j_shreyans wrote:If abc ≠0, is a < b < c ?
(1) 1/c < 1/b < 1/a
(2) c > a
Statement 1: 1/c < 1/b < 1/a
In light of statement 2, test one case where c>a and one case where a<c.
Case 1: c=3, b=2, and a=1
The constraint that 1/c < 1/b < 1/a is satisfied:
1/3 < 1/2 < 1.
In this case, a<b<c.
Case 2: c=-1, b=2, and a=1
The constraint that 1/c < 1/b < 1/a is satisfied:
-1 < 1/2 < 1.
In this case, c<a<b.
INSUFFICIENT.
Statement 2: c > a
No information about b.
INSUFFICIENT.
Statements combined:
a < c and 1/a > 1/c.
Implication:
a and c must have the SAME SIGN, as in the following cases:
a=1, c=3.
a=1/3, c=1.
a=-3, c=-1.
a=-1, c=-1/3.
Since 1/c < 1/b < 1/a -- and a and c must have the same sign -- ALL 3 VALUES must have the same sign, as in the following cases:
a=1, b=2, c=3.
a=1/3, b=1/2, c=1.
a=-3, b=-2, c=-1.
a=-1, b=-1/2, c=-1/3.
In every case, a<b<c.
SUFFICIENT.
The correct answer is
C.
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