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If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:00
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If \(xy > 0\), is \(\frac{y}{x} + \frac{x}{y} > 2\)? (1) \(y = x  1\) (2) \(x = 2y\)
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If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:10
\(\frac{x^2+y^2}{xy}>2\)
Since xy>0, we can multiply the stem by xy maintaining the same inequality and we know that neither x nor y can be 0
So, \(x^2+y^2>2xy\) or \(x^2+y^22xy>0\) or \((xy)^2>0\)
So we need to find whether \((xy)^2>0\), this will be true when xy is not equal to 0
(1) \(y=x1\)
So, \(xy=1\) => We know that xy is non zero
So Sufficient
(2) \(x=2y\)
Or, \(xy=y\)
Since y is non zero, \((xy)^2 > 0\)
Sufficient
Answer is (D)




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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:14
xy > 0 > Both x>0 & y>0 (OR) x<0 & y<0
(1) y = x  1 Since both x & y are of same sign, Irrespective of their values, x/y + y/x will always be > 2
eg: x = 2, y = 1 > 2/1 + 1/2 = 2.5 > 2
x = 5, y = 4 > 5/4 + 4/5 = 1.25 + 0.8 = 2.05 > 2
Sufficient
(2) x = 2y > x/y = 2 > x/y + y/x = 2 + 1/2 = 2.5 > 2
Sufficient
IMO Option D
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:15
xy > 0 which means x and y should be of same sign 1) y= x 1 Substituting in the x/y + y/x we get (2x^22x+1) / x(x1) which can be further reduced to (2x1)(x1)/x(x1) = (2x  1) / x So is (2x1) / x > 2? (2x1) / x = 2  (1/x) For all x >0 the equation will always be < 2 And for all x <0 the equation will always be > 2. Hence 1 is Not Sufficient. 2) x = 2y Substituting in the equation we get y / 2y + 2y / y = 2.5 [note: x and y are of same sign always as given in prompt ] . Hence the equation is >2 So statement 2 alone is sufficient. Hence option B.
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:18
Question Stem \(\frac{y}{x}+\frac{x}{y}>2\) \(y^2+x^2>2xy\) \(y^2+x^22xy>0\) \((xy)^2>0\)
From Statement 1 \(xy=1\) > square both sides \((xy)^2=1\) sufficient
From Statement 2 \(x=2y\) \(x/y=2\) \(y/x=1/2\) \(\frac{x}{y}+\frac{y}{x}=2+\frac{1}{2}\)
Which is greater than 2, hence sufficient
Answer D



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:18
Given, xy > 0 means neither x nor y can be ZERO



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:20
If xy>0, is y/x+x/y>2 (1) y=x−1 (2) x=2y y/x + x/y = (x^2 + y^2)/xy => y/x+ x/y 2 >0 => (x^2 + y^2  2xy)/xy>0 => (xy)^2/xy>0 => xy>0 since (xy)^2 is always >0 if x<>y and it is given that xy>0 The expression x/y+y/x>2 except when x=y 1. y = x 1 => x<>y (xy)^2 = 1 > 0 Sufficient. 2. y = 2x (xy)^2 = x^2 > 0 since x<>0 Sufficient. Each statement alone is sufficient. IMO D
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If xy > 0, is y/x + x/y > 2?
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Updated on: 03 Jul 2019, 08:38
D, each independently sufficient. Let's start with easy statement#2, x=2y So, y/x + x/y = {y^2 + x^2}/xy Substituting x=2y {y^2 + 4*y^2}/2y^2, cancelling y^2 from numerator and denominator (1+4)/2 > 2 sufficient Statement#1 x=y+1, for y=1, x=2 2 + .5 is > 2 Also, for y=.1, x=1.1 11 + 1/11 is always >2 But for y=7, x=6 7/6 + 6/7 >2,sufficient.
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Originally posted by LeoN88 on 03 Jul 2019, 08:20.
Last edited by LeoN88 on 03 Jul 2019, 08:38, edited 2 times in total.



