On a hypothetical scale x, the ice point is ${40}^0$and the steam point is${120}^0$For another scale y, the ice point and steam point are$-{30}^0 and {130}^0$respectively, if X reads${50}^0$, then y wouldread

$$\begin{gathered} X = L.F.P = 40 ; U.F.P = 120 \\ Y = L.F.P = – 30 ; U.F.P. = {130}^0 \\ We know \frac{x-40}{120-40}=\frac{y-(-30)}{130-(-30)} \\ \therefore \frac{x-40}{120-40}=\frac{y+30}{160} \\ now x^0={50}^0;y^0=? \\ \therefore \frac{50-40}{80}=\frac{y-(-30)}{130+30} \\ \Rightarrow \frac{1}{8}=\frac{y+30}{160} \\ \Rightarrow 20=y+30\Rightarrow y=-{10}^{0}C \end{gathered}$$