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# Is x > y ?

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Is x > y ?  [#permalink]

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26 Jul 2017, 08:53
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Difficulty:

45% (medium)

Question Stats:

66% (01:41) correct 34% (01:45) wrong based on 654 sessions

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Is x > y ?

(1) x + y > x − y

(2) 3x > 2y

(DS04468)

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Re: Is x > y ?  [#permalink]

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26 Jul 2017, 09:07
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.

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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26 Jul 2017, 21:14
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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15 Sep 2018, 14:39
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
Probus
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Re: Is x > y ?  [#permalink]

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16 Sep 2018, 04:13
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)

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Re: Is x > y ?  [#permalink]

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22 Oct 2018, 12:07
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.

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Re: Is x > y ?  [#permalink]

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16 Sep 2019, 12:18

Posted from my mobile device
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Re: Is x > y ?  [#permalink]

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18 Sep 2019, 05:20
carcass wrote:
Is x > y ?

(1) x + y > x − y

(2) 3x > 2y

(DS04468)

Ques: Is x > y?

(1) x + y > x − y

This inequality is independent of x. It gives us y > -y (when we subtract x from both sides). This means y is positive.
But we don't know whether x is greater than y.

(2) 3x > 2y
We know that 3 times of x is greater than 2 times of y. This will be true when x is less than y, equal to y or greater than y.
x = 3, y = 4
x = 3, y = 3
x = 3, y = 2

Using both statements, y is positive so dividing both sides of second inequality by 3y we get
x/y > 2/3

Now, x/y could be 3/4 in which case x < y or x/y could be 1 in which case both are equal or x/y could be 4/3 in which case x > y.

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Re: Is x > y ?   [#permalink] 18 Sep 2019, 05:20
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