Answer:
C. 10/9 gallons
Explanation:
Given that
V= 20 gallons
Volume of ethanol = 5% of V = 0.05 x 20 = 1 gallons
Volume of the gasoline = 95 % of V = 19 gallons
Lets take x gallons of ethanol is added to achieve optimum performance.
(1+x)/(20+x) = 10 %
(1+x)/(20+x) =0.1
1+ x = 0.1 ( 20+ x)
1+ x = 2 + 0.1 x
1 = 0.9 x
x= 10 /9 gallons
So the option C is correct.
C. 10/9 gallons