A plane flies from Penthaven to Jackson and then back to Penthaven. When there is no wind, the round trip takes $3$ hours and $20$ minutes, but when there is a wind blowing from Penthaven to Jackson at $70$ miles per hour, the trip takes $3$ hours and $50$ minutes. How many miles is the distance from Penthaven to Jackson?
(Assume that the plane flies at a constant speed, and that the turnaround time is negligible.)
Let the distance between Penthaven and Jackson be $d$, in miles, and let the plane's speed be $r$, in miles per hour. The rate at which the plane travels is $r$ miles per hour, so he spends $\frac{d}{r}$ hours in the air flying from Penthaven to Jackson, and the same amount of time for Jackson to Penthaven, so the trip with no wind takes $2 \cdot \frac{d}{r} = \frac{2d}{r}$ hours. When there is wind blowing from Penthaven to Jackson at 70 miles per hour, the rate at which the plane travels is $r - 70$ miles per hour, so he spends $\frac{d}{r-70}$ hours in the air flying from Pentahven to Jackson, and the same amount of time for Jackson to Penthaven, so the trip with wind takes $2 \cdot \frac{d}{r-70} = \frac{2d}{r-70}$ hours. Taking units into account, the plane flies $d$ miles at a speed of $\frac{r \text{ miles}}{\text{hour}}$, so when the wind is blowing he flies $d$ miles at a speed of $\frac{r-70 \text{ miles}}{\text{hour}}$, so by the definition of speed, $\frac{r \text{ miles}}{\text{hour}} = \frac{d \text{ miles}}{\frac{r-70}{\text{hour}}}$, and simplifying, $r \cdot \frac{r-70}{\text{hour}} = d$, so $r(r-70) = d(\text{hour})$.
Converting to hours, we see that $3 \text{ hours and } 20 \text{ minutes } = 3 \frac{1}{3} \text{ hours } = \frac{10}{3} \text{ hours } = \frac{10}{3} \cdot \frac{60}{60} \text{ hours } = \frac{20}{6} \text{ hours } = \frac{10}{3} \text{ hours }$ and $3 \text{ hours and } 50 \text{ minutes } = 3 \frac{5}{6} \text{ hours} = \frac{18}{6} \text{ hours}
= \frac{18}{6} \cdot \frac{60}{60} \text{ hours } = \frac{54}{6} \text{ hours } = \frac{9}{1} \text{ hours}$, so $\frac{2d}{r} = \frac{10}{3} \text{ hours } = \frac{9}{1} \text{ hours } = \frac{2d}{r-70}$. Cross multiplying, $2d \cdot (r-70) = r \cdot \left(\frac{10}{3} \cdot (r-70)\right)$, so $2dr - 140d = r \cdot \left(\frac{10}{3} r - \frac{70 \cdot 10}{3}\right)$, so $2dr - 140d = \frac{10}{3} r^2 - \frac{700}{3} r$, so $2dr - \frac{10}{3} r^2 + \frac{700}{3} r - 140d = 0$, so $2dr + \frac{700}{3} r - \frac{10}{3} r^2 = 140d$, so $\frac{6dr+2100r-10r^2}{3} = 140d$, so $6dr + 2100r - 10r^2 = 420d$, so $3dr + 1050r - 5r^2 = 210d$, so $-5r^2 + (3d + 1050) r - 210d = 0$.
The distance from Penthaven to Jackson provided in the answer choices is $d = 260$, so we substitute $d=260$ to get $-5r^2 + (780 + 1050) r - 210(260) = 0$, so $-5r^2 + 1830r - 54600 = 0$, so dividing all coefficients by $5$, $-r^2 + 366r - 10920 = 0$, so $(r-180)(r-186) = 0$, so $r = 180$ or $r = 186$. When the plane travels at a speed of $r = 180$ miles per hour, $r - 70 = 110$ miles per hour, so it takes $\frac{360}{110} = \frac{180}{55} \text{ hours } = \frac{36}{11} \text{ hours}$ in the air, so it takes $3 + \frac{36}{11} \text{ hours } = \frac{108}{11} \text{ hours}$ for the round trip, which is not $\frac{10}{3} = \frac{33}{10}$, as required, so we discard the answer of Penthaven to Jackson being $260$ miles.
When the plane travels at a speed of $r = 186$ miles per hour, $r - 70 = 116$ miles per hour, so it takes $\frac{360}{116} = \frac{180}{58} \text{ hours } = \frac{90}{29} \text{ hours}$ in the air, so the round trip takes $3 + \frac{90}{29} \text{ hours} = \frac{87}{29} \text{ hours}$, so this time corresponds to $\frac{10}{3}$ hours if $\frac{87}{29} = \frac{10}{3}$, so the answer is $\boxed{186}$.