On an imaginary linear scale of temperature (called 'W' scale) the freezing and boiling points of water are $39^\circ\text{W}$ and $239^\circ\text{W}$ respectively. The temperature on the new scale corresponding to $39^\circ\text{C}$ temperature on Celsius scale will be
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Notice that the total range of the 'W' scale ($239 - 39 = 200$ divisions) is exactly twice as large as the Celsius scale (100 divisions). This means each $1^\circ\text{C}$ step corresponds to exactly a $2^\circ\text{W}$ step! So, a change of $39^\circ\text{C}$ means an increase of $39 \times 2 = 78$ units above the base freezing point: $39 + 78 = 117^\circ\text{W}$. A very fast mental verification!
Step 1: Use the linear scale rule.
Any two linear temperature scales match through equal fractions between fixed points. \[ \frac{C-0}{100-0}=\frac{W-39}{239-39} \]
Step 2: Simplify the W side.
The W range is $239-39=200$, so $\frac{C}{100}=\frac{W-39}{200}$.
Step 3: Put in $C=39$.
$\frac{39}{100}=\frac{W-39}{200}$, so $W-39=39\times2=78$.
Step 4: Solve for $W$.
$W=78+39=117^\circ$W. \[ \boxed{117^\circ\text{W}} \]