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If x + y + z > 0, is z > 1?

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If x + y + z > 0, is z > 1?  [#permalink]

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New post 12 Feb 2014, 02:11
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A
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E

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x + y + z > 0, is z > 1?

(1) z > x + y + 1
(2) x + y + 1 < 0

Data Sufficiency
Question: 89
Category: Algebra Inequalities
Page: 158
Difficulty: 650


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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 12 Feb 2014, 02:12
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SOLUTION

Note that:
You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Back to the original question:

If x + y + z > 0, is z > 1?

(1) z > x + y + 1 --> as the signs are in the same direction we can add two inequalities: \((x+y+z)+z>x+y+1\) --> \(2z>1\) --> \(z>\frac{1}{2}\), so z may or may not be more than 1. Not sufficient.

(2) x + y + 1 < 0 --> as the signs are in the opposite direction we can subtract two inequalities: \((x+y+z)-(x+y+1)>0\) --> \(z-1>0\) --> \(z>1\). Sufficient.

Answer: B.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 12 Feb 2014, 12:02
If x + y + z > 0, is z > 1?

(1) z > x + y + 1
(2) x + y + 1 > 0

Sol: st1 says z>x+y+1 and x+y+z>0

If x=-2, y=-3,z=6 then z>1
But if x=-0.1,y=-0.1 thus z>0.8 so it may or may not be greater than 1
Not sufficient. A and D ruled out

St 2 says x+y+1>0
Again x=-1/2,y=1/2 then z can take any value greater than or less than 1 but if x=1000,y=1 then again z can take any value so clearly
Not sufficient

On combining we get that z>0 but z may or may not be greater than1

Ans E

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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 12 Feb 2014, 17:47
If x + y + z > 0, is z > 1?

(1) z > x + y + 1
(2) x + y + 1 > 0

Sol: st1 says z>x+y+1 and x+y+z>0

Using stmt 1 : z-x-y>1, therefore z will be >1.

Therefore stmt A is sufficient.
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 12 Feb 2014, 20:32
X017in wrote:
If x + y + z > 0, is z > 1?

(1) z > x + y + 1
(2) x + y + 1 > 0

Sol: st1 says z>x+y+1 and x+y+z>0

Using stmt 1 : z-x-y>1, therefore z will be >1. --->x, y can be negative then conclusion fails

Therefore stmt A is sufficient.


Ans-A is not correct I guess
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 13 Feb 2014, 08:04
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syamen wrote:
X017in wrote:
If x + y + z > 0, is z > 1?

(1) z > x + y + 1
(2) x + y + 1 > 0

Sol: st1 says z>x+y+1 and x+y+z>0

Using stmt 1 : z-x-y>1, therefore z will be >1. --->x, y can be negative then conclusion fails

Therefore stmt A is sufficient.


Ans-A is not correct I guess


OK. Taking a different approach :

x + y + z > 0
- x + y + 1 > 0
= z-1>0
= z>1

So B is sufficient?
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 13 Feb 2014, 17:08
Stmt 1 : z > x + y + 1 i.e x+y-z > -1
Is z>1 ?
yes, put z = 2 & x+y =3 so x+y - z =1, which is > -1 ... also we maintain the given condition x+y+z>0
no, put z = 0 & x+y =3 so x+y - z =3 , which is > -1 again we pick nos such that x+y+z> is not violated

A is insuff ..so A and D eliminated

Stmt 2 : x + y + 1 > 0 i.e. x+y > -1
Is z> 1 ?
yes i.e pick any value for x+y >-1 so that x+y+z>0 is not violated .. So at the least x+y =-1 , then z has to be z > 1 for x+y + z > 0..

no, z = -2 and x+y = 5 then x+y+z = 3 i.e. > 0

B is insuff

So OA is E
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 13 Feb 2014, 23:14
1
X017in wrote:
syamen wrote:
X017in wrote:
If x + y + z > 0, is z > 1?

(1) z > x + y + 1
(2) x + y + 1 > 0

Sol: st1 says z>x+y+1 and x+y+z>0

Using stmt 1 : z-x-y>1, therefore z will be >1. --->x, y can be negative then conclusion fails

Therefore stmt A is sufficient.


Ans-A is not correct I guess


OK. Taking a different approach :

x + y + z > 0
- x + y + 1 > 0
= z-1>0
= z>1

So B is sufficient?


You cannot subtract inequalities.

Consider:

4>2 ....(i)
3>-2 ...(ii)

(i) - (ii) gives
1 > 4 (not true)

You can add them if they have the same inequality sign. If they have opposite signs, multiply one of them by -1, flip the inequality sign and then add them.
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 17 Feb 2014, 02:14
2
SOLUTION

Note that:
You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Back to the original question:

If x + y + z > 0, is z > 1?

(1) z > x + y + 1 --> as the signs are in the same direction we can add two inequalities: \((x+y+z)+z>x+y+1\) --> \(2z>1\) --> \(z>\frac{1}{2}\), so z may or may not be more than 1. Not sufficient.

(2) x + y + 1 < 0 --> as the signs are in the opposite direction we can subtract two inequalities: \((x+y+z)-(x+y+1)>0\) --> \(z-1>0\) --> \(z>1\). Sufficient.

Answer: B.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 23 Mar 2016, 11:01
Statement II is quite obvious.

If x + y + z > 0
(2) x + y + 1 < 0

We are initially told that x + y + z > 0, but statement II tells us that x + y + 1 < 0. Hence, given the question prompt, z must be greater than 1.
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 03 May 2016, 22:02
My approach to this problem
st #1-- z>x+y+1
Take x = -3 , y= -4 then this implies that z> -6 . NS y

st#2 -- x+y+1<0
ok. from the prompt x+y+z>0 , therefore x+y+z > x+y+1 . This is since postive is always greater than negative.
Then we can take away x and y from both sides by adding -x and -y to both sides one by one.
what remains is z>1 . Sufficient
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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New post 30 Jun 2016, 05:32
Given x + y + z > 0
(1) says z > x + y + 1

Adding the two:

x + y + 2z > x + y + 1
z > 0.5

So, we can't say for sure that z > 1. Not sufficient.

(2) x + y + 1 < 0
x + y < -1

Multiplying both sides by -1
-x - y > 1

Question says: x + y + z > 0

Adding the two:
z > 1
So, (2) is sufficient.
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Re: If x + y + z > 0, is z > 1?  [#permalink]

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Re: If x + y + z > 0, is z > 1? &nbs [#permalink] 25 Aug 2017, 13:09
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