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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 08 Apr 2006, 13:51
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A
B
C
D
E

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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.]
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 [#permalink] New post 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.

So answer is E
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 [#permalink] New post 08 Apr 2006, 15:09
1 is definitely sufficient and I think 2 is as well. Therefore D
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 [#permalink] New post 08 Apr 2006, 15:10
Please give explanations to your answers; anyways one of them is wrong
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simply A [#permalink] New post 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

so the answer is A.
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Re: simply A [#permalink] New post 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

so the answer is A.


Thanks;
How did you come up with this 1+z^2 >0: does it mean then when x/y<0, x>0
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Re: simply A [#permalink] New post 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

so the answer is A.


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
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Re: simply A [#permalink] New post 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

so the answer is A.


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..
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Post subject: DS a and b [#permalink] New post 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 :shock:
Idea :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.

So answer choice is

D

Heman & the master of the universe

Last edited by heman on 19 Jul 2006, 05:02, edited 1 time in total.
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Re: Post subject: DS a and b [#permalink] New post 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 :shock:
Idea :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.

So answer choice is

D

Heman & the master of the universe


Thanks Heman that was awesome..
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Re: Post subject: DS a and b [#permalink] New post 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 :oops:

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

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

Sorry...
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Re: Post subject: DS a and b [#permalink] New post 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.
_
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Re: Post subject: DS a and b [#permalink] New post 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.
Hence A answer.


jodeci, Are you sure about OA.
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Post subject: DS a and b [#permalink] New post 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)

With your example

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.

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


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Re: Post subject: DS a and b [#permalink] New post 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).
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Re: Post subject: DS a and b [#permalink] New post 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.

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Re: Post subject: DS a and b [#permalink] New post 12 Apr 2006, 13:21
I am sorry I meant

ccax , gmat_crack's solution Which is A is right.
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Re: Post subject: DS a and b [#permalink] New post 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
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 [#permalink] New post 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.. :) )
  [#permalink] 13 Apr 2006, 07:32
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