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

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Is (A/B)^3<(AB)^3 1) A>0 2) AB>0 [#permalink] New post 05 Sep 2011, 01:18
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

78% (01:40) correct 22% (01:28) wrong based on 9 sessions
Is (A/B)^3<(AB)^3

1) A>0
2) AB>0
[Reveal] Spoiler: OA
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Re: Inequalities!! [#permalink] New post 05 Sep 2011, 05:33
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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Re: Inequalities!! [#permalink] New post 05 Sep 2011, 07:33
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..
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Re: Inequalities!! [#permalink] New post 05 Sep 2011, 16:36
\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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Re: Inequalities!! [#permalink] New post 05 Sep 2011, 22:06
Plug in numbers (1/2 ,1 ) and check
(-1/2,-1) .You will get " E "
+1 for E
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Re: Inequalities!! [#permalink] New post 05 Sep 2011, 23:08
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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Re: Inequalities!! [#permalink] New post 09 Sep 2011, 07:23
E. You dont know if either a or b is a fraction which could change your result
Re: Inequalities!!   [#permalink] 09 Sep 2011, 07:23
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