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Re: Number Prop DS [#permalink]
14 Aug 2010, 07:15

Expert's post

rxs0005 wrote:

Bunuel u mistook the statement

what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion

This means that I wrote new question:

If r is an integer and |r|!=1 is r even?

(1) r is not positive.

(2) 5r>-2.

Answer: C.

As for original question:

If |r|\neq{1} is r=even?

|r|\neq{1} --> r\neq{1} and r\neq{-1}.

(1) r is not positive --> Clearly insufficient, r can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) 2r>-5 --> r>-\frac{5}{2}=-2.5 --> again r can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) r is not positive and r>-2.5 --> r can be -2, -1, or 0. But as given that r\neq{-1} then only valid solutions for r are -2 and 0, both are even. Sufficient.

Re: Number Prop DS [#permalink]
14 Aug 2010, 07:33

Bunuel, If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped? and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E

I just saw, (2) was a typo. Very rare you make a mistake!

In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works. _________________

Re: Number Prop DS [#permalink]
14 Aug 2010, 08:07

Expert's post

mainhoon wrote:

Bunuel, If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped? and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E

I just saw, (2) was a typo. Very rare you make a mistake!

In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works.

No mistake there. Nothing was flipped. Look carefully, I consider in my previous post two questions:

MINE: If r is an integer and |r|!=1 is r even?

(1) r is not positive.

(2) 5r>-2.

Answer: C.

AND ORIGINAL:

If |r|\neq{1} is r=even?

(1) r is not positive.

(2) 2r>-5.

Answer: C.

Totally in my previous posts we have 3 questions. First 2 I "invented" because rxs0005 meant by | r | != 1 |r|\neq{1} and I understood it as it was written |r|!={1}. So two questions I consider in my first post are those I "invented".

Question 1 (mine):

If r is an integer and |r|!=1 is r even?

First of all factorial of a number equals to 1 in two cases: number equals to 0 --> 0!=1 or number equals to 1 --> 1!=1.

|r|!=1 --> |r|=0 or |r|=1, so r can take 3 values: 0, 1, and -1, out of which only 0 is even. So basically questions asks: is r=0?

(1) r is not positive --> r can be 0 or -1. Not sufficient.

(2) 2r>-5 --> r>-\frac{5}{2}=-2.5 --> r can be 0, 1 or -1. Not sufficient.

(1)+(2) still two values of r are possible 0 and -1. Not sufficient.

Answer: E.

Question #2 (mine):

If r is an integer and |r|!=1 is r even?

(1) r is not positive --> r can be 0 or -1. Not sufficient.

(2) 5r>-2 --> r>-\frac{2}{5}=-0.4 --> r can be 0 or 1. Not sufficient.

(1)+(2) only one value of r is possible: 0, hence r is even. Sufficient.

Answer: C.

Question #3 (original):

If |r|\neq{1} is r=even?

|r|\neq{1} --> r\neq{1} and r\neq{-1}.

(1) r is not positive --> Clearly insufficient, r can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) 2r>-5 --> r>-\frac{5}{2}=-2.5 --> again r can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) r is not positive and r>-2.5 --> r can be -2, -1, or 0. But as given that r\neq{-1} then only valid solutions for r are -2 and 0, both are even. Sufficient.

Answer: C.

BUT: if the second statement were as you wrote answer still would be C.

Question #4 (your's):

If |r|\neq{1} is r=even?

|r|\neq{1} --> r\neq{1} and r\neq{-1}.

(1) r is not positive --> Clearly insufficient, r can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) 5r>-2 --> r>-\frac{2}{5}=-0.4 --> r can be 0 or 2, 3, 4, ... Not sufficient.

(1)+(2) Only one solution is possible for r is 0, which is even. Sufficient.

Re: Number Prop DS [#permalink]
14 Aug 2010, 08:22

Bunuel wrote:

rxs0005 wrote:

Bunuel u mistook the statement

what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion

This means that I wrote new question:

If r is an integer and |r|!=1 is r even?

(1) r is not positive.

(2) 5r>-2.

Answer: C.

As for original question:

If |r|\neq{1} is r=even?

|r|\neq{1} --> r\neq{1} and r\neq{-1}.

(1) r is not positive --> Clearly insufficient, r can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) 2r>-5 --> r>-\frac{5}{2}=-2.5 --> again r can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) r is not positive and r>-2.5 --> r can be -2, -1, or 0. But as given that r\neq{-1} then only valid solutions for r are -2 and 0, both are even. Sufficient.

Answer: C.

Bunuel - I see that there was an original question and you had modified it. I got it. Thanks for the clarification. As I said, very rare you make a mistake _________________

Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]
22 Apr 2014, 05:57

jlgdr wrote:

OA on original question should be E. Please correct

Thanks

Cheers! J

Hi jlgdr, The answer for the original question is indeed E , if "!" meant as factorial. However if the same is mentioned as "not equal to", then the answer should be C. I believe the question is originally posted for "!" as nt equal to. In that case "C" is the correct answer as mentioned in the OA. Hope that helps.

Thanks.

gmatclubot

Re: if | r | != 1 is integer r even 1 r is not positive 2 2r
[#permalink]
22 Apr 2014, 05:57