Linear Inequalities — Class 11 MCQs with Answers

Linear Inequalities — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Solve 2x > 6:

   Ax > 3Bx < 3Cx = 3Dx ≥ 3
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   Answer: x > 3

   Explanation: Dividing both sides by the positive number 2 keeps the direction: 2x > 6 → x > 3. The sign only flips when you divide by a negative number, not here.

2. Q2 of 6Easy

   Solve x + 3 > 5:

   Ax ≥ 2Bx > 2Cx < 2Dx = 2
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   Answer: x > 2

   Explanation: Subtract 3 from both sides of x + 3 > 5 to get x > 2 — a strict inequality, since no equality was introduced, so x = 2 itself is not included.

3. Q3 of 6Medium

   Boundary of x − y ≤ 0 passes through origin and has slope:

   A1B−1C0D∞
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   Answer: 1

   Explanation: The boundary of x − y ≤ 0 is the line x − y = 0, i.e. y = x — it passes through the origin with slope 1, the ratio of rise to run being 1:1.

4. Q4 of 6Medium

   Assertion (A): When x is a natural number, the solution set of 30x < 200 is {1, 2, 3, 4, 5, 6}. Reason (R): Dividing both sides of 30x < 200 by 30 gives x > 20/3.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is true but R is false

   Explanation: A is true: dividing by 30 gives x < 20/3, which is about 6.67, so the natural numbers that work are 1 to 6. R is false: 30 is positive, so the sign stays the same and the result is x < 20/3.

5. Q5 of 6Hard

   Number of solutions to: x ≥ 0, y ≥ 0, x + y ≤ 4 with x, y integers:

   A15B16C10D9
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   Answer: 15

   Explanation: Count integer pairs with x,y ≥ 0 and x+y ≤ 4 row by row: 5 pairs when x=0, 4 when x=1, down to 1 when x=4 — total 5+4+3+2+1 = 15.

6. Q6 of 6Hard

   Assertion (A): A student scored 84, 91, 88 and 95 in four examinations of 100 marks each. To get an average of at least 90 over five examinations, the student needs at least 92 marks in the fifth. Reason (R): An average of at least 90 over five examinations means that the total of the five marks must be at least 450.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: A is true: 84 + 91 + 88 + 95 = 358, and 358 + x ≥ 450 gives x ≥ 92. R is true, and it is the step that turns the average condition into 358 + x ≥ 450, so R explains A.

Linear Inequalities notes: key concepts (Class 11 Mathematics)

Inequalities describe ranges rather than single values. This chapter solves linear inequalities in one variable algebraically and marks the answers on the number line, then moves to two variables where each inequality defines a half-plane. Systems of inequalities give a feasible region — the foundation of linear programming.

Linear inequality

A statement like ax+b<0 (or ≤, >, ≥). Unlike an equation, its solution is a range of values — an interval — rather than a single number.

Rule of signs

Adding or subtracting the same quantity keeps the sign; multiplying or dividing both sides by a NEGATIVE number reverses the inequality.

Strict vs non-strict

< and > are strict and exclude the boundary; ≤ and ≥ include it. This decides open versus closed endpoints on the number line.

Solution on the number line

The solution set of a one-variable inequality is an interval, drawn with an open circle for strict and a filled circle for inclusive endpoints.

Interval notation

Answers are written as intervals: (a,b) excludes the ends, [a,b] includes them, and ∞ or −∞ is always paired with a round bracket.

Inequalities with fractions

Multiply out by a denominator only when its sign is known; if the sign is unknown, the inequality direction cannot be fixed safely.

Two-variable inequality

ax+by≤c represents a half-plane. The line ax+by=c is the boundary, drawn solid for ≤, ≥ and dashed for <, >.

Deciding the shaded side

Substitute a test point (usually the origin) into the inequality; if it holds, shade that side, otherwise shade the opposite side.

Graphical solution

Plot each boundary line, shade the correct half-plane for every inequality, and the overlap of all the shadings is the solution region.

Systems of inequalities

Several inequalities considered together; their common region is the intersection of the individual half-planes, bounded by the relevant lines.

Feasible region

The set of points satisfying every constraint at once. In linear programming an optimum is sought at its corner (vertex) points.

Word problems

Real constraints — cost, time, resources — translate into linear inequalities; non-negativity such as x≥0, y≥0 is often implied.