Question

How many gallons of 90% antifreeze must be mixed with 60 gallons of 15% antifreeze to get a mixture that is 80%?

Answer 1

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"

`90 \% * x + 15 \% * 60 = 80 \% * (x+60)\\\\(90)/(100) * x + (15)/(100) * 60 = (80)/(100) * (x+60)\\\\0.9x + 0.15 * 60 = 0.8(x + 60)\\\\0.9x + 9 = 0.8x + 48\\\\0.9x - 0.8x = 48 - 9\\\\0.1x = 39\\\\ x = 390`

Thus 390 gallons of 90 % antifreeze must be mixed