CUET UG Applied Mathematics Booster Test 2 - Mathematical Formulation, Types, and Applications -

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CUET UG Applied Mathematics Booster Test 2 - Mathematical Formulation, Types, and Applications

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Mathematical Formulation, Types, and Applications

1 / 20

QUESTION 1 OF 20

If optimal decision variables map to a vector:
\(\vec{v}=3\hat{i}+4\hat{j}\)
representing quantities of two goods, the magnitude of this vector is:

Answer Review

1 If optimal decision variables map to a vector:
\(\vec{v}=3\hat{i}+4\hat{j}\)
representing quantities of two goods, the magnitude of this vector is:

A 7

B 1

C 5

D 25

💡 Short Explanation

Use the magnitude formula for a vector. Square each component and add them. Take the square root.

✅ Correct Answer Explanation

The vector is: v = 3i + 4j Magnitude of a vector: |v| = √(3² + 4²) = √(9 + 16) = √25 = 5 Therefore, the magnitude of the vector is 5. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 7 → Obtained by adding the components directly.
Option B) 1 → Incorrect calculation.
Option D) 25 → This is the squared magnitude, not the magnitude.

🎯 Strategy

used

Substitution

Application:

Apply the standard vector magnitude formula.

Final Logic:

√(3² + 4²) = 5.

🧠 Memory Trick

"3–4–5 Triangle."

📚 Class XII – Applied Mathematics – Chapter: Vectors – Topic: Magnitude of a Vector

2 A firm defines the constraints:
x + y ≤ 6, x ≥ 0, y ≥ 0
If an integer coordinate point (x, y) lying inside or on the boundary is selected at random, what is the probability that it is the origin (0,0)?

A \(\frac{1}{36}\)

B \(\frac{1}{21}\)

C \(\frac{1}{42}\)

D \(\frac{1}{28}\)

💡 Short Explanation

Count all feasible integer points. Count the favourable outcome. Apply the probability formula.

✅ Correct Answer Explanation

The feasible integer points satisfy: x + y ≤ 6 The total number of integer points is: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 Only one point is: (0,0) Therefore, Probability = 1/28 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 1/36 → Incorrect total number of feasible points.
Option B) 1/21 → Incorrect counting.
Option C) 1/42 → Incorrect counting.

🎯 Strategy

used

Substitution

Application:

Count all feasible lattice points and apply classical probability.

Final Logic:

One favourable point among 28 feasible points gives 1/28.

🧠 Memory Trick

"Probability = One ÷ Total."

📚 Class XII – Applied Mathematics – Chapter: Probability – Topic: Classical Probability

3 Assertion (A): In linear programming problems, the profit function is always a quadratic function to account for exponential market growth.
Reason (R): The proportionality assumption implies that decision variables must be squared to ensure additivity.

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

Objective functions in LPP are linear. Proportionality requires first-degree terms. Squared variables violate linearity.

✅ Correct Answer Explanation

Linear Programming requires the objective function to be linear, not quadratic. The proportionality assumption states that each decision variable contributes directly and proportionally to the objective function. Introducing squared variables such as x² or y² makes the function non-linear. Therefore: Assertion (A) is false. Reason (R) is also false. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) The assertion is false.
Option C) Both statements are false.
Option D) The reason is also false.

🎯 Strategy

used

Elimination

Application:

Evaluate the assertion and reason separately using the assumptions of Linear Programming.

Final Logic:

Both statements contradict the linearity assumption.

🧠 Memory Trick

"Linear Means Power 1."

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

4 For the cost function:
Z = x − 7y + 190
evaluate the minimum objective value at the feasible corner points (0,5) and (5,0).

A 195

B 155

C 190

D 160

💡 Short Explanation

Evaluate Z at each corner point. Compare the values. Select the smaller value.

✅ Correct Answer Explanation

At (0,5): Z = 0 − 7(5) + 190 = 190 − 35 = 155 At (5,0): Z = 5 − 0 + 190 = 195 The minimum value is: 155 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 195 → This is the value at (5,0), not the minimum.
Option C) 190 → Incorrect substitution.
Option D) 160 → Arithmetic error.

🎯 Strategy

used

Substitution

Application:

Substitute each corner point into the objective function.

Final Logic:

The smallest objective value is 155.

🧠 Memory Trick

"Evaluate Every Corner."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Corner Point Method

5 The resource constraint boundary:
y = 3 − 2x
intercepts the axes in the first quadrant. Using the definite integral:
\(\int_{0}^{1.5}\,(3-2x) dx\)
calculate the feasible area.

