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  • On a temperature scale 'X', the boiling point of water is and the freezing

On a temperature scale 'X', the boiling point of water is 65^{0}\mathrm{X} and the freezing point is -15^{0} \mathrm{X}. Assume that the
X scale is linear. The equivalent temperature corresponding to -95^{0} \mathrm{X} on the Farenheit scale would be :
 

Option: 1

-63^{0}\mathrm{F}


Option: 2

-148^{0}\mathrm{F}


Option: 3

-48^{0}\mathrm{F}


Option: 4

-112^{0}\mathrm{F}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \frac{X_T-X_L}{X_H-X_L}=\frac{T_F-32}{212-32} \\ \end{aligned}}

\begin{aligned} & \frac{-95^{\circ}-\left(-15^{\circ}\right)}{65^{\circ}-\left(-15^{\circ}\right)}=\frac{T_F-32}{180} \\ \end{aligned}

\begin{aligned} & \frac{-80^{\circ}}{80^{\circ}}=\frac{T_F-32}{180^{\circ}} \\ \end{aligned}

\begin{aligned} & -180=T_F-32 \\ & T_F=-180+32=-148^{\circ} \mathrm{F} \end{aligned}

Posted by

Pankaj

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