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A chess player won 25 percent of the first 20 games [#permalink]

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11 May 2013, 00:48

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A chess player won 25 percent of the first 20 games he played and all of his remaining games. What is the ratio of the number of games he won to the number of the games he lost?

(1) If the player had won 25 percent of the total games he played, he would have lost 30 more games than he actually did.

(2) The player won 75 percent of the games he played.

Re: A chess player won 25 percent of the first 20 games [#permalink]

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11 May 2013, 00:59

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A chess player won 25 percent of the first 20 games he played and all of his remaining games. What is the ratio of the number of games he won to the number of the games he lost?

In the first 20 games, 5 W and 15 L. Overall he won \(5+R\) games and lost \(15\). We need to find R

(1) If the player had won 25 percent of the total games he played, he would have lost 30 more games than he actually did. "If the player had won 25 percent of the total games he played" = lost 75% \((20+R)*75%=30+15\) the first part is the total losses, the second part is the is 30 more games than he actually did lose (15)

\(0.75R=30\)

\(R=40\) Suffcient

(2) The player won 75 percent of the games he played. \(5+R=75%(20+R)\) the first part is the number of games he won, the second part is the transaltion of "won 75 percent of the games he played"

\(R=40\) Sufficient D _________________

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Re: A chess player won 25 percent of the first 20 games [#permalink]

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11 May 2013, 03:35

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skamal7 wrote:

A chess player won 25 percent of the first 20 games he played and all of his remaining games. What is the ratio of the number of games he won to the number of the games he lost?

(1) If the player had won 25 percent of the total games he played, he would have lost 30 more games than he actually did.

(2) The player won 75 percent of the games he played.

A chess player won 25 percent of the first 20 games he played and all of his remaining games. What is the ratio of the number of games he won to the number of the games he lost?

(1) If the player had won 25 percent of the total games he played, he would have lost 30 more games than he actually did. The player won 25% of his first 20 games and 100% of the remaining games, in order to win 25% of total matches he should have won 25% of the remaining games (instead of 100%, so 75% less). So 75% losses in the remaining games result in 30 more losses: 0.75*R=30, where R is # of the remaining games. We have only one unknown R, hence we can solve for it and thus we'll have all information needed to get the ratio. Sufficient.

(2) The player won 75 percent of the games he played --> 0.25*20+1*R=0.75*(20+R). The same here: we have only one unknown R, hence we can solve for it and thus we'll have all information needed to get the ratio. Sufficient.

Re: A chess player won 25 percent of the first 20 games [#permalink]

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11 May 2013, 01:26

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skamal7 wrote:

A chess player won 25 percent of the first 20 games he played and all of his remaining games. What is the ratio of the number of games he won to the number of the games he lost? (1) If the player had won 25 percent of the total games he played, he would have lost 30 more games than he actually did. (2) The player won 75 percent of the games he played.

I approached the problem in this way, Let x be the number of games played after the first 20 games. Now we already know, that out of the 20 games, 5 were won by the player and 15 lost.

Also, that since he doesnt loose any more games, the total number of losses remain to be 15. But the games won can be assumed as 5 + x Hence the desired ratio is 5+x:15

To calculate value of x we use the given statements: Stamement 1. if the player had won 25% of the total games i.e. games lost = 75/100(20 + x) = 15 + 3/4x. => 15+3/4x -15 = 30 => 3/4x = 30 => x = 4/3 * 30 = 40. Hence the ratio can be found as 45:15.

For Statement 2. If the player had won 75 of total games i.e. 5+x/20+x = 75/100 => x = 40 Hence the ratio is again found to be 45:15

The answer for me would be [D]**edited..typo! , both the statments are individually sufficient.

Re: A chess player won 25 percent of the first 20 games [#permalink]

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11 May 2013, 03:39

Bunnel, I have a query. If the player had won 25 percent of the total games he played" = lost 75%----- why do we want to convert into the opposite of how much % lost..can we get the answer without converting this? _________________

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Re: A chess player won 25 percent of the first 20 games [#permalink]

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11 May 2013, 21:21

skamal7 wrote:

Bunnel, I have a query. If the player had won 25 percent of the total games he played" = lost 75%----- why do we want to convert into the opposite of how much % lost..can we get the answer without converting this?

Since the RHS in the statement 1 is 30 i.e. "the player would have lost 30 more games than he actually did", which is the number of games lost, its important to frame the equation in terms of number of games lost by the player. Statement 1 in general does not speak about the exact number of wins, but then again we can derive the answer from the wins as well.

Let x be the number of games played after the first 20 games. Now we already know, that out of the 20 games, 5 were won by the player and 15 lost. The total number of games won can be assumed as 5 + x and the losses as of now are 15. Total games played is 20 + x. If the player had won 25% of the total games i.e. 25/100 * (20 + x) games won. Now as mentioned, the player would loose 30 more games. Hence the number of wins would reduce by 30 i.e. 5 + x -30 is the total number of wins. Equating the same, 25/100*(20 + x) = 5 + x - 30 =>5 + x/4 = x - 25 => x = 40

The above values are concurrent with our other findings. Although this procedure does provide the answer, but would not be the best of the methods to use given the question. Please correct me if I am wrong Bunnel! Im sorry I went ahead over this even though the query was directed at you

Re: A chess player won 25 percent of the first 20 games [#permalink]

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Re: A chess player won 25 percent of the first 20 games [#permalink]

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