Quant Lesson 20: Mixture Word Problems • GMATCourse.mba

1

Setting Up Mixture Problems

Mixture problems involve combining two or more substances with different concentrations or values. The key is that the total amount of each substance must be conserved.

Mixture Table Framework

Solution	Amount	Concentration	Pure Substance
A	$a$	$p\%$	$a \times p/100$
B	$b$	$q\%$	$b \times q/100$
Mixture	$a+b$	$c\%$	$(a+b) \times c/100$

Key equation: Column 3 of A + Column 3 of B = Column 3 of Mixture

2

The Alligation Method (Weighted Average Shortcut)

Alligation is a shortcut for mixing two solutions to get a target concentration. It gives the ratio of amounts to mix.

Alligation Cross Diagram

Solution A: $p\%$ Solution B: $q\%$

$\backslash$ $/$

Target: $c\%$

$/$ $\backslash$

$(q-c)$ parts of A $(c-p)$ parts of B

Ratio of A to B = $(q-c) : (c-p)$. Use when mixing two concentrations to get a target.

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Concentration Change Problems

Adding Water (Dilution)

Adding pure water (0% concentration) to a solution:

New concentration = $\dfrac{\text{original pure substance}}{\text{original volume + water added}}$

Removing and Replacing

Remove $x$ liters of mixture (concentration $c\%$), replace with water:

Pure substance after = original − $x \times c/100$

New concentration = $\dfrac{\text{pure substance after}}{\text{total volume}}$

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Combining Mixture with Other Topics

Mixture + Value Problems (Coins, Tickets)

Same framework: replace "concentration" with "value per unit." Total value column = amount × unit value. Set up the table and solve.

Mixture Equation Template

$a \cdot c_1 + b \cdot c_2 = (a+b) \cdot c_{\text{mix}}$

Where $a$, $b$ = amounts and $c_1$, $c_2$, $c_{mix}$ = concentrations (or values).

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10 Mixture Traps

1. Concentration applies to the component, not total

30% salt solution: 30% of the volume is salt, not the total weight.

2. Adding water dilutes; concentration must decrease

Verify your final concentration is between the lower bound and the target.

3. Alligation gives ratio, not amounts

The alligation method gives the ratio of A to B. Multiply by a common factor to get actual amounts.

4. Mixture concentration is between the two components

$c_{mix}$ must lie strictly between $c_1$ and $c_2$ (unless you mix equal concentrations).

5. Volume/weight units must match

Don't mix liters and gallons, or grams and ounces, without converting.

6. Removing mixture removes substance proportionally

If a 30% solution has 1L removed, you remove 0.3L of pure substance — not 1L.

7. Two-variable mixture: set up two equations

Amount equation and substance equation. Two unknowns need two equations.

8. Value mix: unit value × quantity = total value

Coins: nickels worth $0.05 each. Total value = $0.05 \times \text{count}$.

9. "How much to add" problems: let $x$ = amount added

Set up the substance equation with $x$ as the unknown, then solve.

10. Weighted average ≠ simple average

Mixed concentration closer to the larger-quantity component. Never just average.

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10 GMAT Practice Questions

Q1 PS Difficulty: 650

A chemist mixes 10 liters of a 30% acid solution with 20 liters of a 60% acid solution. What is the concentration of the resulting mixture?

(C) 50%. Acid: $10(0.30) + 20(0.60) = 3 + 12 = 15$ liters. Total: 30 liters. Concentration = $15/30 = 50\%$.

Q2 PS Difficulty: 700

How many liters of a 20% saline solution must be added to 5 liters of a 60% saline solution to produce a 30% saline solution?

(B) 15 liters. Let $x$ = liters of 20% solution. $0.20x + 0.60(5) = 0.30(x+5)$. $0.20x + 3 = 0.30x + 1.5$. $1.5 = 0.10x$. $x = 15$.

Q3 PS Difficulty: 650

A 40-liter solution is 25% alcohol. If 10 liters of pure alcohol are added, what is the new concentration?

(C) 40%. Original alcohol: $40 \times 0.25 = 10$ L. After adding 10L pure: $10+10=20$ L alcohol in $40+10=50$ L total. $20/50 = 40\%$.

Q4 PS Difficulty: 700

In a mixture of nuts, almonds cost $4 per pound and cashews cost $7 per pound. If a 10-pound mixture costs $5.50 per pound on average, how many pounds of cashews are in the mixture?

(D) 5 pounds. Let $c$ = cashews. $4(10-c) + 7c = 5.50(10)$. $40-4c+7c=55$. $3c=15$. $c=5$.

Q5 PS Difficulty: 700

10 liters of a 30% salt solution has some water evaporated until the concentration becomes 50%. How many liters of water were evaporated?

(C) 4 liters. Salt = $10 \times 0.30 = 3$ L (unchanged). New volume $v$ has 50%: $3/v = 0.50$ → $v=6$ L. Water evaporated = $10-6=4$ L.

Q6 PS Difficulty: 750

From 12 liters of a 20% milk solution, 3 liters are removed and replaced with pure milk. What is the new concentration?

(B) 35%. Original milk: $12 \times 0.20 = 2.4$ L. Remove 3 L (containing $3 \times 0.20 = 0.6$ L milk). Remaining milk: $2.4 - 0.6 = 1.8$ L. Add 3 L pure milk: $1.8+3=4.8$ L milk in 12 L total. $4.8/12 = 40\%$. (C) 40% is correct.

Q7 PS Difficulty: 700

Solution A is 40% alcohol and solution B is 10% alcohol. To make a 30% alcohol mixture, in what ratio should A and B be combined?

(C) 2:1. Alligation: $A:B = (30-10):(40-30) = 20:10 = 2:1$.

Q8 DS Difficulty: 700

Is the concentration of the final mixture greater than 40%?

(1) 5 liters of a 60% solution are mixed with 10 liters of a 30% solution.
(2) The final volume is 15 liters.

(A) Statement (1) alone sufficient. (1): $5(0.60)+10(0.30)=3+3=6$ liters of solute in 15 liters. $6/15=40\%$. NOT greater than 40%. Definitively NO. Sufficient. (2): Volume alone doesn't determine concentration. NOT sufficient. Answer: (A).

Q9 PS Difficulty: 750

A jar contains a mixture of juice and water in the ratio 3:1. If 8 liters of juice are added, the ratio becomes 5:1. How many liters of water are in the jar?

(A) 4 liters. Let water = $w$. Juice = $3w$. After adding 8 L juice: $(3w+8)/w = 5/1$ → $3w+8=5w$ → $2w=8$ → $w=4$.

Q10 PS Difficulty: 700

A store sells two grades of coffee: Grade A at $12/lb and Grade B at $8/lb. The store wants to create a 20-pound blend that sells for $9.50/lb. How many pounds of Grade A should be used?

(D) 7.5 pounds. Let $a$ = pounds of A. $12a + 8(20-a) = 9.50(20)$. $12a+160-8a=190$. $4a=30$. $a=7.5$.

Lesson Summary — Key Takeaways

Mixture equation: a·c₁ + b·c₂ = (a+b)·c_mix

Substance in A + substance in B = substance in mixture. Set up and solve.

Alligation: ratio = (c_mix−c₂) : (c₁−c_mix)

Quick ratio shortcut — cross-subtract the concentrations.

Dilution: adding water reduces concentration

Water = 0% concentration. New conc = original substance / new total volume.

Value mixtures use same framework

Replace concentration with price/value per unit. Same equation structure.