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  4. Let (C/5) = (F - 32/9). If C lies between 10 and 20, then :

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Q. Let $ \frac{C}{5} = \frac{F - 32}{9}$. If C lies between 10 and 20, then :

Linear Inequalities

Solution:

Given: $\frac{ C }{5}=\frac{ F -32}{9}$ and $10< C <20$
$\Rightarrow C =\frac{5 F -(32) 5}{9}$
Since, $ 10< C <20$
$\Rightarrow 10<\frac{5 F -160}{9}<20$
$\Rightarrow 90<5 F -160<180$
$\Rightarrow 90+160<5 F <180+160 $
$\Rightarrow 250<5 F <340 $
$\Rightarrow \frac{250}{5}< F <\frac{340}{5} $
$\Rightarrow 50< F <68$