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05 Jun 2020, 08:52
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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:
65%
(hard)
Question Stats:
56% (02:51) correct
44%
(02:27)
wrong
based on 114
sessions
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If x and y are non-zero numbers, what is the value of x/y?
(1) \(\frac{7y}{x }= \frac{2}{1 + \frac{y}{x}} - 12\)
(2) \(25x^2 - y^2 > 0\)
Are You Up For the Challenge: 700 Level Questions
(1) \(\frac{7y}{x }= \frac{2}{1 + \frac{y}{x}} - 12\)
(2) \(25x^2 - y^2 > 0\)
Are You Up For the Challenge: 700 Level Questions
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
05 Jun 2020, 09:58
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Bunuel
Statement 1:
Let \(\frac{y}{x} = z\), then we would like to find the value of z for sufficiency.
The equation is now: \(7z = \frac{2}{{1 + z} } - 12\)
\(7z + 12 =\frac{2}{{1 + z}} \)
\((7z + 12)(1 + z) = 2\)
\(7z^2 + 19z + 10 = 0\)
\((7z + 5)(z + 2) = 0\)
Since we have two solutions this is insufficient.
Statement 2:
\(25x^2 > y^2\)
\(25 > \frac{y^2 }{ x^2}\) (\(x^2\) must be positive so we are allowed to move it around)
\(|z| < 5\).
Insufficient.
Combined:
Since both of our answers from statement 1 are smaller than 5 in magnitude, combined it is still insufficient.
Ans: E
05 Jun 2020, 10:11
Kudos
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Bunuel
Asked: If x and y are non-zero numbers, what is the value of x/y?
Let x/y = 1/t; 1/t=?
(1) \(\frac{7y}{x }= \frac{2}{1 + \frac{y}{x}} - 12\)
7t = 2/(1+t) - 12
7t(1+t) = 2 - 12(1+t)
7t + 7t^2 = 2 - 12 - 12t
7t^2 + 19t + 10 = 0
7t^2 + 14t + 5t + 10 = 0
(7t+5)(t+2) = 0
t = -5/7 or -2
x/y = - 7/5 or -1/2
NOT SUFFICIENT
(2) \(25x^2 - y^2 > 0\)
(5x + y)(5x - y) > 0
25y^2(x/y + 1/5)(x/y - 1/5) > 0
x/y > 1/5 or x/y < -1/5
NOT SUFFICIENT
(1) + (2)
(1) \(\frac{7y}{x }= \frac{2}{1 + \frac{y}{x}} - 12\)
7t = 2/(1+t) - 12
7t(1+t) = 2 - 12(1+t)
7t + 7t^2 = 2 - 12 - 12t
7t^2 + 19t + 10 = 0
7t^2 + 14t + 5t + 10 = 0
(7t+5)(t+2) = 0
t = -5/7 or -2
x/y = - 7/5 = - 1.4 or -1/2 = - .5
(2) \(25x^2 - y^2 > 0\)
(5x + y)(5x - y) > 0
25y^2(x/y + 1/5)(x/y - 1/5) > 0
x/y > 1/5 = .2 or x/y < -1/5 = - .2
Since x/y is either - 1.4 or -.5 and both numbers are < -.2
NOT SUFFICIENT
IMO E
