Mark is playing poker at a casino. Mark starts playing with 120 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?  Choose the correct choice from the following answers
A) $1,960  B) $1,680  C) $1,540  D) $3,080  E) $2,640
Total chips 120   20% of $100 chips = 24 chips * $100 = $2,400   80% of $20 chips = 96 chips * $20 = $1,920   If x is the number of chips bet and y is the amount of chips remaining:   x + y = 120   y = 120 - x ~ (1)   [First round bet] + [Second round bet] = $2,400 + $1,920   [0.1x * $100 + 0.9x * $20] + [0.3y * $100 + 0.7y * $20] = $4,320   10x + 18x + 30y + 14y = 4320   28x + 44y = 4320 ~ (2)   Substitute (1) into (2) to solve for x:   28x + 44(120 - x) = 4320   28x + 5280 - 44x = 4320   16x = 960   x = 60   Substituting x:   [First round bet] = [0.1x * $100 + 0.9x * $20] = 0.1(60)($100)+0.9(60)($20) = $600 + $1,080 = $1,680   Answer is $1,680
<result>B</result>
Mark is playing poker at a casino. Mark starts playing with 120 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?