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?
Answer
12 dark bars and 12 milk bars
14 dark bars and 18 milk bars
10 dark bars and 15 milk bars
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?
Answer
		12 dark bars and 12 milk bars
		14 dark bars and 18 milk bars
		10 dark bars and 15 milk bars
		6 dark bars and 21 milk bars

anonymous · Answer

just plug in and see which ones dont work lol

ash2326 · Answer

d= no. of dark chocolates
m= no. of milk chocolates
We have
$$6d+4m=120$$
@JWALKER295 we have 4 options.  I'll check the last one and rest all you try
6 dark and 21 milk
so d=6, m=21
$$6\times 6+4\times 12=120$$
$$36+84=120$$
$$120=120$$
So this is a possible combination. Now you check the other options

anonymous · Answer

the second one

anonymous · Answer

i got is the second one thank you

anonymous · Answer

ur welcome!!!

anonymous · Answer

14 dark and 18 milk