Question

Two hikers start from the same point and hike in opposite directions around a lake whose shoreline is 15 mi long. One hiker walks 0.5 mph faster than the other hiker. How fast did each hiker walk if they meet in 2 h?

Answer 1

Answer:

Step-by-step explanation:

Let the distance the slower hiker travels=d,

the distance the faster hiker travels=15-d (Since, the sum of the distances they walk is 15mi).

Let the speed of the slower hiker in mi/hr=s,

the speed of the faster hiker=s+0.5

Now, equation for the slow hiker:

$d=s(2)$

$d=2s$                                                            (1)

and equation for the faster hiker:

$15-d=(s+0.5)(2)$

$15-d=2s+1$

$2s+d=14$                                                       (2)

Now, substituting the equation (1) in equation (2),

$2s+2s=14$

$4s=14$

$s=3.5$

and $s+0.5=3.5+0.5=4.0$

The speed of the slower hiker is $3.5\frac{mi}{hr}$and The speed of the faster hiker is $4\frac{mi}{hr}$.