CUET UG Applied Mathematics Booster Test 1 - Feasible Regions, Optimal Solutions, and Graphical Methods -

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CUET UG Applied Mathematics Booster Test 1 - 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 1 - Feasible Regions, Optimal Solutions, and Graphical Methods

1 / 20

QUESTION 1 OF 20

Match the linear constraints with the x-intercept of their boundary lines:

List I	List II
1. 2x + 3y = 12 → x = 6	a. 2
2. 4x + y = 8 → x = 2	b. 5
3. 5x + 5y = 25 → x = 5	c. 6
4. 3x + 2y = 6 → x = 2	d. 3

Answer Review

1 Match the linear constraints with the x-intercept of their boundary lines:

List I	List II
1. 2x + 3y = 12 → x = 6	a. 2
2. 4x + y = 8 → x = 2	b. 5
3. 5x + 5y = 25 → x = 5	c. 6
4. 3x + 2y = 6 → x = 2	d. 3

A 1–c, 2–a, 3–b, 4–a

B 1–b, 2–a, 3–c, 4–d

C 1–c, 2–d, 3–b, 4–a

D 1–a, 2–b, 3–c, 4–d

💡 Short Explanation

The x-intercept is obtained by putting y = 0. Solve each equation after substituting y = 0. Match the calculated x-intercepts with List II.

✅ Correct Answer Explanation

The x-intercept of a straight line is the point where the line cuts the x-axis. At every x-intercept, the value of y is zero. For each equation: 1. 2x + 3y = 12 Put y = 0. 2x = 12 x = 6 Therefore, 1 → c. 2. 4x + y = 8 Put y = 0. 4x = 8 x = 2 Therefore, 2 → a. 3. 5x + 5y = 25 Put y = 0. 5x = 25 x = 5 Therefore, 3 → b. 4. 3x + 2y = 6 Put y = 0. 3x = 6 x = 2 Therefore, 4 → a. Thus, the correct matching is: 1 → c 2 → a 3 → b 4 → a Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 1–b, 2–a, 3–c, 4–d
Incorrect because the first equation has x-intercept 6, not 5. The third equation has x-intercept 5, not 6. The fourth equation has x-intercept 2, not 3.
Option C) 1–c, 2–d, 3–b, 4–a
Incorrect because the second equation has x-intercept 2, not 3.
Option D) 1–a, 2–b, 3–c, 4–d
Incorrect because the x-intercepts of the first, second, third, and fourth equations are all incorrectly matched.

🎯 Strategy

used

Substitution

Application:

To find the x-intercept of a linear equation, substitute y = 0 into each equation and solve for x. Then compare the obtained values with the options.

Final Logic:

Setting y = 0 gives x-intercepts 6, 2, 5, and 2 respectively, which matches Option A.

🧠 Memory Trick

"x-axis ⇒ y = 0; y-axis ⇒ x = 0."

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

2 Which coordinate pairs lie within the feasible region defined by:
x + y ≤ 8,
x + y ≥ 4,
x ≤ 5,
y ≤ 5,
x ≥ 0,
y ≥ 0

A (0, 0)

B (6, 2)

C (4, 4), (3, 5)

D (5, 5)

💡 Short Explanation

A feasible solution satisfies every constraint simultaneously. Substitute each coordinate pair into all inequalities. Only the points in Option C satisfy every condition.

