68 free MCQs on Linear Inequalities with worked answers and explanations. Solving linear inequalities in one and two variables algebraically and graphically, and representing solutions on a number line or as a region in a plane.
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One-variable inequalities use open (excluded) or closed (included) circles on a number line; two-variable inequalities use a dashed (strict) or solid (non-strict) boundary line, with the solution being the entire shaded half-plane on one side of it.
Q1.
Which symbol represents a strict inequality?
Answer: C. strictly less than (<)
Why: Strict inequalities use < or > only, since the boundary value itself is excluded from the solution.
Q2.
Solve for x: 3x - 5 < 7
Answer: A. x < 4
Why: 3x < 12, so x < 4 by dividing both sides by the positive number 3.
Q3.
If x > 5, what happens to the inequality if both sides are multiplied by -1?
Answer: B. It reverses: -x < -5
Why: Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign.
Q4.
On a number line, how is the boundary point of x > 3 represented?
Answer: B. An open (hollow) circle at 3
Why: Since the inequality is strict (x > 3), the point 3 itself is not included, shown with an open circle.
Q5.
Solve: 5x - 3 > 3x + 1
Answer: A. x > 2
Why: 5x - 3x > 1 + 3, so 2x > 4, giving x > 2.
Q6.
When graphing 2x + 3y <= 12, what kind of boundary line is drawn?
Answer: B. A solid line
Why: Since the inequality includes equality (<=), the boundary line is solid, meaning points on the line are included.
Q7.
To decide which side of a boundary line to shade, what is commonly used?
Answer: B. A test point such as the origin
Why: A convenient test point (often the origin, if not on the line) is substituted into the inequality to see which half-plane satisfies it.
Q8.
Which of these is a linear inequality in two variables?
Answer: B. 3x + 2y > 6
Why: 3x + 2y > 6 has both variables to the first power only, making it linear.
Q9.
The solution region of a linear inequality in two variables is called a:
Answer: B. Half-plane
Why: A linear inequality in two variables divides the plane into two half-planes; the solution set is one of these half-planes.
Q10.
Solve: -3x + 12 >= 0
Answer: B. x <= 4
Why: -3x >= -12. Dividing by -3 flips the sign: x <= 4.
Q11.
If the origin (0,0) satisfies the inequality 4x + 3y <= 12, which side is shaded?
Answer: B. The side containing the origin
Why: Since the test point (0,0) satisfies the inequality (0 <= 12), the half-plane containing the origin is shaded.
Q12.
The solution of the inequality x + 3 > 5 is:
Answer: A. x > 2
Why: Subtracting 3 from both sides gives x > 2.
Q13.
When both sides of an inequality are multiplied or divided by a negative number, the inequality sign:
Answer: A. reverses
Why: Multiplying or dividing by a negative number reverses the direction of the inequality.
Q14.
The solution of 2x < 10 is:
Answer: A. x < 5
Why: Dividing both sides by 2 gives x < 5.
Q15.
The symbol ≥ means:
Answer: A. greater than or equal to
Why: The symbol ≥ denotes "greater than or equal to".
Q16.
Which value satisfies the inequality x > 3?
Answer: A. 4
Why: Only 4 is strictly greater than 3.
Q17.
The solution of x − 2 ≤ 0 is:
Answer: A. x ≤ 2
Why: Adding 2 to both sides gives x ≤ 2.
Q18.
On a number line, the graph of x > 2 uses an open circle at 2 and shading to the:
Answer: A. right
Why: x > 2 means all values greater than 2, so the shading goes to the right.
Q19.
The solution of the inequality 3x ≥ 9 is:
Answer: A. x ≥ 3
Why: Dividing both sides by 3 gives x ≥ 3.
Q20.
A closed (filled) circle on a number line shows that the endpoint is:
Answer: A. included
Why: A filled circle means the endpoint value is part of the solution set (≤ or ≥).
Q21.
Solve: 2(x - 3) <= 3x - 2
Answer: A. x >= -4
Why: 2x - 6 <= 3x - 2, so -6 + 2 <= 3x - 2x, giving -4 <= x, i.e. x >= -4.
