitution

Application:

Write the vectors in coordinate form and apply the magnitude formula.

Final Logic:

The magnitude of (4, −4) is 4√2; therefore, Option B is correct.

🧠 Memory Trick

"|a − b| = √[(Δx)² + (Δy)²]."

📚 Class XI – Mathematics – Chapter: Introduction to Three-Dimensional Geometry – Topic: Position Vectors and Distance

10 Which statements are correct?
1. The optimal value always lies at the origin.
2. Optimal solutions occur at corner points.
3. A bounded region has both maximum and minimum values.

A 1 and 2 only

B 1 and 3 only

C 2 and 3 only

D 1, 2, and 3

💡 Short Explanation

The origin is not always optimal. Optimal solutions occur at corner points. A bounded feasible region has finite maximum and minimum values.

✅ Correct Answer Explanation

Evaluate each statement: Statement 1: "The optimal value always lies at the origin." This is false because the optimum may occur at any corner point depending on the objective function. Statement 2: "Optimal solutions occur at corner points." This is true according to the Corner Point Theorem. Statement 3: "A bounded region has both maximum and minimum values." This is true because the feasible region is closed and bounded. Therefore, Statements 2 and 3 are correct. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 1 and 2 only
Incorrect because Statement 1 is false.
Option B) 1 and 3 only
Incorrect because Statement 1 is false.
Option D) 1, 2, and 3
Incorrect because all three statements are not true.

🎯 Strategy

used

Elimination

Application:

Assess each statement independently using the Corner Point Theorem and the properties of bounded regions.

Final Logic:

Only Statements 2 and 3 are correct; therefore, Option C is correct.

🧠 Memory Trick

"Corner points decide the optimum."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Point Method and Bounded Feasible Region

11 Given points:
A(0, 3), B(2, 4), C(5, 0)
For:
Z = 10x + 5y
Find the median of the evaluated values.

A 15

B 25

C 50

D 40

💡 Short Explanation

Evaluate Z at each given point. Arrange the values in ascending order. The middle value (median) is 40.

✅ Correct Answer Explanation

The objective function is: Z = 10x + 5y Evaluate Z at each point. At A(0, 3): Z = 10(0) + 5(3) = 15 At B(2, 4): Z = 10(2) + 5(4) = 20 + 20 = 40 At C(5, 0): Z = 10(5) + 5(0) = 50 The evaluated values are: 15, 40, 50 Arranging them in ascending order: 15, 40, 50 The median is 40. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 15
Incorrect because it is the smallest value, not the median.
Option B) 25
Incorrect because it is not one of the evaluated values.
Option C) 50
Incorrect because it is the largest value.

🎯 Strategy

used

Substitution

Application:

Evaluate the objective function at each point and arrange the results to identify the middle value.

Final Logic:

The values are 15, 40, and 50; therefore, the median is 40.

🧠 Memory Trick

"Median = Middle after arranging."

📚 Class XII – Mathematics – Chapter: Statistics – Topic: Median

12 If the maximum value of Z occurs at two adjacent vertices, what is the probability that any point on the joining line segment is optimal?

A 0

B 1

C 0.5

D 0.2

💡 Short Explanation

This is the case of alternate optimal solutions. Every point on the joining line segment gives the same maximum value. Hence, the probability is 1.

✅ Correct Answer Explanation

If the maximum value of the objective function occurs at two adjacent corner points, then the objective function is parallel to the corresponding boundary of the feasible region. As a result, every point on the line segment joining these two vertices gives the same optimal value. Therefore, if a point is selected from this line segment, it will always be optimal. Probability = 1 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 0
Incorrect because every point on the segment is optimal.
Option C) 0.5
Incorrect because the probability is certain.
Option D) 0.2
Incorrect because no fraction less than 1 applies.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recall the property of alternate optimal solutions in linear programming.

Final Logic:

Every point on the joining line segment is optimal; therefore, the probability is 1.

