If \(\dfrac{a-b}{c}<0\), is \(a>b?\)
(1) \(c < 0\)
(2) \(a + b < 0\)
[spoiler]OA=A[/spoiler]
Source: Manhattan GMAT
(1) \(c < 0\)
(2) \(a + b < 0\)
[spoiler]OA=A[/spoiler]
Source: Manhattan GMAT
Given: \(\dfrac{a-b}{c}<0\)M7MBA wrote:If \(\dfrac{a-b}{c}<0\), is \(a>b?\)
(1) \(c < 0\)
(2) \(a + b < 0\)
[spoiler]OA=A[/spoiler]
Source: Manhattan GMAT
We have to determine whether \(a>b\).
Since \(\dfrac{a-b}{c}<0\), we see that \(\dfrac{a-b}{c}\) is a negative number. It can happen in two ways:
1. a > b and c < 0, making \(\dfrac{a-b}{c}\) a negative number.
or
2. a < b and c > 0, making \(\dfrac{a-b}{c}\) a negative number.
Let's take each statement one by one.
(1) \(c < 0\)
Since we have \(c < 0\); thus, Case 1 is as discussed above is applicable and a > b. Sufficient.
(2) \(a + b < 0\)
Both cases discussed above are applicable; thus, we cannot conclude that a > b. Insufficient.
The correct answer: A
Hope this helps!
-Jay
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