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?

1 Answer
Feb 7, 2018

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