Number Properties - Please Help!!! - DS

[SaurabhP260]

Is |a| + |b| > |a + b| ?

(1) a(square) > b(square)

(2) |a| × b < 0

Source: https://www.manhattanprep.com/gmat/onlineexams
Answer Provided: E (Statements (1) and (2) TOGETHER are NOT sufficient.)

My Contention: B (Statement 2 alone is sufficient)

Explanation: Both cannot be negative as that will tend to turn the original equation equal.

|-28| + |-27| > |(-28) + (-27)|
28+27 = l -55 l
55=55

Hence a and b cannot be both negative or positive.
Hence at least one of them shall be negative.

See Statement 2,

(2) |a| × b < 0

It specifies that b has to be negative.

Now, as explained above both cannot be negative. So a is positive.

Hence statement 2 is SUFFICIENT.

Thus, I believe answer shall be B (Statement 2 alone is sufficient).

Please guide.

[RobertoB400]

Hi! Just saw your inquiry with no replies so I thought of giving you a hand.

I struggled with absolute values before as I tended to tackle each problem algebraically or with tons of with different possible scenarios, until I realized that focusing on the question stem will save time and guide you to the right answer. So forget about looking at statm 1 and 2 until you know what exactly you need to get this problem right.

By separating the absolute values into blocks |a| + |b| and |a + b|, you know that on one side of the inequality you will have both terms a and b in their absolute values (so it won't matter if they were initially +ve or -ve), while the right side of the inequality will put them together in their raw initial form and only after adding/subtracting them you will be able to obtain their absolute value. So unless you know BOTH signs of a and b, you won't be able to tell whether |a + b| will be less or equal than |a| + |b| and thus answering the problem with sufficiency.

This observation is quite wordy but it will enable you to evaluate each statement within 10 seconds.

(1) NS: You don't care of squaring the terms by an even value as it will manipulate the signs (which you're looking for) anyways.
(2) NS: The initial sign of a won't affect such operation. You're happy to know b is negative, but the problem is hiding the sign of a from you.

Together they're worthless as the sign of a is still unknown, thus go with (E)

[RonPurewal]

the problem doesn't seem to be anything relating to absolute values. rather, it seems that saurabh, the original poster, doesn't have a firm grip on exactly how data sufficiency works.

evidence (emphasis mine):

My Contention: B (Statement 2 alone is sufficient)

Explanation: Both cannot be negative as that will tend to turn the original equation equal.

^^ here, it seems, you're not aware that "the original equation" is a QUESTION. it's a question whose answer can be either "yes" or "no".
if both answers are possible, then that's "not sufficient".
therefore, the bold underlined words represent exactly what you should be trying to do!

i.e., for each statement, you should be actively trying to find...
...a case that gives a "yes" answer to THE QUESTION;
AND
...a case that gives a "no" answer to THE QUESTION.

you're dismissing cases in which the two sides of the equation are equal, when in fact you should treasure those cases (they're the "no" cases, which constitute 50% of what you're looking for).

[RonPurewal]

@saurabh
in other words, ironically, your work (above) actually proves that statement 2 is NOT sufficient!
* you found a case (two negatives) that gives a "no" to THE QUESTION;
* presumably, you can also find a case (opposite signs) that gives a "yes" to THE QUESTION.
* ... so, statement 2 is not sufficient.

[RonPurewal]

you should go back to the basics of DS and make sure that you are 100.000000% comfortable with, and competent at, the basic process.

one activity that will be helpful:
* write two DS problems whose answer is "A": one in which THE QUESTION is a yes/no question (like this one), and one in which THE QUESTION asks for a quantity ("What is x?")
* write two DS problems whose answer is "C": same two types.
* same for answers "D" and "E".
(there's no need to write problems whose answer is "B", since those are functionally the same as the ones whose answer is "A"--you can just swap the two statements.)