SOLUTION: A car dealer stocks mid-size and full size cars on her lot, which holds 60 cars. On the average, she borrows $13,000 to purchase a mid-size car and $18,000 to purchase a full size
Algebra ->
Coordinate Systems and Linear Equations
-> Lessons
-> SOLUTION: A car dealer stocks mid-size and full size cars on her lot, which holds 60 cars. On the average, she borrows $13,000 to purchase a mid-size car and $18,000 to purchase a full size
Log On
Question 1149813: A car dealer stocks mid-size and full size cars on her lot, which holds 60 cars. On the average, she borrows $13,000 to purchase a mid-size car and $18,000 to purchase a full size car. How many cars of each type could she stock if her total debt is 1 million. How many cars of each type could she stock if her total debt is 1 million? Set up and solve a system using either elimination or substitution. Answer by ikleyn(52784) (Show Source):
Let M be the number of the mid-size cars and
let F be the number of the full size cars.
Then you have these 2 equations in 2 unknowns
M + F = 60 (1) (counting cars)
13*M + 18*F = 1000 (2) (counting money in thousand dollars)
Using substitution method, from equation (1) express M = 60 - F and substitute it into equation (2). You will get
13*(60-F) + 18F = 1000. (3)
Simplify; express F and calculate
F = = 44.
ANSWER. 44 full size cars, and the rest, 60-44 = 16 mid-size cars.
CHECK. 44*18 + 16*13 = 1000 thousands dollars. ! Precisely correct !
Solved.
--------------
Buying tickets; buying shirts and socks; buying roses and violets; counting calories and grams of fats in combined food;
buying oranges and apples --- all these problems are of the same type and can be solved by the same method.