Q22.
Find the solution set of -5 <= 2x - 3 < 7.
Answer: A. -1 <= x < 5
Why: Add 3 to all parts: -2 <= 2x < 10. Divide by 2: -1 <= x < 5.
Q23.
For the inequality x + y <= 5 with x >= 0, y >= 0, what shape is the feasible region?
Answer: B. A triangle with vertices (0,0), (5,0), (0,5)
Why: The constraints x>=0, y>=0, x+y<=5 bound a triangular region with corners at the origin and the two intercepts (5,0) and (0,5).
Q24.
Solve the system graphically: x + y >= 4 and x + y <= 8 (with x, y >= 0). The feasible region lies:
Answer: B. Between the two parallel boundary lines, in the first quadrant
Why: The region satisfying both x+y>=4 and x+y<=8 in the first quadrant is the band between the two parallel lines x+y=4 and x+y=8.
Q25.
A student scored 62 and 48 in two tests. What minimum score in a third test (out of 100) gives an average of at least 60?
Answer: A. 70
Why: Let third score be x. (62+48+x)/3 >= 60, so 110+x >= 180, x >= 70.
Q26.
Which point satisfies both x - y <= 2 and x + y <= 6 (along with x,y >= 0)?
Answer: B. (2, 2)
Why: At (2,2): x-y=0<=2 true, x+y=4<=6 true. Both satisfied. Check (5,5): x+y=10>6 fails.
Q27.
Solve: |2x - 3| < 5 is equivalent to which compound inequality (ignoring absolute value notation)?
Answer: A. -5 < 2x - 3 < 5
Why: An inequality of the form |A| < k (k > 0) is equivalent to -k < A < k, here giving -5 < 2x-3 < 5, i.e. -1 < x < 4.
Q28.
The graph of 4x + 3y <= 12 in the first quadrant is bounded by the line through which two intercepts?
Answer: A. (3,0) and (0,4)
Why: Setting y=0 gives x=3; setting x=0 gives y=4. The intercepts are (3,0) and (0,4).
Q29.
Solve: 7x + 3 < 5x + 9
Answer: A. x < 3
Why: 7x - 5x < 9 - 3, so 2x < 6, giving x < 3.
Q30.
Solve the inequality: (x-2)/3 >= (x-4)/5, for real x.
Answer: A. x >= -1
Why: Multiplying both sides by 15 gives 5(x-2) >= 3(x-4), i.e. 5x - 10 >= 3x - 12, so 2x >= -2, giving x >= -1.
Q31.
The solution of the inequality −3x > 6 is:
Answer: A. x < −2
Why: Dividing by −3 reverses the sign, giving x < −2.
Q32.
The solution of 2x + 1 ≤ 7 is:
Answer: A. x ≤ 3
Why: Subtract 1 then divide by 2: 2x ≤ 6, so x ≤ 3.
Q33.
The solution of the inequality |x| < 3 is:
Answer: A. −3 < x < 3
Why: |x| < 3 means x lies within 3 of zero, i.e. −3 < x < 3.
Q34.
The solution of the inequality |x| > 2 is:
Answer: A. x > 2 or x < −2
Why: |x| > 2 means x is more than 2 from zero, so x > 2 or x < −2.
Q35.
The solution set of 5 − 2x > 1 is:
Answer: A. x < 2
Why: −2x > −4; dividing by −2 reverses the sign to give x < 2.
Q36.
In the coordinate plane, the graph of the inequality x + y ≤ 4 is a:
Answer: A. half-plane
Why: A linear inequality in two variables represents a half-plane bounded by the line x + y = 4.
Q37.
The solution of 3(x − 1) < 6 is:
Answer: A. x < 3
Why: Expand: 3x − 3 < 6, so 3x < 9 and x < 3.
Q38.
The common solution of x > 1 and x < 5 is:
Answer: A. 1 < x < 5
Why: The values satisfying both conditions lie between 1 and 5.
Q39.
