Answer: B

|a | > |d|
A: g(|a| * b * e)
B. g(b * e * |d|)

|a|*b^3 * c^2 * |d| * e^5 * f^6 * g <0
We analyze the above expression,
|a|, c^2, |d|, f^6 are definitely positive,
Odd numbers(1 or 3) of other factors(b^3, e^5, g) are negative. Then either one of (b, e, g) are negative or the three of them

A: g(|a| * b * e)
As explained in above, either the three of (b, e, g) are negative or one of them
1: three of them are negative then we will have :
g(|a| * b * e) = g(positive * negative * negative) = g(positive) = negative (because g is considered as negative)
2.one of them is negative ( for example e):
g(|a| * b * e) = g(positive * positive * negative) = g(negative) = negative (because g is considered as positive here)
[the same situation happens when g or b is negative)

B. g(b * e * |d|)
As explained in above, either the three of (b, e, g) are negative or one of them
1: three of them are negative then we will have :
g(|d| * b * e) = g(positive * negative * negative) = g(positive) = negative (because g is considered as negative)
2.one of them is negative ( for example e):
g(|d| * b * e) = g(positive * positive * negative) = g(negative) = negative (because g is considered as positive here)
[the same situation happens when g or b is negative)

So we see both of them are negative, the one which is less is bigger, as |a| > |d| then B is bigger than A.

Ok..
Too many variables.. What do we do? - we try to remove as many variables as possible..

\(|a| * b^3 * c^2 * |d| * e^5 * f^6 * g < 0\)
Of course we cannot find the values of variables but it can tell us what all lead to a NEGATIVE value..
so discard the positive terms as they do not affect the equation..
\(|a| * b^3 * c^2 * |d| * e^5 * f^6 * g < 0...........b^3*e^5*g<0\)
Now whatever be the values of b, g and e, \(b*g*e<0\)

With this information let us see if we can compare A and B.
\(g( |a| * b * e)\))(\(g(b * e * |d|\))
Now both have three terms same, so we compare |a| and |d| and we know |a|>|d|, so the numeric value ||a|*b*g*e|>||d|*b*g*e|
BUT since b*g*e<0 both A and B are NEGATIVE..

We know larger the negative value, the smaller it is.
so |A|>|B| hence A<B..

B

Hello!

Fast question...

Supposing that g is -ve and b and e are +ve´s.

Can we cancel out from both statements?

Regards!

No, unless you do know their magnitude.

Thank you

g*b*e is negative, so when you divide by this, you have to inverse your relation. If you get quantity A bigger, then actual answer is B

|a|>|d|

|a|∗b3∗c2∗|d|∗e5∗f6∗g<0 - We can clearly infer that b3*∗e5*g <0 => b*e*g<0

Also given |a|>|d| => - |a| < -|d|
pick option B

A is greater than D.

Both A and D are positive. Let's assume that b, c, e, f, and g are negative.

b^3= still negative
c^2= now positive
e^5= still negative
f^6= now positive
g= still negative
Now gather it all up and multiply
(+)(+)(+)(+)(-)(-)(-)<0
A C D F Positive B E G Negative

Cancel out b and e from both Quantity A and B and all you're left with is:

g(|a|) and g(|d|)

and we know that g is negative and that A is the larger number so Answer Choice B is correct

This is far easier to do using a placeholder and keeping in mind some simple sign flipping rules.
|a|>|d|
g(|a|∗b∗e) ? g(b∗e* |d|)
g <0, so cancel and flip the sign, let the flipped sign be ?'
|a|*b *e ?' b∗e* |d|
Cancel out b and e, both are <0 so sign is flipped twice and maintains its direction, hence it is still ?'
|a| ?' |d|
As per the question, |a| > |d|, hence the sign ?' is >, so the flipped sign is <, which is the answer, hence B. C isn't possible, because we cannot have zeroes.

A

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i think i could post a better solution

we have

|a| beg

|d| beg

from the given info we can derive that b^3 * e^5 * g < 0

here g has to be -ve regardless cuz b and e have odd powers

if b is +ve then e and g have to be -ve and +ve
if e is +ve then b and g have to be -ve and +ve
if g is +ve then b and e have to be -ve and +ve

now g has no power so lets put g as +ve
then either b or e is +ve
then
|a| beg < |d| beg since |a| > |d|

if g is -ve then its gonna be the same

so B is the option