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04 Jun 2019, 11:01
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gmatcracker2010
If x ≠ -y is \(\frac{x-y}{x+y}>1\)?
(1) x > 0
(2) y < 0
(1) x > 0
(2) y < 0
I'm actually going to go straight to plugging in numbers here, without rephrasing. That's different from usual, but there are a few reasons I'm going to approach it like that:
- I can't just cross-multiply, because I don't know whether the denominator of the fraction is positive or negative. So the easiest way to rephrase won't work at all.
- Also, this looks like an algebra problem, not a positive/negative type problem. That's because the inequality is "greater than 1", not "greater than 0" (positive) or "less than 0" (negative). So, we actually care about what the specific numbers are, and we can't necessarily just use logical reasoning and figure out whether they're positive or negative.
So, let's start with the statements!
Statement 1 x is positive.
Let's try x = 1 to make things easy.
is \(\frac{1-y}{1+y}>1\)?
Let's see what happens when we use different values for y...
if y = 1, then \(\frac{1-y}{1+y}=\frac{1-1}{1+1}\)= 0, which is NOT greater than 1.
How can we make the fraction bigger than 1? We'd like the top to be bigger than the bottom. Well, the only way for that to happen is by subtracting a NEGATIVE number on the top. Let's try making y negative:
if y = -2, then \(\frac{1-y}{1+y}=\frac{1-(-2)}{1+(-2)} = \frac{3}{-1}\)= -3, which is NOT greater than 1.
Interesting. To make it greater than 1, we need to keep the denominator from becoming negative. Let's make y a smaller negative number:
if y = -0.5, then \(\frac{1-y}{1+y}=\frac{1-(-0.5)}{1+(-0.5)} = \frac{1.5}{0.5}\)= 3, which IS greater than 1.
Since it can go either way, the statement is insufficient.
Statement 2: We can reuse the two cases we just tested! We found a case where y = -2, and the result was NOT greater than 1. We also found a case where y = -0.5, and the result WAS greater than 1. This statement is insufficient as well.
Statements 1 and 2 together Since we used the exact same cases to prove both statements insufficient individually, we can also use those same cases when we put the statements together. Therefore, the statements are insufficient together and the answer is E.
06 Mar 2020, 05:14
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Bunuel
If \(x\neq{-y}\) is \(\frac{x-y}{x+y}>1\)?
Is \(\frac{x-y}{x+y}>1\)? --> Is \(0>1-\frac{x-y}{x+y}\)? --> Is \(0>\frac{x+y-x+y}{x+y}\)? --> Is \(0>\frac{2y}{x+y}\)?
(1) \(x>0\) --> Not sufficient.
(2) \(y<0\) --> Not sufficient.
(1)+(2) \(x>0\) and \(y<0\) --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.
Answer: E.
The problem with your solution is that when you are then writing \(x-y>x+y\), you are actually multiplying both sides of inequality by \(x+y\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x+y>0\) you should write \(x-y>x+y\) BUT if \(x+y<0\), you should write \(x-y<x+y\) (flip the sign when multiplying by negative expression).
So again: given inequality can be simplified as follows: \(\frac{x-y}{x+y}>1\) --> \(0>1-\frac{x-y}{x+y}<0\) --> \(0>\frac{x+y-x+y}{x+y}\) --> \(0>\frac{2y}{x+y}\) --> we can drop 2 and finally we'll get: \(0>\frac{y}{x+y}\).
Now, numerator is negative (\(y<0\)), but we don't know about the denominator, as \(x>0\) and \(y<0\) can not help us to determine the sign of \(x+y\). So the answer is E.
Hope it helps.
Hi Bunuel, am I correct in stating that the answer would still have been E if the question was?Is \(\frac{x-y}{x+y}>1\)? --> Is \(0>1-\frac{x-y}{x+y}\)? --> Is \(0>\frac{x+y-x+y}{x+y}\)? --> Is \(0>\frac{2y}{x+y}\)?
(1) \(x>0\) --> Not sufficient.
(2) \(y<0\) --> Not sufficient.
