Question

There are two scales of measuring temperature, namely, Fahrenheit (F) and Celsius (C).
The relation between the two scales is
$F=\frac{9}{5}C+32$
(i) Draw the graph of the given linear equation taking C along the x-axis and F along the y-axis.
Fill in the blanks given below:
(ii) 0°C = (......)°F
(iii) 95°F = (......)°C
(iv) 0°F = (......)°C
(v) Find the temperature which is numerically the same in both (F) and (C).

Answer 1

The given equation is $F=\frac{9}{5}C+32$.
On a graph paper, take $C$ along the $\text{x}$-axis and $F$ along the $y$-axis.
Along the $\text{x}$-axis, take 10 small divisions = $1°C$
Along the $y$-axis, take 10 small divisions = $10°F$
Putting $C=0$ in the given equation, we get: $F=32$
Putting $C=5$ in the given equation, we get: $F=41$
Thus, we get the following table:

$\text{x}$	0	5
$y$	32	41

Plot the points $A(0,32),B(5,41)$ on the graph paper. Join these points and extend them in both the directions.
(ii) Putting $C=0$ in the given equation, we get $F=32$
$0°C=32°F$

(iii) Putting $F=95°$ in the given equation, we get:
$\frac{9C}{5}=95-32$
⇒ $\frac{9}{5}C=63$
⇒ $C=\frac{63\times 5}{9}=35$
∴ $95°F=35°C$

(iv) Putting $F=0$ in the given equation, we get:
$\frac{9}{5}C=-32$
⇒ $C=-\frac{5\times 32}{9}=-\frac{160}{9}=-17.8$
$0°F=-17.8°C$

(v) Putting $C=F$ in the given equation, we get:
$F-\frac{9}{5}F=32$
⇒ $4F=-160$
⇒ $F=-\frac{160}{4}==-40$
∴ $-40°F=-40°C$