The Fahrenheit temperature $F$ decreased by $32$ is equal to nine-fifths of the centigrade temperature $C$ . The correct formula framed for the above statement will be :

F + 32 = \frac{5}{9} C

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F - 32 = \frac{5}{9} C

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F + 32 = \frac{9}{5} C

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F - 32 = \frac{9}{5} C

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Solution

The correct option is C $F - 32 = \frac{9}{5} C$
One-fifth of the centigrade temperature is equal to one-ninth of the difference between Fahrenheit $(F)$ temperature and $32$ .
$\frac{C}{5} = \frac{1}{9} (F - 32)$
$F - 32 = \frac{9}{5} C$

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Similar questions

Q. Frame a formula for the following statement :
The Fahrenheit temperature F decreased by 32 is equal to nine-fifths of the centigrade temperature C.

F - 32 = \frac{9}{5} C

;

x^{\circ}

is a temperature at which centigrade and Fahrenheit read the same. Find x.

Q. The formula

F = \frac{9}{5} C + 32

relates any Centigrade reading to its corresponding Fahrenheit reading. For what temperature is the reading the same on both scales?

F - 32 = \frac{9}{5} C

Change the subject of the above formula to :
(a) C, (b) F.

A solution is to be kept between $68^{\circ} F$ and $77^{\circ} F$ . What is the range of temperature in degree Celsius (C) if the Celsius /Fahrenheit (F) convension formula is given by $F = \frac{9}{5} C + 32$ ?

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The Fahrenheit temperature F decreased by 32 is equal to nine-fifths of the centigrade temperature C. The correct formula framed for the above statement will be :

The correct option is C F−32=95COne-fifth of the centigrade temperature is equal to one-ninth of the difference between Fahrenheit (F) temperature and 32. C5=19(F−32)F−32=95C

The Fahrenheit temperature $F$ decreased by $32$ is equal to nine-fifths of the centigrade temperature $C$ . The correct formula framed for the above statement will be :

The correct option is C $F - 32 = \frac{9}{5} C$
One-fifth of the centigrade temperature is equal to one-ninth of the difference between Fahrenheit $(F)$ temperature and $32$ .
$\frac{C}{5} = \frac{1}{9} (F - 32)$
$F - 32 = \frac{9}{5} C$