Achemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. the premium antifreeze solution contains 80% pure antifreeze. the company wants to obtain 240 gallons of a mixture that contains 60% pure antifreeze. how many gallons of water and how many gallons of the premium antifreeze solution must be mixed?

Volume of 80% pure antifreeze used = 180 gallons.

Volume of water to be used = 240 - 180 gallons = 60 gallons

Explanation:

Let the volume of 80% pure antifreeze used to make 60% pure antifreeze = x gallons  

The volume of 60% pure antifreeze = 240 gallons

For 80% pure antifreeze:

C₁ = 80% , V₁ = x gallons

For 60% pure antifreeze :

C₂ = 60% , V₂ = 240  gallons

Applying,

C₁V₁ = C₂V₂

80×x = 60×240

x = 180  gallons

Thus , volume of 80% pure antifreeze used = 180 gallons.

Volume of water to be used = 240 - 180 gallons = 60 gallons

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6