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Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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18 Mar 2015, 19:31
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Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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18 Mar 2015, 22:24
ColdSushi wrote: Is z>0? (1) (Z+1)(Z)(Z1) < 0 (2) Z< 1 Is Z > 0 (1) (Z+1)(Z)(Z1) < 0 There are 3 transition points 1, 0 and 1. The expression will be negative if Z < 1 or 0 < Z < 1. If you are not sure how we got this, check: http://www.veritasprep.com/blog/2012/06 ... efactors/So Z may be positive or negative in this range. Not sufficient. (2) Z< 1 1 < Z < 1 Z may be positive or negative in this range. Not sufficient. Which range of Z satisfies both inequalities? 0 < Z < 1 In this range, Z is always positive. Sufficient. Answer (C)
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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18 Mar 2015, 19:42
ColdSushi wrote: Is z>0? (1) (Z+1)(Z)(Z1) < 0 (2) Z< 1 Could anyone share their workings for this one? I started with (2) because it'd easier to rule out. For (1) I started by expanding the formula out then picked smart numbers. The OA, however, did it by just picking smart numbers without any prework. I understand how to pick the right smart numbers BUT what I can't work out is what should I look for to determine whether I need to expand or leave the equation as is before I plug in smart numbers?



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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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18 Mar 2015, 22:20
Hi ColdSushi, This DS question is loaded with Number Property rules; you'll actually see Number Properties on many DS questions, so it's a good area to spend some extra time on. We're asked if X > 0. This is a YES/NO question. Fact 1: (Z+1)(Z)(Z1) < 0 This looks like a great opportunity to TEST VALUES. Before randomly TESTing a value though, I want to note the Number Property patterns in this question. Since the product is LESS than 0, we cannot have a product that equals 0. Looking at the three parentheses, we can see that the product will equal 0 if Z is 1, 0 or 1. This means that we CANNOT TEST any of those values. The first TEST should be something fairly obvious. We need a negative product, so let's choose a negative value. IF... Z = 2 (1)(2)(3) = 6, which is < 0 The answer to the question is NO. Notice how in the first Test we had the product of 3 negative numbers. We can also get a negative product with just 1 negative number, so we should look for that option next. IF.... Z = 1/2 (3/2)(1/2)(1/2) = 3/8, which is < 0 The answer to the question is YES. Fact 1 is INSUFFICIENT Fact 2: Z < 1 This means that 1 < Z < 1 IF... Z = 1/2 The answer to the question is YES IF... Z = 1/2 The answer to the question is NO Fact 2 is INSUFFICIENT Combined, we know.... (Z+1)(Z)(Z1) < 0 1 < Z < 1 Since Z is greater than 1, there's no way for (Z+1) to be negative. By extension, we can't end up with the product of 3 negative numbers, so the only way to satisfy the first Fact is if we have JUST 1 negative number (which will be the (Z1) piece of the product. Z itself must be positive....which will give us... (Z+1)(Z)(Z1) (+)(+)() The answer to the question is ALWAYS YES Combined, SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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18 Mar 2015, 22:30
ColdSushi wrote: ColdSushi wrote: Is z>0? (1) (Z+1)(Z)(Z1) < 0 (2) Z< 1 Could anyone share their workings for this one? I started with (2) because it'd easier to rule out. For (1) I started by expanding the formula out then picked smart numbers. The OA, however, did it by just picking smart numbers without any prework. I understand how to pick the right smart numbers BUT what I can't work out is what should I look for to determine whether I need to expand or leave the equation as is before I plug in smart numbers? By the way, establishing a statement using "smart numbers" isn't the best idea. It is easy to negate a statement using numbers but establishing isn't always advisable. Say, I am given that Z< 1. Is Z always positive? I can negate it by plugging in a single value Z = 1/2. Here Z is negative but 1/2 < 1 holds. So I know that Z is not always positive in this case. Great. But if I want to establish using both statements that Z must always be positive, I may miss out on testing some important numbers since it must hold for all values of Z. So be careful when you use "smart numbers" in DS questions.
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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19 Mar 2015, 05:29
ColdSushi wrote: Is z > 0?
(1) (z + 1)(z)(z  1) < 0 (2) z < 1 Option A: (z + 1)(z)(z  1) < 0 implies 0<Z<1 OR Z<1 NOT SUFFICIENT. Option B: z < 1 implies 1<Z<1 NOT SUFFICIENT. Combining Together 0<Z<1 SUFFICIENT. ANSWER C.
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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02 Apr 2015, 22:24
Thanks very much guys  really appreciate your explanations.



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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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11 Apr 2017, 07:26
Is z > 0?
(1) (z + 1)(z)(z  1) < 0 (2) z < 1
answer is C
1 not suff z=0.5 and z =2 2 not suff 1<z<1
combining we get sufficient in the range of 0<z<1



