Question

An athlete cycles 8 mi/hr faster than the run. Every morning they cycle 4 miles and run 2.5 miles for a total of one hour of exercise. How fast do they run?

Answer 1

Let the speed for running be x

Let the speed for cycling be y

Given that the athlete cycles 8 miles/hr faster than he runs, we have

$y\text{ = (x +8) ------ equation 1}$

Every morning, the athlete cycles 4 miles and runs 2.5 miles for a total of 1 hour.

This implies that the total time for cycling and running is 1 hour.

$\begin{gathered} \text{speed}=\frac{\text{distance}}{\text{time}} \\ \Rightarrow time=\frac{dis\tan ce}{speed} \end{gathered}$

Time for cycling:

$t_{\text{cycling}}=\frac{4\text{miles}}{x\text{ miles/hr}}$

Time for running:

$t_{cycling}=\frac{2.5\text{ miles}}{y\text{ miles/hr}}$$\begin{gathered} t_{cycling}+t_{running}=1\text{ hour} \\ \frac{4}{x}+\frac{2.5}{y}=1-------\text{ equation 2} \end{gathered}$

Substitute equation 1 into equation 2

$\begin{gathered} \frac{4}{x}+\frac{2.5}{y}=1 \\ \frac{4}{x}+\frac{2.5}{(x+8)}=1 \\ \text{LCM = x(x+8)} \\ \frac{4(x+8)+2.5x}{x(x+8)}=1 \\ open\text{ the brackets} \\ \frac{4x+32+2.5x_{}}{x^2+8x}=1 \\ \frac{6.5x+32}{x^2+8x}=1 \\ \text{cross}-mul\text{tiply} \\ 6.5x+32=x^2+8x \\ \text{collect like terms} \\ x^2+8x-6.5x-32=0 \\ x^2+1.5x-32=0 \\ \end{gathered}$

solving the quadratic equation by factorization, we have

$\begin{gathered} x^2+1.5x-32=0 \\ (x-4.956)(x+6.456)=0 \\ x-4.956=0 \\ x\Rightarrow4.956 \\ x+6.456=0 \\ \Rightarrow x=-6.456 \end{gathered}$

since x and y are the speeds of running and cycling, their values cannot be negative,

Thus, x= 4.956 miles/hr

Hence, they run at 4.956 miles/hr