✅ Correct Answer Explanation

A feasible solution is a point that satisfies all the given linear inequalities simultaneously. Check each option: Option A: (0, 0) x + y = 0 + 0 = 0 Since 0 < 4, it violates the condition x + y ≥ 4. Therefore, this point is not feasible. Option B: (6, 2) x + y = 6 + 2 = 8 ✓ But x = 6, which violates the condition x ≤ 5. Hence, this point is not feasible. Option C: (4, 4) and (3, 5) For (4, 4): x + y = 4 + 4 = 8 8 ≤ 8 ✓ 8 ≥ 4 ✓ x = 4 ≤ 5 ✓ y = 4 ≤ 5 ✓ x ≥ 0 and y ≥ 0 ✓ Therefore, (4, 4) is feasible. For (3, 5): x + y = 3 + 5 = 8 8 ≤ 8 ✓ 8 ≥ 4 ✓ x = 3 ≤ 5 ✓ y = 5 ≤ 5 ✓ x ≥ 0 and y ≥ 0 ✓ Therefore, (3, 5) is also feasible. Thus, both points satisfy all the constraints. Option D: (5, 5) x + y = 5 + 5 = 10 Since 10 > 8, it violates the condition x + y ≤ 8. Hence, this point is not feasible. Therefore, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) (0, 0)
Incorrect because it does not satisfy the inequality x + y ≥ 4.
Option B) (6, 2)
Incorrect because x = 6 violates the condition x ≤ 5.
Option D) (5, 5)
Incorrect because x + y = 10 exceeds the maximum permitted value of 8.

🎯 Strategy

used

Substitution

Application:

Substitute each coordinate pair into every inequality. Eliminate any point that violates even one constraint. A feasible point must satisfy all inequalities simultaneously.

Final Logic:

Only the coordinate pairs (4, 4) and (3, 5) satisfy every constraint, making Option C the correct answer.

🧠 Memory Trick

"Feasible means ALL conditions true."

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

3 If the vector:
\(\vec{v}=60\hat{i}+20\hat{j}\)
represents the optimal point (x, y) for the objective function:
Z = 15x + 10y
the maximum value of Z is:

A 1100

B 900

C 1000

D 1200

💡 Short Explanation

The vector represents the optimal point (60, 20). Substitute x = 60 and y = 20 into the objective function. The calculated value of Z is 1100.

✅ Correct Answer Explanation

The given vector v⃗ = 60i + 20j represents the point (60, 20). The objective function is: Z = 15x + 10y Substitute x = 60 and y = 20. Z = 15(60) + 10(20) = 900 + 200 = 1100 Since the point (60, 20) is stated to be the optimal point, the value obtained is the maximum value of the objective function. Therefore, the maximum value of Z is 1100. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 900
Incorrect because it considers only the contribution of x and ignores the contribution of y.
Option C) 1000
Incorrect because it is not obtained by correctly substituting x = 60 and y = 20 into the objective function.
Option D) 1200
Incorrect because the calculated value of Z is 1100, not 1200.

🎯 Strategy

used

Substitution

Application:

Identify the coordinates represented by the vector and substitute them directly into the objective function.

Final Logic:

Substituting x = 60 and y = 20 gives Z = 1100, so Option A is correct.

🧠 Memory Trick

"Optimal point → Substitute directly into Z."

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

4 For:
Z = x + 2y
the minimum value exists at all points on a line segment AB. The probability that a randomly chosen point on segment AB yields this minimum value is:

A 0

B 0.5

C 0.8

D 1

💡 Short Explanation

Every point on line segment AB gives the same minimum value. A randomly selected point from AB will always be optimal. Therefore, the probability is 1.

✅ Correct Answer Explanation

In linear programming, multiple optimal solutions occur when the objective function is parallel to one of the boundary lines of the feasible region. If the minimum value of Z exists at every point on the line segment AB, then each point on AB is an optimal solution. Since every point on the segment produces the same minimum value, selecting any point on AB will always give the minimum value. Therefore, Probability = (Number of favourable points) / (Total number of points) = All points on AB / All points on AB = 1 Hence, the probability is 1. Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 0
Incorrect because every point on AB gives the minimum value.
Option B) 0.5
Incorrect because the probability is not one-half; it is certain.
Option C) 0.8
Incorrect because there is no partial success. Every point on AB is optimal.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

The phrase "all points on line segment AB" indicates certainty. Therefore, the probability must be equal to 1.

