CUET UG Applied Mathematics Booster Test 1 - Mathematical Formulation, Types, and Applications -

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CUET UG Applied Mathematics Booster Test 1 - 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 1 - Mathematical Formulation, Types, and Applications

1 / 20

QUESTION 1 OF 20

A firm defines its maximum variable limits for 4 consecutive production runs as:
12, 14, 16, 18
What is the 2-period moving average for the first available subset?

Answer Review

1 A firm defines its maximum variable limits for 4 consecutive production runs as:
12, 14, 16, 18
What is the 2-period moving average for the first available subset?

A 12

B 13

C 15

D 17

💡 Short Explanation

Take the first two observations. Compute their arithmetic mean. This gives the first 2-period moving average.

✅ Correct Answer Explanation

The first available subset consists of: 12 and 14 The 2-period moving average is: (12 + 14) / 2 = 26 / 2 = 13 Therefore, the first 2-period moving average is 13. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 12 → This is the first observation, not the average.
Option C) 15 → This is the average of the second and third observations.
Option D) 17 → This is the average of the third and fourth observations.

🎯 Strategy

used

Substitution

Application:

Select the first two observations and calculate their average.

Final Logic:

The average of 12 and 14 is 13.

🧠 Memory Trick

"Moving Average = Sum ÷ Number of Terms."

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

2 Match List I with List II:

List I	List II
1. At least	a. ≤
2. At most	b. =
3. Exactly	c. ≥
4. Not equal to	d. ≠

A 1–c, 2–a, 3–b, 4–d

B 1–a, 2–b, 3–c, 4–d

C 1–b, 2–c, 3–a, 4–d

D 1–c, 2–b, 3–a, 4–d

💡 Short Explanation

"At least" means greater than or equal to. "At most" means less than or equal to. "Exactly" means equality.

✅ Correct Answer Explanation

The correct mathematical meanings are: At least → ≥ At most → ≤ Exactly → = Not equal to → ≠ Thus, the correct matching is: 1 → c 2 → a 3 → b 4 → d Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B → Incorrectly matches "At least" and "At most."
Option C → Incorrectly matches "Exactly."
Option D → Incorrectly matches "At most."

🎯 Strategy

used

Option Grouping

Application:

Recall the standard meanings of common mathematical inequality terms.

Final Logic:

Only Option A correctly matches every inequality.

🧠 Memory Trick

"At Least ≥, At Most ≤."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Linear Constraints

3 A profit feasible area is defined under the constraint:
y = 10 − 2x
Using the definite integral:
\(\int_{0}^{5}\,(10-2x) dx\)
Find the area defining the profit region.

A 15

B 20

C 30

D 25

💡 Short Explanation

Integrate the given function. Apply the limits 0 and 5. The definite integral equals the required area.

✅ Correct Answer Explanation

Given: ∫ from 0 to 5 of (10 − 2x) dx Integrate: ∫ (10 − 2x) dx = 10x − x² Apply the limits: At x = 5: = 10(5) − 5² = 50 − 25 = 25 At x = 0: = 0 Therefore, Area = 25 − 0 = 25 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 15 → Incorrect integration.
Option B) 20 → Arithmetic error.
Option C) 30 → Does not satisfy the definite integral.

🎯 Strategy

used

Substitution

Application:

Integrate the function and substitute the limits.

Final Logic:

The definite integral evaluates to 25.

🧠 Memory Trick

"Integrate → Apply Limits."

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

4 A cost minimization problem has an enclosed rectangular feasible region bounded by:
x = 0, y = 0, x = 5, y = 4
The area of this region is:

A 20 sq. units

B 15 sq. units

C 9 sq. units

D 25 sq. units

💡 Short Explanation

The region is a rectangle. Area equals length × breadth. Multiply the side lengths.

