is a>0 ?
1. a^3 - a <0
2. 1 -a^2 > 0
I have TWO issues with this question
Answer is 'C' - but I seem to be getting 'A'
and by solving (1) - via substitution and via equation I seem to be getting TWO ranges for A. Can someone help me figure out this flaw?
1) a^3 - a < 0
= a(a^2 - a) < 0
= a <0 or a^2 - 1<0 ( -1 < a < 1)
taking the limiting of the two I get
-1 < a < 0 THUS PROVING A
However, if I use the values of a in (1) via this result set - I DONT get a^3-a < 0!!!
a=-0.9
a^3-a = -0.729 - (-0.900) = +0.171 which is > 0!!!
How is the answer C? and HOW is it that the result set of the inequality equation I've solved is not representative of the original inequality?
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Data Sufficiency problem
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Balrog1978
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Anurag@Gurome
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This is an OR case! You're considering this as an AND case!Balrog1978 wrote:1) a^3 - a < 0
= a(a^2 - a) < 0
= a <0 or a^2 - 1<0 ( -1 < a < 1)
taking the limiting of the two I get
-1 < a < 0 THUS PROVING A
Also you're missing "something" very important.
Either a is negative (in which case (a² - 1) must be positive) or (a² - 1) is negative (in which case a must be positive).
Thus first condition reduces to a < -1 and second condition reduces to 0 < a < 1.
Therefore either a < 1 or 0 < a < 1.
For the solution of the question, see below
Statement 1: (a³ - a) < 0Balrog1978 wrote:is a>0 ?
1. a^3 - a <0
2. 1 -a^2 > 0
=> a(a² - 1) < 0
=> a(a + 1)(a - 1) < 0
Consider the number line with -1, 0 and 1 marked on it. In each of these points the expression will change its sign. For any number on the left of -1, the expression is negative. Therefore the expression is negative for a < -1 or 0 < a < 1
Not sufficient.
Statement 2: (1 - a²) < 0
=> a² > 1
=> a < -1 or a > 1
Not sufficient.
1 & 2 Together: Only range of a that satisfies both the condition is a < -1
Sufficient
The correct answer is C.
Last edited by Anurag@Gurome on Wed Dec 29, 2010 7:30 am, edited 2 times in total.
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anshumishra
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Balrog1978 wrote:is a>0 ?
1. a^3 - a <0
2. 1 -a^2 > 0
I have TWO issues with this question
Answer is 'C' - but I seem to be getting 'A'
and by solving (1) - via substitution and via equation I seem to be getting TWO ranges for A. Can someone help me figure out this flaw?
1) a^3 - a < 0
= a(a^2 - a) < 0
= a <0 or a^2 - 1<0 ( -1 < a < 1)
taking the limiting of the two I get
-1 < a < 0 THUS PROVING A
However, if I use the values of a in (1) via this result set - I DONT get a^3-a < 0!!!
a=-0.9
a^3-a = -0.729 - (-0.900) = +0.171 which is > 0!!!
How is the answer C? and HOW is it that the result set of the inequality equation I've solved is not representative of the original inequality?
The question asks is a > 0, that means , is it always greater than 0 ?1) a^3 - a < 0
= a(a^2 - a) < 0
= a <0 or a^2 - 1<0 ( -1 < a < 1)
taking the limiting of the two I get
-1 < a < 0 THUS PROVING A
When you combine the statements 1 and 2 :
a(a^2-1) < 0 and (1-a^2) > 0
=> a(a^2-1) < 0 and (a^2-1) < 0
=> a > 0. So We can definitely say that a is greater than 0 (always - and it lies in (0,1)).
Hence C.
Last edited by anshumishra on Wed Dec 29, 2010 10:15 am, edited 3 times in total.
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
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Geva@EconomistGMAT
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Balrog1978 wrote:is a>0 ?
1. a^3 - a <0
2. 1 -a^2 > 0
I have TWO issues with this question
Answer is 'C' - but I seem to be getting 'A'
and by solving (1) - via substitution and via equation I seem to be getting TWO ranges for A. Can someone help me figure out this flaw?
1) a^3 - a < 0
= a(a^2 - 1) < 0
For the product of tw factors to be a negative number, you need them to have different signs: one positive, the other negative, or vice versa. your two scenarios are therefore not separate for each factor, but integrated:
a. Either a>0 AND a^2-1<0 (the intersection of which is 0<a<1)
OR
b. a<0 and a^2-1>0 (the intersection of which is any a<-1).
Both these ranges of a will satisfy the inequality: either positive fractions or negative numbers smaller than -1. Since a can be positive or negative, (1) is insufficient.
