, 2012

Since the 200 g solution is 25% salt by mass, then $\frac{1}{4}$ of the mass (or 50 g) is salt and the rest (150 g) is water.

When water is added, the mass of salt does not change. Therefore, the 50 g of salt initially in the solution becomes 10% (or $\frac{1}{10}$) of the final solution by mass.

Therefore, the total mass of the final solution is $10 \times 50 = 500$ g.

Thus, the mass of water added is $500 - 200 = 300$ g.
We are told that $F = \frac{9}{5}C+32$.

From the given information $f = 2C+30$.

We determine an expression for the error in terms of $C$ by first determining when $\text{f lt;F}$.

The inequality $\text{f lt;F}$ is equivalent to $\text{2C+30 lt; 9 5 C+32}$ which is equivalent to $\text{1 5 C lt; 2}$ which is equivalent to $\text{C lt;10}$.

Therefore, $\text{f lt;F}$ precisely when $\text{C lt;10}$.

Thus, for $\text{-20 C lt; 10}$, the error equals $F - f = (\frac{9}{5}C+32) - (2C+30)=2-\frac{1}{5}C$.

Also, for $10 \leq C \leq 35$, the error equals $f-F = (2C+30)-(\frac{9}{5}C+32) = \frac{1}{5}C-2$.

When $\text{-20 C lt; 10}$, the error in terms of $C$ is $2-\frac{1}{5}C$ which is linear with negative slope, so is decreasing as $C$ increases. Thus, the maximum value of error in this range for $C$ occurs when $C$ is smallest, that is, when $C=-20$. This gives an error of $2-\frac{1}{5}(-20)=2+4=6$.

When $10\leq C \leq 35$, the error in terms of $C$ is $\frac{1}{5}C-2$ which is linear with positive slope, so is increasing as $C$ increases. Thus, the maximum value of error in this range for $C$ occurs when $C$ is largest, that is, when $C=35$. This gives an error of $\frac{1}{5}(35)-2=7-2=5$.

Having considered the two possible ranges for $C$, the maximum possible error that Gordie would make is 6.