Mixture problems: complete GMAT guide

For a solution, the amount of solute is

If 200 milliliters of a solution is 30% acid, the amount of acid is $0.30 \times 200=60$ milliliters. The remaining 140 milliliters is solvent or other liquid.

When two solutions are mixed, the total solute is the sum of the solutes, and the total volume is the sum of the volumes:

This equation is the foundation for nearly every mixture problem. Keep concentrations as decimals or fractions, and keep volumes in the same units.

You can also solve by comparing deviations from the target. The 20% solution is 10 percentage points below 30%, and the 50% solution is 20 points above. The amounts must be in the inverse ratio of the deviations: 20 parts of the 20% solution for every 10 parts of the 50% solution, or 2:1. Since there are 8 liters of the 50% solution, there are 16 liters of the 20% solution. This is the alligation method.

Alligation arranges the two concentrations and the target concentration:

The ratio of A to B is the ratio of the differences, taken crosswise: $20:10=2:1$. Alligation is fast when the numbers are clean, but the conservation equation is more reliable when they are not.

Replacement problems remove some mixture and replace it with a pure substance or a different concentration. Track the amount after each step. If a container starts with $V$ liters at concentration $c$, and $x$ liters are removed, the amount of solute removed is $cx$. The remaining solute is $c(V-x)$. Then add the replacement and recalculate.

If the process is repeated, the concentration multiplies by the fraction of volume that remains each time. After one replacement, the fraction is $(V-x)/V$. After $n$ replacements, the concentration is $c\left(\frac{V-x}{V}\right)^n$. This formula applies when the replacement liquid contains none of the solute.

A pure substance has concentration 100%, and a pure solvent has concentration 0%. For example, mixing 3 liters of pure juice with 5 liters of water produces 8 liters at $3/8=37.5\%$ juice. Treat pure juice as a 100% solution and water as a 0% solution in the conservation equation.

Evaporation removes solvent but not solute. The amount of solute stays the same while the volume decreases, so the concentration increases. If 20 liters of a 10% solution loses 5 liters of water, the solute remains 2 liters and the new volume is 15 liters, giving a concentration of $2/15 \approx 13.3\%$.

A table keeps the amounts organized:

Fill in the known values, set the sum of the first two solute amounts equal to the final solute amount, and solve for the unknown.

A mixture can contain more than two ingredients. The conservation equation still works: the total amount of each component before mixing equals the total after mixing. If a blend contains three types of coffee at different prices, multiply each quantity by its price, add the results, and divide by the total quantity to find the average price per pound.

For concentration problems with three solutions, write one equation for the total volume and one equation for the total solute. If there are three unknown volumes, you need a third relationship, such as a fixed ratio between two of the volumes. Without enough relationships, the problem is underdetermined.

A mixture answer should satisfy three quick checks. First, the final concentration must lie between the lowest and highest concentrations of the ingredients. If a 20% solution and a 50% solution produce a 30% mixture, the answer is plausible. A result of 60% would be impossible.

Second, the final volume must equal the sum of the volumes, unless some liquid was removed or evaporated. Third, the amount of solute must be conserved. Substitute your answer back into the conservation equation and confirm that both sides match.

When a problem gives a ratio rather than a volume, assign a multiplier. A mixture with ingredients in the ratio 2:3:5 has volumes $2k$, $3k$, and $5k$. The total volume is $10k$, and the concentration of each ingredient is its part divided by 10. This keeps the arithmetic consistent even when the actual volumes are not given.

1. Identify the solute and the solvent.
2. Convert every percentage to a decimal or fraction.
3. Write the amount of solute in each container before mixing.
4. Set the total solute before mixing equal to the total solute after mixing.
5. For replacement, subtract the solute removed before adding the new liquid.
6. Check that the final concentration lies between the two starting concentrations.

• Adding percentages directly instead of weighting by volume.
• Forgetting to subtract the solute removed during replacement.
• Using the final volume as the denominator for the amount removed.
• Treating pure water as 100% instead of 0%.
• Mixing units such as liters and milliliters.
• Assuming the final concentration is the simple average of the two concentrations.