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If a>b>0

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If a>b>0  [#permalink]

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New post 01 Oct 2018, 20:38
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Question Stats:

34% (01:39) correct 66% (02:09) wrong based on 47 sessions

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If a>b>0 , is b<2?

1. \(\frac{1}{a} >\frac{1}{2}\)
2. \(\frac{1}{a} + \frac{a}{b} = 1\)
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If a>b>0  [#permalink]

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New post 01 Oct 2018, 22:02
rheabiswal wrote:
If a>b>0 , is b<2?

1. \(\frac{1}{a} >\frac{1}{2}\)
2. \(\frac{1}{a} + \frac{a}{b} = 1\)


From the given,
a, b are both +ve , a!=b!= 0

1. 1/a >1/2 ==> a< 2 , hence b <2 . Sufficient.
2. let a=1 , 1+1/b >1 , hence a<1 and b < a < 1 .Sufficient.

Hence D.
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Re: If a>b>0  [#permalink]

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New post 01 Oct 2018, 23:42
If 0<a<1 then the first term 1/a is already more than one, so how can this work? 1/a + a/b = 1 since be cant be negative. (a>b>0)
thx.
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Re: If a>b>0  [#permalink]

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New post 02 Oct 2018, 03:48
Phibo wrote:
If 0<a<1 then the first term 1/a is already more than one, so how can this work? 1/a + a/b = 1 since be cant be negative. (a>b>0)
thx.



Right....I also have same doubt.
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Re: If a>b>0  [#permalink]

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New post 02 Oct 2018, 03:58
If a>b>0 , is b<2?

1. \(\frac{1}{a} >\frac{1}{2}\)
\(\frac{2}{a>1}.....a<2\),
And since a>b>0, b is also less than 2
Sufficient

2. \(\frac{1}{a} + \frac{a}{b} = 1\)
Three cases...
a) a<1 ..... then 1/a >1 and so a/b should be less than 0 to negate excess in 1/a..
Means b<0.... But b>0, so not possible
b) a=1....then 1/a=1/1=1 and so a/b that is 1/b , should be 0...Not possible
c) a>1... Then 1/a<1 and a/b is also<1....
But a/b<1 means a<b....but we are given a>b>0
9
So none of the value can fit in..

FLAWED question
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If a>b>0  [#permalink]

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New post 02 Oct 2018, 05:13
Statement 1 is sufficient but statement 2 is flawed. kindly recheck the OA.
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Re: If a>b>0  [#permalink]

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New post 02 Oct 2018, 06:05
ReadyPlayerOne wrote:
Statement 1 is sufficient but statement 2 is flawed. kindly recheck the OA.



As per MathRevolution diagnostic test OA is D

Not sure how.
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Re: If a>b>0 &nbs [#permalink] 02 Oct 2018, 06:05
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