1. Joined
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    01 Sep '18 05:03
    Hint:
    total distance = average_speed * total_time
  2. Joined
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    26771
    02 Sep '18 16:141 edit
    Second hint. All you need are the equations given, and the two deducible facts:
    1 None of the speeds can be less than zero
    2 The third speed s3 is less than the first two speeds s1 & s2, so we can conclude that s3<(s1+s2)/2, because if it was not then it would be equal to or greater than one of them.
  3. Joined
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    26771
    04 Sep '18 08:23
    Shoul post up the answer? Is anyone still thinking about it?
  4. Joined
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    06 Sep '18 08:49
    http://www.wolframalpha.com/input/?i=x%2By%3D1;+(2%2F3)x%2B(2%2F3)y%2Bz%3Dr;+0%3Cz%3C(2%2F3)x%3C(2%2F3)y;+r%3D1
  5. Joined
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    06 Sep '18 20:211 edit
    ? That says no solutions exist. Would you like the solution?
  6. Joined
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    26771
    07 Sep '18 20:274 edits
    The answer
    Sorry, there is no easy way of hiding a multiple line post.

    The first journey takes 1 hour and is driven at two speeds s1 and s2
    We are given an equation for the average speed of this journey
    A =(s1+ s2)/2

    Since the time is 1 hour, if our speeds are in mph then
    Distance in miles = D = (s1 + s2)/2
    (We can also work out that the times driven at each speed must be equal, but that doesn’t help us)

    For the second journey we are given
    A2 = (s1 + s2 + s3)/3

    Assuming the time for this journey is T we have another equation for distance travelled
    D = T(s1 + s2 + s3)/3
    So T(s1 + s2 + s3)/3 = (s1 + s2) /2
    Rearranging
    T = 3/2 * (s1 + s2)/(s1 + s2 + s3)

    Assuming all speeds are positive, the minimum T happens with maximum s3, and the maximum T happens with minimum S3
    For the minimum s3, all we can assume is s3>0, this gives
    maxT < 3/2 * (s1 + s2)/(s1 + s2 + 0)
    Max t < 3/2
    For the minimum s3, since s3 is lower than s1 and s2, s3 MUST be less than the average of s1 and s2, whatever their values are.
    s3 < (s1 + s2)/2
    So
    MinT > 3/2 * (s1 +s2)/(s1 + s2 + (s1 + s2)/2)
    This rearranges to
    MinT > 1

    So my wife’s second journey must have taken longer than 1 hour, but less than 1 hour and 30 minutes.
  7. Subscribervenda
    Dave
    S.Yorks.England
    Joined
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    08 Sep '18 15:421 edit
    Originally posted by @iamatiger
    The answer
    Sorry, there is no easy way of hiding a multiple line post.

    The first journey takes 1 hour and is driven at two speeds s1 and s2
    We are given an equation for the average speed of this journey
    A =(s1+ s2)/2