Final Logic:

Since every point on AB is optimal, the probability of choosing an optimal point is 1, making Option D correct.

🧠 Memory Trick

"All points optimal ⇒ Probability = 1."

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

5 For the system:
4x + 6y ≤ 360,
3x ≤ 180,
5y ≤ 200
the intersection point of:
4x + 6y = 360 and 3x = 180
is:

A (60, 0)

B (0, 40)

C (60, 20)

D (30, 40)

💡 Short Explanation

Convert the inequalities into boundary equations. Solve the two equations simultaneously. The intersection point is (60, 20).

✅ Correct Answer Explanation

To find the intersection point of two lines, solve the equations simultaneously. The given boundary 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 Hence, the point of intersection is: (60, 20) Therefore, the correct answer is Option C. This point represents the common point where both boundary lines intersect and is an important concept in determining corner points of the feasible region in linear programming.

❓ Why Other Options Are Incorrect

Option A) (60, 0)
Incorrect because substituting (60, 0) into 4x + 6y = 360 gives:
4(60) + 6(0) = 240 ≠ 360
Option B) (0, 40)
Incorrect because it does not satisfy the equation 3x = 180.
Option D) (30, 40)
Incorrect because it does not satisfy the equation 3x = 180, where x must be 60.

🎯 Strategy

used

Substitution

Application:

First solve the simpler equation to obtain x. Then substitute its value into the other equation to calculate y.

Final Logic:

Using x = 60 in 4x + 6y = 360 gives y = 20, so the intersection point is (60, 20).

🧠 Memory Trick

"Solve one, substitute into the other."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Points and Graphical Solution

6 The area of the rectangular region bounded by:
x ≤ 5,
y ≤ 5,
x ≥ 0,
y ≥ 0
can be evaluated using:
∫₀⁵ 5 dx
The value of the area is:

A 10

B 25

C 20

D 15

💡 Short Explanation

The given region is a square with side length 5 units. The definite integral calculates the area under the constant function y = 5. The area of the region is 25 square units.

✅ Correct Answer Explanation

The given inequalities define the region: 0 ≤ x ≤ 5 0 ≤ y ≤ 5 This forms a square in the first quadrant with side length 5 units. The area is evaluated using the definite integral: Area = ∫₀⁵ 5 dx Since the integrand is constant, Area = 5 × (5 − 0) = 25 Alternatively, Area of square = Side × Side = 5 × 5 = 25 square units. Therefore, the value of the area is 25. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 10
Incorrect because it is not the area obtained from the given integral.
Option C) 20
Incorrect because multiplying the side length incorrectly gives this value.
Option D) 15
Incorrect because it does not represent the area of a square with side length 5.

🎯 Strategy

used

Substitution

Application:

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

Final Logic:

The integral ∫₀⁵ 5 dx equals 25, so Option B is correct.

🧠 Memory Trick

"Constant × Interval = Area."

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

7 Nutritional data indicates cost ₹0.60 per lb for Food 1 (x) and ₹1.00 per lb for Food 2 (y). The objective function to minimize cost is:

A Z = 10x + 4y

B Z = 5x + 5y

C Z = x + 0.60y

D Z = 0.60x + 1.00y

💡 Short Explanation

The objective function represents the total cost. Multiply each variable by its respective unit cost. Add the costs to obtain the objective function.

✅ Correct Answer Explanation

In a linear programming problem, the objective function is formed by multiplying each decision variable by its corresponding unit cost. Given: Cost of Food 1 = ₹0.60 per lb Cost of Food 2 = ₹1.00 per lb Let: x = pounds of Food 1 y = pounds of Food 2 Therefore, Total Cost (Z) = (0.60 × x) + (1.00 × y) or Z = 0.60x + 1.00y This objective function correctly represents the total expenditure on both foods and is minimized to obtain the least possible cost. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Z = 10x + 4y
Incorrect because the coefficients do not match the given costs.
Option B) Z = 5x + 5y
Incorrect because equal coefficients are not supported by the given data.
Option C) Z = x + 0.60y
Incorrect because the coefficients of x and y are interchanged and do not represent the stated costs.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Identify the cost associated with each variable and assign it directly as the coefficient in the objective function.

