Tom and his brother Harry had a foot race from their home to school, which was an integral number of yards away. Harry ran faster than Tom, but less than twice as fast, and arrived at the school when Tom was still a 2-digit integer of yards behind.

The next day they ran the race again, but this time Harry started farther away from the school by the same amount as the winning margin the previous day, while Tom still started the same time as Harry, but again from home. Of course, since Harry still runs faster, the same ratio as the day before, he finished the difference in distances in less time than it would have taken Tom, but this time the gap at the end was reduced so that when Harry reached school, the amount by which Tom was behind had the two digits reversed from the preceding day.

How far was it from their home to school, and what was their gap at the end of each run when Harry arrived at school?

let d be distance from home to school
let h be rate of henry
let t be rate of tom
let n be gap during first race
let r be reverse of n

then we have
(1) d(h-t)/h=n from first race
and in the second race we have
d-((d+n)t/h)=r
(dh-dt-nt)/h=r
(d(h-t)/h)-(nt/h)=r
n-(nt/h)=r
n-r=nt/h
h=nt/(n-r)
putting this into (1) we get
d=n^2/r
so we need n to be a 2 digit number with n^2/r a integer.
this is only true for 72 and 84.

Now we also need that t<h<2t and this gives us
t<nt/(n-r)<2t
t(n-r)<nt<2t(n-r)
(n-r)<n<2(n-r)
and of 72 and 84 this is only true for 72
so we and say that the distance from home to school is
72^2/27=192 and the 2 gaps are 72 and 27

whats interesting is that this is independent of the speeds of henry and tom.

Posted by Daniel on 2009-05-20 12:12:17