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

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03 Nov 2011, 14:19

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C

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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.

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

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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]

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?

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/ _________________

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

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21 Jan 2015, 11:22

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

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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..

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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