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Is x > y ?

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Is x > y ?  [#permalink]

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New post 26 Jul 2017, 07:53
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15
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

67% (01:37) correct 33% (01:43) wrong based on 543 sessions

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Re: Is x > y ?  [#permalink]

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New post 26 Jul 2017, 08:07
4
2
Is x > y ?

(1) x + y > x − y

y > -y

2y > 0

y > 0. Since no information is given on x, then this statement is not sufficient.

(2) 3x > 2y.

If x = y = 1, then the answer would be NO but if x = 2 and y = 1, then the answer would be YES. Not sufficient.

(1)+(2) The examples we used for (2) are still valid, so even combined the statements are not sufficient.

Answer: E.

P.S. To elaborate: from (1) y > 0, so from (2) we can get that x/y > 2/3 but the question asks whether x/y > 1. So, still insufficient.
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Re: Is x > y ?  [#permalink]

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New post 26 Jul 2017, 20:14
I messed up on how i interpreted statement two, the fact that x=y=1 would create the right condition for statement 2 to be correct, also means that when answering the prompt, x>y, we get the answer NO; trying another set of numbers- x=2 and y=1- where statement two meets the condition, 2>1/3, means that the answer to the question x>y is YES.

Bunuel explains this but I didn't quite understand the process because I wasn't registering the prompt and statement criteria clearly.
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Re: Is x > y ?  [#permalink]

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New post 15 Sep 2018, 13:39
pingala wrote:
I messed up on how i interpreted statement two, the fact that x=y=1 would create the right condition for statement 2 to be correct, also means that when answering the prompt, x>y, we get the answer NO; trying another set of numbers- x=2 and y=1- where statement two meets the condition, 2>1/3, means that the answer to the question x>y is YES.

Bunuel explains this but I didn't quite understand the process because I wasn't registering the prompt and statement criteria clearly.


Hipingala,

here is how i would approach this question

we are asked that IS x>y
so is x-y>0
what could be the possible senarios
x>0, y>0 and |x|>|y|
x<0, y<0 and |y|>|x|
x>0, y<0
x<0, y>0 and |y|>|x|

in above all scenarios we would get the answer to our question.
Now statement A:
It says y>0. It could be x>0 or x<0, but still we don't about |x| and |y|
So insufficient.

Now Statement B:
3x>2y
or we can have
\(x>\frac{2}{3} y\)
if we read along it says x is greater than two-thirds of y.
We will have two senarios
y>0 even then we can't conclude that x> y. why say y>1 then x can be \(>\frac{2}{3}\) . So its possible that the range of of x is \(\frac{2}{3}<x<\infty\) and range of y would be \(1<y< \infty\) . Now if you imagine a number line with these two ranges then you can either conclude that x<y or x>y

Now if y<0 even then we can't conclude that x> y. Then x can be \(>-\frac{2}{3}\) . So its possible that the range of of x is \(-\frac{2}{3}<x<\infty\) and range of y would be \(-\infty<y<0\). Now if you imagine a number line with these two ranges then you can either conclude that x<y or x>y

Now even if we combine both the statements still we don't have any new information to conclusively answer if x>y

hence option E
let me know if this helps
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Re: Is x > y ?  [#permalink]

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New post 16 Sep 2018, 03:13
4
St1 :- x + y > x - y = 2y > 0 ( here we can only conclude that Y is positive. As no information is given about X, St1 is not sufficient )

St2 :- 3x > 2y = 1.5x > y
The above statement can be hold true in following case :-
X = 2, and Y = 2,
X =3, and Y = 2, or
X = 1, and Y = 2

As there's multiple value of x, this statement is also sufficient

Combining St1, and St2
by adding both Statement, we get 3x>0 (here we can only conclude that X, and Y are positive. But can't conclude that Whether X is greater than Y)

Hence, Answer E
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Re: Is x > y ?  [#permalink]

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New post 22 Oct 2018, 11:07
carcass wrote:
Is x > y ?

(1) x + y > x − y

(2) 3x > 2y


(DS04468)


This can be rearranged in different ways x > y or x - y > 0 or x/y > 1 (if y is positive)

statement 1) x + y > x - y can be rearranged to 2y > 0 then y > 0, insufficient we have no information on x

Eliminate A/D

Statement 2) 3x > 2y

if x = 1 and y = 1 then 3 > 2 then back to the stem 1 > 1 is not true.

if x = 3 and y = 1, then 9 > 2 then back to the stem 3 > 1 is true

Insufficient

Now try combined

given that y > 0

then we can divide by y, \(\frac{x}{y} > \frac{2}{3}\) we don't know for sure whether it is larger than 1? it could be anything larger than 2/3

So insufficient.

Answer choice E
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Re: Is x > y ?   [#permalink] 22 Oct 2018, 11:07
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