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If x, y and z are integers and x – y – z < 0, is

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If x, y and z are integers and x – y – z < 0, is [#permalink]

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  95% (hard)

Question Stats:

30% (01:09) correct 70% (01:38) wrong based on 33 sessions

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If x, y and z are integers and x – y – z < 0, is z > 1?

(1) x - y > 1 - z
(2) y - x < -2

*kudos for all correct solutions

ASIDE: Given the issues with my first posting of the question, I should upgrade this question to 800!!

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Originally posted by GMATPrepNow on 21 Mar 2017, 07:49.
Last edited by GMATPrepNow on 22 Mar 2017, 10:52, edited 2 times in total.
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Re: If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 21 Mar 2017, 08:48
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GMATPrepNow wrote:
If x, y and z are integers and x – y – z < 0, is z > 1?

(1) x - y > 1 - z
(2) y - x < -2

*kudos for all correct solutions


We have \(x<y+z\)

(1) \(x-y > 1 - z \implies x > y +1 -z \implies y+z > y+ 1 -z \implies 2z > 1 \implies z > \frac{1}{2}\).

Hence we have \(z=1\), \(z<1\) or \(z>1\). Insufficient.

(2) \(y-x<-2 \implies y+2 < x < y+ z \implies z > 2 > 1\). Sufficient.

The answer is B.
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If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 22 Mar 2017, 03:46
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If x, y and z are integers and x – y – z < 0, is z > 1?

(1) x - y > 1 - z

Rearranging:

x - y +z > 1
x – y – z < 0
---------------- Subtracting 2 inequality AS LONG AS the signs are in OPPOSITE directions.
0+0 +2z > 1

z>1/2......

z CAN'T be 1 as it violates one of the two conditions. Test other INTEGER numbers GREATER THAN 1

z might be 2, 3,4,..etc

Answer is always YES.........Please read below posts to check proof.

Sufficient

(2) y - x < -2

x – y – z < 0
y - x < -2
---------------- Adding two inequalities AS LONG AS the signs in SAME directions
0 + 0- z< -2

-z <-2 ....... multiply by -1
z > 2>1

Sufficient

Answer: D
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If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 22 Mar 2017, 04:53
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If x, y and z are integers and x – y – z < 0, is z > 1?

Alternative method by plugging some values. It is somehow cumbersome for statement 1.

Starting with Statement 2:

2) y - x < -2

Let y-x=-2.1 < -2

Apply in equation in the stem:

-2.1 - z <0............z>2.1>1...........Answer is Yes . You spot that if we increase the magnitude of (y-x), z is getting bigger. If you do not notice, check other numbers.

Let y-x = -3 < -2

Apply in equation in the stem:

-3 - z <0............z>3>1...........Answer is Yes. You can check other points and you will find the same.

Sufficient

(1) x - y > 1 - z


Let x-y=3.... We need to maintain the inequality in the stem true also

3>1 - z

z might be 4.... check in inequality in stem 3-4 <0...........So z > 1

z might be INTEGER GREATER than 1. So it could be 2, 3,4,..etc

Answer is always YES.........Please read below posts to check proof.

Sufficient

Answer: D

Note: It is really cumbersome to plug in numbers especially for number 1
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Re: If x, y and z are integers and x – y – z < 0, is [#permalink]

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Mo2men wrote:

z might be 1/2....We need to maintain the inequality in the stem true also -3/4 <0.......Sp z <1



Good solution Mo2men. Only one small glitch z cannot be 1/2, since x, y and z are integers.

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Re: If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 22 Mar 2017, 08:16
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GMATPrepNow wrote:
Mo2men wrote:

z might be 1/2....We need to maintain the inequality in the stem true also -3/4 <0.......Sp z <1



Good solution Mo2men. Only one small glitch z cannot be 1/2, since x, y and z are integers.

Cheers,
Brent



Dear Brent,

Good catch from you. But I really could not come up with numbers to prove insufficient. All what we know that z> 1/2.

If z =1 & x-y=3

3 – 1 < 0........Not valid

3 + 1> 1 ........Valid

If z =1 & x-y=- 3

-3 -1 <0....valid

- 3 + 1 > 1...Not valid

Z can't be viable number to test as it violates either conditions in the stem of fact 1

If z =4 & x-y=3

3 – 4 < 0........valid

3 + 4> 1 ........Valid

If z =5 & x-y=-3

- 3 – 5 < 0........valid

- 3 +5> 1 ........Valid

so Z must be greater than 1 So It seems A is sufficient too.

