How to Solve Mixture Problems on the GRE/GMAT

How to Solve Mixture Problems on the GRE/GMAT

Introduction

"Mixture problems appear 4-6 times on every GRE and 3-5 times on the GMAT—master them, and you’ll gain 20+ points by avoiding careless errors and saving 30+ seconds per question."

WHAT THIS QUESTION TYPE IS ACTUALLY TESTING

The GRE/GMAT isn’t testing your ability to solve algebra—it’s testing: 1. Precision under pressure – Can you set up the right equation without mixing up variables? 2. Logical structure – Can you translate words into math without overcomplicating? 3. Trap detection – Can you spot when the question is asking for a ratio vs. an absolute quantity?

ANATOMY OF THE QUESTION

Structure Breakdown

Representative Example Question

"A chemist has 20 liters of a 30% salt solution. She adds 10 liters of a 50% salt solution. What is the concentration of salt in the final mixture?"

Breakdown: - Stem: Two solutions (30% and 50%) being mixed. - Conditions: 20L of 30% + 10L of 50%. - Question: Final concentration (not quantity). - Answer Choices: (A) 35% (B) 36.67% (C) 40% (D) 43.33% (E) 50%

THE DECISION FRAMEWORK (Step-by-Step)

Run this every time—no exceptions.

1. Identify the components.
2. What is being mixed? (e.g., salt, alcohol, acid)

3. What are the concentrations? (e.g., 30% salt = 0.3 salt per liter)

4. Write the total quantity equation.

5. Final volume = Initial volume + Added volume – Removed volume.

6. Example: 20L + 10L = 30L final.

7. Write the "solute" equation.

8. Total solute = (Initial volume × Initial %) + (Added volume × Added %) – (Removed volume × Removed %).

9. Example: (20 × 0.3) + (10 × 0.5) = 6 + 5 = 11L salt.

10. Calculate the final concentration.

11. Final % = (Total solute / Final volume) × 100.

12. Example: (11 / 30) × 100 ≈ 36.67%.

13. Match to answer choices.

14. Eliminate options that don’t fit (e.g., 35% is too low, 40% is too high).

15. Check for traps.

16. Did the question ask for a ratio instead of a percentage? (e.g., "What is the ratio of salt to water?")
17. Did you account for removal (e.g., evaporation)?

Worked Examples

Example 1 – Straightforward

"A 40% alcohol solution is mixed with a 60% alcohol solution to create 10 liters of a 50% solution. How many liters of the 40% solution were used?"

Step-by-Step: 1. Components: Alcohol in two solutions (40% and 60%). 2. Let x = liters of 40% solution.
- Then (10 – x) = liters of 60% solution. 3. Solute equation:
- 0.4x + 0.6(10 – x) = 0.5 × 10
- 0.4x + 6 – 0.6x = 5
- -0.2x = -1 → x = 5. 4. Answer: 5 liters (Option C if given).

Elimination: - If you guessed 6L (60% solution), the final % would be 48% (too low). - If you guessed 4L, the final % would be 52% (too high).

Example 2 – Common Trap (Removal)

"A 25% sugar solution has 5 liters of water evaporated. The remaining solution is 30% sugar. What was the original volume?"

Trap: Students forget to account for the removed water.

Step-by-Step: 1. Let V = original volume. 2. Solute before evaporation: 0.25V. 3. Volume after evaporation: V – 5. 4. Solute after evaporation: 0.3(V – 5). 5. Equation: 0.25V = 0.3(V – 5) → 0.25V = 0.3V – 1.5 → -0.05V = -1.5 → V = 30L.

Elimination: - If you ignored evaporation, you’d get 25L (wrong). - If you reversed the percentages, you’d get 60L (wrong).

Example 3 – Hard Variant (Three Solutions)

"Solution A is 10% acid, Solution B is 20% acid, and Solution C is 40% acid. If 2 liters of A, 3 liters of B, and x liters of C are mixed to form a 25% acid solution, what is x?"

Step-by-Step: 1. Total volume: 2 + 3 + x = 5 + x. 2. Solute equation:
- 0.1(2) + 0.2(3) + 0.4x = 0.25(5 + x)
- 0.2 + 0.6 + 0.4x = 1.25 + 0.25x
- 0.8 + 0.4x = 1.25 + 0.25x
- 0.15x = 0.45 → x = 3.

Elimination: - If you miscalculated the total volume as 5L (ignoring x), you’d get x = 1.25 (wrong). - If you reversed the percentages, you’d get x = 6 (wrong).

WRONG ANSWER PATTERNS

Common Mistakes

TIME STRATEGY

• Target time: 1:30 per question.
• When to skip: If you’re stuck after 2 minutes, mark and return.
• Minimum work:
• Write the solute equation.
• Plug in numbers.
• Eliminate 2-3 options before solving.

BACKSOLVING AND SHORTCUTS

1. Plug in answer choices (if variables are in the options).
2. Example: If the question asks for liters of 40% solution, test Option C first (middle value).
3. Use alligation (for two-solution problems).
4. Example: 30% and 50% → difference is 20%. Ratio is 20:10 → 2:1.
5. If total is 30L, 20L of 30% and 10L of 50%.
6. Eliminate extremes (e.g., if final % must be between 30% and 50%, eliminate 25% and 55%).

1-Minute Recap

"Here’s the exact process to solve any mixture problem in under 90 seconds:

1. Identify the components—what’s being mixed? Salt? Alcohol? Acid?
2. Write the total volume equation—initial + added – removed.
3. Write the solute equation—(volume × %) for each part.
4. Solve for the target—final %, quantity, or ratio.
5. Eliminate wrong answers—check for reversed ratios, ignored removal, or unit errors.

Most students mess up Step 3—they forget to multiply volume by %. Don’t be one of them. Write the equation, plug in the numbers, and move on. You’ve got this."

Final Notes

• Practice with a timer. Mixture problems should take ≤1:30.
• Always define variables. Never assume equal volumes.
• Check units. % vs. decimals, liters vs. grams.

Next step: Do 5 timed problems using this framework. Track how many you get right in under 90 seconds. Adjust as needed.