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Is |xy| > xy?

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Is |xy| > xy?  [#permalink]

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New post 13 Sep 2017, 21:48
2
6
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (01:29) correct 43% (01:35) wrong based on 238 sessions

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Is |xy| > xy?  [#permalink]

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New post 13 Sep 2017, 22:17
4
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Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


If both \(x\) and \(y\) are positive or negative then \(|xy|=xy\) and if either of the two is negative and other is positive then \(|xy|>xy\)

Statement 1: LHS \(= \sqrt{(xy)^2}\) \(= |xy|\)
so we have \(|xy|=xy\). Hence we have a NO for our question stem. Sufficient

Statement 2: this implies magnitude of both \(x\) & \(y\) equals summation of both \(x\) & \(y\). Hence both \(x\) & \(y\) are positive. Hence \(|xy|=xy\)
Hence we have a NO for our question stem. Sufficient

Option \(D\)
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Re: Is |xy| > xy?  [#permalink]

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New post 26 Sep 2017, 22:14
niks18 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


If both \(x\) and \(y\) are positive or negative then \(|xy|=xy\) and if either of the two is negative and other is positive then \(|xy|>xy\)

Statement 1: LHS \(= \sqrt{(xy)^2}\) \(= |xy|\)
so we have \(|xy|=xy\). Hence we have a NO for our question stem. Sufficient

Statement 2: this implies magnitude of both \(x\) & \(y\) equals summation of both \(x\) & \(y\). Hence both \(x\) & \(y\) are positive. Hence \(|xy|=xy\)
Hence we have a NO for our question stem. Sufficient

Option \(D\)



if either of the two is negative and other is positive then \(|xy|>xy\) - Can you explain this further?
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Re: Is |xy| > xy?  [#permalink]

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New post 26 Sep 2017, 22:18
1
Manveetha wrote:
niks18 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


If both \(x\) and \(y\) are positive or negative then \(|xy|=xy\) and if either of the two is negative and other is positive then \(|xy|>xy\)

Statement 1: LHS \(= \sqrt{(xy)^2}\) \(= |xy|\)
so we have \(|xy|=xy\). Hence we have a NO for our question stem. Sufficient

Statement 2: this implies magnitude of both \(x\) & \(y\) equals summation of both \(x\) & \(y\). Hence both \(x\) & \(y\) are positive. Hence \(|xy|=xy\)
Hence we have a NO for our question stem. Sufficient

Option \(D\)



if either of the two is negative and other is positive then \(|xy|>xy\) - Can you explain this further?


Hi Manveetha

Take few examples, lets assume \(x=-2\) and \(y =2\), then \(xy=-2*2=-4\)

but \(|xy|=|-4|=|4|\)

hence \(|xy|>xy\) in this case
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Re: Is |xy| > xy?  [#permalink]

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New post 27 Sep 2017, 21:54
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Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


Bunuel, I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?
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Is |xy| > xy?  [#permalink]

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New post 27 Sep 2017, 22:26
We have to just figure out whether the signs of x and y are same or different.
If they have the same signs, then the two sides would be equal. If one of the two values or both are zero, then the two sides are equal and not less than or greater than.

Both the statements answer this question: The signs are same.
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Re: Is |xy| > xy?  [#permalink]

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New post 27 Sep 2017, 23:09
RMD007 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


Bunuel, I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?


Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.
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Re: Is |xy| > xy?  [#permalink]

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New post 01 Oct 2017, 13:59
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)



Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

The first step of VA (Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.

The question \(|xy| > xy\) is equivalent to \(xy < 0\).

Condition 1)
\(|xy| = xy\) is equivalent ot \(xy \ge 0\).
Since the answer is no, this condition is sufficient by CMT(Common Mistake Type) 1.

Condition 2)
\(|x| + |y| = x + y\)
=> \((|x| + |y|)^2 = (x + y)^2\)
<=> \(|x|^2 + 2|x||y| + |y|^2 = x^2 + 2xy + y^2\)
<=> \(x^2 + 2|x||y| + y^2 = x^2 + 2xy + y^2\)
<=> \(2|x||y| = 2xy\)
<=> \(|xy| = xy\)
<=> \(xy \ge 0\)
Then the answer is no and "no" is also an answer.
By CMT 1, this condition is also sufficient.

Therefore, D is the answer.
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Is |xy| > xy?  [#permalink]

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New post 06 Apr 2018, 09:49
Bunuel wrote:
RMD007 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


Bunuel, I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?


Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.


Hi Bunuel.

I understood Statement-2 but confused with statement-1, my logic is as below

From question we can figure out that if xy>0, then lXYl= XY ; so XY>XY so we have to neglect this option
if xy<0, then l XY l = -(xy), So question becomes Is -(xy)>xy, this is what we have to look out from the question set.

So, -(xy)>xy is possible when we have different signs of x and Y (one + and another -); then we can definitely say "Yes" to the question Is -(xy)>xy
If X and Y have same signs then -(xy)<xy ; then a definite "NO" (to our question which is -(xy)>xy)
now in statement-1

X and Y both can be positive/negative; or X negative and Y positive (or vice versa) as the square will negate the effect of positivity/negativity of X/Y
So we can't be sure from statement 1 whether X/Y have same or different signs. Hence insufficient

I hope you understood my thinking and plz help to correct where me going wrong with my understanding
Thanks
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Re: Is |xy| > xy?  [#permalink]

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New post 06 Apr 2018, 12:22
cruiseav wrote:
Bunuel wrote:
Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.


Hi Bunuel.

I understood Statement-2 but confused with statement-1, my logic is as below

From question we can figure out that if xy>0, then lXYl= XY ; so XY>XY so we have to neglect this option
if xy<0, then l XY l = -(xy), So question becomes Is -(xy)>xy, this is what we have to look out from the question set.

So, -(xy)>xy is possible when we have different signs of x and Y (one + and another -); then we can definitely say "Yes" to the question Is -(xy)>xy
If X and Y have same signs then -(xy)<xy ; then a definite "NO" (to our question which is -(xy)>xy)
now in statement-1

X and Y both can be positive/negative; or X negative and Y positive (or vice versa) as the square will negate the effect of positivity/negativity of X/Y
So we can't be sure from statement 1 whether X/Y have same or different signs. Hence insufficient

I hope you understood my thinking and plz help to correct where me going wrong with my understanding
Thanks


For \(\sqrt{x^2y^2}=xy\) to be true x and y cannot have different signs because if they do, then the right hand side would become negative: \(xy=negative\) and it cannot equal to the square root of any number (\(\sqrt{x^2y^2}\)), which is always positive or 0.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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: Is |xy| > xy? &nbs [#permalink] 06 Apr 2018, 12:22
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