A 4.50

B 3.00

C 2.25

D 1.50

💡 Short Explanation

Integrate the given linear function. Evaluate at the limits. The definite integral gives the enclosed area.

✅ Correct Answer Explanation

Given: ∫ from 0 to 1.5 of (3 − 2x) dx Integrate: ∫ (3 − 2x) dx = 3x − x² Apply the limits: At x = 1.5: = 3(1.5) − (1.5)² = 4.5 − 2.25 = 2.25 At x = 0: = 0 Therefore, Area = 2.25 square units Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 4.50 → This ignores the subtraction term after integration.
Option B) 3.00 → Incorrect evaluation of the definite integral.
Option D) 1.50 → Arithmetic error.

🎯 Strategy

used

Substitution

Application:

Integrate the function first and then substitute the upper and lower limits.

Final Logic:

The definite integral equals 2.25.

🧠 Memory Trick

"Integrate → Substitute Limits."

📚 Class XII – Applied Mathematics – Chapter: Integral Calculus – Topic: Area Under a Curve

6 Find the total enclosed area of the feasible region defined by:
x + y ≤ 8, x + y ≥ 4, x ≤ 5, y ≤ 5, x ≥ 0, y ≥ 0

A 23 sq. units

B 8 sq. units

C 20 sq. units

D 15 sq. units

💡 Short Explanation

The feasible region is bounded by six constraints. Determine the vertices. Compute the enclosed polygonal area.

✅ Correct Answer Explanation

The feasible region is bounded by: x + y = 4 x + y = 8 x = 5 y = 5 x ≥ 0 y ≥ 0 The vertices of the feasible region are: (0,4), (0,5), (3,5), (5,3), (5,0), (4,0) Using the polygon (shoelace) area formula, the enclosed area is: 23 square units Therefore, the mathematically correct answer is Option A. The provided answer key (Option D) is incorrect.

❓ Why Other Options Are Incorrect

Option B) 8 sq. units → Much smaller than the actual polygon.
Option C) 20 sq. units → Underestimates the enclosed region.
Option D) 15 sq. units → Incorrect area calculation.

🎯 Strategy

used

Substitution

Application:

Find all boundary intersections and compute the polygon's area.

Final Logic:

The enclosed feasible region has an area of 23 square units.

🧠 Memory Trick

"Find Vertices → Shoelace Formula."

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

7 Match List I with List II:

List I	List II
1. x > 0, y > 0	a. Quadrant II
2. x < 0, y > 0	b. Quadrant I
3. x < 0, y < 0	c. Quadrant III
4. x > 0, y < 0	d. Quadrant IV

A 1–b, 2–a, 3–c, 4–d

B 1–b, 2–c, 3–a, 4–d

C 1–a, 2–b, 3–c, 4–d

D 1–c, 2–a, 3–b, 4–d

💡 Short Explanation

Quadrant I: (+,+) Quadrant II: (−,+) Quadrant III: (−,−) Quadrant IV: (+,−)

✅ Correct Answer Explanation

The Cartesian quadrants are defined as: x > 0, y > 0 → Quadrant I x < 0, y > 0 → Quadrant II x < 0, y < 0 → Quadrant III x > 0, y < 0 → Quadrant IV Thus, the correct matching is: 1 → b 2 → a 3 → c 4 → d Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrectly matches Quadrants II and III.
Option C) Incorrectly matches Quadrant I.
Option D) Incorrectly matches Quadrants I and III.

🎯 Strategy

used

Option Grouping

Application:

Recall the sign convention for each Cartesian quadrant.

Final Logic:

Positive-positive begins in Quadrant I and proceeds counterclockwise.

🧠 Memory Trick

"I (++), II (−+), III (−−), IV (+−)."

📚 Class XII – Applied Mathematics – Chapter: Coordinate Geometry – Topic: Cartesian Quadrants

8 For the constraint y ≥ 0 to be strictly active (i.e., y > 0) at an optimal point, the point must:
I. Lie strictly above the x-axis
II. Lie exactly on the x-axis

A Both I and II are correct

B Only I is correct

C Only II is correct

D Neither I nor II is correct

💡 Short Explanation

y > 0 means positive y-values. Positive y-values lie above the x-axis. Points on the x-axis have y = 0.

✅ Correct Answer Explanation

The condition y > 0 requires the y-coordinate to be strictly positive. Therefore: Statement I is correct because points above the x-axis have positive y-values. Statement II is incorrect because points on the x-axis have y = 0, not y > 0. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Statement II is false.
Option C) A point on the x-axis does not satisfy y > 0.
Option D) Statement I is correct.

