Question

On a particular day, the wind added 5 miles per hour to Alfonso's rate when he was cycling with the wind and subtracted 5 miles per hour from his rate on his return trip. Alfonso found that in the same amount of time he could cycle 60 miles with the wind, he could go only 30 miles against the wind. What is his normal bicycling speed with no wind?

Answer 1

now, this is pretty much the same as the one before it with Jaime, so I'll do this without much fuss.

recall d = rt.

a = Alfonso's rate

with the wind his speed is a + 5, against it is a - 5, 60 miles with it and 30 miles against it, all in the same time of t hours.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{with the wind}&60&a+5&t\\ \textit{against the wind}&30&a-4&t \end{array} \\\\\\ \begin{cases} 60=(a+5)t\implies \frac{60}{a+5}=\boxed{t}\\\\ 30=(a-4)t\\ -------------\\ 30=(a-4)\left( \boxed{\frac{60}{a+5}} \right) \end{cases} \\\\\\ 30(a+5)=(a-4)60\implies 30a+150=60a-240 \\\\\\ 390=30a\implies \cfrac{390}{30}=a\implies 13=a$