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Re: Awkward DS.... [#permalink]
26 Nov 2010, 14:15

2

This post received KUDOS

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.

Re: Awkward DS.... [#permalink]
26 Nov 2010, 14:26

3

This post received KUDOS

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. _________________

I appreciate the kudos if you find this post helpful! +1

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. _________________

I appreciate the kudos if you find this post helpful! +1

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.

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.

Re: For all numbers t, let //t// be defined as the greatest [#permalink]
31 Jan 2014, 10:25

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