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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]

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11 Sep 2019, 21:10
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55% (hard)

Question Stats:

73% (01:29) correct 27% (01:25) wrong based on 43 sessions

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Is $$(\frac{A}{B})^3 < (AB)^3$$ ?

(1) $$A > 0$$

(2) $$AB > 0$$

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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Updated on: 11 Sep 2019, 22:08
1

(If the questions mentioned A and B are integers, then answer would have been option B)

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Originally posted by EncounterGMAT on 11 Sep 2019, 21:23.
Last edited by EncounterGMAT on 11 Sep 2019, 22:08, edited 2 times in total.
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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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11 Sep 2019, 21:46
1
Statement 1 is insufficient since we know nothing about B.
When A=3; and b=-2
(A/B)^3= -27/8 > (AB)^3=216
When A=3 and B=2,
(AB)^3 > (A/B)^3

2 is also insufficient bcos AB>0 means either both A and B are greater than 0 or both A and B are negative.
When A=3 and B=2, (AB)^3 > (A/B)^3
When A=-3 and B=-1/2, (A/B)^3 > (AB)^3
Hence insufficient.

1+2 is still insufficient.
Because we know A and B must be positive in order for both conditions to be satisfied. Meanwhile, when A=3 and B=2, as we saw already, (AB)^3 > (A/B)^3
However when A=3 and B=1/2,
(A/B)^3 > (AB)^3

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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11 Sep 2019, 23:56
1
is (A/B)^3 < (AB)^3 ?

STATEMENT (1)- A > 0
but we don't have any information about B
if A = 4 B =$$\frac{1}{4}$$ then (A/B)^3 > (AB)^3---is (A/B)^3 < (AB)^3 ? --NO

if A = 4 B = $$\frac{-1}{4}$$ then (A/B)^3 < (AB)^3---is (A/B)^3 < (AB)^3 ? --YES

from here we can't get a definite answer
so, INSUFFICIENT

STATEMENT (2)- AB > 0
from here we know A and B are either positive or negative
if A = 4 B = $$\frac{1}{4}$$ then (A/B)^3 > (AB)^3---is (A/B)^3 < (AB)^3 ? --NO

if A = 4 B = 2 then (A/B)^3 < (AB)^3---is (A/B)^3 < (AB)^3 ? --YES

from here we can't get a definite answer
so, INSUFFICIENT

combining both statements together
we know AB>0 and A>0 this tells us B>0

if A = 4 B = $$\frac{1}{4}$$ then (A/B)^3 > (AB)^3---is (A/B)^3 < (AB)^3 ? --NO

if A = 4 B = 2 then (A/B)^3 < (AB)^3---is (A/B)^3 < (AB)^3 ? --YES

from here we can't get a definite answer
so, INSUFFICIENT

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 00:04
(A/B)^3 < (AB)^3 ?

(1) A>0
Let's A = 1, B = -2; insert A,B value, we've got -1/8 < -8. NO
But A = 1, B = 2; 1/8 < 8. Yes

So, A is sufficient

(2) AB>0
Let's A = 1, B = 2; insert A,B value, we've got 1/8 < 8. Yes
But A = 2, B = 1; 8 < 8 No.
So, 2 alone is not suffcient

(1) + (2); A>0 and AB >0, thus both A,B > 0
If A = 1, B =2; 1/8 < 8 Yes.
If A =2 , B =1 ; 8 < 8 No.

Not sufficient. Therefore E is the answer
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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 00:19
Is $$(\frac{A}{B})^3 < (AB)^3$$ ?

No relation between A and B is given so many possibilities exist.

(1) $$A > 0$$
Here either $$B > A$$ for both integer value and non-integer value of positive B.
Example: Let A = 2 & B = 3 then
$$(\frac{2}{3})^3 < (2*3)^3$$ YES

OR

$$A > B$$ for either positive value of B or negative value of B.
Example: Let A = 2 & B = 1 then
$$(\frac{2}{1})^3 < (2*1)^3$$ NO

INSUFFICIENT.

(2) $$AB > 0$$

Two cases are possible here:

(a) Both A and B are +ve
Take A = 2 & B = 3 then
$$(\frac{2}{3})^3 < (2*3)^3$$ YES

(b) Both A and B are -ve
Take $$A = \frac{-1}{2}$$ & $$B = \frac{-1}{3}$$ then
$$((\frac{-1}{2})/(\frac{-1}{3}))^3 < ((\frac{-1}{2})*(\frac{-1}{3}))^3$$ NO

INSUFFICIENT.

