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# a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ?

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a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink]

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26 Nov 2010, 20:07
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a is an odd integer and a ≠ -b. Is $$\frac{a^2}{|b+a|}> a-b ?$$

(1) $$ab=0$$

(2) $$a^3>a$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Aug 2017, 08:24, edited 1 time in total.
Renamed the topic and edited the question.

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Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink]

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26 Nov 2010, 21:56
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a is an odd integer and a <>-b

Is $$\frac{a^2}{|b+a|}> a-b ?$$

(1) ab=0
(2)$$a^3>a$$

PLease explain the shortest way to solve this

Thanks

I have no clue what "a <> -b" means, but that doesn't seem to be overly important to solve.

Let's dive into the statements:

1) if a is an odd integer, then (1) must mean that b=0. So, the question becomes:

Is a^2/|a|> a?

Let's pick a positive and negative a and see what happens.

If a = 1, we get:

Is 1/1 > 1? NO

If a = -1, we get:

Is 1/1 > -1? YES

Can get a yes and a no: insufficient.

2) if a is an odd integer and a^3 > a, then a must be positive. However, we know nothing about b. If we pick a big negative value of b then a-b will be negative and the left side will be positive, giving us a NO answer. If we pick a small positive value for b then both sides will be positive and we can play with the value of a to produce a YES answer: insufficient.

Together: if a is positive and b=0, the two sides will be equal, give us a "no" answer: sufficient, choose (C)!

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Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink]

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27 Nov 2010, 12:35
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Expert's post
a is an odd integer and a <>-b

Is $$\frac{a^2}{|b+a|}> a-b ?$$

(1) ab=0
(2)$$a^3>a$$

PLease explain the shortest way to solve this

Thanks

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that $$\frac{a^2}{|b+a|}$$ is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Stmnt 2: $$a^3>a$$. This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 16925 [1], given: 230 Intern Joined: 18 Feb 2010 Posts: 28 Kudos [?]: 73 [0], given: 5 Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink] ### Show Tags 27 Nov 2010, 12:56 Thanks Karishma and Stuart. Yes <> means "not equal to". It had a long solution and wanted to know from experts like you if this resembles a GMAT Question. I got it from GMATfix question bank. Kudos [?]: 73 [0], given: 5 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7615 Kudos [?]: 16925 [0], given: 230 Location: Pune, India Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink] ### Show Tags 27 Nov 2010, 13:39 nades09 wrote: Thanks Karishma and Stuart. Yes <> means "not equal to". It had a long solution and wanted to know from experts like you if this resembles a GMAT Question. I got it from GMATfix question bank. Yes, this question could resemble an actual GMAT question. Definitely interesting! Though, it is a higher level question, let us say 49 and above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink]

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27 Nov 2010, 19:47
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VeritasPrepKarishma wrote:
a is an odd integer and a <>-b

Is $$\frac{a^2}{|b+a|}> a-b ?$$

(1) ab=0
(2)$$a^3>a$$

PLease explain the shortest way to solve this

Thanks

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that $$\frac{a^2}{|b+a|}$$ is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Thats a wrong assumption. Question tells us a is an odd integer.

Stmnt 2: $$a^3>a$$. This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.

In equation
$$\frac{a^2}{|b+a|} + b-a > 0$$,
the $$b+a$$ part can be either positive or negetive, so we must consider both to remove the MOD. The two conditions we get, after applying a little algebra, are:
$$\frac{b^2}{b+a} > 0$$ --------------------------------E1
and
$$\frac{b^2 - 2a^2}{b+a} > 0$$-------------------------E2
any statement must satisfy both the equations to be considered good.

S1 gives us b=0 (since a is an odd integer), putting that in E1 and E2 above we get:
E1: LHS = 0, RHS = 0. Answers the question.
E2: $$-2a>0 ?$$. Does not answer the question.
S1 Not sufficient

S2 gives us a>0, putting that in E1 and E2 above does not answer the question.
S2 Not sufficient

Putting b=0 and a>0 in E1 and E2 above:
E1: LHS = 0, RHS = 0. Answers the question.
E2: $$-2a>0 ?$$(where a is positive int). Answers the question.

Both together are sufficient. C
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Vaibhav

PS: Correct me if I am wrong.

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Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink]

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30 Nov 2010, 04:38
VeritasPrepKarishma wrote:
a is an odd integer and a <>-b

Is $$\frac{a^2}{|b+a|}> a-b ?$$

(1) ab=0
(2)$$a^3>a$$

PLease explain the shortest way to solve this

Thanks

Alternate approach:
I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.
First thing that strikes me about the question stem is that $$\frac{a^2}{|b+a|}$$ is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.
If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.
Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Stmnt 2: $$a^3>a$$. This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.

Hi I have a question on solving $$a^3>a$$. I see that you do it directly, but then if I need to solve it ->
$$a^3>a$$ => $$a^3-a>0$$ =>$$a(a^2-1)>0$$ => thats means either a>0 (i) or $$a^2>1$$ -> a>+/- 1 => a>1 or a<-1 (ii)
Final value for a from (i) and (ii) => a<-1 and a>1

@Karishma -Please confirm whether this solution to a is correct by inequality method.
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Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ? [#permalink]

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30 Nov 2010, 06:27
oldstudent wrote:
Hi I have a question on solving $$a^3>a$$. I see that you do it directly, but then if I need to solve it ->
$$a^3>a$$ => $$a^3-a>0$$ =>$$a(a^2-1)>0$$ => that means either a>0 (i) or $$a^2>1$$ -> a>+/- 1 => a>1 or a<-1 (ii)
Final value for a from (i) and (ii) => a<-1 and a>1

@Karishma -Please confirm whether this solution to a is correct by inequality method.

The problem lies in your interpretation as shown by the colored part.
Think ab > 0. When will this happen? When either both a and b are positive or both a and b are negative.
Similarly when ab < 0, we say that either a < 0 and b > 0 or a > 0 and b < 0 i.e. one and only one of a and b is negative.
Only when ab = 0, can we say that either a = 0 or b = 0 or both are 0.

Here is how you will solve the inequality:
$$a^3>a$$ => $$a^3-a>0$$ =>$$a(a^2-1)>0$$=>$$(a-1)a(a + 1)>0$$
For $$(a-1)a(a + 1)$$ to be positive, either all (a - 1), a and (a + 1) should be positive, or exactly two of them should be negative. Then the entire product will be positive.
To deal with this easily, take them on the number line. Plot 1, 0 and -1 on the number line. Now your number line is divided into 4 sections. Start from the rightmost section. That will always be positive. The next section will be negative, next positive again, next negative again and so on...
Attachment:

Ques1.jpg [ 3.59 KiB | Viewed 1455 times ]

The green portions show where the expression will be positive and blue portions show where it will be negative.
So you get x > 1 or -1 < x < 0.
For detailed explanation of why this method works, check out my earlier post:
http://gmatclub.com/forum/inequalities-trick-91482.html#p804990
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Re: a is an odd integer and a ≠ -b. Is a^2/|b+a| > a-b ?   [#permalink] 30 Nov 2010, 06:27
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