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# When William received 10x coins, he then had 5y + 1 times as

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Manager
Joined: 22 Sep 2008
Posts: 117

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02 Oct 2008, 21:17
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?

A) 10x(5y + 1)

B) (5y+1)/10x

C) 2x/y

D) 10/(5y + 1)

E) None of the above

i got consufed when this example comes in exam..... had no idea how to solve

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VP
Joined: 30 Jun 2008
Posts: 1004

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02 Oct 2008, 21:41
vr4indian wrote:
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?

A) 10x(5y + 1)

B) (5y+1)/10x

C) 2x/y

D) 10/(5y + 1)

E) None of the above

i got consufed when this example comes in exam..... had no idea how to solve

Say William had C coins initially, he received 10x more coins. So totally C + 10x. The question says after he received 10x coins he had 5y+1 TIMES as many coins he had originally (originally he had C coins). Times means multiplication, So he had C(5y+1) coins.

so C + 10x = c(5y+1) = 5yC + C
C - C - 5yC = -10x
5yC = 10x
C = 2x/y

The key in these problems is read EACH and EVERY word carefully and converting the words into equations. I agree that it gets really confusing at times. It pays to spend a bit more time in understanding the question properly in word translation problems.
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Manager
Joined: 27 Sep 2008
Posts: 76

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02 Oct 2008, 22:02
Agree with (C)

z+10x = z(5y+1)

z+10x = 5zy+z

10x = 5zy

2x/y = z

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Re: coiiiin: &nbs [#permalink] 02 Oct 2008, 22:02
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# When William received 10x coins, he then had 5y + 1 times as

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