Question

two cyclists leave town 250 miles apart at the same time and Travel toward each other. what cyclist travels 7 mi/h slower than the other. if they meet in 5 hours, what is the rate of each cyclist? rate of the slower cyclists: __mi/hrate of the faster cyclist: __mi/h

Answer 1

let

a = rate of the faster cyclist

a - 7 = rate of the slower cyclist

$\begin{gathered} \text{rate}=\frac{dis\tan ce}{\text{time}} \\ dis\tan ce=\text{rate}\times time \end{gathered}$

Therefore,

$\begin{gathered} \text{distance covered by the faster cyclist=5a} \\ \text{distance covered by the slower cyclist=5(a-7)}=5a-35 \end{gathered}$$\begin{gathered} 215=5a+5a-35 \\ 215=10a-35 \\ 215+35=10a \\ 250=10a \\ a=\frac{250}{10} \\ a=25\text{ mile per hour} \end{gathered}$$\begin{gathered} \text{rate of the slower cyclist = 25-7=18 miles per hour} \\ \text{rate of the faster cyclist=25 miles per hour} \end{gathered}$