5.3 Mixture Problems and Weighted Averages

5.3 Mixture Problems and Weighted Averages

Quick Summary: Mixture and weighted average problems on the GMAT Focus Quantitative Reasoning measure test your ability to determine the equilibrium point when two or more components with differing concentrations, prices, or values are combined. Rather than setting up laborious simultaneous algebraic equations, high-scoring candidates exploit the Teeter-Totter (Balance Point) principle and the Alligation Cross. These proportional shortcuts capitalize on the core mathematical truth that the ratio of component quantities is strictly inverse to the ratio of their distances from the weighted average.

Whether blending chemical solutions of varying acidities, combining coffee beans of different price points, or calculating composite test scores across student cohorts, the underlying mathematics is identical. On an exam where you have approximately two minutes per question and no calculator, mastering proportional balance methods allows you to solve mixture questions in 30 to 45 seconds while avoiding computational errors.

---

The Algebraic Foundations of Weighted Averages

The simple arithmetic mean of two numbers, $\frac{A + B}{2}$, assumes that both numbers carry equal weight. When combining groups of unequal sizes, you must compute a weighted average:

$\bar{X} = \frac{w_1 X_1 + w_2 X_2 + \dots + w_n X_n}{w_1 + w_2 + \dots + w_n} = \frac{\sum w_i X_i}{\sum w_i}$

where $X_i$ represents the value, percentage, or price of component $i$, and $w_i$ represents the weighting factor (volume, weight, quantity, or head count).

The Bounded Value Principle

The weighted average of any two distinct components $X_1$ and $X_2$ (where $X_1 < X_2$) must lie strictly between the two values:

$X_1 < \bar{X} < X_2$

• If $w_1 = w_2$, $\bar{X}$ lies exactly at the numerical midpoint: $\frac{X_1 + X_2}{2}$.
• If $w_1 > w_2$, $\bar{X}$ is pulled closer to $X_1$.
• If $w_2 > w_1$, $\bar{X}$ is pulled closer to $X_2$.

On the GMAT, you can immediately eliminate answer choices that fall outside the $[X_1, X_2]$ range or lie on the wrong side of the midpoint when relative weights are known.

---

The Teeter-Totter (Balance Point) Method

Think of a weighted average as a physical balance beam or teeter-totter in mechanical equilibrium. The fulcrum sits at the weighted average $\bar{X}$, with weight $w_1$ positioned at $X_1$ and weight $w_2$ positioned at $X_2$.

Weight w₁                                    Weight w₂
  [ w₁ ]                                       [ w₂ ]
====▲======================▲=====================▲====
   X₁                  Fulcrum (X_avg)           X₂
    |<--- Distance d₁ --->| |<--- Distance d₂ --->|

For the system to balance, torque on both sides must be equal:

$\text{Weight}_1 \times \text{Distance}_1 = \text{Weight}_2 \times \text{Distance}_2$

$w_1 (\bar{X} - X_1) = w_2 (X_2 - \bar{X})$

Rearranging this identity yields the Golden Proportion of Weighted Averages:

$\frac{w_1}{w_2} = \frac{X_2 - \bar{X}}{\bar{X} - X_1} = \frac{d_2}{d_1}$

The Golden Law: The ratio of the quantities (weights) is the INVERSE of the ratio of their distances from the weighted average. A component that is twice as close to the average must carry twice the weight!

Conceptual Example of the Teeter-Totter

Suppose a solution of 10% acid is mixed with a solution of 40% acid to yield a 20% acid mixture:

• Distance from 10% to 20%: $d_1 = 20 - 10 = 10$.
• Distance from 40% to 20%: $d_2 = 40 - 20 = 20$.
• Distance ratio: $\frac{d_1}{d_2} = \frac{10}{20} = \frac{1}{2}$.
• Inverse weight ratio: $\frac{w_1}{w_2} = \frac{d_2}{d_1} = \frac{2}{1}$. There must be twice as much 10% solution as 40% solution. If the total mixture is 30 liters, the 10% solution constitutes $\frac{2}{3} \times 30 = 20$ liters, and the 40% solution constitutes $\frac{1}{3} \times 30 = 10$ liters.

---

The Alligation Cross: Rapid Visual Layout

The Alligation method is a structured tabular arrangement of the teeter-totter equation that requires no algebraic setup:

Component 1 Value (X₁)                   Parts of Component 1 = |X₂ - X_avg|
                           Target (X_avg)
Component 2 Value (X₂)                   Parts of Component 2 = |X₁ - X_avg|

How to Execute Alligation in 15 Seconds

1. Write the two component concentrations or prices in the left column.
2. Write the desired target mixture average in the center.
3. Subtract diagonally (always taking the positive absolute difference):
   • $\text{Parts of Component 1} = |X_2 - \bar{X}|$
   • $\text{Parts of Component 2} = |X_1 - \bar{X}|$
4. The resulting numbers represent the ratio of parts of Component 1 to Component 2.
5. Divide the total mixture quantity by the sum of parts to determine the volume or mass per part.

