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# ds: abs value

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06 Apr 2005, 13:35
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pls explain
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06 Apr 2005, 14:39
E?
from stat 1=> a^2 = b^2.....insuff take a=1 b=1 and answer is a=|b| yes
take a=-1 and b=1 and answer is a=|b| no
from stat 2 => b=|a| doesn't mean that a=|b|
combine 1 and 2 (seems to me that the statements are saying nothing substantially different)
hope it's E (I'm a bit tired to combine the statements,maybe it will be better if I go to bed )

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Director
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06 Apr 2005, 15:54
would go with A

Can you please share the source of the questions you ask? They are good

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06 Apr 2005, 18:03
Given that, a^2-b^2=b^2-c^2
a^2+c^2=2b^2

from (i), b=lcl, which means b^2=c^2. if so, then
a^2+c^2=2b^2
a^2+b^2=2b^2
a^2 =b^2
a = + or - b, which means a = lbl.............. sufficient.

from (ii), b= lal, which also means b^2=a^2
a^2+c^2=2b^2
b^2+c^2=2b^2
c^2=b^2. then again the same process as in (i)
a^2+c^2=2b^2
a^2+b^2=2b^2
a^2 =b^2
a = + or - b, which means a = lbl.............. sufficient.

for me, it seems D.

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06 Apr 2005, 18:13
E for me....

how on earth can statement 1 be sufficient??

from 1)

a^2-b^2=b^2-c^2

b=|c| we get
a^2-b^2=0 therefore a^2=b^2

how on earth can you extrapolate a=|b| from the above equation??

from 2)

b= |a|

a^2-b^2=b^2-c^2

a^2-a^2=b^2-c^2, therefore 0=b^2-c^2 which can be written as
b^2=c^2

E it is...

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06 Apr 2005, 19:40
Can reduce question to (a+b)(a-b) = (b+c)(b-c)

For 1)

b=c, or b=-c
positive case:
LHS becomes (a+c)(a-c) which is = RHS
negative case:
LHS becomes (a-c)(a+c) which is = RHS

so we know a=b and so a=|b|

Sufficient.

For 2)

b=a, or b=-a
We can put them as a=b, a=-b and so a=|b| hold for both.

2 is sufficient as well.

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06 Apr 2005, 19:59
I think E.
a lot of combination involved here

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06 Apr 2005, 20:03
ywilfred wrote:
Can reduce question to (a+b)(a-b) = (b+c)(b-c)

For 1)

b=c, or b=-c
positive case:
LHS becomes (a+c)(a-c) which is = RHS
negative case:
LHS becomes (a-c)(a+c) which is = RHS

so we know a=b and so a=|b|

Sufficient.

For 2)

b=a, or b=-a
We can put them as a=b, a=-b and so a=|b| hold for both.

2 is sufficient as well.

Hi ywilfred. need clarification

In stmt 2

b= |a| is not the same as a = |b|

b=|a| => b=a if a >0 , b=-a if a<0
This means b is always positive

In a=|b|, a= b if b>0 , a = -b if b<0
This means b can be positive or negative while a is always positive

=> the two are not equivalent.
Hence Stmt 2 is not sufficient

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07 Apr 2005, 08:25
A) !

1) because...b=|c|...b^2-c^2=0 => because...a^2-b^2=b^2-c^2 => a^2-b^2 has to be 0 as well => a=b or a=|b| => suff

2) b=|a|...a is +ve or -ve...if +ve a=|b| but if -ve...-a<>|b| => insuff
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07 Apr 2005, 08:28
christoph wrote:
A) !

1) because...b=|c|...b^2-c^2=0 => because...a^2-b^2=b^2-c^2 => a^2-b^2 has to be 0 as well => a=b or a=|b| => suff

2) b=|a|...a is +ve or -ve...if +ve a=|b| but if -ve...-a<>|b| => insuff

if a^2=b^2 has 2 solutions -> a=b AND a=-b
this is a typical ETS trick

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07 Apr 2005, 08:47
thearch wrote:
christoph wrote:
A) !

