Problem Solving

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?

[spoiler]Source MGMAT fractions book
OA 25[/spoiler]

How is this a weighted average problem ?
Can someone explain how to do it as per this way
.....x.....

some weighted average trick i saw on btg forum by an instructor.

thanks

Hi !

What method are you exactly looking for?

Here's how I solved this :

40% of 25 i.e. 10 games have been already won.

Now, let X be the total no of games played.

After winning 10 games, his chances of winning becomes 80% of the times he plays, so,

10 + .8(X-25) = .6X

10 + .8X - 20 = .6X

X = 50 Total no of games played is 50 and as 25 games have already been played, 50 - 25 = 25 games are

left to be played.

Let me know what exactly were you looking for.
Thanks.

bhumika.k.shah 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?

[spoiler]Source MGMAT fractions book
OA 25[/spoiler]

How is this a weighted average problem ?
Can someone explain how to do it as per this way
.....x.....

some weighted average trick i saw on btg forum by an instructor.

thanks

If he needs to win x more games, then

40% of 25 + 80% of x = 60% of (25 + x)

10 + (4 x/5) = 15 + (3 x/5)

x/5 = 5, so x = [spoiler]25[/spoiler].

I didn't see a need of discussing weighted average here, though it's a case like that. Never mind jumping down if the staircases are not functioning.

bhumika.k.shah 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?

[spoiler]Source MGMAT fractions book
OA 25[/spoiler]

How is this a weighted average problem ?
Can someone explain how to do it as per this way
.....x.....

some weighted average trick i saw on btg forum by an instructor.

thanks

Note: This question is very simple and needs no calculation
Just check here that 60 is the average of 40 and 80 .. so for winning a total of 60% he need to play exactly the same number of games i.e. 25.. (just a 2 sec problem)
though u can go thru process also (will take 30 sec)...

Actually I did not understand the question in the first place.

According to me, the gambler is now winning 10 (40%) games out of 25 games so far. Suddenly he starts winning 20 (80%) of 25 games. So to end up winning 60% of the time why should he play more games in first place? I don't seem to get it...

Thanks

No need to determine number of games... Please use alligation here

Spirit has 40% alcohal how many liters of pure spirit or 80% spirt shall be added to make it 60%

equal quantities is the answer.. a = 40, b =80 ..equal quantities give = (40+80)/2 = 60

HSPA wrote:No need to determine number of games... Please use alligation here

Spirit has 40% alcohal how many liters of pure spirit or 80% spirt shall be added to make it 60%

equal quantities is the answer.. a = 40, b =80 ..equal quantities give = (40+80)/2 = 60

Hi HSPA,

Please explain how did u use Alligation here. Pls show the calculation.

25 games indeed.

25 is indeed the right answer.

This is how i solved it.

10 out of 25 already won.
If he plays x more games...he will win 0.8x out of them.


( 10 + 0.8x ) / (25 + X ) = 0.6

X = 25

force5 wrote:25 games indeed.

you can solve this through allegation:

Ratio
a) 40% wins 80-60= 20=1
b) 80% wins 60-40= 20=1 or 1:1

c) 60% wins - we already know that the number of 40% winning games = 25 = number of 80% winning games