Question

On a certain scale of temperature x, the ice points is ${40}^{\circ }$ and the steam point is ${160}^{\circ }$ . On another scale of temperature y, the ice points is $-{20}^{\circ }$ and the steam point is ${100}^{\circ }$ . Convert ${50}^{\circ }$ in x to the y-scale.

• A. $-{25}^{\circ }$
• B. $-{20}^{\circ }$
• C. $-{10}^{\circ }$
• D. $-{15}^{\circ }$

Answer 1

The correct option is C $-{10}^{\circ }$

Range of x: $R_x=160-40=120$
Range of y: $R_y=100-(-20)=120$
The given temp is ${50}^{\circ }$ in scale x and let it be t in scale y.
$\therefore \frac{50-40}{R_x}=\frac{t-(-20)}{R_y}$

$\therefore \frac{10}{120}=\frac{t+20}{120}$
$\Rightarrow t=-{10}^{\circ }$