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Is z > 0?

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Is z > 0? [#permalink]

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New post 18 Mar 2015, 20:31
3
8
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A
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C
D
E

Difficulty:

  65% (hard)

Question Stats:

43% (01:20) correct 57% (01:07) wrong based on 173 sessions

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Is z > 0?

(1) (z + 1)(z)(z - 1) < 0
(2) |z| < 1
Intern
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Joined: 29 Oct 2014
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Re: Is z > 0? [#permalink]

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New post 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? [#permalink]

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New post 18 Mar 2015, 23:20
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

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Re: Is z > 0? [#permalink]

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New post 18 Mar 2015, 23:24
4
1
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? [#permalink]

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New post 18 Mar 2015, 23:30
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? [#permalink]

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New post 19 Mar 2015, 06:29
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? [#permalink]

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New post 02 Apr 2015, 23:24
Thanks very much guys - really appreciate your explanations.
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Re: Is z > 0? [#permalink]

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New post 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
Re: Is z > 0?   [#permalink] 11 Apr 2017, 08:26
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