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Math Expert V
Joined: 02 Sep 2009
Posts: 55266
M20-30  [#permalink]

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1
2 00:00

Difficulty:   45% (medium)

Question Stats: 66% (01:12) correct 34% (01:12) wrong based on 188 sessions

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Is $$x \lt y$$ ?

(1) $$x^3 \lt y^3$$

(2) $$(x + y)(x - y) \lt 0$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re M20-30  [#permalink]

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Official Solution:

Is $$x \lt y$$ ?

(1) $$x^3 \lt y^3$$. Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from $$x^3 \lt y^3$$ we'll get $$x \lt y$$. Sufficient.

(2) $$(x+y)(x-y) \lt 0$$. This statement tells that $$x^2-y^2 \lt 0$$ or $$x^2 \lt y^2$$, which is not sufficient to answer whether $$x \lt y$$, consider $$x=1$$ and $$y=2$$ for an YES answer and $$x=1$$ and $$y=-2$$ for a NO answer.

Answer: A
_________________
Intern  Joined: 09 Sep 2015
Posts: 5
Location: Grenada
GMAT 1: 630 Q46 V31 GPA: 3.53
WE: Education (Education)
Re: M20-30  [#permalink]

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I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 12 Jan 2013
Posts: 2
Re: M20-30  [#permalink]

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Bunuel wrote:
Official Solution:

(1) $$x^3 \lt y^3$$. Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from $$x^3 \lt y^3$$ we'll get $$x \lt y$$. Sufficient.

(2) $$(x+y)(x-y) \lt 0$$. This statement tells that $$x^2-y^2 \lt 0$$ or $$x^2 \lt y^2$$, which is not sufficient to answer whether $$x \lt y$$, consider $$x=1$$ and $$y=2$$ for an YES answer and $$x=1$$ and $$y=-2$$ for a NO answer.

Answer: A

Hi, why can we not consider that either (x+y)<0 or (x-y)<0? If we do this then we get X<-Y and X<Y. Thx
Intern  B
Joined: 22 May 2017
Posts: 4
Re: M20-30  [#permalink]

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Hi,

Say x = -1/2 and y = -1/3 then this solution of your don't work.

In such case x to the power of 3 is greater than y to the power of 3, but x < y is false.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M20-30  [#permalink]

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omtraders wrote:
Hi,

Say x = -1/2 and y = -1/3 then this solution of your don't work.

In such case x to the power of 3 is greater than y to the power of 3, but x < y is false.

Did you try to use a calculator?

(-1/2)^3 = -1/8 < (-1/3)^3 = -1/27
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Intern  B
Joined: 18 Apr 2015
Posts: 16
Re: M20-30  [#permalink]

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Bunuel wrote:
Official Solution:

(1) $$x^3 \lt y^3$$. Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from $$x^3 \lt y^3$$ we'll get $$x \lt y$$. Sufficient.

(2) $$(x+y)(x-y) \lt 0$$. This statement tells that $$x^2-y^2 \lt 0$$ or $$x^2 \lt y^2$$, which is not sufficient to answer whether $$x \lt y$$, consider $$x=1$$ and $$y=2$$ for an YES answer and $$x=1$$ and $$y=-2$$ for a NO answer.

Answer: A

Hi Bunuel,

For 2, why can't I write -y<x<y ? Is it because we don't really know the sign of y?

Regards
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M20-30  [#permalink]

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sliceoflife wrote:
Bunuel wrote:
Official Solution:

(1) $$x^3 \lt y^3$$. Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from $$x^3 \lt y^3$$ we'll get $$x \lt y$$. Sufficient.

(2) $$(x+y)(x-y) \lt 0$$. This statement tells that $$x^2-y^2 \lt 0$$ or $$x^2 \lt y^2$$, which is not sufficient to answer whether $$x \lt y$$, consider $$x=1$$ and $$y=2$$ for an YES answer and $$x=1$$ and $$y=-2$$ for a NO answer.

Answer: A

Hi Bunuel,

For 2, why can't I write -y<x<y ? Is it because we don't really know the sign of y?

