Mark is playing poker at a casino. Mark starts playing with 110 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:
A) $1,960	B) $1,740	C) $1,540	D) $3,080	E) $2,640
Total chips 110   20% of $100 chips = 22 chips * $100 = $2,200   80% of $20 chips = 88 chips * $20 = $1,760   If x is the number of chips bet and y is the amount of chips remaining:   x + y = 110   y = 110 - x ~ (1)   [First round bet] + [Second round bet] = $2,200 + $1,760   [0.1x * $100 + 0.9x * $20] + [0.3y * $100 + 0.7y * $20] = $3,960   10x + 18x + 30y + 14y = 3960   28x + 44y = 3960 ~ (2)   Substitute (1) into (2) to solve for x:   28x + 44(110 - x) = 3960   28x + 4840 - 44x = 3960   16x = 880   x = 55   Substituting x:   [First round bet] = [0.1x * $100 + 0.9x * $20] = 0.1(55)($100)+0.9(55)($20) = $550 + $990 = $1,540   Answer is $1,540
<result>C</result>
Mark is playing poker at a casino. Mark starts playing with 110 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?