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05 Aug 2016, 15:45
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In the following 700+ Level question, when I solve it with an algebra method, I get answer E (Both statements together not sufficient). When I plug in the values, I get C (Both together sufficient). Can someone please point out where is the flaw in my algebra method?
Is z>0?
(1) (z+1)(z)(z-1) -1
Since we have two possible answers, Statement 2 is INSUFFICIENT.
Looking at both together, we have one common solution: z 0, i.e is Z a positive number? z NEGATIVE
Test Case: z = 1
(1+1)(1)(1-1) 0. -----> POSITIVE
Test Case: z = \(\frac{-1}{2}\)
\((\frac{-1}{2}+1)(\frac{-1}{2})(\frac{-1}{2}-1)\)
\((\frac{1}{2})(\frac{-1}{2})(-\frac{3}{2})\)
\(\frac{3}{8}\) NOT NEGATIVE
Test Case: z = \(\frac{1}{2}\)
|\(\frac{1}{2}\)| = [m]\frac{1}{2} 0 ---> POSITIVE
INSUFFICIENT.
Combining two statements, the only test case that worked for both the statements is z = 1/2. Thus must be z > 0. SUFFICIENT. Thus answer is (C). Both statements together are SUFFICIENT.
What am I missing? Why the discrepancy?
Is z>0?
(1) (z+1)(z)(z-1) -1
Since we have two possible answers, Statement 2 is INSUFFICIENT.
Looking at both together, we have one common solution: z 0, i.e is Z a positive number? z NEGATIVE
Test Case: z = 1
(1+1)(1)(1-1) 0. -----> POSITIVE
Test Case: z = \(\frac{-1}{2}\)
\((\frac{-1}{2}+1)(\frac{-1}{2})(\frac{-1}{2}-1)\)
\((\frac{1}{2})(\frac{-1}{2})(-\frac{3}{2})\)
\(\frac{3}{8}\) NOT NEGATIVE
Test Case: z = \(\frac{1}{2}\)
|\(\frac{1}{2}\)| = [m]\frac{1}{2} 0 ---> POSITIVE
INSUFFICIENT.
Combining two statements, the only test case that worked for both the statements is z = 1/2. Thus must be z > 0. SUFFICIENT. Thus answer is (C). Both statements together are SUFFICIENT.
What am I missing? Why the discrepancy?
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05 Aug 2016, 17:06
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From what I see, Algebra method 1 should be:
\((z+1)(z)(z-1) = z(z^2-1)\)
Therefore we have two cases:
(a) \(z\) is negative and \(z^2 - 1\) is positive.
(b) \(z\) is positive and \(z^2 -1\) is negative.
Obviously \(z\) may be negative or positive, so it is insufficient. But we can go further into the cases.
(a) requires that \(z < 0\) and \(z^2 > 1\)
This resolves to \(z < -1\)
(b) requires that \(z^2 - 1\) is negative and \(z > 0\).
This resolves to \(0 < z < 1\)
Therefore, considering (2) we can remove case (a), we are left with case (b). This is sufficient.
\((z+1)(z)(z-1) = z(z^2-1)\)
Therefore we have two cases:
(a) \(z\) is negative and \(z^2 - 1\) is positive.
(b) \(z\) is positive and \(z^2 -1\) is negative.
Obviously \(z\) may be negative or positive, so it is insufficient. But we can go further into the cases.
(a) requires that \(z < 0\) and \(z^2 > 1\)
This resolves to \(z < -1\)
(b) requires that \(z^2 - 1\) is negative and \(z > 0\).
This resolves to \(0 < z < 1\)
Therefore, considering (2) we can remove case (a), we are left with case (b). This is sufficient.
05 Aug 2016, 18:07
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Does it say in the original problem that z is an Int? That would seem to make a difference here. Here is how I would attack it assuming z is Int. I think you would need logic, more than algebra for this problem which seems to be testing your knowledge of number properties.
Restating the problem we see that it is asking, is Z positive?
Statement 1. The form z-1, z, z+1 should be recognize able as the sequence of three numbers. [this is what makes me think the problem may have specified z is Int]
Three numbers when multiplied will be negative if all three are negative, or only one is.
Because this is a sequence. The only option where one number is positive would mean z=0. {-1, 0, 1}
(-1)(0)(-1) cannot be less than 0, as statement one requires, thus the only possibility is that all three numbers are less than 0. So...
Z+1<0
z<0
z-1<0
Statement 1 seems to be telling us, z cannot be 0, and z cannot be -1, and z is negative
Therefore sufficient, we can say for certain z is not greater than 0
Statement 2 seems a bit easier. The only integer for which |z| < 1 is true is 0.
Thus, if z = 0 then sufficient. And the answer is D.
It looks like in your algebra for statement 1, what you didn't consider was that none of the terms can equal 0 because that would make a false statement. 0 multiplied by anything is 0, and 0 cannot be <0
Your algebra in statement two, I think you forgot to bound the set.
|z| < 1
Breaks down into...
z < 1
z > -1
The set of answers for this inequality are only numbers between and not including 1 and -1. If we are dealing with Ints, then the only option is 0 and we have sufficiency
Sent from my iPhone using GMAT Club Forum mobile app
Restating the problem we see that it is asking, is Z positive?
Statement 1. The form z-1, z, z+1 should be recognize able as the sequence of three numbers. [this is what makes me think the problem may have specified z is Int]
Three numbers when multiplied will be negative if all three are negative, or only one is.
Because this is a sequence. The only option where one number is positive would mean z=0. {-1, 0, 1}
(-1)(0)(-1) cannot be less than 0, as statement one requires, thus the only possibility is that all three numbers are less than 0. So...
Z+1<0
z<0
z-1<0
Statement 1 seems to be telling us, z cannot be 0, and z cannot be -1, and z is negative
Therefore sufficient, we can say for certain z is not greater than 0
Statement 2 seems a bit easier. The only integer for which |z| < 1 is true is 0.
Thus, if z = 0 then sufficient. And the answer is D.
It looks like in your algebra for statement 1, what you didn't consider was that none of the terms can equal 0 because that would make a false statement. 0 multiplied by anything is 0, and 0 cannot be <0
Your algebra in statement two, I think you forgot to bound the set.
|z| < 1
Breaks down into...
z < 1
z > -1
The set of answers for this inequality are only numbers between and not including 1 and -1. If we are dealing with Ints, then the only option is 0 and we have sufficiency
Sent from my iPhone using GMAT Club Forum mobile app
