Linear InequalitiesMCQMTP May 19Question 1141 of 146
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On solving the inequalities 5x+y100,x+y60,x0\displaystyle 5x+y \le 100, x+y \le 60, x \ge 0 and y0\displaystyle y \ge 0, we get the following situation

Options

A(0,0),(20,0),(10,50) and (0,60)\displaystyle (0, 0), (20, 0), (10, 50) \text{ and } (0, 60)
B(0,0),(60,0),(10,50) and (0,60)\displaystyle (0, 0), (60, 0), (10, 50) \text{ and } (0, 60)
C(0,0),(20,0),(0,100) and (10,50)\displaystyle (0, 0), (20, 0), (0, 100) \text{ and } (10, 50)
Dnone of these
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Correct Answer

Option a(0,0),(20,0),(10,50) and (0,60)\displaystyle (0, 0), (20, 0), (10, 50) \text{ and } (0, 60)

All Options:

  • A(0,0),(20,0),(10,50) and (0,60)\displaystyle (0, 0), (20, 0), (10, 50) \text{ and } (0, 60)
  • B(0,0),(60,0),(10,50) and (0,60)\displaystyle (0, 0), (60, 0), (10, 50) \text{ and } (0, 60)
  • C(0,0),(20,0),(0,100) and (10,50)\displaystyle (0, 0), (20, 0), (0, 100) \text{ and } (10, 50)
  • Dnone of these

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Detailed Solution & Explanation

Let's find the corner points (vertices) of the bounded feasible region defined by the inequalities:
1) 5x+y100\displaystyle 5x + y \le 100
2) x+y60\displaystyle x + y \le 60
3) x0\displaystyle x \ge 0, y0\displaystyle y \ge 0

First, let's find the intercepts of the boundary lines on the coordinate axes:
- For the line 5x+y=100\displaystyle 5x + y = 100: intercepts are (20,0)\displaystyle (20, 0) and (0,100)\displaystyle (0, 100).
- For the line x+y=60\displaystyle x + y = 60: intercepts are (60,0)\displaystyle (60, 0) and (0,60)\displaystyle (0, 60).

Next, we find the intersection point of 5x+y=100\displaystyle 5x + y = 100 and x+y=60\displaystyle x + y = 60:
Subtracting the second equation from the first:
(5x+y)(x+y)=10060    4x=40    x=10(5x + y) - (x + y) = 100 - 60 \implies 4x = 40 \implies x = 10
Substituting x=10\displaystyle x = 10 into x+y=60\displaystyle x + y = 60:
10+y=60    y=5010 + y = 60 \implies y = 50
So the intersection point is (10,50)\displaystyle (10, 50).

The boundary points defining the feasible region in the first quadrant are:
- Origin: (0,0)\displaystyle (0, 0)
- x\displaystyle x-axis boundary: (20,0)\displaystyle (20, 0) (since it lies inside x+y60\displaystyle x + y \le 60)
- Intersection point: (10,50)\displaystyle (10, 50)
- y\displaystyle y-axis boundary: (0,60)\displaystyle (0, 60) (since it lies inside 5x+y100\displaystyle 5x + y \le 100)

Therefore, the corner points are (0,0)\displaystyle (0, 0), (20,0)\displaystyle (20, 0), (10,50)\displaystyle (10, 50), and (0,60)\displaystyle (0, 60).

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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