Jimmy's Chocolate shop sells two different fudge bars. The dark chocolate fudge bar sells for $6.00 per pound and the milk chocolate fudge bar sells for $4.00 per pound. You purchased a combination of these bars for all of your family and friends for the holidays and your bill came to $120. This scenario can be modeled by the equation 6d + 4m = 120. Which of the following is not a possible combination of dark chocolate bars and milk chocolate bars?

A. 10 dark bars and 15 milk bars
B. 12 dark bars and 12 milk bars
C. 14 dark bars and 18 milk bars
D. 6 dark bars and 21 milk bars

Question

Jimmy's Chocolate shop sells two different fudge bars. The dark chocolate fudge bar sells for $6.00 per pound and the milk chocolate fudge bar sells for $4.00 per pound. You purchased a combination of these bars for all of your family and friends for the holidays and your bill came to $120. This scenario can be modeled by the equation 6d + 4m = 120. Which of the following is not a possible combination of dark chocolate bars and milk chocolate bars?
 	
	A. 10 dark bars and 15 milk bars
	B. 12 dark bars and 12 milk bars
	C. 14 dark bars and 18 milk bars
	D. 6 dark bars and 21 milk bars

anonymous · Answer

replace in the equation

anonymous · Answer

the answer is c because if you plug in the numbers from c, you will get this:
6*14+18*4=
84+72=
156
156 doesnt equal 120 so it is not a possible combination of dark chocolate and milk chocolate bars.

anonymous · Answer

?

anonymous · Answer

the answer is c because you purchased only 120 and 156 is more than 120