Question 104438This question is from textbook College Algebra
: A new automobile engine can run on a mixture of gasoline and a substitute fuel. If gas costs $1.50 per gallon and the substitute fuel costs 40 cents per gallon, what percent of a mixture must be substitute fuel to bring the cost down to $1.00 per gallon?
This question is from textbook College Algebra
Answer by stormy(4)   (Show Source):
You can put this solution on YOUR website! Let g = the number of gallons of gas and s = the number of gallons of substitute fuel in the mixture. Then:
The price of g gallons of gas = 1.50 g
The price of s gallons of substitute fuel = 0.40 s
Therefore, the price of the mixture is: 1.50 g + 0.40 s
To get a price of 1.00 per gallon, you have to mix the two types. The price of this mixture is $1 multiplied the total number of gallons in the mixture:
1.00 ( g + s )
Now set the two expressions for the price of the mixture equal to each other and solve for "g":
1.50 g + 0.40 s = 1.00 ( g + s )
1.50 g + 0.40 s = 1.00 g + 1.00 s
1.50 g - 1.00 g = 1.00 s - 0.40 s
0.50 g = 0.60 s
g = (0.60 / 0.50) s
g = 1.2 s
From this we see that we need 1.2 gallons of gas for every gallon of substitute fuel.
We can now calculate the percentage of substitute fuel in the mixture:
s / (g + s) = s / (1.2s + s) = s / 2.2s = 0.4545, or 45.45%
To check our solution, verify that 45.45% of the substitute fuel price added to 100 - 45.45 = 54.55% of the gas price equals $1.00:
0.5455($1.50) + 0.4545($0.40) = $1.00
It does!
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