This is a simple weighted average question.

Say the gambler must play x more games, then:

40% of 25 games plus 80% of x games should equal to 60% of total number of games, which is (25+x) --> 0.4*25+0.8*x=0.6*(25+x) --> x=25.

Hope it's clear.
_________________

If he plays X more games, the equation becomes:
games with 40% chance + Games with 80% chance = Total games with 60% chance
.40*25 + .80*X=.60*(25+X)
Solving for X = 25

Responding to a pm:

Yes, you can use the weighted averages method.

w1/w2 = (80 - 60)/(60 - 40) = 1:1

'The number of games out of which he won 40% games' is equal to 'the number of games out of which he must win 80% of the games'. He won 40% of the games out of 25 games.
So he must win 80% of the games out of another 25 games.
_________________

Hi,

Initially the person has played 25 games with probablity of winning as 40%
Thus, number of games won = 10

Later when the probability of winning changes to 80% and assuming x games were played by him:

Total number of win, 60% of all games \(= 0.6(x+25) = 10 + 0.8x\)
\(0.2x=5\)
\(x=25\)
so, he has to play 25 games.

Alternatively, this could be solved by allegation method.
B = 25games.

Regards,

Hi, can you explain the equation please ?

Check this: a-professional-gambler-has-won-40-of-his-25-poker-games-for-133738.html#p1092310

Hope it helps.
_________________

Using alligation method

40 80
60
20 20 ==> 1:1

Must play another 25 games

Ans:B

ok, the question says that 40% of the games 25 games are won => 10 games.
then the question says that 60% of all the games are won => 60% ( x + 25)
it tells you that 80% of the remaining games are won => 80% (x)

the equation is => 40%(25) + 80%(x) = 60% (x+25) => 10+ .80x = .6x + 15 => .2x = 10 => x = 25.

Bunuel - this is the same formula I derived- do we need to necessarily plug in and test each of the answer choices? Or can we just simplify the equation?

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 have:

(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
_________________

The "aggressive-because-the-occasion-permits" style:

60% is the average of 40% and 80%, hence 25 games with 40% winning percentage and another 25 games with 80% winning percentage does the trick!

The solution is in red.

Regards,
Fabio.
_________________

Given:
1. A professional gambler has won 40% of his 25 poker games for the week so far.
2. All of a sudden, his luck changes and he begins winning 80% of the time.

Asked: How many more games must he play to end up winning 60% of all his games for the week?

Let him play x more games
Games won = .4 * 25 + .8 x = 10 + .8x
Game played = 25 + x
.6 (25 +x) = 10 + .8x
15 + .6x = 10 + .8x
5 = .2x
x = 5/.2 = 25

25 more games must he play to end up winning 60% of all his games for the week.

IMO B
_________________

@EMPOWERGmat If we apply triage here, then won't the correct option be eliminated?