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**running-gag** Let d = 16 miles

Let $\displaystyle t_d = 20 $ minutes

Let $\displaystyle t_u = 30 $ minutes

The speed of the boat when it goes downstream is x+y

Therefore $\displaystyle \frac{d}{x+y} = t_d$

Inverse gives (i) $\displaystyle \frac{x}{d} + \frac{y}{d} = \frac{1}{t_d}$

The speed of the boat when it goes upstream is x-y

Therefore $\displaystyle \frac{d}{x-y} = t_u$

Inverse gives (ii) $\displaystyle \frac{x}{d} - \frac{y}{d} = \frac{1}{t_u}$

Equation (i) - equation (ii) gives

$\displaystyle 2 \frac{y}{d} = \frac{1}{t_d} - \frac{1}{t_u}$

$\displaystyle y = \frac{d}{2} (\frac{1}{t_d} - \frac{1}{t_u})$

y = 8 mph