---

Solution Chemistry: Solute, Solvent, and Pure Additions

In liquid mixture questions, solutions consist of a solute (such as acid, salt, sugar, or alcohol) dissolved in a solvent (typically water):

$\text{Concentration } (C) = \frac{\text{Quantity of Solute}}{\text{Total Quantity of Solution}} = \frac{\text{Solute}}{\text{Solute} + \text{Solvent}}$

The Pure Substance Rules (Crucial for GMAT Modeling)

When a problem states that water or pure solute is added to an existing solution, apply two absolute mathematical conventions:

1. Pure Water / Solvent has 0% Solute: If water is added to dilute an acid solution, the concentration of the added liquid is strictly 0% acid ($X_1 = 0$).
2. Pure Solute has 100% Solute: If pure alcohol or pure acid is added to strengthen a mixture, the concentration of the added substance is strictly 100% solute ($X_2 = 100$).

Dilution and Evaporation Dynamics

• Dilution (Adding Water): Total solution volume increases while absolute solute quantity remains constant. Concentration decreases.
• Evaporation (Removing Water): Heat evaporates pure water (0% solute). Total solution volume decreases while absolute solute quantity remains unchanged. Concentration increases.

$\text{Initial Volume} \times C_{\text{initial}} = \text{Final Volume} \times C_{\text{final}} = \text{Constant Solute Mass}$

---

Commodity Blending and Weighted Cost Analysis

Mixture principles apply directly to commercial word problems where different grades of goods (coffee beans, nuts, grain, metal alloys) are blended to achieve a target selling price or metal purity:

$\text{Blended Price per Pound} = \frac{P_A Q_A + P_B Q_B}{Q_A + Q_B}$

Where $P$ represents the unit price and $Q$ represents the weight or quantity.

By treating price per pound as the value variable ($X$) and total pounds as the weighting variable ($w$), the teeter-totter method instantly reveals the exact proportion of each commodity needed without solving a 2-variable linear system.

---

Successive Dilution and Replacement Dynamics

A distinct, high-difficulty GMAT mixture problem features a container of pure liquid from which a fixed volume is removed, replaced with water, stirred, and the process repeated multiple times.

The Compounding Replacement Formula

If a container initially holds volume $V$ of pure substance, and volume $r$ is drawn off and replaced with pure water $n$ times successively, the quantity of the original substance remaining is:

$Q_{\text{final}} = Q_{\text{initial}} \times \left(1 - \frac{r}{V}\right)^n$

Why This Compounding Formula Works

In the first extraction, pure liquid is removed. But in the second extraction, the liquid removed is already diluted, so less pure liquid is lost. In each subsequent cycle, the fraction of remaining original liquid is scaled by $\left(1 - \frac{r}{V}\right)$, analogous to compound interest decay.

Example: A 100-liter cask is full of pure wine. 10 liters are drawn off and replaced with water; this process is repeated a second time. How much pure wine remains?
$Q_{\text{wine}} = 100 \times \left(1 - \frac{10}{100}\right)^2 = 100 \times (0.90)^2 = 100 \times 0.81 = 81 \text{ liters}$

---

Worked Problem Solving Examples

Example 1: Acid Solution Blending via Alligation

Problem: A laboratory technician has access to a 20% hydrochloric acid solution and a 50% hydrochloric acid solution. How many liters of each solution must be mixed together to produce exactly 30 liters of a 40% hydrochloric acid solution?

Step-by-Step Solution:

1. Identify Values and Target:
   • Lower concentration $X_1 = 20%$.
   • Higher concentration $X_2 = 50%$.
   • Desired target concentration $\bar{X} = 40%$.
   • Total target volume $= 30 \text{ liters}$.
2. Apply the Teeter-Totter Distance Principle:
   • Distance from lower concentration to target: $d_1 = 40 - 20 = 20$.
   • Distance from higher concentration to target: $d_2 = 50 - 40 = 10$.
3. Compute the Inverse Ratio of Parts: $\frac{\text{Volume of } 20\%}{\text{Volume of } 50\%} = \frac{d_2}{d_1} = \frac{10}{20} = \frac{1}{2}$
4. Apportion the Total Target Volume (30 Liters):
   • Total parts $= 1 + 2 = 3 \text{ equal parts}$.
   • Volume per part $= \frac{30}{3} = 10 \text{ liters}$.
   • Volume of 20% solution $= 1 \times 10 = 10 \text{ liters}$.
   • Volume of 50% solution $= 2 \times 10 = 20 \text{ liters}$.
5. Verification via Solute Balance:
   • Solute from 20% solution: $10 \times 0.20 = 2 \text{ liters of acid}$.
   • Solute from 50% solution: $20 \times 0.50 = 10 \text{ liters of acid}$.
   • Total acid $= 2 + 10 = 12 \text{ liters}$.
   • Final concentration $= \frac{12}{30} = 40%$. Exact match.

