SATHEE: Chapter 5 Linear Inequalities Miscellaneous Exercise

Chapter 5 Linear Inequalities Miscellaneous Exercise

Miscellaneous Exercise on Chapter 5

Solve the inequalities in Exercises $1$ to $6.$

1. $2 \leq 3 x-4 \leq 5$

Answer :

$2 \leq 3 x-4 \leq 5$

$\Rightarrow 2+4 \leq 3 x-4+4 \leq 5+4$

$\Rightarrow 6 \leq 3 x \leq 9$

$\Rightarrow 2 \leq x \leq 3$

Thus, all the real numbers, $x$, which are greater than or equal to $2$ but less than or equal to $3 ,$ are the solutions of the given inequality. The solution set for the given inequalityis $[2,3]$.

2. $6 \leq-3(2 x-4) < 12$

Answer :

$6 \leq -3(2 x-4) < 12$

$\Rightarrow 2 \leq -(2 x-4) < 4$

$\Rightarrow-2 \geq 2 x-4 > -4$

$\Rightarrow 4 - 2 \geq 2 x > 4 - 4$

$\Rightarrow 2 \geq 2 x > 0$

$\Rightarrow 1 \geq x > 0$

Thus, the solution set for the given inequalityis $(0,1]$.

3. $-3 \leq 4-\dfrac{7 x}{2} \leq 18$

Answer :

$-3 \leq 4-\dfrac{7 x}{2} \leq 18$

$\Rightarrow-3-4 \leq-\dfrac{7 x}{2} \leq 18-4$

$\Rightarrow-7 \leq-\dfrac{7 x}{2} \leq 14$

$\Rightarrow 7 \geq \dfrac{7 x}{2} \geq-14$

$\Rightarrow 1 \geq \dfrac{x}{2} \geq-2$

$\Rightarrow 2 \geq x \geq-4$

Thus, the solution set for the given inequalityis $[-4, 2].$

4. $-15 < \dfrac{3(x-2)}{5} \leq 0$

Answer :

$-15 < \dfrac{3(x-2)}{5} \leq 0$

$\Rightarrow - 75 < 3(x - 2) \leq 0$

$\Rightarrow - 25 < x - ~ 2 \leq 0$

$\Rightarrow - 25+2 < x \leq 2$

$\Rightarrow -23 < x \leq 2$

Thus, the solution set for the given inequalityis $(-23, 2]$

5. $-12 < 4-\dfrac{3 x}{-5} \leq 2$

Answer :

$-12 < 4-\dfrac{3 x}{-5} \leq 2$

$\Rightarrow-12-4 < \dfrac{-3 x}{-5} \leq 2-4$

$\Rightarrow-16 < \dfrac{3 x}{5} \leq-2$

$\Rightarrow-80 < 3 x \leq-10$

$\Rightarrow \dfrac{-80}{3} < x \leq \dfrac{-10}{3}$

Thus, the solution set for the given inequalityis $\bigg(\dfrac{-80}{3}, \dfrac{-10}{3}\bigg]$.

6. $7 \leq \dfrac{(3 x+11)}{2} \leq 11$.

Answer :

$7 \leq \dfrac{(3 x+11)}{2} \leq 11$

$\Rightarrow 14 \leq 3 x+11 \leq 22$

$\Rightarrow 14-11 \leq 3 x \leq 22-11$

$\Rightarrow 3 \leq 3 x \leq 11$

$\Rightarrow 1 \leq x \leq \dfrac{11}{3}$

Thus, the solution set for the given inequalityis $\bigg[1, \dfrac{11}{3}\bigg]$.

Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

7. $5 x+1 > -24,5 x-1 < 24$

Answer :

$5 x+1 > -24$

$\Rightarrow 5 x > -25$

$\Rightarrow x > -5$

$5 x-1 < 24\qquad …(1) $

$\Rightarrow 5 x < 25$

$\Rightarrow x < 5\qquad …(2) $

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is $(-5,5)$. The solution of the given system of inequalities can be represented on number line as

8. $2(x-1) < x+5,3(x+2) > 2-x$

Answer :

$2(x-1) < x+5$

$\Rightarrow 2 x-2 < x+5$

$\Rightarrow 2 x-x < 5+2$

$\Rightarrow x < 7 \qquad \ldots$ $(1)$

$3(x+2) > 2-x$

$\Rightarrow 3 x+6 > 2-x$

$\Rightarrow 3 x+x > 2-6$

$\Rightarrow 4 x > -4$

$\Rightarrow x > -1 \qquad\ldots$ $(2)$

From $(1)$ and $(2),$ it can be concluded that the solution set for the given system of inequalities is $(-1, 7).$ The solution of the given system of inequalities can be represented on number line as

