1. Subscribervenda
    Dave
    S.Yorks.England
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    30 May '19 19:35
    At one time I could do this type of thing easily.
    Now apparently not because recently I came across this and I can't work out how to get to the answer so I must be going wrong somewhere.The given answer doesn't agree with what I get it to.
    I'm sure someone here can solve the puzzle and show me the working out please:-
    A man walks from one town to another.
    On the first day he covers one third of the total distance
    The next day he covers a quarter of what is left
    The following day he covers two fifths of the remainder and on the fourth day half of the remaining distance.He now has 14 miles left.
    How far has he travelled?
  2. Joined
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    30 May '19 19:473 edits
    @venda said
    At one time I could do this type of thing easily.
    Now apparently not because recently I came across this and I can't work out how to get to the answer so I must be going wrong somewhere.The given answer doesn't agree with what I get it to.
    I'm sure someone here can solve the puzzle and show me the working out please:-
    A man walks from one town to another.
    On the first day h ...[text shortened]... urth day half of the remaining distance.He now has 14 miles left.
    How far has he travelled?
    1/3 means 2/3 left. 1/4 of 2/3 is 1/6. So now he has traveled 1/3+1/6 =1/2.

    2/5 of half is 1/5 So now he has traveled 7/10 leaving 3/10 of the trip to go.

    So if half if 3/10 is 14 miles the entire 3/10 is 28 miles. 3/10 of what number is 28?
    280/3 miles. 93 and 1/3 miles.

    Now subtract the 14 he has not traveled...79 and 1/3 miles traveled.

    I think tearing it apart is easier than using variables and number crunching.
  3. Joined
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    30 May '19 21:22
    I think the only thing I was not explicit about was the half. Since he traveled half way, he had half remaining. The half remaining was what was used to continue the rest.
  4. Subscribervenda
    Dave
    S.Yorks.England
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    31 May '19 07:40
    @eladar said
    1/3 means 2/3 left. 1/4 of 2/3 is 1/6. So now he has traveled 1/3+1/6 =1/2.

    2/5 of half is 1/5 So now he has traveled 7/10 leaving 3/10 of the trip to go.

    So if half if 3/10 is 14 miles the entire 3/10 is 28 miles. 3/10 of what number is 28?
    280/3 miles. 93 and 1/3 miles.

    Now subtract the 14 he has not traveled...79 and 1/3 miles traveled.

    I think tearing it apart is easier than using variables and number crunching.
    Exellent thank you.
    79.333 was the answer given so well done.
    I think I got 126 or something like that.
    Must have missed a stage
  5. Joined
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    05 Jun '19 23:19
    @venda

    Total distance to go = X
    Distance travelled so far = X - 14
    X * 2/3 * 3/4 * 3/5 * 1/2 = 14
    X = 14*20/3
    Distance travelled so far = 14*20/3 - 14
    = 280/3 - 14
    = 93 1/3 - 14
    = 79 1/3

    I think the trick is you are suppose to forget to take 14 away at the end.
  6. Subscribervenda
    Dave
    S.Yorks.England
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    07 Jun '19 13:38
    @iamatiger said
    @venda

    Total distance to go = X
    Distance travelled so far = X - 14
    X * 2/3 * 3/4 * 3/5 * 1/2 = 14
    X = 14*20/3
    Distance travelled so far = 14*20/3 - 14
    = 280/3 - 14
    = 93 1/3 - 14
    = 79 1/3

    I think the trick is you are suppose to forget to take 14 away at the end.
    Thanks for that.
    As they say there's more than one way to skin a cat(or something like that)
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    07 Jun '19 17:11
    @venda

    Yep that was a great algebraic approach.

    He just found what was left over. A much simpler approach of the algebraic I was going to take.

