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
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