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

Question

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.

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

Kudos for a correct  solution .

balamoon · Accepted Answer

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

GMATinsight · Answer

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

IanStewart · Answer

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.

AmoyV · Answer

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

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

Answer: B

mejia401 · Answer

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}

Thanks,

ENGRTOMBA2018 · Answer

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

kelvind13 · Answer

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?

kanigmat011 · Answer

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

Bunuel · Answer

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.

rahulms94 · Answer

Doesn't  one half    as many losses mean 1 1/2 = 3/2?????

Bunuel · Answer

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).

bumpbot · Answer

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

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