Results 1 to 7 of 7

Math Help - a problem in confusion

  1. #1
    Member
    Joined
    May 2009
    Posts
    91

    Talking a problem in confusion

    Hi guys,

    Now, this is in my opinion a very tricky puzzle! It comes from an old book with answers but no working. I have been fiddling with it for ages! The wording is very difficult to understand. Can anyone help?

    When Bert was just one year younger than Bill was when Ben was half as old as Bill will be 3 years from now, Ben was twice as old as Bill was when Ben was 1/3 as old as Bert was 3 years ago. But, when Bill was twice as old as Bert, Ben was 1/4 as old as Bill was one year ago. Ignoring odd months and considering that Bert is over 50 years old, it should be no problem to find out how old these three friends are.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    In general for this kind of problem, a line representing the ages would help you. But I hate drawing, even a line, so I'll try to explain it with words

    Let BertNow be the age of Bert now XD. Let BillNow and BenNow be the age of Bill and Ben now (do you follow for the moment ? lol). I call them this way in order not to confuse with a,b,c when reading the problem.

    All the green sentences will refer to the "condition" : something happened when blablabla

    When Bert was just one year younger than Bill was when Ben was half as old as Bill will be 3 years from now
    I think the red was is actually an "is"... referring to the age of Bill now. Otherwise, this piece of information can't be used further.

    Suppose that the green sentence happened X years ago.
    Transcription of the green sentence gives : BenNow-X=(BillNow+3)/2 ---> X=BenNow-(BillNow+3)/2
    Transcription of the black sentence gives : BertNow-X=BillNow-1
    Substitution gives : BertNow-BenNow+BillNow/2+3/2=BillNow-1 ---> BertNow-BenNow-BillNow/2+5/2=0

    Ben was twice as old as Bill was when Ben was 1/3 as old as Bert was 3 years ago
    Suppose it was Y years ago.
    The green sentence gives : BenNow-Y=(BertNow-3)/3 ---> Y=BenNow-(BertNow-3)/3
    And the black sentence : BenNow-Y=2*(BillNow-Y) --> BenNow+Y=2*BillNow
    Substitution : BenNow+BenNow-(BertNow-3)/3=2*BillNow ---> 2*BenNow-BertNow/3-2*BillNow+1=0

    when Bill was twice as old as Bert, Ben was 1/4 as old as Bill was one year ago.
    Suppose it was Z years ago.
    Green sentence : (BillNow-Z)=2*(BertNow-Z) ---> Z=2*BertNow-BillNow
    Black sentence : BenNow-Z=(BillNow-1)/4
    Substitution : BenNow-2*BertNow+BillNow=BillNow/4-1/4 ---> BenNow-2*BertNow+3*BillNow/4+1/4=0


    Let a=BertNow, b=BenNow, c=BillNow.

    Thus we have the system :

    \begin{cases} a-b-\frac c2+\frac 52=0 \\ -\frac a3+2b-2c+1=0 \\ -2a+b+\frac{3c}{4}+\frac 14=0 \end{cases}


    I hope it's clear !
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    91
    Moo!

    Sorry for the delay in replying. I congratulate you on supplying an answer to this puzzle. It is not a typical one but it was interesting and I realised when I read your solution where I was going wrong. So a BIG thank you! There is just one little problem... the answers come out wrong!! I thought it might be me, so I solved the equations using different methods but get the same result:

    a= 117/31; b = 131/31; c = 127/31
    a = 3.7 b = 4.2 c= 4.10

    I checked the logic again expecting a little slip somewhere, but it seems flawless! How can such beautiful logic not lead to the correct answer? I will continue looking and if you can find anything, it will be great to hear from you.

