Question 1112756: The management of hartman rent a car has allocated $1.5 million to buy a new fleet of automobiles consisting of compact intermediate size cars an full size cars. Compact costs $12000 each intermediate size cars cost $18000 each and full sized cars cost $24000 each. If hartman purchases twice as many compacts as intermediate size cars and the total amount of cars to be purchased is 100, determine how many cars of each type will be used
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let c = number of compact size cars.
let i = number of intermediate size cars.
let f = number of full size cars.
total number of cars = 100, therefore:
c + i + f = 100 (original equation 1)
total cost of cars is 1,500,000, therefore:
12,000 * c + 18,000 * i + 24,000 * f = 1,500,000 (original equation 2)
number of compact size cars is twice the number of intermediate size cars, therefore:
c = 2 * i
replace c with 2 * i in both original equations to get:
3 * i + f = 100 (first equation)
42,000 * i + 24,000 * f = 1,500,000 (second equation)
solve for f in first equation to get:
f = 100 - 3 * i
replace f with 100 - 3 * i in second equation to get:
42,000 * i + 24,000 * (100 - 3 * i) = 1,500,000
simplify to get:
42,000 * i + 2,400,000 - 72,000 * i = 1,500,000
solve for i to get:
i = 30
c = 2 * i, therefore c = 60
this makes f = 10, because c + i + f = 100
you have:
c = 60
i = 30
f = 10
c + i + f = 100 becomes 60 + 30 + 10 = 100 which becomes 100 = 100, which is true.
12,000 * c + 18,000 * i + 24,000 * f = 1,500,000 becomes 12,000 * 60 + 18,000 * 30 + 24,000 * 10 = 1,500,000, which becomes 720,000 + 540,000 + 240,000 = 1,500,000, which becomes 1,500,000 = 1,500,000, which is true.
both original equations are true, so solution looks good.
solution is 60 compact cars and 30 intermediate size cars and 10 full size cars will be used.
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