Let's "write equations for all this."

Part A - the original problem.

I know, it's one of those horrible word problem things. Let's take a look at the original problem.

My car's cooling system holds 12 quarts. It has four quarts of antifreeze, and eight quarts of water. How much of this mixture do I have to drain out and replenish with straight antifreeze in order to get a 50/50 mixture?

So, where do we start? All the algebra problems I ever saw had an x in them, so we need an x. What shall x stand for? It could be water, antifreeze, or something else entirely. When in doubt, let x stand for the thing you want to find. So:

Let x = "how much of this mixture do I have to drain out."

Next, take a look at the following "equation."

start - remove + add = finish

Think apples. If you start with 4, take away 2, and add 3, you end up with 5. Simple, right? It works exactly the same way with fluids in the cooling system. The only thing going both in and out of the cooling system is antifreeze, so let's look at that.

We are told in the problem we have 4 quarts of antifreeze at the start. So,

start = 4.

How much antifreeze do we want in the system AFTER we have taken x of the mixture OUT and put x of the antifreeze back IN? Well, we want a 50/50 mixture, so we want 6 quarts of antifreeze and 6 quarts of water in the 12-quart system. So,

finish = 6.

Remember, we let x stand for the number of quarts of MIXUTRE we removed. So, x must also equal the number of quarts of ANTIFREEZE we are adding back. So,

add = x.

So far, we have the following:

4 - removed + x = 6

So, how many quarts of ANTIFREEZE did we remove when we removed x quarts of MIXTURE? Well, there are 4 quarts of antifreeze and 8 quarts of water, so each drop (or pint or quart) is 1/3 antifreeze and 2/3 water. So the amount of antifreeze we removed is (1/3) times the amount of mixture we removed. So,

remove = (1/3)x.

Now we have a real equation we can solve. Yahoo! Those who hate algebra, just take a deep breath. I will make it as painless as possible.

4 - (1/3)x + x = 6
4 - (1/3)x + (3/3)x = 6 since 1 = (3/3)
4 + (2/3)x = 6 you have 3 thirds of something and remove 1 third
(2/3)x = 2 subtract 4 from both sides
2x = 6 multiply both sides by 3
x = 3 divide both sides by 2

So, we think the answer is 3. Stop! We need to check our work! Oh, wait, Ray already did that. Thanks, Ray.

Now, the rest of the problems should be, dare I say it, easy. Right?

Part B - replacing with 50/50 mixture.

Let's begin by restating Part B of the original puzzler.

My car's cooling system holds 12 quarts. It has four quarts of antifreeze, and eight quarts of water. How much of this mixture do I have to drain out and replenish with a 50/50 mixture in order to get a 50/50 mixture?

Remember our starting idea.

start - remove + add = finish

What has changed? Just the fact that we now add a 50/50 mixture, so instead of adding x quarts of antifreeze, we are adding (1/2)x quarts. OK, let's plug that in and find our answer.

4 - (1/3)x + (1/2)x = 6
4 - (2/6)x + (3/6)x = 6 since 1/3 = 2/6 and 1/2 = 3/6
4 + (1/6)x = 6 you have 3 sixths of something and remove 2 sixths
(1/6)x = 2 subtract 4 from both sides
x = 12 multiply both sides by 6

Tricky, tricky. Don't mess with my head, Ray.

Challenge problems.

Remember my challenge problems?

Suppose you add a mixture that is 2/3 antifreeze instead of pure antifreeze, and want to end up with a 50/50 mixture as before?
Suppose you add pure antifreeze as before, but we want to end up with a mixture that is 2/3 antifreeze instead of a 50/50 mixture?

Part 1 should be especially easy, based on the discussion of Ray's Part B. Now, instead of adding (1/2)x we are adding (2/3)x.

4 - (1/3)x + (2/3)x = 6
4 + (1/3)x = 6 you have 2 thirds of something and remove 1 third
(1/6)x = 2 subtract 4 from both sides
x = 6 multiply both sides by 3

Part 2 is not difficult. Now, instead of ending up with (1/2) of 12 = 6 we end up with (2/3) of 12 = 8. So, only the right side changes from the original.

4 - (1/3)x + x = 8
4 - (1/3)x + (3/3)x = 8 since 1 = (3/3)
4 + (2/3)x = 8 you have 3 thirds of something and remove 1 third
(2/3)x = 4 subtract 4 from both sides
2x = 12 multiply both sides by 3
x = 6 divide both sides by 2

Congratulations. You did it.

Too easy, you say?

If you are still curious, how about tackling a real hard core problem?

How to Solve It

Differential Equations

Last modified:

Peter C. Scott

`4 - (1/3)x + x = 6`
`4 - (1/3)x + (3/3)x = 6`	`since 1 = (3/3)`
`4 + (2/3)x = 6`	`you have 3 thirds of something and remove 1 third`
`(2/3)x = 2`	`subtract 4 from both sides`
`2x = 6`	`multiply both sides by 3`
`x = 3`	`divide both sides by 2`

`4 - (1/3)x + (1/2)x = 6`
`4 - (2/6)x + (3/6)x = 6`	`since 1/3 = 2/6 and 1/2 = 3/6`
`4 + (1/6)x = 6`	`you have 3 sixths of something and remove 2 sixths`
`(1/6)x = 2`	`subtract 4 from both sides`
`x = 12`	`multiply both sides by 6`

`4 - (1/3)x + (2/3)x = 6`
`4 + (1/3)x = 6`	`you have 2 thirds of something and remove 1 third`
`(1/6)x = 2`	`subtract 4 from both sides`
`x = 6`	`multiply both sides by 3`

`4 - (1/3)x + x = 8`
`4 - (1/3)x + (3/3)x = 8`	`since 1 = (3/3)`
`4 + (2/3)x = 8`	`you have 3 thirds of something and remove 1 third`
`(2/3)x = 4`	`subtract 4 from both sides`
`2x = 12`	`multiply both sides by 3`
`x = 6`	`divide both sides by 2`