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  • Unknown's avatar

    Jim Randell 8:28 am on 17 March 2024 Permalink | Reply
    Tags: by: C M Gates   

    Brain-Teaser 948: Journey north-east 

    From The Sunday Times, 21st September 1980 [link]

    For the purpose of my problem I have to choose two towns 26 miles apart. I might have chosen Oxford and Newbury, but it would be more appropriate for me as a Scotsman to go much farther north and choose Kingussie and Grantown-on-Spey, where the roads are somewhat less busy.

    Alf, Bert and Charles start off at the same time from Kingussie to make their way north-eastwards to Grantown-on-Spey, 26 miles distant.

    Alf walks at a constant speed of four miles per hour. Bert and Charles drive together in a car. After a certain time, Bert leaves the car, and walks forward at the same rate as Alf, while Charles drives back to meet Alf.

    Alf gets Into the car with Charles, and they continue to drive to Grantown-on-Spey, arriving there just as Bert does.

    On each stretch Charles averages 40 miles per hour.

    What is the time (in minutes) taken for them all to travel from Kingussie to Grantown-on-Spey?

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994).

    [teaser948]

     
    • Jim Randell's avatar

      Jim Randell 8:29 am on 17 March 2024 Permalink | Reply

      See: Teaser 3140, where we determined:

      If there are k pedestrians to be transported a distance d, and each walks a distance x at velocity w and is transported a distance (d − x) at velocity v, and the total time taken is t, then we have:

      n = 2k − 1
      x = dw(n − 1) / (v + wn)
      t = d(w + vn) / v(v + wn)

      We can plug the numbers for this puzzle in and calculate the result:

      Run: [ @replit ]

      from enigma import (fdiv, printf)
      
      # initial conditions
      k = 2   # number of walkers
      d = 26  # distance
      w = 4   # walking speed
      v = 40  # driving speed
      
      # calculate walking distance per person
      n = 2 * k - 1
      x = fdiv(d * w * (n - 1), v + w * n)
      
      # calculate time taken
      t = fdiv(x, w) + fdiv(d - x, v)
      
      # output solution
      printf("t = {t:g} hours (= {m:g} min) [x={x:g} miles]", m=t * 60)
      

      Solution: The total time taken is 93 minutes.

      Alf walks the first 4 miles (in 60 minutes), and is driven the remaining 22 miles (in 33 minutes).

      Bert is driven 22 miles first, and walks the last 4 miles.

      Charles drives 22 miles to drop off Bert, returns 18 miles to collect Alf, and then 22 miles to the destination, a total of 62 miles (in 93 minutes).

      So each arrives at the destination after 93 minutes.

      Like

    • Lise Andreasen's avatar

      Lise Andreasen 4:22 pm on 4 April 2024 Permalink | Reply

      Alf walks x miles. For symmetry reasons, Bert also walks x miles. In the middle we have y miles. 26 = x + y + x.
      They all spend the same amount of time.
      x/4 + (x+y)/40 = (2x+3y)/40.
      Solve.

      Like

  • Unknown's avatar

    Jim Randell 9:32 am on 17 September 2023 Permalink | Reply
    Tags: by: C M Gates   

    Brain-Teaser 919: Golden age of steam 

    From The Sunday Times, 2nd March 1980 [link]

    Ever since the dawn of history, mankind has looked back to a Golden Age when everybody and everything was so much better.

    As far as our railway system is concerned, the Golden Age was the prewar era of the great steam locomotives. In spite of their noise, steam, and smoke pollution, (or perhaps even because of these characteristics), the older generation look back with nostalgia to that golden era. Let us therefore pose a not-too-difficult problem concerning those wonderful days.

    Two steam locomotives, each travelling at 50 mph, are driven by Tom and Dick. They are approaching Harry’s signal box from opposite directions.

    Tom sounds his whistle. Immediately upon hearing it, Dick replies. Harry hears both whistles with an interval of seven seconds between them. The two engines later pass the signal box at the same instant.

    You may assume that the speed of sound was 1100 feet per second relative to the ground. One mile has 5280 feet.

    How far (in feet) was Tom from the box when he sounded his whistle?

    This puzzle is included in the book The Sunday Times Book of Brainteasers (1994). The puzzle text above is taken from the book.

    [teaser919]

     
    • Jim Randell's avatar

      Jim Randell 9:33 am on 17 September 2023 Permalink | Reply

      The trains are travelling at the same speed, and (presumably on different tracks) pass the signal box at the same time, so they must start the same distance from the signal box.

      Working in feet per second, a speed of 50 mph = 220/3 ft/s.

      Suppose at time t = 0 both T and D are a distance x feet from H, and T sounds his whistle.

      At a time t = x / 1100 H hears T’s whistle.

      And if D hears the whistle when he is a distance y from H, then he hears T’s whistle at time t = (x + y) / 1100.

      D instantaneously responds (still at distance y), and H hears D’s whistle at time t = (x + 2y) / 1100.

      H hears the whistles 7 seconds apart, so:

      (x + 2y) / 1100 = 7 + x / 1100
      y = (7/2)1100 = 3850

      So D sounds his whistle when he is 3850 ft from H at time t = (x + 3850) / 1100.

      And at this time T must also be 3850 ft from H.

      Hence:

      x = 3850 + (220/3)(x + 3850) / 1100
      ⇒ x = 4400

      Solution: Tom was 4400 ft from the signal box when he sounded his whistle.

      Like

    • Hugo's avatar

      Hugo 9:49 am on 17 September 2023 Permalink | Reply

      It must have been a cold day: the speed of sound is about 1100 ft/s at 5°C and 60% relative humidity
      (about 3 ft/s less in dry air at that temperature).

      Like

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