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# IF a,b,c are distinct non zero nos, is (a+b)^2 * (b-c)

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Joined: 11 May 2008
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IF a,b,c are distinct non zero nos, is (a+b)^2 * (b-c) [#permalink]

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06 Aug 2008, 05:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

Last edited by arjtryarjtry on 06 Aug 2008, 09:40, edited 1 time in total.
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Joined: 29 Aug 2007
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06 Aug 2008, 06:51
arjtryarjtry wrote:
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^2

1) a > b
2) b > c

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^2

or, (b-c)/(a-b)^3 >=0?

1: not suff as no value for c.
2: not suff as no value for a.

from 1 and 2: since a > b > c, (b-c) and (a-b)^3 both are +ve. suff..

C.
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Senior Manager
Joined: 06 Apr 2008
Posts: 428

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06 Aug 2008, 06:58
arjtryarjtry wrote:
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^2

1) a>b
2)b>c

The numerator will be +ve when b>c and denominator will be +ve when a>b

This is satisfied by both statements together so its C)
Manager
Joined: 27 Mar 2008
Posts: 64

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06 Aug 2008, 07:25
Should be C, as nothing can be cancelled out from the fraction, and we need to know the relationship b/w (a& b) AND (b&c) to determine what will be the sign of the result
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Joined: 07 Nov 2007
Posts: 1789
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06 Aug 2008, 08:50
arjtryarjtry wrote:
hmm.. wrong...

I got it..
it should be E then
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Joined: 11 May 2008
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06 Aug 2008, 09:39
x2suresh wrote:
arjtryarjtry wrote:
hmm.. wrong...

I got it..
it should be E then

sorry guys, slight mistake in the problem.. it should be cube, not square in the second term of Dr i.e (b-c)^3
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06 Aug 2008, 10:00
arjtryarjtry wrote:
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

1/(a-b)>=0?

1) a>b
a-b>0 1/(A-B)>0

A is suffcient.

2) not sufficient .

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06 Aug 2008, 10:25
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

Problem can be written as :

(a+b)^2>=0 ?
(a-b)^3 * (b-c)^2

S1. we do not know whether a and b are + or -ve.
numerator will always be +ve
(b-c)^2 will be +ve even though we do not know the value of c

NOW
even if a is +ve and b is -ve we will still get (a-b)^3 as positive ex: (2- (-5))^3. Even if a is -ve and a>b we will get a +ve value for (a-b)^3 ex: (-1 -(-2)). And even if both are =ve we will ge a +ve value for (a-b)^3. SUFF. BCE are out adn AD stays

S2. Tells you about b and c only and not about the relationship b/w a and b

(a+b)^2_ will always be +ve and so will (b-c)^2
However, (a-b)^3 can be +ve or -ve
ex. (2-3)^3 = -ve
(3 -2)^ 3 = +ve
hence MAYBE

Ans A
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Joined: 17 Jun 2008
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06 Aug 2008, 17:33
arjtryarjtry wrote:
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

the equation above depends on sign of a-b only since odd power of a-b hence (1) is SUFFI
(2) is not

IMO A
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Joined: 17 Jun 2008
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06 Aug 2008, 17:36
spriya wrote:
arjtryarjtry wrote:
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

the equation above depends on sign of a-b only since odd power of a-b hence (1) is SUFFI
(2) is not

IMO A

oops I misread the question again .. it depends on a-b as well as b-c
hence both (1) and (2) are important to solve
hence IMO C
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Joined: 15 Jul 2008
Posts: 205

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07 Aug 2008, 03:41
I think it should be C

You also need to ensure that a <> b and b <> c. The question statement only states that they all are non zero.

1) Ensures that a-b is +ve, but says nothing about b<>c. The possible options are > 0 and undefined (if b=c)
2) makes sure that b-c <> 0. Now we know that there is only one option >0
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Joined: 07 Nov 2007
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07 Aug 2008, 05:15
bhushangiri wrote:
I think it should be C

You also need to ensure that a <> b and b <> c. The question statement only states that they all are non zero.

1) Ensures that a-b is +ve, but says nothing about b<>c. The possible options are > 0 and undefined (if b=c)
2) makes sure that b-c <> 0. Now we know that there is only one option >0

Question says.. distinct non zero numbers
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07 Aug 2008, 05:29
arjtryarjtry wrote:
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

(a+b)^2 must be >=0
(b-c)/[(b-c)^3] = 1/[(b-c)^2] must be >0

So only the term (a-b)^3 will determine whether the fraction > or < = 0

(1) sufficient:
a > b
then (a-b)^3 > 0
then the fraction > 0

(2) NOT sufficient
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Joined: 15 Jul 2008
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07 Aug 2008, 06:50
x2suresh wrote:
bhushangiri wrote:
I think it should be C

You also need to ensure that a <> b and b <> c. The question statement only states that they all are non zero.

1) Ensures that a-b is +ve, but says nothing about b<>c. The possible options are > 0 and undefined (if b=c)
2) makes sure that b-c <> 0. Now we know that there is only one option >0

Question says.. distinct non zero numbers

that was a good one...
Re: distinct nos..   [#permalink] 07 Aug 2008, 06:50
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# IF a,b,c are distinct non zero nos, is (a+b)^2 * (b-c)

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