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Math Expert V
Joined: 02 Sep 2009
Posts: 55266
M19-36  [#permalink]

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16 00:00

Difficulty:   95% (hard)

Question Stats: 37% (01:45) correct 63% (01:44) wrong based on 249 sessions

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If $$[x]$$ denotes the largest integer smaller than $$x$$, is $$[x] \gt [-x]$$?

(1) $$x = [x] + 1$$

(2) $$x + 1 \gt 0$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re M19-36  [#permalink]

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Official Solution:

Statement (1) by itself is insufficient. S1 only tells that $$x$$ is an integer.

Statement (2) by itself is insufficient.

Statements (1) and (2) combined are insufficient. Adding S2 to S1 will give that $$x$$ is an integer bigger than -1. However, if $$x$$ is 0, the answer is "no" ($$ = $$); if $$x$$ is positive, the answer is "yes".

Answer: E
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Intern  Joined: 11 Nov 2014
Posts: 3
GPA: 3.3
WE: Information Technology (Computer Software)
Re: M19-36  [#permalink]

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Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 only tells that $$x$$ is an integer.

Statement (2) by itself is insufficient.

Statements (1) and (2) combined are insufficient. Adding S2 to S1 will give that $$x$$ is an integer bigger than -1. However, if $$x$$ is 0, the answer is "no" ($$ = $$); if $$x$$ is positive, the answer is "yes".

Answer: E

Hi Bunuel,

I think 1st statement is wrong, How can x= [x]+1 ?
because x = [x] + {x}, Where {x} is fractional part, which is always less than 1.
so x = [x] +1 can't be true in any circumstances.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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2
2
prasadk wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 only tells that $$x$$ is an integer.

Statement (2) by itself is insufficient.

Statements (1) and (2) combined are insufficient. Adding S2 to S1 will give that $$x$$ is an integer bigger than -1. However, if $$x$$ is 0, the answer is "no" ($$ = $$); if $$x$$ is positive, the answer is "yes".

Answer: E

Hi Bunuel,

I think 1st statement is wrong, How can x= [x]+1 ?
because x = [x] + {x}, Where {x} is fractional part, which is always less than 1.
so x = [x] +1 can't be true in any circumstances.

x = [x] + 1 is true for all integers. For example, if x = 3, then  = 2 (the largest integer smaller than 3), so 3 = 2 + 1.
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Intern  Joined: 11 Nov 2014
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Re: M19-36  [#permalink]

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Bunuel wrote:
prasadk wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 only tells that $$x$$ is an integer.

Statement (2) by itself is insufficient.

Statements (1) and (2) combined are insufficient. Adding S2 to S1 will give that $$x$$ is an integer bigger than -1. However, if $$x$$ is 0, the answer is "no" ($$ = $$); if $$x$$ is positive, the answer is "yes".

Answer: E

Hi Bunuel,

I think 1st statement is wrong, How can x= [x]+1 ?
because x = [x] + {x}, Where {x} is fractional part, which is always less than 1.
so x = [x] +1 can't be true in any circumstances.

x = [x] + 1 is true for all integers. For example, if x = 3, then  = 2 (the largest integer smaller than 3), so 3 = 2 + 1.

Oh Sorry, My bad.

It's "smaller than" , I took it for greatest integer function.
You removed "equal to".
Intern  Joined: 07 Dec 2014
Posts: 9
Re: M19-36  [#permalink]

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Hi Bunuel,

I have only one doubt.. when we combined two statements.. (1) & (2)..

We understood that "x is an integer" and "x>-1"

So my Q is when x = 0 then [x] = -1; [-x] = 1; Hence [x]<[-x] and for all other cases [x]>[-x]..

Hence insufficient to draw a conclusion, so option E is correct.

But as per you when x is 0 ( = ).. it is not as per the context of question which says (largest integer smaller than x)

Am I right? let me know.. thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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arunmandapaka wrote:
Hi Bunuel,

I have only one doubt.. when we combined two statements.. (1) & (2)..

We understood that "x is an integer" and "x>-1"

So my Q is when x = 0 then [x] = -1; [-x] = 1; Hence [x]<[-x] and for all other cases [x]>[-x]..

Hence insufficient to draw a conclusion, so option E is correct.

But as per you when x is 0 ( = ).. it is not as per the context of question which says (largest integer smaller than x)

Am I right? let me know.. thanks

If x = 0, then  = -1. Since -0 = 0, then [-0] = -1 too.
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Intern  B
Joined: 08 Apr 2014
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M19-36  [#permalink]

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can the question stem be further solved into asking if "x >2?"

if [x] > [-x], (x-1) > -(x-1), x > 2?
Intern  B
Joined: 04 May 2014
Posts: 3
GMAT 1: 720 Q50 V37 Re: M19-36  [#permalink]

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Hi Bunuel,

With all due respect, I want to point out that there is something wrong with the question and the explanation you have provided to one of the fellow students.

