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A professional gambler has won 40% of his 25 poker games for
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03 Jun 2012, 11:37
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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
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Re: A professional gambler has won 40% of his 25 poker games for
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04 Jun 2012, 00:41
farukqmul wrote: 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 began winning 80% of the time, how many more games must he play to end up winning 60% of all his games for the week? explanation is given in MGMAT book but I didn't understand it...Can anyone please explain? 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.
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A professional gambler has won 40% of his 25 poker games for
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03 Jun 2012, 23:54
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



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Re: A professional gambler has won 40% of his 25 poker games for
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04 Jun 2012, 02:52
farukqmul wrote: 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 began winning 80% of the time, how many more games must he play to end up winning 60% of all his games for the week? explanation is given in MGMAT book but I didn't understand it...Can anyone please explain? 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.
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Re: A professional gambler has won 40% of his 25 poker games for
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04 Jun 2012, 05:21
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,
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Re: Percentage problem
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04 Jun 2012, 07:31
davidandcompany wrote: farukqmul wrote: 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 began winning 80% of the time,how many more games must he play to end up winning 60% of all his games for the week? explanation is given in MGMAT book but I didn't understand it...Can anyone please explain? Let x = number of games more must he play 0.4(25) + 0.8x = 0.6(25+x) < You should be able to grasp this idea. 10 + 0.8x = 15 + 0.6x 0.8x  0.6x = 15  10 0.2x = 5 x = 25 Hi, can you explain the equation please ?



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Re: Percentage problem
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04 Jun 2012, 07:33
farukqmul wrote: davidandcompany wrote: farukqmul wrote: 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 began winning 80% of the time,how many more games must he play to end up winning 60% of all his games for the week? explanation is given in MGMAT book but I didn't understand it...Can anyone please explain? Let x = number of games more must he play 0.4(25) + 0.8x = 0.6(25+x) < You should be able to grasp this idea. 10 + 0.8x = 15 + 0.6x 0.8x  0.6x = 15  10 0.2x = 5 x = 25 Hi, can you explain the equation please ? Check this: aprofessionalgamblerhaswon40ofhis25pokergamesfor133738.html#p1092310Hope it helps.
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Re: A professional gambler has won 40% of his 25 poker games for
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27 Aug 2013, 18:58
Using alligation method
40 80 60 20 20 ==> 1:1
Must play another 25 games
Ans:B



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Re: A professional gambler has won 40% of his 25 poker games for
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17 Apr 2016, 08:56
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.



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Re: A professional gambler has won 40% of his 25 poker games for
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07 Jun 2017, 20:53
Bunuel wrote: farukqmul wrote: 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 began winning 80% of the time, how many more games must he play to end up winning 60% of all his games for the week? explanation is given in MGMAT book but I didn't understand it...Can anyone please explain? 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. 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?



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Re: A professional gambler has won 40% of his 25 poker games for
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10 Jun 2017, 08:34
farukqmul wrote: 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 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
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A professional gambler has won 40% of his 25 poker games for
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18 Oct 2018, 11:13
farukqmul wrote: 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
The " aggressivebecausetheoccasionpermits" 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.
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Re: A professional gambler has won 40% of his 25 poker games for
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22 Aug 2019, 08:58
farukqmul wrote: 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 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
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Re: A professional gambler has won 40% of his 25 poker games for
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