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

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

Data Sufficiency
Question: 120
Category: Arithmetic; Algebra Arithmetic operations; Inequalities
Page: 161
Difficulty: 650


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Re: Is 1/(a - b) > b - a ? [#permalink]

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SOLUTION

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 ? [#permalink]

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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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



Statement 1: a < b
Thus, a-b<0, implying that b-a>0.
Therefore, LHS is negative and RHS is positive. Which implies LHS<RHS. Therefore Sufficient.

Statement 2: 1 < |a - b|
This statement implies 1<(a-b)<-1

Substitute a-b = 2 and a-b= -2 we get different answers for both cases. INSUFFICIENT.


Therefore answer is A.
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Is 1/(a - b) > b - a ? [#permalink]

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SOLUTION

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 YES 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. Not sufficient.

Answer: A.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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Bunuel wrote:
SOLUTION

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.



for the second statement;

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

AND

a-b < -1 --> 1< b-a therefore LHS will be negative and RHS will be positive -->negative<positive. Sufficient.


what is wrong with my approach ?
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Re: Is 1/(a - b) > b - a ? [#permalink]

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lool wrote:
Bunuel wrote:
SOLUTION

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.



for the second statement;

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

AND

a-b < -1 --> 1< b-a therefore LHS will be negative and RHS will be positive -->negative<positive. Sufficient.


what is wrong with my approach ?


From \(1 < |a - b|\) we can have two cases:

A. \(1 < a-b\) --> \((\frac{1}{a - b}=positive) > (b - a=negative)\) --> answer YES.

B. \(a-b<-1\) --> \((\frac{1}{a - b}=negative) < (b - a=positive)\) --> answer NO.

Two different answers, hence insufficient.

Hope it's clear.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 11 Mar 2014, 02:24
I did it like this:

We can reduce the given statement as:

1/(a-b)>(b-a) -> Multiplying both sides by (a-b), we get: 1>(a-b)(b-a)

Taking negative out from RHS: 1<(a-b)(a-b), which is what we have to prove.

Now, Statement 1: a<b -> a-b<0. Squaring both sides: (a-b)(a-b)<0. Hence, the answer would be No. Thus, sufficient.

Statement 2: 1<|a-b| -> 1<(a-b) or 1>(a-b). Thus insufficient.

Hence, answer is A.

Please do let me know if this is a good way to proceed with such questions.

Thanks.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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I did it like this:

We can reduce the given statement as:

1/(a-b)>(b-a) -> Multiplying both sides by (a-b), we get: 1>(a-b)(b-a)

Taking negative out from RHS: 1<(a-b)(a-b), which is what we have to prove.

Now, Statement 1: a<b -> a-b<0. Squaring both sides: (a-b)(a-b)<0. Hence, the answer would be No. Thus, sufficient.

Statement 2: 1<|a-b| -> 1<(a-b) or 1>(a-b). Thus insufficient.

Hence, answer is A.

Please do let me know if this is a good way to proceed with such questions.

Thanks.


Unfortunately, most of it is wrong.

We cannot multiply 1/(a-b)>(b-a) by a-b, because we don't know its sign.
If a-b is positive, then we would have 1 > (b-a)(a-b);
If a-b is negative, then we would have 1 < (b-a)(a-b): flip the sign when multiplying by negative value.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

Next, you cannot square a-b<0 and write (a-b)^2<0. This is obviously wrong: the square of a number cannot be less than zero.

We can only raise both parts of an inequality to an even power if we know that both parts of the inequality are non-negative (the same for taking an even root of both sides of an inequality).

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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Darn it! :(
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 03 Jun 2014, 18:58
Is 1/(a - b) > b - a ?

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

From stmt1 : a<b
(a-b)<0; (negative) ->( b-a)>0 (positive);

So sufficient;

Stmt2 : 1 < |a - b|
let x= |a - b|;
then 1<x<-1 -> x>1 {2,3,4,5,6}(or) x<-1{-2,-3,-4,-5,-6} ; Hence Insufficient.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

Data Sufficiency
Question: 120
Category: Arithmetic; Algebra Arithmetic operations; Inequalities
Page: 161
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
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Thank you!


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

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

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

\(\frac{1+(a-b)^2}{(a-b)}\) > 0 ? we can see that numerator is always +ive . all we need to know is a>b?

