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Senior Manager
Joined: 30 Aug 2003
Posts: 322
Location: dallas , tx
Followers: 1

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28 Dec 2003, 16:48
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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
_________________

shubhangi

Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

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28 Dec 2003, 16:52
shubhangi wrote:
working??

intial = P

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

ans: 2X/Y
Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405
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Kudos [?]: 25 [0], given: 0

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28 Dec 2003, 23:05
Hey, dj--

The algebra took me seven steps and three minutes!

Would you mind terribly showing your work?
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

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28 Dec 2003, 23:16
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
Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405
Followers: 1

Kudos [?]: 25 [0], given: 0

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28 Dec 2003, 23:22
I didn't even think to un-factor it. Stupid me...

Worn out, indeed...

Thank you.
28 Dec 2003, 23:22
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