The GIF value of an integer I is the integer I itself and not I-1, something that you have stated and the explanation implies.
It can be understood by the definition and graph plot of the GIF function.
[x] : the greatest integer less than or equal to x. Equal when I is an integer of course.
[6.4]=6
=6
[-6.4]=-7
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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siddharthasthana2212 wrote:
Hi Bunuel,

With all due respect, I want to point out that there is something wrong with the question and the explanation you have provided to one of the fellow students.

The GIF value of an integer I is the integer I itself and not I-1, something that you have stated and the explanation implies.
It can be understood by the definition and graph plot of the GIF function.
[x] : the greatest integer less than or equal to x. Equal when I is an integer of course.
[6.4]=6
=6
[-6.4]=-7

Not sure I understand what you mean but the function is defined as "the largest integer smaller than..." not "smaller than or equal to...". There are many other functions, this one is as it is.

So, for example, [6.4]=6 and [-6.4]=-7 but =5, not 6 because the largest integer smaller than 6 is 5.
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GMAT 1: 720 Q50 V37 Re: M19-36  [#permalink]

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Hi Bunuel,

This is exactly what I want to point out. The GIF is indeed "the greatest integer less than or EQUAL to x."
=6 and not 5.
I am unable to copy the link for some reason, but you can check the definition and graph of the function anywhere.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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siddharthasthana2212 wrote:
Hi Bunuel,

This is exactly what I want to point out. The GIF is indeed "the greatest integer less than or EQUAL to x."
=6 and not 5.
I am unable to copy the link for some reason, but you can check the definition and graph of the function anywhere.

It's not that function. It's a different function using same concept and notation.
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Intern  Joined: 25 Sep 2017
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Re: M19-36  [#permalink]

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X +1 > 0
-X
CROSS MULTIPLY
= - X × 0 > X + 1
= 0 > × + 1
Manager  B
Joined: 04 Jul 2017
Posts: 59
Location: India
Concentration: Marketing, General Management
GPA: 1
WE: Analyst (Consulting)
Re: M19-36  [#permalink]

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Bunuel wrote:
If $$[x]$$ denotes the largest integer smaller than $$x$$, is $$[x] \gt [-x]$$?

(1) $$x = [x] + 1$$

(2) $$x + 1 \gt 0$$

This question is part of which topic?
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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2
aashishagarwal2 wrote:
Bunuel wrote:
If $$[x]$$ denotes the largest integer smaller than $$x$$, is $$[x] \gt [-x]$$?

(1) $$x = [x] + 1$$

(2) $$x + 1 \gt 0$$

This question is part of which topic?

This is a rounding functions question.

Check other Rounding Functions Questions in our Special Questions Directory.
_________________
Manager  B
Joined: 04 Jul 2017
Posts: 59
Location: India
Concentration: Marketing, General Management
GPA: 1
WE: Analyst (Consulting)
Re: M19-36  [#permalink]

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Bunuel wrote:
aashishagarwal2 wrote:
Bunuel wrote:
If $$[x]$$ denotes the largest integer smaller than $$x$$, is $$[x] \gt [-x]$$?

(1) $$x = [x] + 1$$

(2) $$x + 1 \gt 0$$

This question is part of which topic?

This is a rounding functions question.

Check other Rounding Functions Questions in our Special Questions Directory.

What these types "[x]" mean? Whenever such variables inside square brackets appear, I become clueless on to handle them or what they mean.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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aashishagarwal2 wrote:

What these types "[x]" mean? Whenever such variables inside square brackets appear, I become clueless on to handle them or what they mean.

I think practising the questions from that link should help.
_________________
Manager  B
Joined: 04 Jul 2017
Posts: 59
Location: India
Concentration: Marketing, General Management
GPA: 1
WE: Analyst (Consulting)
Re: M19-36  [#permalink]

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Bunuel wrote:
aashishagarwal2 wrote:

What these types "[x]" mean? Whenever such variables inside square brackets appear, I become clueless on to handle them or what they mean.

I think practising the questions from that link should help.

Thanks Bunuel.
SVP  V
Joined: 26 Mar 2013
Posts: 2176
Re: M19-36  [#permalink]

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Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 only tells that $$x$$ is an integer.

Statement (2) by itself is insufficient.

Statements (1) and (2) combined are insufficient. Adding S2 to S1 will give that $$x$$ is an integer bigger than -1. However, if $$x$$ is 0, the answer is "no" ($$ = $$); if $$x$$ is positive, the answer is "yes".

Answer: E

Can you elaborate the highlighted part pls? it is not clear enough.
Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M19-36  [#permalink]

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Mo2men wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. S1 only tells that $$x$$ is an integer.

Statement (2) by itself is insufficient.

Statements (1) and (2) combined are insufficient. Adding S2 to S1 will give that $$x$$ is an integer bigger than -1. However, if $$x$$ is 0, the answer is "no" ($$ = $$); if $$x$$ is positive, the answer is "yes".

Answer: E

Can you elaborate the highlighted part pls? it is not clear enough.
Thanks

Check here: https://gmatclub.com/forum/if-denotes-t ... 01744.html
_________________ Re: M19-36   [#permalink] 02 Oct 2017, 00:45

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