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If y≠0 and y≠-1, is x/y > x/(y+1)?

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If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 24 Feb 2011, 12:50
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E

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  55% (hard)

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63% (01:13) correct 37% (01:20) wrong based on 131 sessions

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Project DS Butler: Day 53: Data Sufficiency (DS106)


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If y≠0 and y≠-1, is x/y > x/(y+1)?

(1) x≠0
(2) x > y
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 24 Feb 2011, 13:19
banksy wrote:
If y≠0 and y≠-1, which is greater, x/y or x/(y+1)?
(1) x≠0
(2) x > y


I think question should be:

If y≠0 and y≠-1, is x/y > x/(y+1)?

Is \(\frac{x}{y}>\frac{x}{y+1}\)? --> is \(\frac{x}{y}-\frac{x}{y+1}>0\)? --> is \(\frac{xy+x-xy}{y(y+1)}>0\)? --> is \(\frac{x}{y(y+1)}>0\)?

(1) x≠0. Not sufficient.
(2) x > y. Not sufficient.

(1)+(2) Still not sufficient, for example: if x>0>(y=-1/2) answer is NO but if x>y>0 answer is YES.

Answer: E.
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 24 Feb 2011, 13:23
If y≠0 and y≠-1, which is greater, x/y or x/(y+1)?
(1) x≠0
(2) x > y

Sol:

(1) x!=0 doesn't tell us anything.

x=10
y=1

\(\frac{x}{y}=10\)
\(\frac{x}{y+1}=\frac{10}{2}=5\)
\(\frac{x}{y}>\frac{x}{y+1}\)

x=-10
y=-2

\(\frac{x}{y}=\frac{-10}{-2}=5\)
\(\frac{x}{y+1}=\frac{-10}{-1}=10\)
\(\frac{x}{y}<\frac{x}{y+1}\)

(2) It tells us that x and y have same sign

x > y

x/y > 1 (same sign)
Use sample set from statement 1.
Not sufficient.

Combining both;
Use sample set from statement 1.
Not Sufficient.

Ans: "E"
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 24 Feb 2011, 15:42
So the bottom line is that we need to know the sign of both x and y, to answer the question. And neither provides that.
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 24 Feb 2011, 15:52
bellcurve wrote:
So the bottom line is that we need to know the sign of both x and y, to answer the question. And neither provides that.


Just signs won't be enough.

\(\frac{x}{y(y+1)}>0\) holds true if:

A. \(x>0\) and \(y>0\) or \(y<-1\);
B. \(x<0\) and \(-1<y<0\).

Check the image below:
Attachment:
MSP4919ebfi83e60e30b700002i4ed3dd94aec949.gif
MSP4919ebfi83e60e30b700002i4ed3dd94aec949.gif [ 3.03 KiB | Viewed 4357 times ]

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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 12 Apr 2014, 10:47
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Hi experts, Why is this question tagged 'poor quality'? I found this question quite tough.
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 13 Apr 2014, 05:39
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 13 Apr 2014, 07:29
Bunuel wrote:
MensaNumber wrote:
Hi experts, Why is this question tagged 'poor quality'? I found this question quite tough.


Edited the tags. Thank you.


Thanks Bunuel! Just a quick follow up question- though, you edited it for 600-700 level now, you still retained 'poor quality' tag? Is that intentional or an error?
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 13 Apr 2014, 08:05
MensaNumber wrote:
Bunuel wrote:
MensaNumber wrote:
Hi experts, Why is this question tagged 'poor quality'? I found this question quite tough.


Edited the tags. Thank you.


Thanks Bunuel! Just a quick follow up question- though, you edited it for 600-700 level now, you still retained 'poor quality' tag? Is that intentional or an error?


That tag is gone now.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If y≠0 and y≠-1, is x/y > x/(y+1)?  [#permalink]

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New post 28 Dec 2018, 07:48
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Re: If y≠0 and y≠-1, is x/y > x/(y+1)? &nbs [#permalink] 28 Dec 2018, 07:48
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