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

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Joined: 02 Sep 2009
Posts: 51215
Re: If a and b are positive integers, is a/b < 9/11 ?  [#permalink]

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26 Jul 2018, 09:39
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.
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If a and b are positive integers, is a/b < 9/11 ?  [#permalink]

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12 Aug 2018, 12:09
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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12 Aug 2018, 12:21
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.
Re: If a and b are positive integers, is a/b < 9/11 ? &nbs [#permalink] 12 Aug 2018, 12:21

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