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M10 Q35

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M10 Q35 [#permalink] New post 21 Dec 2008, 10:13
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The scoring system in a certain football competition goes as follows: 3 points for victory, 1 point for a draw, and 0 points for defeat. Each team plays 20 matches. If the team scored 9 points after 5 games, what is the least number of the remaining matches it has to win to reach a 40-point mark by the end of the tournament?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Is n't the answer 11?
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Re: M10 Q35 [#permalink] New post 21 Dec 2008, 13:28
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No the answer is C:) 8...

Remaining matches: 15
40-9= 31 points in the remaining matches
8 wins X3= 24
7 draws x1=7
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Re: M10 Q35 [#permalink] New post 21 Dec 2008, 13:47
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It was n't quite apparent then but it makes sense now. And I guess for the same reason 11 was not one of the choices!!
Thanks Mbadreamer4ever.
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Re: M10 Q35 [#permalink] New post 24 Jul 2010, 16:22
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gk2k2 wrote:
The scoring system in a certain football competition goes as follows: 3 points for victory, 1 point for a draw, and 0 points for defeat. Each team plays 20 matches. If the team scored 9 points after 5 games, what is the least number of the remaining matches it has to win to reach a 40-point mark by the end of the tournament?

A: 6
B: 7
C: 8
D: 9
E: 10

Is n't the answer 11?


Remaining matches = 15
Remaining points = 40=9 = 31

let x be the number of matches won

3*x + (15-x)*1 >= 31
2x >= 16 => x=8
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Re: M10 Q35 [#permalink] New post 29 Jul 2010, 06:22
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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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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 05:15
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tricky question...

i picked D~10. because i think 3win*10+1draw=31, but i forgot there is 15 games altogether to be played, and the question asks about AT LEAST...
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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 05:26
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3 for Victory; 1 for Draw; 0 for Loss
20 matches total

After 5 games of total of 9 points, the team has 15 games left to gain 31 points to reach 40.

To find the least number of wins, we need to assume that the team doesn't lose and can get points from drawing the games too.

With no loss: L = 0
Eq 1, With 15 games left: V + D = 15
Eq 2, With 31 points to go: 3*V + 1*D = 31

Solve for D with equation 1: D = 15 - V
Solve for V with equation 2: V = 8

Number of victories needed = 8
Ans C
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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 06:34
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This question can be solved in a much easier fashion, if we use the answer options..
Ans: C
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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 06:45
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.
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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 06:46
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very tricky question.
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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 19:36
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Ans C 8*3=24 , 7*1=7 and 0
so 31 points can be earn
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Re: M10 Q35 [#permalink] New post 08 Sep 2010, 19:53
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I believe that there is something wrong in the question formation. I donno, I may be wrong.. Are there anybody with me?
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Re: M10 Q35 [#permalink] New post 09 Sep 2010, 00:45
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To get the least number of required wins, we need the most optimistic scenario for the remaining matches which means draws for each of these matches.

Let's x be the number of wins and y the number of draws:
x+y=15 (1)

The number of points to reach is 40 while current score is 9, hence we need 31 points. We can write the following equation

3x+1y=31
or using (1) 3x+(15-x)=31-->2x=16-->x=8
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Re: M10 Q35 [#permalink] New post 09 Sep 2010, 05:23
Francois wrote:
To get the least number of required wins, we need the most optimistic scenario for the remaining matches which means draws for each of these matches.

Let's x be the number of wins and y the number of draws:
x+y=15 (1)

The number of points to reach is 40 while current score is 9, hence we need 31 points. We can write the following equation

3x+1y=31
or using (1) 3x+(15-x)=31-->2x=16-->x=8


Atta Boy!..that is some way to do it..
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Re: M10 Q35 [#permalink] New post 09 Sep 2010, 09:41
3 points for win (w); 1 point for draw (d)
20 total - 5 played = 15 games left = # wins + # draws --> w + d = 15
points to make = 40 total - 9 so far = 31 points --> 3w + 1d = 31
subtract:
-(w+d)-15

-w - d = -15
3w + d = 31
___________ +
2w = 16
w = 8

Check: 8 + d = 15 ; d=7; 3(8) +1(7) = 24 + 7 = 31

Answer is 8 wins.
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Re: M10 Q35 [#permalink] New post 14 Sep 2010, 07:55
Make a note of what we have to find out. We need to find how many winning matches are needed to get 40 points?

Since 9 points are achieved in 5 matches we are left with 31 points that should be filled with less winning matches. If you try with 8; you will have 8*3(win)+7*1(draw)=31.

Hence answer is 8.
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Re: M10 Q35 [#permalink] New post 13 Sep 2011, 07:25
picking numbers solved in under 1min30sec
40 - 9 = 31 points need
Points from Draws + Points from Winning
15 * 1 + 3*0 (0) = 15
14 *1 + 3*1 (3) = 17
and so on
13 6
12 9
11 12
10 15
9 18
8 21
until
7 *1 + 3*8 (24) = 31
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Re: M10 Q35 [#permalink] New post 13 Sep 2011, 22:15
nice explanation by gurpreetsingh.
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Re: M10 Q35 [#permalink] New post 13 Sep 2012, 04:33
Very tricky question got 11 as the poster ..

Knew i had missed something and 10 could not be the answer but did not the logic
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Re: M10 Q35 [#permalink] New post 13 Sep 2012, 05:03
Apply-"hit and trial" , start with the middle number(when you see the ACs in order such as 6,7,8,9,10 ,because it will tell you whether you have to go up or down . in this question one might get lucky because C=8 is the right answer.GMAT loves to ask such questions, because in the middle of the test GMAT wants you to boggle with simple yet silly manual calculation.
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Re: M10 Q35   [#permalink] 13 Sep 2012, 05:03
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