Saturday, February 17, 2007

an interesting physics/mathematics problem on distances, speeds and time (and my solution to it)

minutes back when i logged in to my gmail account, i saw a very interesting problem sent to all his orkut friends by my junior cousin. i couldn't resist the temptation to get my hands on it. since i could not find a piece of paper nearby, i opened microsoft paint and tried solving it there (very stupid thing). i hurriedly declared victory, having thought that i had solved it, and mailed the png format image to my cousin, only to be struck by a thought that made me feel my solution was incorrect. i hastily replied again to him that previous solution is incorrect. and i set out to solve it again, this time finding a piece of torn paper. i solved it, and will soon tell him the solution too. here are the resources

below is part of mail/puzzle he sent to me
this is a real brain teaser..
WARNING:the answer may seem logically correct as the figures here have been arbitarilly chosen.. there's this girl who reaches the station at 6am everyday.Her driver comes and picks her up and they reach home by 7am.
Now,one day the girl reaches the station 5am and decides to walk towards her house.She meets the driver on the way and that day they reach home by 6.40 am..
The ques is..for how long was the girl walking??..and NO the answer is NOT 20 mins....
Best of luck folks!

[red][b]please inform me the ans and the way u found it[/red][/b]

and here is original incorrect solution i sent to my cousin

here is final (hopefully correct) solution
and here is the logic
  • let distance from station to home be x km
  • since driver picks up girl at six (obviously he would travel from home such that he picks her immediately) and reaches at seven, x km traveled in 1 hr. driver speed= x km/hr
  • since it takes him 1 hr from station to home, it takes him 1 hr from home to station. so he starts from home at 5 am
  • girl too started at 5 am that day. let distance [from station] at which they meet be z km
  • let girl speed be y km/hr
  • girl travels z km at y km/hr, so time = z/y hr
  • driver travels x-z km at x km/hr, so time= (x-z)/x
  • equate time by girl to time by driver to get first equation
  • since girl started at 5 and reached at 6:40, total time [girl on foot+ girl in car] = 1 hour 40 minutes. 2nd equation obtained
  • since driver started at 5 and reached home at 6:40, his journey to meeting point took half of 1 hour 40 minutes i.e. 50 minutes. his distance till meeting place = x-z and time= 50 minutes and his speed as already established is x km/hr. 3rd equation obtained. 3 variables 3 equations the problem is solved!


  1. Given: Car takes 1hr to reach home from station
    => same 1hr is taken by it for coming from home to station
    => the car also starts from home at 5am
    => Total time taken by car from home to back home is 2hrs in general

    Let distance between home & station be 20 Km

    Now, Distance traveled in 2 hrs by car= 40 Km
    Distance traveled in 1hr 40 min (5/3 hrs) by car= (40/2)* 5/3=100/3= 33.33 Km

     Distance traveled by girl is 40-(33.33)= 6.67 Km

    but since the car and girl meet after this distance, so car till that time would have traveled 20-(6.67)=13.33 Km

    Now, 40 km traveled by car in = 2 hrs
    therefore,13.33 Km in =(2/40)* 13.33=0.6665 hrs = 39.99min =40 min (app)

    Answer up to 2 decimal places if we consider different distances,
    Distance Time taken by girl
    10 km 40.02 min
    45 km 40.00 min
    23 km 39.99 min
    20 km 39.99 min

    So answer is verified.
    Hence, time for which the girl walked= 40 min.

  2. I didn't go through ur solutions b4 solving it...but now on reading ur solution it seems to be more systematic. But...anyhow I got to the answer which is correct according to me. Does it match ur answer?

  3. I never calculated the final answer! As you can see, I left it when I got three equations...