CA Foundation P3 · Chapter 3

Linear Inequalities MCQs with Answers

8 multiple-choice questions on Linear Inequalities for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   The solution set of the inequality 4x − 7 < 2x + 5 is:

   • A) x > 6
   • B) x < 1
   • C) x < 12
   • D) x < 6
   Show answer & explanation

   Answer: D) x < 6

   4x − 2x < 5 + 7, so 2x < 12 and x < 6. Dividing by a positive number does not reverse the inequality. Forgetting to divide by 2 gives x < 12.

2. Question 2

   The solution of −3x + 9 ≥ 0 is:

   • A) x ≤ 3
   • B) x ≥ 3
   • C) x ≤ −3
   • D) x ≥ −3
   Show answer & explanation

   Answer: A) x ≤ 3

   −3x ≥ −9. Dividing both sides by −3 reverses the inequality sign, so x ≤ 3. Students who forget to reverse the sign get x ≥ 3.

3. Question 3

   Which of the following points lies in the region defined by 2x + 3y ≤ 12, x ≥ 0, y ≥ 0?

   • A) (2, 3)
   • B) (3, 2)
   • C) (4, 2)
   • D) (0, 5)
   Show answer & explanation

   Answer: B) (3, 2)

   Substitute each point into 2x + 3y: (3, 2) gives 6 + 6 = 12 ≤ 12, so it lies on the boundary and is included. (2, 3) gives 13, (4, 2) gives 14 and (0, 5) gives 15, all greater than 12.

4. Question 4

   Subject to x + y ≤ 6, x ≤ 4, x ≥ 0, y ≥ 0, the maximum value of Z = 5x + 4y is:

   • A) 24
   • B) 20
   • C) 28
   • D) 30
   Show answer & explanation

   Answer: C) 28

   The corner points of the feasible region are (0, 0), (4, 0), (4, 2) and (0, 6). Z takes values 0, 20, 5(4) + 4(2) = 28 and 24 respectively. The maximum is 28 at (4, 2). The value 30 comes from the point (6, 0), which violates x ≤ 4.

5. Question 5

   For the constraints x + 2y ≤ 8, 3x + y ≤ 9, x ≥ 0, y ≥ 0, which of the following is a corner point of the feasible region?

   • A) (2, 3)
   • B) (3, 2)
   • C) (8, 0)
   • D) (0, 9)
   Show answer & explanation

   Answer: A) (2, 3)

   The corner where the two lines meet: from x = 8 − 2y, 3(8 − 2y) + y = 9 gives 24 − 5y = 9, so y = 3 and x = 2. (3, 2) violates 3x + y ≤ 9 (gives 11), (8, 0) violates 3x + y ≤ 9 and (0, 9) violates x + 2y ≤ 8.

6. Question 6

   A student buys x pens at ₹15 each and y notebooks at ₹40 each with a budget of ₹500, and must buy at least 4 notebooks. The maximum number of pens she can buy is:

   • A) 33
   • B) 23
   • C) 20
   • D) 22
   Show answer & explanation

   Answer: D) 22

   Constraints: 15x + 40y ≤ 500, y ≥ 4. Pens are maximised when y takes its minimum value 4, so 15x ≤ 500 − 160 = 340 and x ≤ 22.67. Since x must be a whole number, the maximum is 22 (rounding down). Ignoring the notebook condition gives 500/15 = 33.

7. Question 7

   The graph of the inequality x ≥ 0 on the XY-plane is:

   • A) The half-plane above the x-axis, including the x-axis
   • B) The half-plane to the right of the y-axis, including the y-axis
   • C) The half-plane to the left of the y-axis
   • D) The y-axis only
   Show answer & explanation

   Answer: B) The half-plane to the right of the y-axis, including the y-axis

   x = 0 is the equation of the y-axis. All points with non-negative x-coordinates lie on or to the right of it, so x ≥ 0 represents the right half-plane together with the y-axis. y ≥ 0 would give the upper half-plane.

8. Question 8

   The number of integer values of x satisfying both 2x − 3 > 5 and 3x + 1 ≤ 22 is:

   • A) 4
   • B) 2
   • C) 7
   • D) 3
   Show answer & explanation

   Answer: D) 3

   From 2x − 3 > 5: 2x > 8, so x > 4. From 3x + 1 ≤ 22: 3x ≤ 21, so x ≤ 7. Hence 4 < x ≤ 7 and the integers are 5, 6, 7, i.e. 3 values. Including 4 by mistake (treating > as ≥) gives 4.