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# Is a<0?

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Is a<0?  [#permalink]

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Updated on: 05 Aug 2013, 22:29
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Difficulty:

75% (hard)

Question Stats:

57% (02:16) correct 43% (02:22) wrong based on 183 sessions

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Is a<0?

(1) $$a^3<a^2+2a$$
(2) $$a^2>a^3$$

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Originally posted by bagdbmba on 05 Aug 2013, 22:28.
Last edited by mau5 on 05 Aug 2013, 22:29, edited 1 time in total.
Edited the Q
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Re: Is a<0?  [#permalink]

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05 Aug 2013, 22:37
1
1
bagdbmba wrote:
Is a<0?

(1) $$a^3<a^2+2a$$
(2) $$a^2>a^3$$

From F.S 1, add 1 on both sides : $$a^2+2a+1>a^3+1 \to (a+1)^2-(a+1)(a^2+1-a)>0 \to (a+1)[(a+1)-(a^2+1-a)]>0 \to (a+1)(2a-a^2)>0$$
Thus, we get a(a+1)(a-2)<0. Either 0<a<2 OR a<-1. Insufficient.

From F.S 2, we can divide by$$a^2$$ on both sides and we get a<1. Insufficient.

Taking both together, we know that a<1. Thus, a could be 0<a<1 OR a<-1. Insufficient.

E.
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Re: Is a<0?  [#permalink]

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05 Aug 2013, 22:42
bagdbmba wrote:
Is a<0?

(1) $$a^3<a^2+2a$$
(2) $$a^2>a^3$$

STMNT 1:

$$a^3<a^2+2a$$
$$a^3-a^2-2a < 0$$
a(a+1)(a-2)<0
when a = -2 expression above is= -8 which is < 0
when a = 1 expression above is = -2 which is <0
hence a can be -ve /+ve insufficient

STMNT 2:
$$a^2$$$$> a^3$$
$$a^2(a-1)<0$$
clearly this satisfies for a<1
hence a can be -ve /+ve insufficient

combining both
still both statement satisfies for a = 0.5 and a =-2
hence insufficient

hence E
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Re: Is a<0?  [#permalink]

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05 Aug 2013, 22:50
bagdbmba wrote:
Is a<0?

(1) $$a^3<a^2+2a$$
(2) $$a^2>a^3$$

st(1) , add 1 to both sides. then a= -1 and a= 1/2. you will have a double case.
st(2), a can be any positive fraction here and a can be any negative integer too such as a= 1/2 and a = -2 . a double case too.
so both statement depict the same answers, which are both a double case.
so Answer is (E)
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Re: Is a<0?  [#permalink]

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08 Aug 2013, 00:45
1
kumar23badgujar wrote:
Consider,

Statement (1): $$a^3 < a^2 + 2a$$

$$a^3 - a^2 - 2a < 0$$
$$a(a^2 - a - 2) < 0$$
$$a(a -2)(a + 1) < 0$$

i.e. a = 0 or a = 2 or a = -1, hence not sufficient.

Statement (2): $$a^2 > a^3$$
This simply means,
a < 1, hence not sufficient.

Combining both statements, we get a = 0 or a = -1, Hence both statements together not sufficient.

Correct Ans: E

Even though you have the correct answer, I am sorry but it is not what it means(the highlighted part). It is not an equality, rather an in-equality.Please refer through the above posts, for the correct method.
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Is a < 0 ?  [#permalink]

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Updated on: 20 Sep 2013, 10:48
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Is a < 0 ?

(1) a^3 < a^2 + 2a
(2) a^2 > a^3

Originally posted by EnemyInside on 20 Sep 2013, 10:45.
Last edited by Bunuel on 20 Sep 2013, 10:48, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is a < 0 ?  [#permalink]

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21 Sep 2013, 03:33
1
1
pjagadish27 wrote:
Statement 1 -> a^3-a^2-2a<0.
a(a-2)(a+1)<0=>a<0 or a<2 or a<-1. Not Sufficient.

Statement 2->a^2-a^3>0=>a^2(1-a)>0 => a>0 or a<1.Not Sufficient.

1&2,still a<2 exists which does not answer the question is a<0. So E.

The answer is E, but the ranges are not correct.

Is a < 0 ?

(1) a^3 < a^2 + 2a --> $$(a+1)a(a-2)<0$$ --> $$a<-1$$ or $$0<a<2$$. Not sufficient.