Final Logic:

The coefficients must equal the unit costs, giving Z = 0.60x + 1.00y, which is Option D.

🧠 Memory Trick

"Coefficient = Unit Cost."

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

8 Assertion (A): Unbounded regions always possess a distinct maximum value.
Reason (R): Unbounded feasible regions are empty and contain no points.

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

An unbounded feasible region does not necessarily have a maximum or minimum value. An unbounded region is not empty; it contains infinitely many feasible points. Therefore, both the Assertion and the Reason are false.

✅ Correct Answer Explanation

Assertion (A): Unbounded regions always possess a distinct maximum value. This statement is false. An unbounded feasible region extends indefinitely in one or more directions. Whether a maximum or minimum value exists depends on the direction of the objective function, not merely on the nature of the feasible region. Therefore, an unbounded region may have an optimal solution, may have multiple optimal solutions, or may have no finite optimum. Reason (R): Unbounded feasible regions are empty and contain no points. This statement is also false. An unbounded feasible region is not empty. It consists of infinitely many feasible points and extends indefinitely in one or more directions. An empty feasible region is called an infeasible region, which is entirely different from an unbounded region. Hence, both the Assertion and the Reason are false. Therefore, 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. An unbounded feasible region does not always have a distinct maximum value.
Option C) Both A and R are true, and R is the correct explanation of A
Incorrect because both statements are false.
Option D) A is false, R is true
Incorrect because the Reason is also false. An unbounded feasible region contains infinitely many feasible points.

🎯 Strategy

used

Elimination

Application:

Evaluate the truth of the Assertion and the Reason independently. Since both statements are false, eliminate all options containing a true statement.

Final Logic:

Both statements are false; therefore, Option A is correct.

🧠 Memory Trick

"Unbounded ≠ Empty."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Types of Feasible Regions

9 Arrange the values of:
Z = 5x + 3y
at the points:
1. \(\left(0\ ,\ 3\right)\)
2. \(\left(2\ ,\ 0\right)\)
3. \(\left(\frac{20}{19}\ ,\ \frac{45}{19}\right)\)
in descending order.

A 1, 2, 3

B 2, 1, 3

C 3, 1, 2

D 3, 2, 1

💡 Short Explanation

Evaluate Z at each given point. Compare the three values obtained. Arrange them from highest to lowest.

✅ Correct Answer Explanation

The objective function is: Z = 5x + 3y Evaluate Z at each point. Point 1: (0, 3) Z = 5(0) + 3(3) = 0 + 9 = 9 Point 2: (2, 0) Z = 5(2) + 3(0) = 10 + 0 = 10 Point 3: (20/19, 45/19) Z = 5(20/19) + 3(45/19) = 100/19 + 135/19 = 235/19 ≈ 12.37 Thus, Point 3 → 12.37 Point 2 → 10 Point 1 → 9 Hence, the descending order is: 3, 2, 1 Therefore, Option D is mathematically correct. Note: The provided answer key is incorrect.

❓ Why Other Options Are Incorrect

Option A) 1, 2, 3
Incorrect because it arranges the values in ascending order rather than descending order.
Option B) 2, 1, 3
Incorrect because Point 3 has the highest value of Z.
Option C) 3, 1, 2
Incorrect because Point 2 gives a greater value than Point 1.

🎯 Strategy

used

Substitution

Application:

Substitute each coordinate into the objective function and compare the numerical values obtained.

Final Logic:

Since 12.37 > 10 > 9, the descending order is 3, 2, 1, making Option D the correct answer.

🧠 Memory Trick

"Evaluate first, arrange later."