---

Example 2: Dilution with Pure Water

Problem: A pharmacist has 80 ounces of a 35% alcohol solution. How many ounces of pure water must be added to dilute the solution to a concentration of 20% alcohol?

Step-by-Step Solution:

1. Model the Components:
   • Component 1: Added pure water $\implies 0%$ alcohol concentration ($X_1 = 0%$).
   • Component 2: Existing solution $\implies 35%$ alcohol concentration ($X_2 = 35%$), weight $w_2 = 80$ ounces.
   • Target concentration: $\bar{X} = 20%$.
2. Calculate Distance to Target:
   • Distance for pure water: $d_1 = 20 - 0 = 20$.
   • Distance for 35% solution: $d_2 = 35 - 20 = 15$.
3. Apply the Inverse Weight-to-Distance Ratio: $\frac{w_{\text{water}}}{w_{\text{solution}}} = \frac{d_2}{d_1} = \frac{15}{20} = \frac{3}{4}$
4. Solve for Water Weight: $w_{\text{water}} = \frac{3}{4} \times w_{\text{solution}} = \frac{3}{4} \times 80 = 60 \text{ ounces}$
5. Verification via Algebraic Solute Conservation:
   • Initial alcohol $= 80 \times 0.35 = 28 \text{ ounces}$.
   • After adding 60 oz of water, total volume $= 80 + 60 = 140 \text{ ounces}$.
   • Final concentration $= \frac{28}{140} = \frac{2}{10} = 20%$. Verified in 20 seconds.

---

Example 3: Commodity Blending with Unknown Proportions

Problem: A specialty grocer blends Grade A Arabica beans costing $14 per pound with Grade B Robusta beans costing $7 per pound to create a 50-pound batch of custom blend that costs exactly $9 per pound. How many pounds of Grade A Arabica beans are in the blend?

Step-by-Step Solution:

1. Identify Given Prices and Target:
   • Grade B (lower): $P_B = $7$ per pound.
   • Grade A (higher): $P_A = $14$ per pound.
   • Target blend price: $\bar{P} = $9$ per pound.
   • Total batch weight $= 50$ pounds.
2. Calculate Price Distances from Target:
   • Distance for Grade B: $d_B = 9 - 7 = $2$.
   • Distance for Grade A: $d_A = 14 - 9 = $5$.
3. Determine Inverse Weight Ratio: $\frac{\text{Weight of Grade B}}{\text{Weight of Grade A}} = \frac{d_A}{d_B} = \frac{5}{2}$
4. Distribute 50 Pounds Across Total Parts:
   • Total ratio parts $= 5 + 2 = 7 \text{ parts}$.
   • Weight of Grade A beans:
     $\text{Weight of Grade A} = \frac{2}{7} \times 50 = \frac{100}{7} = 14\frac{2}{7} \text{ pounds}$
   • Weight of Grade B beans:
     $\text{Weight of Grade B} = \frac{5}{7} \times 50 = \frac{250}{7} = 35\frac{5}{7} \text{ pounds}$ (Notice how the inverse ratio ensures that Grade B, being only $2 away from the target, receives the lion's share of the total weight).

---

High-Frequency GMAT Traps & Pacing Strategies

• Trap 1: Direct Averaging of Percentages: When mixing 10 liters of a 20% solution with 20 liters of a 50% solution, candidates frequently compute $\frac{20 + 50}{2} = 35%$. The true average is $40%$ because twice as much 50% liquid was added. Never take the arithmetic mean of concentrations unless volumes are identical.
• Trap 2: The Ratio Inversion Mistake: When using alligation or the teeter-totter method, candidates frequently associate distance $d_1$ with weight $w_1$. Always remember: distance to average is inversely proportional to weight ($w_1 \propto d_2$). The component closer to the average has the larger volume.
• Trap 3: Confusing Solute Amount with Total Solution Volume: The denominator of concentration is always $\text{Solute} + \text{Solvent}$. If a problem says "40 grams of salt are added to 160 grams of water," the total solution is $40 + 160 = 200$ grams, making the concentration $\frac{40}{200} = 20%$, not $\frac{40}{160} = 25%$.
• Trap 4: Missing the 0% or 100% Boundary for Pure Substances: When pure water is added, treat its concentration as $0%$. When pure solute is added, treat its concentration as $100%$. Forgetting to assign these values stalls algebraic progress.