Regards

$$x^2 \lt y^2$$ means that |x| < |y|, so y is further from 0 than x is. We can have the following cases:

-----------0---x---y---
-------x---0-------y---
---y---x---0-----------
---y-------0---x------

As you can see, for the first two cases x < y and for the remaining two cases x > y.
_________________
Senior Manager  S
Joined: 08 Jun 2015
Posts: 424
Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33
Re: M20-30  [#permalink]

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+1 for A.
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" The few , the fearless "
Intern  B
Joined: 15 Aug 2013
Posts: 41
Re: M20-30  [#permalink]

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Hi brunel,
Cannt we take x=1 and y=1 in that case statement 1 fails??

Thanks
Abhi
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M20-30  [#permalink]

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1
1
abhinashgc wrote:
Hi brunel,
Cannt we take x=1 and y=1 in that case statement 1 fails??

Thanks
Abhi

It's Bunuel, not brunel.

When testing numbers for DS, we should choose so that these numbers satisfy the statement we are examining. Does x = y = 1 satisfy x^3 < y^3?

3. Strategies and Tactics for DS Section

For more check below:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
_________________
Intern  B
Joined: 15 Aug 2013
Posts: 41
Re: M20-30  [#permalink]

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Bunuel wrote:
abhinashgc wrote:
Hi brunel,
Cannt we take x=1 and y=1 in that case statement 1 fails??

Thanks
Abhi

It's Bunuel, not brunel.

When testing numbers for DS, we should choose so that these numbers satisfy the statement we are examining. Does x = y = 1 satisfy x^3 < y^3?

3. Strategies and Tactics for DS Section

For more check below:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Thank you Bunuel Manager  B
Joined: 26 Feb 2018
Posts: 77
Location: United Arab Emirates
GMAT 1: 710 Q47 V41 GMAT 2: 770 Q49 V47 Re: M20-30  [#permalink]

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another way to think of it is that statement 2 tells us that the absolute value of x is more than the absolute value of y. This means that x could be more or less than y, so (II) is insufficient. We don't need to test any cases for (II)

Good question but too easy to be a Q50/51 question. More like Q45-48. Would classify as medium
Intern  B
Joined: 19 Nov 2012
Posts: 28
Re: M20-30  [#permalink]

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Bunuel wrote:
sliceoflife wrote:
Bunuel wrote:
Official Solution:

(1) $$x^3 \lt y^3$$. Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from $$x^3 \lt y^3$$ we'll get $$x \lt y$$. Sufficient.

(2) $$(x+y)(x-y) \lt 0$$. This statement tells that $$x^2-y^2 \lt 0$$ or $$x^2 \lt y^2$$, which is not sufficient to answer whether $$x \lt y$$, consider $$x=1$$ and $$y=2$$ for an YES answer and $$x=1$$ and $$y=-2$$ for a NO answer.

Answer: A

Hi Bunuel,

For 2, why can't I write -y<x<y ? Is it because we don't really know the sign of y?

Regards

$$x^2 \lt y^2$$ means that |x| < |y|, so y is further from 0 than x is. We can have the following cases:

-----------0---x---y---
-------x---0-------y---
---y---x---0-----------
---y-------0---x------

As you can see, for the first two cases x < y and for the remaining two cases x > y.

Hi Bunuel, chetan2u,

Bunuel: I read your explanation but still have one doubt. x^2 - y^2 < 0 is quadratic and we do use the wavy line method to solve quadratic. With great difficulty I understood and use the wavy line method for many quadratic inequality questions, so now, why can't this quadratic expression be solved using wavy line method ? Just to explain, wavy line method was explained on gmatclub posts itself. Basically, in that method, we factorize this expression to get zero points such as -y and y and choose that range where the value is negative ( since x^2 - y^2 < 0) which in this case indeed works out to be -y < x < y. so how is it not correct? OR is there way forward to get the ans using this wavy line method?

Thanks a lot. Re: M20-30   [#permalink] 25 Nov 2018, 04:22
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# M20-30

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