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If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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Updated on: 04 Oct 2013, 00:01
Question Stats:
77% (01:20) correct 23% (01:58) wrong based on 167 sessions
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If k > 0, x + k = y, and y + 3k = z, what is the ratio between z  x and y  x ? (A) 1 to 4 (B) 1 to 2 (C) 2 to 1 (D) 3 to 1 (E) 4 to 1
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Originally posted by manishuol on 14 May 2013, 21:44.
Last edited by Bunuel on 04 Oct 2013, 00:01, edited 1 time in total.
Edited the question.



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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14 May 2013, 21:54
manishuol wrote: If k > 0, x + k = y, and y + 3k = z, what is the ratio between z � x and y � x ?
(A) 1 to 4 (B) 1 to 2 (C) 2 to 1 (D) 3 to 1 (E) 4 to 1 The question is not clear, all I can see is a question mark: z � x and y � x Can you repost this? Or is this z?x and y?x the actual question?



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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14 May 2013, 23:23
psychout wrote: manishuol wrote: If k > 0, x + k = y, and y + 3k = z, what is the ratio between z � x and y � x ?
(A) 1 to 4 (B) 1 to 2 (C) 2 to 1 (D) 3 to 1 (E) 4 to 1 The question is not clear, all I can see is a question mark: z � x and y � x Can you repost this? Or is this z?x and y?x the actual question? This is How the Question is ............
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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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15 May 2013, 01:29
manishuol wrote: If k > 0, x + k = y, and y + 3k = z, what is the ratio between z � x and y � x ?
(A) 1 to 4 (B) 1 to 2 (C) 2 to 1 (D) 3 to 1 (E) 4 to 1 Hi, From the answer, I could only assume that "z � x" is zx. So, (zx)/(yx) = 4k/k = 4:1 thus (E) Regards,



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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02 Oct 2013, 12:07
If k > 0, x + k = y, and y + 3k = z, what is the ratio between z � x and y � x ? (A) 1 to 4 (B) 1 to 2 (C) 2 to 1 (D) 3 to 1 (E) 4 to 1 Picking numbers is probably the fastest solution to solve this problem. First, k>0 = k could be ANY positive number. For the purposes of the explanation K= 2. x + 2 = y and y + 3(2) = z. Rearrange the latter equation as to isolate y. X + 2 = Y Z  6 = Y Now pick Any number for Y. For the sake of our example Y will be 10. Now the equation looks like: X + 2 = 10 Z  6 = 10 Solve for X. Solve for Z. X= 8 and Z = 16 what is the ratio between z � x and y � x 16 to 8 as 10 to 8 8 2 8/2 = 4 2/2=1 Thus answer choice E is correct 4 to 1. Kudos please



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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03 Oct 2013, 21:11
cmclaurin wrote: what is the ratio between z � x and y � x 16 to 8 as 10 to 8 8 2 8/2 = 4 2/2=1
Thus answer choice E is correct 4 to 1.
Could you please elaborate on how you got 8 2 and further? thanks



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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03 Oct 2013, 21:27
16  8 = 8
10  8 = 2
I hope this is helpful now. Lol (I'm being a troll). Thanks



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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03 Oct 2013, 23:51
manishuol wrote: If k > 0, x + k = y, and y + 3k = z, what is the ratio between z � x and y � x ?
(A) 1 to 4 (B) 1 to 2 (C) 2 to 1 (D) 3 to 1 (E) 4 to 1 Assuming that the required ratio is (z  x)/(y  x), x + k = y y  x = k y + 3k = z x + k + 3k = z z  x = 4k (z  x)/(y  x) = 4/1
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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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04 Oct 2013, 23:31
(zx )/ (yx) == (y + 3k x )/ (yx) where k = yx
hence == (yx) + 3(yx) / ( yx) = 4(yx)/(yx) == 4/1 !!



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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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04 Mar 2015, 17:49
Hi All, This question is perfect for TESTing VALUES. We're given 3 "rules" that we have to follow (regarding the values we choose to TEST): 1) K > 0 2) X + K = Y 3) Y + 3K = Z Let's keep the values small: X = 1 K = 2 1 + 2 = 3 = Y 3 + 3(2) = 9 = Z The question asks for the ratio of (ZX) to (YX). Using the values we chose.... (91) to (31) = 8 to 2 = 4 to 1 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If k > 0, x + k = y, and y + 3k = z, what is the ratio betwe [#permalink]
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