On a new scale of temperature (which is linear) and called the `W` scale. The freezing and boiling points of water are `39^(@)W` and `239^(@)W` respectively. What will be the temperature on the new scale, corresponding to a temperature of `39^(@)C` on the Celsius scale?

Options

A. `200^@ W`
B. `139^@ W`
C. `78^(@) W`
D. `117^(@)W`

Answer

D

Explanation

`(T_C)/(100) = (W-39)/(239-39) rArr (39)/(100) = (W-39)/(200)`
`rArr 39 xx 2 + 39 = W rArr W = 117^@`.