Linear InequalitiesPYQ Jan 26Question 4255 of 73
All Questions

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
For any discrepancies in this question, email contact@cadada.in

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.

More Questions from Linear Inequalities

On solving the inequalities 5x+y≤100\displaystyle 5x + y \le 100, x+y≤60\displaystyle x + y \le 60, x≥0\displaystyle x \ge 0, y≥0\displaystyle y \ge 0, we get the following solution:

An employer recruits experienced (x)\displaystyle (x) and fresh workmen (y)\displaystyle (y) for his under the condition that he cannot employ more than 11\displaystyle 11 people x\displaystyle x and y\displaystyle y can be related by the inequality.

The solution set of the equations x+2>0\displaystyle x + 2 > 0 and 2x−6>0\displaystyle 2x - 6 > 0 is

On solving the inequalities; we get 6x+y≥18\displaystyle 6x + y \ge 18, x+4y≥12\displaystyle x + 4y \ge 12, 2x+y≥10\displaystyle 2x + y \ge 10

Solve for x\displaystyle x of the inequalities 2≤3x−25≤4\displaystyle 2 \le \frac{3x - 2}{5} \le 4 where x∈N\displaystyle x \in N

The common region in the graph of the inequalities x+y≤4\displaystyle x + y \le 4, x−y≤4\displaystyle x - y \le 4, x≥2\displaystyle x \ge 2 is

Ready to Master Linear Inequalities?

Practice all 73 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free