(1)+(2) \(x>0\) and \(y<0\) --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.
Answer: E.
The problem with your solution is that when you are then writing \(x-y>x+y\), you are actually multiplying both sides of inequality by \(x+y\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x+y>0\) you should write \(x-y>x+y\) BUT if \(x+y<0\), you should write \(x-y<x+y\) (flip the sign when multiplying by negative expression).
So again: given inequality can be simplified as follows: \(\frac{x-y}{x+y}>1\) --> \(0>1-\frac{x-y}{x+y}<0\) --> \(0>\frac{x+y-x+y}{x+y}\) --> \(0>\frac{2y}{x+y}\) --> we can drop 2 and finally we'll get: \(0>\frac{y}{x+y}\).
Now, numerator is negative (\(y<0\)), but we don't know about the denominator, as \(x>0\) and \(y<0\) can not help us to determine the sign of \(x+y\). So the answer is E.
Hope it helps.
Is (x - y) / (x + y) > 0 ?
06 Mar 2020, 05:17
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OjhaShishir
Bunuel
If \(x\neq{-y}\) is \(\frac{x-y}{x+y}>1\)?
Is \(\frac{x-y}{x+y}>1\)? --> Is \(0>1-\frac{x-y}{x+y}\)? --> Is \(0>\frac{x+y-x+y}{x+y}\)? --> Is \(0>\frac{2y}{x+y}\)?
(1) \(x>0\) --> Not sufficient.
(2) \(y<0\) --> Not sufficient.
(1)+(2) \(x>0\) and \(y<0\) --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.
Answer: E.
The problem with your solution is that when you are then writing \(x-y>x+y\), you are actually multiplying both sides of inequality by \(x+y\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x+y>0\) you should write \(x-y>x+y\) BUT if \(x+y<0\), you should write \(x-y<x+y\) (flip the sign when multiplying by negative expression).
So again: given inequality can be simplified as follows: \(\frac{x-y}{x+y}>1\) --> \(0>1-\frac{x-y}{x+y}<0\) --> \(0>\frac{x+y-x+y}{x+y}\) --> \(0>\frac{2y}{x+y}\) --> we can drop 2 and finally we'll get: \(0>\frac{y}{x+y}\).
Now, numerator is negative (\(y<0\)), but we don't know about the denominator, as \(x>0\) and \(y<0\) can not help us to determine the sign of \(x+y\). So the answer is E.
Hope it helps.
Hi Bunuel, am I correct in stating that the answer would still have been E if the question was?Is \(\frac{x-y}{x+y}>1\)? --> Is \(0>1-\frac{x-y}{x+y}\)? --> Is \(0>\frac{x+y-x+y}{x+y}\)? --> Is \(0>\frac{2y}{x+y}\)?
(1) \(x>0\) --> Not sufficient.
(2) \(y<0\) --> Not sufficient.
(1)+(2) \(x>0\) and \(y<0\) --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.
Answer: E.
The problem with your solution is that when you are then writing \(x-y>x+y\), you are actually multiplying both sides of inequality by \(x+y\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x+y>0\) you should write \(x-y>x+y\) BUT if \(x+y<0\), you should write \(x-y<x+y\) (flip the sign when multiplying by negative expression).
So again: given inequality can be simplified as follows: \(\frac{x-y}{x+y}>1\) --> \(0>1-\frac{x-y}{x+y}<0\) --> \(0>\frac{x+y-x+y}{x+y}\) --> \(0>\frac{2y}{x+y}\) --> we can drop 2 and finally we'll get: \(0>\frac{y}{x+y}\).
Now, numerator is negative (\(y<0\)), but we don't know about the denominator, as \(x>0\) and \(y<0\) can not help us to determine the sign of \(x+y\). So the answer is E.
Hope it helps.
Is (x - y) / (x + y) > 0 ?
Absolutely. The numerator (x - y) will be positive but the denominator (x + y) could be positive as well as negative, so the whole fraction ((x - y)/(x + y)) could be positive as well as negative.
