Question 221397
Let {{{x}}} = gallons of 10% mixture to be added
given:
Iso in 20% mix = {{{.2*60 = 12}}}
Iso in 10% mix = {{{.1x}}}
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In words
(iso in 10% mix)+(iso in 20% mix)/(total gallons of mix) = 16%
{{{(12 + .1x)/(60 + x) = .16}}}
{{{12 + .1x = .16*(60 + x)}}}
{{{12 + .1x = 9.6 + .16x}}}
{{{.06x = 2.4}}}
{{{x = 40}}}
He must add 40 gallons of the iso mix
check answer
{{{(12 + .1x)/(60 + x) = .16}}}
{{{(12 + .1*40)/(60 + 40) = .16}}}
{{{(12 + 4)/100 = .16}}}
{{{16/100 = .16}}}
{{{.16 = .16}}}
OK