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

10 Minimize:
Z = 3x + 5y
subject to:
x + 3y ≥ 3
x + y ≥ 2
At the point (1.5, 0.5), the value of Z is:

A 7

B 9

C 10

D 8

💡 Short Explanation

Substitute the given coordinates into the objective function. Compute the value of Z directly. The calculated value is 7.

✅ Correct Answer Explanation

The objective function is: Z = 3x + 5y The given point is: (x, y) = (1.5, 0.5) Substitute these values into the objective function: Z = 3(1.5) + 5(0.5) = 4.5 + 2.5 = 7 Thus, the value of the objective function at the point (1.5, 0.5) is 7. The given point also satisfies the constraints: For x + 3y ≥ 3: 1.5 + 3(0.5) = 1.5 + 1.5 = 3 ✓ For x + y ≥ 2: 1.5 + 0.5 = 2 ✓ Therefore, the point is feasible, and the value of the objective function at this point is 7. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 9
Incorrect because substituting x = 1.5 and y = 0.5 gives Z = 7, not 9.
Option C) 10
Incorrect because it is not obtained from the given objective function and coordinates.
Option D) 8
Incorrect because the correct calculation yields Z = 7.

🎯 Strategy

used

Substitution

Application:

Substitute the given coordinates directly into the objective function and perform the arithmetic carefully.

Final Logic:

Substituting x = 1.5 and y = 0.5 gives Z = 7; therefore, Option A is correct.

🧠 Memory Trick

"Given point → Substitute into Z."

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

11 The number of corner points of the feasible region defined by:
x + y ≤ 8,
x + y ≥ 4,
x ≤ 5,
y ≤ 5,
x ≥ 0,
y ≥ 0
is:

A 4

B 6

C 5

D 3

💡 Short Explanation

The feasible region is formed by the intersection of six linear constraints. The corner points are obtained by the intersections of the boundary lines. The feasible region has six vertices.

✅ Correct Answer Explanation

The given constraints are: x + y ≤ 8 x + y ≥ 4 x ≤ 5 y ≤ 5 x ≥ 0 y ≥ 0 These inequalities form a closed feasible region in the first quadrant. The corner points are obtained from the intersections of the boundary lines: (0, 4) (0, 5) (3, 5) (5, 3) (5, 0) (4, 0) Thus, the feasible region contains 6 corner points. In linear programming, these corner points are important because the optimum value of the objective function, if it exists, always occurs at one or more corner points of the feasible region. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 4
Incorrect because the feasible region has more than four vertices after considering all the constraints.
Option C) 5
Incorrect because one corner point has been omitted.
Option D) 3
Incorrect because the feasible region is a hexagon, not a triangle.

🎯 Strategy

used

Substitution

Application:

Find the intersection points of the boundary lines and identify those satisfying all the constraints. Count the feasible vertices.

Final Logic:

The feasible region has six valid corner points; therefore, Option B is correct.

🧠 Memory Trick

"Corner points = Feasible vertices."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Corner Points of a Feasible Region

12 The point of intersection of:
2x + y = 3,
x + 2y = 6
is:

A (1, 1)

B (2, 2)

C (0, 3)

D (3, 0)

💡 Short Explanation

Solve the two linear equations simultaneously. Find the common values of x and y. The intersection point is (0, 3).

✅ Correct Answer Explanation

The equations are: 2x + y = 3 x + 2y = 6 From the first equation, y = 3 − 2x Substitute this into the second equation: x + 2(3 − 2x) = 6 x + 6 − 4x = 6 −3x = 0 x = 0 Substitute x = 0 into the first equation: 2(0) + y = 3 y = 3 Therefore, the point of intersection is: (0, 3) Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) (1, 1)
Incorrect because it satisfies the first equation but not the second.
Option B) (2, 2)
Incorrect because it does not satisfy either equation.
Option D) (3, 0)
Incorrect because it does not satisfy the second equation.

🎯 Strategy

used

Substitution

Application:

Express one variable from one equation and substitute it into the other equation to obtain the common solution.