9. $3 x-7 > 2(x-6), 6-x > 11-2 x$

Answer :

$3 x - 7 > 2 (x $ - 6 $ ) $

$\Rightarrow 3 x - 7 > 2 x - 12$

$\Rightarrow 3 x - 2 x > - 12+7$

$\Rightarrow x > - 5\qquad … $ $(1)$

$\Rightarrow 6 - x > 11 - 2 x$

$\Rightarrow - x+2 x > 11- 6$

$\Rightarrow x > 5 \qquad …$ $(2)$

From $(1)$ and $(2),$ it can be concluded that the solution set for the given system of inequalities is $(5, \infty)$. The solution of the given system of inequalities can be represented on number line as

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10. $5(2 x-7)-3(2 x+3) \leq 0,2 x+19 \leq 6 x+47$.

Answer :

$5(2 x-7)-3(2 x+3) \leq 0$

$\Rightarrow 10 x-35-6 x-9 \leq 0$

$\Rightarrow 4 x-44 \leq 0$

$\Rightarrow 4 x \leq 44$

$\Rightarrow x \leq 11 \qquad\ldots$ $(1)$

$\Rightarrow 2 x+19 \leq 6 x+47$

$\Rightarrow 19-47$ $\leq x-2 x$

$\Rightarrow-28 \leq x $

$\Rightarrow-7 \leq x\qquad …$ $(2)$

From $(1)$ and $(2),$ it can be concluded that the solution set for the given system of inequalities is $[-7, 11].$ The solution of the given system of inequalities can be represented on number line as

11. A solution is to be kept between $68^{\circ} F$ and $77^{\circ} F$. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by $ F=\dfrac{9}{5} C+32 ? $

Answer :

Since the solution is to be kept between $68^{\circ} F$ and $77^{\circ} F$,

$68 < F < 77$

Putting $F=\dfrac{9}{5} C+32$, we obtain

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

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

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

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

$\Rightarrow 20 < C < 25$

Thus, the required range of temperature in degree Celsius is between $20^{\circ} C$ and $25^{\circ} C$.

12. A solution of $8 %$ boric acid is to be diluted by adding a $2 %$ boric acid solution to it. The resulting mixture is to be more than $4 %$ but less than $6 %$ boric acid. If we have 640 litres of the $8 %$ solution, how many litres of the $2 %$ solution will have to be added?

Answer :

Let $x$ litres of $2 %$ 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.

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

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

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

$\Rightarrow \dfrac{2}{100} x+\dfrac{8}{100}(640) > \dfrac{4}{100}(x+640)$

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

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

$\Rightarrow 5120 - 2560 > 2 x$

$\Rightarrow 2560 > 2 x$

$\Rightarrow 1280 > x$

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

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

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

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

$\Rightarrow 1280 < 4 x$

$\Rightarrow 320 < x$

$\therefore \ \ 320 < x < 1280$

Thus, the number of litres of $2 %$ of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres.

13. 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?

Answer :

Let $x$ litres of water is required to be added.

Then,total mixture $=(x+1125)$ litres

It is evident that the amount of acid contained in the resulting mixture is $45 %$ of 1125 litres.

This resulting mixture will contain more than $25 %$ but less than $30 %$ acid content.

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

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

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

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

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

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

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

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

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

$25 %$ of $(1125+x) < 45 %$ of $1125$

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

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

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

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

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

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

$\therefore \ \ 562.5 < x < 900$

Thus, the required number of litres of water that is to be added will have to be more than $562.5$ but less than $900$ .

14. IQ of a person is given by the formula

$ IQ=\dfrac{MA}{CA} \times 100 $

where MA is mental age and CA is chronological age. If $80 \leq IQ \leq 140$ for a group of $12 $ years old children, find the range of their mental age.

Answer :

It is given that for a group of $12$ years old children, $80 \leq IQ \leq 140 \ldots (i)$

For a group of $12$ years old children, $CA=12$ years

$ IQ=\dfrac{MA}{12} \times 100 $

Putting this value of IQ in $(i),$ we obtain

$ \begin{aligned} & 80 \leq \dfrac{MA}{12} \times 100 \leq 140 \\ \\ & \Rightarrow 80 \times \dfrac{12}{100} \leq MA \leq 140 \times \dfrac{12}{100} \\ \\ & \Rightarrow 9.6 \leq MA \leq 16.8 \end{aligned} $

Thus, the range of mental age of the group of $12$ years old children is $9.6 \leq MA \leq 16.8$.

Chapter 5 Linear Inequalities EXERCISE 5.1