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Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b

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Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 30 Jun 2006, 05:49
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Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-1-a-b-b-a-168100.html
[Reveal] Spoiler: OA
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 30 Jun 2006, 09:41
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likar wrote:
Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|


To start with a,b belong to a set of real numbers, since nothing else is known.

Let b=a+k (where k is a +ve or -ve real)
=>k = b-a

To prove

1/(a-b) < b-a
=>1/(a-a-k)<a+k-a

=> -1/k < k ....(1)

From 1:
a<b.
=>0<b-a
=>0<k
Therefore k is +ve. Therefore 1 is sufficient.

From 2:
1<|a-b|
=> (a-b)>1 OR -(a-b)>1
=> -k>1 OR k>1
=> k<-1 OR k>1
If k<-1 say -2, then (1) becomes:-
1/2 < -1 which means the inequality does not hold. So 2 is insufficient.

I'd say A.
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 30 Jun 2006, 10:18
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A

1/(a-b) < b-a ?

St1: a-b <0. This means left hand side term will always be -ve and right hand side term will always be +ve.: SUFF

St2: 1 < |a-b| we can't say anything about the sign of (a-b) so either side can be either positive or negative. : INSUFF
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 02 Jul 2006, 07:31
likar wrote:
Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|


AD-BCE

rephrasing the Q stem,

1/(a-b) < b-a => 1 < (a-b)(b-a) => 1 < - (a-b)^2,

(a-b)^2 will always be +ve regardless of whether a < b or not, so INSUFF

therefore, BCE are choices.

taking second condition,
1 < |a-b| => -(a-b)^2 is definitely < 1 therefore B is SUFF

(B)

but i'm not too sure of my answer...now let me see others solutions :)
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 10 Feb 2011, 11:36
guys im concerned with (2)

im getting:

1/a-b< b-a?
1/a-b< - (a-b) ?

1st scenario if a-b >
1< - (a-b)^2

2nd scenario:
of a-b<0
a
1< - (a-b)^2

why im geting the same answer? could somebody advice whats wrong?
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 10 Feb 2011, 12:14
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tinki wrote:
guys im concerned with (2)

im getting:

1/a-b< b-a?
1/a-b< - (a-b) ?

1st scenario if a-b >
1< - (a-b)^2

2nd scenario:
of a-b<0
a
1< - (a-b)^2


why im geting the same answer? could somebody advice whats wrong?


How did you even find this thread?

Anyway: when you consider 2nd scenario (a-b<0) and multiply the inequality by this expression then you must flip the sign as you multiply be the negative value --> 1>- (a-b)^2.

But you don't actually need to manipulate this way.

Is 1/(a-b) < b-a ?

(1) a<b --> we can rewrite this as: \(a-b<0\) so LHS is negative, also we can rewrite it as: \(b-a>0\) so RHS is positive --> negative<positive. Sufficient.

(2) 1<|a-b| --> if \(a-b=2\) (or which is the same \(b-a=-2\)) then LHS>0 and RHS<0 and in this case the answer will be NO if \(a-b=-2\) (or which is the same \(b-a=2\)) then LHS<0 and RHS>0 and in this case the answer will be YES. Not sufficient.

Answer: A.
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 10 Feb 2011, 12:33
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1/(a-b) < b-a ?

1/(a-b) - (b-a) < 0
1/(a-b) + (a-b) < 0
(1+(a-b)^2)/(a-b) < 0

(1+(a-b)^2)= +ve

a-b < 0

Q: Is a<b?

(1) a<b. Sufficient.

(2) 1<|a-b|

|a-b| > 1

(a-b) > 1
a > 1+b
Answer: No.

or

a-b < -1
a < b-1
Answer: Yes
Not sufficient.

Ans: "A"
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 11 Feb 2011, 02:29
Bunuel wrote:
tinki wrote:
guys im concerned with (2)

im getting:

1/a-b< b-a?
1/a-b< - (a-b) ?

1st scenario if a-b >
1< - (a-b)^2

2nd scenario:
of a-b<0
a
1< - (a-b)^2


why im geting the same answer? could somebody advice whats wrong?


How did you even find this thread?

Anyway: when you consider 2nd scenario (a-b<0) and multiply the inequality by this expression then you must flip the sign as you multiply be the negative value --> 1>- (a-b)^2.

But you don't actually need to manipulate this way.

Is 1/(a-b) < b-a ?

(1) a<b --> we can rewrite this as: \(a-b<0\) so LHS is negative, also we can rewrite it as: \(b-a>0\) so RHS is positive --> negative<positive. Sufficient.

(2) 1<|a-b| --> if \(a-b=2\) (or which is the same \(b-a=-2\)) then LHS>0 and RHS<0 and in this case the answer will be NO if \(a-b=-2\) (or which is the same \(b-a=2\)) then LHS<0 and RHS>0 and in this case the answer will be YES. Not sufficient.

Answer: A.


Thanks Bunuel, u re my Math angels, always there when help is needed.
1 more request : could you specify definition for abbreviations for LHS and RHS ?
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 11 Feb 2011, 02:37
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 11 Feb 2011, 02:40
thanks bunuel , thanks fluke
+kudos to both
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 23 Feb 2011, 20:19
Agree with A.

