On an imaginary linear scale of temperature (called ' $\mathrm{W}$ ' scale) the freezing and boiling points…
On an imaginary linear scale of temperature (called ' $\mathrm{W}$ ' scale) the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. The temperature on the new scale corresponding to $39^{\circ} \mathrm{C}$ temperature on Celsius scale will be
$139^{\circ} \mathrm{W}$
$78^{\circ} \mathrm{W}$
$117^{\circ} \mathrm{W}$
$200^{\circ} \mathrm{W}$
Solution
In Celsius scale the freezing and boiling point are $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$. In the given imaginary scale the freezing and boiling points are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$. Hence, we can write
$\frac{\mathrm{C}-0}{100}=\frac{\mathrm{W}-39}{200}$
For $\mathrm{C}=39^{\circ}, \frac{39}{100}=\frac{\mathrm{W}-39}{200}$
Solving, $\mathrm{W}=117$