    The published answers are: Bert 72, Ben 80 and Bill 85.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    I'm glad you understood the logic.
    But I'm sorry I can't tell you where the flaw is... I still have a problem with the first part of the text : changing is into was doesn't change anything to the problem, it's like we're assuming that there is always one year of difference between the two.
    And I solved the system as if it were BertNow=BillNow-1, but it doesn't give the solution

    I don't know if the original text was in English, maybe there's a problemin the translation :s in particular if you mistranslated something into "ago" instead of "before" or something else...
    If the text was in French (since you're Swiss ), I can have a look !
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2009
    Posts
    91
    Hi Moo,
    I think it is the age of the book not the language that may be the problem. It was published in English in 1966 when even a calculator was a bit avant garde! I have found another case with an arithmetic error. For now we will have to assume the book is wrong unless someone else on the forum can come up with something. I didn't think of making the change from "was" to "is" which is one reason I didn't get anywhere with my solution. I agree that without this assumption, the information given is redundant and I just can't believe it's wrong! So, thanks again. Your solution was very instructive. I think these puzzles are great because they really test your ability to formulate questions mathematically. It's not like reeling off standard textbook stuff.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by s_ingram View Post
    Hi guys,

    Now, this is in my opinion a very tricky puzzle! It comes from an old book with answers but no working. I have been fiddling with it for ages! The wording is very difficult to understand. Can anyone help?

    When Bert was just one year younger than Bill was when Ben was half as old as Bill will be 3 years from now, Ben was twice as old as Bill was when Ben was 1/3 as old as Bert was 3 years ago. But, when Bill was twice as old as Bert, Ben was 1/4 as old as Bill was one year ago. Ignoring odd months and considering that Bert is over 50 years old, it should be no problem to find out how old these three friends are.
    I spent way too much time on this complicated problem, but I think I finally solved it.

    Let
    Bert's age = a
    Bill's age = b
    Ben's age = c

    We adopt the convention that today's date is time zero, times in the future are positive numbers, and times in the past are negative, so a date t years in the past, for example, is time -t.

    (1) Consider Bill's age "when Ben was 1/3 as old as Bert was 3 years ago"; this is
    b - c + (1/3)(a-3).

    (2) "When Ben was half as old as Bill will be 3 years from now" is time
    (1/2)(b+3) - c.

    (3) Bill's age then was
    b + (1/2)(b+3) - c.

    (4) The date when Bert was 1 year younger than this is
    b + (1/2)(b+3) - c - 1 - a.

    (5) On this date, "Ben was twice as old as Bill was when Ben was 1/3 as old as Bert was 3 years ago"; so using (1),
    c + b + (1/2)(b+3) - c - 1 - a = 2 [b - c + (1/3)(a-3)].

    (6) Let's say the date "when Bill was twice as old as Bert" was time t.
    Then b + t = 2(a + t), so
    t = b - 2a.

    (7) On this date, "Ben was 1/4 as old as Bill was one year ago"; so
    b - 2a + c = (1/4)(b - 1).

    The problem statement boils down to the two equations (5) and (7). Since we have 2 equations and 3 unknowns, there seem to be infinitely many solutions. I think "ignoring odd months" means that we require integer values for a, b, and c, so let's see if we can find an integer solution. (5) and (7) are equivalent (after some algebra) to

    (8) 10a + 3b -12c = 15
    (9) -8a + 3b + 4c = -1

    (8) - (9) yields
    (10) 9a - 8c = 8

    Since all the terms in (10) are divisible by 8 except for 9a, a must be a multiple of 8; say
    (11) a = 8n.
    Then substituting in (10) and solving for c,
    (12) c = 9n - 1.
    Substituting (11) and (12) in (8), we find
    (13) b = 1 + (28/3) n.

    Since a > 50, n >= 7. It is apparent from (13) that n must be a multiple of 3, so we try n = 9, with the result
    a = 72
    b = 85
    c = 80
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2009
    Posts
    91

    Thumbs up

    Well, I don't know in which way you are awkward, but mathematically you sure are not! What a beautiful solution you have provided! As I said before, an ability to apply maths in unusual situations is the measure of real mathematical ability. I haven't been through the solution in detail yet. It will take me a little time. I just wanted to say thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. confusion with simple trig problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 30th 2011, 12:46 PM
  2. Word problem confusion
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 21st 2010, 12:52 PM
  3. Problem with Roots. Confusion.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 15th 2009, 03:09 PM
  4. venn diagram problem little confusion
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 10th 2009, 05:15 AM
  5. Confusion regarding problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 1st 2008, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum