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Let ] represent the average of the greatest integer less [#permalink]
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17 Sep 2011, 01:07
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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) \(0< [[a]]<1\)?
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Re: Toughie [#permalink]
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17 Sep 2011, 02:05
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) \(0< [[a]]<1\)?
Please explain your answer. Will award the most comprehensive answer with kudos! 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]] = (xk1+x+k2)/2 = (2x+ k2k1)/2 = x + (k2k1)/2 (1) [[x]]  x = (k2k1)/2 Or a=(k2k1)/2 Because 0=< k1,k2 <=1 > 0=<  (k2k1)/2  <=1 or 0=<a<=1 Hence, (1) is suff. (2) Similar to [[x]] : [[a]] = a+ (r2r1)/2 in which 0=<r1,r2<=1 0<[[a]]<1 0< (ar1+a+r2)/2 <1 0< ar1+a+r2 <2 Because ar1 and a+ r2 are interger, ar1+a+r2 =1 > 2a = 1 (r2r1) > a = 1/2(r2r1)/2 0=<r1,r2<=1 Hence, 0=<a<=1 (2) is suff ==> The answer is D



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Re: Toughie [#permalink]
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17 Sep 2011, 02:35
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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]
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19 Sep 2011, 09:00
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]
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20 Sep 2011, 08:38
1)Simpliy the statement Greatest int <=x is x Least int >=x is x Hence [[x]] = x 2)Evaluate 1. [[x]]  x = a xx = 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]
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20 Sep 2011, 11:37
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]
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21 Sep 2011, 22:54
check for x= 0.4  1.6 0.4  1.7 a for 0.4, 0.50.4=0.1 Y ; for 1.6, 1.51.6= 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]
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29 Sep 2011, 13:03
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 2.52.3=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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Re: Let [[x]] represent the average of the greatest integer less [#permalink]
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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. So the answer is D
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Re: Let [[x]] represent the average of the greatest integer less [#permalink]
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19 May 2012, 09:45
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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 ?
(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]
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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]
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Updated on: 04 Jul 2012, 11:55
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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Originally posted by EvaJager on 04 Jul 2012, 10:28.
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]
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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]
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02 Aug 2013, 13:52
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. So the answer is D 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. 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]
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02 Aug 2013, 14:06
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
Then the answer for [[x]] is 4 not 4.5. Please explain. 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 ] represent the average of the greatest integer less [#permalink]
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20 Nov 2014, 05:21
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



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Re: Let ] represent the average of the greatest integer less [#permalink]
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