32 gallons of 25 % antifreeze should be combined with 8 gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze
From given,
Final mixture is 40 gallon
Let "x" be the gallon of 25 % antifreeze
Then, (40 - x) is the gallon of 50 % antifreeze
Therefore, according to question,
"x" gallons of 25 % antifreeze should be combined with (40 - x) gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze
25 % of x + 50 % of (40 - x) = 30 % of 40
Solve for "x"
[tex]\frac{25}{100} \times x + \frac{50}{100} \times (40-x) = \frac{30}{100} \times 40\\\\0.25x + 0.5(40-x) = 12\\\\0.25x + 20 - 0.5x = 12\\\\0.25x - 0.5x = 12 - 20\\\\-0.25x = -8\\\\0.25x = 8\\\\Divide\ both\ sides\ by\ 0.25\\\\x = 32[/tex]
Thus, 32 gallons of 25 % antifreeze is used
Then, (40 - x) = (40 - 32) = 8
Thus 8 gallons of 50 % antifreeze is used
