Question

How many gallons of 90% antifreeze must be mixed with 60 gallons of 10% antifreeze to get mixture that is 80% antifreeze?

Answer 1

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:

`x * 90 \% + 60 * 10 \% = (x + 60) * 80 \%`

Solve the above equation for "x"

`x * (90)/(100) + 60 * (10)/(100) = (x + 60) * (80)/(100)\\\\0.9x + 6 = (x+60) * 0.8\\\\0.9x + 6 = 0.8x + 48\\\\\text{Combine the like terms }\\\\0.9x - 0.8x = 48 - 6\\\\0.1x = 42\\\\x = (42)/(0.1) = 420`

Thus, 420 gallons of 90% antifreeze is used