On solving the inequalities $2x+5y ≤ 20, 3x+2y ≤ 12, x ≥ 0, y ≥ 0$, we get the following situation

Option a: (0, 0), (0, 4), (4, 0) and (20/11, 36/11). Solution: Let's find the corner points (vertices) of the bounded feasible region defined by the inequalities: 1) $2x + 5y ≤ 20 2) $3x + 2y ≤ 12 3) x ≥ 0 , y ≥ 0 First, let's find the intercepts of the boundary lines on the axes: - For the line $2x + 5y = 20 : intercepts are (10, 0) and (0, 4) . - For the line $3x + 2y = 12 : intercepts are (4, 0) and (0, 6) . Next, we find the intersection of the two boundary lines: Multiply the first equation by 2 and the second by 5 : 4x + 10y = 40 (1) 15x + 10y = 60 (2) Subtracting (1) from (2): 11x = 20 x = 20/11 Substitute x = 20/11 into $3x + 2y = 12 : 3(20/11) + 2y = 12 60/11 + 2y = 12 2y = 12 - 60/11 = 72/11 y = 36/11 So the intersection point is (20/11, 36/11) . Now we identify the corner points forming the boundary of the feasible region in the first quadrant: - Origin: (0, 0) - y -axis boundary: (0, 4) (since it lies inside $3x + 2y ≤ 12 ) - x -axis boundary: (4, 0) (since it lies inside $2x + 5y ≤ 20 ) - Intersection point: (20/11, 36/11) Therefore, the corner points are (0, 0) , (0, 4) , (4, 0) , and (20/11, 36/11)$. This matches Option A. Hence, Option A is the correct answer.

On solving the inequalities $2x+5y ≤ 20, 3x+2y ≤ 12, x ≥ 0, y ≥ 0$, we get the following situation

(0, 0), (0, 4), (4, 0) and (20/11, 36/11)

(0, 0), (10, 0), (0, 6) and (20/11, 36/11)

(0, 0), (0, 4), (4, 0) and (2, 3)

(0, 0), (10, 0), (0, 6) and (2, 3)

On solving the inequalities $2x+5y ≤ 20, 3x+2y ≤ 12, x ≥ 0, y ≥ 0$, we get the following situation

The correct answer is Option a: (0, 0), (0, 4), (4, 0) and (20/11, 36/11). Let's find the corner points (vertices) of the bounded feasible region defined by the inequalities: 1) $2x + 5y ≤ 20 2) $3x + 2y ≤ 12 3) x ≥ 0 , y ≥ 0 First, let's find the intercepts of the boundary lines on the axes: - For the line $2x + 5y = 20 : intercepts are (10, 0) and (0, 4) . - For the line $3x + 2y = 12 : intercepts are (4, 0) and (0, 6) . Next, we find the intersection of the two boundary lines: Multiply the first equation by 2 and the second by 5 : 4x + 10y = 40 (1) 15x + 10y = 60 (2) Subtracting (1) from (2): 11x = 20 x = 20/11 Substitute x = 20/11 into $3x + 2y = 12 : 3(20/11) + 2y = 12 60/11 + 2y = 12 2y = 12 - 60/11 = 72/11 y = 36/11 So the intersection point is (20/11, 36/11) . Now we identify the corner points forming the boundary of the feasible region in the first quadrant: - Origin: (0, 0) - y -axis boundary: (0, 4) (since it lies inside $3x + 2y ≤ 12 ) - x -axis boundary: (4, 0) (since it lies inside $2x + 5y ≤ 20 ) - Intersection point: (20/11, 36/11) Therefore, the corner points are (0, 0) , (0, 4) , (4, 0) , and (20/11, 36/11)$. This matches Option A. Hence, Option A is the correct answer.

Linear InequalitiesMCQMTP Mar 20, ICAI SMQuestion 1145 of 146

All Questions

On solving the inequalities $2x+5y $\displaystyle \le$ 20, 3x+2y $\displaystyle \le$ 12, x $\displaystyle \ge$ 0, y $\displaystyle \ge$ 0$, we get the following situation

Select Your Answer (Keyboard: A, B, C, D)

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

Correct Answer

✅ Option a — $\displaystyle (0, 0), (0, 4), (4, 0)$

All Options:

A $\displaystyle (0, 0), (0, 4), (4, 0)$
B $\displaystyle (0, 0), (10, 0), (0, 6)$
C $\displaystyle (0, 0), (0, 4), (4, 0)$
D $\displaystyle (0, 0), (10, 0), (0, 6)$

Detailed Solution & Explanation

Let's find the corner points (vertices) of the bounded feasible region defined by the inequalities:
1) $2x + 5y

\displaystyle \le

\displaystyle \le

…" style="color:#cc0000">\displaystyle <br>2) $3x + 2y

\displaystyle \le

12
3)

\displaystyle x

\displaystyle y

First, let's find the intercepts of the boundary lines on the axes:
- For the line $2x + 5y = 20$: intercepts are $(10, 0)

\displaystyle and

(0, 4)

\displaystyle .<br>- For the line &#36;3x + 2y = 12

: intercepts are

\displaystyle (4, 0)

and

\displaystyle (0, 6)

.

Next, we find the intersection of the two boundary lines:
Multiply the first equation by

\displaystyle 2

and the second by

\displaystyle 5

4x + 10y = 40 \quad (1)

15x + 10y = 60 \quad (2)

Subtracting (1) from (2):

11x = 20 \implies x =

Substitute

\displaystyle x =

into $3x + 2y = 12

\displaystyle :<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>3</mn><mrow><mo fence="true">(</mo><mfrac><mn>20</mn><mn>11</mn></mfrac><mo fence="true">)</mo></mrow><mo>+</mo><mn>2</mn><mi>y</mi><mo>=</mo><mn>12</mn><mtext>  </mtext><mo>⟹</mo><mtext>  </mtext><mfrac><mn>60</mn><mn>11</mn></mfrac><mo>+</mo><mn>2</mn><mi>y</mi><mo>=</mo><mn>12</mn></mrow><annotation encoding="application/x-tex">3\left(

\left(

\displaystyle \frac{20}{11}

\displaystyle \frac{36}{11}

\right)

\displaystyle .<br><br>Now we identify the corner points forming the boundary of the feasible region in the first quadrant:<br>- Origin:

(0, 0)

\displaystyle <br>-

\displaystyle -axis boundary:

(0, 4)

\displaystyle \le

…" style="color:#cc0000">\displaystyle (since it lies inside $3x + 2y

\displaystyle \le

12)
-

\displaystyle x

-axis boundary:

\displaystyle (4, 0)

(since it lies inside $2x + 5y

\displaystyle \le

\displaystyle )<br>- Intersection point:

\left(

\displaystyle \frac{20}{11}

\displaystyle \frac{36}{11}

\right)

\displaystyle <br><br>Therefore, the corner points are

(0, 0)

\displaystyle ,

(0, 4)

\displaystyle ,

(4, 0)

\displaystyle , and

\left(

\displaystyle \frac{20}{11}

\displaystyle \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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On solving the inequalities $2x+5y ≤\displaystyle \le≤ 20, 3x+2y ≤\displaystyle \le≤ 12, x ≥\displaystyle \ge≥ 0, y ≥\displaystyle \ge≥ 0$, we get the following situation

Correct Answer

All Options:

Detailed Solution & Explanation

About This Chapter: Linear Inequalities

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On solving the inequalities $2x+5y $\displaystyle \le$ 20, 3x+2y $\displaystyle \le$ 12, x $\displaystyle \ge$ 0, y $\displaystyle \ge$ 0$, we get the following situation