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 $[2,3]$.

Answer :

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

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

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

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

$\Rightarrow 2 \geq 2 x>0$

$\Rightarrow 1 > = x>0$

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

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].

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]

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 $(\dfrac{-80}{3}, \dfrac{-10}{3}]$.

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 $[1, \dfrac{11}{3}]$.

Answer :

$5 x+1>-24$

$\Rightarrow 5 x>-25$

$\Rightarrow x>-5$

$5 x-1<24$

$\Rightarrow 5 x<25$

$\Rightarrow x<5$

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

Answer :

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

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

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

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

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

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

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

$\Rightarrow 4 x>-4$

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

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

Answer :

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

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

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

$\Rightarrow x>$- 5

6 - $x>11$ - $2 x$

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

$\Rightarrow x>5$.

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

Answer :

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

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

$\Rightarrow 4 x-44 \leq 0$

$\Rightarrow 4 x \leq 44$

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

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

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

$\Rightarrow-28 \leq x $

$\Rightarrow-7 \leq x$…

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

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 $25^{\circ} C$.

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.

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 .

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 $9.6 \leq MA \leq 16.8$.