The owner of Dennis's garage wants to fill a 55 gallon drum with a 20% winter mixture of antifreeze for his customers. Denis wants to mix 100% antifreeze with a 10% antifreeze mix to get a 20% mixture. How many gallons of 100% antifreeze and 10% antifreeze does he need in order to fill the drum

6.1 gallons of 100% antifreeze and 48.9 gallons of 10% antifreeze

Step-by-step explanation:

Let the amount of 100% antifreeze be x gallon

let the amount of 10% antifreeze be y gallon

total amount to drum = 55 gallons

therefore x + y = 55 Eqn 1

    100%(x) + 10(y) = 20%(55)

     x + 0.1y = 11Eqn 2

Both equations are solved simultaneously

From eqn 1 x = 55-y

Substitute x = 55-y into eqn 2

55-y+0.1y = 11

collecting like terms

44 = 0.9y

y = 48.88889

x = 55- 48.88889 = 6.11111 gallons

Therefore 6.1 gallons of 100% antifreeze and 48.9 gallons of 10% antifreeze

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!