# 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?

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?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Ezra Herbert
Let x be the amount of 90% gasoline mixture so that 1200-x is the amount of 75% gasoline mixture, both in liters.
In terms of percentage, we write: 0.90(x)+0.75(1200-x)=0.85(1200)
Solve for x: 0.90x+900-0.75x=1020
0.15x+900=1020
0.15x=120
$x=\frac{120}{0.15}$
x=800
Hence, we need 1800 liters of the 90% gasoline mixture and 400 liters of the 75% gasoline mixture.
Jeffrey Jordon