Linear InequalitiesMCQMTP Mar 20, ICAI SMQuestion 1145 of 146
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On solving the inequalities 2x+5y20,3x+2y12,x0,y0\displaystyle 2x+5y \le 20, 3x+2y \le 12, x \ge 0, y \ge 0, we get the following situation

Options

A(0,0),(0,4),(4,0) and (20/11,36/11)\displaystyle (0, 0), (0, 4), (4, 0) \text{ and } (20/11, 36/11)
B(0,0),(10,0),(0,6) and (20/11,36/11)\displaystyle (0, 0), (10, 0), (0, 6) \text{ and } (20/11, 36/11)
C(0,0),(0,4),(4,0) and (2,3)\displaystyle (0, 0), (0, 4), (4, 0) \text{ and } (2, 3)
D(0,0),(10,0),(0,6) and (2,3)\displaystyle (0, 0), (10, 0), (0, 6) \text{ and } (2, 3)
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Correct Answer

Option a(0,0),(0,4),(4,0) and (20/11,36/11)\displaystyle (0, 0), (0, 4), (4, 0) \text{ and } (20/11, 36/11)

All Options:

  • A(0,0),(0,4),(4,0) and (20/11,36/11)\displaystyle (0, 0), (0, 4), (4, 0) \text{ and } (20/11, 36/11)
  • B(0,0),(10,0),(0,6) and (20/11,36/11)\displaystyle (0, 0), (10, 0), (0, 6) \text{ and } (20/11, 36/11)
  • C(0,0),(0,4),(4,0) and (2,3)\displaystyle (0, 0), (0, 4), (4, 0) \text{ and } (2, 3)
  • D(0,0),(10,0),(0,6) and (2,3)\displaystyle (0, 0), (10, 0), (0, 6) \text{ and } (2, 3)

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Detailed Solution & Explanation

Let's find the corner points (vertices) of the bounded feasible region defined by the inequalities:
1) 2x+5y20\displaystyle 2x + 5y \le 20
2) 3x+2y12\displaystyle 3x + 2y \le 12
3) x0\displaystyle x \ge 0, y0\displaystyle y \ge 0

First, let's find the intercepts of the boundary lines on the axes:
- For the line 2x+5y=20\displaystyle 2x + 5y = 20: intercepts are (10,0)\displaystyle (10, 0) and (0,4)\displaystyle (0, 4).
- For the line 3x+2y=12\displaystyle 3x + 2y = 12: intercepts are (4,0)\displaystyle (4, 0) and (0,6)\displaystyle (0, 6).

Next, we find the intersection of the two boundary lines:
Multiply the first equation by 2\displaystyle 2 and the second by 5\displaystyle 5:
4x+10y=40(1)4x + 10y = 40 \quad (1)
15x+10y=60(2)15x + 10y = 60 \quad (2)
Subtracting (1) from (2):
11x=20    x=201111x = 20 \implies x = \frac{20}{11}
Substitute x=2011\displaystyle x = \frac{20}{11} into 3x+2y=12\displaystyle 3x + 2y = 12:
3(2011)+2y=12    6011+2y=123\left(\frac{20}{11}\right) + 2y = 12 \implies \frac{60}{11} + 2y = 12
2y=126011=7211    y=36112y = 12 - \frac{60}{11} = \frac{72}{11} \implies y = \frac{36}{11}
So the intersection point is (2011,3611)\displaystyle \left(\frac{20}{11}, \frac{36}{11}\right).

Now we identify the corner points forming the boundary of the feasible region in the first quadrant:
- Origin: (0,0)\displaystyle (0, 0)
- y\displaystyle y-axis boundary: (0,4)\displaystyle (0, 4) (since it lies inside 3x+2y12\displaystyle 3x + 2y \le 12)
- x\displaystyle x-axis boundary: (4,0)\displaystyle (4, 0) (since it lies inside 2x+5y20\displaystyle 2x + 5y \le 20)
- Intersection point: (2011,3611)\displaystyle \left(\frac{20}{11}, \frac{36}{11}\right)

Therefore, the corner points are (0,0)\displaystyle (0, 0), (0,4)\displaystyle (0, 4), (4,0)\displaystyle (4, 0), and (2011,3611)\displaystyle \left(\frac{20}{11}, \frac{36}{11}\right).

This matches 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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