Final Logic:

The simultaneous solution gives x = 0 and y = 3; therefore, Option C is correct.

🧠 Memory Trick

"Common solution = Intersection point."

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

📄 Passage

A company uses an objective function to measure its overall performance based on two decision variables, xxx and yyy. After identifying a feasible solution during a linear programming analysis, the management needs to determine the corresponding value of the objective function at that point. Evaluating the objective function at a given coordinate helps compare different feasible solutions and identify the most suitable one for decision-making.

Based on the above passage, answer the following question:
Evaluating the objective function:
Z = x − 7y + 190
at the point (0, 4), the value of Z is:

A 162

B 190

C 155

D 194

💡 Short Explanation

Substitute x = 0 and y = 4 into the objective function. Perform the arithmetic carefully. The value obtained is 162.

✅ Correct Answer Explanation

The objective function is: Z = x − 7y + 190 Substitute x = 0 and y = 4. Z = 0 − 7(4) + 190 = 0 − 28 + 190 = 162 Thus, the value of the objective function at (0, 4) is 162. Therefore, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 190
Incorrect because it ignores the term −7y.
Option C) 155
Incorrect because it is not obtained by correct substitution.
Option D) 194
Incorrect because it results from incorrect arithmetic.

🎯 Strategy

used

Substitution

Application:

Replace x and y in the objective function with the given coordinates and simplify step by step.

Final Logic:

Substituting (0, 4) gives Z = 162; therefore, Option A is correct.

🧠 Memory Trick

"Point given → Substitute into Z."

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

📄 Passage

A manufacturing firm evaluates different feasible solutions in a linear programming problem using an objective function. Each feasible point represents a possible decision, and the value of the objective function at that point indicates the effectiveness of that decision. By substituting the coordinates of a feasible point into the objective function, the firm can compare outcomes and identify the most appropriate solution.

Based on the above passage, answer the following question:
Evaluating the function:
Z = x − 7y + 190
at the point (0, 5), the value of Z is:

A 190

B 162

C 155

D 195

💡 Short Explanation

Substitute x = 0 and y = 5 into the objective function. Simplify the expression. The resulting value is 155.

✅ Correct Answer Explanation

The objective function is: Z = x − 7y + 190 Substitute x = 0 and y = 5. Z = 0 − 7(5) + 190 = 0 − 35 + 190 = 155 Thus, the value of the objective function at the point (0, 5) is 155. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 190
Incorrect because it ignores the contribution of −7y.
Option B) 162
Incorrect because it is the value obtained when y = 4, not y = 5.
Option D) 195
Incorrect because it is not obtained from the given objective function.

🎯 Strategy

used

Substitution

Application:

Insert the given coordinates into the objective function and evaluate the numerical value.

Final Logic:

Substituting (0, 5) gives Z = 155; therefore, Option C is correct.

🧠 Memory Trick

"Substitute carefully, then simplify."

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

15 Maximize:
Z = –x + 2y
subject to:
–0.5x + y ≤ 2
x – y ≤ –1
x ≥ 0
y ≥ 0
The maximum value of Z is:

A 2

B 4

C 6

D 8

💡 Short Explanation

Determine the feasible region from the given constraints. Evaluate the objective function at the corner points. The maximum value obtained is 4.

✅ Correct Answer Explanation

The constraints are: –0.5x + y ≤ 2 x – y ≤ –1 x ≥ 0 y ≥ 0 Rewrite the constraints: y ≤ 2 + 0.5x y ≥ x + 1 The corner points of the feasible region are: (0, 1) (0, 2) (2, 3) Evaluate the objective function Z = –x + 2y. At (0, 1): Z = –0 + 2(1) = 2 At (0, 2): Z = –0 + 2(2) = 4 At (2, 3): Z = –2 + 2(3) = –2 + 6 = 4 Thus, the maximum value of Z is 4. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 2
Incorrect because although Z = 2 at one corner point, it is not the maximum value.
Option C) 6
Incorrect because no corner point produces a value of 6.
Option D) 8
Incorrect because the objective function never attains this value within the feasible region.

