Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z|

ColdSushi

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

18 Mar 2015, 20:42

ColdSushi wrote:

Is z>0?
(1) (Z+1)(Z)(Z-1) < 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 [#permalink]

18 Mar 2015, 23:30

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ColdSushi wrote:

Is z>0?
(1) (Z+1)(Z)(Z-1) < 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 [#permalink]

19 Mar 2015, 06:29

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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 [#permalink]

02 Apr 2015, 23:24

Thanks very much guys - really appreciate your explanations.

S

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

11 Apr 2017, 08:26

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 [#permalink]

04 Sep 2018, 12:25

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Quote:

Is z > 0?

(1) (z + 1)(z)(z - 1) < 0
(2) |z| < 1

Statement 1:

-1 0 1
From 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<1
in statement 1 .
These 2 cases are satisfied in both statements and two possible solutions.
Please help me where I am going wrong
Bunuel

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

04 Sep 2018, 13:09

Expert Reply

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 [#permalink]

04 Sep 2018, 13:16

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 deduce-1<z<0 from statement 1 ?

L

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

04 Sep 2018, 13:38

Expert Reply

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 [#permalink]

21 Nov 2018, 12:02

Question: Is z>0?
(1) (z+1)(z)(z-1)<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.

G

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

21 Nov 2018, 18:43

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aj3001 wrote:

Question: Is z>0?
(1) (z+1)(z)(z-1)<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 (z-1) (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 [#permalink]

21 Nov 2018, 23:00

Salsanousi wrote:

aj3001 wrote:

Question: Is z>0?
(1) (z+1)(z)(z-1)<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 (z-1) (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

Thank you so much. Your explanation makes a lot of sense. There is a typo in the book. Thanks a lot!

L

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

03 Aug 2020, 20:33

ColdSushi wrote:

Is z > 0?

(1) (z + 1)(z)(z - 1) < 0
(2) |z| < 1

target is z>0
#1
(z + 1)(z)(z - 1) < 0
possible when z=1/2 , -2,-3
insufficient
#2
|z| < 1
range of values of z
-1<z<1
insufficient clearly
from 1 &2
z = 1/2 only possible
hence sufficient option c

G

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

21 Sep 2021, 04:37

VeritasKarishma wrote:

ColdSushi wrote:

Is z>0?
(1) (Z+1)(Z)(Z-1) < 0
(2) |Z|< 1

Is Z > 0

(1) (Z+1)(Z)(Z-1) < 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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/06 ... e-factors/

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)

Hi VeritasKarishma
Thank you for the explanation.
I have a small doubt, though.
Taking two statement together, I chose option E. (Please don't mind the grammatical error in my sentence.)
I could not internalize this part: (Which range of Z satisfies both inequalities? 0 < Z < 1). Why did we not took other unsatisfied range into consideration and thus chose option E. Please shed some light. Thank you.

L

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

21 Sep 2021, 22:28

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Mck2023 wrote:

VeritasKarishma wrote:

ColdSushi wrote:

Is z>0?
(1) (Z+1)(Z)(Z-1) < 0
(2) |Z|< 1

Is Z > 0

(1) (Z+1)(Z)(Z-1) < 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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/06 ... e-factors/

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)

Hi VeritasKarishma
Thank you for the explanation.
I have a small doubt, though.
Taking two statement together, I chose option E. (Please don't mind the grammatical error in my sentence.)
I could not internalize this part: (Which range of Z satisfies both inequalities? 0 < Z < 1). Why did we not took other unsatisfied range into consideration and thus chose option E. Please shed some light. Thank you.

This has to do with understanding the way DS questions are formulated.
You are asked: Is Z > 0?

Hint 1 gives to you is: (Z+1)(Z)(Z-1) < 0
You can rewrite this as Z < -1 or 0 < Z < 1.
So you know now that either Z is less than -1 or between 0 and 1. Can you say whether Z is positive? No. It may or may not be.

Hint 2 given to you is: |Z|< 1
You can rewrite this as -1 < Z < 1.
So you know now that Z lies between -1 and 1. Can you say whether Z is positive? No. It may or may not be.

Using both the hints (statements), you know that "either Z is less than -1 or between 0 and 1" and also that "Z lies between -1 and 1". Both statements need to be satisfied.
"Z must be either less than -1 or between 0 and 1" AND "Z must lie between -1 and 1".
Both can hold only if Z lies between 0 and 1 (e.g. say if Z = 0.5, it works. If Z = -0.5 or -2 etc, it doesn't work. One or the other condition is not met.)

So Z must lie between 0 and 1 only.

B

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

27 Oct 2021, 12:29

(1) Could we simplify: (z+1)(z)(z-1)<0
=> (z^2-1)(z)<0
=> z^3-z<0
=> z^3<z
this way we know that z could be either positive fraction or a negative number more than 1?

(2) |z|<1
z could be positive and negative fractions.

(1) and (2)
from (1) could be positive fraction or a negative number more than 1
from (2) could be positive or negative fractions.
The only thing that overlaps is positive fraction.

Therefore, the answer is C

Could this way of solving work?

L

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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

27 Oct 2021, 21:05

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mgan wrote:

(1) Could we simplify: (z+1)(z)(z-1)<0
=> (z^2-1)(z)<0
=> z^3-z<0
=> z^3<z
this way we know that z could be either positive fraction or a negative number more than 1?

(2) |z|<1
z could be positive and negative fractions.

(1) and (2)
from (1) could be positive fraction or a negative number more than 1
from (2) could be positive or negative fractions.
The only thing that overlaps is positive fraction.

Therefore, the answer is C

Could this way of solving work?

Yes you can provided you know and understand the relations between a number and its squares, cubes, square roots, absolute values etc.
How do you arrive at "either a positive fraction or a negative number less than -1"? If you understand these number properties well, it's great!

gmatclubot

Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1 [#permalink]

27 Oct 2021, 21:05

GMAT Data Sufficiency (DS) Questions

Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1