If abc ≠0, is a < b < c ?
(1) 1/c < 1/b < 1/a
(2) c > a
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a < b < c
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j_shreyans
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GMATGuruNY
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Statement 1: 1/c < 1/b < 1/aj_shreyans wrote:If abc ≠0, is a < b < c ?
(1) 1/c < 1/b < 1/a
(2) c > 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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Another approach!
Start with the easier statement. S2 tells us nothing about b, so it's clearly INSUFFICIENT.
S1 gives us 1/c < 1/b < 1/a, so we know that 1/c < 1/a.
Now let's consider two possibilities.
#1: a > c
If a > c and 1/a > 1/c, then a > 0 and 0 > c. In this case, c > b > a is impossible.
#2: c > a
In this case, c > b > a IS possible.
Since we could have either a > c or c > a, S1 is INSUFFICIENT.
Now let's take the two together.
If 1/a > 1/c and c > a, we must have either c > a > 0 or 0 > c > a.
In the first case, all three numbers are positive. If 1/a > 1/b > 1/c > 0, then c > b > a > 0.
In the second case, all three numbers are negative. If 0 > 1/a > 1/b > 1/c, then 0 > c > b > a.
So in both cases, c > b > a, and the two statements together are SUFFICIENT.
Start with the easier statement. S2 tells us nothing about b, so it's clearly INSUFFICIENT.
S1 gives us 1/c < 1/b < 1/a, so we know that 1/c < 1/a.
Now let's consider two possibilities.
#1: a > c
If a > c and 1/a > 1/c, then a > 0 and 0 > c. In this case, c > b > a is impossible.
#2: c > a
In this case, c > b > a IS possible.
Since we could have either a > c or c > a, S1 is INSUFFICIENT.
Now let's take the two together.
If 1/a > 1/c and c > a, we must have either c > a > 0 or 0 > c > a.
In the first case, all three numbers are positive. If 1/a > 1/b > 1/c > 0, then c > b > a > 0.
In the second case, all three numbers are negative. If 0 > 1/a > 1/b > 1/c, then 0 > c > b > a.
So in both cases, c > b > a, and the two statements together are SUFFICIENT.
