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# If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1

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If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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06 Jul 2017, 06:58
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If a > b > 0, is b < 2?

(1) 1/a > 1/2
(2) 1/a + 1/b = 1
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Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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06 Jul 2017, 12:07
1
1) 1/a >1/b
This is possible if and only if a < 2.
Since we have relation a>b>0, and b <2 is definitely true. Sufficient

2) (1/a) + (1/b) =1
For relation a>b>0,
this is possible when b=3/2 and a=3, or when b=4/3 and a=4..... and so on.
At b=2, a must be equal to 2, which is not possible as a>b>0.
Thereafter, for every value of b, we will not be able to find a such that a>b. Sufficient (Option D)
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Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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16 Jul 2017, 04:05
roastedchips wrote:
If a>b>0, is b <2?
1) 1/a >1/2
2) (1/a) + (1/b) =1

I think the question should mention for statement 2 that "a" is not equal to 1, since at that point b becomes infinite or undefined.Rest is fine.
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Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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18 Jul 2017, 04:33
If a>b>0, is b <2?

1) 1/a >1/2

As 'a' is positive then we can safely reverse the fraction but flip the inequality sign

The a < 2. For certain we know that b < 2 as a > b.

Sufficient

2) (1/a) + (1/b) =1

$$\frac{1}{a}$$ + $$\frac{1}{b}$$ =1

$$\frac{ab}{b+a}$$= 1

a +b = ab

Let a =4.......then b= 4/3 < 2

let a =10...... then b= 10/9 < 2

let a = 12......then b = 12/11 <2

There is a pattern. Always the numerator is greater than denominator by 1 making the fraction 1.xy <2

Sufficient

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Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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30 Jul 2017, 17:27
2
roastedchips wrote:
If a>b>0, is b <2?
1) 1/a >1/2
2) (1/a) + (1/b) =1

We are given that a > b > 0 and need to determine whether b < 2.

Statement One Alone:

1/a > 1/2

Taking the reciprocal of each side and reversing the direction of the inequality, we get:

a < 2

We see that a is less than 2, and thus b must also be less than 2. Statement one is sufficient to answer the question.

Statement Two Alone:

(1/a) + (1/b) = 1

Multiplying the entire inequality by ab (which is non-zero because a > b > 0), we have:

b + a = ab

b = ab - a

b = a(b - 1)

a = b/(b - 1)

Substituting for a in the inequality given in the question stem, a > b > 0, we have:

b/(b - 1) > b > 0

Since b > 0, we can divide each side of the inequality by b without changing the inequality sign:

1/(b - 1) > 1

Taking the reciprocal of each side and reversing the direction of the inequality, we get:

b - 1 < 1

b < 2

Statement two is sufficient to answer the question.

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Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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30 Jul 2018, 11:13
ScottTargetTestPrep

Bunuel

I did the same thing for statement 2 but I replaced b.

a + b= ab
a = ab - b = b(a-1)
b = a/(a-1)

Now we have a > a/(a-1) > 0
a > a/(a-1)
1 > 1/(a-1)
1 < a-1
a > 2

But this conflicts with what we got in statement 1 i.e. a < 2. Doesn't GMAT have the same answers when both statements are sufficient on their own?
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Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1  [#permalink]

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06 Sep 2018, 15:53
roastedchips wrote:
If a > b > 0, is b < 2?

(1) 1/a > 1/2
(2) 1/a + 1/b = 1

In the version of the question I encountered, and the version that was archived here, (2) is the following: 1/a+a/b=1. (Difference is a/b and not 1/b). Does this change the analysis?
Re: If a > b > 0, is b < 2? (1) 1/a > 1/2 (2) 1/a + 1/b = 1 &nbs [#permalink] 06 Sep 2018, 15:53
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