The solution of x/2 ≥ 3 is:
Answer: A. x ≥ 6
Why: Multiplying both sides by 2 gives x ≥ 6.
Q40.
For a system of linear inequalities, the solution region is the ___ of the individual regions:
Answer: A. the intersection
Why: A system is satisfied only where all the individual regions overlap, i.e. their intersection.
Q41.
Find all x such that (x-1)(x-2) >= 0 using the sign-rule method for inequalities (not just linear, as an extension exercise).
Answer: A. x <= 1 or x >= 2
Why: Sign-rule method: critical points x=1 and x=2 divide the number line into three intervals. Test each: x<1: (-)(-) =+≥0 ✓; 1<x<2: (+)(-) =-<0 ✗; x>2: (+)(+)=+≥0 ✓. Include endpoints (equality holds). Answer: x ≤ 1 or x ≥ 2
Q42.
A company makes two products. Constraints from a related LP setup give x >= 0, y >= 0, x + 2y <= 10, and 2x + y <= 10. Which of these is NOT a corner point of the feasible region?
Answer: C. (10, 10)
Why: (10,10) violates x + 2y <= 10 since 10 + 20 = 30 is far greater than 10, so it lies well outside the feasible region. The true corners are (0,0), (5,0), (0,5), and the intersection (10/3, 10/3).
Q43.
For what values of k does the system x + y <= k and x - y <= 2 (with x, y >= 0) have a feasible region that includes the point (3, 1)?
Answer: A. k >= 4
Why: At (3,1): x+y=4 and x-y=2. For (3,1) to be feasible under x+y<=k, we need k >= 4 (the second constraint x-y<=2 is already satisfied with equality).
Q44.
Solve for x: (1/(x-2)) > 0, where x is a real number not equal to 2.
Answer: A. x > 2
Why: Since the numerator is 1 (positive constant), the fraction 1/(x-2) is positive iff the denominator (x-2) is positive. So x-2 > 0, which gives x > 2. Check: x=3: 1/1=1>0 ✓; x=1: 1/(-1)=-1<0 ✗. Answer: x > 2
Q45.
Two linear inequalities 3x + 4y <= 24 and x + y <= 7 (with x, y >= 0) are graphed together. What are the coordinates of their boundary lines' intersection point?
Answer: A. (4, 3)
Why: Solve 3x+4y=24 and x+y=7 together: from the second, y=7-x. Substitute: 3x+4(7-x)=24, 3x+28-4x=24, -x=-4, x=4, y=3. Intersection: (4,3).
Q46.
If x is a natural number satisfying 5x - 3 < 3x + 7, how many positive integer solutions are there?
Answer: A. 4
Why: Solve: 5x-3<3x+7 ⟹ 5x-3x<7+3 ⟹ 2x<10 ⟹ x<5. Natural numbers less than 5: {1, 2, 3, 4}. Count = 4. (x=0 excluded as 0 is not a natural number in NCERT convention.) Answer: 4 positive integer solutions
Q47.
A shopkeeper wants the cost price x of an item (in rupees) to satisfy: at least Rs 20 and the selling price (x + 0.25x) to not exceed Rs 100. What is the valid range for x?
Answer: A. 20 <= x <= 80
Why: Two conditions: (1) x ≥ 20. (2) Selling price = x+0.25x = 1.25x ≤ 100, so x ≤ 100/1.25 = 80. Intersection of both: 20 ≤ x ≤ 80. Answer: 20 ≤ x ≤ 80
Q48.
The solution of the system 2x + y > 4 and x - y < 1 includes which of these points?
Answer: A. (2, 3)
Why: At (2,3): 2(2)+3 = 7 > 4 is true, and 2-3 = -1 < 1 is true, so both conditions hold. The other points all fail the first condition, since 2x+y is at most 3 for each of them.
Q49.
For the system 3x + 2y <= 18, x + 2y <= 10, x >= 0, y >= 0, what are the coordinates of the corner point where both boundary lines intersect (away from the axes)?
Answer: A. (4, 3)
Why: Subtracting x + 2y = 10 from 3x + 2y = 18 gives 2x = 8, so x = 4. Then 4 + 2y = 10 gives y = 3. The intersection point is (4, 3).
