Question

A solution is to be kept between ${86}^{\circ }and{95}^{\circ }F$.What is the range of temperature in degree Celsius,if the Celsius(C)/Fahrenheit (F) conversion formula is given by $F=\frac{9}{5}C+32.$.

Answer 1

We have,

$F_1={86}^{\circ }F{F}_1=\frac{9}{5}{C}_1+32[\because F=\frac{9}{5}C+32]\Rightarrow 86=\frac{9}{5}{C}_1+32\Rightarrow 86-32=\frac{9}{5}{C}_1\Rightarrow 54=\frac{9}{5}{C}_1\Rightarrow 9C_1=5\times 54\Rightarrow C_1=\frac{5\times 54}{9}\Rightarrow C_1=5\times 6={30}^{\circ }CNow{F}_2={95}^{\circ }F\therefore F_2=\frac{9}{5}{C}_2+32\Rightarrow 95=\frac{9}{5}{C}_2+32\Rightarrow 95-32=\frac{9}{5}{C}_2\Rightarrow 63=\frac{9}{5}{C}_2\Rightarrow 9C_2=63\times 5\Rightarrow C_2=\frac{63\times 5}{9}\Rightarrow C_2=7\times 5={35}^{\circ }C$

$\therefore$ The range of temperature of the solution is from ${30}^{\circ }Cto{35}^{\circ }C$.