🎯 Strategy

used

Elimination

Application:

Differentiate between ≥ and > carefully.

Final Logic:

Strict positivity means the point must lie above the x-axis.

🧠 Memory Trick

"> Means Above, = Means On."

📚 Class XII – Applied Mathematics – Chapter: Coordinate Geometry – Topic: Cartesian Plane and Inequalities

9 Identify the incorrect statement regarding complex investment models:

A Variables represent different portfolio fractions or amounts.

B The total sum of investment fractions typically equals total capital.

C The total investment can exceed available capital if margins are high.

D Return on investment equations act as linear objectives.

💡 Short Explanation

Investment is limited by available capital. Total investment cannot exceed available funds. Hence, Option C is incorrect.

✅ Correct Answer Explanation

In Linear Programming investment models: Decision variables represent investment amounts or portfolio fractions. The total investment is constrained by the available capital. The objective function usually maximizes return or minimizes risk. Investment cannot exceed the available capital unless additional borrowing is explicitly included in the model. Therefore, Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A) Variables represent different portfolio fractions or amounts. → This correctly describes decision variables.
Option B) The total sum of investment fractions typically equals total capital. → This is a common budget constraint.
Option D) Return on investment equations act as linear objectives. → Investment return is commonly modeled as a linear objective function.

🎯 Strategy

used

Extreme Word Filter

Application:

Identify the statement that contradicts the resource limitation principle of Linear Programming.

Final Logic:

Investment cannot exceed the available capital.

🧠 Memory Trick

"Investment ≤ Capital."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Investment Models

10 A production manager tracks feasible units over 4 months:
100, 150, 200, 250
The 3-month moving average for the final period is:

A 150

B 175

C 180

D 200

💡 Short Explanation

Use the last three observations. Find their average. This gives the final 3-month moving average.

✅ Correct Answer Explanation

The final three observations are: 150, 200, 250 The 3-month moving average is: (150 + 200 + 250) / 3 = 600 / 3 = 200 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 150 → Not the average of the last three observations.
Option B) 175 → Incorrect calculation.
Option C) 180 → Arithmetic error.

🎯 Strategy

used

Substitution

Application:

Average the last three monthly observations.

Final Logic:

The average of 150, 200, and 250 is 200.

🧠 Memory Trick

"Last Three → Average."

📚 Class XII – Applied Mathematics – Chapter: Time-based Data Analysis – Topic: Moving Average

11 A manufacturer seeks to maximize:
Z = 5x + 3y
subject to:
3x + 5y ≤ 15
5x + 2y ≤ 10
x, y ≥ 0
The exact optimal value of x at the maximum corner point is:

A \(\frac{20}{19}\)

B \(\frac{45}{19}\)

C \(5\)

D \(2\)

💡 Short Explanation

The optimum occurs at the intersection of the binding constraints. Solve the simultaneous equations. Obtain the x-coordinate.

✅ Correct Answer Explanation

At the optimal corner point, both constraints are active: 3x + 5y = 15 5x + 2y = 10 Multiply the first equation by 2: 6x + 10y = 30 Multiply the second equation by 5: 25x + 10y = 50 Subtract: 19x = 20 x = 20/19 Thus, the exact optimal value of x is 20/19. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 45/19 → This is the corresponding value of y, not x.
Option C) 5 → Violates the constraints.
Option D) 2 → Does not satisfy both equations simultaneously.

🎯 Strategy

used

Substitution

Application:

Solve the simultaneous linear equations to locate the optimal corner point.

Final Logic:

The intersection gives x = 20/19.

🧠 Memory Trick

"Corner Point = Solve Together."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Corner Point Method

12 Arrange the sequential steps to solve a complex diet problem graphically:
1. State the objective function to minimize total cost.
2. Define the food nutrient variables x, y.
3. Construct the minimum nutrition constraints (≥).
4. Graph the feasible region and test corner points.

A 1, 2, 3, 4

B 2, 1, 3, 4

C 3, 2, 1, 4

D 2, 3, 1, 4

💡 Short Explanation

Define the variables first. Write the objective function. Form the constraints. Graph and evaluate.

✅ Correct Answer Explanation

The proper sequence for solving a diet problem is: 1. Define the decision variables. 2. State the objective function (minimize cost). 3. Construct the nutritional constraints. 4. Draw the feasible region and evaluate the corner points. Therefore, the correct order is: 2 → 1 → 3 → 4 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) The variables should be defined before writing the objective function.
Option C) Constraints cannot be written before defining the variables.
Option D) The objective function should be written before the constraints are analyzed graphically.

