Question 1003627: A car dealer stocks mid-size and full size cars on her lot, which holds 75 cars. On the average she borrows $10,000 to purchase a mid-size car and $15,000 to purchase a full-size car. How many cars of each type could she stock if her total debt is 1 million? please include steps. thanks!!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! m = numbr of midsized cars.
f = number of full sized cars.
m + f = 75
10,000m + 15,000f = 1,000,000
you need to solve both these equations simultaneously to get a solution that is common to both.
first equation says total number of midsize and fullsize cars is equal to 75.
second equation says total amount of money paid for midsize cars and full size cars is 1,000,000.
multiply both sides of the first equation by 10,000 and bring down the second equation unchanged to get:
10,000m + 10,000f = 750,000
10,000m + 15,000f = 1,000,000
subtract the first equation form the second equation to get:
5,000f = 250,000
divide both sides of this equation by 5,000 to get:
f = 50
since m + f = 75, then m = 25
you have m = 25 and f = 50
25 + 50 = 75
10,000 * 25 = 250,000
15,000 * 50 = 750,000
total = 1,000,000
solution is:
she could stock 25 midsize cars and 50 full size cars.
|
|
|
