Word problem

Question

Antifreeze is a compound added to water to reduce the freezing point of a mixture. In extreme cold (less than -35 F ) one car manufacturer recommends that a mixture of 65% antifreeze be used. How much 50% antifreeze solution should drained from a 4-gal tank and replaced with pure antifreeze to produce a 65% antifreeze mixture?

Answer 1

Let x = number of gal of 50% solution that is drained 

         = number of gal of 100% solution that is added

 

Antifreeze content of original solution - Antifreeze content of solution that is drained + Antifreeze content of solution that is added = Antifreeze content of mixture

 

0.50(4) - 0.50x + 1.00x = 0.65(4)

 

2 + 0.50x = 2.60

 

       0.50x = 0.60

 

              x = 1.2 gal

 

Drain 1.2 gallons of 50% solution and replace it with 1.2 gallons of pure antifreeze to obtain 4 gallons of solution that is 65% antifreeze.  

Answer 2

You have a mixture that is 4 gallons at 50% antifreeze = 2 gallons of antifreeze in 2 gallons water. 

You want the mixture to be 65% antifreeze = 2.6 gallons antifreeze in 1.4 gallons of water. You need to drain 1.2 gallons from the tank, taking away 0.6 gallons each of water and antifreeze, replacing it with 1.2 gallons of pure antifreeze. 

2 gallons - 0.6 gallons + 1.2 gallons = 2.6 gallons of antifreeze in the radiator with 2 - 0.6 = 1.4 gallons of water.