Data Sufficiency

Let [[x]] represent the average of the greatest integer less than or equal to x and the least integer greater than or equal to x. Is 0 <= |a| <= 1 ?

1) [[x]] - x = a
2) 0 < [[a]] < 1

OA: D (*Source: Knewton GMAT CAT Test #3)
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The explanation provided by Knewton is really complicated and long since it's testing cases using PIN - I'm wondering if there's a simpler and faster way to solve this.

During the test, I was already lost at the question stem. *discouraged* Help is deeply appreciated!

i_have_no_cool_username wrote:Let [[x]] represent the average of the greatest integer less than or equal to x and the least integer greater than or equal to x. Is 0 <= |a| <= 1 ?

1) [[x]] - x = a
2) 0 < [[a]] < 1

OA: D (*Source: Knewton GMAT CAT Test #3)
-------------------------------------------------------------------------------------------

The explanation provided by Knewton is really complicated and long since it's testing cases using PIN - I'm wondering if there's a simpler and faster way to solve this.

During the test, I was already lost at the question stem. *discouraged* Help is deeply appreciated!

Hi,
This is a good one.

Let's try to understand the meaning of [[x]].
If x = 4.1, [[x]] = (4 + 5)/2 = 4.5
If x = 3.9, [[x]] = (3 + 4)/2 = 3.5

So, if x = Integer + d (where 'I' is its decimal part whereas 'd' is its decimal part),
then [[x]] = I + 0.5

Statement(1):
[[x]] - x = (I + 0.5) - (I + d) = 0.5 - d
|0.5 - d| will always be between 0 and 1, since d is between 0 and 0.999...
SUFFICIENT.

Statement(2):
0< [[a]] < 1
0 < I + d < 1
Since the decimal value is always between 0 and 1, the integral value of a = 0
So,
0 < a < 1
or
0 <= |a| <= 1
SUFFICIENT.

[spoiler](D)[/spoiler] is correct.

aneesh.kg wrote:
Statement(2):
0< [[a]] < 1
0 < I + d < 1
Since the decimal value is always between 0 and 1, the integral value of a = 0
So,
0 < a < 1
or
0 <= |a| <= 1
SUFFICIENT.

Hi Aneesh! Thanks for the explanation - I think I understand the prompt and Statement 1 better now. It seems like the important part is to spot a pattern for x and [[x]] after testing some numbers... Also did you mean that "I" is the integer part (not the decimal part) and "d" is the decimal part? If so, then it makes sense to me.

I still don't really understand Statement 2 though. Since the statement states 'a' in the inequality, are you using the same rules stated in the prompt for x to evaluate 'a' in statement 2? If so, 0 < d + i < 1, and if 0 < i < 0.999, then like you said d=a=0.
In this case, a IS equal to 0, not between 0 and 1...can you explain a little bit more on how you got from a=0 to 0 < a < 1, please?

i_have_no_cool_username wrote:

aneesh.kg wrote:
Statement(2):
0< [[a]] < 1
0 < I + d < 1
Since the decimal value is always between 0 and 1, the integral value of a = 0
So,
0 < a < 1
or
0 <= |a| <= 1
SUFFICIENT.

Hi Aneesh! Thanks for the explanation - I think I understand the prompt and Statement 1 better now. It seems like the important part is to spot a pattern for x and [[x]] after testing some numbers... Also did you mean that "I" is the integer part (not the decimal part) and "d" is the decimal part? If so, then it makes sense to me.

I still don't really understand Statement 2 though. Since the statement states 'a' in the inequality, are you using the same rules stated in the prompt for x to evaluate 'a' in statement 2? If so, 0 < d + i < 1, and if 0 < i < 0.999, then like you said d=a=0.
In this case, a IS equal to 0, not between 0 and 1...can you explain a little bit more on how you got from a=0 to 0 < a < 1, please?

Yes, sorry for the typo. 'I' is the integral part.

I'm using the same prompt for Statement(2) but I am thinking a little harder along with using the prompt.
0 < [[a]] < 1
0 < [[I + d]] < 1
0 < I + 0.5 < 1
If, in the above inequality I is anything except 0, the inequality will get violated. For example, I = -1 means that I + 0.5 = - 0.5, I = 1 means that I + 0.5 = 1.5. In other words, since I is an integer, 0.5 is the only value of (I + 0.5) that lies between 0 and 1.
So, I (or the integral part of a) has to be 0.

'a' is now left with only its decimal part.. i.e. a = 0 + d = d.
Since 0 < d < 0.999.., 0 < a < 0.999.. also.

Please let me know if this makes sense.