(2) a^2 > a^3 --> $$a^2(1-a)>0$$ --> $$a<0$$ or $$0<a<1$$. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$a<-1$$ or $$0<a<1$$. Not sufficient.

Answer: E.

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Re: Is a < 0 ?  [#permalink]

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09 Oct 2013, 00:49
Bunuel why can't we factorize a^2>a^3 as 1>a or a<1.
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Re: Is a < 0 ?  [#permalink]

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09 Oct 2013, 01:00
1
mohnish104 wrote:
Bunuel why can't we factorize a^2>a^3 as 1>a or a<1.

a<1 implies that a can be 0. But a=0 does not satisfy a^2>a^3, so the correct ranges for which this inequality holds true is a<0 or 0<a<1 (the same range as you have excluding 0).

Hope it helps.
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Re: Is a < 0 ?  [#permalink]

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18 Jan 2014, 09:58
Bunuel wrote:
mohnish104 wrote:
Bunuel why can't we factorize a^2>a^3 as 1>a or a<1.

a<1 implies that a can be 0. But a=0 does not satisfy a^2>a^3, so the correct ranges for which this inequality holds true is a<0 or 0<a<1 (the same range as you have excluding 0).

Hope it helps.

Hi,

There's a concept related to inequalities that I fail to understand. Could you please tell me what I'm doing wrong?

For statement 2 we have "a^2 > a^3". Depending on how I solve this, I'm getting two complete different ranges, both of which are incorrect.

First case:

a^2 - a^3 > 0
a^2*(1-a)>0------> this gives me the critical points 0 and 1.

Hence I have 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1.

As the inequality has the ">0" sign I took only the 1st and 3rd range and got "a<0 or a>1".

Second case:

0>a^3 - a^2
0>a^2*(a-1)------> this also gives me the critical points 0 and 1.

Hence I have the same 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1.

However, this time we have "0>" sign, so I took only the second range: 0<a<1.

In any case, both are false. What's odd is that I used the same method to find the range for statement 1 but I got the correct answer.

Thanks a lot for your help.

Aurèle
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Re: Is a < 0 ?  [#permalink]

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19 Jan 2014, 09:34
Aurele wrote:
Bunuel wrote:
mohnish104 wrote:
Bunuel why can't we factorize a^2>a^3 as 1>a or a<1.

a<1 implies that a can be 0. But a=0 does not satisfy a^2>a^3, so the correct ranges for which this inequality holds true is a<0 or 0<a<1 (the same range as you have excluding 0).

Hope it helps.

Hi,

There's a concept related to inequalities that I fail to understand. Could you please tell me what I'm doing wrong?

For statement 2 we have "a^2 > a^3". Depending on how I solve this, I'm getting two complete different ranges, both of which are incorrect.

First case:

a^2 - a^3 > 0
a^2*(1-a)>0------> this gives me the critical points 0 and 1.

Hence I have 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1.

As the inequality has the ">0" sign I took only the 1st and 3rd range and got "a<0 or a>1".

Second case:

0>a^3 - a^2
0>a^2*(a-1)------> this also gives me the critical points 0 and 1.

Hence I have the same 3 ranges: (1st) a<0, (2nd) 0<a<1, and (3rd) a>1.

However, this time we have "0>" sign, so I took only the second range: 0<a<1.

In any case, both are false. What's odd is that I used the same method to find the range for statement 1 but I got the correct answer.

Thanks a lot for your help.

Aurèle

0 is not a critical point. The squared terms (basically even powers) must be ignored because they cannot be negative and hence doesn't affect the sign.
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Re: Is a < 0 ?  [#permalink]

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21 Jan 2014, 06:42
Bunuel wrote:

0 is not a critical point. The squared terms (basically even powers) must be ignored because they cannot be negative and hence doesn't affect the sign.

Great, thanks for the reply. Just to be sure that I understand the concept fully, could you tell me please if I'm solving the following equation correctly (from Manhattan books):

$$x^6 - x^7 > x^5 - x^6$$

I factor the equation and get:

$$x^5*(x-1)^2 < 0$$

Here, If I understand your explanation well, $$x^5$$ is raised to an odd power; hence, it should be considered, as it can yield a negative or positive result. Conversely, $$(x-1)^2$$ is raised to an even power; thus, we ignore it.

