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Absolute Value Confusion

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by [email protected] » Fri Nov 15, 2013 6:18 am
If a and b are integers, and |a| > |b|, is a · |b| < a - b?

(1) a < 0

(2) ab >= 0

anybody have a solution for this (under 2 minute). testing value takes way too long here.


E
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Source: — Data Sufficiency |
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by theCodeToGMAT » Fri Nov 15, 2013 10:55 am
|a| > |b|

if a is +ve ==> a>b
if a is -ve ==> a<-b

To find: a |b| < a - b

Statement 1:
a < 0
that means 'a' is negative
a<-b ==> a - (-b) < 0
"a" & "b" can have any value and also maybe "b" is 0
So, we cannot say whether a|b| is zero
INSUFFICIENT

Statement 2:
ab >= 0
a & b have same sign or either one of them is "0"
if a is positive, (a-b) > 0
if a is negative, (a - (-b)) < 0
we cannot say whether a|B| is "0"
INSUFFICIENT


Combining...
a<0
ab >=0
-- a is negative
-- b is negative or "0"
So, a<-b ==> a - (-b) < 0
again, we cannot say a|b| is "0" or not.
INSUFFICIENT

Answer [spoiler]{E}[/spoiler]
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by [email protected] » Fri Nov 15, 2013 1:57 pm
Hi shibsriz,

This DS question is perfect for TESTing Values. You just have to keep your tests SIMPLE (and by simple, I mean that you should TEST small values whenever possible, including, -1, 0, 1):

We're told that A and B are integers and |A| > |B|. We're asked: Is A|B| < A - B? This is a YES/NO question.

Fact 1: A < 0

A = -1
B = 0
The answer to the question is NO

A = -2
B = -1
The answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: AB >= 0
**Notice how BOTH TEST cases from Fact 1 fit into Fact 2??? This is a HUGE shortcut**

A = -1
B = 0
The answer to the question is NO

A = -2
B = -1
The answer to the question is YES
Fact 2 is INSUFFICIENT

Combined, the same sets of Values fit BOTH Facts, so we have a NO and a YES answer.
Combined, INSUFFICIENT

Final Answer: E

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by Mathsbuddy » Mon Nov 18, 2013 10:06 am
[email protected] wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?

(1) a < 0

(2) ab >= 0

anybody have a solution for this (under 2 minute). testing value takes way too long here.


E
You could try this, but it still takes a few minutes:


As |a| > |b| and |a| < 0, then |b| < 0 too


Now we can limit our testing:

If we let |a| = p + k, |b| = p, then a = -(p + k) and b = -p where p > 0, k > 0 are unknown.

So, looking at the question "is a*|b| < a - b?"

The LHS is a · |b| = -(p + k)*p = -p(p + k)
and the RHS is a - b = -(p + k)--p = -k

So the question becomes: is -p(p + k) < -k ?

Add k to both sides:

-p^2 - kp + k < 0

p^2 + kp - k > 0 (changing sign changes the inequality symbol)

Solve:

p -> [-k +/- SQRT(k^2 + 4k)]/2 which is real and solvable; so it is possible. Integer solutions are findable.

However, the quadratic curve will intersect the p axis to define acceptable and unacceptable ranges.

Therefore it is unsufficient.
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