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Re: M04 Q21  soccer coach bike ride [#permalink]
16 Oct 2013, 08:28
A.
A good place to start is with organizing the given information into a Distance = Rate x Time chart.
D = R x T
We are trying to find D, the distance. We are given a set of clues to include the coach's rate of travel and variations of rate and time for the same distance.
We know that the dude is traveling at 24 km/hr.
D = R x T ?? = 24 x ??
We know that he was 5 minutes late. Let T = time in minutes. T+5. D = R x T ?? = 24 x (T+5)
We know that if he traveled 25% faster (1 + 25/100 = 125/100 or 125%), he would be 4 minutes early (T4). Make a new line in your chart. D = R x T ?? = 24 x (T+5) ?? = 24(125%) x (T4)
Here it would be useful to convert everything into the same units (minutes not hours). 24 k/hr converted into kilometers/minute = 24k/60min = 2km/5min. Also, 125% of 2/5 is 1/2.
Now you have (Let D = the distance the coach traveled): D = R x T D = 2/5 x (T+5) D = 1/2 x (T4)
Since the distance is the same in both the late and the early scenarios, you can set the equations equal to each other and solve for Time, your only variable. 2/5 x (T+5) = 1/2 x (T4) Which gives you T = 40 minutes.
Plug 40 minutes into either equation to find the Distance. D = 2/5 x (40+5) D = 18 km. And this is the answer.
Long algebraic way to do it, I'm sure there are more efficient ways.
