Is 1/ ( a-b ) < b-a? 1. a<b 2.1<|a-b : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 15:26

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is 1/ ( a-b ) < b-a? 1. a<b 2.1<|a-b

Author Message
Manager
Joined: 21 Dec 2005
Posts: 102
Followers: 1

Kudos [?]: 4 [0], given: 0

Is 1/ ( a-b ) < b-a? 1. a<b 2.1<|a-b [#permalink]

### Show Tags

08 Apr 2006, 13:51
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

1. a<b
2.1<|a-b|

[Edited by HongHu. Please change it back if it is not correct editing. Thanks.]
Senior Manager
Joined: 24 Jan 2006
Posts: 252
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

08 Apr 2006, 14:16
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

For 1. a<b=> a-b<1. Now, if 0<a-b<1 then 1/a-b!<b-a and if a-b<0 then
1/a-b<b-a So we can't realy say from 1

For 2. |a-b|>1. We can't say again.

Manager
Joined: 21 Mar 2006
Posts: 89
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

08 Apr 2006, 15:09
1 is definitely sufficient and I think 2 is as well. Therefore D
Manager
Joined: 21 Dec 2005
Posts: 102
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

08 Apr 2006, 15:10
Manager
Joined: 31 Mar 2006
Posts: 68
Followers: 1

Kudos [?]: 32 [0], given: 0

### Show Tags

08 Apr 2006, 15:22
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

put a-b = Z, then the question can be described as "Is 1/Z < -Z?"
Do a little caculation as below.

1/Z + Z < 0 => (1+Z^2)/Z < 0 here we know that 1+z^2 >0 so that we only need to find out whether Z < 0, that is a - b < 0.

from 1. a < b => a - b < 0 this is sufficent.
from 2. a-b < -1 or a - b > 1 this isn't sufficent

Manager
Joined: 21 Dec 2005
Posts: 102
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

08 Apr 2006, 15:36
antiant wrote:
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

put a-b = Z, then the question can be described as "Is 1/Z < -Z?"
Do a little caculation as below.

1/Z + Z < 0 => (1+Z^2)/Z < 0 here we know that 1+z^2 >0 so that we only need to find out whether Z < 0, that is a - b < 0.

from 1. a < b => a - b < 0 this is sufficent.
from 2. a-b < -1 or a - b > 1 this isn't sufficent

Thanks;
How did you come up with this 1+z^2 >0: does it mean then when x/y<0, x>0
Manager
Joined: 31 Mar 2006
Posts: 68
Followers: 1

Kudos [?]: 32 [0], given: 0

### Show Tags

08 Apr 2006, 21:27
jodeci wrote:
antiant wrote:
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

put a-b = Z, then the question can be described as "Is 1/Z < -Z?"
Do a little caculation as below.

1/Z + Z < 0 => (1+Z^2)/Z < 0 here we know that 1+z^2 >0 so that we only need to find out whether Z < 0, that is a - b < 0.

from 1. a < b => a - b < 0 this is sufficent.
from 2. a-b < -1 or a - b > 1 this isn't sufficent

Thanks;
How did you come up with this 1+z^2 >0: does it mean then when x/y<0, x>0

1>0 and z^2 >0 => 1+z^2 >0
Manager
Joined: 21 Dec 2005
Posts: 102
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

08 Apr 2006, 21:32
antiant wrote:
jodeci wrote:
antiant wrote:
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

put a-b = Z, then the question can be described as "Is 1/Z < -Z?"
Do a little caculation as below.

1/Z + Z < 0 => (1+Z^2)/Z < 0 here we know that 1+z^2 >0 so that we only need to find out whether Z < 0, that is a - b < 0.

from 1. a < b => a - b < 0 this is sufficent.
from 2. a-b < -1 or a - b > 1 this isn't sufficent

Thanks;
How did you come up with this 1+z^2 >0: does it mean then when x/y<0, x>0

1>0 and z^2 >0 => 1+z^2 >0

You changed the sign
1/Z + Z < 0 => (1+Z^2)/Z < 0 here we know that 1+z^2 >0

How could this be..
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

Post subject: DS a and b [#permalink]

### Show Tags

09 Apr 2006, 14:13
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

Alright First off : Here is an important rule with absolute values
if |x| < a then it is always true that -a<x<a
if |x| > a then it is always true that x>a & x <-a

Memorize the two important statments above. It will apply to all problems with absolute values.

If you to know how we derived it check the link below
http://www.purplemath.com/modules/absineq.htm

Now getting to the problem

Is 1/a-b < b-a?
The question asks us if we can answer the above inequality with a resounding Yes or a resounding No.

