Board Exam-Pattern Questions

How This Bank Works

Note: These are original practice questions written in the CBSE/State Board pattern — matched to the style, difficulty and mark-distribution of recent board papers, not copied from any specific year's paper. Attempt each question on paper before reading the solution.

Linear Programming is the board paper's most predictable chapter: a guaranteed 5-mark graphical problem (draw, tabulate, conclude — sometimes with an unbounded twist or a tie), supported by MCQs and 2-3-mark items on the vocabulary, corner evaluation, and region identification.

Presentation rules that earn marks:

1. Draw a labelled graph — axes named, lines labelled with their equations, feasible region shaded and named. The graph itself carries marks.
2. Tabulate corner points against $Z$-values with an arrow at the optimum. The table format is expected, not optional.
3. State boundedness explicitly ("the feasible region is bounded, so by Theorem 2…") — and for unbounded regions, write the half-plane test before concluding.
4. Conclude in a sentence: "the maximum value of $Z$ is … attained at …" — mirroring the question's wording.

2-Mark Questions

Q1. Write the corner points of the feasible region determined by $x + y \leq 3$, $x \geq 0$, $y \geq 0$.

Solution:

1. The region is the triangle cut from the first quadrant by the line $x + y = 3$.

Answer: $(0, 0)$, $(3, 0)$ and $(0, 3)$.

Q2. The corner points of a feasible region are $(0, 0)$, $(2, 1)$ and $(0, 3)$. Find the maximum of $Z = 7x + 3y$.

Solution:

1. Evaluate: $Z = 0$; $Z = 14 + 3 = 17$; $Z = 9$.

Answer: maximum $Z = 17$ at $(2, 1)$.

Q3. State the theorem that justifies checking only the corner points of a feasible region.

Solution:

1. Theorem 1: when the objective function $Z = ax + by$ has an optimal value over the feasible region, that value must occur at a corner point (vertex) of the region.

Answer: as stated — with the companion fact (Theorem 2) that a bounded region guarantees both a maximum and a minimum exist at corners.

Q4. Is the region determined by $x \geq 2$, $y \geq 3$ bounded or unbounded? Justify.

Solution:

1. Test with a circle: points like $(100, 100)$ satisfy both constraints, so no circle can enclose the region.

Answer: unbounded — it extends indefinitely up and to the right from the corner $(2, 3)$.

3-Mark Questions

Q5. Maximise $Z = 4x + 3y$ subject to $2x + y \leq 10$, $x + 3y \leq 15$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(5,0)$, $(3,4)$ (solving $2x + y = 10$ with $x + 3y = 15$), $(0,5)$.
2. Table: $Z = 0, 20, 24, 15$.

Answer: maximum $Z = 24$ at $(3, 4)$.

Q6. Minimise $Z = 2x + 3y$ subject to $x + y \geq 6$, $2x + y \geq 8$, $x, y \geq 0$.

Solution:

1. Corners of the unbounded region: $(0, 8)$, $(2, 4)$ (solving the lines), $(6, 0)$.
2. Table: $24, 16, 12$; smallest $m = 12$.
3. Half-plane test: $2x + 3y < 12$ shares no point with the region (on the $x$-axis the region needs $x \geq 6$, where $2x \geq 12$; adding $y$ only raises $Z$).

Answer: minimum $Z = 12$ at $(6, 0)$ — note the winner is an intercept corner, not the interior intersection; the table must contain all three.

Q7. Maximise $Z = x + y$ subject to $x + y \leq 5$, $x, y \geq 0$, and state the number of optimal solutions.

Solution:

1. Corners: $(0,0), (5,0), (0,5)$ with $Z = 0, 5, 5$.
2. The tie: the objective $x + y$ is parallel to the boundary $x + y = 5$, so the maximum is shared along the whole edge.

Answer: maximum $Z = 5$ at every point of the segment joining $(5, 0)$ and $(0, 5)$ — infinitely many optimal solutions.

Q8. Maximise $Z = 5x + 2y$ subject to $x - y \leq 2$, $3x + 2y \leq 12$, $x, y \geq 0$.

Solution:

1. Corners: $(0,0)$, $(2,0)$, $\left(\dfrac{16}{5}, \dfrac{6}{5}\right)$ (solving $x - y = 2$ with $3x + 2y = 12$), $(0,6)$.
2. Table: $0, 10, \dfrac{92}{5}, 12$.

Answer: maximum $Z = \dfrac{92}{5}$ at $\left(\dfrac{16}{5}, \dfrac{6}{5}\right)$ — fractional corners must be kept as exact fractions in the table.

5-Mark Questions

Q9. Solve graphically: maximise $Z = 22x + 18y$ subject to $x + y \leq 20$, $3x + 2y \leq 48$, $x, y \geq 0$.

Solution:

1. Draw both lines and shade the common region — bounded, with corners $(0,0)$, $(16,0)$, $(8,12)$, $(0,20)$; the interior corner solves $x + y = 20$ with $3x + 2y = 48$.
2. Table:

3. Bounded region — Theorem 2 applies directly.

Answer: maximum $Z = 392$ at $(8, 12)$.

Q10. Solve graphically: minimise $Z = x + y$ subject to $3x + 2y \geq 12$, $x + 3y \geq 11$, $x, y \geq 0$.

Solution:

1. Corners of the unbounded region: $(0, 6)$, $(2, 3)$ (solving the lines), $(11, 0)$.
2. Table: $6, 5, 11$; smallest $m = 5$.
3. Half-plane test: $x + y < 5$ has no point in common with the feasible region.

Answer: minimum $Z = 5$ at $(2, 3)$.

Q11. Show that $Z = x + y$ has no maximum subject to $x - y \leq 1$, $x + y \geq 3$, $x, y \geq 0$.

Solution:

1. Corners: $(0, 3)$ and $(2, 1)$, both giving $Z = 3$; largest table value $M = 3$.
2. The region is unbounded (it extends without limit above both lines).
3. Half-plane test: $x + y > 3$ plainly shares points with the region — e.g. $(0, 10)$ satisfies every constraint and gives $Z = 10$.

Answer: since the open half-plane $x + y > 3$ meets the feasible region, $Z$ exceeds every bound — no maximum exists. The table's $M$ was a false summit; only the test reveals it.

Q12. Solve graphically: maximise $Z = 4x + 6y$ subject to $x + y \leq 8$, $2x + 3y \leq 21$, $x, y \geq 0$, and describe all optimal solutions.

Solution:

1. Corners: $(0,0)$, $(8,0)$, $(3,5)$ (solving the lines), $(0,7)$.
2. Table: $0, 32, 42, 42$ — a tie at $(3,5)$ and $(0,7)$.
3. Why the tie: $Z = 4x + 6y = 2(2x + 3y)$ is proportional to the second constraint, so the objective line is parallel to that edge.

Answer: maximum $Z = 42$ at every point of the segment joining $(3, 5)$ and $(0, 7)$ — the complete description of the optimal set is required for full marks.