$$\begin{aligned} &\begin{aligned} \text { We have, } \quad & -5 \leq \frac{2-3 x}{4} \\ \Rightarrow \quad& -20 \leq 2-3 x \quad \text { [multiplying by } 4 \text { on both sides] }\\ \Rightarrow \quad& 3 x \leq 2+20 \\ \Rightarrow \quad& 3 x \leq 22 \\ \Rightarrow \quad& x \leq \frac{22}{3} \\ \text { and }\quad & \frac{2-3 x}{4} \leq 9 \end{aligned}\\ \end{aligned}$$

$$\begin{array}{lc} \Rightarrow & 2-3 x \leq 36 \\ \Rightarrow & -3 x \leq 36-2 \\ \Rightarrow & -3 x \leq 34 \\ \Rightarrow & 3 x \geq-34 \\ \Rightarrow & x \geq-\frac{34}{3} \\ \Rightarrow & -\frac{34}{3} \leq x \leq \frac{22}{3} \\ \Rightarrow & x \in\left[\frac{-34}{3}, \frac{22}{3}\right] \end{array}$$

We have,

$$\begin{aligned} & 4 x+3 \geq 2 x+17 \\ \Rightarrow \quad & 4 x-2 x \geq 17-3 \Rightarrow 2 x \geq 14 \\ \Rightarrow \quad & x \geq \frac{14}{2} \\ \Rightarrow \quad & x \geq 7 \quad \supset\text{(i)}\\ \text { Also, we have } & \\ & 3 x-5<-2 \\ \Rightarrow\quad & 3 x<-2+5 \Rightarrow 3 x<3 \\ \Rightarrow \quad & x<1\quad \supset\text{(ii)} \end{aligned}$$

On combining Eqs. (i) and (ii), we see that solution is not possible because nothing is common between these two solutions. (i.e., $x<1, x \geq 7$ ).

$$\begin{aligned} \text{Cost function, }\quad C(x) =26000+30 x \\ \text{and revenue function, }\quad R(x) =43 x \\ \text{For profit,}\quad R(x) >C(x) \\ \Rightarrow \quad 26000+30 x <43 x \\ \Rightarrow \quad 30 x-43 x <-26000 \\ \Rightarrow \quad -13 x <-26000 \\ \Rightarrow \quad 13 x >26000 \\ \Rightarrow \quad x >\frac{26000}{13} \\ \therefore \quad x >2000 \end{aligned}$$

Hence, more than 2000 cassettes must be produced to get profit.

Given, $\quad\text{ first pH value} = 8.48$

and $\quad \text{second pH value}$ = 8.35

Let third pH value be $x$.

Since, it is given that average pH value lies between 8.2 and 8.5 .

$$\begin{aligned} & \therefore \quad 8.2< \frac{8.48+8.35+x}{3}< 8.5 \\ & \Rightarrow \quad 8.2 < \frac{16.83+x}{3} < 8.5 \\ & \Rightarrow \quad 3 \times 8.2 < 16.83+x < 8.5 \times 3 \\ & \Rightarrow \quad 24.6 < 16.83+x < 25.5 \\ & \Rightarrow \quad 24.6-16.83 < x< 25.5-16.83 \\ & \Rightarrow \quad 7.77< x<8.67 \end{aligned}$$ Thus, third pH value lies between 7.77 and 8.67 .

Let $x \mathrm{~L}$ of $3 \%$ solution be added to 460 L of $9 \%$ solution of acid.

Then, total quantity of mixture $=(460+x) \mathrm{L}$

Total acid content in the $(460+x)$ L of mixture

$$=\left(460 \times \frac{9}{100}+x \times \frac{3}{100}\right)$$

It is given that acid content in the resulting mixture must be more than $5 \%$ but less than $7 \%$ acid.

$$ \begin{aligned} & \text { Therefore, } \quad 5 \% \text { of }(460+x)<460 \times \frac{9}{100}+\frac{3 x}{100}<7 \% \text { of }(460+x) \\ & \Rightarrow \quad \frac{5}{100} \times(460+x)<460 \times \frac{9}{100}+\frac{3}{100} x<\frac{7}{100} \times(460+x) \\ & \Rightarrow \quad 5 \times(460+x)<460 \times 9+3 x<7 \times(460+x)\quad \text{[multiplying by 100]} \end{aligned}$$

$$ \begin{array}{lc} \Rightarrow & 2300+5 x<4140+3 x<3220+7 x \\ \text { Taking first two inequalities, } & 2300+5 x<4140+3 x \\ \Rightarrow & 5 x-3 x<4140-2300 \\ \Rightarrow & 2 x<1840 \\ \Rightarrow & x<\frac{1840}{2} \\ \Rightarrow & x<920\quad \text{... (i)} \end{array}$$

$$ \begin{array}{lr} \text { Taking last two inequalities, } & 4140+3 x<3220+7 x \\ \Rightarrow & 3 x-7 x<3220-4140 \\ \Rightarrow & -4 x<-920 \\ \Rightarrow & 4 x>920 \\ \Rightarrow & x>\frac{920}{4} \\ \Rightarrow & x>230\quad \text{... (ii)} \end{array}$$

Hence, the number of litres of the 3% solution of acid must be more than 230 L and less than 920 L.