<PRE><B>Question 159816</B>
An outboard engine uses a gasoline-oil fuel mixture in the ratio of 15 to 1. How much gasoline must be mixed with a gasoline-oil mixture, which is 75% gasoline, to make 8.0L of the mixture for the outboard engine?
:
Find the percent of gasoline in a 15:1 mixture
% gasoline = {{{15/((15+1))}}} * 100 = 93.75% gasoline
:
Let x = amt of gasoline to accomplish this
Then
(8-x) = amt of 75% mixtures used
:
.75(8-x) + 1.0x = .9375(8)
:
6 - .75x + 1x =  7.5
:
-.75x + 1x = 7.5 - 6
:
.25x = 1.5
:
x = {{{1.5/.25}}}
x = 6 liters of gas required (plus 2 liters of the 75% mixture)
;
:
Check our solution :
.75(8-6) + 6 = .9375(8)
.75(2) + 6 = 7.5
1.5 + 6 = 7.5
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