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  4. On a hypothetical scale X, the ice point is 40° and the steam point is 120°. For another scale Y, the ice point and steam point are -30° and 130° respectively. If X reads 50°, then Y would read :

Q. On a hypothetical scale $X$, the ice point is $40^{\circ}$ and the steam point is $120^{\circ}$. For another scale $Y$, the ice point and steam point are $-30^{\circ}$ and $130^{\circ}$ respectively. If $X$ reads $50^{\circ}$, then $Y$ would read :

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Solution:

We use
$\Rightarrow \frac{50-40}{120-40} =\frac{Y-(-30)}{130-(-30)}$
$\Rightarrow \frac{10}{80} =\frac{Y+30}{160}$
$\Rightarrow \frac{1}{8} =\frac{Y+30}{160}$
$\Rightarrow Y+30 =20$
$\Rightarrow Y =-10^{\circ}$

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