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?

30

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25

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15

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20

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Solution

The correct option is B $25$
This is a weighted averages problem. You can set up a table to calculate the number of games the gambler must play to obtain a weighted average win rate of $60$ %:
Poker Games First $25$ Games Remaining Games Total
Wins $(0.4) 25 = 10$ $(0.8) x$ $(0.6) (25 + x)$
Losses
Total $25$ $x$ $25 + x$
Thus, $10 + 0.8 x = (0.6) (25 + x), 10 + 0.8 x = 15 + 0.6 x, 0.2 x = 5, x = 25$ .

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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 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?