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655-705 Level|   Arithmetic|      
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rxs0005
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if | r | != 1 is integer r even 1 r is not positive 2 2r Updated on: 16 Apr 2014, 01:13
Originally posted by rxs0005 on 14 Aug 2010, 04:38.
Last edited by WoundedTiger on 16 Apr 2014, 01:13, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

50% (01:40) correct 50% (01:34) wrong based on 203 sessions

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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) \(5r>-2\).
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C
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Bunuel
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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 05:18
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rxs0005
if | r | != 1 is integer r even

1 r is not positive

2 2r > -5


OA is C but i am not convinced
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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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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 07:53
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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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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 08:15
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rxs0005
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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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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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 08:33
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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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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 08:55
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true, the condition for statement 2 has been flipped.
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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 09:07
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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.
Show more

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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if | r | != 1 is integer r even 1 r is not positive 2 2r 14 Aug 2010, 09:22
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Bunuel
rxs0005
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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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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if | r | != 1 is integer r even 1 r is not positive 2 2r 15 Apr 2014, 14:01
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OA on original question should be E.
Please correct

Thanks

Cheers!
J :twisted:
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if | r | != 1 is integer r even 1 r is not positive 2 2r 16 Apr 2014, 01:14
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jlgdr
OA on original question should be E.
Please correct

Thanks

Cheers!
J :twisted:
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Question formatting done.
OA is correct
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if | r | != 1 is integer r even 1 r is not positive 2 2r 22 Apr 2014, 06:57
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jlgdr
OA on original question should be E.
Please correct

Thanks

Cheers!
J :twisted:
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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.
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if | r | != 1 is integer r even 1 r is not positive 2 2r 21 Dec 2017, 14:06
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