Problem Solving

AAPL wrote:Manhattan Prep

A professional gambler has won 40% of his 25 poker games for the week so far. If, all of a sudden, his luck changes and he begins winning 80% of the time, how many more games must he play to end up winning 60% of all his games for the week?

A. 20
B. 25
C. 30
D. 35
E. 40

OA B

Say he plays x more games after he has played 25 games.

Thus,

Number of games won out of the first 25 games = 40% of 25 = 10;
Number of games to be won out of the x games = 80% of x = 0.8x;

Number of games won out of the all (x + 25) games = 60% of (x + 25) = 3(x + 25)/5

Thus, 10 + 0.8x = 3(x + 25)/5;

50 + 4x = 3x + 75

x = 25 games

The correct answer: B

Hope this helps!

-Jay
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AAPL wrote:Manhattan Prep

A professional gambler has won 40% of his 25 poker games for the week so far. If, all of a sudden, his luck changes and he begins winning 80% of the time, how many more games must he play to end up winning 60% of all his games for the week?

A. 20
B. 25
C. 30
D. 35
E. 40

We are given that a poker player has won 0.4 x 25 = 10 poker games. If he starts winning 80% of his games, we need to determine how many more games must be played to have a winning percentage of 60% for the week. We can let x = the number of additional games played, and we set up the proportion:

(10 + 0.8x)/(25 + x) = 60/100

(10 + 0.8x)/(25 + x) = 3/5

5(10 + 0.8x) = 3(25 + x)

50 + 4x = 75 + 3x

x = 25

Answer: B