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M70-20

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M70-20  [#permalink]

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New post Updated on: 16 Oct 2018, 01:16
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

29% (00:35) correct 71% (00:29) wrong based on 21 sessions

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Originally posted by Bunuel on 03 Sep 2018, 04:01.
Last edited by Bunuel on 16 Oct 2018, 01:16, edited 3 times in total.
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Re M70-20  [#permalink]

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New post 03 Sep 2018, 04:01
Official Solution:


We’ll go for LOGICAL because there is a logic to understanding the operator.

Since \([\frac{a}{b}]\) is defined as an integer either equal to or less than \(\frac{a}{b}\), no possible value of \((a,b)\) can ever make \([\frac{a}{b}]\) necessarily greater than any integer. This means a definitive answer to the question stem, if there is sufficient information, can only be ‘NO!’ – when \([\frac{a}{b}] < {1}\). This would be the case if \(b\) is greater than \(a\). (1) gives us no such information, and (2) gives us the opposite: the greater integer \(b\) is, integer \(a\) becomes even greater. Thus, combining the two we’ll still get \(a > b\). Therefore, (E) is correct.


Answer: E
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Re: M70-20  [#permalink]

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New post 12 Dec 2018, 09:10
someone please give mathematical solution
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Re: M70-20  [#permalink]

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New post 18 Dec 2018, 13:39
if u combine two, b^3=64 then b=4. and a=16, thus sufficient. Am I wrong?
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Re: M70-20  [#permalink]

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New post 18 Dec 2018, 17:48
Bunuel wrote:
Official Solution:


We’ll go for LOGICAL because there is a logic to understanding the operator.

Since \([\frac{a}{b}]\) is defined as an integer either equal to or less than \(\frac{a}{b}\), no possible value of \((a,b)\) can ever make \([\frac{a}{b}]\) necessarily greater than any integer. This means a definitive answer to the question stem, if there is sufficient information, can only be ‘NO!’ – when \([\frac{a}{b}] < {1}\). This would be the case if \(b\) is greater than \(a\). (1) gives us no such information, and (2) gives us the opposite: the greater integer \(b\) is, integer \(a\) becomes even greater. Thus, combining the two we’ll still get \(a > b\). Therefore, (E) is correct.


Answer: E


Hello chetan2u Bunuel

OS above is not clear to me...

Can you explain with some other approach...

I find OA dubious....

Thanks.
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M70-20  [#permalink]

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New post 18 Dec 2018, 18:09
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Bunuel wrote:
\(a\) and \(b\) are integers. \([x]\) is an integer less than or equal to \(x\). Is \([\frac{a}{b}] \geq {1}\)?



(1) \(ab = 64\)

(2) \(a=b^2\)


Harshgmat
Yes the question is a bit confusing because we have been dealing with such questions with a bit different wordings.

Had it been \([x]\) is the GREATEST integer less than or equal to \(x\), the answer would be YES. Because a/B would be 4 and thus >1.

However here GREATEST is missing, so the value could be anything but not greater than a/B..
So if a/b=4, [a/b] can be 4,3,2,1,0,-1,-2.....
So it could be any integer as shown above and \(\frac{a}{b}\leq{4}\)

The difference is the word GREATEST not being there.

Hope it helps.
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: M70-20  [#permalink]

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New post 18 Dec 2018, 19:23
chetan2u wrote:
Bunuel wrote:
\(a\) and \(b\) are integers. \([x]\) is an integer less than or equal to \(x\). Is \([\frac{a}{b}] \geq {1}\)?



(1) \(ab = 64\)

(2) \(a=b^2\)


Harshgmat
Yes the question is a bit confusing because we have been dealing with such questions with a bit different wordings.

Had it been \([x]\) is the GREATEST integer less than or equal to \(x\), the answer would be YES. Because a/B would be 4 and thus >1.

However here GREATEST is missing, so the value could be anything but not greater than a/B..
So if a/b=4, [a/b] can be 4,3,2,1,0,-1,-2.....
So it could be any integer as shown above and \(\frac{a}{b}\leq{4}\)

The difference is the word GREATEST not being there.

Hope it helps.


chetan2u

Yes it is helpful.

Thanks. Kudos.

Posted from my mobile device
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Re: M70-20 &nbs [#permalink] 18 Dec 2018, 19:23
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