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We had many correct answers to this
problem, including some well argued solutions from Thomas Wong
(Reed High School, Sparks, Nevada, USA) and Stelios Serghiou
(G.C.School of Careers).
Peter Simpson from Castell Alun School
argued as follows:
The number must be odd because the holiday makers split into
two group, with the car group one less than the walkers.
If N was 7 (or greater) the time delays would be greater than 59
minutes, the time for the car to travel to B (the walkers arrive in
60 minutes - N miles in an hour, the car arrives 1 minute
earlier).
If N was 3 the distance would be 22km, but this would mean the
holiday makers would split into 2 walkers and 1 car traveller, and
as the problem states "they" for the car travellers the number of
people in the car must be greater than 1.
Therefore N is 5, and the distance by road is 20km.
Momtchil Iliev from Drayton Manor School
argued in a very similar way and added an explanation for how he
calculated the distance by road:
Calculating Distance of road travelled by car
Time taken driving = 59 - (5) ² - 2(5)
= 59 - 25 - 10 = 24 minutes = 0.4 hours
Speed of travelling by car = 10N = 10(5) = 50 km/h
Distance of road = Speed of car x time taken travelling
= 50 x 0.4 = 20km
Harry also sent in his work on this
problem:
Because each group has at least 2 people in it, and there is one
more walker than person in the car, there must be at least 5
people, so $N\geq 5$.
The walkers walk $N$ km at $N$ km/hr, so take 1 hour, which is 60
minutes.
The people in the car drive $d$ km at $10N km/hr$, which takes
$\frac{60d}{10N}$ minutes, and stop for $N+N^2+N$ minutes. This
takes 60-1=59 minutes. So $\frac{60d}{10N}+N+N^2+N=59$, so
$\frac{6d}{N}+2N+N^2=59$, so $d=\frac{59N-2N^2 -N^3}{6}$. If $N=6$,
this gives 11/6, which is less than $N$, so that's not allowed, and
since it's just going to get smaller and smaller, we must have
$N< 6$. But $N\geq 5$, so $N=5$. So $d=20km$.
Well done to you
all