CBSE Maths › Class 12 PYQs › Linear Programming

Linear Programming: 1 mark Questions (CBSE Class 12)

8 different 1 mark questions on Linear Programming from CBSE Class 12 Maths board exams 2026, newest first.

Q171 markMCQLinear Programming CBSE 2026 · 65/1/1

The feasible region of a linear programming problem with objective function $Z = 5 x + 7 y$ is shown below :
The maximum value of Z – minimum value of Z is

Diagram for CBSE 2026 Class 12 Maths question 17

(A)8
(B)29
(C)35
(D)43

Show answer & solution

Answer: (D) 43

Corner points: O(0, 0), (0, 2), (3, 4), (7, 0).
Z at these: 0, 14, 15 + 28 = 43, 35.
Maximum $= 43$ , minimum $= 0$ .
Maximum $-$ minimum $= 43$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q181 markMCQLinear Programming CBSE 2026 · 65/1/1

The degree of an objective function of a linear programming problem is

(A)0
(B)1
(C)2
(D)Any natural number

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Answer: (B) 1

The objective function of an LPP is linear, e.g. $Z = a x + b y$ .
So its degree is 1.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q161 markMCQLinear Programming CBSE 2026 · 65/2/1

In a linear programming problem, the linear function which has to be maximized or minimized is called

(A)a feasible function
(B)an objective function
(C)an optimal function
(D)a constraint

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Answer: (B) an objective function

The linear function to be optimised in an LPP is called the objective function.

Also asked in: 2026 65/2/2, 2026 65/2/3

Q171 markMCQLinear Programming CBSE 2026 · 65/2/1

For the feasible region shown below, the non-trivial constraints of the linear programming problem are

(A) $x + y \leq 5$ , $x + 3 y \leq 9$
(B) $x + y \leq 5$ , $x + 3 y \geq 9$
(C) $x + y \geq 5$ , $x + 3 y \leq 9$
(D) $x + y \geq 5$ , $3 x + y \leq 9$

Show answer & solution

Answer: (C)

x + y \geq 5

x + 3 y \leq 9

The line through $(0, 5)$ and $(5, 0)$ is $x + y = 5$ ; the line through $(0, 3)$ and $(9, 0)$ is $x + 3 y = 9$ .
The shaded region (vertices $(3, 2)$ , $(5, 0)$ , $(9, 0)$ ) lies on the side of $x + y = 5$ away from the origin: $x + y \geq 5$ .
It lies on the origin side of $x + 3 y = 9$ : $x + 3 y \leq 9$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q141 markMCQLinear Programming CBSE 2026 · 65/4/1

The region represented by the system of inequations $3 x + y \geq 3$ , $2 x - y \geq - 5$ , $x, y \geq 0$ is :

(A)unbounded in 1 $^{s t}$ quadrant
(B)bounded in 1 $^{s t}$ quadrant
(C)unbounded in 2 $^{n d}$ quadrant
(D)bounded in 2 $^{n d}$ quadrant

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Answer: (A) unbounded in 1

^{s t}

quadrant

$x, y \geq 0$ keeps the region in the first quadrant.
$3 x + y \geq 3$ keeps it away from the origin; $2 x - y \geq - 5$ means $y \leq 2 x + 5$ .
As x increases, points such as $(x, 0)$ with $x \geq 1$ satisfy all inequations, so the region extends without limit.
Hence it is unbounded in the first quadrant.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q151 markMCQLinear Programming CBSE 2026 · 65/4/1

In the graph, the feasible region representing the Linear Programming Problem for maximising objective function $Z = p x + q y$ , $p, q > 0$ is shaded. If all points on segment AB give max (Z), then which of the following is true ?

Diagram for CBSE 2026 Class 12 Maths question 15

(A) $p = 2 q$
(B) $p = 3 q$
(C) $q = 3 p$
(D) $q = 2 p$

Show answer & solution

Answer: (C)

q = 3 p

At A(0, 5): $Z = 5 q$ . At B(3, 4): $Z = 3 p + 4 q$ .
Maximum at every point of AB means $Z (A) = Z (B)$ : $5 q = 3 p + 4 q$ .
So $q = 3 p$ . (Then $Z (C) = 5 p < 15 p = Z (A)$ , consistent.)

Also asked in: 2026 65/4/2, 2026 65/4/3

Q151 markMCQLinear Programming CBSE 2026 · 65/5/1

The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40) (60, 20) and (60, 0). If the objective function of an LPP is $Z = 4 x + 3 y$ , then the maximum value is :

(A)200
(B)300
(C)240
(D)120

Show answer & solution

Answer: (B) 300

Z at the corners: (0, 0): 0; (0, 40): 120; (20, 40): 200; (60, 20): 300; (60, 0): 240.
Maximum value is 300 at (60, 20).

Also asked in: 2026 65/5/2, 2026 65/5/3

Q201 markAssertion–ReasonLinear Programming CBSE 2026 · 65/5/1

Assertion (A): Consider a Linear Programming Problem with minimise $Z = x + 2 y$ subject to constraints $2 x + y \geq 3$ , $x + 2 y \geq 6$ , $x, y \geq 0$ which gives minimum Z at infinitely many points. The corner points of feasible region are (0, 3) and (6, 0).
Reason (R): If two corner points produce the same minimum value of the objective function, then every point on the line segment joining the points will give the same minimum value.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (A) Both A and R are true and R is the correct explanation of A.

Lines $2 x + y = 3$ and $x + 2 y = 6$ meet at (0, 3); the feasible (unbounded) region has corner points (0, 3) and (6, 0).
$Z (0, 3) = 6$ and $Z (6, 0) = 6$ ; since $x + 2 y \geq 6$ in the region, the minimum is 6.
Every point on the segment joining (0, 3) and (6, 0) gives $Z = 6$ , so the minimum occurs at infinitely many points.
R is a true property and is exactly why A holds.

Also asked in: 2026 65/5/2, 2026 65/5/3

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