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# GRE Weekly Challenge #4

Author Message
Senior Manager
Joined: 25 Jul 2011
Posts: 298

Kudos [?]: 233 [0], given: 286

Location: India
Concentration: Strategy, General Management
GMAT 1: 730 Q50 V40
GPA: 3
WE: Operations (Non-Profit and Government)

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23 Sep 2011, 03:52
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:
$$| a | > | d |$$
$$| a |*b^3*c^2*| d |*e^5*f ^6*g < 0$$

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

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.

Edit: This challenge is now closed

Solution

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 A Column 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 A Column 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 winner of this Week's Challenge is... (drum roll).... Ardarandir. Congratulations! Please send me a pm with your shipping address, and choice of Manhattan GRE Guide

Details of next week's competition will be posted in the Weekly Challenge Master thread... Stay tuned!
_________________

Kudos [?]: 233 [0], given: 286

Joined: 31 Dec 1969

Kudos [?]: [0], given:

Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)
Re: GRE Weekly Challenge #4 [#permalink]

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23 Sep 2011, 04:48
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 |)

Kudos [?]: [0], given:

Manager
Joined: 24 Nov 2010
Posts: 203

Kudos [?]: 90 [0], given: 7

Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Re: GRE Weekly Challenge #4 [#permalink]

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23 Sep 2011, 06:31
| 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

Kudos [?]: 90 [0], given: 7

Intern
Status: single
Joined: 13 May 2011
Posts: 19

Kudos [?]: 22 [0], given: 6

Location: India
Concentration: Entrepreneurship, Strategy
GPA: 3
WE: Manufacturing and Production (Other)
Re: GRE Weekly Challenge #4 [#permalink]

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23 Sep 2011, 06:35
Quantity B is greater.

Explanation:

$| a |*b^3*c^2*| d |*e^5*f ^6*g < 0$
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|.

Kudos [?]: 22 [0], given: 6

Manager
Joined: 30 Sep 2009
Posts: 118

Kudos [?]: 34 [0], given: 183

Re: GRE Weekly Challenge #4 [#permalink]

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23 Sep 2011, 10:12
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|

Kudos [?]: 34 [0], given: 183

Intern
Joined: 25 Sep 2011
Posts: 1

Kudos [?]: [0], given: 0

Re: GRE Weekly Challenge #4 [#permalink]

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25 Sep 2011, 06:56
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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Intern
Joined: 06 Feb 2012
Posts: 5

Kudos [?]: [0], given: 0

Re: GRE Weekly Challenge #4 [#permalink]

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09 Feb 2012, 10:52
Wow great topic to discuss and have a chance to win.

Kudos [?]: [0], given: 0

Re: GRE Weekly Challenge #4   [#permalink] 09 Feb 2012, 10:52
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