    Usually in math there are easier or more difficult roads, but the best road is the one you see and understand. Half the time I find the easier road.
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    07 Jun '19 17:16
    @venda said
    Exellent thank you.
    79.333 was the answer given so well done.
    I think I got 126 or something like that.
    Must have missed a stage
    Do you remember you line of thought?
  9. Subscribervenda
    Dave
    S.Yorks.England
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    08 Jun '19 15:01
    @eladar said
    Do you remember you line of thought?
    The same way as your solution.
    I think I just got mixed up with the distances travelled and distances remaining.
    After I had seen the solution I went thro' my fractions again but couldn't see where the error was.
    Sometimes I find an easy way to do problems is to make up an "easy sum" to make sure the algebra is correct.
    For this problem for instance you could say the distance was 128 miles and make the distances 1/2 for each day.
    Another type of problem I always used to have trouble with is:-
    If "a" gives "b" £7 "b" has the same as "a"
    If "b" gives "a" £7 "a" has 5 times more than "b"
    How much does each have to start with?
    I used to get a mental block when trying to visualise the equations required.
    The solution once you get your mind round the equations is simple to work out.
    It's just simultaneous equations.
    One problem I have NEVER found an algebraic solution to is the wrong cheque thing.I think there must be a trial and error element to it
    Here goes:-
    A man recieves a cheque and realises that the pounds and pence have been transposed by mistake.
    Seeing it is to his advantage he cashes the cheque and spends £3.97
    Looking at his change he sees that he now has 8 times what the cheque should have been worth
    How much sholud the cheque have been for?
    In summary 8x + 3.97 =y
    How do you form another equation?
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    08 Jun '19 15:151 edit
    @venda

    Yeah well although I think British monetary units are cool and know I should save my tuppens, I do not understand the units.

    Your x and y need to be the number of each unit.

    For instance the number 17 would be 1 times 10 plus 7 or as a variable representation..10x + y

    X=1 and y=7
  11. Subscribervenda
    Dave
    S.Yorks.England
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    08 Jun '19 19:51
    @eladar said
    @venda

    Yeah well although I think British monetary units are cool and know I should save my tuppens, I do not understand the units.

    Your x and y need to be the number of each unit.

    For instance the number 17 would be 1 times 10 plus 7 or as a variable representation..10x + y

    X=1 and y=7
    £3.97 reversed would be £97.30 if the pounds and the pence were reversed.
  12. Joined
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    08 Jun '19 20:402 edits
    @venda

    So given x and y are whole numbers 100x + y would be the amount the Check should have read, but now the value is 100y+x.
    y>x and that 1 <y <99 So 0 <x <98

    We also know that 8 (100y+x-397)=100x+y

    Kind of stuck here.

    The difference in value of the checks is 99y-99x
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    09 Jun '19 00:231 edit
    Work in pence
    Cheque correct val. = 100x + y
    We know x and y are integers >= 0
    also y < 100
    Man gets 100y + x
    (Likely x < 100 or the transposition wouldn’t make sense)
    Then we have
    100y + x - 397 = 800x + 8y
    92y - 397 = 799x
    So we are looking for a multiple of 92 that is 397 more than a multiple of 799
    Because everything is integers there may be a unique solution
    Tried 13*92 = 1196
    1196 - 397 = 799, so that is an answer
    Cheque is 1.13, gets 13.01, spends 3.97, has 9.04 which is 8*1.13
    Is this unique?
    Another answer would have to be
    (13+c)*92-397
    Where c*92 is a multiple of 799
    799 and 92 have no common factors, so the next answer would be (13 +799)*92-397=93*799
    Which gives y=812, x=93
    But as y must be less that 100 it is no good
    So the answer above is the only one that works,
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    09 Jun '19 00:262 edits
    @iamatiger

    So it was a guess and check possible solution problem. And you are right I did have that backwards.

    To make myself feel better I did have it right on my paper lol.
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    09 Jun '19 07:581 edit
    @Eladar

    No probs eladar, I like these puzzles venda!
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