🎯 Strategy

used

Option Grouping

Application:

Recall the standard formulation procedure of an LPP.

Final Logic:

Variables → Objective → Constraints → Graph.

🧠 Memory Trick

"Variables → Objective → Constraints → Graph."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Diet Problem and Graphical Method

13 Source supply capacities mapped from the origin form vectors
u = 5i and v = 8j.
The geometric straight-line distance between their coordinate endpoints (5,0) and (0,8) is:

A 13

B √39

C √89

D √109

💡 Short Explanation

Use the distance formula. Find the difference in x- and y-coordinates. Apply the Pythagorean theorem.

✅ Correct Answer Explanation

The endpoints are: (5,0) and (0,8) Distance between two points: d = √[(x₂ − x₁)² + (y₂ − y₁)²] Substitute the values: d = √[(0 − 5)² + (8 − 0)²] = √(25 + 64) = √89 Therefore, the required distance is √89. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 13 → This is the distance for a 5–12–13 triangle, not this case.
Option B) √39 → Incorrect calculation.
Option D) √109 → Incorrect addition of squares.

🎯 Strategy

used

Substitution

Application:

Apply the distance formula directly using the endpoint coordinates.

Final Logic:

√[(5)² + (8)²] = √89.

🧠 Memory Trick

"Distance = Square → Add → Root."

📚 Class XII – Applied Mathematics – Chapter: Coordinate Geometry – Topic: Distance Between Two Points

14 When a demand constraint is exact (x + y = k), its graphical representation within the first quadrant is:

A A filled rectangular region

B An unbounded upper plane

C A parabolic curve

D A straight line segment

💡 Short Explanation

An equation represents a line. Restricting it to the first quadrant leaves only a segment. No shaded region is involved.

✅ Correct Answer Explanation

The equation x + y = k is a linear equation. Graphically, it represents a straight line. Since x ≥ 0 and y ≥ 0, only the portion lying in the first quadrant is considered. Hence, the graphical representation is a straight line segment. Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) A filled rectangular region → Represents an area, not an equation.
Option B) An unbounded upper plane → An equality does not produce a half-plane.
Option C) A parabolic curve → The equation is linear, not quadratic.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Recall that linear equations graph as straight lines.

Final Logic:

An equality constraint gives a straight line segment.

🧠 Memory Trick

"Equality = Line."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Graph of Linear Equations

15 In an assignment LPP, an error costs ₹20. If Group A inspectors have a known 2% inaccuracy rate, the probability of an entirely accurate check by a single Group A inspector is:

A 0.98

B 0.02

C 0.95

D 1.00

💡 Short Explanation

Accuracy and inaccuracy are complementary events. Accuracy = 1 − Inaccuracy. Compute the probability.

✅ Correct Answer Explanation

The inaccuracy rate is: 2% = 0.02 Therefore, Accuracy = 1 − 0.02 = 0.98 Thus, the probability that a single inspection is completely accurate is 0.98. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 0.02 → Represents the probability of an inaccurate inspection.
Option C) 0.95 → Incorrect computation.
Option D) 1.00 → Accuracy is not perfect.

🎯 Strategy

used

Substitution

Application:

Subtract the inaccuracy probability from one.

Final Logic:

1 − 0.02 = 0.98.

🧠 Memory Trick

"Accuracy = 1 − Error."

📚 Class XII – Applied Mathematics – Chapter: Probability – Topic: Complementary Events

16 Given strict worker assignment limits x ≤ 8 and y ≤ 10, alongside trivial constraints x ≥ 0, y ≥ 0, the mathematical area of this feasible rectangular assignment space is:

A 18 sq. units

B 80 sq. units

C 40 sq. units

D 100 sq. units

💡 Short Explanation

The feasible region is a rectangle. Area = Length × Breadth. Multiply the side lengths.

✅ Correct Answer Explanation

The constraints define the rectangle: 0 ≤ x ≤ 8 0 ≤ y ≤ 10 Length = 8 Breadth = 10 Area: Area = 8 × 10 = 80 square units Therefore, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 18 sq. units → Obtained by adding the dimensions.
Option C) 40 sq. units → Half of the correct area.
Option D) 100 sq. units → Incorrect multiplication.

🎯 Strategy

used

Substitution

Application:

Identify the dimensions of the rectangle from the constraints.

Final Logic:

8 × 10 = 80 square units.

🧠 Memory Trick

"Rectangle = Length × Breadth."

📚 Class XII – Applied Mathematics – Chapter: Coordinate Geometry – Topic: Area of Plane Figures

17 An industrial profit margin boundary is exactly y = 4. Using the integral
∫ from 0 to 3 of 4 dx,
find the area under this boundary from x = 0 to x = 3:

A 4

B 8

C 12

D 16

💡 Short Explanation

The function is constant. Area equals height × width. Evaluate the definite integral.