    Since the time is 1 hour, if our speeds are in mph then
    Distance in miles = D = (s1 + s2)/2
    (We can also work out that the times ...[text shortened]... wife’s second journey must have taken longer than 1 hour, but less than 1 hour and 30 minutes.
    Interesting stuff.The Algebra is a bit too complicated for me but I can just about see how it works.
    I've got a book somewhere with some algebra problems that are way beyond my ability(I'm virtually self taught ,I couldn't be bothered at school and failed Maths).I only became interested in later life when I got this book.
    Would you like me to find it and post one of the problems?
    I assure you it'll be more challenging than the bloke filling a bath with a hole in it type of problem.The answers are in the back so I can provide an answer but I wouldn't be able to show you how to get to it!!
  8. Joined
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    08 Sep '18 16:42
    Sounds challenging!
  9. Subscribervenda
    Dave
    S.Yorks.England
    Joined
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    83451
    08 Sep '18 19:12
    Originally posted by @iamatiger
    Sounds challenging!
    Right then.I remember spending hours on this one , but never worked out how to do it.Here goes:-
    An army 25 miles long starts on a journey of 50 miles just an an orderly at the rear starts to deliver a message to the General at the front.The orderly, travelling at uniform speed delivers his message and returns to the rear ,arriving there just as the army finishes the journey.
    Q- How far does the orderly travel?
    I remember thinking at the time that the question was a bit vague.Does it mean when the General at the front finishes , or when the orderly at the back finishes(assuming they are walking in one long column).?
    Anyway,as i said, I know the answer but even with that knowledge I never worked out how it was arrived at.
    Good luck.
  10. Joined
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    26771
    08 Sep '18 23:49
    Looks like the back of the army has to travel 75 miles
    Assume the army travels at speed a, and the messenger travels at ka
    Army total time = 75/a
    Messenger time to front = 25/(ka-a)
    Messanger time to back = 25/(ka+a)
    Messenger total time = (25/a)(1/(k-1)+1/(k+1))
    Messenger time = army time so
    75/a = (25/a)(1/(k-1)+1/(k+1))
    Divide by 25/a
    3 = 1/(k-1)+1/(k+1)
    Multiply by (k-1)(k+1)= k^2 -1
    3(k^2-1)= 2k
    k^2 - 2k/3 = 1
    k^2 - 2k/3 + 1/9 = 10/9
    (k-1/3)^2 = 10/9
    k=sqrt(10)/3 + 1/3 = (1+sqrt(10))/3
    Orderly distance = orderly speed x time = ka*75/a=75a
    Orderly distance = 25*(1+sqrt(10))=104.0569 miles
  11. Joined
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    26771
    09 Sep '18 04:08
    Sorry, second to last line should be
    Orderly distance = orderly speed x time = ka*75/a=75k
  12. Subscribervenda
    Dave
    S.Yorks.England
    Joined
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    83451
    09 Sep '18 08:231 edit
    Originally posted by @iamatiger
    Looks like the back of the army has to travel 75 miles
    Assume the army travels at speed a, and the messenger travels at ka
    Army total time = 75/a
    Messenger time to front = 25/(ka-a)
    Messanger time to back = 25/(ka+a)
    Messenger total time = (25/a)(1/(k-1)+1/(k+1))
    Messenger time = army time so
    75/a = (25/a)(1/(k-1)+1/(k+1))
    Divide by 25/a
    3 = 1/( ...[text shortened]... distance = orderly speed x time = ka*75/a=75a
    Orderly distance = 25*(1+sqrt(10))=104.0569 miles
    The answer in the book isn't that far but as I said, the question is a bit vague and the book doesn't give any working out.I'll let you think about it and maybe make other assumptions and then give the quoted answer when you(or someone else) has had another go.
  13. Joined
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    26771
    09 Sep '18 21:00
    Ok, let's assume that the army stops when the general has gone 50 miles.
    This changes the calcs as follows:
    Assume the army travels at speed a, and the messenger travels at ka
    Army total time = 50/a
    Messenger time to front = 25/(ka-a)
    Messanger time to back = 25/(ka+a)
    Messenger total time = (25/a)(1/(k-1)+1/(k+1))
    Messenger time = army time so
    50/a = (25/a)(1/(k-1)+1/(k+1))
    Divide by 25/a
    2 = 1/(k-1)+1/(k+1)
    Multiply by (k-1)(k+1)= k^2 -1
    2(k^2-1)= 2k
    k^2 - k = 1
    k^2 - k + 1/4 = 5/4
    (k-1/2)^2 = 5/4
    k=sqrt(5)/2 + 1/2 = (1+sqrt(5))/2
    Orderly distance = orderly speed x time = ka*50/a=50k
    Orderly distance = 25*(1+sqrt(5))= 80.9017 miles
    is that right?
  14. Subscribervenda
    Dave
    S.Yorks.England
    Joined
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    83451
    10 Sep '18 07:54
    Originally posted by @iamatiger
    Ok, let's assume that the army stops when the general has gone 50 miles.
    This changes the calcs as follows:
    Assume the army travels at speed a, and the messenger travels at ka
    Army total time = 50/a
    Messenger time to front = 25/(ka-a)
    Messanger time to back = 25/(ka+a)
    Messenger total time = (25/a)(1/(k-1)+1/(k+1))
    Messenger time = army time ...[text shortened]... ly speed x time = ka*50/a=50k
    Orderly distance = 25*(1+sqrt(5))= 80.9017 miles
    is that right?
    It is!!.
    Well done!!
    Your Algebraic talents are impressive.
    I'll study your solution in detail when I have a bit of time and see what I can learn.
    Perhaps when I'm trying to fill a bath with the plug out!!
  15. Joined
    26 Apr '03
    Moves
    26771
    11 Sep '18 02:374 edits
    Originally posted by @venda
    It is!!.
    Well done!!
    Your Algebraic talents are impressive.
    I'll study your solution in detail when I have a bit of time and see what I can learn.
    Perhaps when I'm trying to fill a bath with the plug out!!
    Thanks 😀
    Got any other tricky ones?
    I think the key was trying to solve for messenger_speed = k*army_speed.
    At first I tried messenger_speed = k+army_speed
    But there is no solution for k then, because the army speed does not cancel out.
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