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PS: Always negative?

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02 Mar 2009, 15:33
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If |A| < |B| , which of the following numbers is always negative?

a) A/B - B/A

b) A-B/A+B

c) (A^B)-(B^A)

d) A (B/A-B)

e) B-A/B

Is there a fast way of solving these, other than plugging and chugging numbers?

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08 Mar 2009, 14:19
If |A| < |B| , which of the following numbers is always negative?

a) A/B - B/A
b) A-B/A+B
c) (A^B)-(B^A)
d) A (B/A-B)
e) B-A/B

Is there a fast way of solving these, other than plugging and chugging numbers?

I use POE not plugging-in nor chugging-in.

Given that: |A| < |B|

The clue: which of the following is always -ve?
So lets try finding +ve one.

a) A/B - B/A: If A and B both are -ve, the expression is +ve. out.
b) (A-B)/(A+B): It is in any case -ve...
c) (A^B)-(B^A): If A is -ve integer, and B is +ve integer, the expression is +ve. out.
d) A (B/(A-B)): If A and B both are -ve, the expression is +ve. out.
e) B - (A/B): If A is -ve (or even +ve) and B is +ve, the expression is +ve. out.

So B.
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Retired Moderator
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09 Mar 2009, 13:48
I really don't understand this part:

Given that: |A| < |B|
The clue: which of the following is always -ve?
So lets try finding +ve one.

GMAT TIGER wrote:
If |A| < |B| , which of the following numbers is always negative?

a) A/B - B/A
b) A-B/A+B
c) (A^B)-(B^A)
d) A (B/A-B)
e) B-A/B

Is there a fast way of solving these, other than plugging and chugging numbers?

I use POE not plugging-in nor chugging-in.

Given that: |A| < |B|

The clue: which of the following is always -ve?
So lets try finding +ve one.

a) A/B - B/A: If A and B both are -ve, the expression is +ve. out.
b) (A-B)/(A+B): It is in any case -ve...
c) (A^B)-(B^A): If A is -ve integer, and B is +ve integer, the expression is +ve. out.
d) A (B/(A-B)): If A and B both are -ve, the expression is +ve. out.
e) B - (A/B): If A is -ve (or even +ve) and B is +ve, the expression is +ve. out.

So B.

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Joined: 07 Nov 2007
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Location: New York

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11 Mar 2009, 22:22
If |A| < |B| , which of the following numbers is always negative?

a) A/B - B/A

b) A-B/A+B

c) (A^B)-(B^A)

d) A (B/A-B)

e) B-A/B

Is there a fast way of solving these, other than plugging and chugging numbers?

A-B/A+B

= {(A/B)-1 } / {(A/B)+1}

|A| < |B| --> means magnitude of A/B <1

A/B lies between -1 and 1
i.e -1<A/B<1

= {(A/B)-1 } / {(A/B)+1}

clearly neumarotr -ve for that range
denominator +ve for that range.. it (B) is always negative.

B
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Name: Ronak Amin
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14 Mar 2009, 01:54
Hi GmatTiger,

I could not understand: a) A/B - B/A: If A and B both are -ve, the expression is +ve. out.

if A and B are both -ve, A/B is +ve, and B/A is +ve. Now,
A/B - B/A should be -ve because...magnitude of A is less than B, which means A/B < B/A !!

Similarly,
c) (A^B)-(B^A): If A is -ve integer, and B is +ve integer, the expression is +ve. out.

Here, A^B can be -ve or +ve depending upon B is odd or even,
B^A will be 1/B^A (as A is -ve) , How can u conclude that the expression is +ve ??

GMAT TIGER wrote:
If |A| < |B| , which of the following numbers is always negative?

a) A/B - B/A
b) A-B/A+B
c) (A^B)-(B^A)
d) A (B/A-B)
e) B-A/B

Is there a fast way of solving these, other than plugging and chugging numbers?

I use POE not plugging-in nor chugging-in.

Given that: |A| < |B|

The clue: which of the following is always -ve?
So lets try finding +ve one.

a) A/B - B/A: If A and B both are -ve, the expression is +ve. out.
b) (A-B)/(A+B): It is in any case -ve...
c) (A^B)-(B^A): If A is -ve integer, and B is +ve integer, the expression is +ve. out.
d) A (B/(A-B)): If A and B both are -ve, the expression is +ve. out.
e) B - (A/B): If A is -ve (or even +ve) and B is +ve, the expression is +ve. out.

So B.

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15 Mar 2009, 06:56
That is because if we find all possible +ves, we can rule out those +ves so that the remaining is -ve.
I really don't understand this part:

Given that: |A| < |B|
The clue: which of the following is always -ve?
So lets try finding +ve one.

_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 845 [0], given: 19

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15 Mar 2009, 08:07
Thanks for pointing out that A and C are not complete in my previous posts. I thought correctly but wrote incorrectly. A has typo but C should be revised.

Economist wrote:
Hi GmatTiger,

I could not understand: a) A/B - B/A: If A and B both are -ve, the expression is +ve. out.

if A and B are both -ve, A/B is +ve, and B/A is +ve. Now,
A/B - B/A should be -ve because...magnitude of A is less than B, which means A/B < B/A !!

Similarly,
c) (A^B)-(B^A): If A is -ve integer, and B is +ve integer, the expression is +ve. out.

Here, A^B can be -ve or +ve depending upon B is odd or even,
B^A will be 1/B^A (as A is -ve) , How can u conclude that the expression is +ve ??

If |A| < |B| , which of the following numbers is always negative?

a) A/B - B/A
b) A-B/A+B
c) (A^B)-(B^A)
d) A (B/A-B)
e) B-A/B

Is there a fast way of solving these, other than plugging and chugging numbers?

Given that: |A| < |B|

The clue: which of the following is always -ve?
So lets try finding +ve one.

a) A/B - B/A: If one is +ve and the other is -ve, the expression is +ve. out.
b) (A-B)/(A+B): It is in any case -ve...
c) (A^B)-(B^A): If A is +ve odd integer, and B is -ve integer, the expression is +ve. out.

For ex: A = 3 and B = -4:
(A^B)-(B^A) = 3^(-4) - (-4)^3 = 1/81 - (-256) = 256 approx.

d) A (B/(A-B)): If A and B both are -ve, the expression is +ve. out.
e) B - (A/B): If A is -ve (or even +ve) and B is +ve, the expression is +ve. out.

So B.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

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15 Mar 2009, 08:32
I think the quickest and safest way is by plugging in numbers:

Consider 4 possibilities for each of them:

A=-1, B=-5
A=1, B=5
A=-1, B=5
A=1, B=-5

Just plug them in for each of them and you will find B gives a negative answer for each of the cases.

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Re: PS: Always negative?   [#permalink] 15 Mar 2009, 08:32
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