SOLUTION: In testing an engine, various mixtures of gasoline and methanol are being tried. How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200 L of an 85% gasol

Algebra ->  Equations -> SOLUTION: In testing an engine, various mixtures of gasoline and methanol are being tried. How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200 L of an 85% gasol      Log On


   

Question 289131: In testing an engine, various mixtures of gasoline and methanol are being tried. How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200 L of an 85% gasoline mixture?
Found 2 solutions by brucewill, stanbon:
Answer by brucewill(101) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = L of 90% gas
Then (1200 - x) would be the # of L of 75% gas.
We will end up with 1020L (85% * 1200 L) of overall 100% gas.
.9x + .75(1200 - x) = 1020
.9x + 900 - .75x = 1020
.15x = 120
highlight%28x+=+800%29
Prove:
800 L of 90% gas (720 L pure)
400 L of 75% gas (300 L pure).

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How much of a 90% gasoline mixture and a 75% gasoline mixture are needed for 1200 L of an 85% gasoline mixture?
--------------
Equation:
gas + gas = gas
0.90x + 0.75(1200-x) = 0.85*1200
----
Multiply thru by 100 and solve for "x":
90x + 75*1200 - 75x = 85*1200
15x = 10*1200
x = (2/3)1200
x = 800 L (amt. of 90% gasoline needed in the mixture)
---
1200-x = 400 L (amt. of 75% gasoline needed in the mixture)
====================================
Cheers,
Stan H.