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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 (T-4). Make a new line in your chart. D = R x T ?? = 24 x (T+5) ?? = 24(125%) x (T-4)
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 (T-4)
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 (T-4) 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.
An increase in speed by 25% means reduction in time by 20% (percentage change graphic) which is 9. so actual time is 9x5=45 mins. At the speed of 24km/hr, time taken to reach is 3/4 hrs. hence distance = 24x3/4 =18km (A one-liner if you'r through with the basics of percentages)
Re: M04 Q21 - soccer coach bike ride
17 Oct 2013, 03:18