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Mgmat: Absolute value DS

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by Pharo » Sun Apr 08, 2012 12:40 pm
Since none of the statements are giving you information about the the relationship between a and b; the solution is E :)
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by pemdas » Sun Apr 08, 2012 1:52 pm
shubhamkumar wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0

(2) ab > 0

OA :E.Looking for solutions taking less than 2 mins.
I found an answer a here
|a| > |b| could be translated into a^2 > b^2 and a^2-b^2>0 or (a+b)(a-b)>0
Either a-b>0 or a-b<0 (cannot be zero). As we see a*|b| could be Either +ve or -ve.

st(1) a<0 implies a-b<0 or a<b. Sufficient
st(2) ab>=0 Insufficient as the signs of a and b not known
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by Shalabh's Quants » Sun Apr 08, 2012 8:40 pm
shubhamkumar wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0

(2) ab > 0

OA :E.Looking for solutions taking less than 2 mins.
=>|a| > |b| ; Is a · |b| < a - b; Taking few random values of a & b and analysing.


=>Case 1; |-2|>|0| ; is a · |b| < a - b => -2.|0| < -2 - 0 => 0 < -2 => No.

=>Case 2; |-2|>|1| ; is a · |b| < a - b => -2.|1| < -2 - 1 => -2 < -3 => No.

=>Case 3; |-2|>|-1| ; is a · |b| < a - b => -2.|-1| < -2 + 1 => -2 < -1 => Yes.


Taking Stat.1...

a<0 ; This statement does not help as in above all cases a = -2 is less than 0. Stat. 1 is insufficient.

Taking Stat.2...

ab>0

In cases 1 & 3 ab>0 with contradicting results. Stat. 2 is insufficient.

Taking Stat.1 & 2 together...


Combining stat. 1 & 2 will make case 2 not applicable, but still cases 1 & 3 holds with contradicting results. Hence Stat. 1 & 2 together are insufficient.
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by GMATGuruNY » Sun Apr 08, 2012 9:06 pm
shubhamkumar wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0

(2) ab > 0

OA :E.Looking for solutions taking less than 2 mins.
To save time, try to plug in combinations that satisfy both statements.
Since ab≥0 in statement 2, we should consider b=0.

Let b=0 and a=-1.
|a| > |b|, and both statements are satisfied.
Plugging these values into a · |b| < a - b, we get:
-1 * |0| < -1-0
0 < -1.
NO.

Let b=-1 and a=-2.
|a| > |b|, and both statements are satisfied.
Plugging these values into a · |b| < a - b, we get:
-2 * |-1| < -2-(-1)
-2 < -1.
YES.

Since in the first case the answer is NO, and in the second case the answer is YES, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Shalabh's Quants » Sun Apr 08, 2012 9:12 pm
pemdas wrote:
shubhamkumar wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0

(2) ab > 0

OA :E.Looking for solutions taking less than 2 mins.
I found an answer a here
|a| > |b| could be translated into a^2 > b^2 and a^2-b^2>0 or (a+b)(a-b)>0
Either a-b>0 or a-b<0 (cannot be zero). As we see a*|b| could be Either +ve or -ve.

st(1) a<0 implies a-b<0 or a<b. Sufficient
st(2) ab>=0 Insufficient as the signs of a and b not known
Hi,

Just to check with you. How can we conclude a · |b| < a - b if a-b<0 or a<b ?

Try these... say (a,b)=(-2,-1) => It says a · |b| < a - b

& (a,b)=(-2,1) => It says a · |b| > a - b

In both the cases a<0 and a<b.
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by shubhamkumar » Mon Apr 09, 2012 9:05 am
GMATGuruNY wrote:
shubhamkumar wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0

(2) ab > 0

OA :E.Looking for solutions taking less than 2 mins.
To save time, try to plug in combinations that satisfy both statements.
Since ab≥0 in statement 2, we should consider b=0.

Let b=0 and a=-1.
|a| > |b|, and both statements are satisfied.
Plugging these values into a · |b| < a - b, we get:
-1 * |0| < -1-0
0 < -1.
NO.

Let b=-1 and a=-2.
|a| > |b|, and both statements are satisfied.
Plugging these values into a · |b| < a - b, we get:
-2 * |-1| < -2-(-1)
-2 < -1.
YES.

Since in the first case the answer is NO, and in the second case the answer is YES, the two statements combined are INSUFFICIENT.

The correct answer is E.
That makes sense and the approach is out of the box.Are you suggesting that if we use both statements and prove that the stem cannot be solved with both statements, then answer would be E.
Traditionally we are advised to eliminate each statement before using the two together. :?
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by GMATGuruNY » Mon Apr 09, 2012 9:38 am
shubhamkumar wrote:
GMATGuruNY wrote:
shubhamkumar wrote:If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0

(2) ab > 0

OA :E.Looking for solutions taking less than 2 mins.
To save time, try to plug in combinations that satisfy both statements.
Since ab≥0 in statement 2, we should consider b=0.

Let b=0 and a=-1.
|a| > |b|, and both statements are satisfied.
Plugging these values into a · |b| < a - b, we get:
-1 * |0| < -1-0
0 < -1.
NO.

Let b=-1 and a=-2.
|a| > |b|, and both statements are satisfied.
Plugging these values into a · |b| < a - b, we get:
-2 * |-1| < -2-(-1)
-2 < -1.
YES.

Since in the first case the answer is NO, and in the second case the answer is YES, the two statements combined are INSUFFICIENT.

The correct answer is E.
That makes sense and the approach is out of the box.Are you suggesting that if we use both statements and prove that the stem cannot be solved with both statements, then answer would be E.
Traditionally we are advised to eliminate each statement before using the two together. :?
When we plug values into statement 1, the ONLY requirement is that we satisfy statement 1 and any conditions in the question stem.
But we should also have our eye on statement 2.
If we can find combinations that satisfy BOTH statements, we can evaluate both statements at the same time.
In the problem at hand, I could see that it would be relatively easy to plug a value into statement 1 (a<0) that would also satisfy statement 2 (ab≥0).
Since the two combinations that I plugged in satisfied both statements and yielded different answers to the question stem, the correct answer here is E.

If the two combinations HAD NOT yielded different answers to the question stem, then I would have had to consider values that satisfy statement 1 but DO NOT satisfy statement 2 (such as a=-1 and b=2).
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