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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
1
Kudos
Statement 1 says that x>y. Let's try some values. If both are positive, then the answer is yes.
x=3, y=1 says that 3/1>3/1
However, if both values are negative, the answer is no and the same conditions apply when one is positive and one is negative.
x=-1, y=-3 says that 1/3>3/1 or x=3, y=-2 says that 3/-2>-2/3 (both of these are not true)
Insufficient

Statement 2 says that xy>0, meaning that they both have the same sign. This doesn't help as we don't know their relative values. Insufficient

Combining also doesn't help as both having the same sign with x being greater than y is still insufficient to determine whether the inequality holds true.

Choose E
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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
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Hi Awli,

This is another great opportunity to TEST VALUES.

We're told that neither X nor Y can equal 0. We're asked if X/Y > Y/X. This is a YES/NO question.

Fact 1: X > Y

IF....
X = 2
Y = 1
2/1 > 1/2 and the answer to the question is YES

IF...
X = 2
Y = -1
2/-1 is NOT > -1/2 and the answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: XY > 0

This Fact tells us that X and Y have the SAME SIGN.

IF....
X = 2
Y = 1
2/1 > 1/2 and the answer to the question is YES

IF....
X = 1
Y = 2
1/2 is not > 2/1 and the answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know...
X > Y
X and Y have the SAME sign

IF....
X = 2
Y = 1
2/1 > 1/2 and the answer to the question is YES

IF....
X = -1
Y = -2
(-1)/(-2) is NOT > (-2)/(-1) and the answer to the question is NO
Combined, INSUFFICIENT

Final Answer:

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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
Given \(x\neq{0}\) and \(y\neq{0}\)

\(is \frac{x}{y} > \frac {y}{x} ?\)
it will be True when

Case (i): Both \(x\) and \(y\) positive
\(x > y\)

Case (ii): Both \(x\) and \(y\) negative
\(|x| > |y|\)

Case (iii): \(x\) and \(y\) opposite signs
\(|x| < |y|\)

From Statement 1 and 2, there is no definitive information about signs of \(x\) or \(y\)

Answer E
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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
Bunuel wrote:

Tough and Tricky questions: Inequalities.



If x and y cannot be equal to zero, is x/y > y/x ?

(1) x > y
(2) xy > 0

Kudos for a correct solution.

Source: Chili Hot GMAT


from stem

x/y>y/x = x/y-y/x = y(x-y) / yx = 1-y/x , question asks is y/x<1 ( if yx has different signs or same sign and /y/</x/)

from 1

x-y>0 .... insuff

from 2

yx has same sign but we dont know whether or not /y/</x/....... insuff

both

usingy: y(x-y) /yx = (unknown sign*+Ve ) / +ve.... still insuff

E
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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
Hi,

I am not sure how E is the answer

I solved it by below method

x/y>y/x

x/y-y/x>0

(x^2-y^2)/xy>0

1.x>y

This means x^2-y^2 is positive. But we don't know abt sign of xy. So Insufficient

2. xy>0

This tells us xy is positive. But we dont know abt sign of x^2-y^2.So insufficient

Taking 1 and 2 together

1.x^2-y^2 is positive
2.xy is positive

Therefore both together are sufficient

Can anyone help understand what am I missing here.

Thanks in Advance !!
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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
1
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Expert Reply
Bunuel wrote:

Tough and Tricky questions: Inequalities.



If x and y cannot be equal to zero, is x/y > y/x ?

(1) x > y
(2) xy > 0


One approach:

\(\frac{x}{y} > \frac{y}{x}?\)
\(\frac{x}{y} - \frac{y}{x} > 0?\)
\(\frac{x^2 - y^2}{xy} > 0?\)

Statement 2: xy > 0
Case 1: x=2 and y=1
In this case, \(\frac{x^2 - y^2}{xy} = \frac{4-1}{2} = \frac{3}{2}\), so the answer to the question stem is YES.

Case 2: x=-1 and y=-2
In this case, \(\frac{x^2 - y^2}{xy} = \frac{1-4}{2} = -\frac{3}{2}\), so the answer to the question stem is NO.

Cases 1 and 2 satisfy BOTH STATEMENTS.
Since the answer is YES in Case 1 but NO in Case 2, the two statements combined are INSUFFICIENT.

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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
So, we have x≠0 and y≠0 for x/y and y/x to be possible divisions. The question is x/y > y/x
=> x/y - y/x > 0
=> (x^2-y^2)/xy > 0
=> ((x-y)(x+y))/xy > 0
To know the final sign of the equation, we need to know the signs of all individual factors, i.e. x+y, x-y, and x*y
(1) x > y => x – y > 0 => INSUFFICIENT
When x > y or x – y > 0, we cannot conclude anything about x + y or x*y.
Example:
x = 2, y = 1 => x + y = 3 > 0, x*y = 2 > 0
x = 2, y = -3 => x + y= -1 < 0, x*y = -6 < 0
(2) x*y > 0 => INSUFFICIENT
Similarly, when x*y>0, it only tells you that x and y either both negative or both positive (same sign). We cannot conclude anything about x - y and x + y.
Example:
x = 4, y = 5, x - y = -1 < 0, x + y = 9 > 0
x = -4, y = -5, x - y = 1 > 0, x + y = -9 < 0
Together (1) + (2) -> STILL INSUFFICIENT because we do not know anything about x + y.
So, E is the final answer.
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Re: If x and y cannot be equal to zero, is x/y > y/x ? [#permalink]
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