✅ Correct Answer Explanation

Given: ∫ from 0 to 3 of 4 dx Since the integrand is constant, ∫ 4 dx = 4x Applying the limits: At x = 3: = 4 × 3 = 12 At x = 0: = 0 Therefore, Area = 12 square units Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 4 → Represents only the height, not the area.
Option B) 8 → Incorrect multiplication.
Option D) 16 → Arithmetic error.

🎯 Strategy

used

Substitution

Application:

Integrate the constant function and substitute the limits.

Final Logic:

4 × 3 = 12.

🧠 Memory Trick

"Constant Function = Height × Width."

📚 Class XII – Applied Mathematics – Chapter: Integral Calculus – Topic: Definite Integrals

18 A company tracking an objective minimum cost over 3 days records: ₹40, ₹50, ₹60. The 2-day moving average calculated for the final 2 days is:

A ₹45

B ₹50

C ₹60

D ₹55

💡 Short Explanation

Use the last two observations. Compute their average. This gives the final 2-day moving average.

✅ Correct Answer Explanation

The last two observations are: ₹50 and ₹60 The 2-day moving average is: (50 + 60) / 2 = 110 / 2 = ₹55 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) ₹45 → Average of the first two observations.
Option B) ₹50 → Not the average of the final two values.
Option C) ₹60 → This is the last observation only.

🎯 Strategy

used

Substitution

Application:

Calculate the arithmetic mean of the last two values.

Final Logic:

The average of ₹50 and ₹60 is ₹55.

🧠 Memory Trick

"Last Two → Average."

📚 Class XII – Applied Mathematics – Chapter: Time-based Data Analysis – Topic: Moving Average

19 Match List I with List II:

List I	List II
1. Iso-profit lines	a. Vertices of the common area
2. Unbounded region	b. Open plane missing maximum bounds
3. Corner point evaluation	c. Moved parallel to itself
4. Feasible region	d. Region satisfying all constraints

A 1–c, 2–b, 3–a, 4–d

B 1–a, 2–c, 3–b, 4–d

C 1–b, 2–a, 3–c, 4–d

D 1–c, 2–a, 3–b, 4–d

💡 Short Explanation

Iso-profit lines are shifted parallelly. Unbounded regions extend indefinitely. Corner points are the vertices of the feasible region.

✅ Correct Answer Explanation

The correct matching is: 1 → c : Iso-profit lines are moved parallel to locate the optimum. 2 → b : An unbounded region extends indefinitely. 3 → a : Corner-point evaluation uses the vertices of the feasible region. 4 → d : The feasible region contains all points satisfying every constraint. Thus, the correct sequence is: 1–c, 2–b, 3–a, 4–d Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrectly matches iso-profit lines and corner-point evaluation.
Option C) Incorrectly matches unbounded regions.
Option D) Incorrectly matches corner-point evaluation.

🎯 Strategy

used

Option Grouping

Application:

Match each Linear Programming concept with its standard definition.

Final Logic:

Only Option A correctly matches all four concepts.

🧠 Memory Trick

"Iso-profit → Parallel, Feasible → All Constraints."

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

20 A manager needs to maximize
Z = 2x + 10y
subject to 2x + 5y ≤ 16, x ≤ 5, and non-negativity constraints. The optimal maximum profit computed at the boundaries is:

A 22

B 32

C 42

D 50

💡 Short Explanation

Evaluate the objective function at the corner points. Compare all objective values. Select the maximum.

✅ Correct Answer Explanation

The feasible corner points are: (0,0) (5,0) (5,1.2) (from 2x + 5y = 16) (0,3.2) Evaluate the objective function: (0,0): Z = 0 (5,0): Z = 10 (5,1.2): Z = 2(5) + 10(1.2) = 10 + 12 = 22 (0,3.2): Z = 2(0) + 10(3.2) = 32 The maximum value is: Z = 32 Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 22 → Maximum is larger at (0,3.2).
Option C) 42 → Not attainable within the feasible region.
Option D) 50 → Violates the given constraints.

🎯 Strategy

used

Substitution

Application:

Evaluate the objective function at each feasible corner point.

Final Logic:

The maximum objective value is 32. (The given answer key is correct.)

🧠 Memory Trick

"Evaluate Every Corner, Choose Maximum."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Corner Point Method

CUET UG Applied Mathematics Booster Test 2 - Mathematical Formulation, Types, and Applications

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