Riding Against the Wind Puzzle

Problem: A horse rider went a mile in 5 minutes with the wind and returned in 7 minutes against the wind. How fast could he ride a mile if there was no wind?

horse-riding puzzle

Solution:

Most of us will proceed like that if a rider goes a mile in 2 minutes with the wind, and returns against the wind in 3 minutes, that 2 and 3 equal 5, should give a correct average, so that time taken should be two and half minutes. We find this answer to be incorrect, because the wind has helped him for only 2 minutes, while it has worked adversely for 3 minutes.

If he could ride a mile in 2 minutes with the wind, it is clear that he could go a 1.5 mile 3 minutes, and 1 mile in 3 minutes against the wind.

Therefore 2.5 miles in 6 minutes gives his actual speed, because the wind helped him just as much as it has retarded him, so his actual speed for a single mile without any wind would be (2.5)/6 = 5/12 miles/sec

11 Thoughts on “Riding Against the Wind Puzzle

  1. (H+w) *5=(H-W)*7=1 w is speed of wind and h is speed of horse gives speed of horse H=6/35 therefore answer is 6/35 miles per minute

    • Rishabh kumar gaur on September 2, 2015 at 9:58 am said:

      nice work dude…..hats off…

    • And to complete the answer, given that the question is “How fast could he ride a mile if there was no wind?”, we must invert that (because time = 1 mile / speed), to get 35/6 minutes, which is 5 minutes and 50 seconds.

    • shivam shukla on May 29, 2016 at 10:37 am said:

      you are very correct dude!!!
      Even i was confused by the answer of puzzle.
      Maths did its job.. good work .

  2. Nitin Gupta on April 28, 2015 at 12:19 pm said:

    Distance / Time = speed
    So
    5(X+Y) = 1
    7(X-Y) = 7
    12X = 2
    x = 1/6 {Speed}
    Time = 6 Minutes

  3. Sumit Sharma on May 12, 2015 at 4:17 am said:

    1/r + 1/w=5/60

    1/r – 1/w=7/60

    2/r=12/60

    r/2=60/12 r=60*2/12= 10 miles /hr

    r=rider, w=wind

  4. harshitha on June 23, 2015 at 7:38 pm said:

    can you please clearly explain

    • Nitik on July 30, 2015 at 2:53 am said:

      Rider rides 1 mile in 5 min with wind
      => speed with wind = 1/5 miles per min
      i.e R + W = 1/5 equation 1
      Rider rides 1 mile in 7 min against wind
      => speed against wind = 1/7 miles per min
      i.e R – W = 1/7 equation 2

      ADDING 2 EQUATIONS
      R + W = 1/5 where : R = speed of rider
      + W = speed of wind
      R – W = 1/7
      => 2R = 12/35
      hence R = 6/35 miles/min

  5. Horse speed with the wind = 1mile in 5 minutes.
    So he was going 1/5 miles per minute.

    Horse speed against the wind = 1 mile in 7 minutes.
    So he was going 1/7 miles per minute.

    average of speeds = ( 1/5 + 1/7 ) ÷ 2 = 6/35 miles per minute.

  6. a. W+H = 12 Miles/Hr (Since 5 minutes/mile with wind; 12 *5 = 60)
    b. H-W = 8.57 Miles/Hr (Since 7 minutes/mile against wind; 8.57 * 7 = 60)

    a+b = 2H = (12+8.57) Miles/Hr
    = 2H = 20.57 Mi/Hr
    = H = 10.285 Mi/Hr

    and 1 mile would take 60 min/10.285 Miles = 5.8337 min; or 5 min 50 seconds

  7. Irebel on January 3, 2021 at 3:14 am said:

    how about the simplest solution of all. 5 min. + 7 min. = 12 min divide by 2 miles = 6 mile with no wind.

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