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# M28-37

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Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135189 [0], given: 12671

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16 Sep 2014, 01:31
Expert's post
2
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

36% (00:37) correct 64% (00:46) wrong based on 42 sessions

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If $$x$$ is not equal to 0 and $$x^y=1$$, then which of the following must be true?
I. $$x=1$$

II. $$x=1$$ and $$y=0$$

III. $$x=1$$ or $$y=0$$

A. $$I$$ only
B. $$II$$ only
C. $$III$$ only
D. $$I$$ and $$III$$ only
E. $$None$$
[Reveal] Spoiler: OA

_________________

Kudos [?]: 135189 [0], given: 12671

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135189 [0], given: 12671

### Show Tags

16 Sep 2014, 01:31
Official Solution:

If $$x$$ is not equal to 0 and $$x^y=1$$, then which of the following must be true?
I. $$x=1$$

II. $$x=1$$ and $$y=0$$

III. $$x=1$$ or $$y=0$$

A. $$I$$ only
B. $$II$$ only
C. $$III$$ only
D. $$I$$ and $$III$$ only
E. $$None$$

Notice that if $$x=-1$$ and $$y$$ is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

_________________

Kudos [?]: 135189 [0], given: 12671

Intern
Joined: 02 Aug 2014
Posts: 4

Kudos [?]: [0], given: 14

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16 Jul 2015, 04:25
Bunuel wrote:
Official Solution:

If $$x$$ is not equal to 0 and $$x^y=1$$, then which of the following must be true?
I. $$x=1$$

II. $$x=1$$ and $$y=0$$

III. $$x=1$$ or $$y=0$$

A. $$I$$ only
B. $$II$$ only
C. $$III$$ only
D. $$I$$ and $$III$$ only
E. $$None$$

Notice that if $$x=-1$$ and $$y$$ is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Dear Bunuel,

Please explain why C is incorrect. Even that satisfies the must condition.

Kudos [?]: [0], given: 14

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135189 [0], given: 12671

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16 Jul 2015, 04:38
shanti47 wrote:
Bunuel wrote:
Official Solution:

If $$x$$ is not equal to 0 and $$x^y=1$$, then which of the following must be true?
I. $$x=1$$

II. $$x=1$$ and $$y=0$$

III. $$x=1$$ or $$y=0$$

A. $$I$$ only
B. $$II$$ only
C. $$III$$ only
D. $$I$$ and $$III$$ only
E. $$None$$

Notice that if $$x=-1$$ and $$y$$ is any even number, then $$(-1)^{even}=1$$, thus none of the options must be true.

Dear Bunuel,

Please explain why C is incorrect. Even that satisfies the must condition.

III says $$x=1$$ or $$y=0$$. This is NOT necessarily true: consider x = -1 and y = 2 --> (-1)^2 = 1.
_________________

Kudos [?]: 135189 [0], given: 12671

Current Student
Joined: 12 Aug 2015
Posts: 298

Kudos [?]: 592 [0], given: 1474

Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

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12 Feb 2016, 00:22
_________________

KUDO me plenty

Kudos [?]: 592 [0], given: 1474

Intern
Joined: 12 Nov 2016
Posts: 29

Kudos [?]: 23 [0], given: 27

Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
WE: Account Management (Accounting)

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18 Jun 2017, 03:33
Hi Bunuel, I didn't understand why B is incorrect. When x=1 and y=0, then the answer is always 1. Please clarify.

Kudos [?]: 23 [0], given: 27

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135189 [0], given: 12671

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18 Jun 2017, 03:37
Romannepal wrote:
Hi Bunuel, I didn't understand why B is incorrect. When x=1 and y=0, then the answer is always 1. Please clarify.

The question asks which of the following MUST be true? Not COULD be true.

II says: $$x=1$$ and $$y=0$$.

This is not necessarily true. Consider x = -1 and y = 0.
_________________

Kudos [?]: 135189 [0], given: 12671

Intern
Joined: 30 Nov 2013
Posts: 7

Kudos [?]: 25 [0], given: 7

Location: India
GMAT 1: 770 Q51 V42
GPA: 3.4

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05 Oct 2017, 07:58
Hi Bunuel, why is C incorrect. It says X=1 or Y=0. Now if X=1, x^y will always be 1 irrespective of Y and if Y = 0, then x^y will be 1 as well since X cannot be 0. Can you please explain.

Kudos [?]: 25 [0], given: 7

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135189 [0], given: 12671

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05 Oct 2017, 08:00
mansi20081990 wrote:
Hi Bunuel, why is C incorrect. It says X=1 or Y=0. Now if X=1, x^y will always be 1 irrespective of Y and if Y = 0, then x^y will be 1 as well since X cannot be 0. Can you please explain.

This is explained above:

III says $$x=1$$ or $$y=0$$. This is NOT necessarily true: consider x = -1 and y = 2 --> (-1)^2 = 1.
_________________

Kudos [?]: 135189 [0], given: 12671

Re: M28-37   [#permalink] 05 Oct 2017, 08:00
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# M28-37

Moderators: chetan2u, Bunuel

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