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M19-36

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M19-36 [#permalink]

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New post 16 Sep 2014, 01:07
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" (\([0] = [0]\)); if \(x\) is positive, the answer is "yes".


Answer: E
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Re: M19-36 [#permalink]

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New post 14 Mar 2015, 05:08
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" (\([0] = [0]\)); 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.

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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" (\([0] = [0]\)); 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 [3] = 2 (the largest integer smaller than 3), so 3 = 2 + 1.
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Re: M19-36 [#permalink]

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New post 14 Mar 2015, 06:43
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" (\([0] = [0]\)); 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 [3] = 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".

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Re: M19-36 [#permalink]

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New post 01 Nov 2015, 00:42
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 ([0] = [0]).. it is not as per the context of question which says (largest integer smaller than x)

Am I right? let me know.. thanks

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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 ([0] = [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 [0] = -1. Since -0 = 0, then [-0] = -1 too.
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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?

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Re: M19-36 [#permalink]

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New post 20 Aug 2017, 19:19
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
[-6.4]=-7

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New post 21 Aug 2017, 03:34
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
[-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 [6]=5, not 6 because the largest integer smaller than 6 is 5.
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Re: M19-36 [#permalink]

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New post 21 Aug 2017, 04:58
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]=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.

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New post 21 Aug 2017, 05:04
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]=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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New post 25 Sep 2017, 05:45
X +1 > 0
-X
CROSS MULTIPLY
= - X × 0 > X + 1
= 0 > × + 1

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Re: M19-36 [#permalink]

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New post 30 Sep 2017, 02:17
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?

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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.
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Re: M19-36 [#permalink]

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New post 01 Oct 2017, 09:48
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.

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Re: M19-36 [#permalink]

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New post 01 Oct 2017, 10:24
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.

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Re: M19-36 [#permalink]

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New post 02 Oct 2017, 00:41
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" (\([0] = [0]\)); if \(x\) is positive, the answer is "yes".


Answer: E


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

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Re: M19-36 [#permalink]

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New post 02 Oct 2017, 00:45
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" (\([0] = [0]\)); 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
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Re: M19-36   [#permalink] 02 Oct 2017, 00:45
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