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:23
Question stem =
[(x^2+y^2)/ xy] > 2
Using 2 we can easily get 2.5 by substituting y = x1
using 1 we get (y^2/y^2+1) +1 that may or may not be >2 hence not sufficient



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:24
Ans D
1) Substitute the value of y in the eq(n)
(x1/x) + (x/x1) x cannot be 0 and 1 since denominator cannot be zero
Put x=2, value= 2.5 Put x=1 or 2 value of eq(n) still >2 So sufficient
2) Substitute the value of x in terms of y
(y/2y)+(2y/y) gives 2+1/2 which is greater than 2. So sufficient
Therefore Ans D



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:28
Quote: If xy>0, is y/x+x/y>2?
(1) y=x−1
(2) x=2y If xy>0, i.e. signs of both x and y are identical Now, y/x+x/y will be equal to 2 only if both x and y are equal otherwise it will always be greater than 2? So question becomes Is x = y?(1) y=x−1 i.e. x and y are not equal hence SUFFICIENT (2) x=2y i.e. x and y are NOT equal hence SUFFICIENT Answer: Option D
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:29
IMO D
If xy > 0, is y/x + x/y > 2? Let's reorder the question: y/x + x/y > 2 => x^2 + y^2 > 2xy (as xy > 0, we can easily multiply both sides by xy) => (xy)^2 > 0 => xy > 0 => x>y Thus the question asks is x>y ?
St1: y=x1 => if x=2, y=1 (note x or y cannot be 0 as given xy>0) That means x>y, sufficient
St2: x=2y => if y=1, x=2 That means x>y, sufficient
Answer D
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:32
Statements 1 and 2 together are sufficient to answer the question. Hence option C would be the answer. Posted from my mobile device
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:33
Either both x and y are negative or positive. Lets start with second statement. Direct substitution will give us 2.5> 2 Statement 1 is a bit tricky. First simplify what the question is asking. y/x+x/y>2 LHS take common denominator and take it to RHS y^2 + x^2 > 2xy This is basically simplified to (x  y)^2 > 0 Now replace statement 1 .... x  (x  1) > 0 which is 1 > 0 Hence answer is D
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:34
Given xy>0 implies that x and y have same sign. Either both are +ve or both are ve.
y/x+x/y>2 It is possible when x and y have sign, and when x and y are distinct real numbers. We already know that x and y have same sign. If we know that x and y are distinct, we can give our answer.
Statement 1 x=y1 x and y are distinct, hence statement is sufficient.
Statement 2 x=2y Again we can see x and y are distinct, statement is sufficient
IMO D



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:34
Required is (x^2+y^2)/xy>0 when xy>0 (1) y=x−1 ; Insufficient as when substituted we get 2+1/x (2) x=2y sufficient; 2+1/2 is greater than 2 IMO B
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:41
From the xy>0 >> x>0; y>0 or x<0; y<0
1) y=x1 (x1)/x+x/(x1)>2 Not sufficient to answer the question
2) x=2y y/2y+2y/y>2 1/2+2>2 Sufficient
Answer: B



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:41
If xy>0, is y/x+x/y>2?
(1) y=x−1 ==>xy = 1 ; (xy)^2 = 1 ==> x^2 2xy + y^2 = 1 > 0 ==> x^2+y^2 > 2xy => (x^2 + y^2)/xy > 2 (because xy > 0) ==> y/x + x/y > 2 > yes (2) x=2y ==> x/y = 2 & y/x = 1/2, so y/x+x/y = 2.5 > 2 > yes
Answer: D



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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:41
Hi, Given xy>0 means either x> 0, y>0 or x<0, y<0. Now, the question: \(\frac{(y)}{(x)}\) + \(\frac{(x)}{(y)}\) >2
=> (\(x^2\) + \(y^2\) 2*x*y) >0 => \((xy)^2\) >0 => (xy) >0 => x > y ?
1. y = x1 => x = y+1 if x> 0 , y>0 then x > y. if x<0 , y<0 then also x>y. Sufficient.
2. x =2y if x>0, y>0 then x>y. if x<0, y<0 then x<y. Insufficient.
The answer is A.
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Re: If xy > 0, is y/x + x/y > 2?
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03 Jul 2019, 08:44
the given says that x,y ≠ zero, and have the same signthe question can be rephrased to: is \(\frac{y}{x} + \frac{x}{y} > 2\) \(\frac{{y^2 + x^2}}{{xy}} > 2\) \(y^2 + x^2 > 2xy\) \(x^2  2xy + y^2 > 0\) \((xy)^2 > 0?\) is \(x≠y\)?? statement 1 says that (xy) = 1, so yes \((xy)^2 > 0\) > sufficientstatement 2 says that xy = y, so \((xy)^2 > 0\) > \((y)^2 > 0\) which is correct as y ≠ zero > sufficientD
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Re: If xy > 0, is y/x + x/y > 2?
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