Question

A solution is to be kept between ${40}^{o}C$ and ${45}^{o}C$ . What is the range of temperature in degree farenheit, if the conversion formula is $F=?$

Answer 1

We have,

$F=\frac{9}{5}C+32$

$C=\frac{5F-160}{9}$

Since, temperature is degree Celsius lies between ${40}^{0}C$ to ${45}^{0}C$, we get

$40<\frac{5F-160}{9}<45$

$360<5F-160<405$

$360+160<5F<405+160$

$520<5F<565$

$104<F<133$

So, the range is ${104}^{0}F$ to ${113}^{0}F$.

Hence, this is the answer.