🧠 Memory Trick

"Two optimal vertices ⇒ Whole edge optimal."

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

13 For:
Z = 30x + 20y
What is true about iso-profit lines?

A Slope changes with constraints

B All iso-profit lines are parallel

C They intersect only at origin

D Slope increases with profit

💡 Short Explanation

Iso-profit lines differ only in the value of Z. Their slope remains constant. Hence, all iso-profit lines are parallel.

✅ Correct Answer Explanation

For the objective function: Z = 30x + 20y Rearranging, y = -(30/20)x + Z/20 The slope is: -30/20 = -3/2 Since the coefficients of x and y remain unchanged, every iso-profit line has the same slope. Only the intercept changes as Z changes. Therefore, all iso-profit lines are parallel. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Slope changes with constraints
Incorrect because the slope depends only on the objective function.
Option C) They intersect only at origin
Incorrect because parallel lines never intersect.
Option D) Slope increases with profit
Incorrect because changing Z changes only the intercept, not the slope.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Rewrite the objective function in slope-intercept form and observe that only the intercept changes.

Final Logic:

The slope remains constant for all values of Z; therefore, Option B is correct.

🧠 Memory Trick

"Different Z, same slope."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-profit Lines

14 Match the arbitrary objective function with its mathematically derived slope \(m=-\frac{a}{b}\)obtained from the slope-intercept form:
\(y=-\frac{a}{b}x+\frac{Z}{b}\)

List I	List II
1. Z = 2x + 4y	a. 1
2. Z = 5x + 2y	b. -2/3
3. Z = x − y	c. -1/2
4. Z = 4x + 6y	d. -5/2

A 1–b, 2–c, 3–d, 4–a

B 1–a, 2–d, 3–c, 4–b

C 1–c, 2–d, 3–a, 4–b

D 1–d, 2–c, 3–b, 4–a

💡 Short Explanation

Rewrite each objective function in slope-intercept form. Calculate the slope using m = -a/b. Match the slopes correctly.

✅ Correct Answer Explanation

For each objective function: 1. Z = 2x + 4y Slope = -2/4 = -1/2 Therefore, 1 → c 2. Z = 5x + 2y Slope = -5/2 Therefore, 2 → d 3. Z = x - y Here, a = 1 b = -1 Slope = -1/(-1) = 1 Therefore, 3 → a 4. Z = 4x + 6y Slope = -4/6 = -2/3 Therefore, 4 → b Thus, the correct matching is: 1–c, 2–d, 3–a, 4–b Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the slopes for the first and third equations are mismatched.
Option B) Incorrect because the first equation has slope -1/2, not 1.
Option D) Incorrect because several equations are paired with incorrect slopes.

🎯 Strategy

used

Substitution

Application:

Use the formula m = -a/b for each objective function and compare the results with the given slopes.

Final Logic:

The calculated slopes match Option C exactly.

🧠 Memory Trick

"Slope of Z = -Coefficient of x / Coefficient of y."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-profit Lines

15 In a manufacturing minimization problem governed by the linear objective function:
Z = 400x + 450y
an arbitrary iso-cost line is initially drawn on the graph. This line is then shifted parallelly toward the origin in order to locate the minimum cost. During this shifting process, the iso-cost line first makes contact with the feasible bounded region at the corner point:
B(6, 4)
Determine the evaluated minimum cost corresponding to this optimal point.

A ₹2400

B ₹1800

C ₹4500

D ₹4200

💡 Short Explanation

Substitute the coordinates into the objective function. Evaluate the total cost. The minimum cost is ₹4200.

✅ Correct Answer Explanation

The objective function is: Z = 400x + 450y Substitute the optimal point: x = 6 y = 4 Therefore, Z = 400(6) + 450(4) = 2400 + 1800 = ₹4200 Since this is the first point touched by the iso-cost line during minimization, it gives the minimum cost. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) ₹2400
Incorrect because it includes only the cost contribution of x.
Option B) ₹1800
Incorrect because it includes only the cost contribution of y.
Option C) ₹4500
Incorrect because it is not obtained by substituting the given coordinates.

