NCERT Solutions · Class 11 Mathematics Linear Inequalities

Substitute the formula into the double inequality.The temperature is required to satisfy \[68 < \mathrm{F} < 77. \]Replacing \(\displaystyle \mathrm{F} \) by \(\displaystyle \dfrac{9}{5}\mathrm{C} + 32 \), \[68 < \frac{9}{5}\mathrm{C} + 32 < 77. \]A double inequality may be solved by performing the same operation on all three members. Subtract \(\displaystyle 32 \) throughout: \[36 < \frac{9}{5}\mathrm{C} < 45. \]Multiply throughout by \(\displaystyle \dfrac{5}{9} \). Since \(\displaystyle \dfrac{5}{9} > 0 \), the inequality signs do not reverse: \[36 \times \frac{5}{9} < \mathrm{C} < 45 \times \frac{5}{9}, \] \[20 < \mathrm{C} < 25. \]Check. At \(\displaystyle \mathrm{C} = 20 \), \(\displaystyle \mathrm{F} = \frac{9}{5}(20)+32 = 68 \); at \(\displaystyle \mathrm{C} = 25 \), \(\displaystyle \mathrm{F} = \frac{9}{5}(25)+32 = 77 \) — the two given end temperatures, so the endpoints correspond correctly and are excluded in both scales.The solution must be kept between $\displaystyle 20$° C and $\displaystyle 25$° C.

Form the mixture inequality on the quantity of acid.Let \(\displaystyle x \) litres of the $\displaystyle 2$% solution be added to the $\displaystyle 640$ litres of $\displaystyle 8$% solution. Then The mixture must contain more than $\displaystyle 4$% but less than $\displaystyle 6$% boric acid, so the acid present must lie between $\displaystyle 4$% and $\displaystyle 6$% of the total volume: \[4\% \text{ of } (640+x) \;<\; 51.2 + \frac{2x}{100} \;<\; 6\% \text{ of } (640+x). \]Left inequality. \[\frac{4}{100}(640+x) < 51.2 + \frac{2x}{100} \] \[25.6 + 0.04x < 51.2 + 0.02x \] \[0.02x < 25.6 \quad\Longrightarrow\quad x < 1280. \]Right inequality. \[51.2 + \frac{2x}{100} < \frac{6}{100}(640+x) \] \[51.2 + 0.02x < 38.4 + 0.06x \] \[12.8 < 0.04x \quad\Longrightarrow\quad x > 320. \]Both conditions must hold together, so \(\displaystyle 320 < x < 1280 \).Check. With \(\displaystyle x = 320 \) the strength is \(\displaystyle \dfrac{51.2 + 6.4}{960}\times 100 = 6\% \), and with \(\displaystyle x = 1280 \) it is \(\displaystyle \dfrac{51.2 + 25.6}{1920}\times 100 = 4\% \) — exactly the excluded limits, so the open interval is right.More than $\displaystyle 320$ litres but less than $\displaystyle 1280$ litres of the $\displaystyle 2$% boric acid solution must be added.

Form the mixture inequality on the quantity of acid.Let \(\displaystyle x \) litres of water be added. Water carries no acid, so the amount of acid is unchanged while the volume grows: The mixture must be more than $\displaystyle 25$% but less than $\displaystyle 30$% acid: \[25\% \text{ of } (1125+x) \;<\; 506.25 \;<\; 30\% \text{ of } (1125+x). \]Left inequality. \[\frac{25}{100}(1125+x) < 506.25 \] \[281.25 + 0.25x < 506.25 \] \[0.25x < 225 \quad\Longrightarrow\quad x < 900. \]Right inequality. \[506.25 < \frac{30}{100}(1125+x) \] \[506.25 < 337.5 + 0.3x \] \[168.75 < 0.3x \quad\Longrightarrow\quad x > 562.5. \]Both must hold, so \(\displaystyle 562.5 < x < 900 \).Check. At \(\displaystyle x = 562.5 \) the strength is \(\displaystyle \dfrac{506.25}{1687.5}\times 100 = 30\% \); at \(\displaystyle x = 900 \) it is \(\displaystyle \dfrac{506.25}{2025}\times 100 = 25\% \). Adding more water dilutes further, which is why the larger \(\displaystyle x \) gives the weaker mixture.More than $\displaystyle 562.5$ litres but less than $\displaystyle 900$ litres of water must be added.

Substitute the formula into the double inequality.The children are $\displaystyle 12$ years old, so \(\displaystyle \mathrm{CA} = 12 \) and \[\mathrm{IQ} = \frac{\mathrm{MA}}{12}\times 100 = \frac{100\,\mathrm{MA}}{12}. \]The given condition \(\displaystyle 80 \le \mathrm{IQ} \le 140 \) becomes \[80 \le \frac{100\,\mathrm{MA}}{12} \le 140. \]Multiply throughout by \(\displaystyle \dfrac{12}{100} \), a positive number, so the inequality signs are preserved: \[80 \times \frac{12}{100} \le \mathrm{MA} \le 140 \times \frac{12}{100}, \] \[9.6 \le \mathrm{MA} \le 16.8. \]Check. \(\displaystyle \mathrm{MA} = 9.6 \) gives \(\displaystyle \mathrm{IQ} = \frac{9.6}{12}\times 100 = 80 \), and \(\displaystyle \mathrm{MA} = 16.8 \) gives \(\displaystyle \mathrm{IQ} = \frac{16.8}{12}\times 100 = 140 \); the endpoints are attained, so they are included.The mental age of the children lies in the range \(\displaystyle 9.6 \le \mathrm{MA} \le 16.8 \) years.