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‎[x] denotes to be the least integer no less than x. Is [2d]

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‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 17 Apr 2012, 19:53
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‎[x] denotes to be the least integer no less than x. Is [2d] = 0?

(1) [d] = 0
(2) [3d] = 0
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 17 Apr 2012, 21:17
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‎[x] denotes to be the least integer no less than x. Is [2d] = 0?

[x] denotes to be the least integer no less than x, means that some function [] rounds UP a number to the nearest integer, for example:
\([1.5] = 2\) since 2 is the least integer which no less than 1.5;
\([-0.5]=0\) since 0 is the least integer which no less than -0.5;
\([1] = 1\) since 1 itself is the least integer which no less than 1;
...

(1) [d] = 0 --> \(-1<d\leq{0}\). Now, if \(d=0\) then \([2d]=[0]=0\) but if \(d=-0.5\) then \([2d]=[-1]=-1\). Not sufficient.

Or: \(-1<d\leq{0}\) --> \(-2<d\leq{0}\) --> \([2d]=0\) (if \(-1<2d\leq{0}\)) or \([2d]=-1\) (if \(-2<2d\leq{-1}\)). Not sufficient.

(2) [3d] = 0 --> \(-1<3d\leq{0}\) --> \(-\frac{1}{3}<d\leq{0}\). Even if \(d=-\frac{1}{3}\) then \([2d]=[-\frac{2}{3}]=0\) (again since 0 is the least integer which no less than -2/3). Sufficient.

Or: \(-1<3d\leq{0}\) --> \(-\frac{2}{3}<2d\leq{0}\) --> \([2d]=0\). Sufficient.

Answer: B.

Similar questions to practice:
if-denotes-the-greatest-integer-less-than-or-equal-to-x-94687.html
for-all-z-denotes-the-least-integer-greater-than-or-110561.html

Hope it helps.
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Re: Integers  [#permalink]

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New post Updated on: 17 Apr 2012, 23:43
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Smita04 wrote:
‎[x] denotes to be the least integer no less than x. Is [2d] = 0?
(1) [d] = 0
(2) [3d] = 0


'no less than' indicated [x]\(>=\) x....with the equality holding when x is an integer

st1: [d]=0 indicates -1<d<0 (eg. -0.5) => -2<2d<0(2x -0.5=-1) =>[2d]=-1
but if d= -0.3......2d= -0.6.....[2d]= 0 (not -1 as -1< -0.6)
NOT Sufficient

st2: -1<3d<0 => -1/3<d<0
d can be -0.65...then [2d]=0..... d can be [-0.01]....[d]=0....
(taking extreme values )
Suffecient

Answer:B

Originally posted by Dreaming on 17 Apr 2012, 20:49.
Last edited by Dreaming on 17 Apr 2012, 23:43, edited 1 time in total.
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 21 Jun 2013, 02:51
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 22 Jun 2013, 03:56
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Smita04 wrote:
‎[x] denotes to be the least integer no less than x. Is [2d] = 0?

(1) [d] = 0
(2) [3d] = 0


I did not jump into the equations, but tried to solve logically. Please let me know if wrong.

We need to find out if [2d] = 0?

Again its a YES or NO, question, getting Either in the statements will help us to the solutions.

(1) [d] = 0

Now according to the question [x] = integer no less than x

Hence, considering two extremes for evaluating this particular case, Lets take
[0.12] = 0 as well as [0.99] = 0 (both are valid for [d] =0)

therefore, [2d], i.e. taking the same values as above
[0.24] = 0, but [1.98] = 1
so we cant say most definitely that [2d] =0

Not Sufficient.

(2) [3d] = 0

again considering etremes here,

lets take
Minimum value => 3d = 0.12 (you can take less then this as well, i am just taking this for easy divisibility purpose, as anything less then 0.12 will also result in 0 we already know that)
and
Maximum value => 3d = 0.99
[3d] = 0 holds for both the values.

now, [2d]
minimum= [0.8] =0
maximum= [0.22] =0

hence, no matter what range we take for 3d, 2d will always be less then that.
Thus, if [3d]= 0, holds true, then [2d] =0 should hold true as well.

Hence, B
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 23 Jun 2013, 06:53
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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Hi Bunuel,

Please let me know if the above approach is correct?
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 24 Jun 2013, 01:59
kpali wrote:
Smita04 wrote:
‎[x] denotes to be the least integer no less than x. Is [2d] = 0?

(1) [d] = 0
(2) [3d] = 0


I did not jump into the equations, but tried to solve logically. Please let me know if wrong.

We need to find out if [2d] = 0?

Again its a YES or NO, question, getting Either in the statements will help us to the solutions.

(1) [d] = 0

Now according to the question [x] = integer no less than x

Hence, considering two extremes for evaluating this particular case, Lets take
[0.12] = 0 as well as [0.99] = 0 (both are valid for [d] =0)

therefore, [2d], i.e. taking the same values as above
[0.24] = 0, but [1.98] = 1
so we cant say most definitely that [2d] =0

Not Sufficient.

(2) [3d] = 0

again considering etremes here,

lets take
Minimum value => 3d = 0.12 (you can take less then this as well, i am just taking this for easy divisibility purpose, as anything less then 0.12 will also result in 0 we already know that)
and
Maximum value => 3d = 0.99
[3d] = 0 holds for both the values.

now, [2d]
minimum= [0.8] =0
maximum= [0.22] =0

hence, no matter what range we take for 3d, 2d will always be less then that.
Thus, if [3d]= 0, holds true, then [2d] =0 should hold true as well.

Hence, B


No, that's not correct. Function [] rounds UP a number to the nearest integer, thus [0.12] = 1, not 0 and [0.99] = 1, not 1.
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 13 Sep 2014, 00:37
Meaning that even -0.99 will round UP to 0 and not -1??
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 13 Sep 2014, 07:16
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 22 Dec 2017, 07:02
Smita04 wrote:
‎[x] denotes to be the least integer no less than x. Is [2d] = 0?

(1) [d] = 0
(2) [3d] = 0



Given [x] LI no less than x i.e [x]>=x
Find Is [2d]=0

Now [2d]=0 possible if 2d satisfies the inequality/range -1<2d<=0 ---equ(a)

Statement 1 [d]=0
=> Since [d]=0 so d lies in -1<d<=0
=> Multiplying the above inequality by 2 we have -2<2d<=0. So this range makes 2 cases
=> Case 1 When -2<2d<=-1 then [2d]=-1
=> Case 2 When -1<2d<=0 then [2d]=0
=> Since no unique sol. So NOT sufficient

Statement 2 [3d]=0
Since [3d]=0 so 3d lies in -1<3d<=0
=> Multiplying the above inequality by 2/3 we have -2/3<2d<=0.
=> So the above range satisfy equ (a)
=> Therefore [2d]=0
=> Since unique sol. So SUFFICIENT

Therefore "B"

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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]  [#permalink]

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New post 29 Dec 2018, 12:49
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Re: ‎[x] denotes to be the least integer no less than x. Is [2d]   [#permalink] 29 Dec 2018, 12:49
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