What do you think?
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Re: If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 22 Mar 2017, 10:45
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Mo2men wrote:
But I really could not come up with numbers to prove insufficient. All what we know that z> 1/2.

I think you're right.

Given: x – y – z < 0

Target question: Is z > 1

(1) x - y > 1 - z

From statement 1 (and the given inequality), we learn that z > 1/2. So, we might (incorrectly) conclude that it could be the case that z = 1 or z = 2, in which case, we get different answers to the target question.
HOWEVER, if we try to come up with values for x, y and z that demonstrate this, we find that we have a problem.

If z = 1, then we can plug this value into our two inequalities.
For the statement 1 inequality, we get x - y > 1 - 1
Simplify to get: x - y > 0

For the given inequality, we get x – y – 1 < 0
Simplify to get: x - y < 1

When we combine the two inequalities, we get: 0 < x - y < 1
In other words, the difference between x and y is a fractional value BETWEEN 0 and 1.
This is IMPOSSIBLE, since it's given that x and y are integers.
So, it cannot be the case that z = 1

Since we already know that z > 1/2, we can conclude that it's possible that z = 2, z = 3, z = 4, etc.
For example, consider these situations:
Case a: x = 0, y = 0 and z = 2. In this case, z IS greater than 1
Case b: x = 0, y = 0 and z = 3. In this case, z IS greater than 1
Case c: x = 0, y = 0 and z = 4. In this case, z IS greater than 1
Case d: x = 0, y = 0 and z = 5. In this case, z IS greater than 1
etc..

So, it turns out that the correct answer is actually D (both statements are sufficient)

Sorry for not knowing the correct answer when I first posted the question. It's even harder than I first imagined!!

Cheers,
Brent
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Re: If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 22 Mar 2017, 10:52
3
GMATPrepNow wrote:
Mo2men wrote:
But I really could not come up with numbers to prove insufficient. All what we know that z> 1/2.

I think you're right.

Given: x – y – z < 0

Target question: Is z > 1

(1) x - y > 1 - z

From statement 1 (and the given inequality), we learn that z > 1/2. So, we might (incorrectly) conclude that it could be the case that z = 1 or z = 2, in which case, we get different answers to the target question.
HOWEVER, if we try to come up with values for x, y and z that demonstrate this, we find that we have a problem.

If z = 1, then we can plug this value into our two inequalities.
For the statement 1 inequality, we get x - y > 1 - 1
Simplify to get: x - y > 0

For the given inequality, we get x – y – 1 < 0
Simplify to get: x - y < 1

When we combine the two inequalities, we get: 0 < x - y < 1
In other words, the difference between x and y is a fractional value BETWEEN 0 and 1.
This is IMPOSSIBLE, since it's given that x and y are integers.
So, it cannot be the case that z = 1

Since we already know that z > 1/2, we can conclude that it's possible that z = 2, z = 3, z = 4, etc.
For example, consider these situations:
Case a: x = 0, y = 0 and z = 2. In this case, z IS greater than 1
Case b: x = 0, y = 0 and z = 3. In this case, z IS greater than 1
Case c: x = 0, y = 0 and z = 4. In this case, z IS greater than 1
Case d: x = 0, y = 0 and z = 5. In this case, z IS greater than 1
etc..

So, it turns out that the correct answer is actually D (both statements are sufficient)

Sorry for not knowing the correct answer when I first posted the question. It's even harder than I first imagined!!

Cheers,
Brent


Dear Brent,

You do not need to be sorry. You always provide us with genuine question and answers.

Thanks for your keen support and help

I have never asked before to get kudos but I need a kudos from you :wink:
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If x, y and z are integers and x – y – z < 0, is [#permalink]

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New post 22 Mar 2017, 10:58
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Mo2men wrote:

Dear Brent,

You do not need to be sorry. You always provide us with genuine question and answers.

Thanks for your keen support and help

I have never asked before to get kudos but I need a kudos from you :wink:


You deserve a bucket of kudos for catching that!!
:-D

Cheers,
Brent
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If x, y and z are integers and x – y – z < 0, is   [#permalink] 22 Mar 2017, 10:58
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