$4 x+3 \geq 2 x+17,3 x-5<-2$
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., $x<1, x \geq 7$ ).
A company manufactures cassettes. Its cost and revenue functions are $C(x)=26000+30 x$ and $R(x)=43 x$, respectively, where $x$ is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit?
$$\begin{aligned} \text{Cost function, }\quad C(x) =26000+30 x \\ \text{and revenue function, }\quad R(x) =43 x \\ \text{For profit,}\quad R(x) >C(x) \\ \Rightarrow \quad 26000+30 x <43 x \\ \Rightarrow \quad 30 x-43 x <-26000 \\ \Rightarrow \quad -13 x <-26000 \\ \Rightarrow \quad 13 x >26000 \\ \Rightarrow \quad x >\frac{26000}{13} \\ \therefore \quad x >2000 \end{aligned}$$
Hence, more than 2000 cassettes must be produced to get profit.
The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5 . If the first two pH readings are 8.48 and 8.35 , then find the range of pH value for the third reading that will result in the acidity level being normal.
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 .
A solution of $9 \%$ acid is to be diluted by adding $3 \%$ acid solution to it. The resulting mixture is to be more than $5 \%$ but less than $7 \%$ acid. If there is 460 L of the $9 \%$ solution, how many litres of $3 \%$ solution will have to be added?
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.
A solution is to be kept between $40^{\circ} \mathrm{C}$ and $45^{\circ} \mathrm{C}$. What is the range of temperature in degree fahrenheit, if the conversion formula is $F=\frac{9}{5} C+32 ?$
Let the required temperature be $x \Upsilon\mathrm{F}$.
$$\begin{array}{ll} \text { Given that, } & F=\frac{9}{5} C+32 \\ \Rightarrow & 5 F=9 C+32 \times 5 \\ \Rightarrow & 9 C=5 F-32 \times 5 \\ \therefore & C=\frac{5 F-160}{9} \end{array}$$
Since, temperature in degree calcius lies between $40^{\circ} \mathrm{C}$ to $45^{\circ} \mathrm{C}$.
Therefore,
$$40<\frac{5 F-160}{9}<45$$
$$\Rightarrow \quad 40<\frac{5 x-160}{9}<45$$
$$\begin{array}{lrl} \Rightarrow & 40 \times 9 & <5 x-160<45 \times 9 \quad \text{[multiplying throughout by 9]}\\ \Rightarrow & 360 & <5 x-160<405 \quad \text{[adding 160 throughout]}\\ \Rightarrow & 360+160 & <5 x<405+160 \\ \Rightarrow & 520 & <5 x<565 \\ \Rightarrow & \frac{520}{5} & < x<\frac{565}{5}\quad \text{[divide throughout by 5]} \\ \Rightarrow & 104 & < x < 113 \end{array}$$
Hence, the range of temperature in degree fahrenheit is 104$^\circ$F to 113$^\circ$F.