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# For all numbers t, let //t// be defined as the greatest

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Manager
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For all numbers t, let //t// be defined as the greatest [#permalink]  26 Nov 2010, 14:03
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For all numbers t, let //t// be defined as the greatest integer less than or equal to t. Is //k// evenly divisible by 2?

(1) 5 < k < 6
(2) //k + 2.3// = 7

Source: Knewton

Solution to follow...interested to see how people interrupt this question.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 01 Feb 2012, 12:56, edited 2 times in total.
Edited the question
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Manager
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Re: Awkward DS.... [#permalink]  26 Nov 2010, 14:15
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Statement A: Tells you that k is somewhere between 5 and 6. Based on the definition of the question stem /k/ is thus 5, which is not evently divisible by 2.

=> Sufficient

Statement B:Tells you that the following inequality has to hold: 7 < k + 2.3 < 8 => 4.7 < k < 5.7 => /k/ has to equal 5, which is not evenly divisible.

=> Sufficient

Hence, solution D is correct.
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Re: Awkward DS.... [#permalink]  26 Nov 2010, 14:26
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Thanks for the reply stanford2012.

Looking at statement b...
Slight correction in your inequality: 4.7 ≤ k < 5.7. With this inequality, we see that k can be 4.7, 4.8, 4.9, etc. Hence, if k < 5 then //k// = 4, which is divisible by 2.
However, if k = 5.2, then //k// = 5, which is not divisible by 2.

Hence insufficient.

Correct answer is A. Updated post with OA.
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Re: Awkward DS.... [#permalink]  26 Nov 2010, 22:25
krishnasty wrote:
martie11 wrote:
Source: Knewton

For all numbers t, let //t// be defined as the greatest integer less than or equal to t. Is //k// evenly divisible by 2?

a) 5 < k < 6
b) //k + 2.3// = 7

Solution to follow...interested to see how people interrupt this question.

Can somebody please explain me what '//' symbol implies over here??

-----------------------------------------------------------------
Consider KUDOS if you like my post!

It's my interpretation that this is equivalent to the "floor" of the number.

In this case, say you have x = 11.3. Now the question tries to confuse you: // x // is equal to the greatest integer less than or equal to x, in other words:

// x // --> the largest integer <= x

In this case x = 11.3, so the largest number that is <= 11.3 is 11. So // x // = 11. Note, if x = 11, then by definition, // x // = 11.

I think the concept itself is fairly straightforward, it's the wording that is awkward....

HTHs.
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Re: Awkward DS.... [#permalink]  27 Nov 2010, 01:15
Expert's post
krishnasty wrote:
martie11 wrote:
Source: Knewton

For all numbers t, let //t// be defined as the greatest integer less than or equal to t. Is //k// evenly divisible by 2?

a) 5 < k < 6
b) //k + 2.3// = 7

Solution to follow...interested to see how people interrupt this question.

Can somebody please explain me what '//' symbol implies over here??

-----------------------------------------------------------------
Consider KUDOS if you like my post!

Stem defines some function, represented by the symbol $////$, as the function which rounds down any number to an integer value:

$//3.4//=3$, $//2//=2$, $//-7.5//=-8$, ...

Check similar question for practice: how-x-become-fraction-if-its-been-said-an-integer-94687.html

Hope it helps.
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Re: Awkward DS.... [#permalink]  27 Nov 2010, 07:59
hmmmmmm...........

much clear now....thanks!!
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Re: Awkward DS.... [#permalink]  27 Nov 2010, 19:39
martie11 wrote:
Source: Knewton

For all numbers t, let //t// be defined as the greatest integer less than or equal to t. Is //k// evenly divisible by 2?

a) 5 < k < 6
b) //k + 2.3// = 7

Solution to follow...interested to see how people interrupt this question.

S1: 5< k < 6 means //k// = 5. Not divisible by 2. Sufficient.
S2: //k+2.3// = 7 means 8 > k+2.3 >= 7. therefore 5.7 > k >=4.7 and //k// = 4 or 5. Not Sufficient.

Answer: A
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Re: Awkward DS.... [#permalink]  31 Dec 2010, 10:11
Sat.1- 5<k<6 mean any integer n.m. between 5 to 6. 5 is possible integer value.and 5 isn't evenly divisible by 2. suffucient
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Re: For all numbers t, let //t// be defined as the greatest [#permalink]  18 Dec 2012, 00:29
Rule: //t// is the greatest integer less than or equal to t
Problem: Is //k// evenly divisible by 2?

(1) 5 < k < 6
//k// = 5 which is not divisible by 2
SUFFICIENT.
(2) //k + 2.3// = 7
8 > k + 2.3 >= 7
5.7 > k > 4.7
if k = 4, YES
if not, NO
INSUFFICIENT.

Answer: A
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Re: For all numbers t, let //t// be defined as the greatest [#permalink]  31 Jan 2014, 10:25
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Re: For all numbers t, let //t// be defined as the greatest [#permalink]  01 Feb 2014, 03:28
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Re: For all numbers t, let //t// be defined as the greatest [#permalink]  11 Sep 2014, 18:11
Can somebody explain to me why statement B //k + 2.3// = 7 become this 8 > k + 2.3 >= 7 ?

I am confused ...
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Re: For all numbers t, let //t// be defined as the greatest [#permalink]  12 Sep 2014, 03:51
Expert's post
shelrod007 wrote:
Can somebody explain to me why statement B //k + 2.3// = 7 become this 8 > k + 2.3 >= 7 ?

I am confused ...

The function rounds down any number to an integer value:

$//3.4//=3$, $//2//=2$, $//-7.5//=-8$, ...

So, //x// = 7, means that 7 <= x < 8. Any number from this range when rounded down gives 7.

Check other Rounding Functions Questions in our Special Questions Directory.
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Re: For all numbers t, let //t// be defined as the greatest   [#permalink] 12 Sep 2014, 03:51
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# For all numbers t, let //t// be defined as the greatest

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