🎯 Strategy

used

Substitution

Application:

Determine the corner points of the feasible region and substitute each point into the objective function to identify the maximum value.

Final Logic:

The largest value of Z at the corner points is 4; therefore, Option B is correct.

🧠 Memory Trick

"Maximum occurs at a corner point."

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

16 Minimize:
Z = 4x – 2y
subject to:
x + y ≤ 14
2x + y ≤ 24
3x + 2y ≥ 14
x ≥ 0
y ≥ 0
The minimum value of Z is:

A 14

B 24

C 0

D 20

💡 Short Explanation

Determine the corner points of the feasible region. Evaluate the objective function at each corner point. The minimum value obtained is 0.

✅ Correct Answer Explanation

The objective function is: Z = 4x – 2y The feasible region is obtained from the given constraints. The relevant corner points are: (0, 7) (0, 14) (10, 4) (12, 0) Evaluate the objective function at these points. At (0, 7): Z = 4(0) – 2(7) = –14 At (0, 14): Z = 4(0) – 2(14) = –28 This point does not satisfy 3x + 2y ≥ 14? It does satisfy the constraint. At (10, 4): Z = 40 – 8 = 32 At (12, 0): Z = 48 Among all feasible corner points, the smallest value is –28. Since –28 is not given among the options, the provided options are inconsistent with the mathematical solution. Therefore, none of the listed options is correct. Note: The answer key provided (Option D) is incorrect.

❓ Why Other Options Are Incorrect

Option A) 14
Incorrect because the objective function never attains this minimum value.
Option B) 24
Incorrect because it is not obtained at any feasible corner point.
Option C) 0
Incorrect because values smaller than 0 are attainable within the feasible region.
Option D) 20
Incorrect because the minimum value is not 20.

🎯 Strategy

used

Substitution

Application:

Find all feasible corner points and substitute each into the objective function to determine the smallest value.

Final Logic:

The mathematical minimum is –28; therefore, the question contains an incorrect answer key and inconsistent options.

🧠 Memory Trick

"Min/Max → Check every corner point."

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

17 For:
Z = 5x + 4y
subject to:
x – 2y ≤ 1
x + 2y ≤ 6
x – y ≥ 3
identify the correct conclusion:

A Maximum value is 10

B No solution (infeasible or unbounded region)

C Maximum value is 20

D Maximum value is 5

💡 Short Explanation

Analyze whether the constraints have a common feasible region. If no common region exists, the problem has no feasible solution. Therefore, no optimum value exists.

✅ Correct Answer Explanation

The constraints are: x – 2y ≤ 1 x + 2y ≤ 6 x – y ≥ 3 Graphically, these inequalities do not produce a common feasible region satisfying all three constraints simultaneously. Since there is no feasible region, no corner points exist for evaluating the objective function. Without a feasible region, the linear programming problem has no optimal solution. Therefore, the correct conclusion is that the problem has no feasible solution. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Maximum value is 10
Incorrect because no feasible solution exists.
Option C) Maximum value is 20
Incorrect because the objective function cannot be evaluated without a feasible region.
Option D) Maximum value is 5
Incorrect because the feasible region itself does not exist.

🎯 Strategy

used

Elimination

Application:

First determine whether a feasible region exists. If not, eliminate all options claiming a maximum value.

Final Logic:

Since no common feasible region exists, Option B is correct.

🧠 Memory Trick

"No feasible region = No optimum."

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

18 To verify whether a minimum value m exists for an unbounded feasible region, consider:
ax + by < m
If this half-plane has no common point with the feasible region, then:

A Z has no solution

B Z is automatically maximum

C The problem is bounded

D m is the minimum value

💡 Short Explanation

The inequality ax + by < m represents a half-plane. If this half-plane has no common point with the feasible region, no feasible solution has an objective value less than m. Therefore, m is the minimum value.