✅ Correct Answer Explanation

The rectangle has: Length = 5 units Breadth = 4 units Area of the rectangle: Area = Length × Breadth = 5 × 4 = 20 square units Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 15 sq. units → Incorrect multiplication.
Option C) 9 sq. units → Not the area of the rectangle.
Option D) 25 sq. units → Uses only one side length squared.

🎯 Strategy

used

Substitution

Application:

Identify the side lengths from the boundary equations and apply the rectangle area formula.

Final Logic:

Area = 5 × 4 = 20 square units.

🧠 Memory Trick

"Rectangle = Length × Breadth."

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

5 A firm's resource limits are given by:
x ≤ 4, y ≤ 3
The displacement vector from the origin (0,0) to the maximum capacity point (4,3) is:

A \(3\hat{i}+4\hat{j}\)

B \(4\hat{i}+3\hat{j}\)

C \(7\hat{i}+7\hat{j}\)

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

💡 Short Explanation

The displacement vector equals the terminal point minus the initial point. Initial point is the origin. Read the coordinates directly.

✅ Correct Answer Explanation

The displacement vector from: (0,0) to (4,3) is (4−0)i + (3−0)j = 4i + 3j Thus, the vector has an x-component of 4 and a y-component of 3. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 3i + 4j → The x and y components are interchanged.
Option C) 7i + 7j → Obtained by incorrectly adding the coordinates.
Option D) i + j → Does not represent the displacement from the origin to (4,3).

🎯 Strategy

used

Substitution

Application:

Subtract the coordinates of the initial point from those of the terminal point.

Final Logic:

Displacement vector = 4i + 3j.

🧠 Memory Trick

"Vector = Destination − Start."

📚 Class XII – Applied Mathematics – Chapter: Vectors – Topic: Position and Displacement Vectors

6 A factory requires 10 man-hours for each unit of x and 6 man-hours for each unit of y. The maximum weekly capacity is 1000 hours.
If 50 units of y are produced, the maximum possible production of x is:

A 50

B 60

C 70

D 80

💡 Short Explanation

Form the capacity constraint. Substitute the value of y. Solve for the maximum value of x.

✅ Correct Answer Explanation

The capacity constraint is: 10x + 6y ≤ 1000 Given: y = 50 Substitute: 10x + 6(50) ≤ 1000 10x + 300 ≤ 1000 10x ≤ 700 x ≤ 70 Thus, the maximum possible production of x is 70 units. Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 50 → Does not satisfy the maximum available capacity.
Option B) 60 → Underestimates the allowable production.
Option D) 80 → Requires more than the available 1000 man-hours.

🎯 Strategy

used

Substitution

Application:

Substitute the given value of y into the capacity constraint and solve for x.

Final Logic:

After producing 50 units of y, the maximum value of x is 70.

🧠 Memory Trick

"Substitute First, Solve Later."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Linear Constraints

7 Assertion (A): Decision variables in linear programming problems can take negative values to represent debt.
Reason (R): The condition x ≥ 0, y ≥ 0 is called the non-negativity restriction.

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

Standard LPP does not allow negative decision variables. Non-negative restrictions require variables to be zero or positive. Thus, the assertion is false.

✅ Correct Answer Explanation

In a standard Linear Programming Problem, decision variables satisfy: x ≥ 0 y ≥ 0 These are called the non-negativity restrictions. Therefore: Assertion (A) is false because decision variables cannot take negative values in a standard LPP. Reason (R) is true because x ≥ 0, y ≥ 0 are indeed the non-negative restrictions. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Both A and R are false → The reason is true.
Option B) A is true, R is false → The assertion is false.
Option C) Both A and R are true → The assertion is incorrect.

🎯 Strategy

used

Elimination

Application:

Evaluate the assertion and reason independently.

Final Logic:

Decision variables are non-negative; therefore, only the reason is true.

🧠 Memory Trick

"LPP Variables Never Negative."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Non-negative Restrictions

8 Which of the Cartesian quadrants satisfy the condition:
y ≥ 0
I. Quadrant I
II. Quadrant II
III. Quadrant III

A I and II only

B I and III only

C II and III only

D I, II, and III

💡 Short Explanation

Positive y-values occur above the x-axis. Quadrants I and II satisfy y ≥ 0. Quadrant III has negative y-values.

