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# M28-37

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Math Expert
Joined: 02 Sep 2009
Posts: 55266

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16 Sep 2014, 01:31
1
9
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Difficulty:

75% (hard)

Question Stats:

30% (00:35) correct 70% (00:41) wrong based on 244 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. $$None$$

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Math Expert
Joined: 02 Sep 2009
Posts: 55266

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16 Sep 2014, 01:31
1
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.

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Intern
Joined: 02 Aug 2014
Posts: 4

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 55266

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

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12 Feb 2016, 00:22
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KUDO me plenty
Intern
Joined: 12 Nov 2016
Posts: 29
Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
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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.
Math Expert
Joined: 02 Sep 2009
Posts: 55266

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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.
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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.
Math Expert
Joined: 02 Sep 2009
Posts: 55266

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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.
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Joined: 31 Mar 2018
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28 Nov 2018, 06:34
Hi,
Can someone explain why A isn't correct. When X=1, then 1 to the power of any except 0 is 1 right?
Math Expert
Joined: 02 Sep 2009
Posts: 55266

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28 Nov 2018, 06:40
swathi27 wrote:
Hi,
Can someone explain why A isn't correct. When X=1, then 1 to the power of any except 0 is 1 right?

The question asks: which of the following must be true? NOT which of the following could be true?

I says $$x=1$$. This is NOT necessarily true: consider x = -1 and y = 2 --> (-1)^2 = 1.
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Intern
Joined: 01 Jan 2016
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28 Nov 2018, 08:20
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.

In the question, its already mentioned that x=1, so why are we taking x=-1 as other possibility ?
Math Expert
Joined: 02 Sep 2009
Posts: 55266

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28 Nov 2018, 08:22
bhateja 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.

In the question, its already mentioned that x=1, so why are we taking x=-1 as other possibility ?

The stem says that $$x$$ is not equal to 0. Where does it say that x = 1?
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28 Nov 2018, 10:38
Bunuel, why we are considering x=-1 when its nowhere mentioned in options.
Intern
Joined: 24 Apr 2016
Posts: 30

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28 Nov 2018, 18:25
dine5207 wrote:
Bunuel, why we are considering x=-1 when its nowhere mentioned in options.

I'm not Bunuel but going to answer this one.

The question asks "which of the following MUST be true?"
In order to show that the answer is "E. None", we must present a solution in which we can have $$x,y$$ such that $$x^y = 1$$,and $$x\neq 0$$ and all of I, II, and III are all not true.

The example Bunuel gives here is $$x=-1$$ and $$y={\text{any even number}}$$. Suppose $$y = 2$$. Then $$x^y = (-1)^2 = 1$$ and yet all of I, II, and III are all not true.
Intern
Joined: 24 Apr 2016
Posts: 30

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28 Nov 2018, 18:27
Bunuel wrote:
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. $$None$$

Could someone confirm that the set specified by III also includes where $$x=1 \text{ AND } y = 0$$?
Intern
Joined: 17 Oct 2018
Posts: 1

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18 Dec 2018, 16:30
I think this is a poor-quality question and I don't agree with the explanation. the statements that are given clearly say that x=1 and not x<1 or x>1. I don't understand why are we considering x=-1 when each statement clearly gives a value to x and/or y and not a range of values
Math Expert
Joined: 02 Sep 2009
Posts: 55266

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18 Dec 2018, 22:11
FrankMartin wrote:
I think this is a poor-quality question and I don't agree with the explanation. the statements that are given clearly say that x=1 and not x<1 or x>1. I don't understand why are we considering x=-1 when each statement clearly gives a value to x and/or y and not a range of values

I think you did not understand the question.

The stem says: $$x$$ is not equal to 0 and $$x^y=1$$.

Then we are given three options, and are asked to determine which of them MUST be true.

The solution shows that none of the options MUST be true (none of the options is NECESSARILY true).

The questions is 100% correct.
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Re: M28-37   [#permalink] 18 Dec 2018, 22:11
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# M28-37

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