Mixture Problem: Antifreeze Solution

What's Really Going On Here

The tank stays full at 13 quarts the whole time, yet its antifreeze content changes. That is what makes this problem work. The percentages are a distraction; the real bookkeeping happens in quarts of antifreeze, and the fluid you drain takes some of it with it.

• How to convert a percentage into an actual amount of a pure ingredient, and why that amount is the quantity worth tracking.
• How to model a "drain and replace" operation as amount before − amount removed + amount added = amount after.
• Why drained fluid still counts: it carries the old concentration out with it, so each swapped quart gains less than a full quart.
• How to solve the same problem three ways (algebra, net-gain-per-quart, alligation) and recognize that they are one idea in three costumes.
• How to handle an answer like 39/7 without panic. It is a perfectly respectable result.

Solution: Method 1 — Tracking the Pure Antifreeze

The cleanest way into any mixture problem is to ask: what quantity is conserved or accounted for? Here it is the antifreeze itself. The total volume never changes, but the antifreeze does, so we write an equation about antifreeze.

Step 1 — Convert both percentages into quarts

At the start, 30% of 13 quarts is antifreeze. At the finish, 60% of the same 13 quarts should be antifreeze.

So the swap must raise the antifreeze content by 7.8 − 3.9 = 3.9 quarts.

Step 2 — Name the unknown

Let x be the number of quarts drained and then replaced with pure antifreeze. Because we put back exactly what we took out, the radiator holds 13 quarts at the end, just as at the start.

Step 3 — Follow the antifreeze through the swap

Draining x quarts of the well-mixed fluid removes 30% of x as antifreeze. Pouring in x quarts of pure antifreeze adds all of x. A table keeps the bookkeeping honest:

Step 4 — Set the last row equal to the goal and solve

Notice that −0.3x + x combines to 0.7x. That 0.7 is the heart of the problem: each quart swapped nets only 0.7 quart of new antifreeze, because 0.3 quart of antifreeze goes down the drain with it.

Step 5 — Translate the fraction into something usable

39/7 = 5 4/7 ≈ 5.57 quarts. In kitchen units, 4/7 of a quart is about 2.3 cups, so you would drain a little over five and a half quarts.

Solution: Method 2 — The Net-Gain-Per-Quart Approach

Method 1 is algebra on autopilot. This method is the same fact told as a story, and it lets you solve the problem without writing an equation at all.

One quart swapped: you remove a quart that is 30% antifreeze (losing 0.3 quart of antifreeze) and add a quart that is 100% antifreeze (gaining 1 quart). The net gain per quart swapped is

The gap to close: we need 7.8 quarts of antifreeze and have 3.9, a shortfall of 3.9 quarts.

How many swaps? It is a "how many 0.7s fit into 3.9" question:

This is exactly the equation 0.7x = 3.9 from Method 1, reached by thinking instead of manipulating. If you ever blank on an exam, ask "what does one swapped quart accomplish, and how big is the gap?"

Solution: Method 3 — The Alligation (Ratio) Approach

Alligation is an old merchants' and pharmacists' technique. It becomes available once you notice a reframing: after you drain x quarts, the radiator holds 13 − x quarts of 30% fluid, and you are blending that with x quarts of 100% antifreeze. A drain-and-replace problem is a two-ingredient blend in disguise.

To blend a 30% ingredient with a 100% ingredient into a 60% mixture, subtract diagonally through the target:

Each ingredient's parts come from the other ingredient's distance to the target. The weaker ingredient (30%) must be the bigger share because the target is closer to 30% than to 100%? Not quite. The target is 30 points from 30% but 40 points from 100%, so it sits closer to 30%, which is why 30%-fluid gets the larger share, 40 parts. The blend is 40 : 30 = 4 : 3 old fluid to pure antifreeze, so the pure antifreeze makes up 3 of every 7 parts:

Same answer, no equation. A useful bonus: the ratio 3/7 says that 60% sits 3/7 of the way from 30% to 100%. We will turn that observation into a general formula below.

Verification

Substitute x = 39/7 back into the antifreeze bookkeeping, keeping fractions exact.

Independent check: Method 1 (equation), Method 2 (gap ÷ rate) and Method 3 (3/7 of 13) all land on 39/7. Three different routes giving the same number is strong evidence.

Does This Seem Reasonable?

Before trusting 5.57 quarts, pin it between two extremes. Because the concentration changes linearly with the amount swapped, the boundary cases tell you where the answer must live:

The target of 60% is 3/7 of the way from 30% to 100%, so the swap should be about 3/7 of the tank, a little less than half. 5.57 out of 13 is 42.9%, and that fits. The tempting "halfway" intuition, 6.5 quarts, overshoots to 65%.

Three Mistakes That Are Easy to Make

The General Formula Behind This

Everything above generalizes. Suppose a container holds volume V at concentration c, you swap x of it for fluid at concentration r, and the target is t. The balance equation is

For this problem: x = 13 × (0.60 − 0.30)/(1.00 − 0.30) = 13 × 3/7 = 39/7. This is the 3/7 from alligation, the "gap ÷ rate" from Method 2, and the algebra from Method 1, all in one line.

On the "ugly" answer: 39/7 is a repeating decimal (5.571428…) because the tank size (13), the percentages (30% and 60%) and the net rate (0.7) do not share nice factors. That is normal in real problems. Keep the fraction until the very end, then round to a sensible precision, since nobody can pour 5.571428 quarts anyway.

If You See These Words...

• "drained and replaced with" or "removed and replaced": the total volume is unchanged, and you will have a −cx + rx pattern.
• "pure" means concentration 1 (100%), which makes the algebra simplest.
• "so that the new mixture is …%" means the target concentration applies to the whole final volume.
• A full tank, flask or container that cannot hold more. That implies you must remove before you add.

Disguised twin: "A 40-liter tank of 25% saltwater is partly drained and refilled with fresh water to dilute it to 10%." It is the same structure with a replacement concentration of r = 0.

Beyond the Textbook

• Engine coolant: a mechanic who finds coolant too weak does exactly this calculation. Many manufacturers recommend staying near a 50/50 mix and not much above 70% antifreeze, because too much antifreeze actually lowers heat transfer and freeze protection.
• Pharmacy and laboratories: adjusting a stock solution's strength by removing some and topping up with a stronger one uses the same balance equation.
• Swimming pools and aquariums: partial water changes dilute a chemical level, which is the same math with r = 0.

Push Further

Try each one on paper first, then open the solution to compare. They build in difficulty, and the last one challenges an assumption you may not know you made.