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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 \(<\)) --> \(a-c>b-d\) (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.

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 _________________

"Wherever you go, go with all your heart" - Confucius

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 re-write 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.

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 z-1>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.

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 \(<\)) --> \(a-c>b-d\) (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.

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)>0-0 (as a>b -c<d = a-b>c-d) Help me for my understanding. _________________

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 _________________

......................................................................... +1 Kudos please, if you like my post

Re: If x+y+z > 0, is z > 1 ? [#permalink]
15 Sep 2015, 12:32

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