🎯 Strategy

used

Substitution

Application:

Replace x and y with the coordinates of the optimal point and evaluate the objective function.

Final Logic:

Substituting (6, 4) gives ₹4200; therefore, Option D is correct.

🧠 Memory Trick

"Optimal point → Substitute into Z."

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

16 When performing an extensive minimum search using the graphical Iso-cost method, identify the specific geometric point that is strictly considered optimal.

A The point on the feasible region that is touched by the iso-cost line closest to the origin

B The point on the feasible region where the iso-cost line is farthest from the origin

C The central intersection point of all constraint boundary lines

D The origin itself, irrespective of whether it satisfies the constraints

💡 Short Explanation

For minimization, the iso-cost line is shifted toward the origin. The first point of contact with the feasible region is optimal. This point gives the minimum cost.

✅ Correct Answer Explanation

In the graphical Iso-cost method, the objective function is represented by an iso-cost line. For a minimization problem: An arbitrary iso-cost line is drawn. It is shifted parallel to itself toward the origin. The first point where the line touches the feasible region gives the minimum value of the objective function. Thus, the optimal point is the point on the feasible region touched by the iso-cost line closest to the origin. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because moving farthest from the origin is associated with maximization, not minimization.
Option C) Incorrect because the intersection of all constraint lines is generally not a feasible or optimal point.
Option D) Incorrect because the origin may not satisfy all constraints.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recall the graphical procedure used in the Iso-cost method for minimization.

Final Logic:

The first point touched by the iso-cost line nearest the origin is optimal; therefore, Option A is correct.

🧠 Memory Trick

"Minimum → Nearest to origin."

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

17 Which specific graphical situation strictly and directly indicates that a Linear Programming Problem has no feasible region?

A The constraints form a bounded polygonal region

B There is no common region that satisfies all the given linear constraints and non-negativity conditions simultaneously

C The feasible region lies entirely in the second quadrant

D The iso-profit line is parallel to one of the constraints

💡 Short Explanation

A feasible region is the common region satisfying all constraints. If no common region exists, the LPP has no feasible solution. Hence, there is no feasible region.

✅ Correct Answer Explanation

A feasible region is the common area that satisfies: All linear constraints. Non-negativity constraints. If these constraints do not overlap, then no point satisfies all the conditions simultaneously. Consequently, the Linear Programming Problem has no feasible region and therefore no feasible solution. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because a bounded polygon represents a valid feasible region.
Option C) Incorrect because a feasible region can exist outside the first quadrant if non-negativity constraints are absent.
Option D) Incorrect because parallel iso-profit lines may indicate alternate optimal solutions, not infeasibility.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recall the definition of a feasible region in Linear Programming.

Final Logic:

No common region means no feasible solution; therefore, Option B is correct.

🧠 Memory Trick

"No common region = No feasible solution."

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

18 Consider the minimization problem:
Z = x + 2y
subject to the constraints:
2x + y ≥ 3
x + 2y ≥ 6
x ≥ 0
y ≥ 0
The resulting corner points of the feasible region are:
A(6, 0) and B(0, 3)
Evaluating the objective function at these points gives:
Z(A) = 6
Z(B) = 6
Since the feasible region is unbounded, we further analyze the inequality:
x + 2y < 6
Based on this analysis, determine the correct conclusion regarding the solution.

A The minimum value is 6, attained at infinitely many points lying on the line segment joining A(6, 0) and B(0, 3)

B There is no minimum value for the objective function

C The minimum value is 3

D The unique minimum value occurs at the origin

💡 Short Explanation

Both corner points give the same objective value. No feasible point satisfies x + 2y < 6. Therefore, every point on the joining line segment is optimal.

