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if | r | != 1 is integer r even 1 r is not positive 2 2r

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if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink] New post 14 Aug 2010, 03:38
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

44% (01:57) correct 55% (00:52) wrong based on 59 sessions
Orginiallly Posted Question

if | r | != 1 is integer r even

1 r is not positive

2 2r > -5


OA is C but i am not convinced

CORRECT QUESTION

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

(1) r is not positive.

(2) 5r>-2.
[Reveal] Spoiler: OA

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Last edited by WoundedTiger on 16 Apr 2014, 00:13, edited 1 time in total.
Question Formatting done
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Re: Number Prop DS [#permalink] New post 14 Aug 2010, 04:18
Expert's post
rxs0005 wrote:
if | r | != 1 is integer r even

1 r is not positive

2 2r > -5


OA is C but i am not convinced


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 basicalyy 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.

If statement (2) were 5r>-2, then answer would be C: 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.
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Re: Number Prop DS [#permalink] New post 14 Aug 2010, 06:53
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
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Re: Number Prop DS [#permalink] New post 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.

Answer: C.
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Re: Number Prop DS [#permalink] New post 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.
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Re: Number Prop DS [#permalink] New post 14 Aug 2010, 07:55
true, the condition for statement 2 has been flipped.
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Re: Number Prop DS [#permalink] New post 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.

Answer: C.

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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: Number Prop DS [#permalink] New post 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 :)
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink] New post 15 Apr 2014, 13:01
OA on original question should be E.
Please correct

Thanks

Cheers!
J :twisted:
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink] New post 16 Apr 2014, 00:14
jlgdr wrote:
OA on original question should be E.
Please correct

Thanks

Cheers!
J :twisted:



Question formatting done.
OA is correct
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink] New post 22 Apr 2014, 05:57
jlgdr wrote:
OA on original question should be E.
Please correct

Thanks

Cheers!
J :twisted:


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