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GRE Weekly Challenge #4

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GRE Weekly Challenge #4 [#permalink] New post 23 Sep 2011, 02:52
GMAT Club invites you to test your GRE knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GRE Strategy guide. What are you waiting for? Get out your scrap paper and start solving! Click here to view contest & prize details

This week's question:
| a | > | d |
| a |*b^3*c^2*| d |*e^5*f ^6*g < 0

g(| a |*b*e)g(b*e*| d |)

Compare Quantity A and Quantity B using the information given above, and select one of the following answer choices:
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Please post your answer, along with the explanation, below. Get cracking! :)

Edit: This challenge is now closed


First, let’s look at the common information. If |a| > |d|, then a is further from 0 than d is, but we don’t know their signs.
Since |a|b3c2|d|e5f6g and we know that |a|, c2, |d|, and f6 must be positive (any nonzero number becomes positive when put in an absolute value sign or given an even exponent), then whatever is “left over” must be causing the entire expression to be negative.
Therefore, b3e5g < 0.
Since an odd exponent merely preserves the sign of the original number (that is, a negative number to a power of 1, 3, 5, etc. will still be negative, and a positive number to a power of 1, 3, 5, etc. will still be positive), we can ignore the exponents for the purposes of evaluating the signs of our variables:
beg < 0
For three numbers multiplied together to yield a negative, there are two possibilities:
neg x neg x neg = neg
pos x pos x neg = neg
Of course, we have no way to know which of b, e, and g are negative, but we know that either all of them are negative, or exactly one of them is negative.
Now that we have fully processed the common information, we are ready to proceed to Columns A and B. In any problem of this type, it is important to draw logical conclusions from the common information before approaching the Columns.
First, note that order doesn’t matter when multiplying, so the parentheses are meaningless, and the Columns can be re-ordered to look more similar to one another:

Column AColumn B
|a| beg|d| beg

Although b, e, and g appear on both sides, we cannot simply ignore them—importantly, the term beg is negative. We cannot “divide out” a term from both sides unless it is known to be positive.
Let’s rephrase what we now know about the Columns:

Column AColumn B
(big absolute value)(negative)(small absolute value)(negative)

Multiplying the negative term beg by a big absolute value will make it bigger on the negative side. Multiplying this same negative term beg by a smaller absolute value will result in a negative, but a negative closer to zero than Column A's value. Still confused? Let's illustrate with real numbers.
For instance, imagine that beg is equal to –2. The terms a and d could be something like 5 and 3, or –7 and 4, or –100 and –50, etc. (as long as the absolute value of a is bigger than the absolute value of d). So, if a and d were originally –6 and 3, for instance, then their absolute values would be 6 and 3, and when we multiplied them by –2, we’d get that Column A = –12 and Column B = –6. Since –6 is closer to zero, it is larger and the answer is B. This will work for any numbers you select that follow the rules of the problem (that is, beg is negative and |a| > |d|).
The answer is B.

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Re: GRE Weekly Challenge #4 [#permalink] New post 23 Sep 2011, 03:48
The correct answer is B) Quantity B is greater.

Since in : | a |*b^3*c^2*| d |*e^5*f ^6*g < 0
|a|, |d|, c^2, f^6 can not be negative. One (or all three) of b, e, g must be negative and since | a | > | d |,

==> g(| a |*b*e) must be lower than g(b*e*| d |)
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Re: GRE Weekly Challenge #4 [#permalink] New post 23 Sep 2011, 05:31
| a |*b^3*c^2*| d |*e^5*f ^6*g < 0

from the above we know that b^3*e^5*g = -ve. since all others are either even powers or absolute values so they cannot be -ve.
therefore b*e*g = -ve

A = g(| a |*b*e)
B = g(b*e*| d |)

Since b*e*g = -ve, we know that both A and B = -ve. However, we also know that |a|>|d|. therefore A<B

Answer B.
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Re: GRE Weekly Challenge #4 [#permalink] New post 23 Sep 2011, 05:35
Quantity B is greater.


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] New post 23 Sep 2011, 09:12
| 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

the answer willl be B
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Re: GRE Weekly Challenge #4 [#permalink] New post 25 Sep 2011, 05:56
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] New post 09 Feb 2012, 09:52
Wow great topic to discuss and have a chance to win.
Re: GRE Weekly Challenge #4   [#permalink] 09 Feb 2012, 09:52
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GRE Weekly Challenge #4

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