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If x+y+z > 0, is z > 1 ? [#permalink]
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05 Feb 2011, 09:33
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If x+y+z > 0, is z > 1 ? (1) z > x + y + 1 (2) x + y + 1 < 0 This is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough. Thanks, Eddy
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Re: is z > 1 ? [#permalink]
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05 Feb 2011, 09:45
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eddyed911 wrote: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y + 1 < 0 this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough. Thanks, Eddy Welcome to Gmat Club Eddy! 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 \(<\)) > \(ac>bd\) (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. Hope it's clear.
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Re: is z > 1 ? [#permalink]
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05 Feb 2011, 09:52
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Statement 1: Given: z>x+y+1 < Equ. 1 & x+y+z>0 Now add z to both sides of Equ. 1 2z>x+y+z+1 Since x+y+z > 0 2z>GT0+1 (GT0 means a quantity greater than 0) 2z>GT1 z>GT(1/2) So we cannot conclude that z>1 Therefore, insufficient! Statement 2: x+y+1<0 Adding z to both sides of the Equation x+y+z+1<Z GT0+1<z or GT1<z Therefore, z>1 Sufficient! Ans: 'B' I hope the explanation is clear enough
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Re: is z > 1 ? [#permalink]
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05 Feb 2011, 09:56
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If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 z  1 > x+ y and x+y > z z1 > z 2z>1 z>1/2 There are infinite real numbers between 1/2 and 1. Not sufficient. 2) x + y + 1 < 0 x+y < 1 To make x+y+z>0; z should be greater than 1 because x+y is less than 1. Sufficient. Ans: "B"
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Re: is z > 1 ? [#permalink]
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05 Feb 2011, 09:58
Thanks guys ! these explanations really clarify things



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Re: is z > 1 ? [#permalink]
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07 Feb 2011, 07:43
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thanks for the expalanation bunuel



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Re: is z > 1 ? [#permalink]
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16 Feb 2011, 19:41
eddyed911 wrote: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y + 1 < 0 this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough. Thanks, Eddy In case you get confused with addition/subtraction of inequalities, you can stick with addition only, if you like. Just rewrite one inequality to give them the same sign. Remember that you can add inequalities only when they have the same sign. Ques: Is z > 1? Given: x+y+z > 0 .......... (I) Stmnt 1: z > x + y + 1 .........(II) (I) and (II) both have the same inequality sign '>' so we can add them. x + y + 2z > x + y + 1 We get, z > (1/2) z may or may not be greater than 1. Not sufficient. Stmnt 2: x + y + 1 < 0 We can rewrite this as 0 > x + y + 1 .......(III) Now, (I) and (III) have the same sign '>' so you can add them. x + y + z > x + y + 1 We get z > 1. Sufficient. Answer (B)
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Re: is z > 1 ? [#permalink]
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17 Feb 2011, 00:36
eddyed911 wrote: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y + 1 < 0 this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough. Thanks, Eddy Statement 1: just tells us z1>x+y. Tells us nothing about the signs of any variable. Insufficient. Statement 2: tells us x+y<1. Now for x+y+z>0, minimum quantity we have to add has to be greater than one. Test numbers. Sufficient. Answer B.
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Re: is z > 1 ? [#permalink]
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03 Mar 2011, 07:13
Bunuel wrote: eddyed911 wrote: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y + 1 < 0 this is a question from OG 10, could someone please explain this. The explanation in the OG is not clear enough. Thanks, Eddy Welcome to Gmat Club Eddy! 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 \(<\)) > \(ac>bd\) (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. Hope it's clear. is the option (1) is like : x+y+z+z>0+x+y+1 (as a>b+c>d= a+c>b+d), and option 2 (2) x+y+z (x+y+z)>00 (as a>b c<d = ab>cd) Help me for my understanding.
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Re: DS inequalities [#permalink]
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12 Jan 2012, 00:44
Smita04 wrote: If x + y + z > 0, is z > 1?
(1) z > x + y + 1 (2) x + y + 1 < 0 stmnt1: adding z to both sides of the equation 2z >x+y+z+1 2z>1 or z> 0.5. this can give us any value for z (0.6,1,1.5). taking numeric values assume x=  0.2 y=  0.3 z= + 0.6 z> x+y+1 but z<1 for x= 1, y= 2 and z= 5 we have x+y+z>0 and z> x+y+1 and z>1. Hence insuff stmnt2: adding z to either sides x+y+1+z<0+z or z>1 also given x+y+1<0 or x+y < 1 we have x+y+z>0 or 1 + z >0 or z>1 hence suff so B



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Re: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y [#permalink]
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19 Jan 2012, 09:02
Thanks Bunuel...bcoz of the property u told this Q took 30 seconds to solve...



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Re: If x+y+z > 0, is z > 1 ? 1) z > x + y + 1 2) x + y [#permalink]
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20 Jan 2012, 05:54
Thanks Bunuel, makes the questions clear and easy to solve!



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Re: If x+y+z > 0, is z > 1 ? [#permalink]
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22 Jul 2014, 16:42
Entwistle wrote: Statement 1: Given: z>x+y+1 < Equ. 1 & x+y+z>0 Now add z to both sides of Equ. 1 2z>x+y+z+1 Since x+y+z > 0 2z>GT0+1 (GT0 means a quantity greater than 0) 2z>GT1 z>GT(1/2) So we cannot conclude that z>1 Therefore, insufficient! Statement 2: x+y+1<0 Adding z to both sides of the Equation x+y+z+1<Z GT0+1<z or GT1<z Therefore, z>1 Sufficient! Ans: 'B' I hope the explanation is clear enough Yeah, Add z to the both sides of the equation is the fastest way
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