A Man & His Faithful Dog
A Man & His Faithful Dog --- One day, the dog waited patiently by the door, to see which way the man was going, and when the man started along a familiar road, at that very instant, the dog raced along to the end of the road, immediately returning to the man; again racing to the end of the road and again returning. He did this four times in all, at a uniform speed, and then walked at the man's side the remaining distance, which according to the man's paces measured 81 feet. Afterwards the man measured the distance from his door to the end of the road and found it to be 625 feet. Now, if the man walked at exactly 4 miles per hour at a uniform speed, what was the speed of his dog in mph when racing to and fro? This puzzle is from Henry Ernest Dudeney (10 April 1857 – 23 April 1930),
The answer is 16.
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1 solution
Assume Vm = Speed of man and Vd = Speed of dog. Let Vd/Vm = k (>1) & d be the distance between the door and road-end. After the dog runs once to the road-end and returns to the man in 't' units of time, we can say that: Vm t + Vd t = 2d or t = 2d/(Vd+Vm) and the distance walked by the man in time 't' = 2 d Vm/(Vd+Vm) = 2 d/(k+1). Now the new distance between the man and road-end, d1=d - 2d/(k+1) = d(k-1)/(k+1). Likewise, d2 = d[(k-1)/(k+1)]², d3 = d[(k-1)/(k+1)]³ and lastly after the 4th meeting, d4 = d [(k-1)/(k+1)]^4. We're given that d4 = 81 ft. while d = 625 ft. and Vm = 4 mph. Thus, 81 = 625[(k-1)/(k+1)]^4 or (3/5)^4=[(k-1)/(k+1)]^4 or (k-1)/(k+1)=3/5, which yields k = 4 or Vd = 4*4 = 16 mph.