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Re: GRE Weekly Challenge #4 [#permalink]
Quantity B is greater.

Explanation:


in this,
|a|, |d|, c^2, f^6 are always positive.
therefore b*e*g < 0 (b^3 = b if we consider contribution of + or - sign) ..............(i)

also |a| > |d|...............(ii)

# either one of b, e and g is negetive or all three are negetive.

Case 1 - g is negetive then b*e is positive.
so, (|a|*b*e) > (b*e*|d|). .........using (ii)
when a negetive g is multiplied, g(|a|*b*e) < g(b*e*|d|).


Case 2 - all three are negetive
so like in previous case g(|a|*b*e) < g(b*e*|d|). [as b*e is positive; (-)x(-)= (+)]

Case 3 - g is positive, any one of b or e is negetive.
so, (|a|*b*e) < 0 and (b*e*|d|) < 0 but (|a|*b*e) is more negetive because |a|> |d|.
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Re: GRE Weekly Challenge #4 [#permalink]
given:
| a | > | d | -->eqn 1
as the above has modulus on both side we know |a| and |b| are positive.

| a |*b^3*c^2*| d |*e^5*f ^6*g < 0 -->eqn 2
from eqn 2 , it is seen that all the even powers will be positive i.e c^2 and f^6,
hence, b^3*e^5*g<0
from the above we can deduce b*e*g <0 (or a -ve value)

the question asked to compare the two terms:
first term :g(| a |*b*e) -->g*b*e <0 and |a| >0
2nd term : g(b*e*| d |) --->g*b*e <0 and |b| >0

hence ,
from eqn 1, we know |a| >|b|. therefore
-|a|<-|b|

the answer willl be B
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Re: GRE Weekly Challenge #4 [#permalink]
Quantity B is greater.
according to the second statement b,e,g have odd power....... so either all of them or any one of them is negative...
so therefore b*g*e < 0.......
since |a|>|d|......... therefore magnitude of A is greater than Magnitude of B.........
And due to minus sign B will be greater...
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Re: GRE Weekly Challenge #4 [#permalink]
Wow great topic to discuss and have a chance to win.
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Re: GRE Weekly Challenge #4 [#permalink]

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