On a temperature scale ‘X’, the boiling point of water is 65°X and the freezing point is –15°X. Assume that the X scale is linear. The equivalent temperature corresponding to –95°X on the Fahrenheit scale would be [2023]

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Solution

Q.
On a temperature scale ‘X’, the boiling point of water is 65°X and the freezing point is –15°X. Assume that the X scale is linear. The equivalent temperature corresponding to –95°X on the Fahrenheit scale would be [2023]

1 –63°F

2 –112°F

3 –48°F

4 –148°F

Ans.
(4)

$\frac{X - X_{freez}}{X_{boil} - X_{freez}} = \frac{t - 32}{212 - 32}$

$\frac{- 95 - (- 15)}{65 - (- 15)} = \frac{t - 32}{180}$

$\Rightarrow \frac{- 80}{80} = \frac{t - 32}{180}$

$t = - 180 + 32$

$\Rightarrow t = - 148 ° F$

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