DS - GMATPrep (negative or positive problem)

Question

Any ideas, guys?....

SysAdmin · Answer

Since 1 and 2 suggests that any number before and after 1/k is greater than 0, isn't 1/k always greater than 0? 

How about D?

Fig · Answer

For me (A)

1/k > 0 => k > 0 ? Notice that: o k could be any real number. o k-1 < k < k+1 From 1 1/(k-1) > 0 => k-1 > 0 So, 0 < k-1 < k < k+1 SUFF. From 2 1/(k+1) > 0 => k+1 > 0 Here, we cannot conclude because we could have: o 0 < k-1 < k < k+1 o k-1 < 0 < k < k+1 o k-1 < k < 0 < k+1 To be sure, try to pick numbers: o k = -0,5 : 1/(-0,5+1) > 0 but k < 0 so is 1/k. o k = 0,5 : 1/(0,5+1) > 0 and 1/k > 0. INSUFF.

querio · Answer

Another A)


Stmt 1)
From the condition 1/(k-1)>0 we know that k must be positive, therefore 1/k>0 --> Suff

Stmt 2)
the condition allows positive/negative values for k. The trick here is to realize that k DOES NOT NECESSARILY NEED TO BE AN INTEGER.
if k=-0.5, then 1/(-0.5+1)>0 (satisfies the condition) and 1/k<0
if k=8, then 1/(8+1)>0 (satisfies the condition) but 1/k>0
---> Insuff

Andr359 · Answer

That 1/k>0 and k is not 0, 1, or -1 implies that k is in the range:0< <1 U 1<. Therefore

(1) k is in the range 1< <2 U 2<, which is a subset of the original range of the question, then (1) is suff; A or D.

(2) Similarly to (1): k is in the range -1< 0 U 0<. In this case, k may not fall in the range required by the question, then (2) is insuff and the answer is A.

nick_sun · Answer

Very good point, Fig! Indeed we need not forget about fractions.. :wink:

DS - GMATPrep (negative or positive problem)

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