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If a and b are positive integers, is a/b < 9/11 ?

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Re: If a and b are positive integers, is a/b < 9/11 ?  [#permalink]

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New post 26 Jul 2018, 10:39
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If a and b are positive integers, is a/b < 9/11 ?  [#permalink]

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New post 12 Aug 2018, 13:09
Bunuel wrote:
kablayi wrote:
If understand correctly,

a/b < 9/11 then b/a>9/11

I guess this rule holds only when both sides of the inequality have the same sign, am I correct?


When a and b have the same sign.

Also, while @kablayi's inequality holds true in this specific case, generally if \(a>0, b>0\) and \(\frac{a}{b} < \frac{9}{11}\), then \(\frac{b}{a} > \frac{11}{9}\). It's simple if you try to manipulate the inequality step-by-step.
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Re: If a and b are positive integers, is a/b < 9/11 ?  [#permalink]

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New post 12 Aug 2018, 13:21
On a different note, I disagree with the first few posters who argued that you needn't calculate anything, because you could obviously derive a solution from either statement (1) or (2). Maybe I misunderstood you guys, but things aren't that clear in my opionion.

Assume, for the sake of the argument, that statement (1) instead says \(\frac{a}{b} < 0.82\). This would be insufficient, because \(\frac{a}{b}\) could be \(0.819 > \frac{9}{11}\), but it could also be less than \(\frac{9}{11}\).

For me, the take-away here is to remember my fractions, because the answer isn't as clear-cut as some might make it seem.
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Re: If a and b are positive integers, is a/b < 9/11 ?  [#permalink]

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New post 18 Feb 2019, 01:56
the condition 2 is what
a/b=1/1.123
or
a/b<1/1.123

what is the question?
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Re: If a and b are positive integers, is a/b < 9/11 ?   [#permalink] 18 Feb 2019, 01:56

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