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# Let [[x]] represent the average of the greatest integer less

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Let [[x]] represent the average of the greatest integer less [#permalink]  18 May 2012, 05:44
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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
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Re: Let [[x]] represent the average of the greatest integer [#permalink]  18 May 2012, 06:45
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Smita04 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 ?
Hi Smita

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

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.
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  19 May 2012, 09:45
Smita04 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

consider 1: 0<a<1/2
consider 2: --> a=1/2,
in both cases , sufficient,
thus D
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  22 May 2012, 10:55
[[x]] = x when x is integer.==> a=0

[[x]]= [x]+.5 when x is a not integer.==>a=[x]+.5 -x = -{x}+.5 which always lies in between -.5 to .5
so |a|<1.
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  22 May 2012, 10:55
yeah D, take some cases to solve before 2 minutes
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  04 Jul 2012, 04:14
Bunuel can you please explain this question, esp. the second part. Thanks in advance
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  04 Jul 2012, 10:28
riteshgupta wrote:
Bunuel can you please explain this question, esp. the second part. Thanks in advance

I can try to explain it:
Any real number is situated between two consecutive integers. So, there is an integer k such that k \leq a < k+1.
Then [[a]] = k+0.5, and combining with (2), we get that 0 < k+0.5 < 1, which means -0.5 < k < 0.5. Being an integer,
it follows that k must be 0. Therefore, 0 \leq a < 1, and (2) is sufficient.
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Last edited by EvaJager on 04 Jul 2012, 11:55, edited 1 time in total.
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  04 Jul 2012, 10:42
EvaJager wrote:
riteshgupta wrote:
Bunuel can you please explain this question, esp. the second part. Thanks in advance

I can try to explain it:
Any real number is situated between two consecutive integers. So, there is an integer k such that k \leq a \leq k+1.
Then [[a]] = k+0.5, and combining with (2), we get that 0 < k+0.5 < 1, which means -0.5 < k < 0.5. Being an integer,
it follows that k must be 0. Therefore, 0 \leq a \leq 1, and (2) is sufficient.

Thanks EvaJager...
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  04 Jul 2012, 11:57
riteshgupta wrote:
EvaJager wrote:
riteshgupta wrote:
Bunuel can you please explain this question, esp. the second part. Thanks in advance

I can try to explain it:
Any real number is situated between two consecutive integers. So, there is an integer k such that k \leq a \leq k+1.
Then [[a]] = k+0.5, and combining with (2), we get that 0 < k+0.5 < 1, which means -0.5 < k < 0.5. Being an integer,
it follows that k must be 0. Therefore, 0 \leq a \leq 1, and (2) is sufficient.

Thanks EvaJager...

Very welcome.

Just slight corrections: it should be k \leq a < k+1 and 0 \leq a < 1.
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Re: Let [[x]] represent the average of the greatest integer less   [#permalink] 04 Jul 2012, 11:57
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