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If x and y are integers and xy ≠ 0, is x - y > 0?

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If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 24 Mar 2015, 03:46
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 30 Mar 2015, 03:32
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Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) \(\sqrt{x}= x\) and \(\sqrt{y} = y\)

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
isx-ypositive_explanation.png
isx-ypositive_explanation.png [ 36.76 KiB | Viewed 2300 times ]


Frequently Asked Questions

Q: For statement 1, why can't we multiply both sides by y to get x < 1/2y? Wouldn't that be sufficient on its own?

A: You have good intuition, and this might work for an equation, but there's a crucial difference between equations and inequalities: if you divide (or multiply) both sides by a negative number, then you have to change the direction of the inequality sign.

If I have -5x > 10

and divide both sides by -5, I need to flip the inequality so I have:

x < -2

Let's see why with actual numbers:

Equation:

8 = 8

divide both sides by - 2:

-4 = -4 <--- still true!

Inequality:

16 > 8

divide both sides by -2

-8 > -4 ....no! We changed the relationship when dividing by -2. So we need to flip the sign:

-8 < -4

Now this is correct.

In this problem, we have to ask "Is y negative?" If it isn't, then we can rework the equation as you did. If it is negative, then we have to flip the sign. But we don't know if it's negative or not! We can't multiply by y.

So, we cannot multiply across an inequality with variables unless we know for certain whether they're positive or negative.

Q: Can you please explain to me how I know that the √y² = y, is positive? Why can't it be positive and negative? For example, √4 is +2 or -2.

A: The radical sign (more commonly known as the square root sign) actually means the POSITIVE square root of a given number. This positive square root is often referred to as the principal square root. Anytime we see the radical sign, we want the principal (positive) square root.

Also, from Wikipedia :)

"Every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by [square root symbol], where [square root symbol] is called radical sign.

For example, the principal square root of 9 is 3, denoted √9 = 3, because 3² = 3 × 3 = 9 and 3 is non-negative."

So if x²= 4, then x = 2 or -2

But if √4 = x, then x = 2 only

√4 is the positive root of 4 only.

When you see a "√" sign with something inside, there is only one possible output.

So √(x^2) = |x|

Also check out this blog on square roots for more information.
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 24 Mar 2015, 05:26
Answer should be B.
Given that non of a and y is 0.
Statement 1 is not enough, as relationship between x and Y are different for +ve and -ve values of x and y.
Statement 2 is enough. With this information x = y =1, as they can not be 0. SUFFICIENT.
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 24 Mar 2015, 05:42
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) \(\sqrt{x}= x\) and \(\sqrt{y} = y\)

Kudos for a correct solution.


1) x/y < 1/2
At X=1,Y=3 above inequality holds but is x>y ? NO
At X=-1,Y=-4 above inequality holds but is x>y ? YES
NOT SUFFICIENT.

(2) \(\sqrt{x}= x\) and \(\sqrt{y} = y\)
As X and Y are integers, we can infer X=Y=1
SUFFICIENT.

ANSWER :B
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 24 Mar 2015, 11:26
St 1) x/y < 1/2
At X=1,Y=3 above inequality holds but is x>y ? NO
At X=-1,Y=-4 above inequality holds but is x>y ? YES
NOT SUFFICIENT.

St 2) \sqrt{x}= x and \sqrt{y} = y
As X and Y are integers, we can infer X=Y=1
SUFFICIENT.
Ans: B
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 24 Mar 2015, 17:00
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) \(\sqrt{x}= x\) and \(\sqrt{y} = y\)

Kudos for a correct solution.


Statement 1: x/y < 1/2
test numbers
If x=-3, y=2 No
if x=3 y=-2 yes
Insufficient

Statement 2:
x, y = 1
sufficient

Answer: B
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 24 Mar 2015, 23:21
x &y integers & x*y not=0; is x>y?

1. Resolve the equation and 2x<y; this will hold true for +ve value but not hold true for -ve values; Not sufficient

2. Square of number = number itself only for 1, hence x=y=1; sufficient

Hence answer is B

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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 08 Apr 2015, 04:42
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Bunuel could you please explain this more briefly as why B is not the answer as the above fluff is bouncing all above the head.
if x^2=x and y^2=y this implies that x=1 and y=1. what am i missing here?????
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 08 Apr 2015, 04:52
anshul2014 wrote:
Bunuel could you please explain this more briefly as why B is not the answer as the above fluff is bouncing all above the head.
if x^2=x and y^2=y this implies that x=1 and y=1. what am i missing here?????