More on part B, if :

1< |a-b|, this gives two inequalities

1< a-b or

1< -(a-b) = 1< b-a = -1> a-b

This gives two inequalities, which allow (a-b) to be anywhere on interval (-Infinite, -1) or (1, infininty)

If a-b is positive, then b-a is negative....proof a-b>0.....a>b......0>b-a

Therefore 1/positive < negative is false

If a-b is negative, then b-a is positive

Therefore 1/negative< positive is true

Cannot determine based on B

Answer is A
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 17 Jun 2013, 07:06
I want to know where is the error in following method for statement 2

1 < la-bl
la-bl > 1

then

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since both sides are positive for first possibility

1/(a-b) < 1 or 1 < b-a ( i think problem is here) :x

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 17 Jun 2013, 08:34
Expert's post
GMAT98 wrote:
I want to know where is the error in following method for statement 2

1 < la-bl
la-bl > 1

then

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since both sides are positive for first possibility

1/(a-b) < 1 or 1 < b-a ( i think problem is here) :x

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow


Please elaborate your question.
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 18 Jun 2013, 08:35
Bunuel wrote:
GMAT98 wrote:
I want to know where is the error in following method for statement 2

1 < la-bl
la-bl > 1

then

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since both sides are positive for first possibility

1/(a-b) < 1 or 1 < b-a ( i think problem is here) :x

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore yes... i know i am wrong somewhere.. but where i dontknow


Please elaborate your question.


Hi Bunuel

Pl help me with following problem

Is 1/ a-b < b-a
(1) a<b
(2) 1<|a-b|

My thought process is as follows
(1) a <b
a-b < 0 or 0 < b-a
1/(a-b) < 0 or 0 < b-a
negative will always be less than positive, therefore, statement one is sufficient


(2) 1 < la-bl
la-bl > 1

then opening absolute bracket

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since a-b > 1 it implies a-b is positive and 1 is also positive therefore taking reciprocal

1/(a-b) < 1 or b-a > 1

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore statement two is also sufficient Therefore answer D but answer is A
... i know i am wrong somewhere.. but where i dont know
can you please go through my solution for second statement and pin-point the error
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 21 Jun 2013, 10:18
Hi Bunuel

Pl help me with following problem

Is 1/ a-b < b-a
(1) a<b
(2) 1<|a-b|

My thought process is as follows
(1) a <b
a-b < 0 or 0 < b-a
1/(a-b) < 0 or 0 < b-a
negative will always be less than positive, therefore, statement one is sufficient


(2) 1 < la-bl
la-bl > 1

then opening absolute bracket

a-b > 1 or a-b< -1

a-b > 1 or b-a> 1

since a-b > 1 it implies a-b is positive and 1 is also positive therefore taking reciprocal

1/(a-b) < 1 or b-a > 1

combining

1/(a-b) < 1 < b-a

1/(a-b)< b-a

therefore statement two is also sufficient Therefore answer D but answer is A
... i know i am wrong somewhere.. but where i dont know
can you please go through my solution for second statement and pin-point the error
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 27 Jun 2013, 12:19
Is 1/(a-b) < b-a ?

(1) a<b
(2) 1<|a-b|

(1) a<b

If a<b then (b-a) will always be a positive number. Therefore, a-b will always be negative. Even if b-a turned out to be a fraction, a-b will negative.
SUFFICIENT

(2) 1<|a-b|

a-b could be 4-2 = 2 or it could be b-a = 2-4 = -2. In other words, we don't know if it's positive or negative meaning we cannot solve for (a-b) or (b-a)
INSUFFICIENT

(A)
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 27 Aug 2014, 02:29
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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 08 May 2015, 00:40
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Going through the solutions posted above, I realized that this question is a good illustration of the importance of first analyzing the question statement before moving to St. 1 and 2 in a DS question.

Most students dived straight into Statement 1 and then tried to draw inferences from St. 1 to determine whether the inequality given in the question statement was true or not.

Imagine if you had done this instead before going to St. 1:

The question is asking if:
\(\frac{1}{a-b} < b-a\)

Case 1: a - b is positive (that is, a > b)

Multiplying both sides of an inequality with with the positive number (a-b) will not change the sign of inequality.

So, the question simplifies to: Is 1 < (b-a)(a-b)

Now, b - a will be negative.

So, the question simplifies to: Is \(1 < -(a-b)^2\) . . . (1)

\((a-b)^2\), being a square term, will be >0 (Note: a - b cannot be equal to zero because then the fraction given in the question: 1/a-b becomes undefined)

So, \(-(a-b)^2\) will be < 0

Therefore, the question simplifies to: Is 1 < (a negative number?)

And the answer is NO.

Case 2: a - b is negative (that is, a < b)

Multiplying both sides of an inequality with the negative number (a-b) will change the sign of inequality.

So, the question simplifies to: Is 1 > (b-a)(a-b)

Now, b - a will be positive.

So, the question simplifies to: Is \(1 > -(a-b)^2\) . . . (2)

Again, by the same logic as above, we see that the question simplifies to: Is 1 > (a negative number?)

And the answer is YES

Thus, from the question statement itself, we've inferred that:

If a > b, the answer to the question asked is NO
If a < b, the answer to the question asked is YES

Thus, the only thing we need to find now is whether a > b or a < b.

Please note how we are going to Statement 1 now with a much simpler 'To Find' task now. :-D

One look at St. 1 and we know that it will be sufficient.

One look at St. 2 and we know that it doesn't give us a clear idea of which is greater between a and b, and so, is not sufficient.

To sum up this discussion, spending time on analyzing the question statement before going to the two statements usually simplifies DS questions a good deal.

Hope this helped!

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Re: Is 1/(a-b) < b-a ? (1) a<b (2) 1<|a-b [#permalink] New post 08 May 2015, 02:49
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