Lets simplify the question stem first
Rewrite the question stem as (1/a-b) < (b-a/1)
Therefore cross multiplying we get
1/(a-b)(b-a) < 1

Now the new question stem -> Is 1/(a-b)(b-a) <1

Good Now lets look at Statement 1
a<b

if a< b --> a-b is -ve and b-a is +ve
If you are not sure plug in
a=2, b=3 implies 2-3(i.e a-b) is -ve and (3-2)(i.e b-a) is +ve
a=-4, b=-3 implies -4-(-3)(i.e a-b) is -ve and -3-(-4) which is b-a is +ve

What do we have in the Denominator of the new question stem we have one a-b term and one b-a term. So when a<b a-b is -ve and b-a is +ve so the product of the two is always -ve.

So in our new question stem the Left hand side is always negative which answers the question that it is < 1.
So statement 1 is good.

Ok now White wash your memory so that you forget statement 1

Lets looks at what we have in statement 2.
1<|a-b|
Absolute values and inequalities
Idea Remember what we said about absolute values above
So from statement 2 --> a-b > 1 and a-b < -1 . In otherwords we have two ranges of solutions. Lets take them one at a time

a-b>1 implies a>b. So in In our new question stem a-b is +ve and b-a is -ve . Therefore the product of a-b x b-a is -ve . hence the answer is less than 1.

Take a-b< -1 . Lets multiply both sides by -1> remember to flip the sign of the inequality when multiplying by -ve so we have
-(a-b) > 1 which is b-a > 1 which is b> a.
So with this in our new question stem b-a is +ve and a-b is -ve. So the product of a-b x b-a is -ve. Hence the answer is -ve which is less than 1.

Therefore statement 2 is sufficient as well.

D

Heman & the master of the universe

Last edited by heman on 19 Jul 2006, 05:02, edited 1 time in total.
Manager
Joined: 21 Dec 2005
Posts: 102
Followers: 1

Kudos [?]: 4 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

09 Apr 2006, 14:19
heman wrote:
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

Alright First off : Here is an important rule with absolute values
if |x| < a then it is always true that -a<x<a
if |x| > a then it is always true that x>a & x <-a

Memorize the two important statments above. It will apply to all problems with absolute values.

If you are a curious cat and want to know how we derived it check the link below
http://www.purplemath.com/modules/absineq.htm

Now getting to the problem

Is 1/a-b < b-a?
The question asks us if we can answer the above inequality with a resounding Yes or a resounding No.

Lets simplify the question stem first
Rewrite the question stem as (1/a-b) < (b-a/1)
Therefore cross multiplying we get
1/(a-b)(b-a) < 1

Now the new question stem -> Is 1/(a-b)(b-a) <1

Good Now lets look at Statement 1
a<b

if a< b --> a-b is -ve and b-a is +ve
If you are not sure plug in
a=2, b=3 implies 2-3(i.e a-b) is -ve and (3-2)(i.e b-a) is +ve
a=-4, b=-3 implies -4-(-3)(i.e a-b) is -ve and -3-(-4) which is b-a is +ve

What do we have in the Denominator of the new question stem we have one a-b term and one b-a term. So when a<b a-b is -ve and b-a is +ve so the product of the two is always -ve.

So in our new question stem the Left hand side is always negative which answers the question that it is < 1.
So statement 1 is good.

Ok now White wash your memory so that you forget statement 1

Lets looks at what we have in statement 2.
1<|a-b|
Absolute values and inequalities
Idea Remember what we said about absolute values above
So from statement 2 --> a-b > 1 and a-b < -1 . In otherwords we have two ranges of solutions. Lets take them one at a time

a-b>1 implies a>b. So in In our new question stem a-b is +ve and b-a is -ve . Therefore the product of a-b x b-a is -ve . hence the answer is less than 1.

Take a-b< -1 . Lets multiply both sides by -1> remember to flip the sign of the inequality when multiplying by -ve so we have
-(a-b) > 1 which is b-a > 1 which is b> a.
So with this in our new question stem b-a is +ve and a-b is -ve. So the product of a-b x b-a is -ve. Hence the answer is -ve which is less than 1.

Therefore statement 2 is sufficient as well.

D

Heman & the master of the universe

Thanks Heman that was awesome..
Manager
Joined: 20 Nov 2004
Posts: 108
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 09:20
heman wrote:
if |x| > a then it is always true that x>a & x <-a

Memorize the two important statments above.

I hope nobody will memorize this statement, since it's
certainly not true

Just consider e. g. x = -10 and a = 1

Then |x| > a, but x<a.

