KCC's quiz AQQ274 about 2 sportsmen running and walking

Take: w the walking speed, r the running speed, d the total distance, t1 the time taken byt the first sportsman and t2, taken by the second sportsman.

First sportsman:  t1 = (d/2)/w  + (d/2)/r     so  t1=(d/2) ( 1/w +1/r)  = (d/2) (w+r)/(wr)
Second sportman:  w(t2/2) + r(t2/2) = d    so  t2 =                              2d / (w+r)

t1/t2  =  1/4  (w+r)^2  / wr.  If that value is 1.0, they both reach the end at the same time.  Less that 1, t1 is less than t2 so the first sportsman arrives first, and if greate than 1, the second sportsman arrives first.



Clearly, that depends on the ratio of w/r.  If w/r  =1,  the expression leads us to conclude that they both arrive at the same time.
If we assume running is faster than walking, the expression leads us to conclude that the first sportsman arrives last.

Have  a = r/w, t1/t2=1/4  (1+a)^2  / a.  When equals to 1, we have   4a = (1+a)^2, leading to a = 1, that is, if running equals walking, both arrive at the same time.
If we assume running is faster than walking, as with a=2 for example, the expression lead us  9/8, so to conclude that the first sportsman arrives last.  (its time, t1, is larger than t2) 

Take: w the walking speed, r the running speed, d the total distance, t1 the time taken byt the first sportsman and t2, taken by the second sportsman.

First sportsman:  t1 = (d/2)/w  + (d/2)/r     so  t1=(d/2) ( 1/w +1/r)  = (d/2) (w+r)/(wr)
Second sportman:  w(t2/2) + r(t2/2) = d    so  t2 =                              2d / (w+r)

t1/t2  =  1/4  (w+r)^2  / wr.  If that value is 1.0, they both reach the end at the same time.  Less that 1, t1 is less than t2 so the first sportsman arrives first, and if greate than 1, the second sportsman arrives first.



Clearly, that depends on the ratio of w/r.  If w/r  =1,  the expression leads us to conclude that they both arrive at the same time.
If we assume running is faster than walking, the expression leads us to conclude that the first sportsman arrives last.

Have  a = r/w, t1/t2=1/4  (1+a)^2  / a.  When equals to 1, we have   4a = (1+a)^2, leading to a = 1, that is, if running equals walking, both arrive at the same time.
If we assume running is faster than walking, as with a=2 for example, the expression lead us  9/8, so to conclude that the first sportsman arrives last.  (its time, t1, is larger than t2) 

Congratulations vanderghast , you get it; and with a crystal clear explanation!

We are dearly missing your "Quote of the Day" in the latest quiz  :-)

Yes, I will reconsider to put it again... I made a pause because their selection requires more verification (i.e.  they have to be reviewed by our marketing to be sure they are not hurting people sensibility....). Some of them can appear little bit sarcastic...