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24 Jun 2017, 07:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (01:35) correct
23%
(01:43)
wrong
based on 148
sessions
History
Date
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Not Attempted Yet
If \(x < y < –1\), which of the following must be true?
(A) \(\frac{x}{y} > xy\)
(B) \(\frac{y}{x} > x + y\)
(C) \(\frac{y}{x} > xy\)
(D) \(\frac{y}{x} < x + y\)
(E) \(\frac{y}{x} > \frac{x}{y}\)
(A) \(\frac{x}{y} > xy\)
(B) \(\frac{y}{x} > x + y\)
(C) \(\frac{y}{x} > xy\)
(D) \(\frac{y}{x} < x + y\)
(E) \(\frac{y}{x} > \frac{x}{y}\)
Source: Nova GMAT
Difficulty Level: 600
Difficulty Level: 600
ydmuley
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24 Jun 2017, 11:39
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\(x < y < –1\)
This means that both x and y are NEGATIVE numbers
(A) \(\frac{x}{y} > xy\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(x/y = 1.5\\
xy = 6\)
So \(x/y < xy\) ====> Hence, FALSE
(B) \(\frac{y}{x} > x + y\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(x+y = -3+(-2) = -3-2 = -5\)
So, \(y/x > x+y\) ===> TRUE
(C) \(\frac{y}{x} > xy\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(xy = 6\)
So\(y/x < xy\) ===> Hence, FALSE
(D) \(\frac{y}{x} < x + y\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(x+y = -3+(-2) = -3-2 = -5\)
So, \(y/x > x+y\) ===> Hence, FALSE
(E) \(\frac{y}{x} > \frac{x}{y}\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(x/y = -3/-2 = 1.5\)
So, \(y/x < x/y\) ===> Hence, FALSE
Hence, the Answer is B
This means that both x and y are NEGATIVE numbers
(A) \(\frac{x}{y} > xy\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(x/y = 1.5\\
xy = 6\)
So \(x/y < xy\) ====> Hence, FALSE
(B) \(\frac{y}{x} > x + y\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(x+y = -3+(-2) = -3-2 = -5\)
So, \(y/x > x+y\) ===> TRUE
(C) \(\frac{y}{x} > xy\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(xy = 6\)
So\(y/x < xy\) ===> Hence, FALSE
(D) \(\frac{y}{x} < x + y\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(x+y = -3+(-2) = -3-2 = -5\)
So, \(y/x > x+y\) ===> Hence, FALSE
(E) \(\frac{y}{x} > \frac{x}{y}\)
Take \(x = -3\) and \(y = -2\) - when we add these values in the above equation we get
\(y/x = -2/-3 = 0.66\)
\(x/y = -3/-2 = 1.5\)
So, \(y/x < x/y\) ===> Hence, FALSE
Hence, the Answer is B
24 Jun 2017, 11:50
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Bunuel
it is clear that x & y are negative numbers. So summation of x & y will be negative (for eg. -3+-2=-5)
So, x+y<0 and
y/x will be positive (as negative signs will cancel out. for eg. -3/-2 = 3/2 = 1.5)
So, y/x >0
Obviously Positive number is greater than negative number
So, \(\frac{y}{x} > x + y\)
Hence Option B