Sorry...
Manager
Joined: 20 Nov 2004
Posts: 108
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 10:06
heman wrote:
Rewrite the question stem as (1/a-b) < (b-a/1)
Therefore cross multiplying we get
1/(a-b)(b-a) < 1

Now the new question stem -> Is 1/(a-b)(b-a) <1

Sorry, but what are you doing?

First, the question reads 1/a-b < b-a and NOT 1/(a-b) < b-a.
And even if the latter were true, there's nothing to multiply with
since you don't now if b-a is negative. If it were, you had to swop
the inequality sign from < to >.

And D is definitely not true.

If we take the question as it's written, it is

Is 1/a-b < b-a?

Consider a = 3 and b = 5:
(1) and (2) are satisfied and the answer is yes.

Consider a=0.01 and b=5:
(1) and (2) are satisfied and the answer is no.

So it is E.

==

If the question were 1/(a-b) < b-a:

(1)
If a<b, then a-b<0 and therefore 1/(a-b) negative.
If a<b, then b-a is positive.

So (1) is sufficient.
_

(2)
Consider a = 1 and b = 5:
(2) is satisfied and the answer is yes.

Consider a = 7 and b = 5:
(2) is satisfied and the answer is no.
--> Not sufficient

So in this case, the answer were A.
_
Senior Manager
Joined: 22 Nov 2005
Posts: 476
Followers: 2

Kudos [?]: 20 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 11:35
heman wrote:
Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

Alright First off : Here is an important rule with absolute values
if |x| < a then it is always true that -a<x<a
if |x| > a then it is always true that x>a & x <-a

I agree that St. 1 is correct

For the second here is the my version

1/(a-b) < b-a

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

Now from st. two 1 <|a -b|
---> |a -b| > 1

------> a-b < -1 and a-b > 1

case 1 - (a - b) < -1
then from (A) (a-b) < -1 hence A will be -ve which is true.

Case 2--- (a -b) > 1
then (A) will be +ve, which is False.

hence B is not sufficient.

jodeci, Are you sure about OA.
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 12:31
ccax
I said
if |x| > a then it is always true that x>a & x < -a(read negative of a)

Just consider e. g. x = -10 and a = 1

Then |x| > a, ------->you got that right

but x<a. -----> I said x < -a which is -10 < -1 which is true.

Heman
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

Post subject: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 12:46
Jodeci since ccax brought up the point

Is the ?

1/(a-b) < b-a

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

Heman
Manager
Joined: 20 Nov 2004
Posts: 108
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 13:00
heman wrote:
but x<a. -----> I said x < -a which is -10 < -1 which is true.

Heman,

you said that if
|x| > a then it is always true that x>a & x < -a

but in the case of e. g. x = -10 and a = 1 (where |x | > a),
then x < -a (as you said) BUT x NOT > a since x < a
(and that's the opposite of what you said).
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 13:19
ccax Now I see the point

Can we say this : If |x| > a then x has 2 range of solutions
Either x > a or x < -a

Also I after reviewing my inequalities I see that I should not have cross multiplied. In that case

ccax , gmat_crack's solution Which is B is right.

Heman
Manager
Joined: 20 Mar 2006
Posts: 200
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

12 Apr 2006, 13:21
I am sorry I meant

ccax , gmat_crack's solution Which is A is right.
Manager
Joined: 20 Nov 2004
Posts: 108
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: Post subject: DS a and b [#permalink]

### Show Tags

13 Apr 2006, 06:49
heman wrote:
Can we say this : If |x| > a then x has 2 range of solutions
Either x > a or x < -a

That's true, if you consider the OR in a not-excluding sense (like and/or)
because it's also possible that both are true.

For that case, consider e. g. x = -1 and a = -2:
|x| > a
1 > -2

x > a
-1 > -2

x < -a
-1 < 2
Senior Manager
Joined: 05 Jan 2006
Posts: 382
Followers: 1

Kudos [?]: 84 [0], given: 0

### Show Tags

13 Apr 2006, 07:32
I am getting A as answer...could not justify B...

Is 1/a-b < b-a?

1. a<b
2.1<|a-b|

1 a<b => a-b<0 and -(a-b)>0
so a-b is negative and b-a is positive
=> 1/a-b is negative and b-a is positive
=> 1/a-b < b-a
sufficient

2) 1<|a-b|
=> 1<a-b or 1<-(a-b)
=> 1<a-b or 1>(b-a)

We need to take one euqation at a time and see weather they come to same conclusion

1<a-b => b-a>-1
so a-b can be negative or positive so does b-a for different int or fractional values hence insufficient...

same analysis for 1>b-a

so A sufficient B not sufficient...

(If OA is D I may have to rethink my analysis of B.. )
13 Apr 2006, 07:32

Go to page    1   2    Next  [ 24 posts ]

Display posts from previous: Sort by