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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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04 Sep 2018, 11:25
Quote: Is z > 0?
(1) (z + 1)(z)(z  1) < 0 (2) z < 1 Statement 1:1 0 1From 1 we can deduce Z<1 1<z<0 0<z<1 z>1 Many cases So we cannot say whether Z>0.Not Sufficient Statement 2:From 2 we can get: 1<z<1 positive or negative so not sufficient 1+2 here still 2 cases can be possible right? Because we have 1<z<0 0<z<1in statement 1 . These 2 cases are satisfied in both statements and two possible solutions. Please help me where I am going wrongBunuel
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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04 Sep 2018, 12:09
Hi SonGoku, You made a minor mistake when dealing with Fact 1. There are a couple of different 'ranges' of values that will fit... (Z + 1)(Z)(Z  1) < 0 Z < 1 and 0 < Z < 1 are both possibilities. However, 1 < Z < 0 is NOT an option. With a bit more work, you'll find that the ONLY possibilities that will fit BOTH Facts are when Z is a positive fraction. GMAT assassins aren't born, they're made, Rich
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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04 Sep 2018, 12:16
EMPOWERgmatRichC wrote: Hi SonGoku,
You made a minor mistake when dealing with Fact 1. There are a couple of different 'ranges' of values that will fit...
(Z + 1)(Z)(Z  1) < 0
Z < 1 and 0 < Z < 1 are both possibilities.
However, 1 < Z < 0 is NOT an option. With a bit more work, you'll find that the ONLY possibilities that will fit BOTH Facts are when Z is a positive fraction.
GMAT assassins aren't born, they're made, Rich Thanks for the reply But why can’t we deduce1<z<0 from statement 1 ?
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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04 Sep 2018, 12:38
Hi SonGoku, Consider what would happen to each of the 3 parentheses if 1 < Z < 0 (meaning that Z is a negative fraction).... (Z + 1)(Z)(Z  1) < 0 With any negative fraction in place for Z, we would end up with a product that is... ( + )(  )(  ) = POSITIVE For example, if Z = 0.5..... (0.5 + 1)(0.5)(0.5  1) (+0.5)(0.5)(1.5) = +.375 This does NOT fit what we were told; the product is supposed to be NEGATIVE. Thus, Z cannot be in the range 1 < Z < 0. GMAT assassins aren't born, they're made, Rich
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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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21 Nov 2018, 11:02
Question: Is z>0? (1) (z+1)(z)(z1)<0 (2) z <1
Both the statements are insufficient independently however they are sufficient together. (C)
But in explanation of trying to prove (2) insufficient to strike option B and D, they have tested three cases using 2, ½, ½. On testing each of the three cases they have considered the following: a) 2: 2<1 Not considered b) ½: ½<1 so z>0 Yes c) ½: ½<1 so z>0 No
But my doubt is that in the third case shouldn't the answer be yes as the modulus changes the sign of any value inside it to positive? Wont ½ = ½? But in the answer they have considered ½= ½
Please clarify. Thanks in advance.



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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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21 Nov 2018, 17:43
aj3001 wrote: Question: Is z>0? (1) (z+1)(z)(z1)<0 (2) z <1
Both the statements are insufficient independently however they are sufficient together. (C)
But in explanation of trying to prove (2) insufficient to strike option B and D, they have tested three cases using 2, ½, ½. On testing each of the three cases they have considered the following: a) 2: 2<1 Not considered b) ½: ½<1 so z>0 Yes c) ½: ½<1 so z>0 No
But my doubt is that in the third case shouldn't the answer be yes as the modulus changes the sign of any value inside it to positive? Wont ½ = ½? But in the answer they have considered ½= ½
Please clarify. Thanks in advance. For statement one We have z (z1) (z +1) < 0 3 different values multipled it means either all are negative or only one is negative. This way we cannot tell whether z is negative or z  1 etc.. For statement two z < 1  z < 1 this then becomes z > 1 1 < z < 1 Pick z = 1/2 or  1/2 1/2 = 1/2 < 1 (satisfies the statement) but is z > 0? Yes.  1/2  = 1/2 < 1(satisfies the statement) But is z > 0? No. I think something went off on the explanation. Now combined we know z is between 1 and 1 So for the first statement to be negative z has to be a positive number less than 1 The reason is 1/2 * (1/2  1) * (1/2 + 1) 1/2 * ( 1/2) * (3/2) < 0 If z was negative then it would be 1/2 This way it becomes 1/2 * (3/2) * 1/2 which invalidates the statement. Since z has to be a positive number less than 1. We have an answer. Hope this helps! Posted from my mobile device



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Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1
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21 Nov 2018, 22:00
Salsanousi wrote: aj3001 wrote: Question: Is z>0? (1) (z+1)(z)(z1)<0 (2) z <1
Both the statements are insufficient independently however they are sufficient together. (C)
But in explanation of trying to prove (2) insufficient to strike option B and D, they have tested three cases using 2, ½, ½. On testing each of the three cases they have considered the following: a) 2: 2<1 Not considered b) ½: ½<1 so z>0 Yes c) ½: ½<1 so z>0 No
But my doubt is that in the third case shouldn't the answer be yes as the modulus changes the sign of any value inside it to positive? Wont ½ = ½? But in the answer they have considered ½= ½
Please clarify. Thanks in advance. For statement one We have z (z1) (z +1) < 0 3 different values multipled it means either all are negative or only one is negative. This way we cannot tell whether z is negative or z  1 etc.. For statement two z < 1  z < 1 this then becomes z > 1 1 < z < 1 Pick z = 1/2 or  1/2 1/2 = 1/2 < 1 (satisfies the statement) but is z > 0? Yes.  1/2  = 1/2 < 1(satisfies the statement) But is z > 0? No. I think something went off on the explanation. Now combined we know z is between 1 and 1 So for the first statement to be negative z has to be a positive number less than 1 The reason is 1/2 * (1/2  1) * (1/2 + 1) 1/2 * ( 1/2) * (3/2) < 0 If z was negative then it would be 1/2 This way it becomes 1/2 * (3/2) * 1/2 which invalidates the statement. Since z has to be a positive number less than 1. We have an answer. Hope this helps! Posted from my mobile deviceThank you so much. Your explanation makes a lot of sense. There is a typo in the book. Thanks a lot!




Re: Is z > 0? (1) (z + 1)(z)(z  1) < 0 (2) z < 1 &nbs
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