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actleader
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If x, y, and k are positive numbers such that 35x/(x-y) 30 Oct 2012, 12:46
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

43% (02:45) correct 57% (02:51) wrong based on 184 sessions

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If x, y, and k are positive numbers such that [35x][/(x-y)] +[(70y-70x)][/(x-y)]=k , and if x>y , which of the following could be a value for k?

(A) −85
(B) −70
(C) −65
(D) −35
(E) −20
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E
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If x, y, and k are positive numbers such that 35x/(x-y) 30 Oct 2012, 19:59
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actleader
Hi! Could you help with the following... what is the best strategy?
Thanx


If x, y, and k are positive numbers such that [35x][/(x-y)] +[(70y-70x)][/(x-y)]=k , and if x>y , which of the following could be a value for k?

(A) −85
(B) −70
(C) −65
(D) −35
(E) −20
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Is the OA correct for this? The ans should be E.
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If x, y, and k are positive numbers such that 35x/(x-y) 30 Oct 2012, 21:42
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If x, y and k are all +ve numbers, how could k be any of the answer choices??. I can only get k>-35 and so only E satisfies the requirement. But even then, the requirement that k is a postive number is not satisfied.

Kudos Please... If my post helped.
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If x, y, and k are positive numbers such that 35x/(x-y) 30 Oct 2012, 21:47
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MacFauz
If x, y and k are all +ve numbers, how could k be any of the answer choices??. I can only get k>-35 and so only E satisfies the requirement. But even then, the requirement that k is a postive number is not satisfied.

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Yup, this question is posted second time on forum... and again its incorrect :shock:
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If x, y, and k are positive numbers such that 35x/(x-y) 31 Oct 2012, 12:40
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MacFauz
If x, y and k are all +ve numbers, how could k be any of the answer choices??.
Kudos Please... If my post helped.
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Great notation, I havn't noticed it initially ))
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If x, y, and k are positive numbers such that 35x/(x-y) 02 Nov 2012, 12:50
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Vips0000
MacFauz
If x, y and k are all +ve numbers, how could k be any of the answer choices??. I can only get k>-35 and so only E satisfies the requirement. But even then, the requirement that k is a postive number is not satisfied.

Kudos Please... If my post helped.
Yup, this question is posted second time on forum... and again its incorrect :shock:
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I've brang the question up.
Waiting for the GoGMAT answer.
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If x, y, and k are positive numbers such that 35x/(x-y) 03 Dec 2012, 11:53
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If number k such that k=35x/(x-y)+(70y-70x)/(x-y) for some positive numbers x and y, and if x>y, which of the following could be a value for k?
-85
-75
-65
-35
-20
Explanation:
First of all simplify expression for k:
k=35x/(x-y)+(70y-70x)/(x-y)=(35x+70y-70x)/(x-y)=(70y-35x)/(x-y)=(35y+35y-35x)/(x-y)=35y/(x-y)-35
Since y>0 and x>y, 35y/(x-y)>0 or 35y/(x-y)-35>0-35 or k>-35.
The answer is E.

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If x, y, and k are positive numbers such that 35x/(x-y) 20 Sep 2013, 18:27
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actleader
Hi! Could you help with the following... what is the best strategy?
Thanx


If x, y, and k are positive numbers such that [35x][/(x-y)] +[(70y-70x)][/(x-y)]=k , and if x>y , which of the following could be a value for k?

(A) −85
(B) −70
(C) −65
(D) −35
(E) −20
Show more

The easiest way I found is K = 35/[1-Y/X] - 70
35/[1-Y/X] is always >35 [as the denominator is always between 0 and 1]
therefore K>-35, hence K can only be -20

Please give kudos if the explanation helps!! :P
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If x, y, and k are positive numbers such that 35x/(x-y) 21 Sep 2013, 02:59
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actleader
If x, y, and k are positive numbers such that [35x][/(x-y)] +[(70y-70x)][/(x-y)]=k , and if x>y , which of the following could be a value for k?

(A) −85
(B) −70
(C) −65
(D) −35
(E) −20
Show more

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If x, y, and k are positive numbers such that 35x/(x-y) 05 Mar 2017, 05:27
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x = 10
y = 3

(70*3-35*10)/(10-3) = (210-350)/7 = 20

Answer E
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If x, y, and k are positive numbers such that 35x/(x-y) 09 Apr 2021, 13:22
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\(\frac{35x}{x-y} + \frac{70y-70x}{x-y}=k\)

\(\frac{35x + 70y - 70x}{x-y} = k\)

\(\frac{-35x + 70y}{x-y} = k\)

\(-35x + 70y = kx - ky\)

\(-35x - kx = -70y - ky\)

\(x(35+k) = y(70+k)\)

\(\frac{x}{y} = \frac{70+k}{35+k}\)

As \(\frac{x}{y} > 1\), so, the above euation should also satisfy this with the answer choices.

From A till C, \(\frac{x}{y} < 1\) hence these are not answers

D is coming as \(\frac{35}{0}\) which is NA

E is the the only choice left, further on substituting we see its indeed \(\frac{x}{y} > 1\)
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If x, y, and k are positive numbers such that 35x/(x-y) 30 Jan 2025, 06:21
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