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# Let ] represent the average of the greatest integer less

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Let ] represent the average of the greatest integer less [#permalink]  17 Sep 2011, 00:07
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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) $$0< [[a]]<1$$?
[Reveal] Spoiler: OA

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Re: Toughie [#permalink]  17 Sep 2011, 01:05
akhileshgupta05 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) $$0< [[a]]<1$$?

x - k1 : the greatest integer less than or equal to x
x + k2 : the least integer greater than or equal to x
--> 0=< k1,k2 <=1 and [[x]] = (x-k1+x+k2)/2 = (2x+ k2-k1)/2 = x + (k2-k1)/2

(1) [[x]] - x = (k2-k1)/2 Or a=(k2-k1)/2
Because 0=< k1,k2 <=1 --> 0=< | (k2-k1)/2 | <=1 or 0=<|a|<=1
Hence, (1) is suff.

(2) Similar to [[x]] : [[a]] = a+ (r2-r1)/2 in which 0=<r1,r2<=1
0<[[a]]<1
0< (a-r1+a+r2)/2 <1
0< a-r1+a+r2 <2
Because a-r1 and a+ r2 are interger, a-r1+a+r2 =1
--> 2a = 1- (r2-r1)
--> a = 1/2-(r2-r1)/2
0=<r1,r2<=1 Hence, 0=<|a|<=1
(2) is suff

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Re: Toughie [#permalink]  17 Sep 2011, 01:35
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akhileshgupta05 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) $$0< [[a]]<1$$?

The answer should be D. This is because:

for any value of x, the difference between the greatest integer less than or equal to x AND the least integer greater than or equal to x will always vary between 1 unit. hence
1)
=> -1<= [[x]] - x <=1
or -1<= a <= 1
or 0 <= |a| <=1

hence sufficient

2) again as mentioned above and given:0< [[a]] < 1.

This will be possible only when x is between 0 and 1 so that we get he greatest value as one and least value as 0. This will only result in 0< [[a]] < 1, as [[a]] is av. of greatest and the least integer.
hence:
0 <= a <=1 and hence 0 <= |a| <=1.
sufficient
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Re: Toughie [#permalink]  19 Sep 2011, 08:00
For such daunting questions, instead of looking for a solution visually better write down what is given and what is required.

Suppose the number x = n.m ( if n.m = 2.1 => n = 2 and m =1)

[[x]] = (n+n+1)/2 = n +1/2 because greatest integer is n+1 and smallest is n

1. [[x]] - x = a
=> n+1/2 -n.m = a
=> a = n+1/2 - n - 0.m = 1/2 - 0.m
The absolute value of a will always oscillate between 0 and 1. Hence sufficient.

2. 0< [[a]] <1 => 0 < a+1/2 < 1
=> -1/2 < a < 1/2
=> the value of a oscillates between -1/2 and 1/2 => the absolute value lies between 0 and 1/2.
Hence sufficient.

Hence D
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Re: Toughie [#permalink]  20 Sep 2011, 07:38
1)Simpliy the statement
Greatest int <=x is x
Least int >=x is x
Hence [[x]] = x
2)Evaluate 1.
[[x]] - x = a
x-x = 0 so a = 0
1. is Sufficient
4)Evaluate 2.
0<[[a]]<1
0<a<1
2. is sufficient

Ans D
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Re: Toughie [#permalink]  20 Sep 2011, 10:25
it's not so difficult question
i think [[x]] is nothing but x itself, so dont get confused

My answer - D (both suff.)
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Re: Toughie [#permalink]  20 Sep 2011, 10:37
parulbhatnagar wrote:
it's not so difficult question
i think [[x]] is nothing but x itself, so dont get confused

My answer - D (both suff.)

x=2.3456; [[x]]=2.5
x=0; [[x]=0
x=-2.3456; [[x]]=-2.5

So, your statement is true only if x=integer.
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Re: Toughie [#permalink]  21 Sep 2011, 21:54
check for x= 0.4 | 1.6 |-0.4 | -1.7

a for 0.4, 0.5-0.4=0.1 Y ; for 1.6, 1.5-1.6= -0.1 Y
for -0.4, -0.5+0.4 = 0.1 Y ; for -1.7, -1.5+1.7 = 0.2 Y

thus sufficient.

b satisfies for 0 <= a <= 1 hence sufficient.

D it is.
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Re: Toughie [#permalink]  29 Sep 2011, 12:03
It was hard to understand the question. I picked A initially but then I agreed that D was the answer.

Pick some numbers first and draw a numbers line. x=2.3 then the lower integer= 2, upper integer=3 therefore [[2.3]]=2.5

stm 1: [[2.3]]-2.3=a
2.5-2.3=a
0.2=a....sufficient

stm 2: if [[a]]=average of the upper and lower integers and it is between 0 and 1 then a must be between 0 and 1....sufficient
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Let [[x]] represent the average of the greatest integer less [#permalink]  18 May 2012, 04:44
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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 less [#permalink]  18 May 2012, 05:45
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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, 08:45
1
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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 ?

(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, 09:55
[[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]  04 Jul 2012, 09:28
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, 10:55, edited 1 time in total.
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  04 Jul 2012, 10:57
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]  02 Aug 2013, 12:52
narangvaibhav wrote:
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.

For statement 1 couldn't x = 0? If x=0 or any other integer say 4 then greatest integer less than or equal to x is x which = 4
and the least integer greater than or equal to x is x which is 4

Would the answer then be B? then you know that it has to be .5
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  02 Aug 2013, 13:06
MBA2015hopeful wrote:

For statement 1 couldn't x = 0? If x=0 or any other integer say 4 then greatest integer less than or equal to x is x which = 4
and the least integer greater than or equal to x is x which is 4

Would the answer then be B? then you know that it has to be .5

For statement 1 couldn't x = 0? If x=0 or any other integer say 4 then greatest integer less than or equal to x is x which = 4
and the least integer greater than or equal to x is x which is 4

Then the answer for [[x]] is 4 not 4.5. Please explain.==>this is perfectly fine
now when you apply statement 1:[[x]] - x==>4.5 - 4 = 0.5 which is between 0 and 1 (included)

hope it helps
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]  10 Oct 2014, 20:06
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Re: Let ] represent the average of the greatest integer less [#permalink]  18 Nov 2014, 20:01
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Re: Let ] represent the average of the greatest integer less [#permalink]  20 Nov 2014, 04:21
akhileshgupta05 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) $$0< [[a]]<1$$?

I did it this way... dont know whether this is a right method or not..

[[x]]= avg of [(greatest integer less than equal to x)+(smallest integer greater than or equal to x)]
--> greatest integer less than equal to x = number one unit smaller than x...lets say z, similarly
--> smallest int more than or equal to x = number one unit more than x...lets say y
HENCE
z<= x <=y
THESE 3 NUMBERS ARE EITHER 3 CONSECUTIVE NUMBERS OR THEY ARE SAME (as there is <= sign)
so avg of these three numbers, will be X only, even though they are same or consecutive.
Therefore
[[x]] = x
and this helps in solving the question further..
Key is to identify [[x]] = x
Re: Let ] represent the average of the greatest integer less   [#permalink] 20 Nov 2014, 04:21
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