Question

Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If Yes , find it.

Answer 1

$\begin{array}{c}\frac{{}^{0}f-31}{9}=\frac{{}^{0}C}{5}\\ {}^{0}C{=}^{0}f\\ \frac{{}^{0}f-32}{9}=\frac{{}^{0}C}{5}\\ 5^{0}C-160=9^{0}C\\ -160=4^{0}C\\ {}^{0}C=40\end{array}$

$Therefore{}^{0}Cand{}^{0}fareequalat-{40}^0$