1) because...b=|c|...b^2-c^2=0 => because...a^2-b^2=b^2-c^2 => a^2-b^2 has to be 0 as well => a=b or a=|b| => suff

2) b=|a|...a is +ve or -ve...if +ve a=|b| but if -ve...-a<>|b| => insuff

if a^2=b^2 has 2 solutions -> a=b AND a=-b
this is a typical ETS trick

you are right...its D)...thx!
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07 Apr 2005, 08:51
I still dont get it....Ithink it should be E,

we have 2 possible solutions from each statement that repeat!

christoph wrote:
thearch wrote:
christoph wrote:
A) !

1) because...b=|c|...b^2-c^2=0 => because...a^2-b^2=b^2-c^2 => a^2-b^2 has to be 0 as well => a=b or a=|b| => suff

2) b=|a|...a is +ve or -ve...if +ve a=|b| but if -ve...-a<>|b| => insuff

if a^2=b^2 has 2 solutions -> a=b AND a=-b
this is a typical ETS trick

you are right...its D)...thx!

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07 Apr 2005, 09:14

1 st statement is equivalent to a = +-b so a >0 or <0 we can not conclude.

2st statement leads to this question : is |b|=a equivalent to b=|a|, answer is no.

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07 Apr 2005, 09:31
Don't be fooled by the seemly complexity of this question.

The question asks if a=|b|. With absolute value questions your first relection should be "do we know it's sign?" |b| is non negative. If we aren't able to determine if a is non negative then we can't determine if a=|b|.

Now look at the stem: a^2-b^2=b^2-c^2
and the two choices:
b=|c| and b=|a|
All we know is b is non negative. Do we know anything about a's sign? No!
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12 Apr 2005, 23:48
OA?????????????????????????

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13 Apr 2005, 04:19
HongHu wrote:
Don't be fooled by the seemly complexity of this question.

The question asks if a=|b|. With absolute value questions your first relection should be "do we know it's sign?" |b| is non negative. If we aren't able to determine if a is non negative then we can't determine if a=|b|.

Now look at the stem: a^2-b^2=b^2-c^2
and the two choices:
b=|c| and b=|a|
All we know is b is non negative. Do we know anything about a's sign? No!

It took me just over 60 secs to choose 'E'. Hong's approach reduces this to a less that 10 sec question. Good one, that!

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01 Jun 2005, 06:43
Sorry guys I don't have the OA. I guess Hong is correct. We don't know anything about A's sign. Hence it's E.

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01 Jun 2005, 10:16
HIMALAYA wrote:
Do you have OA?

I vote for E.

The stem reduces to (A^2 + C^2) = 2*B^2

with 1) b^2 = C^2, i.e A^2 = B ^2 No way to conclude that A is +/-.
with 2) B^2 = A^2 i.2 C^2 = B^2 ,Can't conclude if B is +/-

HMTG.

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03 Jun 2005, 13:25
mirhaque wrote:
pls explain

Think it'll be E
as the first statement only proves that a^2 = b^2
for a to equal |b|, we would have to be sure that a is positive which we can't be...
second statement doesn't give any new information
even with both together there is no way of knowing that a is positive

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03 Jun 2005, 13:39
HongHu wrote:
Don't be fooled by the seemly complexity of this question.

The question asks if a=|b|. With absolute value questions your first relection should be "do we know it's sign?" |b| is non negative. If we aren't able to determine if a is non negative then we can't determine if a=|b|.

Now look at the stem: a^2-b^2=b^2-c^2
and the two choices:
b=|c| and b=|a|
All we know is b is non negative. Do we know anything about a's sign? No!

Honghu, I am not sure if your approach to this particular question is right, because the question is really asking us whether a = b irrespective of their signs, so this can be deduced to is a^2 = b^2.

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03 Jun 2005, 13:39

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