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.
The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm , then find the minimum length of the shortest side.
Let the length of shortest side be $x \mathrm{~cm}$.
According to the given information,
$$\begin{aligned} \text { Longest side } & =2 \times \text { Shortest side } \\ & =2 x \mathrm{~cm} \\ \text { and third side } & =2+\text { Shortest side } \\ & =(2+x) \mathrm{cm} \end{aligned}$$
Perimeter of triangle $=x+2 x+(x+2)=4 x+2$
According to the question,
Perimeter $>166 \mathrm{~cm}$
$$\begin{aligned} & \Rightarrow \quad 4 x+2>166 \\ & \Rightarrow \quad 4 x>166-2 \\ & \Rightarrow \quad 4 x>164 \\ & \therefore \quad x>\frac{164}{4}=41 \mathrm{~cm} \end{aligned}$$
Hence, the minimum length of shortest side be 41 cm .
In drilling world's deepest hole it was found that the temperature $T$ in degree celcius, $x \mathrm{~km}$ below the earth's surface was given by $T=30+25(x-3), 3 \leq x \leq 15$. At what depth will the temperature be between $155^{\circ} \mathrm{C}$ and $205^{\circ} \mathrm{C}$ ?
Given that, $$T=30+25(x-3), 3 \leq x \leq 15$$
According to the question,
$$\begin{array}{lc} & 155 < T < 205 \\ \Rightarrow & 155 < 30+25(x-3) < 205 \\ \Rightarrow & 155-30 < 25(x-3)<205-30 \quad \text{[subtracting 30 in whole]}\\ \Rightarrow & 125 < 25(x-3)< 175 \\ \Rightarrow & \frac{125}{25}< x-3 < \frac{175}{25} \quad \text{[dividing by 25 in whole] }\\ \Rightarrow & 5 < x-3 < 7 \\ \Rightarrow & 5+3 < x< 7+3 \\ \Rightarrow & 8< x< 10\quad \text{[adding 3 in whole] } \end{array}$$
Hence, at the depth 8 to 10 km temperature lies between $155^{\circ}$ to $205^{\circ} \mathrm{C}$.