The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by

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$\frac{t_{Q}}{150} = \frac{t_{p} - 180}{100}$

$\frac{t_{P}}{180} = \frac{t_{Q} - 40}{100}$

$\frac{t_{Q}}{100} = \frac{t_{p} - 30}{150}$

$\frac{t_{P}}{100} = \frac{t_{Q} - 180}{150}$

answer is B.

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Detailed Solution

$From graph at t_{Q} = 0, t_{P} = 30^{\circ}$
$\frac{t_{p} - 30^{\circ}}{180^{\circ} - 30^{\circ}} = \frac{t_{q} - 0^{\circ}}{100 - 0} \Rightarrow \frac{t_{p} - 30}{150} = \frac{t_{q}}{100}$

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