Raising the Antifreeze Ratio

Date: 08/07/99 at 11:32:45
From: Steven Robles
Subject: Antifreeze Ratio

I can't understand this problem:

A 5-gallon radiator containing a mixture of water and antifreeze was 
supposed to contain a 50% antifreeze solution. When tested, it was 
found to have only 40% antifreeze. How much must be drained out and 
replaced with pure antifreeze so that the radiator will then contain 
the desired 50% antifreeze solution?

This is a hard one. I set up the equation wrong. Please help.
Thank you.

Date: 08/20/99 at 19:48:12
From: Doctor Jesse
Subject: Re: Antifreeze Ratio

Hey - this *is* a hard one! Good for you, Steven, for not giving up on 
it.

Like most word problems, the hard part is in setting up the equation. 
Let's look at what we know:

  A 5-gallon radiator containing a mixture of water and antifreeze 
  was supposed to contain a 50% antifreeze solution.

Well, that means that the tank is supposed to contain 2.5 gallons of 
antifreeze, and 2.5 gallons of water.

  When tested, it was found to have only 40% antifreeze.

Hmm... That means the tank had (5gal * 0.4) = 2 gallons of antifreeze 
and (5gal * 0.6) = 3 gallons of water in it.

  How much must be drained out and replaced with pure antifreeze so 
  that the radiator will then contain the desired 50% antifreeze 
  solution?

If you could take a half-gallon of water out of the tank and replace 
it with a half-gallon of antifreeze, you'd be done... But you can't do 
that. When you pour out the tank, whatever you pour out will be 40% 
antifreeze and 60% water, which is what makes this problem tricky. But 
think of it this way:

We have 2 gallons of antifreeze in the tank now, and we need to get it 
so that there are 2.5 gallons of antifreeze instead. If we let x be 
the amount of pure antifreeze we need to add to the tank, then we can 
set up an equation:

  2 + x = 2.5

But wait - for every gallon of pure antifreeze we add to the tank, we 
first have to take out an equal quantity of 40% antifreeze... so, 
subtracting that part gives us the equation:

  2 - 0.4x + x = 2.5

Solve for x, and you will know how much 40% antifreeze had to be 
poured out, and how much pure antifreeze added, to get to the 50% 
mixture. I bet you can take it from here.

You also might want to check your answer when you are done. You can do 
this by going to the auto parts store and buying 2.5 gallons of pure 
antifreeze... just kidding! You can check it by plugging x into the 
equation you just solved, and making sure that the answer on the left 
side really does come out to be 2.5 gallons. 

Good luck, Steven!

- Doctor Jesse, The Math Forum
  http://mathforum.org/dr.math/