SOLUTION: a truck can be rented from company a for $80 a day plus $0.60 per mile. company b charges $50 a day plus $0.90 per mile to rent the same truck. how many miles must be d
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-> SOLUTION: a truck can be rented from company a for $80 a day plus $0.60 per mile. company b charges $50 a day plus $0.90 per mile to rent the same truck. how many miles must be d
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Question 1196226: a truck can be rented from company a for $80 a day plus $0.60 per mile. company b charges $50 a day plus $0.90 per mile to rent the same truck. how many miles must be driven in a day to make the rental cost for company a a better deal than company b's? Found 2 solutions by amoresroy, MathTherapy:Answer by amoresroy(361) (Show Source):
You can put this solution on YOUR website! Let X = number of miles the truck must be driven in a day
The equation
80 + 0.6X = 50 + 0.9X
Combine like terms
80 - 50 = 0.9X - 0.6X
30 = 0.3X
X = 30/0.3
X = 100 miles
Answer:
The truck must be driven at least 100 miles in a day to make the rental cost for Company A a better deal than company B's.
You can put this solution on YOUR website!
a truck can be rented from company a for $80 a day plus $0.60 per mile. company b charges $50 a day plus $0.90 per mile to rent the same truck. how many miles must be driven in a day to make the rental cost for company a a better deal than company b's?
With M being number of miles, correct INEQUALITY (NOT equation), is: .6M + 80 < .9M + 50
M, or number of miles driven must be GREATER than 100 in order for A to be a better deal than B.
