The roosevelt middle school band is having a fundraiser. They sold a total of 300 hotdogs and hamburgers. Hot dogs were sold for $2 and hamburgers soldfor $3. They made a total of $780 how many individual hamburgers And hotdogs did they sell.

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Mathematics

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25 April, 04:47

The roosevelt middle school band is having a fundraiser. They sold a total of 300 hotdogs and hamburgers. Hot dogs were sold for $2 and hamburgers soldfor $3. They made a total of $780 how many individual hamburgers And hotdogs did they sell.

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Lena Hurley · Answer 1

Answer: The School band sold a total of; 180 hamburgers and 120 hotdogs.

Step-by-step explanation: For a start we shall represent hotdogs by the letter d and hamburgers by the letter b. If they sold a total of 300 hotdogs and hamburgers, then we can express this as, d + b = 300.

Also if one hotdog was sold for $2 and one hamburger was sold for $3, and altogether realized $780, this can also be expressed as, 2d + 3b = 780.

We now have a pair of simultaneous equations as follows;

d + b = 300 - - - (1)

2d + 3b = 780 - - - (2)

From equation (1), make d the subject of the equation. Therefore d = 300 - b

Substitute for d into equation (2)

2 (300 - b) + 3b = 780

600 - 2b + 3b = 780

Collect like terms

3b - 2b = 780 - 600

b = 180

We can now substitute for the value of b into equation (1)

d + b = 300

d + 180 = 300

Subtract 180 from both sides of the equation

d = 120

Therefore, the school band sold 180 hamburgers and 120 hotdogs

Osvaldo Ware · Answer 2

180 hamburgers

120 hotdogs

Step-by-step explanation:

In this question, we are asked to calculate the number of hamburgers and hotdogs sold by a company given the amount made by them and the total number of these snacks sold

We proceed as follows;

Let the amount of hotdogs sold be x and the amount of hamburgers sold be y.

We have a total of 300 snacks sold, mathematically;

x + y = 300 ... (I)

Now let's look at the prices.

x number of hotdogs sold at $2, this give a total of $2x hotdogs

y number of hamburgers sold at $3, this give a total of $3y.

Adding both to give total, we have;

2x + 3y = 780 ... (ii)

This means we have two equations to solve simultaneously. From equation 1, we can say x = 300 - y

Now let's insert this in the second equation;

2 (300-y) + 3y = 780

600-2y + 3y = 780

y = 780-600 = 180

Recall; x + y = 300

x = 300 - y

x = 300-180 = 120