Together 1) and 2)

We have $$A > 0$$ and $$B > 0$$
Again Take A = a2 & B = 3 then
$$(\frac{2}{3})^3 < (2*3)^3$$ YES

And

Take $$A = \frac{1}{2}$$ & $$B = \frac{1}{3}$$ then
$$((\frac{1}{2}/\frac{1}{3})^3 < ((\frac{1}{2})*(\frac{1}{3}))^3$$ NO

INSUFFICIENT.

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 00:54
1
Is $$(\frac{A}{B})^{3}<(AB)^{3}$$?

$$A^{3}(B^{3}-(\frac{1}{B})^{3}$$)>0

$$A^{3}*(B-\frac{1}{B})*(B^{2}+B*\frac{1}{B}+(\frac{1}{B})^{2}$$)>0

$$(B^{2}+B*\frac{1}{B}+(\frac{1}{B})^{2}$$) is always greater than zero.
--> $$A^{3}*(B-\frac{1}{B})$$>0 ???

(1) A > 0

(2) AB > 0

Statement1:
A > 0
--> (B-$$\frac{1}{B}$$)>0 ???

If B=2, then 2-$$\frac{1}{2}$$>0 (Yes)
If B=$$\frac{1}{2}$$, then $$\frac{1}{2}$$-2>0 (NO)

Insufficient.

Statement2:
AB > 0
--> Both A and B are Positive or Negative:
$$A^{3}*(B-\frac{1}{B})$$>0

if A=B=2, then --> $$2^{3}(2-\frac{1}{2})$$>0 (yes)
if A=B=$$\frac{1}{2}$$, then --> $$(\frac{1}{2})^{3}(\frac{1}{2}-2)$$>0 (NO)

Insufficient.

Taken together 1 and 2,
A>0 and AB>0 --> B>0.

$$A^{3}*(B-\frac{1}{B})$$>0 --> (B-$$\frac{1}{B}$$)>0 ???

if B=2, then --> (2-$$\frac{1}{2}$$)>0 (yes)
if B=$$\frac{1}{2}$$, then -->($$\frac{1}{2}$$-2)>0 (NO)

Insufficient

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 02:28
#1
A>0 so A is +ve but no relation of B with A insufficient
#2
AB>0
so either both AB are +ve or -ve also they can be same value /integer/ fraction no given info
insufficient
from 1 &2
we can say that AB are +ve but whether A=B ; A>B or B<A cannot be determined
IMO E

Is(AB)3<(AB)3 ?

(1) A>0

(2) AB>0
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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 03:34
IMO it's E.
Because even after combining if we take A=100 and B=1 then NO is the answer.
And if A=100 and B=2 then YES is the answer.

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 04:32
Quote:
Is $$(A/B)^3<(AB)^3$$?

(1) A>0
(2) AB>0

rephrase: $$(A/B)^3<(AB)^3…A/B<AB…A/B<AB$$; Is $$A/B<AB$$?

case 1: $$A,B=(1,2)…A/B<AB…1/2<2:true$$
case 2: $$A,B=(10,1)…A/B<AB…10<10:false$$

(1) A>0: case 1 and 2, insufi.
(2) AB>0: case 1 and 2, insufi.
(1&2) A,B>0: case 1 and 2, insufi.

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Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0  [#permalink]

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12 Sep 2019, 07:35
Is (A/B)^3 <(AB)^3 ?

(1) A>0
if A>0, then
$$(A/B)^3<(AB)^3$$
$$A^3/B^3<A^3*B^3$$
$$1/B^3<B^3$$
if B = 1, $$1/1^3<1^3$$ is not possible
if B = 2, $$1/2^3<2^3$$ is possible

(2) AB>0
A>0 & B>0 (I) or A<0 & B<0 (II)
condition (I) is same as (1), so no unique answer
we don't have to check condition (II)

(1) & (2)
A>0 (from 1), so B>0
again no unique solution if we take B = 1 or 2
Re: Is (A/B)^3 < (AB)^3 ? (1) A > 0 (2) AB > 0   [#permalink] 12 Sep 2019, 07:35
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