Question

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

• A. 50 < F < 78
• B. 50 < F < 68
• C. 49 < F < 68
• D. 49 < F < 78

Answer 1

Answer: (B)

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