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# M20-30

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Math Expert
Joined: 02 Sep 2009
Posts: 43292

Kudos [?]: 139153 [0], given: 12777

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16 Sep 2014, 00:09
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Difficulty:

35% (medium)

Question Stats:

70% (01:06) correct 30% (01:09) wrong based on 105 sessions

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Is $$x \lt y$$ ?

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

(2) $$(x + y)(x - y) \lt 0$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139153 [0], given: 12777

Math Expert
Joined: 02 Sep 2009
Posts: 43292

Kudos [?]: 139153 [0], given: 12777

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16 Sep 2014, 00:09
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.

_________________

Kudos [?]: 139153 [0], given: 12777

Intern
Joined: 09 Sep 2015
Posts: 5

Kudos [?]: [0], given: 0

GMAT 1: 630 Q46 V31
GPA: 3.53
WE: Education (Education)

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25 Jul 2016, 09:42
I think this is a high-quality question and I agree with explanation.

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Intern
Joined: 12 Jan 2013
Posts: 2

Kudos [?]: 5 [0], given: 12

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11 May 2017, 11:36
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.

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

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Intern
Joined: 22 May 2017
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08 Jun 2017, 09:03
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.

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Math Expert
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Kudos [?]: 139153 [0], given: 12777

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08 Jun 2017, 14:13
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
_________________

Kudos [?]: 139153 [0], given: 12777

Intern
Joined: 17 Apr 2015
Posts: 13

Kudos [?]: 1 [0], given: 7

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23 Aug 2017, 12:17
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.

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

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Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 139153 [0], given: 12777

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23 Aug 2017, 12:35
Expert's post
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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.

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.
_________________

Kudos [?]: 139153 [0], given: 12777

Senior Manager
Joined: 08 Jun 2015
Posts: 365

Kudos [?]: 27 [0], given: 106

Location: India
GMAT 1: 640 Q48 V29

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24 Oct 2017, 01:51
+1 for A.
_________________

" The few , the fearless "

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Intern
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05 Dec 2017, 20:34
Hi brunel,
Cannt we take x=1 and y=1 in that case statement 1 fails??

Thanks
Abhi

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Math Expert
Joined: 02 Sep 2009
Posts: 43292

Kudos [?]: 139153 [0], given: 12777

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05 Dec 2017, 20:43
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 ! ! !
_________________

Kudos [?]: 139153 [0], given: 12777

Intern
Joined: 15 Aug 2013
Posts: 28

Kudos [?]: [0], given: 167

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06 Dec 2017, 10:46
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 ! ! !

Thank you Bunuel

Kudos [?]: [0], given: 167

Re: M20-30   [#permalink] 06 Dec 2017, 10:46
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# M20-30

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