Question

On a temperature scale '$\mathbf{X}$'. The boiling point of water is $65^{\circ}\mathbf{X}$ and the freczing point is $-15^{\circ}\mathbf{X}$.

Assume that the $\mathbf{X}$ scale is linear. The equivalent temperature corresponding to $-95^{\circ}\mathbf{X}$ on the Farenheit scale would be:

• A. $-63^{\circ}$F
• B. $-112^{\circ}$F
• C. $-48^{\circ}$F
• D. $-148^{\circ}$F

Answer 1

Answer: (D)

The equivalent temperature on the Fahrenheit scale is $-148^{\circ}\text{F}$.

Answer 2

The relationship between two linear temperature scales is given by the formula:

$\frac{T_1 - FP_1}{BP_1 - FP_1} = \frac{T_2 - FP_2}{BP_2 - FP_2}$

where $T$ is the temperature, $FP$ is the freezing point, and $BP$ is the boiling point of water on the respective scales (Scale 1 and Scale 2).

Let Scale 1 be the X scale and Scale 2 be the Fahrenheit scale.

Given: Freezing point on X scale ($FP_X$) = $-15^{\circ}\text{X}$ Boiling point on X scale ($BP_X$) = $65^{\circ}\text{X}$ Freezing point on Fahrenheit scale ($FP_F$) = $32^{\circ}\text{F}$ Boiling point on Fahrenheit scale ($BP_F$) = $212^{\circ}\text{F}$ Given temperature on X scale ($T_X$) = $-95^{\circ}\text{X}$

We need to find the equivalent temperature on the Fahrenheit scale ($T_F$). Using the formula: $\frac{T_X - FP_X}{BP_X - FP_X} = \frac{T_F - FP_F}{BP_F - FP_F}$ Substitute the given values: $\frac{-95 - (-15)}{65 - (-15)} = \frac{T_F - 32}{212 - 32}$ $\frac{-95 + 15}{65 + 15} = \frac{T_F - 32}{180}$ $\frac{-80}{80} = \frac{T_F - 32}{180}$ $-1 = \frac{T_F - 32}{180}$ Multiply both sides by 180: $-1 \times 180 = T_F - 32$ $-180 = T_F - 32$ $T_F = -180 + 32$ $T_F = -148$

Thus, the equivalent temperature corresponding to $-95^{\circ}\text{X}$ on the Fahrenheit scale is $-148^{\circ}\text{F}$.