Answered

A concession stand sells hamburgers (h) for $2 and hotdogs (d) for $1. On Friday night they sold a total of 300 hamburgers and hotdogs and made $420. How many of each did they sell?


What solution is reasonable?

Question 3 options:



200 hamburgers and 50 hotdogs



50 hamburgers and 200 hotdogs



120 hamburgers and 180 hotdogs



100 hamburgers and 180 hotdogs


I think the answer is 120 hamburgers and 180 hotdogs ,but I'm not sure.Thanks to whoever helps me .

Verified Answer :

120 hamburgers and 180 hot dogs.

JeanaShupp

Answer: 120 hamburgers and 180 hotdogs

Step-by-step explanation:

Let 'x' be the number of hamburgers and 'y' be the number of hotdogs sold.

Then, the total number of hamburgers and hotdogs sold on friday will be given by :_

\(x+y=300.....................(1)\)

On the basis of their cost, the total cost will be:-

\(2x+y=420..............................(2)\)

Subtract (1) from (2), we get

\(x=120\)

Substitute the value of x in  (1), we get

\(120+y=300\\\Rightarrow y=300-120\\\Rightarrow y=180\)

Hence, they sold 120 hamburgers and 180 hotdogs.

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