✅ Correct Answer Explanation

In linear programming, the objective function can be written as: Z = ax + by To verify whether m is the minimum value, consider the inequality: ax + by < m This inequality represents all points having an objective function value less than m. If this half-plane has no common point with the feasible region, then no feasible solution can produce a value smaller than m. Hence, the smallest attainable value of the objective function is exactly m. This graphical test is commonly used to verify the existence of a minimum value even when the feasible region is unbounded. Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Z has no solution
Incorrect because the condition confirms the existence of a minimum value rather than indicating the absence of a solution.
Option B) Z is automatically maximum
Incorrect because the question concerns verification of a minimum value, not a maximum value.
Option C) The problem is bounded
Incorrect because the feasible region may still be unbounded even though a minimum value exists.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Interpret the graphical meaning of the inequality and determine what the absence of common points implies for the objective function.

Final Logic:

If no feasible point satisfies ax + by < m, then m is the least attainable value; therefore, Option D is correct.

🧠 Memory Trick

"No value below m ⇒ m is minimum."

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

19 In the iso-profit method, an iso-profit line represents:

A Points with the same value of the objective function

B Boundary of constraints

C Non-negativity constraints

D The origin (0, 0)

💡 Short Explanation

An iso-profit line joins points having an equal objective function value. Every point on the same line gives the same profit. Parallel shifting of the line helps identify the optimum solution.

✅ Correct Answer Explanation

An iso-profit line is a line representing a constant value of the objective function. For example, if Z = 5x + 3y then 5x + 3y = k where k is a constant. Every point lying on this line produces the same value of the objective function. In the graphical method of linear programming, the iso-profit line is shifted parallel to itself until it reaches the last point or edge of the feasible region. That point or edge gives the optimal solution. Therefore, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Boundary of constraints
Incorrect because constraint boundaries define the feasible region, not the objective function.
Option C) Non-negativity constraints
Incorrect because non-negativity constraints are x ≥ 0 and y ≥ 0 and are unrelated to the iso-profit line.
Option D) The origin (0, 0)
Incorrect because an iso-profit line is a straight line and is not necessarily associated with the origin.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recognize the meaning of the term "iso-profit." The prefix "iso" means "equal," indicating equal values of profit.

Final Logic:

An iso-profit line represents equal values of the objective function; therefore, Option A is correct.

🧠 Memory Trick

"Iso = Equal Profit."

📚 Class XII – Mathematics – Chapter: Linear Programming – Topic: Iso-profit (Objective Function) Line

20 If an iso-profit line is parallel to a constraint boundary of the feasible region, the number of optimal solutions is:

A 0

B 1

C Infinite

D 2

💡 Short Explanation

Parallel objective and constraint lines produce multiple optimal solutions. Every point on the common boundary gives the same optimum value. Hence, infinitely many optimal solutions exist.

✅ Correct Answer Explanation

In the graphical method of linear programming, if the objective function (iso-profit line) is parallel to one of the boundary lines of the feasible region and coincides with a boundary segment at the optimum position, then every point on that boundary segment gives the same value of the objective function. Since a line segment contains infinitely many points, there are infinitely many optimal solutions. This situation is known as multiple optimal solutions. Therefore, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 0
Incorrect because optimal solutions do exist.
Option B) 1
Incorrect because more than one point on the boundary is optimal.
Option D) 2
Incorrect because the entire boundary segment is optimal, not just two points.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recognize that parallel objective and boundary lines indicate multiple optimal solutions along a line segment.

Final Logic:

A complete boundary segment becomes optimal, giving infinitely many solutions; therefore, Option C is correct.

🧠 Memory Trick

"Parallel lines ⇒ Infinite optima."

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

CUET UG Applied Mathematics Booster Test 1 - Feasible Regions, Optimal Solutions, and Graphical Methods

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