Q50.
Solve the inequality (2x-1)/(x+3) <= 1, for x not equal to -3.
Answer: B. -3 < x <= 4
Why: Rearranging gives (x-4)/(x+3) <= 0. Testing sign intervals around the critical points x=-3 (excluded) and x=4 shows the expression is negative or zero exactly when -3 < x <= 4.
Q51.
The solution of the inequality (x − 1)(x − 3) < 0 is:
Answer: A. 1 < x < 3
Why: The product is negative only between the roots, so 1 < x < 3.
Q52.
The solution of 2x − 3 < x + 5 is:
Answer: A. x < 8
Why: Subtract x and add 3: x < 8.
Q53.
The solution of the inequality |2x − 1| ≤ 5 is:
Answer: A. −2 ≤ x ≤ 3
Why: −5 ≤ 2x − 1 ≤ 5 gives −4 ≤ 2x ≤ 6, so −2 ≤ x ≤ 3.
Q54.
The number of integer values of x satisfying −2 < x ≤ 3 is:
Answer: A. 5
Why: The integers are −1, 0, 1, 2, 3 - a total of five.
Q55.
The solution region of the system x ≥ 0 and y ≥ 0 lies entirely in the:
Answer: A. first quadrant
Why: Both coordinates non-negative describes the first quadrant.
Q56.
The solution of the inequality (x + 2)/(x − 1) > 0 is:
Answer: A. x < −2 or x > 1
Why: The quotient is positive when both factors share a sign: x < −2 or x > 1.
Q57.
If a < b and c < 0, then multiplying the inequality by c gives:
Answer: A. ac > bc
Why: Multiplying by a negative number reverses the inequality, so ac > bc.
Q58.
The solution of the compound inequality 4 ≤ 2x < 10 is:
Answer: A. 2 ≤ x < 5
Why: Dividing every part by 2 gives 2 ≤ x < 5.
Q59.
The greatest integer value of x satisfying 3x + 2 < 14 is:
Answer: A. 3
Why: 3x < 12 gives x < 4, so the greatest integer value is 3.
Q60.
The solution of the inequality x² < 4 is:
Answer: A. −2 < x < 2
Why: x² < 4 means |x| < 2, i.e. −2 < x < 2.
Q61.
The solution set of |2x − 1| < 3 is:
Answer: A. (−1, 2)
Why: −3 < 2x − 1 < 3 gives −2 < 2x < 4, so −1 < x < 2.
Q62.
The solution set of |x − 2| ≥ 5 is:
Answer: B. x ≤ −3 or x ≥ 7
Why: x − 2 ≤ −5 or x − 2 ≥ 5 gives x ≤ −3 or x ≥ 7.
Q63.
The solution set of (x − 1)/(x − 2) > 0 is:
Answer: B. x < 1 or x > 2
Why: The ratio is positive when both factors share sign: x < 1 or x > 2.
Q64.
The number of integer points (x, y) with |x| + |y| ≤ 2 is:
Answer: B. 13
Why: Counting lattice points in the diamond gives 1 + 4 + 8 = 13.
Q65.
The solution set of the system 3x − 2 < 2x + 1 and x > 0 is:
Answer: A. (0, 3)
Why: The first gives x < 3; combined with x > 0 the solution is (0, 3).
Q66.
The solution set of x² − 5x + 6 < 0 is:
Answer: A. (2, 3)
Why: (x − 2)(x − 3) < 0 holds between the roots, i.e. 2 < x < 3.
Q67.
The solution set of x² − 5x + 6 > 0 is:
Answer: B. x < 2 or x > 3
Why: (x − 2)(x − 3) > 0 holds outside the roots: x < 2 or x > 3.
Q68.
The solution set of 1/(x − 1) < 2 (x ≠ 1) is:
Answer: A. x < 1 or x > 3/2
Why: For x > 1 the inequality gives x > 3/2; for x < 1 the left side is negative and always less than 2. Solution: x < 1 or x > 3/2.