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mirhaque
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disagree with OA 19 Jul 2005, 19:18
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disagree with OA



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Ozmba
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disagree with OA 19 Jul 2005, 20:56
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1. X is an integer ..Nothing can be said

2.X2 < 1 ..Possible only when X is between 0 -1 or x = 0

In either of the case mod{x-6} is always greater than 5 ..

So Answer is B...correct me if i am wrong

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forrestgump
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disagree with OA 20 Jul 2005, 09:51
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although B stands - i would say that the value of x can lie between -1 and 1 ( exclusive of those values)....

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Ozmba2006
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Joined: 14 Feb 2005
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Posted: Tue Jul 19, 2005 7:56 pm Post subject:

--------------------------------------------------------------------------------

1. X is an integer ..Nothing can be said

2.X2 < 1 ..Possible only when X is between 0 -1 or x = 0

In either of the case mod{x-6} is always greater than 5 ..

So Answer is B...correct me if i am wrong

Cheers
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mandy
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disagree with OA 20 Jul 2005, 16:46
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plz what is the OA
I agree that st 1 NOT SUFFICIENT


ST2
x^2 <1
x<sq root of 1
x<- sq root of 1
hence x<-1
x <+1
then i am stuck
can anyone correct me thanks
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FN
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disagree with OA 21 Jul 2005, 09:11
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Hey

this is B....


|x-6| >5?

1) x is an integer than x=4

|4-6|=2 which is <5

x can be 20

|20-6|= 14 which is > 5

insuff

2)

X^2<1

then X is a fraction

say X is 1/2

|1/2-6| = 5.5 which is >5

or X could be 0

|0-6| = 6 which is > 5...

so either way we can that (2) is always sufficient...it is always >5

B it is...
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disagree with OA 21 Jul 2005, 21:05
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B

x>11 or x<1

with x^2 < 1, for sure x < 1
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HowManyToGo
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disagree with OA 22 Jul 2005, 07:06
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mandy
plz what is the OA
I agree that st 1 NOT SUFFICIENT


ST2
x^2 <1
x<sq root of 1
x<- sq root of 1
hence x<-1
x <+1
then i am stuck
can anyone correct me thanks
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so x should be b/w -1 and 1.
Then for any such value | x - 5 | > 6

HMTG.
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HIMALAYA
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disagree with OA 22 Jul 2005, 20:17
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mirhaque
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However, i answerd B as the OA, the question is poorly worded. suppose if statement ii is not sufficient to answer the question, statements i and ii jointly cannot answer the question because these statement contradict each other.



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Hi there,
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