The critical point is then : $$0$$

Because we have the $$<0$$ sign, we'd get $$x<0$$. Would this be correct?
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Re: Is a < 0 ?  [#permalink]

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29 Jan 2015, 21:49
Is a<0?
(1) Insufficient. If a=-5, then -125 < 25-10 --->Yes
If a=1/2, then 1/8 < 1/4+1 --->No

(2) Insufficient. If a=-5, then 25>-125 --->Yes
If a=1/2, then 1/4>1/8 --->No
(1)+(2) Insufficient. We can use the same numbers and we get two different answers
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Is a<0? Help understanding why my solution is not valid.  [#permalink]

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02 Jun 2015, 10:37
Hello! I was looking through these posts, answering questions, and I came across this one:

Is a<0?

(1) a3<a2+2a
(2) a2>a3

from: forum/is-a-157421.html

In the topic above, the answer is that both statements together are insufficient to answer the problem but surely the only case in which $$a^{2}$$ is greater than$$a^{3}$$ is if a is negative?
My answer is that both statements alone can answer the question, as the only way $$a^{3}$$ is smaller than what it is compared to is if a is negative. What am I overlooking?

Thank you.
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Re: Is a<0?  [#permalink]

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02 Jun 2015, 21:02
1
cmcmcm wrote:
Hello! I was looking through these posts, answering questions, and I came across this one:

Is a<0?

(1) a3<a2+2a
(2) a2>a3

from: forum/is-a-157421.html

In the topic above, the answer is that both statements together are insufficient to answer the problem but surely the only case in which $$a^{2}$$ is greater than$$a^{3}$$ is if a is negative?
My answer is that both statements alone can answer the question, as the only way $$a^{3}$$ is smaller than what it is compared to is if a is negative. What am I overlooking?

Thank you.

Hi cmcmcm,

Always be very careful on how you are finding out the range of an inequality. Let me help you out with finding the range of a for both the inequalities.

Statement-I
$$a^3<a^2+2a$$ can be simplified to $$a(a + 1)(a - 2) < 0$$. Using the wavy line method to find the range of $$a$$ with the zero points being 2, 0 and -1.

We can see from the wavy line diagram that the inequality is negative in the range where $$a < -1$$ or $$0 < a < 2$$. Hence, you can't say for sure if $$a < 0$$ using statement-I alone

Statement-II
$$a^2>a^3$$ can be simplified to $$a^2(a - 1) < 0$$. Since $$a^2$$ is always non-negative, for $$a^2(a - 1) < 0$$, $$(a - 1) < 0$$ i.e. $$a < 1$$.
So $$a < 0$$ or $$a > 0$$. Hence using statement-II alone you cant' say for sure if $$a < 0$$.

Combining statement-I & II
Combining statements-I & II will give us the range as $$a < -1$$ or $$0 <a <1$$. Hence, it is not sufficient to tell if $$a < 0$$. Therefore the answer is E.

You can read more about the Wavy line method here.

Hope it's clear .Let me know if you have any doubt in any part of the explanation.

Regards
Harsh
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Re: Is a<0?  [#permalink]

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03 Jun 2015, 14:26
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Hi cmcmcm,

The prompt does NOT state that "A" has to be an integer, so you have to consider the possibility that it's NOT an integer (meaning "A" could be a fraction).

While that level of 'thoroughness' isn't going to be required on that many DS questions, Test Takers who score at the higher levels in the Quant section are more likely to see questions in which fractional answers have to be considered.

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Re: Is a<0?  [#permalink]

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03 Jun 2015, 23:44
E
(1) take a = -1, -2 ---- If a=-1 then eqn gives -1<-1 which is not true and if a=-2 then eqn gives -8<0 which is true ....thats y insuff
(2) a^2>a^3 implies a can be -ve or 0<a<1 thus insuff....
Combined also is insuff because of above reasons
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Re: Is a < 0 ?  [#permalink]

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04 Jun 2015, 00:12
(1) a^3 < a^2 + 2a - --- a=-1 then -1<-1 not true ; a=-2 then -8<0 true insuff
(2) a^2 > a^3 then a<0 or 0<a<1 insuff

If we combine both we cannot answer bc of above reasons
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Re: Is a<0?   [#permalink] 30 Mar 2018, 20:20
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