SOLUTION: A mixture of 90% gas and 10% oil is in a 50 gallon tank. The owner desires a 94% gas and 6% oil mixture. The owner will drain the necessary amount of the mixture and replace it wi
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A mixture of 90% gas and 10% oil is in a 50 gallon tank. The owner desires a 94% gas and 6% oil mixture. The owner will drain the necessary amount of the mixture and replace it wi
Log On
Question 663012: A mixture of 90% gas and 10% oil is in a 50 gallon tank. The owner desires a 94% gas and 6% oil mixture. The owner will drain the necessary amount of the mixture and replace it with gas. what is the amount the owner should drain from the tank and replace with gas?
Please help: I think you start with .9x + .1y = 50, but then what?
You can put this solution on YOUR website! A mixture of 90% gas and 10% oil is in a 50 gallon tank.
The owner desires a 94% gas and 6% oil mixture.
The owner will drain the necessary amount of the mixture and replace it with gas.
what is the amount the owner should drain from the tank and replace with gas?
:
let's do it this way
let x = amt of original mixture drained, and the amt of gas to be added
:
Write an equation based on the amt of gas
:
.90(50-x) + x = .94(50)
45 - .9x + x = 47
.1x = 47 - 45
.1x = 2
x = 2/.1
x = 20 gallons drained and 20 gal of gas added
:
Did this make sense to you? C
