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Source: — Data Sufficiency |
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by Harbinder » Fri Oct 23, 2009 7:37 pm
is it E
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by xcusemeplz2009 » Fri Oct 23, 2009 7:44 pm
IMO C

statement 1) gives x=3,1/3 not suff

statement 2) gives x#3 and x#-3

combining x=1/3

hence mod X<1

OA pls
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by sanjana » Sat Oct 24, 2009 6:37 am
xcusemeplz2009 wrote:IMO C

statement 1) gives x=3,1/3 not suff

statement 2) gives x#3 and x#-3

combining x=1/3

hence mod X<1

OA pls
How does |x-3|#0 imply that x#3 and x#-3??

If x=-3 |x-3| = |-3-3| = |-6| = 6#0

even I think its E.
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by life is a test » Sat Oct 24, 2009 7:08 am
sanjana wrote:
xcusemeplz2009 wrote:IMO C

statement 1) gives x=3,1/3 not suff

statement 2) gives x#3 and x#-3

combining x=1/3

hence mod X<1

OA pls
How does |x-3|#0 imply that x#3 and x#-3??

If x=-3 |x-3| = |-3-3| = |-6| = 6#0

even I think its E.
1) |x + 1| = 2|x - 1|
x+1 = 2x-2 or x+1 = -2x+1 --> x = 3 or 1/3

2) says that x cannot =3 hence it must = 1/3 so IMO C
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by mehravikas » Tue Oct 27, 2009 7:49 pm
1) |x + 1| = 2|x - 1|
x+1 = 2x-2 or x+1 = -2x+1 --> x = 3 or 1/3

If x is -ve. shouldn't the equation be -
-x - 1 = -2x+1 ?????

Please correct if I am wrong?

life is a test wrote:
sanjana wrote:
xcusemeplz2009 wrote:IMO C

statement 1) gives x=3,1/3 not suff

statement 2) gives x#3 and x#-3

combining x=1/3

hence mod X<1

OA pls
How does |x-3|#0 imply that x#3 and x#-3??

If x=-3 |x-3| = |-3-3| = |-6| = 6#0

even I think its E.
1) |x + 1| = 2|x - 1|
x+1 = 2x-2 or x+1 = -2x+1 --> x = 3 or 1/3

2) says that x cannot =3 hence it must = 1/3 so IMO C
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