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A professional gambler has won 40% of his 25 poker games for

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A professional gambler has won 40% of his 25 poker games for  [#permalink]

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New post 03 Jun 2012, 10:37
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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  [#permalink]

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New post 03 Jun 2012, 23:41
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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?

25


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  [#permalink]

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New post 03 Jun 2012, 22:54
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1
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  [#permalink]

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New post 04 Jun 2012, 01:52
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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?

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.
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Re: A professional gambler has won 40% of his 25 poker games for  [#permalink]

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New post 04 Jun 2012, 04:21
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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  [#permalink]

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New post 04 Jun 2012, 06:31
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  [#permalink]

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New post 04 Jun 2012, 06:33
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: a-professional-gambler-has-won-40-of-his-25-poker-games-for-133738.html#p1092310

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A professional gambler has won 40% of his 25 poker games for  [#permalink]

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New post 27 Aug 2013, 17:58
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  [#permalink]

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New post 17 Apr 2016, 07:56
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  [#permalink]

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New post 07 Jun 2017, 19:53
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?

25


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  [#permalink]

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New post 10 Jun 2017, 07:34
1
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  [#permalink]

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New post 18 Oct 2018, 10:13
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 "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.
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A professional gambler has won 40% of his 25 poker games for &nbs [#permalink] 18 Oct 2018, 10:13
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