✅ Correct Answer Explanation

The condition: y ≥ 0 means the point must lie on or above the x-axis. Quadrant I → y is positive ✔ Quadrant II → y is positive ✔ Quadrant III → y is negative ✘ Hence, only Quadrants I and II satisfy the condition. Therefore, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) I and III only → Quadrant III has negative y-values.
Option C) II and III only → Quadrant III does not satisfy y ≥ 0.
Option D) I, II, and III → Includes Quadrant III incorrectly.

🎯 Strategy

used

Elimination

Application:

Recall the signs of coordinates in each Cartesian quadrant.

Final Logic:

Only Quadrants I and II have non-negative y-values.

🧠 Memory Trick

"Above x-axis ⇒ y ≥ 0."

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

9 Identify the incorrect statement regarding mathematical modeling of investment problems:

A The capital available acts as a capacity constraint.

B Investment problems do not require an objective function.

C Risk factors can be mapped into linear inequalities.

D The goal is usually to maximize return on investment.

💡 Short Explanation

Every LPP requires an objective function. Investment models aim to maximize returns or minimize risk. Therefore, the statement denying the objective function is incorrect.

✅ Correct Answer Explanation

An investment problem in Linear Programming consists of: Decision variables representing investment amounts. Objective function representing return, profit, or risk. Constraints representing available capital, risk limits, and other restrictions. Thus: Option A is correct because available capital acts as a resource constraint. Option B is incorrect because every LPP requires an objective function. Option C is correct because risk or policy limits can often be expressed using linear constraints. Option D is correct because investment models generally aim to maximize returns. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) The capital available acts as a capacity constraint. → This is a standard resource constraint in investment models.
Option C) Risk factors can be mapped into linear inequalities. → Many investment constraints are expressed as linear inequalities.
Option D) The goal is usually to maximize return on investment. → This is the usual objective in investment planning.

🎯 Strategy

used

Extreme Word Filter

Application:

Look for statements that contradict the essential components of an LPP.

Final Logic:

Every Linear Programming model must contain an objective function.

🧠 Memory Trick

"No Objective = No LPP."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Mathematical Formulation of LPP

10 For a production LPP constrained by:
3x + 5y ≤ 15, x ≥ 0, y ≥ 0
The corner points of the feasible region are:

A (0,0), (5,0), (0,5)

B (0,0), (3,0), (0,5)

C (0,0), (5,0), (0,3)

D (0,0), (3,0), (0,3)

💡 Short Explanation

Find the x-intercept. Find the y-intercept. Include the origin because of the non-negative restrictions.

✅ Correct Answer Explanation

The boundary line is: 3x + 5y = 15 For the x-intercept: Put y = 0 3x = 15 x = 5 So the point is (5,0). For the y-intercept: Put x = 0 5y = 15 y = 3 So the point is (0,3). Together with the origin, the corner points are: (0,0) (5,0) (0,3) However, none of the given options list these three points correctly. Option C incorrectly gives (0,5). Option D incorrectly gives (3,0). Therefore, the answer key is incorrect, and the correct corner points are (0,0), (5,0), (0,3).

❓ Why Other Options Are Incorrect

Option A) Incorrect y-intercept (0,5).
Option B) Incorrect x-intercept (3,0) and y-intercept (0,5).
Option C) Correct x-intercept but incorrect y-intercept.
Option D) Correct y-intercept but incorrect x-intercept.

🎯 Strategy

used

Substitution

Application:

Set one variable to zero at a time to determine the intercepts.

Final Logic:

The intercepts are (5,0) and (0,3), so the provided options contain an error.

🧠 Memory Trick

"Set y = 0 → x-intercept; Set x = 0 → y-intercept."

