How many gallons of a 50% antifreeze solution must be mixed with 60 gallons of 30% antifreeze to get a mixture that is 42% antifreeze?
+1084 We can form a "proportion" with information. First, let's find out a general "formula":
a% b% c%
---- + ---- = ---------
x y x+y
Multiplying the percentages with the corresponding gallon, we see: ax + by = c(x+y)
From here, we can plug in numbers:
0.5x + 0.3(60) = (0.42)(x + 60)
=> 0.5x + 18 = 0.42x + 25.2
=> 0.5x = 0.42x + 7.2
=> 0.08x = 7.2
=> x = 90
Therefore, we must add 90 gallons.
Edit: Thanks to guest's answer, I corrected my numbers (but my answer remains the same...)
:)
Let the number of gallon of 50% antifrreeze =G
0.50G + 0.30*60 =0.42 *[G + 60]
0.50G + 18 =0.42G + 25.2
0.50G - 0.42G =25.2 - 18
0.08G = 7.2
G = 7.2 / 0.08
G = 90 - gallons of 50% antifreeze must be mixed with 60 gallons of 30% antifreeze to make [90 + 60]=150 gallons of 0.42% of antifreeze.
