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

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

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New post 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  [#permalink]

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New post 18 Mar 2015, 22:24
5
2
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: http://www.veritasprep.com/blog/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)
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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1  [#permalink]

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New post 18 Mar 2015, 19:42
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]

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New post 18 Mar 2015, 22:20
2
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)(Z-1) < 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)(Z-1) < 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 (Z-1) piece of the product. Z itself must be positive....which will give us...

(Z+1)(Z)(Z-1)
(+)(+)(-)
The answer to the question is ALWAYS YES
Combined, SUFFICIENT

Final Answer:

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

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New post 18 Mar 2015, 22:30
1
ColdSushi wrote:
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]

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New post 19 Mar 2015, 05:29
1
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]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 04 Sep 2018, 11:25
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
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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1  [#permalink]

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New post 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.

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

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New post 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.

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Thanks for the reply
But why can’t we deduce-1<z<0 from statement 1 ?
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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1  [#permalink]

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New post 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.

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

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New post 21 Nov 2018, 11:02
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.
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Re: Is z > 0? (1) (z + 1)(z)(z - 1) < 0 (2) |z| < 1  [#permalink]

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New post 21 Nov 2018, 17:43
1
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!

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

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New post 21 Nov 2018, 22:00
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!
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