15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
A bounded feasible region has corner points \((0, 0)\), \((7, 0)\), \((5, 4)\) and \((0, 6)\). The maximum of \(Z = 3x + 4y\) over the region is
Why: Evaluate Z at each corner: 0, 21, 31, 24.
\(Z(0,0) = 0\), \(Z(7,0) = 21\), \(Z(5,4) = 15 + 16 = 31\), \(Z(0,6) = 24\). Maximum \(= 31\) at \((5, 4)\).
Over the same region (corners \((0, 0)\), \((7, 0)\), \((5, 4)\), \((0, 6)\)), the least value of \(Z = x - 2y\) is
Why: Z values: 0, 7, −3, −12.
\(Z(0,0) = 0\), \(Z(7,0) = 7\), \(Z(5,4) = 5 - 8 = -3\), \(Z(0,6) = -12\). Minimum \(= -12\) at \((0, 6)\).
Which point lies in the feasible region of \(2x + y \le 10\), \(x + 3y \le 15\), \(x \ge 0\), \(y \ge 0\)?
Why: (3, 3): 9 ≤ 10 and 12 ≤ 15, with x, y ≥ 0.
\((3,3)\): \(6 + 3 = 9 \le 10\), \(3 + 9 = 12 \le 15\) ✓. \((4,3)\): \(11 \gt 10\); \((1,5)\): \(16 \gt 15\); \((-1,2)\): \(x \lt 0\).
The feasible region of an LPP is bounded, and \(Z = ax + by\) takes its maximum value at two adjacent corner points \(P\) and \(Q\). Then \(Z\) is maximum
Why: Z is linear, so along the edge PQ it takes every value between Z(P) and Z(Q), which are equal.
On the segment \(PQ\), a point is \((1 - t)P + tQ\) with \(0 \le t \le 1\), and since \(Z\) is linear, \(Z = (1 - t)Z(P) + tZ(Q) = Z(P)\). So every point of \(PQ\) gives the same maximum; there are infinitely many optimal solutions.
The corner points of the feasible region given by \(x + y \le 4\), \(x \ge 0\), \(y \ge 0\) are
Why: The region is the triangle cut off by x + y = 4 and the axes.
The boundary lines \(x + y = 4\), \(x = 0\), \(y = 0\) meet in pairs at \((0,0)\), \((4,0)\), \((0,4)\), all feasible: a triangle.
The maximum value of \(Z = 5x + 3y\) subject to \(x + y \le 4\), \(x \ge 0\), \(y \ge 0\) is
Why: Corners (0, 0), (4, 0), (0, 4) give 0, 20, 12.
\(Z(0,0) = 0\), \(Z(4,0) = 20\), \(Z(0,4) = 12\). Maximum \(= 20\) at \((4, 0)\).
If the feasible region of an LPP is bounded, the maximum and minimum of the objective function \(Z = ax + by\) occur
Why: Corner point theorem.
For a bounded feasible region, \(Z\) has both a maximum and a minimum, and each occurs at a corner (vertex) of the region.
The maximum value of \(Z = x + y\) subject to \(x + 3y \le 12\), \(3x + y \le 12\), \(x \ge 0\), \(y \ge 0\) is
Why: Corners (0, 0), (4, 0), (3, 3), (0, 4): Z = 0, 4, 6, 4.
\(x + 3y = 12\) and \(3x + y = 12\) meet at \((3, 3)\). Corners \((0,0), (4,0), (3,3), (0,4)\) give \(Z = 0, 4, 6, 4\). Maximum \(= 6\).
The minimum value of \(Z = 3x + 2y\) subject to \(x + y \ge 6\), \(x \le 4\), \(y \le 5\), \(x \ge 0\), \(y \ge 0\) is
Why: Corners (1, 5), (4, 2), (4, 5): Z = 13, 16, 22.
The feasible region is the triangle with corners \((1, 5)\), \((4, 2)\), \((4, 5)\). \(Z = 13, 16, 22\); minimum \(= 13\) at \((1, 5)\).
The maximum value of \(Z = 2x + 5y\) subject to \(0 \le x \le 3\), \(0 \le y \le 2\) is
Why: The region is a rectangle; Z is largest at (3, 2).
Corners \((0,0), (3,0), (3,2), (0,2)\): \(Z = 0, 6, 16, 10\). Maximum \(= 16\) at \((3, 2)\).
A point \((x, y)\) that satisfies all the constraints of an LPP (including \(x, y \ge 0\)) is called
Why: Feasible = satisfies every constraint; optimal = feasible and best.
Every point of the feasible region is a feasible solution. An optimal solution is a feasible solution that gives the best value of \(Z\); a corner point is a vertex of the region.
A tailor has \(30\) m of cloth. A shirt needs \(2\) m and a pair of trousers needs \(3\) m. If he makes \(x\) shirts and \(y\) pairs of trousers, the cloth constraint is
Why: Cloth used = 2x + 3y and cannot exceed 30.
Cloth used \(= 2x + 3y\) metres, which cannot exceed the \(30\) m available: \(2x + 3y \le 30\).
The corner points of the feasible region of an LPP are \((0, 2)\), \((3, 0)\), \((6, 0)\), \((6, 8)\) and \((0, 5)\). The minimum value of \(Z = 4x + 6y\) is
Why: Z values: 12, 12, 24, 72, 30; the minimum 12 occurs at two corners.
\(Z(0,2) = 12\), \(Z(3,0) = 12\), \(Z(6,0) = 24\), \(Z(6,8) = 72\), \(Z(0,5) = 30\). Minimum \(= 12\), attained at both \((0,2)\) and \((3,0)\), hence at every point of the segment joining them. (\((0,0)\) is not a corner of this region.)
The corner points of a feasible region are \((0, 0)\), \((4, 0)\), \((3, 3)\) and \((0, 5)\). For which objective function is the maximum attained at \((3, 3)\)?
Why: x + y gives 0, 4, 6, 5 — largest at (3, 3).
\(Z = x + y\): \(0, 4, 6, 5\), maximum at \((3,3)\). \(Z = x\) is largest at \((4,0)\); \(Z = y\) at \((0,5)\); \(Z = x - y\) at \((4,0)\).
For the constraints \(x + y \ge 2\), \(x \ge 0\), \(y \ge 0\), the objective function \(Z = x + y\)
Why: The region is unbounded upwards and Z grows without limit along it.
The feasible region is unbounded; e.g. \((n, 0)\) is feasible for every \(n \ge 2\) with \(Z = n\). So \(Z\) has no maximum (its minimum is \(2\), on the segment from \((2,0)\) to \((0,2)\)).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A craftsman makes wooden toys and jewellery boxes. A toy takes \(2\) hours of carving and \(1\) kg of wood; a box takes \(1\) hour and \(2\) kg. Each week he has at most \(40\) hours and \(50\) kg of wood. He earns ₹\(30\) profit on a toy and ₹\(20\) on a box. Let him make \(x\) toys and \(y\) boxes a week.
Ananya plans \(x\) hours of mathematics and \(y\) hours of physics revision a week. She can study at most \(12\) hours in all, wants at most \(8\) hours of mathematics and at least \(2\) hours of physics. Her teacher estimates the gain in test score as \(Z = 5x + 4y\).
This free set is separate from the chapter's question bank. On the Linear Programming chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 37 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.