Linear Inequalities Miscellaneous Ex Class 11 NCERT - Concepts & Practice

Solving Double (Compound) Inequalities

A double inequality is of the form $a \le f(x) \le b$. Solving these requires finding all values of $x$ that satisfy both inequalities simultaneously. The most efficient way to solve $a \le g(x) + c \le b$ is to perform algebraic operations on all three parts of the inequality at once. For example, to isolate $x$, you can add, subtract, multiply, or divide all three sections by the same real number. However, the absolute golden rule of inequalities still applies: if you multiply or divide all three parts of a double inequality by a negative number, you must reverse the direction of both inequality signs. Always express your final solution set in standard interval notation (e.g., open brackets $(a, b)$ or closed brackets $[a, b]$) and verify that your endpoints make physical sense in applied word problems.

Methodology for Solving Mixture Word Problems

1. Define the Variable — Let $x$ represent the volume (usually in litres) of the solution of known concentration that needs to be added to the existing mixture.
2. Express Total Volume and Acid Content — Write algebraic terms for the final volume of the combined mixture, which will be $(\text{existing volume} + x)$. Express the total amount of pure substance (like acid) in this mixture by summing the contributions from both solutions.
3. Set Up the Double Inequality — Translate the constraints given in the problem (e.g., 'resulting mixture must be more than $A\%$ but less than $B\%$') into a double inequality. This takes the form: $A\% \text{ of total volume} < \text{total pure substance} < B\% \text{ of total volume}$.
4. Solve and State the Range — Solve the double inequality system by splitting it into two separate inequalities or solving them simultaneously. The final solution gives the valid range of volume $x$ that satisfies the requirements.

Step-by-Step Worked NCERT Examples

• Example 1 (Double Inequality): Solve $-2 \le \frac{3x - 4}{5} \le 9$.

  Step 1: Multiply all three parts by 5 to clear the denominator:
  $-2 \times 5 \le 3x - 4 \le 9 \times 5$
  $-10 \le 3x - 4 \le 45$

  Step 2: Add 4 to all parts to isolate the term with $x$:
  $-10 + 4 \le 3x \le 45 + 4$
  $-6 \le 3x \le 49$

  Step 3: Divide all parts by 3:
  $\frac{-6}{3} \le x \le \frac{49}{3}$
  $-2 \le x \le \frac{49}{3}$

  Final Solution: The solution set in interval notation is $[-2, \frac{49}{3}]$.

• Example 2 (Mixture Problem): A manufacturer has 600 litres of a $12\%$ solution of acid. How many litres of a $30\%$ acid solution must be added to it so that the acid content in the resulting mixture will be more than $15\%$ but less than $18\%$?

  Step 1: Let $x$ litres of $30\%$ acid solution be added. Total volume of mixture = $(600 + x)$ litres.

  Step 2: Express acid content: Acid in existing solution = $12\%$ of $600 = 72$ litres. Acid in added solution = $30\%$ of $x = 0.3x$ litres. Total acid = $72 + 0.3x$.

  Step 3: Set up the compound inequality according to the limits:
  $15\% \text{ of } (600 + x) < 72 + 0.3x < 18\% \text{ of } (600 + x)$

  Step 4: Solve the left inequality:
  $0.15(600 + x) < 72 + 0.3x$
  $90 + 0.15x < 72 + 0.3x$
  $18 < 0.15x \implies x > 120$

  Step 5: Solve the right inequality:
  $72 + 0.3x < 0.18(600 + x)$
  $72 + 0.3x < 108 + 0.18x$
  $0.12x < 36 \implies x < 300$

  Final Solution: The manufacturer must add more than 120 litres but less than 300 litres of the $30\%$ acid solution.

Board Exam Tips & Common Pitfalls

1. The Sign Reversal Trap: When solving $-x < 5$, multiplying by $-1$ gives $x > -5$. Students frequently forget to reverse the inequality sign, leading to entirely incorrect intervals.
2. Bracket Precision: Use closed brackets $[ ]$ when the inequality is weak ($\le, \ge$) and open brackets $( )$ when it is strict ($<, >$). Representing these properly on a number line (solid dot vs empty circle) is worth 1 critical step mark in CBSE examinations.
3. Word Problem Units: Ensure that you write clear concluding statements with correct physical units (e.g., 'litres', '$^\circ\text{C}$'). Never leave your answer as a raw numerical interval like $(120, 300)$ in application-based questions.

Practice Questions with Solutions

• Q: Solve the double inequality for real $x$: $-12 < 4 - 3x \le 2$.
  A: Step 1: Subtract 4 from all parts of the inequality:
  $-12 - 4 < -3x \le 2 - 4$
  $-16 < -3x \le -2$

  Step 2: Divide all parts by $-3$. Since we are dividing by a negative number, we must reverse the inequality signs:
  $\frac{-16}{-3} > x \ge \frac{-2}{-3}$
  $\frac{16}{3} > x \ge \frac{2}{3}$

  Step 3: Rewrite in standard ascending order:
  $\frac{2}{3} \le x < \frac{16}{3}$

  Final answer: The solution interval is $[\frac{2}{3}, \frac{16}{3})$.

• Q: Solve the system of inequalities and plot the solution on a number line: $5x + 1 > -24$ and $5x - 1 < 24$.
  A: Step 1: Solve the first inequality:
  $5x + 1 > -24$
  $5x > -25 \implies x > -5$

  Step 2: Solve the second inequality:
  $5x - 1 < 24$
  $5x < 25 \implies x < 5$

  Step 3: Combine both solutions to find the intersection:
  $-5 < x < 5$

  Final answer: The solution set is $(-5, 5)$, which on a number line is represented by an open interval between $-5$ and $5$ with empty circles at both endpoints.

• Q: A solution is to be kept between $30^\circ\text{C}$ and $35^\circ\text{C}$. What is the range of temperature in degree Fahrenheit if the conversion formula is given by $\text{C} = \frac{5}{9}(\text{F} - 32)$?
  A: Step 1: Substitute the conversion formula into the given temperature range:
  $30 < \frac{5}{9}(\text{F} - 32) < 35$

  Step 2: Multiply all parts by 9 to clear the fraction:
  $270 < 5(\text{F} - 32) < 315$

  Step 3: Divide all parts by 5:
  $54 < \text{F} - 32 < 63$

  Step 4: Add 32 to all parts to solve for $\text{F}$:
  $54 + 32 < F < 63 + 32$
  $86 < F < 95$

  Final answer: The range of temperature in degree Fahrenheit is between $86^\circ\text{F}$ and $95^\circ\text{F}$.

• Q: A manufacturer has 500 litres of a $10\%$ acid solution. How many litres of a $25\%$ acid solution must be added to it so that the acid content in the resulting mixture will be more than $12\%$ but less than $15\%$?
  A: Step 1: Let $x$ litres of $25\%$ acid solution be added. Total volume of mixture = $(500 + x)$ litres.

  Step 2: Write the expression for total acid content:
  Acid content = $10\%$ of $500 + 25\%$ of $x = 50 + 0.25x$ litres.

  Step 3: Set up the compound inequality:
  $12\% \text{ of } (500 + x) < 50 + 0.25x < 15\% \text{ of } (500 + x)$

  Step 4: Solve the left inequality:
  $0.12(500 + x) < 50 + 0.25x$
  $60 + 0.12x < 50 + 0.25x$
  $10 < 0.13x \implies x > \frac{10}{0.13} \implies x > 76.92 \text{ litres}$

  Step 5: Solve the right inequality:
  $50 + 0.25x < 0.15(500 + x)$
  $50 + 0.25x < 75 + 0.15x$
  $0.10x < 25 \implies x < 250 \text{ litres}$

  Final answer: The manufacturer must add more than $76.92$ litres (or exactly $1000/13$ litres) but less than 250 litres of the $25\%$ acid solution.

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