390 gallons of 90 % antifreeze must be mixed with 60 gallons of 15% antifreeze to get a mixture that is 80%
Solution:
Let "x" be the gallons of 90 % antifreeze
60 gallons of 15 % antifreeze
Then, (x + 60) is the gallons present in final mixture that is 80 %
Therefore, x gallons of 90 % antifreeze is mixed with 60 gallons of 15 % antifreeze to get a mixture that is (x + 60) gallons of 80 %
Thus, a equation is framed:
90 % of x + 15 % of 60 = 80 % of (x + 60)
Solve the above equation for "x"
[tex]90 \% \times x + 15 \% \times 60 = 80 \% \times (x+60)\\\\\frac{90}{100} \times x + \frac{15}{100} \times 60 = \frac{80}{100} \times (x+60)\\\\0.9x + 0.15 \times 60 = 0.8(x + 60)\\\\0.9x + 9 = 0.8x + 48\\\\0.9x - 0.8x = 48 - 9\\\\0.1x = 39\\\\ x = 390[/tex]
Thus 390 gallons of 90 % antifreeze must be mixed