(1) a < b ; then \(\frac{1+(a-b)^2}{(a-b)}\) < 0 . Sufficient.

(2) 1 < |a - b|
case 1 : a-b < -1 or a<b-1 --> a<b
case 2: a-b>1 or a> b+1------> a>b
not sufficient.


Ans : A
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Is 1/(a - b) > b - a ? [#permalink]

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Before going into Statements 1 and 2 here, the question itself can be much simplified, as shown below:

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 YES.

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 NO

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

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

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. :)

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!

Japinder
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 11 Apr 2016, 01:00
[quote="Bunuel"]SOLUTION

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 YES 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. Not sufficient.


Dear,

Please Clarify me Rewrite like this one.I thought rewite variable without considering its Sign is Wrong
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Re: Is 1/(a - b) > b - a ? [#permalink]

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AbdurRakib wrote:
Bunuel wrote:
SOLUTION

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 YES 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. Not sufficient.


Dear,

Please Clarify me Rewrite like this one.I thought rewite variable without considering its Sign is Wrong


We cannot multiply an inequality by a variable if don't know its sign but we can add/subtract a value to both sides. To get a-b<0 from a < b we are subtracting b from both sides and to get b - a > 0 we are subtracting a from both sides.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 30 Jun 2016, 05:19
(1) is sufficient because if a < b, then (a - b) will always be negative, while (b - a) will always be positive. So, LHS will be < RHS.

(2) 1 < |a - b|
This means |a - b| > 1
This doesn't tell us whether a is larger or b is larger. For example a = 1, b = 5 or a = 5, b = 1; for both these cases, |a - b| > 1.

So, not sufficient.

Hence, A.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 10 Oct 2016, 19:34
yezz wrote:
Is 1/(a-b)<b-a?

1. a<b
2. 1< la-bl

I have my own answer but i dont have OA

Please share ur explanation

1) a<b
or a-b<0
then b-a>0
substituting........yes...........suff...

2) 1< la-bl

if a-b = -2
then b-a=2.......yes

but if a-b=2
then b-a= -2.........No.........insuff

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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 11 Oct 2016, 00:35
yezz wrote:
Is 1/(a-b)<b-a?

1. a<b
2. 1< la-bl

I have my own answer but i dont have OA


from stem

-1/(a-b) - (a-b) <0 , -1 - (a-b)^2 / a-b <0 thus -[1+(a-b)^2/(a-b)] <0 ??... we only need sign of (a-b)

from 1

a-b < 0 ... suff

from 2

-1< a-b<1 ... insuff

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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 17 Oct 2016, 08:08
yezz wrote:
yezz wrote:
Is 1/(a-b)<b-a?

1. a<b
2. 1< la-bl

I have my own answer but i dont have OA


from stem

-1/(a-b) - (a-b) <0 , -1 - (a-b)^2 / a-b <0 thus -[1+(a-b)^2/(a-b)] <0 ??... we only need sign of (a-b)

from 1

a-b < 0 ... suff


from 2

-1< a-b<1 ... insuff

A


2nd method

let (a-b) = x , thus b-a = -(a-b) = -x , questions asks is 1/x> -x , ( if x is +ve it is impossible , then question asks whether X<0 or a-b< 0 , i.e. is a<b

from 1

clear.... suff

from 2

insuff

A
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 26 Jan 2017, 19:27
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Statement 1: a < b
Thus, a-b<0, implying that b-a>0.
Is 1/(negative) < positive?
YES.
SUFFICIENT.

Statement 2: 1 < |a - b|
It's possible that a-b = 2, implying that b-a = -2.
Plugging a-b=2 and b-a=-2 into 1/(a-b) < b-a, we get:
1/2 < -2?
NO.

It's possible that a-b = -2, implying that b-a = 2.
Plugging a-b=-2 and b-a=2 into 1/(a-b) < b-a, we get:
1/-2 < 2?
YES.

Since in the first case the answer is NO but in the second case the answer is YES, INSUFFICIENT.

The correct answer is A.
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Re: Is 1/(a - b) > b - a ? [#permalink]

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New post 28 Mar 2018, 03:30
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Re: Is 1/(a - b) > b - a ?   [#permalink] 28 Mar 2018, 03:30
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