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Re: Let [[x]] represent the average of the greatest integer [#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, 03:14

Bunuel can you please explain this question, esp. the second part. Thanks in advance _________________

_______________________________________________________________________________________________________________________________ If you like my solution kindly reward me with Kudos.

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, 09:42

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... _________________

_______________________________________________________________________________________________________________________________ If you like my solution kindly reward me with Kudos.

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 [#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 [#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....