The temperature at which numerical value of the Celsius reading is half that of the Fahrenheit reading

40^{\circ} C

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20^{\circ} C

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160^{\circ} C

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80^{\circ} C

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Solution

The correct option is D $160^{\circ} C$
$\frac{2 x - 32}{180} = \frac{x}{100}$
$0.2 x = 32$
$\Rightarrow x = 160^{o} C$

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Similar questions

\frac{5 F - 160}{9}

A measured temperature on Fahrenheit scale is $200^{\circ} F .$ What will this reading be on celsius scale ?

(a) $40^{\circ} C$ (b) $94^{\circ} C$

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