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Senior Manager
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Joined: 30 Aug 2003
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ps [#permalink] New post 28 Dec 2003, 15:48
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Question Stats:

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When William received 10x coins, he then had 5y + 1 times as many coins as he had originally. In terms of x and y, how many coins did William have originally?

10x(5y + 1)
(5y+1)/10x
2x/y
10/(5y + 1)
none
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shubhangi

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 [#permalink] New post 28 Dec 2003, 15:52
shubhangi wrote:
working??


intial = P

P+10X = (5Y+1)P
solve for P

ans: 2X/Y
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 [#permalink] New post 28 Dec 2003, 22:05
Hey, dj--

The algebra took me seven steps and three minutes!

Would you mind terribly showing your work?
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 [#permalink] New post 28 Dec 2003, 22:16
stoolfi wrote:
Hey, dj--

The algebra took me seven steps and three minutes!

Would you mind terribly showing your work?


suppose, intial amount was P

he for 10X more, thus,
P+10X & this is equal to (5Y+1)P

P+10X=5YP+P
10X=5YP
P=2X/Y

you must be feeling worn out after the holidays.. oh, we are still in the middle :wink:
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 [#permalink] New post 28 Dec 2003, 22:22
I didn't even think to un-factor it. Stupid me...

Worn out, indeed...

Thank you.
  [#permalink] 28 Dec 2003, 22:22
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