Sorry, there was a typo in (2). It should read: \(\sqrt{x^2}= x\) and \(\sqrt{y^2} = y\)

Edited. Thank you.
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 08 Apr 2015, 08:51
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) \(\sqrt{x^2}= x\) and \(\sqrt{y^2} = y\)

Kudos for a correct solution.



Hi,

The way I approached is-

1) x/y< 1/2 => 2x < y

Now if x = -3 and y = -1 => 2x < y => -6 < -1. Then is x - y > 0 - NO.

if x = 1 and y = 3 => 2x < y => 2 < 3. Then is x - y > 0 - NO.

if x = -1/2 and y = -1/6 => 2x < y => -1 < -1/6. Then is x - y > 0 - NO.

if x = 1/6 and y = 1/2 => 2x < y => 1/3 < 1/2. Then is x - y > 0 - NO.

To simply put, shouldn't A be sufficient enough to say that if 2x < y then x - y will never be greater zero??

Please suggest.
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 11 Apr 2015, 10:19
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HI King407,

TESTing VALUES is a great approach for this question, but the problem with your approach is that it does not factor in all of the possibilities that can occur. If you look at your work, you chose to make X and Y either BOTH positive OR both NEGATIVE, but you COULD have 1 positive and 1 negative....

We're told that X/Y < 1/2

IF...
X = 1
Y = -3
then -1/3 < 1/2
and the answer to the question is.....(1) - (-3) = 4 > 0 so the answer is YES.
Fact 1 is INSUFFICIENT

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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 11 Apr 2015, 10:38
EMPOWERgmatRichC wrote:
HI King407,

TESTing VALUES is a great approach for this question, but the problem with your approach is that it does not factor in all of the possibilities that can occur. If you look at your work, you chose to make X and Y either BOTH positive OR both NEGATIVE, but you COULD have 1 positive and 1 negative....

We're told that X/Y < 1/2

IF...
X = 1
Y = -3
then -1/3 < 1/2
and the answer to the question is.....(1) - (-3) = 4 > 0 so the answer is YES.
Fact 1 is INSUFFICIENT

GMAT assassins aren't born, they're made,
Rich


Thanks it is clear now...



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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 08 Nov 2016, 05:01
Hi,

Can someone pls explain why stmt 2 by itself is not correct?

The statement deduces to x=1 and y=1, thus x-y=0.

Am I missing something?
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 08 Nov 2016, 05:33
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 09 Nov 2016, 00:55
Bunuel wrote:
believer700 wrote:
Hi,

Can someone pls explain why stmt 2 by itself is not correct?

The statement deduces to x=1 and y=1, thus x-y=0.

Am I missing something?


Any positive x and y satisfy the second statement, not only x = 1 and y = 1.


Many thanks!!

I messed up on the square part.
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?  [#permalink]

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New post 08 Oct 2018, 10:33
Bunuel wrote:
If x and y are integers and xy ≠ 0, is x - y > 0?

(1) x/y < 1/2

(2) \(\sqrt{x^2}= x\) and \(\sqrt{y^2} = y\)

\(\left( * \right)\,\,\,\,x,y\,\,\, \ne 0\,\,\,\,{\rm{ints}}\,\,\,\,\left( {xy \ne 0} \right)\,\,\,\,\)

\(x\,\,\mathop > \limits^? \,\,y\)

\(\left( 1 \right)\,\,{x \over y} < {1 \over 2}\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1, - 3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,\left\{ \matrix{
\,\left| x \right| = x\,\,\,\, \Rightarrow \,\,\,\,x \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,x > 0 \hfill \cr
\,\left| y \right| = y\,\,\,\, \Rightarrow \,\,\,\,y \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\)

\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
\,\,{x \over y} < {1 \over 2}\,\,\,\,\,\mathop \Rightarrow \limits_{y\, > \,\,0}^{ \cdot \,\,2y} \,\,\,\,\,2x < y \hfill \cr
\,\,\,1 < 2\,\,\,\,\,\mathop \Rightarrow \limits_{\,x\, > \,\,0}^{ \cdot \,\,x} \,\,\,\,\,\,\,x < 2x \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,x < 2x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\)



This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If x and y are integers and xy ≠ 0, is x - y > 0?   [#permalink] 08 Oct 2018, 10:33
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