Amount of water = 260 gallons
Amount of antifreeze = 60 gallons
Explanation:
let the amount of water = w
amount of water + amount of antifreeze will give 320 gallons of mixture
w + amount of antifreeze = 320
amount of antifreeze = 320 - w ...(1)
The concentration for the antifreeze = 80% = 0.8
The concentration of the mixture = 15% = 0.15
From our question, we are mixing antifreeze with pure water (two different things). In our calculation, we can write the concentration in terms of water or in terms of antifreeze.
Writing the concentration in terms of amount of antifreeze
0.8(amount of antifreeze) + percent of antifreeze (amount of water) = concentraton of the mixture
The mixture contains 15% antifreeze
water contains no antifreeze, percent = 0
0.8(320 - w) + 0(w) = 0.15(amount of mixture)
0.8(320 - w) + 0 = 0.15(320)
256 - 0.8w + 0 = 48
256 - 0.8w = 48
collect like terms:
256 - 48 = 0.8w
208 = 0.8w
divide both sides by 0.8:
208/0.8 = w
w = 260
substitute for w in equation (1)
amount of antifreeze = 320 - 260 = 60
Amount of water = 260 gallons
Amount of antifreeze = 60 gallons
