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Apr 6, 2016
 #2
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Mmmmm...I'm not sure about Guest's answer.....it could be correct......but consider

 

It will take Rick 18km/ [9k//hr]  = 2 hours to finish  

 

And runnining at his current pace, it will take D**k 2.5 hours to finish

But...he needs to cut this to 2 hours just to finish even with Rick

 

Call his normal rate per hour, R  and call his increased rate in order to catch Rick, (R + x)

 

And since R * T  = D.......we have this situation......

 

At his current rate, the distance he has left to run = R *2.5

And at  his his increased rate, the distance he has left to  run is the same....and this is given by:

(R + x)*2  = D     .....so.....equating the distance, we have......

 

R * 2.5  =  (R + x) * 2    simplify

 

2.5R  = 2R + 2x       subtract 2R from both sides

 

.5R  = 2x      divide both sides by 2

 

.25R   = x 

 

So.....he needs to run (R + .25R)    = 1.25R   = 5/4 his current rate just to tie Rick   

 

Consider this in light of the Guest's answer.......if 5/4  of D**k's current rate = 11.25km/hr, then his current rate must  just be 4/5 of this......but 4/5 of this = 9km/hr......the very same rate that Rick runs.....and if D**k runs at the same rate as Rick, he would be even with him, not 30 minutes behind....!!!

 

In my opinion.....we actually can't provide an exact numerical answer to this because we would have to know how far D**k has run and the time he has taken to do so to get his current rate.....

 

I may have blown a fuse, here......can some other mathematician look at  this   ????

 

 

cool cool cool

Apr 6, 2016
 #1
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Apr 6, 2016
Apr 5, 2016

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