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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?

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boffeemadrid

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:

[tex]d=s(2)[/tex]

[tex]d=2s[/tex]                                                            (1)

and equation for the faster hiker:

[tex]15-d=(s+0.5)(2)[/tex]

[tex]15-d=2s+1[/tex]

[tex]2s+d=14[/tex]                                                       (2)

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

[tex]2s+2s=14[/tex]

[tex]4s=14[/tex]

[tex]s=3.5[/tex]

and [tex]s+0.5=3.5+0.5=4.0[/tex]

The speed of the slower hiker is [tex]3.5\frac{mi}{hr}[/tex]and The speed of the faster hiker is [tex]4\frac{mi}{hr}[/tex].

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