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 $205^{\circ} \mathrm{C}$.
Solve the following system of inequalities $\frac{2 x+1}{7 x-1}>5, \frac{x+7}{x-8}>2$.
The given system of inequations is
$$\begin{aligned} & \frac{2 x+1}{7 x-1}>5 \quad \text{... (i)}\\ & \text { and } \\ & \frac{x+7}{x-8}>2 \quad \text{... (ii)}\\ & \text { Now, } \\ & \frac{2 x+1}{7 x-1}-5>0 \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & \frac{(2 x+1)-5(7 x-1)}{7 x-1}>0 \\ \Rightarrow \quad & \frac{2 x+1-35 x+5}{7 x-1}>0 \\ \Rightarrow \quad & \frac{-33 x+6}{7 x-1}>0 \Rightarrow\frac{33 x-6}{7 x-1}<0 \\ \Rightarrow \quad & x \in\left(\frac{1}{7}, \frac{6}{33}\right)\quad \text{... (iii)} \end{aligned}$$
$$\begin{aligned} & \text { and } \quad \frac{x+7}{x-8}>2 \Rightarrow \frac{x+7}{x-8}-2>0 \\ & \Rightarrow \quad \frac{x+7-2(x-8)}{x-8}>0 \Rightarrow \frac{x+7-2 x+16}{x-8}>0 \\ & \Rightarrow \quad \frac{-x+23}{x-8}>0 \Rightarrow \frac{x-23}{x-8}<0 \end{aligned}$$
$\Rightarrow \quad x \in(8,23)\quad \text{.... (iv)}$
Since, the intersection of Eqs. (iii) and (iv) is the null set. Hence, the given system of equation has no solution.
Find the linear inequalities for which the shaded region in the given figure is the solution set.
Consider the line $3 x+2 y=48$, we observe that the shaded region and the origin are on the same side of the line $3 x+2 y=48$ and $(0,0)$ satisfy the linear constraint $3 x+2 y \leq 48$. So, we must have one inequation as $3 x+2 y \leq 48$.
Now, consider the line $x+y=20$. We find that the shaded region and the origin are on the same side of the line $x+y=20$ and $(0,0)$ satisfy the constraints $x+y \leq 20$. So, the second inequation is $x+y \leq 20$.
We also notice that the shaded region is above $X$-axis and is on the right side of $Y$-axis, so we must have $x \geq 0, y \geq 0$.
Thus, the linear inequations corresponding to the given solution set are $3 x+2 y \leq 48$, $x+y \leq 20$ and $x \geq 0, y \geq 0$.