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5.3 (continued) — Word Problems with Linear Inequalities

🎓 Class 11 Mathematics CBSE Theory Ch 5 — Linear Inequalities ⏱ ~16 min
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This MCQ module is based on: 5.3 (continued) — Word Problems with Linear Inequalities

This mathematics assessment will be based on: 5.3 (continued) — Word Problems with Linear Inequalities
Targeting Class 11 level in Algebra, with Advanced difficulty.

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5.3 (continued) — Word Problems with Linear Inequalities

Many real-life problems in business, medicine, economics and science can be modelled by linear inequalities and then solved algebraically. The key steps are:

Four-step method
Step 1: Identify the unknown and assign a variable.
Step 2: Translate each condition into an inequality.
Step 3: Solve using Rules 1 and 2.
Step 4: Interpret the answer in the original context and check feasibility.

Miscellaneous Examples

Example 9. Solve \(-8 \leq 5x - 3 < 7\).
Add 3 throughout: \(-5 \leq 5x < 10\). Divide by 5: \(-1 \leq x < 2\). Solution set: \([-1, 2)\).
Example 10. Solve \(-5 \leq \frac{5 - 3x}{2} \leq 8\).
Multiply by 2: \(-10 \leq 5 - 3x \leq 16\). Subtract 5: \(-15 \leq -3x \leq 11\). Divide by \(-3\) and reverse: \(5 \geq x \geq -\frac{11}{3}\) i.e. \(-\frac{11}{3} \leq x \leq 5\).
Example 11. Solve the system: \(3x - 7 > 2(x - 6)\) ...(1) and \(6 - x > 11 - 2x\) ...(2), representing the solution on a number line.
From (1): \(3x - 7 > 2x - 12 \Rightarrow x > -5\). From (2): \(-x + 2x > 11 - 6 \Rightarrow x > 5\). Common solution: \(x > 5\) i.e. \((5, \infty)\).
-505 -505 (1) x > -5 (2) x > 5 — common solution
Fig 5.3 — Common solution of a system of inequalities is the intersection \(x>5\).
Example 12. In an experiment, a solution of hydrochloric acid is to be kept between 30° C and 35° C. What is the range of temperature in degree Fahrenheit (°F) if conversion formula is \(C = \frac{5}{9}(F - 32)\)?
Given \(30 < C < 35\). Put \(C = \frac{5}{9}(F-32)\): \(30 < \frac{5}{9}(F-32) < 35 \Rightarrow 54 < F - 32 < 63 \Rightarrow 86 < F < 95\). Range is between 86°F and 95°F.
Example 13. A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added so that acid content in the resulting mixture will be more than 15% but less than 18%?
Let \(x\) litres of 30% solution be added. Total mixture = \((600+x)\) L. Acid content: 15% of \((600+x) < \frac{30x}{100} + \frac{12 \times 600}{100} < 18\%\) of \((600+x)\). That is, \(\frac{15}{100}(600+x) < \frac{30x+7200}{100} < \frac{18}{100}(600+x)\). Simplifying gives \(9000 + 15x < 30x + 7200\) and \(30x + 7200 < 10800 + 18x\). First: \(1800 < 15x \Rightarrow x > 120\). Second: \(12x < 3600 \Rightarrow x < 300\). So 120 L \(< x <\) 300 L.

Miscellaneous Exercise — Selected Questions

Q11. A solution is to be kept between 68°F and 77°F. What is the range in temperature in degrees Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by \(F = \frac{9}{5}C + 32\)?
\(68 < \frac{9}{5}C + 32 < 77 \Rightarrow 36 < \frac{9}{5}C < 45 \Rightarrow 20 < C < 25\). Range: 20°C to 25°C.
Q13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
Let \(x\) L of water be added. Acid remains \(\frac{45}{100}\times 1125 = 506.25\) L. Need: \(\frac{25}{100}(1125+x) < 506.25 < \frac{30}{100}(1125+x)\). Solving: \(562.5 + 0.25x < 506.25 \Rightarrow\) wait, rearrange: for "more than 25%": \(506.25 > \frac{25}{100}(1125+x) \Rightarrow 1125+x < 2025 \Rightarrow x < 900\); for "less than 30%": \(506.25 < \frac{30}{100}(1125+x) \Rightarrow 1125+x > 1687.5 \Rightarrow x > 562.5\). So 562.5 L \(< x <\) 900 L.

5.4 Linear Inequalities in Two Variables & Graphical Solution

Recall that an inequality of the form \(ax + by < c\) (or with \(>, \leq, \geq\)) where \(a, b \neq 0\) is called a linear inequality in two variables?. A pair \((x_0, y_0)\) is a solution if it satisfies the inequality.

Graphical solution
The graph of \(ax + by = c\) is a straight line that divides the Cartesian plane into two half-planes. Any point in one half-plane satisfies \(ax + by < c\); any point in the other satisfies \(ax + by > c\). Points on the line satisfy \(ax + by = c\).
Shading rules
(i) Draw the boundary line \(ax+by=c\). Use a solid line for \(\leq\) or \(\geq\) (points on line included); use a dashed/broken line for strict \(<\) or \(>\).
(ii) Choose a test point not on the line (often the origin \((0,0)\) if it is not on the line). If the test point satisfies the inequality, shade the half-plane containing it; otherwise shade the other half-plane.

Worked Graphs

Example 14. Solve \(3x + 2y > 6\) graphically.
Draw \(3x + 2y = 6\) as a dashed line through \((2,0)\) and \((0,3)\). Test \((0,0)\): \(3(0)+2(0)=0>6\)? No. So shade the half-plane away from origin (upper-right).
(2,0) (0,3) x y 3x+2y > 6
Fig 5.4 — Dashed line \(3x+2y=6\); shaded half-plane represents \(3x+2y>6\).
Example 15. Solve \(3x - 6 \geq 0\) graphically in two-dimensional plane.
\(3x \geq 6 \Rightarrow x \geq 2\). In 2-D plane, draw the vertical solid line \(x=2\). Test \((0,0)\): \(0 \geq 2\)? No. So shade the half-plane to the right of \(x=2\) (including the line).
Example 16. Solve \(y < 2\) graphically.
Draw dashed horizontal line \(y=2\). Test \((0,0)\): \(0 < 2\)? Yes. Shade the half-plane below \(y=2\), excluding the line.
Activity 5.2 — Test a Point, Pick a Side
Predict: For the inequality \(x + 2y \leq 6\), does the origin \((0,0)\) lie in the solution region?
  1. On graph paper draw the line \(x + 2y = 6\) passing through \((6,0)\) and \((0,3)\). Because the inequality is \(\leq\), make the line solid.
  2. Substitute \((0,0)\) into \(x+2y\): \(0 + 0 = 0\). Is \(0 \leq 6\)? Yes.
  3. Conclude: the half-plane containing the origin is the solution region. Shade it.
  4. Pick any point in the shaded region (say \((2,1)\)) and verify: \(2 + 2(1) = 4 \leq 6\) ✓.
  5. Pick a point outside (say \((5,3)\)): \(5 + 6 = 11 \leq 6\) ✗.
Insight: The "test-point" trick works for any linear inequality because a straight line divides the plane into exactly two half-planes; the inequality is either true in one or the other.

Figure it Out — Exercise 5.2 (Selected)

Q1. \(x + y < 5\) — solve graphically.
Dashed line through \((5,0)\) and \((0,5)\). \((0,0)\): \(0 < 5\) ✓. Shade half-plane containing origin.
Q2. \(2x + y \geq 6\) — solve graphically.
Solid line through \((3,0)\) and \((0,6)\). \((0,0)\): \(0 \geq 6\) ✗. Shade region away from origin, including line.
Q3. \(3x + 4y \leq 12\).
Solid line through \((4,0)\) and \((0,3)\). Origin satisfies the inequality, so shade region containing \((0,0)\).
Q4. \(y + 8 \geq 2x\).
Rewrite: \(2x - y \leq 8\). Line through \((4,0)\) and \((0,-8)\), solid. \((0,0)\): \(0 \leq 8\) ✓. Shade origin side.
Q5. \(x - y \leq 2\).
Solid line through \((2,0)\) and \((0,-2)\). Test \((0,0)\): \(0 \leq 2\) ✓. Shade side containing origin (above the line).
Competency-Based Questions — Half-Planes & Applications
A small workshop makes toys \(T_1\) and \(T_2\). Toy \(T_1\) needs 2 hours and toy \(T_2\) needs 3 hours of labour. The workshop has at most 60 labour-hours per week. Let \(x\) and \(y\) be the number of \(T_1\) and \(T_2\) toys made each week.
Q1. Write the inequality capturing the labour constraint and identify the type of boundary line.
L3 Apply
\(2x + 3y \leq 60\). Because the inequality is \(\leq\), the boundary line \(2x+3y=60\) is solid.
Q2. Is \((10, 15)\) a feasible production plan? Justify using the test-point method.
L4 Analyse
\(2(10) + 3(15) = 20 + 45 = 65 > 60\). Not feasible — the point lies in the rejected half-plane.
Q3. Sketch/describe the feasible region when \(x \geq 0, y \geq 0, 2x + 3y \leq 60\) hold simultaneously.
L5 Evaluate
A triangular region in the first quadrant bounded by the x-axis, y-axis and line \(2x+3y=60\) with vertices \((0,0), (30,0), (0,20)\). All three sides are solid because all inequalities are non-strict.
Q4. Design an additional material constraint so that only the plan \((15,10)\) lies on its boundary while \((30,0)\) becomes infeasible.
L6 Create
Try \(x + y \leq 25\): \(15 + 10 = 25\) (on boundary) ✓; \(30 + 0 = 30 > 25\) (infeasible) ✓. Any valid choice of the form \(ax+by = 15a + 10b\) that rejects \((30,0)\) works.
Assertion–Reason Questions
Assertion (A): The boundary line of \(2x + 3y < 12\) is drawn as a dashed (broken) line.
Reason (R): Strict inequalities \(<\) and \(>\) do not include the points on the boundary line in their solution set.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. Dashed lines visually indicate exclusion of boundary points, which is exactly what strict inequality requires.
Assertion (A): The origin \((0,0)\) always lies in the solution region of \(ax + by \leq c\) whenever \(c \geq 0\).
Reason (R): Substituting \((0,0)\) gives \(0 \leq c\), which is true when \(c \geq 0\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. The test-point algorithm confirms origin is in the half-plane \(ax+by \leq c\) whenever \(0 \leq c\).
Assertion (A): The inequality \(x \geq 2\) has no graphical meaning in two dimensions.
Reason (R): An inequality involving only one variable defines a half-line on the number line.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. A is false — in 2-D, \(x \geq 2\) represents the half-plane to the right of the vertical line \(x=2\). R is true but does not describe the 2-D case.

Summary

  • Two real numbers or two algebraic expressions related by \(<,>,\leq,\geq\) form an inequality.
  • Equal numbers may be added to (or subtracted from) both sides.
  • Both sides may be multiplied (or divided) by the same positive number; if the multiplier is negative, the inequality sign is reversed.
  • Values of \(x\) that make the inequality true form its solution set.
  • To graph \(ax + by \leq c\) in 2-D: draw the boundary line (solid for \(\leq/\geq\), dashed for \(</>\)), test a point not on the line, and shade the appropriate half-plane.

Frequently Asked Questions — Linear Inequalities

What is Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool in NCERT Class 11 Mathematics?

Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool is a key concept covered in NCERT Class 11 Mathematics, Chapter 5: Linear Inequalities. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool step by step?

To solve problems on Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Linear Inequalities?

The essential formulas of Chapter 5 (Linear Inequalities) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool important for the Class 11 board exam?

Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool?

Common mistakes in Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 — Linear Inequalities: Word Problems & Graphical Solutions in Two Variables | Class 11 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

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