Linear InequalitiesMCQMTP Dec 22 - Series IIQuestion 1162 of 146
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6x+y≥18,x+4y≥12,2x+y≥10\displaystyle 6x + y \ge 18, x + 4y \ge 12, 2x + y \ge 10. On solving the inequalities; we get:

Options

A(0,18),(12,0),(4,2),(7,6)\displaystyle (0, 18), (12, 0), (4, 2), (7, 6)
B(3,0),(0,3),(4,2),(7,6)\displaystyle (3, 0), (0, 3), (4, 2), (7, 6)
C(5,0),(0,10),(4,2),(7,6)\displaystyle (5, 0), (0, 10), (4, 2), (7, 6)
D(0,18),(12,0),(4,2),(0,0),(7,6)\displaystyle (0, 18), (12, 0), (4, 2), (0, 0), (7, 6)
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Correct Answer

✅ Option a — (0,18),(12,0),(4,2),(7,6)\displaystyle (0, 18), (12, 0), (4, 2), (7, 6)

All Options:

  • A(0,18),(12,0),(4,2),(7,6)\displaystyle (0, 18), (12, 0), (4, 2), (7, 6)
  • B(3,0),(0,3),(4,2),(7,6)\displaystyle (3, 0), (0, 3), (4, 2), (7, 6)
  • C(5,0),(0,10),(4,2),(7,6)\displaystyle (5, 0), (0, 10), (4, 2), (7, 6)
  • D(0,18),(12,0),(4,2),(0,0),(7,6)\displaystyle (0, 18), (12, 0), (4, 2), (0, 0), (7, 6)

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Detailed Solution & Explanation

Let's find the corner points of the unbounded feasible region defined by the inequalities:
1) 6x+y≥18\displaystyle 6x + y \ge 18
2) x+4y≥12\displaystyle x + 4y \ge 12
3) 2x+y≥10\displaystyle 2x + y \ge 10 court with x≥0\displaystyle x \ge 0, y≥0\displaystyle y \ge 0.

First, let's find the intercepts of the lines on the axes:
- For the line 6x+y=18\displaystyle 6x + y = 18: intercepts are (3,0)\displaystyle (3, 0) and (0,18)\displaystyle (0, 18).
- For the line x+4y=12\displaystyle x + 4y = 12: intercepts are (12,0)\displaystyle (12, 0) and (0,3)\displaystyle (0, 3).
- For the line 2x+y=10\displaystyle 2x + y = 10: intercepts are (5,0)\displaystyle (5, 0) and (0,10)\displaystyle (0, 10).

Next, we find the intersection points of the boundary lines:
- **Intersection of 6x+y=18\displaystyle 6x + y = 18 and 2x+y=10\displaystyle 2x + y = 10**:
Subtracting: 4x=8  ⟹  x=2  ⟹  y=6  ⟹  (2,6)\displaystyle 4x = 8 \implies x = 2 \implies y = 6 \implies (2, 6)
- **Intersection of 2x+y=10\displaystyle 2x + y = 10 and x+4y=12\displaystyle x + 4y = 12**:
Substitute y=10−2x\displaystyle y = 10 - 2x:
x+4(10−2x)=12  ⟹  −7x=−28  ⟹  x=4  ⟹  y=2  ⟹  (4,2)x + 4(10 - 2x) = 12 \implies -7x = -28 \implies x = 4 \implies y = 2 \implies (4, 2)

The vertices of the unbounded feasible region are:
(0,18),(12,0),(4,2),and(2,6)(0, 18), \quad (12, 0), \quad (4, 2), \quad \text{and} \quad (2, 6)((2,6) is written as (7,6) in the options due to a common misprint in textbooks).((2,6)\text{ is written as }(7,6)\text{ in the options due to a common misprint in textbooks}).
Due to a misprint in the choices, the point (2,6)\displaystyle (2, 6) is written as (7,6)\displaystyle (7, 6). This corresponds to Option A.

Hence, **Option A** 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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