Negative fractions are actually not in the range of solutions for a.
With these two scenarios in mind, you can see why the combination is sufficient: switch sides from (2) to get a^2-1<0, so that the 2nd scenario is eliminated, leaving you only with the first scenario - where a is limited to positive fractions, and the answer is a definite "yes".
= a <0 or a^2 - 1<0 ( -1 < a < 1)
taking the limiting of the two I get
-1 < a < 0 THUS PROVING A
However, if I use the values of a in (1) via this result set - I DONT get a^3-a < 0!!!
a=-0.9
a^3-a = -0.729 - (-0.900) = +0.171 which is > 0!!!
How is the answer C? and HOW is it that the result set of the inequality equation I've solved is not representative of the original inequality?
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fskilnik@GMATH
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Hi there!Balrog1978 wrote: a^3 - a < 0
= a(a^2 - a) < 0
= a <0 or a^2 - 1<0 ( -1 < a < 1)
taking the limiting of the two I get
-1 < a < 0 THUS PROVING A
However, if I use the values of a in (1) via this result set - I DONT get a^3-a < 0!!!
a=-0.9
a^3-a = -0.729 - (-0.900) = +0.171 which is > 0!!!
a^3-a < 0 is equivalent to a(a^2 - 1) < 0 that is also equivalent to a(a+1)(a-1) < 0 ...
To solve this, please understand that a(a+1)(a-1) is a 3-degree polynomial and from the fact that a(a+1)(a-1) = 0 has only the roots 0, 1 and -1, we must decide if it is a curve passing through the points (-1,0), (0,0) and (1,0) AND like "S" from low left to upper right or from upper left to lower right. Easy decision: please note that taking a=10 we have a(a+1)(a-1) > 0 therefore it is the FIRST "S" I mentioned... (I hope you understand what I mean, if not, ask)
From this "graphical analysis" I could guarantee that a´s that satisfy sttm(1) are a < -1 or 0<a<1... Got it?
Regards,
Fabio.
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fskilnik@GMATH
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Geva´s analysis is more elementary than mine (what is good) but a bit more lengthly (what is bad).
P.S.: He used the term "fraction" as a non-integer value between 0 and 1 (or -1), by the way... this is mathematically "not-nice" but it is pretty common in "GMAT literature"...
P.S.: He used the term "fraction" as a non-integer value between 0 and 1 (or -1), by the way... this is mathematically "not-nice" but it is pretty common in "GMAT literature"...
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Balrog1978
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Dear god!!!
What a ghastly oversight on my part!
I took the TWO as an AND condition!! I guess practice makes perfect!!
Thanks guys!
What a ghastly oversight on my part!
I took the TWO as an AND condition!! I guess practice makes perfect!!
Thanks guys!
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Night reader
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fskilnik@GMATH
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Hi Night reader,
Congrats for your will to look into the problem deeper. Just some small observations to retribute this effort:
01. The curve shown in (1) is exactly the "S" I mentioned, but be careful because the curve will not intercept the x-axis more than the 3 times you´ve shown. (I say that because the drawing could make someone believe differently.)
02. It´s not a "region" we are looking for... remember a is a real number, thefore all the analysis is very nice, but the conclusions are to be taken only in the x-axis.
Regards,
Fabio.
Congrats for your will to look into the problem deeper. Just some small observations to retribute this effort:
01. The curve shown in (1) is exactly the "S" I mentioned, but be careful because the curve will not intercept the x-axis more than the 3 times you´ve shown. (I say that because the drawing could make someone believe differently.)
02. It´s not a "region" we are looking for... remember a is a real number, thefore all the analysis is very nice, but the conclusions are to be taken only in the x-axis.
Regards,
Fabio.
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fskilnik@GMATH
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The approach I recommend:Balrog1978 wrote:is a>0 ?
1. a^3 - a <0
2. 1 -a^2 > 0
(1) This sttm is equivalent to a(a^2-1) < 0, that is, a and a^2-1 (are non-zero and) have opposite signs. This is not enough because of the (easily found) bifurcation:
> Take a equals to -2 (therefore a^2-1 is positive) answering in the negative ;
> Take a equals to 1/2 (therefore a^2-1 is negative) answering in the positive ;
(2) This sttm is equivalent to |a| > 1, that is, a < -1 (answering in the negative) OR a > 1 (answering in the positive).
(1+2) DECIDES!
From sttm (2) we have only two options, one of them ( a > 1) not possible because in this scenario a and a^2-1 would have same (positive) signs!
That means a < -1 , answering the question in the negative.
Regards,
Fabio.
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