Question

John sells hamburgers ($3) and cheeseburgers ($3.50). One afternoon he sells a total of 24 burger for $79. How many of these were hamburgers and how many cheeseburgers?

Answer 1

John sold 10 hamburgers and 14 cheeseburgers.

Step-by-step explanation:

Given,

Cost of one hamburger = $3

Cost of one cheeseburger = $3.50

Total burgers sold = 24

Total revenue earned = $79

Let,

Number of hamburgers sold = x

Number of cheeseburgers sold = y

According to given statement;

x+y=24   Eqn 1

3x+3.50y=79    Eqn 2

Multiplying Eqn 1 by 3

$3(x+y=24)\\3x+3y=72\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(3x+3.50y)-(3x+3y)=79-72\\3x+3.50y-3x-3y=7\\0.50y=7$

Dividing both sides by 0.50

$\frac{0.50y}{0.50}=\frac{7}{0.50}\\y=14$

Putting y=14 in Eqn 1

$x+14=24\\x=24-14\\x=10$

John sold 10 hamburgers and 14 cheeseburgers.

Keywords: linear equation, elimination method

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