Think of the “chips bet” as “Solution A”
Think of the “chips remaining” as “Solution B”
If we add the following:
(# of chips bet) + (# of chips remaining at end) = Total Beginning Chips
It would be akin to mixing:
(Solution A) + (Solution B) = New Mix
in which the “New Mix” is the Number of Total Beginning Chips
Solution A (“# of chips bet”): has a 90% “concentration” of $20 Chips
Solution B (“# of chips remaining at end”): has a 70% “concentration” of $20 Chips
If we mix all of solution A with all of solution B we will get a NEW Mix (#of BEGINNING Chips) that has ———-> 80% “concentration” of $20 Chips
Since 80% is the Arithmetic Mean of 70% and 90%, the 2 solutions must have EQUAL Weight ———-> in other words, each Solution must make up (1/2) of the Final Mix = Total Number of Beginning Chips
Since there 140 Chips:
He must have bet 70 Chips
And
There must be 70 chips remaining.
Out of 70 chips bet, 10% are $100 chips and 90% are $20 chips
(7) * ($100) + (63) * ($20) =
$1,960 was bet
(A)
christoph wrote:
Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark's remaining chips are $20 chips, how much money did Mark bet?
A. $1,960
B. $1,740
C. $1,540
D. $3,080
E. $2,640
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