Question

The math club sells candy bars and drinks during football games. 60 candy bars and 110 drinks will sell for $265. 120 candy bars and 90 drinks will sell for $270. How much does each candy bar sell for?

Answer 1

See a solution process below:

Candy Bars cost $0.75

Drinks cost $2.00

Explanation:

First, let's define our variables:

• We will call the cost of candy bars: `c`
• We will call the cost of drinks: `d`

We can now write two equations from the information in the problem;

• Equation 1: `60c + 110d = $265`

• Equation 2: `120c + 90d = $270`

Step 1) Solve both equations for `60c`.

Equation 1:

`60c + 110d - color(red)(110d) = $265 - color(red)(110d)`

`60c + 0 = $265 - 110d`

`60c = $265 - 110d`

Equation 2:

`120c + 90d - color(red)(90d) = $270 - color(red)(90d)`

`120c + 0 = $270 - 90d`

`120c = $270 - 90d`

`(120c)/color(red)(2) = ($270 - 90d)//color(red)(2)`

`60c = ($270 )/color(red)(2) - (90d)//color(red)(2)`

`60c = $135 - 45d`

Step 2) Because the left side of both equations are the same we can equation the right side of each equation and solve for `d`:

`$265 - 110d = $135 - 45d`

`$265 - color(blue)($135) - 110d + color(red)(110d) = $135 - color(blue)($135) - 45d + color(red)(110d)`

`$130 - 0 = 0 + (-45 + color(red)(110))d`

`$130 = 0 + 65d`

`$130 = 65d`

`($130)/color(red)(65) = (65d)/color(red)(65)`

`$2 = (color(red)(cancel(color(black)(65)))d)/cancel(color(red)(65))`

`$2 = d`

`d = $2`

Step 3) Substitute `$2` for `d` in either of the equations from Step 1 and solve for `c`:

`60c = $135 - 45d` becomes:

`60c = $135 - (45 xx $2)`

`60c = $135 - $90`

`60c = $45`

`(60c)/color(red)(60) = ($45)/color(red)(60)`

`(color(red)(cancel(color(black)(60)))c)/cancel(color(red)(60)) = $0.75`

The Solution Is:

Candy Bars cost $0.75

Drinks cost $2.00