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Given that x ≠ 0, y ≠ 0, and (x + y) ≠ 0, is y/x > y/(x + y) ? (1) x

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Bunuel
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Given that x ≠ 0, y ≠ 0, and (x + y) ≠ 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   02 Mar 2022, 08:00
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Given that \(x ≠ 0\), \(y ≠ 0\), and \((x + y) ≠ 0\), is \(\frac{y}{x} > \frac{y}{(x + y)}\) ?


(1) \(x > -1\)

(2) \(y > -1\)



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E
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BrentGMATPrepNow
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Re: Given that x ≠ 0, y ≠ 0, and (x + y) ≠ 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   02 Mar 2022, 08:15
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Bunuel wrote:
Given that \(x ≠ 0\), \(y ≠ 0\), and \((x + y) ≠ 0\), is \(\frac{y}{x} > \frac{y}{(x + y)}\) ?


(1) \(x > -1\)

(2) \(y > -1\)


Target question: Is \(\frac{y}{x} > \frac{y}{(x + y)}\) ?

STRATEGY: When I scan the two statements, they both look insufficient, since only one variable is defined for each statement. Also notice that if statement allows for positive and negative values of x and y. Given all of this, I’m pretty sure I can identify some cases with conflicting answers to the target question. So, I’m going to head straight to……

Statements 1 and 2 combined
Statement 1 tells us that \(x > -1\)
Statement 2 tells us that \(y > -1\)
There are several values of \(x\) and \(y\) that satisfy BOTH statements. Here are two:
Case a: \(x = 1\) and \(y = 1\). In this case, our inequality becomes: \(\frac{1}{1} > \frac{1}{(1 + 1)}\), which simplifies to get: \(1 > \frac{1}{2}\) (a true statement), which means the answer to the target question is YES, it's true that \(\frac{y}{x} > \frac{y}{(x + y)}\)

Case b: \(x = 0.3\) and \(y = -0.6\). In this case, our inequality becomes: \(\frac{-0.6}{0.3} > \frac{-0.6}{0.3 + (-0.6)}\), which simplifies to get: \(-2 > 2\) (a false statement), which means the answer to the target question is NO, it's not true that \(\frac{y}{x} > \frac{y}{(x + y)}\)

Answer: E

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Brent
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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   21 Jun 2022, 20:27
Bunuel, is this a 700 level question?
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Bunuel
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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   22 Jun 2022, 01:05
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Aarya20 wrote:
Bunuel, is this a 700 level question?

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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   06 May 2023, 05:26
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I initially approached this question in the following way.

y/x > y/(x + y)
can be simplified to
x/y < (x+y)/y
x/y < (x/y) + (y/y)
x/y < (x/y) + 1

This is true for all numbers, so I thought the answer was D, either suffices.
Can anyone point out to where I'm going wrong?
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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   07 Oct 2023, 21:47
Bunuel wrote:
Given that \(x ≠ 0\), \(y ≠ 0\), and \((x + y) ≠ 0\), is \(\frac{y}{x} > \frac{y}{(x + y)}\) ?

(1) \(x > -1\)

(2) \(y > -1\)

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\(\frac{y}{x} > \frac{y}{(x + y)}\)
=> \(\frac{y}{x} - \frac{y}{(x + y)}\) > 0
=> \(\frac{(yx + y^2 - xy)}{x(x+y)}\) > 0
=> \(\frac{(y^2)}{x(x+y)}\) > 0

Since \(y^2\) is positive \(x(x+y)\) has to be positive
Hence the question is essentially asking if \(x(x+y)\) > 0

Considering statements

(1) \(x > -1\)
Let x be 10 and y be 10 the equation above holds.
But if x is 0 and y is 10 it does not hold

(2) \(y > -1\)
Similar statements as above are applicable.

However even if we combine the statement, we have the below possibilities.
If x = 10 and y be 10 the equation above holds
But if x is 0 and y is 0 it does not hold

Hence E
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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   14 Oct 2023, 17:02
mlindak wrote:
I initially approached this question in the following way.

y/x > y/(x + y)
can be simplified to
x/y < (x+y)/y
x/y < (x/y) + (y/y)
x/y < (x/y) + 1

This is true for all numbers, so I thought the answer was D, either suffices.
Can anyone point out to where I'm going wrong?


IMO here when you simplify you have
y/x > y/(x + y)
(x + y) * y > xy
xy + y^2 > xy

which simplifies to y^2 > 0

This will forever be true because we know that y is different than 0, and a square of a real number can only be 0 or positive
Thus answer choices are irrelevant IMO...
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DanTheGMATMan
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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink] New post   14 Oct 2023, 17:32
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LucienH wrote:
mlindak wrote:
I initially approached this question in the following way.

y/x > y/(x + y)
can be simplified to
x/y < (x+y)/y
x/y < (x/y) + (y/y)
x/y < (x/y) + 1

This is true for all numbers, so I thought the answer was D, either suffices.
Can anyone point out to where I'm going wrong?


IMO here when you simplify you have
y/x > y/(x + y)
(x + y) * y > xy
xy + y^2 > xy

which simplifies to y^2 > 0

This will forever be true because we know that y is different than 0, and a square of a real number can only be 0 or positive
Thus answer choices are irrelevant IMO...


Be careful simplifying the stem when you don't know whether x, y or x+y are negative, for example. You can't actually simplify this because you don't know whether to flip the inequality or not because you don't know whether these values are negative or not. Thanks
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Re: Given that x 0, y 0, and (x + y) 0, is y/x > y/(x + y) ? (1) x [#permalink]
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