SOLUTION: A new edition of a video game cost $60, but its previous version only cost $40. If 15 games are sold for $840, how many copies of the newest game sold?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A new edition of a video game cost $60, but its previous version only cost $40. If 15 games are sold for $840, how many copies of the newest game sold?       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 844196: A new edition of a video game cost $60, but its previous version only cost $40. If 15 games are sold for $840, how many copies of the newest game sold?

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Maybe a mixture problem but also a linear system problem.

n is how many new
p is how many previous versions

Account for total versions sold: n%2Bp=15
Account for sales value: 60n%2B40p=840 which is simplifiable to 3n%2B2p=42

SOLVE THE SYSTEM:
----------------------
n+p=15
----------------------
3n+2p=42
----------------------

Choosing Elimination Method, you can use 2n+2p=30 and 3n+2p=42, and subtract first equation from second equation.
%283n%2B2p%29-%282n%2B2p%29=42-30
3n%2B2p-2n-2p=12
n%2B0=12
highlight%28n=12%29 and this means highlight%28p=3%29