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manhattan challenge problem_DS_is n < 1

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by cramya » Sat May 02, 2009 6:59 am
I am getting E

Stmt I

nx-n<0

n(x-1)<0

n->positive greater than 1 and x-1<0 then NO n is not less than 1

n-> negative and x-1>0 then YES n<1

INSUFF

Stmt II

x-1 = -2

No idea what n could be

TOGETHER:

n (x-1) < 0

x-1 = -2

-2n < 0

All we can say is n>0
n = 10000000 NO
n=1/2 YES


Must be missing a trick if its C

Regards,
CR
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by dumb.doofus » Sat May 02, 2009 7:05 am
Just my opinion..

I think the above problem makes more sense if

a) = nx - x instead of nx - n

if it is nx - x then the answer is definitely C..

Can you confirm the above please?
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by abhinav85 » Sat May 02, 2009 7:21 am
IMO E for me too.

To find out if n<1 we have to find out the value of n?

From A we cannot say that n is positive or negative..becoz we
don't know the value of x.(insufficient)

From B we get the value of x that is -1.(insufficient)

Now when we take them together and put the value of
x in nx-n< 0
= -n-n < 0
=-2n<0

now if we take n as negative that is say -1 we get n> 0.

and if we take n as postive say 1 we get n < 0.
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by nh8404052006 » Sat May 02, 2009 7:26 am
From statement (2) ==>x = &#8211; 1 , but no info abt n ==> insufficient
From statement (1) ==> n (x &#8211; 1 ) < 0 . If n = &#8211; 1 and x =2 , n (x &#8211; 1 ) < 0 but when n = 2 and x = 0 , then n (x &#8211; 1 ) < 0 ==> insufficient
Both statements together ==>if x = &#8211; 1, and n (x &#8211; 1 ) < 0 ==> n > 0 ==>sufficient
Why? when the red portion is negative, the blue portion MUST be positive; otherwise, the inequaility( <0 )does not hold
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