Linear InequalitiesPYQ May 25Question 4008 of 73
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On solving the inequalities 6x+y≥18\displaystyle 6x + y \ge 18, x+4y≥12\displaystyle x + 4y \ge 12, 2x+y≥10\displaystyle 2x + y \ge 10; which of the following are correct solutions?

Options

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

✅ Option a — (0, 18), (12, 0), (4, 2) and (2, 6)

All Options:

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

Detailed Solution & Explanation

To find the correct solution set (which represent the corner points or vertices of the feasible region defined by the given system of inequalities), we analyze the boundary lines and their intersections, assuming the non-negativity constraints x≥0\displaystyle x \ge 0 and y≥0\displaystyle y \ge 0.
The given inequalities are:
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

Let's first find the boundary lines and their corresponding intercepts:
- **Boundary Line 1**: 6x+y=18\displaystyle 6x + y = 18
Intercepts: If x=0\displaystyle x = 0, y=18\displaystyle y = 18 giving (0,18)\displaystyle (0, 18). If y=0\displaystyle y = 0, x=3\displaystyle x = 3 giving (3,0)\displaystyle (3, 0).
- **Boundary Line 2**: x+4y=12\displaystyle x + 4y = 12
Intercepts: If x=0\displaystyle x = 0, y=3\displaystyle y = 3 giving (0,3)\displaystyle (0, 3). If y=0\displaystyle y = 0, x=12\displaystyle x = 12 giving (12,0)\displaystyle (12, 0).
- **Boundary Line 3**: 2x+y=10\displaystyle 2x + y = 10
Intercepts: If x=0\displaystyle x = 0, y=10\displaystyle y = 10 giving (0,10)\displaystyle (0, 10). If y=0\displaystyle y = 0, x=5\displaystyle x = 5 giving (5,0)\displaystyle (5, 0).

Next, let's find the intersection points of these lines that satisfy all inequalities:
- **Intersection of Line 1 (6x+y=18\displaystyle 6x + y = 18) and Line 3 (2x+y=10\displaystyle 2x + y = 10)**:
Subtracting the two equations:
(6x+y)−(2x+y)=18−10  ⟹  4x=8  ⟹  x=2(6x + y) - (2x + y) = 18 - 10 \implies 4x = 8 \implies x = 2
Substitute x=2\displaystyle x = 2 into 2x+y=10\displaystyle 2x + y = 10:
2(2)+y=10  ⟹  y=62(2) + y = 10 \implies y = 6
So the intersection point is (2,6)\displaystyle (2, 6). Let's check if it satisfies all three inequalities:
1. 6(2)+6=18≥18\displaystyle 6(2) + 6 = 18 \ge 18 (True)
2. 2+4(6)=26≥12\displaystyle 2 + 4(6) = 26 \ge 12 (True)
3. 2(2)+6=10≥10\displaystyle 2(2) + 6 = 10 \ge 10 (True)
So, (2,6)\displaystyle (2, 6) is a valid corner point.

- **Intersection of Line 2 (x+4y=12\displaystyle x + 4y = 12) and Line 3 (2x+y=10\displaystyle 2x + y = 10)**:
From Line 2, x=12−4y\displaystyle x = 12 - 4y. Substitute this into Line 3:
2(12−4y)+y=102(12 - 4y) + y = 10
24−8y+y=10  ⟹  24−7y=10  ⟹  7y=14  ⟹  y=224 - 8y + y = 10 \implies 24 - 7y = 10 \implies 7y = 14 \implies y = 2
Then, x=12−4(2)=4\displaystyle x = 12 - 4(2) = 4.
So the intersection point is (4,2)\displaystyle (4, 2). Let's check if it satisfies all three inequalities:
1. 6(4)+2=26≥18\displaystyle 6(4) + 2 = 26 \ge 18 (True)
2. 4+4(2)=12≥12\displaystyle 4 + 4(2) = 12 \ge 12 (True)
3. 2(4)+2=10≥10\displaystyle 2(4) + 2 = 10 \ge 10 (True)
So, (4,2)\displaystyle (4, 2) is a valid corner point.

Now, let's check the boundary points on the axes (under x≥0,y≥0\displaystyle x \ge 0, y \ge 0):
- **On the y-axis (x=0\displaystyle x = 0)**:
The inequalities become y≥18\displaystyle y \ge 18, 4y≥12  ⟹  y≥3\displaystyle 4y \ge 12 \implies y \ge 3, and y≥10\displaystyle y \ge 10. The minimum value of y\displaystyle y that satisfies all three is y=18\displaystyle y = 18. This gives the corner point (0,18)\displaystyle (0, 18).
- **On the x-axis (y=0\displaystyle y = 0)**:
The inequalities become 6x≥18  ⟹  x≥3\displaystyle 6x \ge 18 \implies x \ge 3, x≥12\displaystyle x \ge 12, and 2x≥10  ⟹  x≥5\displaystyle 2x \ge 10 \implies x \ge 5. The minimum value of x\displaystyle x that satisfies all three is x=12\displaystyle x = 12. This gives the corner point (12,0)\displaystyle (12, 0).

Therefore, the corner points (vertices) of the feasible region are (0,18)\displaystyle (0, 18), (12,0)\displaystyle (12, 0), (4,2)\displaystyle (4, 2), and (2,6)\displaystyle (2, 6).
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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