swerve wrote:a, b, and c are three integers such that a < b < c. Is (a + b) less than c?

(1) ab is negative
(2) a + b is negative

The OA is D.

Please, can any expert explain this DS question for me? I don't understand why that is the correct answer. Thanks.

We have to determine whether (a + b) < c.

(1) ab is negative.

Since ab is negative, only one between a and b is negative. Since we know that a < b, the only possibility is that a is negative and b is positive. Let's determine the value of (a + b). Since a is negative and b is positive, we can deduce that (a + b) = b - |a|. Since b < c, we can deduce that (b - |a|) < c. Sufficient.

(2) a + b is negative.

Case 1: a and b both are negative.

Thus, a < b < c can be deduced as -|a| < -|b| < c. For this relationship to hold true, if c is a non-negative number, then a + b = -|a + b| < c. However if c is also a negative number, then a + b = -|a + b| < -|c|.

Case 2: Only one between a and b is negative. Since a < b, only a can be negative and b would be positive. This is the same case as discussed in Statement 1.

Sufficient.

The correct answer: C

Hope this helps!

-Jay
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