✅ Correct Answer Explanation

The objective function is: Z = x + 2y At the given corner points: A(6, 0): Z = 6 + 0 = 6 B(0, 3): Z = 0 + 6 = 6 Both adjacent vertices give the same minimum value. Now consider: x + 2y < 6 This half-plane has no common point with the feasible region. Therefore, no value smaller than 6 is attainable. Hence, every point on the line segment joining A and B also gives the same minimum value. Thus, the minimum value is 6, attained at infinitely many points. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because a minimum value clearly exists.
Option C) Incorrect because no feasible point gives Z = 3.
Option D) Incorrect because the origin does not satisfy the given constraints.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recognize the graphical condition for alternate optimal solutions in an unbounded feasible region.

Final Logic:

Since no feasible point has Z < 6, every point on the joining segment is optimal; therefore, Option A is correct.

🧠 Memory Trick

"Equal Z at adjacent vertices ⇒ Entire edge optimal."

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

19 Assertion (A): In an unbounded graphical region, a maximum value of the objective function Z = ax + by may not exist.
Reason (R): The open half-plane represented by the inequality ax + by > M may contain points that also belong to the feasible region, thereby allowing the value of Z to increase indefinitely.

A is false, but Reason (R) is true

B Assertion

C Both Assertion

D Assertion

💡 Short Explanation

An unbounded region may allow the objective function to increase indefinitely. Hence, a finite maximum may not exist. The Reason correctly explains the Assertion.

✅ Correct Answer Explanation

Assertion (A): An unbounded feasible region is open in one or more directions. If the objective function can continue increasing along that direction, then a finite maximum value does not exist. Therefore, the Assertion is true. Reason (R): The inequality: ax + by > M represents objective function values greater than M. If this region still overlaps with the feasible region, then values larger than M remain attainable. Thus, the objective function can increase indefinitely. The Reason correctly explains why a maximum may not exist. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because both statements are true.
Option B) Incorrect because the Reason is also true.
Option D) Incorrect because the Assertion is true.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

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

Final Logic:

The Reason correctly explains why an unbounded region may not have a maximum value; therefore, Option C is correct.

🧠 Memory Trick

"Unbounded + Increasing Z = No maximum."

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

20 Which comprehensive combination of geometric and mathematical conditions correctly determines the final optimal solution of a Linear Programming Problem?
1. The optimal point must lie within or on the boundary of the feasible region.
2. The optimal point must yield the optimal value (maximum or minimum) of the objective function.
3. The optimal point must satisfy the non-negativity conditions x ≥ 0, y ≥ 0.

A 1 and 2 only

B 2 and 3 only

C 1 and 3 only

D 1, 2, and 3

💡 Short Explanation

The optimal point must satisfy all constraints. It must lie in the feasible region. It must produce the optimum value of the objective function.

✅ Correct Answer Explanation

For a point to be the final optimal solution of a Linear Programming Problem, it must: Satisfy all the given constraints and lie within or on the boundary of the feasible region. Satisfy the non-negativity conditions: x ≥ 0 y ≥ 0 Produce the maximum or minimum value of the objective function. Since all three statements are essential, the correct choice is Option D.

❓ Why Other Options Are Incorrect

Option A) Incorrect because it ignores the non-negativity conditions.
Option B) Incorrect because it omits the requirement that the point must lie in the feasible region.
Option C) Incorrect because it does not mention the objective function.

🎯 Strategy

used

Elimination

Application:

Evaluate each statement independently and retain only those that are necessary conditions for an optimal solution.

Final Logic:

All three conditions must be satisfied; therefore, Option D is correct.

🧠 Memory Trick

"Feasible + Optimal + Non-negative = Final answer."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Graphical Method and Optimal Solution

CUET UG Applied Mathematics Booster Test 1 - Graphical Method, Feasible Region, and Optimal Solution

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