Overview
A concentration is the amount of one ingredient divided by the total amount of mixture. Changes can affect the numerator, denominator or both, so manipulating percentages alone can hide what is happening. We will keep an ingredient ledger: how much is present, how much leaves, and what total remains. This gives one consistent method for blending liquids, adding ingredient-free material and reversing repeated replacement.
Add ingredient amounts, then divide by the combined total
Two liters of a drink contain 40% fruit juice, and three liters contain 80%. The first portion contributes 2·0.40 = 0.8 L of juice; the second contributes 3·0.80 = 2.4 L. Their combined 3.2 L of juice is contained in 5 L of mixture, so the concentration is 3.2/5 = 0.64, or 64%.
The answer 60 averages the two percentages equally, ignoring that the stronger drink contributes more volume. A weighted average can also solve the problem, but the ingredient ledger explains the weights: each percentage acts on its own volume. The result should lie between the original concentrations and be closer to the larger portion’s concentration. Those expectations check the direction and size of the answer without replacing the exact calculation.
Adding ingredient-free material changes only the denominator
A 500 g cereal blend is 20% nuts by mass, so it contains 100 g of nuts. Adding 100 g of nut-free cereal leaves the nuts unchanged but raises the total mass to 600 g. The new nut fraction is therefore 100/600 = 1/6.
The choice 1/5 preserves the old concentration as though numerator and denominator grew together. Here they do not: the added material contains none of the ingredient being tracked. Label the numerator as “nut mass” and the denominator as “total blend mass” before changing either. The same idea applies to dilution by pure water when the tracked substance is salt or dye. It is conservation of the ingredient amount, not conservation of the percentage, that determines the new concentration.
Distinguish removal alone from removal and replacement
A well-mixed liquid contains 18% dye. Pouring out one quarter removes one quarter of the dye and one quarter of the total liquid. Their ratio stays the same, so the remaining liquid is still 18% dye. The answer 13.5% reduces the numerator’s percentage without also accounting for the smaller total.
“Well-mixed” supplies the reason the removed portion has the same composition as the whole. If pure liquid is added afterward, the total is restored while the lost dye is not, and the concentration then falls. These are two different stages. When a problem describes removal and replacement, draw an intermediate line after removal rather than jumping straight to the final percentage. The amount of tracked ingredient and the total volume may change at different moments, so separating the stages prevents an unjustified shortcut.
Work backward through retention factors
A 40 L tank starts at 25% salt, so it contains 10 L of salt. After an unknown removal and water replacement, a second operation removes 10 L of the well-mixed liquid and replaces it with water. That second operation retains 30/40 = 3/4 of the salt then present. The final 12% concentration in 40 L means 4.8 L of salt remains.
Undo the last retention: before it, there were 4.8/(3/4) = 6.4 L of salt. Thus the first operation removed 10 − 6.4 = 3.6 L of salt. Since the original liquid was 25% salt, removing that amount of salt required 3.6/0.25 = 14.4 L of mixture, or 72/5 L. For transfer, choose either a forward ingredient equation or backward retention steps, but keep every concentration attached to its stage. In practice, verify that removed volumes are physically possible and that adding pure water never increases the tracked ingredient amount.
The replacement water restores the tank to forty liters after each operation but does not restore any removed salt. That is why the second removal uses a denominator of forty rather than the temporary volume immediately after the first removal. Reading the full sequence—remove, replace, mix, then remove again—is necessary to choose the retention factor. Omitting the replacement step would describe a different tank history.
Apply the method to a new case
A solution contains 12 grams of salt in 60 grams of solution. Pure water evaporates until the total mass is 40 grams, with no salt lost. Find the final salt mass fraction. The salt amount stays 12 grams, while the water drops from 48 to 28 grams. The final fraction is 12/40 = 3/10, or 30%. Applying the original 20% concentration to the remaining mass would incorrectly imply that salt evaporated in proportion to water. Check conservation: 12 grams salt plus 28 grams water equals the final 40 grams. This is neither dilution nor uniform removal; the process selectively changes one component. Identify what actually leaves the mixture before reusing a concentration formula. The result should increase because the same solute is now distributed through less solvent.