Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]] = (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

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

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.

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

Re: Let [[x]] represent the average of the greatest integer less [#permalink]
18 May 2012, 05:45

3

This post received KUDOS

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 _________________

Best Vaibhav

If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks

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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 04 Jul 2012, 10:55, edited 1 time in total.

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 \(0 \leq a < 1\). _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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. 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

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

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 _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: Let [[x]] represent the average of the greatest integer less [#permalink]
10 Oct 2014, 20:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Let ] represent the average of the greatest integer less [#permalink]
18 Nov 2014, 20:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

gmatclubot

Re: Let ] represent the average of the greatest integer less
[#permalink]
20 Nov 2014, 04:21

Michigan Ross: Center for Social Impact : The Center for Social Impact provides leaders with practical skills and insight to tackle complex social challenges and catalyze a career in...

The Importance of Financial Regulation : Before immersing in the technical details of valuing stocks, bonds, derivatives and companies, I always told my students that the financial system is...