Question

Question 8
In countries like USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
$F=\frac{9}{5}C+32$
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.
(ii) If the temperature is ${30}^{\circ }C$, what is the temperature in Fahrenheit?
(iii) If the temperature is ${95}^{\circ }F$, what is the temperature in Celsius?
(iv) If the temperature is $0^{\circ }C$, what is the temperature in Fahrenheit and if the temperature is $0^{\circ }F$, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer 1

$\left(i\right)F=\frac{9}{5}C+32$
When $C=0$ then $F=32$
also, when $C=5$ then $F=41$

$\begin{array}{ccc}C& 0& 5\\ F& 32& 41\end{array}$


(ii) Putting the value of $C=30$ in $F=\frac{9}{5}C+32$, we get
$F=\frac{9}{5}\times 30+32$
$\Rightarrow F=54+32$
$\Rightarrow F=86$
(iii) Putting the value of $F=95$ in $F=\frac{9}{5}C+32$, we get
$95=\frac{9}{5}C+32$
$\Rightarrow \frac{9}{5}C=95-32$
$\Rightarrow C=63\times \frac{5}{9}$
$\Rightarrow C=35$
(iv) Putting the value of $F=0$ in $F=\frac{9}{5}C+32,$ we get
$0=\frac{9}{5}C+32$
$\Rightarrow \frac{9}{5}C=-32$
$\Rightarrow C=-32\times \frac{5}{9}$
$\Rightarrow C=\frac{-160}{9}$
Putting the value of $C=0$ in $F=\frac{9}{5}C+32,$ we get
$F=\frac{9}{5}\times 0+32$
$\Rightarrow F=32$
(v) Here, we have to find when $F=C.$
Therefore, Putting $F=C$ in $F=\frac{9}{5}C+32,$ we get
$F=\frac{9}{5}F+32$
$\Rightarrow F-\frac{9}{5}F=32$
$\Rightarrow \frac{5F-9F}{5}=32$
$\Rightarrow \frac{-4}{5}F=32$
$\Rightarrow F=\frac{32\times 5}{-4}$
$\Rightarrow F=-40$
Therefore at $-40,$ both Fahrenheit and Celsius are numerically the same.