Mark is playing poker at a casino. Mark starts playing with 130 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?
Pick one:	A) $1,960	B) $1,740	C) $1,820	D) $3,080	E) $2,640
Total chips 130   20% of $100 chips = 26 chips * $100 = $2,600   80% of $20 chips = 104 chips * $20 = $2,080   If x is the number of chips bet and y is the amount of chips remaining:   x + y = 130   y = 130 - x ~ (1)   [First round bet] + [Second round bet] = $2,600 + $2,080   [0.1x * $100 + 0.9x * $20] + [0.3y * $100 + 0.7y * $20] = $4,680   10x + 18x + 30y + 14y = 4680   28x + 44y = 4680 ~ (2)   Substitute (1) into (2) to solve for x:   28x + 44(130 - x) = 4680   28x + 5720 - 44x = 4680   16x = 1040   x = 65   Substituting x:   [First round bet] = [0.1x * $100 + 0.9x * $20] = 0.1(65)($100)+0.9(65)($20) = $650 + $1,170 = $1,820   Answer is $1,820
<result>C</result>
Mark is playing poker at a casino. Mark starts playing with 130 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?