How many gallons of a 50% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 40% antifreeze? Use the six step method.

x = 270 gallons

270 gallons of a 50% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 40% antifreeze

Step-by-step explanation:

Let x represent the number of gallons of the 50% antifreeze solution.

The volume of the 10% antifreeze is = 90 gallons

The volume of the mixture that is 40% antifreeze = 90+x

The amount of antifreeze in the two solutions must be equal to the amount of anti freeze in the mixture;

50% of x + 10% of 90 = 40% of (90+x)

0.50(x) + 0.10(90) = 0.40(90+x)

0.50x + 9 = 36 + 0.40x

Collecting the like terms;

0.50x - 0.40x = 36-9

0.10x = 27

x = 27/0.10

x = 270 gallons

270 gallons of a 50% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 40% antifreeze

imma say 16%. sorry if its wrong. i'm doing the same stuff right now

step-by-step explanation:

step-by-step explanation: