Chapter 5: Linear Inequalities

Ans: Because the solution must be preserved somewhere in the middle, ${68^\circ }{\text{F}}$ and ${77^\circ }{\text{F}},68 < F < 77$ Putting $F = \dfrac{9}{5}C + 32$, we obtain $68 < \dfrac{9}{5}C + 32 < 77$

$\Rightarrow 68 - 32 < \dfrac{9}{5}C < 77 - 32$

$\Rightarrow 36 < \dfrac{9}{5}C < 45$

$\Rightarrow 36 \times \dfrac{5}{9} < C < 45 \times \dfrac{5}{9}$

$\Rightarrow 20 < C < 25$

As a result, the required temperature range in degrees Celsius is between ${20^\circ }C$ and ${25^\circ }C$.

Ans: Let $2\% $ of $x$ litres of boric acid solution is required to be added. Then, total mixture $ = (x + 640)$ litres.

This resulting mixture is to be more than $4\% $ but less than $6\% $ boric acid. 

$\therefore \quad 2 \%$ of $x+8 \%$ of $640>4 \%$ of $(x+640)$

And, $2 \%$ of $x+8 \%$ of $640<6 \%$ of $(x+640)$`

$\Rightarrow 2x + 5120 > 4x + 2560$

$\Rightarrow 5120 - 2560 > 4x - 2x$

$\Rightarrow 5120 - 2560 > 2x$

$\Rightarrow 2560 > 2x$

$\Rightarrow 1280 > x$

$2\% x + 8\% $ of $640 < 6\% $ of $(x + 640)$ $\dfrac{2}{{100}}x + \dfrac{8}{{100}}(640) < \dfrac{6}{{100}}(x + 640)$

$\Rightarrow 2x + 5120 < 6x + 3840$

$\Rightarrow 5120 - 3840 < 6x - 2x$

$\Rightarrow 5120 - 3840 < 6x - 2x$

$\Rightarrow 1280 < 4x$

$\Rightarrow 320 < x$

$\therefore 320 < x < 1280$

As a result, the total amount of boric acid solution to be added must be greater than 320 litres but less than 1280 litres.

Ans: Allow for the addition of x litres of water. The entire mixture is then calculated $ = (x + 1125)$litres It is clear that the amount of acid in the final mixture is excessive. $45\% $ of 1125 litres. The resulting mixture will have a higher concentration of $25\% $ but less than $30\% $ acid content.

$\therefore 30\% $ of $(1125 + x) > 45\% $ of 1125

And, $25\% $ of $(1125 + x) < 45\% $ of 1125 

$\Rightarrow \dfrac{{30}}{{100}}(1125 + x) > \dfrac{{45}}{{100}} \times 1125$

$\Rightarrow 30(1125 + x) > 45 \times 1125$

$\Rightarrow 30 \times 1125 + 30x > 45 \times 1125$

$\Rightarrow 30 > 45 \times 1125 - 30 \times 1125$

$\Rightarrow 30x > (45 - 30) \times 1125$

$\Rightarrow x > \dfrac{{15 \times 1125}}{{30}} = 562.5$

$25\% $ of $(1125 + x) < 45\% $ of 1125 $\Rightarrow \dfrac{{25}}{{100}}(1125 + x) < \dfrac{{45}}{{100}} \times 1125$

$\Rightarrow 25(1125 + x) > 45 \times 1125$

$\Rightarrow 25 \times 1125 + 25x > 45 \times 1125$

$\Rightarrow 25x > 45 \times 1125 - 25 \times 1125$

$\Rightarrow 25x > (45 - 25) \times 1125$

$\Rightarrow x > \dfrac{{20 \times 1125}}{{25}} = 900$

$\therefore 562.5 < x < 900$

As a result, the required number of litres of water must be greater than 562.5 but less than 900.

Ans: It is reported that for a group of twelve-year-olds $80 \leqslant IQ \leqslant 140 \ldots  \ldots (i)$

For a group of 12 years old children, ${\text{CA}} = 12$ years ${\text{IQ}} = \dfrac{{{\text{MA}}}}{{12}} \times 100$

Putting this value of IQ in (i), we obtain $80 \leqslant \dfrac{{{\text{MA}}}}{{12}} \times 100 \leqslant 140$

$\Rightarrow 80 \times \dfrac{{12}}{{100}} \leqslant {\text{MA}} \leqslant 140 \times \dfrac{{12}}{{100}}$

$\Rightarrow 9.6 \leqslant {\text{MA}} \leqslant 16.8$

As a result, the mental age range of the 12-year-olds has widened.$\Rightarrow 9.6 \leqslant {\text{MA}} \leqslant 16.8$.