420 gallons of 90% antifreeze must be mixed with 60 gallons of 10% antifreeze to get mixture that is 80% antifreeze
Solution:
Let "x" be the gallons of 90% antifreeze
The, final mixture would contain (x + 60) gallons
Then by given question, we can say,
"x" be the gallons of 90% antifreeze must be mixed with 60 gallons of 10% antifreeze to get 80 % of (x + 60) gallons
Thus we frame a equation as:
[tex]x \times 90 \% + 60 \times 10 \% = (x + 60) \times 80 \%[/tex]
Solve the above equation for "x"
[tex]x \times \frac{90}{100} + 60 \times \frac{10}{100} = (x + 60) \times \frac{80}{100}\\\\0.9x + 6 = (x+60) \times 0.8\\\\0.9x + 6 = 0.8x + 48\\\\\text{Combine the like terms }\\\\0.9x - 0.8x = 48 - 6\\\\0.1x = 42\\\\x = \frac{42}{0.1} = 420[/tex]
Thus, 420 gallons of 90% antifreeze is used
