Linear InequalitiesPYQ Jan 26Question 4555 of 73
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The common region represented by inequalities: 2x+y≥8,x+y≥12,3x+2y≤34,x≥0\displaystyle 2x + y \ge 8, x + y \ge 12, 3x + 2y \le 34, x \ge 0 and y≥0\displaystyle y \ge 0 is

Options

AUnbounded
BFeasible unbounded
CInfeasible bounded
DFeasible bounded
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Correct Answer

✅ Option d — Feasible bounded

All Options:

  • AUnbounded
  • BFeasible unbounded
  • CInfeasible bounded
  • DFeasible bounded

Detailed Solution & Explanation

Let us analyze the system of inequalities in the first quadrant (x≥0,y≥0\displaystyle x \ge 0, y \ge 0): 1) 2x+y≥8\displaystyle 2x + y \ge 8 2) x+y≥12\displaystyle x + y \ge 12 3) 3x+2y≤34\displaystyle 3x + 2y \le 34
Let us find the boundary lines and their key intersection points: - Line 1 (L1\displaystyle L_1): 2x+y=8\displaystyle 2x + y = 8 - Line 2 (L2\displaystyle L_2): x+y=12\displaystyle x + y = 12 - Line 3 (L3\displaystyle L_3): 3x+2y=34\displaystyle 3x + 2y = 34
Since x≥0\displaystyle x \ge 0 and y≥0\displaystyle y \ge 0, any point satisfying x+y≥12\displaystyle x + y \ge 12 also satisfies: 2x+y=x+(x+y)≥x+12≥12>82x + y = x + (x + y) \ge x + 12 \ge 12 > 8 Thus, the inequality 2x+y≥8\displaystyle 2x + y \ge 8 is redundant under the condition x+y≥12\displaystyle x + y \ge 12 in the first quadrant.
So, the effective region is bounded by: x+y≥12x + y \ge 12 3x+2y≤343x + 2y \le 34 x≥0,y≥0x \ge 0, y \ge 0
Let us find the boundary intersection points: - On the y-axis (x=0\displaystyle x = 0): - From x+y≥12\displaystyle x + y \ge 12, we get y≥12\displaystyle y \ge 12. The lower vertex on the y-axis is (0,12)\displaystyle (0, 12). - From 3x+2y≤34\displaystyle 3x + 2y \le 34, we get 2y≤34  ⟹  y≤17\displaystyle 2y \le 34 \implies y \le 17. The upper vertex on the y-axis is (0,17)\displaystyle (0, 17). - Intersection of L2\displaystyle L_2 (x+y=12\displaystyle x+y=12) and L3\displaystyle L_3 (3x+2y=34\displaystyle 3x+2y=34): Multiply x+y=12\displaystyle x+y=12 by 2: 2x+2y=242x + 2y = 24 Subtract this from 3x+2y=34\displaystyle 3x + 2y = 34: (3x+2y)−(2x+2y)=34−24  ⟹  x=10(3x + 2y) - (2x + 2y) = 34 - 24 \implies x = 10 Substituting x=10\displaystyle x = 10 in x+y=12\displaystyle x+y=12: 10+y=12  ⟹  y=210 + y = 12 \implies y = 2 So, the third vertex of the region is (10,2)\displaystyle (10, 2).
The feasible region is the triangle formed by the vertices (0,12)\displaystyle (0, 12), (0,17)\displaystyle (0, 17), and (10,2)\displaystyle (10, 2). Since all three vertices are finite and the region is closed, the common region is both feasible and bounded (Feasible bounded). Hence, **Option D** is the correct answer.

About This Chapter: Linear Inequalities

Paper

Paper 3: Quantitative Aptitude

Weightage

1-3 Marks

Key Topics

Linear Inequalities in one & two variables

This chapter covers Linear Inequalities in one & two variables and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 1-3 Marks weightage. Focus on understanding core concepts rather than memorizing.

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