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If x + y + z > 0, is z > 1? [#permalink]
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12 Feb 2014, 01:11
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Re: If x + y + z > 0, is z > 1? [#permalink]
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12 Feb 2014, 01:12
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SOLUTIONNote 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). 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: \((x+y+z)(x+y+1)>0\) > \(z1>0\) > \(z>1\). Sufficient. Answer: B. Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.html
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Re: If x + y + z > 0, is z > 1? [#permalink]
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12 Feb 2014, 11:02
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 Posted from my mobile device
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Re: If x + y + z > 0, is z > 1? [#permalink]
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12 Feb 2014, 16:47
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 : zxy>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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12 Feb 2014, 19:32
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 : zxy>1, therefore z will be >1. >x, y can be negative then conclusion fails
Therefore stmt A is sufficient. AnsA is not correct I guess



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Re: If x + y + z > 0, is z > 1? [#permalink]
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13 Feb 2014, 07:04
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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 : zxy>1, therefore z will be >1. >x, y can be negative then conclusion fails
Therefore stmt A is sufficient. AnsA is not correct I guess OK. Taking a different approach : x + y + z > 0  x + y + 1 > 0= z1>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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13 Feb 2014, 09:40
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Re: If x + y + z > 0, is z > 1? [#permalink]
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13 Feb 2014, 16:08
Stmt 1 : z > x + y + 1 i.e x+yz > 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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13 Feb 2014, 22:14
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 : zxy>1, therefore z will be >1. >x, y can be negative then conclusion fails
Therefore stmt A is sufficient. AnsA is not correct I guess OK. Taking a different approach : x + y + z > 0  x + y + 1 > 0= z1>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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17 Feb 2014, 01:14
SOLUTIONNote 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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). 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: \((x+y+z)(x+y+1)>0\) > \(z1>0\) > \(z>1\). Sufficient. Answer: B. Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.html
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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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If x + y + z > 0, is z > 1? [#permalink]
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23 Mar 2016, 10:01
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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03 May 2016, 21:02
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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30 Jun 2016, 04:32
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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