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The "competitive edge" of a baseball team is defined by the formula

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Bunuel
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The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 03:09
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The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 03:34
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Bunuel wrote:
The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


Kudos for a correct solution.


Solution -
GMAT had Last season W wins and L losses and this season 3W wins and L/2 losses.

By substituting in the equation, \(\sqrt{\frac{3W}{(L/2)}}\) = \(\sqrt{6}\)*\(\sqrt{\frac{W}{L}}\).
ANS B
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The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 03:47
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Bunuel wrote:
The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


Kudos for a correct solution.


Competitive Edge = \(\sqrt{\frac{W}{L}}\)

Let, Last Year Losses = 2a
and Last year Wins = b
Last year Competitive Edge = \(\sqrt{\frac{b}{2a}}\)

then This year Losses = a
and This year Wins = 3b
This year Competitive Edge = \(\sqrt{\frac{3b}{a}}\)

This year Competitive Edge = \(\sqrt{\frac{b}{2a}}\) \(*\sqrt{6}\) = Last year Competitive factor \(*\sqrt{6}\)

Answer: Option B
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 04:17
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It's a pure ratio question, so the answer will be the same for any numbers we care to pick. If we assume they won 2 games and lost 2 games last year, then their 'competitive edge' was 1 last year. This year, they'd win 6 games and lose 1 game, so their competitive edge would be √6 this year, which is √6 times last year's value.
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 07:34
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Bunuel wrote:
The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


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Initial = \(\sqrt{\frac{W}{L}}\)

Now = \(\sqrt{\frac{3W}{0.5L}}\)
= \(\sqrt{\frac{6W}{L}}\)

Answer: B
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The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 08:11
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First approach, algebra:
Last year, \(\sqrt{\frac{W}{L}}\)
This year, \(\sqrt{\frac{3W}{.5*L}}=\sqrt{\frac{6W}{L}}\)
Thus, B. \(\sqrt{6}\)

Or second approach, picking numbers:
Number of wins 100
Number of losses 50
Last year, \(\sqrt{\frac{100}{50}}\) = \(\sqrt{2}\)
This year, \(\sqrt{\frac{3*100}{.5*50}}\)=\(\sqrt{\frac{600}{50}}\)
Thus, between the two is B. \(\sqrt{6}\)

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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   03 Jul 2015, 09:10
Bunuel wrote:
The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


Kudos for a correct solution.


\(Y_L\) (Last year)=\(\sqrt{\frac{W}{L}}\)

\(Y_T\) (This year)=\(\sqrt{\frac{3W}{L/2}}\)

Thus, factor of increase = \(Y_L\)/\(Y_T\) = 1/\(\sqrt{6}\). Thus the factor is \(\sqrt{6}\), B is the correct answer
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   04 Jul 2015, 05:31
This is a very good question that tests your skills at going through wording of a statement, I managed to get \sqrt{6} pretty quickly as the answer.

Shouldn't this question be classified for under 600 and not between 600-700?

Bunuel wrote:
The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   04 Jul 2015, 09:00
Team
had Last season W wins and L losses and this season 3W wins and L/2 losses.

By substituting in the equation,
Y1= Sq rt(W/L)
Y2 = Sq rt(3W/(L/2))

So Multiplication factor is SQ rt(6)

Option B is correct
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   06 Jul 2015, 04:51
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Bunuel wrote:
The "competitive edge" of a baseball team is defined by the formula \(\sqrt{\frac{W}{L}}\) where W represents the number of the team's wins, and L represents the number of the team's losses. This year, the GMAT All-Stars had 3 times as many wins and one-half as many losses as they had last year. By what factor did their "competitive edge" increase?

A. \(\sqrt{2}\)
B. \(\sqrt{6}\)
C. \(\sqrt{12}\)
D. 6
E. 12


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Let c = competitive edge

\(c = \sqrt{\frac{W}{L}}\)

Pick numbers to see what happens to the competitive edge when W is tripled and L is halved. If the original value of W is 4 and the original value of L is 2 , the original value of c is \(\sqrt{\frac{4}{2}}=\sqrt{2}\). If W triples to 12 and L is halved to 1, the new value of c is \(\sqrt{\frac{12}{1}}=\sqrt{12}\). The competitive edge has increased from \(\sqrt{2}\) to \(\sqrt{6}\).

\(\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\).

The competitive edge has increased by a factor of \(\sqrt{6}\).

Answer: B.
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   02 Jun 2021, 07:45
Doesn't one half[b][/b] as many losses mean 1 1/2 = 3/2?????
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   02 Jun 2021, 08:14
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rahulms94 wrote:
Doesn't one half[b][/b] as many losses mean 1 1/2 = 3/2?????


w is x times as many as y means w is x multiplied by y. So, "this year, the GMAT All-Stars had one-half as many losses as they had last year." means (this year losses) = 1/2*(last year losses).
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Re: The "competitive edge" of a baseball team is defined by the formula [#permalink] New post   07 Aug 2022, 00:03
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