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A solution is to be kept between $68^{\circ} \mathrm{F}$ and $77^{\circ} \mathrm{F}$. What is the range in temperature in degree Celsius $(C)$ if the Celsius / Fahrenheit (F) conversion formula is given by
$F=\frac{9}{5} C+32 ?$
$F=\frac{9}{5} C+32 ?$
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Verified Answer
We have $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$ But $68^{\circ} < \mathrm{F} < 77^{\circ}$
$\therefore 68^{\circ} < \frac{9}{5} \mathrm{C}+32 < 77^{\circ}$
Subtract 32, we get $68-32 < \frac{9}{5} \mathrm{C}+32-32 < 77-32$ $\Rightarrow \quad 36 < \frac{9}{5} \mathrm{C} < 45$
Multiplying by $\frac{5}{9}, 36 \times \frac{5}{9} < \frac{9}{5} \mathrm{C} \times \frac{5}{9} < 45 \times \frac{5}{9}$
$\Rightarrow 20 < \mathrm{C} < 25$. This is the required range.
$\therefore 68^{\circ} < \frac{9}{5} \mathrm{C}+32 < 77^{\circ}$
Subtract 32, we get $68-32 < \frac{9}{5} \mathrm{C}+32-32 < 77-32$ $\Rightarrow \quad 36 < \frac{9}{5} \mathrm{C} < 45$
Multiplying by $\frac{5}{9}, 36 \times \frac{5}{9} < \frac{9}{5} \mathrm{C} \times \frac{5}{9} < 45 \times \frac{5}{9}$
$\Rightarrow 20 < \mathrm{C} < 25$. This is the required range.
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