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A certain football team played x games last season, of which

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A certain football team played x games last season, of which [#permalink]

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A certain football team played x games last season, of which the team won exactly y games. If tied games were not possible, how many games did the team win last season?

(1) If the team had lost two more of its games last season, it would have won 20 percent of its games for the season.
(2) If the team had won three more of its games last season, it would have lost 30 percent of its games for the season.

[Reveal] Spoiler:
As I don't have an official answer, I will go for C. This is how I did this. Can you guys please help and let me know whether I am correct or not?

Games Lost = x-y

Considering statement 1 --> x-y+2= 0.20x
Considering statement 2 --> y+3=0.3(x-y)

Two variables and two equations and we can solve for y.
[Reveal] Spoiler: OA

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Re: A certain football team played x games last season, of which [#permalink]

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New post 03 Nov 2011, 22:02
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Your approach is correct, though the equations you are getting are incorrect.

Using statement 1: y-2 = 0.20x [i.e. equate the number of games won to 20% of the total games]
Using statement 2: x-y-3 = 0.30x [i.e. equate the number of games lost to 30% of the total games]

The answer should still be (C).
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Re: A certain football team played x games last season, of which [#permalink]

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New post 03 Nov 2011, 23:59
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Expert's post
enigma123 wrote:
A certain football team played x games last season, of which the team won exactly y games. If tied games
were not possible, how many games did the team win last season?
(1) If the team had lost two more of its games last season, it would have won 20 percent of its games for
the season.
(2) If the team had won three more of its games last season, it would have lost 30 percent of its games
for the season.

As I don't have an official answer, I will go for C. This is how I did this. Can you guys please help and let me know whether I am correct or not?

Games Lost = x-y

Considering statement 1 --> x-y+2= 0.20x
Considering statement 2 --> y+3=0.3(x-y)

Two variables and two equations and we can solve for y.


Let me add a little tip about this conclusion. Make sure that the equations you have are distinct. You can solve for y only if you have two equations. If both the equations are the same (even though they may not appear to be same), you cannot solve for y.
Check out this post for an example of this trickery: http://www.veritasprep.com/blog/2010/10 ... er-pulled/
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Re: A certain football team played x games last season, of which [#permalink]

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Re: A certain football team played x games last season, of which [#permalink]

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New post 21 Jan 2015, 14:06
Expert's post
Hi all,

These types of DS questions can be approached with a mix of Algebra and TESTing VALUES.

We're told that a team played X games and won Y of those games (there were NO ties). We're asked for the value of Y.

Fact 1: If the team lost 2 more games, then it would have won 20% of its games.

Algebraically, you can set up the following equation:
Y - 2 = .2X

This equation does not help you to determine the value of Y.

From a practical standpoint, you can also TEST VALUES:

IF...
X = 5 games
Y = 3 wins
Y-2 = 1 win
1/5 = 20% and the answer to the question is 3.

IF....
X = 10 games
Y = 4 wins
Y-2 = 2 wins
2/10 = 20% and the answer to the question is 4
Fact 1 is INSUFFICIENT

Fact 2: If the team won 3 more games, then it would have lost 30% of its games (meaning the team would have WON 70% of its games

Again, we can use algebra:

Y + 3 = .7X

But we have the same problem we had in Fact 1 - the equation does not help you to determine the value of Y.

Here's the proof by TESTing VALUES:

IF....
X = 10 games
Y = 4 wins
Y+3 = 7 wins
7/10 = 70% and the answer to the question is 4.

IF....
X = 20 games
Y = 11 wins
Y+3 = 14 wins
14/20 = 70% and the answer to the question is 11
Fact 2 is INSUFFICIENT

Combined, we know:
Y - 2 = .2X
Y + 3 = .7X

Here, we have a "system" of equations - with 2 variables and 2 unique equations, we CAN solve and get the value of Y.
Combined, SUFFICIENT

Final Answer:
[Reveal] Spoiler:
C


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Re: A certain football team played x games last season, of which [#permalink]

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New post 05 Nov 2015, 07:45
I tried to solve it differently, and yet..I am unsure whether my method works..
from statement 1, we have 2 more lost = 0.2x
lost game: x-y
(x-y+2)/x = 8x/10

on top we have the number of games he lost, divided by the total number of games, and get 80% of X or 8x/10
from this, we cannot solve for y

from statement 2:
y+2 = games would have been 0.3x lost
or
y+3/x = 7x/10 - since 7x/10 => 70% of games the team won
again 2 unknowns. can't solve it

1+2
well...
10x-10y+20 = 8x^2
10y+30 = 7x^2

rewrite second = 10y = 7x^2 - 30

substitute in the first equation:
10x - 7x^2 +30 +20 = 8x^2
50 = 15x^2 - 10x
divide everything by 5
10 = 3x^2 - 2x
factor a x => x(3x-2) = 10

aand here I stopped...this doesn't make any sense..
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Re: A certain football team played x games last season, of which [#permalink]

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New post 05 Nov 2015, 21:25
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Expert's post
mvictor wrote:
I tried to solve it differently, and yet..I am unsure whether my method works..
from statement 1, we have 2 more lost = 0.2x
lost game: x-y
(x-y+2)/x = 8x/10

on top we have the number of games he lost, divided by the total number of games, and get 80% of X or 8x/10
.


This is the problem in your analysis. why will you get 8x/10? Actually, you get 8/10

Games lost = 80% of Total Games

Games lost = (80/100) * Total Games

Games lost / Total Games = 80/100
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Re: A certain football team played x games last season, of which   [#permalink] 05 Nov 2015, 21:25
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