"Brian M. Scott" <b.scott@[EMAIL PROTECTED]
> wrote in message
news:10n8b6g0vf79u.1lzfmxxf2p9pf.dlg@[EMAIL PROTECTED]
> On Mon, 12 Nov 2007 20:34:12 -0500, Marvel
> <Martel@[EMAIL PROTECTED]
> wrote in
> <news:4738fefa$0$19614$4c368faf@[EMAIL PROTECTED]
> in
> alt.algebra.help:
>
>> I cant figure it out any help appreciated!
>
>> I am not smarter than a fifth grader!
>> ------
>
>> Ben had 50 more coins than Joan.
>> After Joan gave 29 of her coins to Ben, she had 1/3 as
>> many coins as Ben.
>
>> How many coins did Ben have at first?
>
> Barb's pointed the way towards an algebraic solution, but
> since it's a fifth grade problem, you might also like a
> non-algebraic solution.
>
> Ben starts with 50 more coins than Joan. When she gives him
> 29 coins, he gains 29, and she loses 29, so his advantage
> increases by 29 + 29 = 58. Thus, he now has 50 + 58 = 108
> coins more than Joan. She has only a third as many coins as
> he has, so his excess 108 coins must be the other two-thirds
> of his present holding. From this it follows easily that
> Joan and Ben now have 54 and 162 coins, respectively, and to
> answer the question you need only undo the 29-coin transfer.
>
> Brian
Thank You for the help it was quite simple that way.
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