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

11 Arrange the logical steps for solving a manufacturing problem using the graphical method:
1. Determine the coordinates of the corner points.
2. Formulate the LPP mathematically.
3. Plot the inequality constraints to identify the feasible region.
4. Evaluate the objective function Z at each corner point.

A 2, 1, 3, 4

B 3, 2, 1, 4

C 2, 4, 3, 1

D 2, 3, 1, 4

💡 Short Explanation

Formulate the problem first. Plot the constraints. Identify the corner points. Evaluate the objective function.

✅ Correct Answer Explanation

The standard graphical method follows these steps: 1. Formulate the Linear Programming Problem. 2. Plot the constraints to obtain the feasible region. 3. Determine the corner points of the feasible region. 4. Evaluate the objective function at each corner point. Therefore, the correct sequence is: 2 → 3 → 1 → 4 Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Corner points cannot be determined before plotting the constraints.
Option B) The LPP must be formulated before graphing.
Option C) The objective function is evaluated only after identifying the corner points.

🎯 Strategy

used

Option Grouping

Application:

Recall the standard sequence of the graphical solution procedure.

Final Logic:

Formulate → Plot → Corner Points → Evaluate.

🧠 Memory Trick

"Model → Graph → Corners → Evaluate."

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

12 The feasible region of a diet problem has exactly 4 distinct corner points. If a corner point is selected uniformly at random, what is the probability that it is the unique optimal minimum point?

A \(\frac{1}{4}\)

B \(\frac{1}{2}\)

C \(\frac{3}{4}\)

D \(1\)

💡 Short Explanation

There are four equally likely corner points. Only one is the unique optimal point. Probability equals favourable outcomes divided by total outcomes.

✅ Correct Answer Explanation

The feasible region has 4 distinct corner points. If exactly one of these is the unique optimal minimum point, then: Probability = Number of favourable corner points / Total corner points = 1 / 4 Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) 1/2 → Assumes two optimal corner points.
Option C) 3/4 → Assumes three optimal corner points.
Option D) 1 → Assumes every corner point is optimal.

🎯 Strategy

used

Substitution

Application:

Use the classical probability formula.

Final Logic:

One favourable outcome among four equally likely corner points gives 1/4.

🧠 Memory Trick

"Probability = Favourable ÷ Total."

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

13 If a transportation firm has Factory P with a capacity of 8 units and Factory Q with a capacity of 6 units, the total supply capacity is:

A 2 units

B 14 units

C 48 units

D 10 units

💡 Short Explanation

Total supply is the sum of capacities. Add the capacities of both factories. This gives the overall supply.

✅ Correct Answer Explanation

Factory capacities are: Factory P = 8 units Factory Q = 6 units Total supply capacity: 8 + 6 = 14 units Thus, the transportation network can supply 14 units in total. Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 2 units → Obtained by subtracting the capacities.
Option C) 48 units → Obtained by multiplying the capacities.
Option D) 10 units → Incorrect addition.

🎯 Strategy

used

Substitution

Application:

Add the capacities of all supply sources.

Final Logic:

Total supply = 8 + 6 = 14 units.

🧠 Memory Trick

"Total Supply = Sum of Sources."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Transportation Problem

14 Match List I with List II for a transportation network:

List I	List II
1. Depot A must receive at least 40% of total available supply of 50 units	a. Total supply ≥ 24
2. Depot B cannot receive more than the surplus after satisfying a compulsory demand of 30 units elsewhere	b. Received units ≤ 20
3. Depot C must receive exactly the difference between total supply (50 units) and combined fixed allocations of 18 and 12 units	c. Received units = 20
4. Total shipment must satisfy a lower bound equal to sum of minimum demands: 6, 8, and 10 units	d. Received units ≥ 20

A 1–d, 2–b, 3–c, 4–a

B 1–c, 2–d, 3–b, 4–a

C 1–d, 2–c, 3–b, 4–a

D 1–b, 2–d, 3–c, 4–a

💡 Short Explanation

"At least 40%" means ≥ 20. Surplus after 30 units from 50 is 20, giving ≤ 20. Remaining allocation is exactly 20.

✅ Correct Answer Explanation

Evaluate each statement: 1. 40% of 50 = 20, so Received units ≥ 20 → d 2. Surplus after allocating 30 units is 20, so Received units ≤ 20 → b 3. Remaining supply = 50 − (18 + 12) = 20, so Received units = 20 → c 4. Minimum demand = 6 + 8 + 10 = 24, so Total supply ≥ 24 → a Thus, the correct matching is: 1–d, 2–b, 3–c, 4–a Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B) Incorrectly matches Statements 1, 2, and 3.
Option C) Incorrectly matches Statements 2 and 3.
Option D) Incorrectly matches Statement 1.

🎯 Strategy

used

Option Grouping

Application:

Compute the numerical values first and then match them with the appropriate inequalities.

Final Logic:

The calculated values match Option A exactly.

🧠 Memory Trick

"40% of 50 = 20; Remaining = 20; Minimum = 24."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Transportation Constraints

15 Group A checks 25 pieces per hour with an accuracy rate of 98%. If an inspector from Group A is assigned a task for exactly 2 hours, how many accurate pieces are produced?

A 45

B 47

C 48

D 49

💡 Short Explanation

Calculate the total pieces checked. Apply the accuracy percentage. The result gives the number of accurate pieces.

✅ Correct Answer Explanation

Pieces checked in 2 hours: 25 × 2 = 50 pieces Accuracy = 98% Accurate pieces: 0.98 × 50 = 49 Therefore, the inspector accurately checks 49 pieces. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 45 → Underestimates the accurate output.
Option B) 47 → Incorrect application of the percentage.
Option C) 48 → Arithmetic error.

🎯 Strategy

used

Substitution

Application:

Compute total production and multiply by the accuracy rate.

Final Logic:

50 × 98% = 49.

🧠 Memory Trick

"Total × Accuracy = Accurate Output."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Assignment Problem

16 A firm assigns workers where Group A has 8 workers and Group B has 10 workers. Let x and y represent the number of inspectors scheduled.
Which limits must inherently apply?
I. Group A inspectors ≤ 8
II. Group B inspectors ≤ 10

A Both I and II

B I only

C II only

D Neither I nor II

💡 Short Explanation

Scheduled workers cannot exceed those available. Both groups have upper limits. Therefore, both conditions are necessary.

✅ Correct Answer Explanation

Available workers: Group A = 8 Group B = 10 Hence, the scheduling variables must satisfy: x ≤ 8 y ≤ 10 Both constraints are essential to ensure that the number of scheduled inspectors does not exceed the available workforce. Therefore, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) I only → Ignores the limit on Group B.
Option C) II only → Ignores the limit on Group A.
Option D) Neither I nor II → Contradicts the given resource limits.

🎯 Strategy

used

Elimination

Application:

Compare the available workers with the scheduling variables.

Final Logic:

Both upper-bound constraints must be satisfied.

🧠 Memory Trick

"Available Workers = Maximum Schedule."

📚 Class XII – Applied Mathematics – Chapter: Linear Programming – Topic: Resource Constraints

17 Identify the incorrect statement related to industrial profit maximization graphs:

A The optimal solution always lies at a corner point.

B The objective function Z has only a minimum value in bounded regions.

C An iso-profit line can be moved parallelly to find the optimal point.

D x, y ≥ 0 restricts the graph to the first quadrant.

💡 Short Explanation

A bounded feasible region may have both maximum and minimum values. Iso-profit lines help locate the optimum. Therefore, the given statement is incorrect.

✅ Correct Answer Explanation

In a bounded feasible region: The objective function may have both a maximum and a minimum value. According to the Fundamental Theorem of Linear Programming, if an optimum exists, it occurs at a corner (extreme) point. The iso-profit line is shifted parallel to itself until the last point of contact with the feasible region. The conditions x ≥ 0, y ≥ 0 restrict the feasible region to the first quadrant. Hence, the statement that the objective function has only a minimum value is incorrect. Therefore, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) The optimal solution always lies at a corner point. → This is a standard result in the graphical method of LPP.
Option C) An iso-profit line can be moved parallelly to find the optimal point. → This is the basis of the iso-profit method.
Option D) x, y ≥ 0 restricts the graph to the first quadrant. → This is the non-negative restriction.

🎯 Strategy

used

Extreme Word Filter

Application:

Notice the word "only". A bounded feasible region can have both maximum and minimum objective values.

Final Logic:

A bounded region does not imply only a minimum value.

🧠 Memory Trick

"Bounded ⇒ Max and Min Possible."

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

18 The area bounded by the cost limit line:
x + 2y = 6
in the first quadrant can be evaluated using:
\(\int_{0}^{6}\,\left(3-\frac{x}{2}\right) dx\)
Calculate this area.

A 18

B 12

C 9

D 6

💡 Short Explanation

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

✅ Correct Answer Explanation

Given: ∫ from 0 to 6 (3 − x/2) dx Integrate: ∫ (3 − x/2) dx = 3x − x²/4 Apply the limits: At x = 6: = 3(6) − 6²/4 = 18 − 36/4 = 18 − 9 = 9 At x = 0: = 0 Therefore, Area = 9 square units Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A) 18 → Ignores the subtraction term.
Option B) 12 → Incorrect integration.
Option D) 6 → Arithmetic error.

🎯 Strategy

used

Substitution

Application:

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

Final Logic:

The definite integral evaluates to 9.

🧠 Memory Trick

"Integrate → Apply Limits → Area."

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

📄 Passage

Most organizations, whether large or small, face the problem of planning and optimizing available resources to achieve maximum production (or maximize profit), or in some cases, to minimize the cost of production. Solving such problems using mathematical techniques is referred to as constrained optimization. Linear programming deals with the optimization of a linear function subject to linear constraints.

Based on the passage, the commercial problems of planning and optimizing available resources using mathematical constraints are technically referred to as:

A Unbounded operations

B Trivial simplifications

C Infinite programming

D Constrained optimization problems

💡 Short Explanation

Organizations optimize limited resources. Mathematical techniques are used for optimization. Such problems are called constrained optimization problems.

✅ Correct Answer Explanation

The passage states that organizations optimize available resources while satisfying various constraints. Problems solved using mathematical techniques under such constraints are known as constrained optimization problems. Linear Programming is one of the most widely used methods for solving these problems. Therefore, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Unbounded operations → This is not the term used in optimization.
Option B) Trivial simplifications → This is unrelated to resource optimization.
Option C) Infinite programming → No such standard term exists in Linear Programming.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Identify the exact terminology used in the passage.

Final Logic:

The passage explicitly refers to constrained optimization problems.

🧠 Memory Trick

"Resources + Constraints = Constrained Optimization."

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

📄 Passage

Based on the passage, in linear programming used by organizations, the objective function that is optimized is always a:

A Linear function

B Quadratic function

C Cubic function

D Non-linear expression

💡 Short Explanation

Linear Programming optimizes a linear objective function. Constraints are also linear. This is the defining feature of LPP.

✅ Correct Answer Explanation

The passage clearly states that Linear Programming deals with the optimization of a linear function subject to linear constraints. Therefore, the objective function must always be linear. Quadratic, cubic, or other non-linear functions belong to different optimization techniques and are not part of standard Linear Programming. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Quadratic function → Used in quadratic programming, not standard LPP.
Option C) Cubic function → Not used in Linear Programming.
Option D) Non-linear expression → Violates the definition of Linear Programming.

🎯 Strategy

used

Contextual/Tonal Matching

Application:

Use the definition of Linear Programming provided in the passage.

Final Logic:

Linear Programming always optimizes a linear function.

🧠 Memory Trick

"Linear Programming → Linear Function."

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

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