Question

How many gallons of a 80% antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze

Answer 1

440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze

Solution:

Let "x" be the gallons of 80 % antifreeze added

Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze

Final mixture is x + 80

Therefore, we can frame a equation as:

"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze

Thus, we get,

x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %

$x \times \frac{80}{100} + 80 \times \frac{15}{100} = (x+80) \times \frac{70}{100}\\\\0.8x + 80 \times 0.15 = (x+80) \times 0.7\\\\0.8x+12 = 0.7x+56\\\\0.8x-0.7x=56-12\\\\0.1x = 44\\\\x = \frac{44}{0.1}\\\\x = 440$

Thus 440 gallons of 80 % antifreeze solution must be mixed