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On a temperature scale ' $X$ '. The boiling point of water is $65^{\circ} \mathrm{X}$ and the freezing point is $-15^{\circ} \mathrm{X}$. Assume that the $\mathrm{X}$ scale is linear. The equivalent temperature corresponding to $-95^{\circ} \mathrm{X}$ on the Farenheit scale would be:
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The correct answer is:
$-148^{\circ} \mathrm{F}$
Boiling point of water, $=$ UFP $65^{\circ}$
Freezing point of water, $=\mathrm{LFP}-15^{\circ}$
$\frac{\mathrm{X}-\mathrm{LFP}}{\mathrm{UFP}-\mathrm{LFP}}=\frac{\mathrm{T}_{\mathrm{F}}-32}{212-32}$
$\Rightarrow \frac{-95-(-15)}{65-(-15)}=\frac{\mathrm{T}_{\mathrm{F}}-32}{180}$
$\Rightarrow \frac{-80}{80}=\frac{T_F-32}{180} \Rightarrow T_F=-180+32$
$\Rightarrow \mathrm{T}_{\mathrm{F}}=-148^{\circ} \mathrm{F}$
Freezing point of water, $=\mathrm{LFP}-15^{\circ}$
$\frac{\mathrm{X}-\mathrm{LFP}}{\mathrm{UFP}-\mathrm{LFP}}=\frac{\mathrm{T}_{\mathrm{F}}-32}{212-32}$
$\Rightarrow \frac{-95-(-15)}{65-(-15)}=\frac{\mathrm{T}_{\mathrm{F}}-32}{180}$
$\Rightarrow \frac{-80}{80}=\frac{T_F-32}{180} \Rightarrow T_F=-180+32$
$\Rightarrow \mathrm{T}_{\mathrm{F}}=-148^{\circ} \mathrm{F}$
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