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M10 Q35

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Manager
Manager
Joined: 07 Sep 2011
Posts: 65
GMAT 1: 660 Q41 V40
GMAT 2: 720 Q49 V39
WE: Analyst (Mutual Funds and Brokerage)
Followers: 0

Kudos [?]: 12 [0], given: 13

Re: M10 Q35 [#permalink] New post 13 Sep 2012, 07:33
40-9= need 31 more points at least
20-5= in 15 games

let x= # wins
let (15-x)= # ties
we want to see min amount of wins with max ties

3x+(15-x)>=31
3x+15-x>=31
2x>=16
x>=8

Answer is C
Intern
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Re: M10 Q35 [#permalink] New post 13 Sep 2012, 10:27
If the answer is 8... then is it necessary rest all the matches will be drawn... not a single match will be defeated and score 0....
Intern
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Joined: 30 Jun 2011
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Re: M10 Q35 [#permalink] New post 14 Sep 2012, 09:18
The Answer is C.
In order to have the least % of victory, the defeat must be 0.
Let x = # of victory after 5 games
y = # of draw after 5 games.

We have:
3x+y=40-9=31 (points)
x+y=15 (games)

Then 2x=16=> x=8
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Re: M10 Q35 [#permalink] New post 16 Sep 2012, 14:54
I think you have to consider the defeated matches for the second equation, even though it does not affect the final result.

magnetictempest wrote:
I set this up in two equations:

1) 3W + D = 31
To minimize the number of wins, then we need to maximize the number of points via draw. Total points add up to the remaining 31 points to make it to 40 incremental to the 9 obtained after 5 games.

2) W + D = 15
Total number of wins + draws should add up to the 15 games to make 20 games on top of the 5 games already played

W = 8
D = 7

Minimum wins is 8 => c)
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Joined: 16 Sep 2012
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Re: M10 Q35 [#permalink] New post 16 Sep 2012, 14:59
hit and trial is very efficient for this type of problem, save lots of time..

debaranjansahoo wrote:
I know answer is 8. By guessing i started with 7 and ended up with 8. Does any one know any direct method to get the answer for these types of questions rather than hit and trial method.
Re: M10 Q35   [#permalink] 16 Sep 2012, 14:59
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