If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? : Data Sufficiency (DS)

2017-01-25T04:16:37-08:00

If yz ≠ 0, is [m][fraction]x–y+z/2z[/fraction] < [fraction]x/2z[/fraction]–[fraction]y/2z[/fraction]–[fraction]x/y[/fraction][/m]? (1) [m][fraction]x/y[/fraction] < -[fraction]1/2[/fraction][/m] (2) xy < 0

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RMD007

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

28 Jan 2017, 04:44

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It should be A.

After simplifying, we get equation x/y< -1/2

Since A confirms that x/y is less than -1/2, (x-y+z)/2z should be less than (x/2z)–(y/2z)–(x/y).

If option A would have been x/y >= -1/2, then also A would be the answer. Because it confirms that (x-y+z)/2z can not be less than (x/2z)–(y/2z)–(x/y).

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mohdabbas5

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

28 Jan 2017, 22:42

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RMD007 wrote:

It should be A.

After simplifying, we get equation x/y< -1/2

Since A confirms that x/y is less than -1/2, (x-y+z)/2z should be less than (x/2z)–(y/2z)–(x/y).

If option A would have been x/y >= -1/2, then also A would be the answer. Because it confirms that (x-y+z)/2z can not be less than (x/2z)–(y/2z)–(x/y).

Thanks a lot for the solution

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

13 Mar 2017, 10:55

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Expert Reply

Moving this to a better forum.

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ameyaprabhu

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

30 Jun 2017, 00:24

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CrackuM7 wrote:

->(x–y+z)/2z < (x/2z)–(y/2z)–(x/y) and neither y nor z =0 as yz ≠ 0
->Expanding LHS -> (x/2z)–(y/2z) + (z/2z) < (x/2z)–(y/2z)–(x/y)
-> x/2z and y/2z cancels from botht he sides as z ≠ 0. And also z/2z = 1/2, as z ≠ 0
->1/2 < -(x/y)
->x/y < -1/2 as when +ve and negaitive signs are reveresed, the lessthan and greater than signs also reverses.

I hope this help..

Hi, I understand that on expanding we get (x/2z)–(y/2z) + (z/2z) < (x/2z)–(y/2z)–(x/y)

But I thought we cannot move the variables, unless we know the sign, so how can we cancel out (X/2z) and (y/2z)?

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

30 Jun 2017, 00:41

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ameyaprabhu wrote:

CrackuM7 wrote:

->(x–y+z)/2z < (x/2z)–(y/2z)–(x/y) and neither y nor z =0 as yz ≠ 0
->Expanding LHS -> (x/2z)–(y/2z) + (z/2z) < (x/2z)–(y/2z)–(x/y)
-> x/2z and y/2z cancels from botht he sides as z ≠ 0. And also z/2z = 1/2, as z ≠ 0
->1/2 < -(x/y)
->x/y < -1/2 as when +ve and negaitive signs are reveresed, the lessthan and greater than signs also reverses.

I hope this help..

Hi, I understand that on expanding we get (x/2z)–(y/2z) + (z/2z) < (x/2z)–(y/2z)–(x/y)

But I thought we cannot move the variables, unless we know the sign, so how can we cancel out (X/2z) and (y/2z)?

We cannot multiply/divide an inequality by the variable if we don't know its sign but we can add/subtract whatever we want to/from both sides of an inequality.

For example, we cannot divide xy > xz by x unless we know the sign of x. If x is positive, then we'll get y > z but if x is negative, then we'll get y < z (flip the sign when multiplying/dividing by negative number). On the other hand we can subtract x from both sides of x + y > x to get y > 0.

Hope it's clear.

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mohdabbas5

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

25 Jan 2017, 22:29

Thanks for the solution :-D

but my question is, after simplifying the equation it gives the result of same as statement 1. then how statement 1 alone is sufficient to answer the question being asked.

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

26 Jan 2017, 03:34

mohdabbas5 wrote:

Thanks for the solution :-D

but my question is, after simplifying the equation it gives the result of same as statement 1. then how statement 1 alone is sufficient to answer the question being asked.

simplifying the equation does not give the result. It only simplifies the question asked. Statement 1 confirms the question asked.

B

mohdabbas5

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

28 Jan 2017, 04:22

sangarajubharadwaj wrote:

mohdabbas5 wrote:

Thanks for the solution :-D

but my question is, after simplifying the equation it gives the result of same as statement 1. then how statement 1 alone is sufficient to answer the question being asked.

Hi mohdabbas5,

I know why you were confused.. Upon simplification of question stem, you get (x/y) <-1/2.... that is the question they are actually asking... if you know that x/y <-1/2 or >-1/2, you can answer the question either with a 'yes' or a 'No'...
' yes' or 'no' is what you need from the choices.. choice 1 directly answers your question... by using choice 1, you can say "yes, the value is less than -1/2"... but only using choice 2 you can't answer the question.... so A is the answer...
Now coming toyour confusion, it's tricky but also silly.. ?Usually we are accustomed to solve the given question stem, then apply the truth from the given choice and get to a common point where the question stem matches your calculation.. but here, ther's no need tocalculate anything.. it's directly given and you were confused....

For e.g.., the question stem is "Am I human?"

Option 1: "I am human"...

The question is so simple that u never encountered such questikn and the answer is too obvious.. so you were confused somehow...

Thanks
Hopethe explanation helps... But please don't yawn..?

Hi sangarajubharadwaj
Many thanks for an excellent explanation. i got the point.

S

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Re: If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

02 Jul 2018, 00:23

mohdabbas5 wrote:

If yz ≠ 0, is \(\frac{x–y+z}{2z} < \frac{x}{2z}–\frac{y}{2z}–\frac{x}{y}\)?

(1) \(\frac{x}{y} < -\frac{1}{2}\)

(2) xy < 0

But why option (2) is incorrect??? please explain.

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If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

24 Sep 2019, 01:36

sajon wrote:

mohdabbas5 wrote:

If yz ≠ 0, is \(\frac{x–y+z}{2z} < \frac{x}{2z}–\frac{y}{2z}–\frac{x}{y}\)?

(1) \(\frac{x}{y} < -\frac{1}{2}\)

(2) xy < 0

But why option (2) is incorrect??? please explain.

The best approach is to break down the question stem before going to Statements I and II. A detailed solution can be found in the attachment.

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Detailed Solution.doc [44 KiB]
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If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)? [#permalink]

24 Sep 2019, 01:36

GMAT Data Sufficiency (DS) Questions

If yz ≠ 0, is (x–y+z)/2z < (x/2z)–(y/2z)–(x/y)?