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Is (A/B)^3<(AB)^3 1) A>0 2) AB>0

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Senior Manager
Joined: 16 Feb 2011
Posts: 258
Is (A/B)^3<(AB)^3 1) A>0 2) AB>0 [#permalink]

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05 Sep 2011, 02:18
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:34) correct 33% (00:29) wrong based on 18 sessions

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Is (A/B)^3<(AB)^3

1) A>0
2) AB>0
[Reveal] Spoiler: OA
Intern
Joined: 11 May 2011
Posts: 22

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05 Sep 2011, 06:33
Hello,

Option E: Both alone are in sufficient and both together are insufficient too.

Consider 1)

A>0 from this alone we cannot say anything.
1. Do not know if they are integers or fractions.
2. Do not know if B is positive or negative.

If fraction then consider and even if both positive:

A= 1/2 and B = 1/3
(A/B)^3 = ((1/2)/(1/3))^3 = (3/2)^3 = 27/8
(AB)^3 = (1/6)^3 = 1/216

Please note that since the power on both sides in ODD. Since ODD powers do not alter the sign we have

In sufficient.

Consider 2)
AB>0
1. Both are positive or both are negative.
2. Do not know if they are integers or fractions.

If fraction then consider:

A= 1/2 and B = 1/3
(A/B)^3 = ((1/2)/(1/3))^3 = (3/2)^3 = 27/8
(AB)^3 = (1/6)^3 = 1/216

In sufficient.

Consider both 1 and 2 :

1. A is positive hence B is also positive
2. Do not know if they are integers or fractions.

If integers
then the equation is true

If fraction then consider:

A= 1/2 and B = 1/3
(A/B)^3 = ((1/2)/(1/3))^3 = (3/2)^3 = 27/8
(AB)^3 = (1/6)^3 = 1/216

Hence both together are insufficient.

Regards
Raghav.V

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Senior Manager
Joined: 16 Feb 2011
Posts: 258

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05 Sep 2011, 08:33
Thanks Raghava..your help is much appreciated..

Is there a simpler way instead of computing random numbers as they are time consuming and i somehow donot seem to select the right set hence, get lost in the statements..
Intern
Joined: 17 May 2009
Posts: 31
Location: United States
GMAT 1: 770 Q51 V44
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05 Sep 2011, 17:36
$$\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{a}{b} < ab \cdot ab \cdot ab$$?

I simplified the expression a bit before carrying on:

$$\frac{a}{b} \cdot \frac{a^2}{b^2} < a^2 \cdot b^2 \cdot ab$$?

$$\frac{a}{b} < ab^5$$?

Statement 1) if $$a > 0$$, then we need to know whether $$\frac{1}{b} < b^5$$. It can clearly be seen that this inequality is violated for $$0< b < 1$$ while it holds for $$b > 1$$.

Statement 2) if $$ab>0$$, then $$a$$ and $$b$$ have the same sign, so $$\frac{a}{b}>0$$ also. We can then simplify the question stem further and discover that we now need to know whether $$b^6 > 1$$. Again, this inequality holds for $$b > 1$$, but not for $$0 < b < 1$$

Combined) I used the same ranges for $$b$$ to demonstrate the same yes/no examples in each statement. The combined statements are insufficient.
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Joined: 13 Apr 2010
Posts: 166
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05 Sep 2011, 23:06
Plug in numbers (1/2 ,1 ) and check
(-1/2,-1) .You will get " E "
+1 for E
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Joined: 20 Dec 2010
Posts: 2010

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06 Sep 2011, 00:08
DeeptiM wrote:
Is (A/B)^3<(AB)^3

1) A>0
2) AB>0

Let's take the back solving approach; Just try to eliminate "E" as the answer:

So,
A>0 & AB>0 are both TRUE. means; A and B are both +ves.

Now, solve the expression
$$(\frac{A}{B})^3<(AB)^3$$
$$B^6>1$$
OR
$$B>1$$

But, we won't be able to tell that using both the statements.

0<B<=1. Is (A/B)^3<(AB)^3? No.
B>1. Is (A/B)^3<(AB)^3? Yes.
Not Sufficient.

Ans: "E"
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Joined: 08 Sep 2011
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09 Sep 2011, 08:23
E. You dont know if either a or b is a fraction which could change your result
Re: Inequalities!!   [#permalink] 09 Sep 2011, 08:23
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