Is (A/B)^3 0 (2) AB 0

Question

Competition Mode Question

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

(1)  A > 0

(2)  AB > 0

EncounterGMAT · Answer

IMO answer is option E

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

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eakabuah · Answer

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

Hence the answer is E.

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shridhar786 · Answer

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

E is the correct answer

LalitaSiri · Answer

(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

unraveled · Answer

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. Answer (E).

lacktutor · Answer

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

The answer is E.

Archit3110 · Answer

#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

Prasannathawait · Answer

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.

exc4libur · Answer

rephrase: (A/B)^3<(AB)^3…A/B0: case 1 and 2, insufi. (2) AB>0: case 1 and 2, insufi. (1&2) A,B>0: case 1 and 2, insufi. Answer (E)

kris19 · Answer

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
so, no unique answer

(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
So, answer is E

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

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