The temperature at which the Fahrenheit and Celsius scales will give numerically equal (but opposite in sign) value

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$- 40^{0} F a n d 40^{0} C$

both (B) and (C)

$11 . 43^{0} F a n d - 11 . 43^{0} C$

$11 . 43^{0} F a n d 11 . 43^{0} C$

answer is C.

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Detailed Solution

We are asked to find the temperatures in faherenheit & Celsius scale numerical equal but opposite in sign.

Let the temperature $= x^{0} C i n c e l s i u s E q u i v a l e n t t o = x^{0} F W e h a v e r e l a t i o n \frac{C}{100} = \frac{F - 32}{180} \Rightarrow \frac{x}{100} = \frac{- x - 32}{180} \Rightarrow \frac{x}{5} = \frac{- x - 32}{9} \Rightarrow 9 x = - 5 x - 160 ∴ 14 x = - 160 ∴ x = \frac{- 160}{14} = - \frac{80}{7} = - 11 . 43^{0} C T h e n i t s - x^{0} F = 11 . 43^{0} F F o r - 11 . 43^{0} C w e g e t F a h r e n h e i t m e a s u r e 11 . 43^{0} F S i m i l a r l y i f w e c o n s i d e r T e m p e r a t u r e = x^{0} F T h e n i t s e q u i v a l e n t i n C e l s i u s s c a l e s h o u l d b e - x^{0} C U \sin g r e l a t i o n \frac{C}{100} = \frac{F - 32}{180} \frac{- x}{100} = \frac{x - 32}{180} \Rightarrow \frac{- x}{5} = \frac{x - 32}{9} - 9 x = 5 x - 160 - 14 x = - 160 ∴ x = 11 . 43^{0} F T h e n i t s C e l s i u s e q u i v a l e n t i s - 11 . 43^{0} F i n b o t h c a s e s h a v i n g s a m e u m e r i c a l v a l u e b u t o p p o s i t e i n s i g n .$