4.2 Ratios, Proportions, Mixtures & Weighted Averages

4.2 Ratios, Proportions, Mixtures & Weighted Averages

Ratio, proportion, and mixture problems are useful quantitative practice; official HEC documents publish broad domain weights rather than subtype frequencies. These concepts evaluate a candidate's capacity to model proportional relationships, solve multi-variable work allocations, balance chemical or physical concentrations, and summarize statistical data using weighted averages. Mastering these algebraic frameworks enables candidates to solve complex quantitative word problems quickly without getting bogged down in tedious arithmetic.

---

Ratio Fundamentals, Scaling & Multi-Part Division

A ratio is a mathematical comparison of two homogeneous quantities expressed as $a:b$ or $\frac{a}{b}$. Ratios express relative magnitudes rather than absolute values.

Scaling and Combining Multi-Part Ratios

When given individual binary ratios that share a common element—such as $A:B$ and $B:C$—they can be combined into a single unified ratio $A:B:C$ by finding the Least Common Multiple (LCM) of the shared terms.

Worked Example: If $A : B = 3 : 4$ and $B : C = 6 : 5$, determine the combined ratio $A : B : C$.

1. The shared term is $B$, which has ratio values of $4$ and $6$.
2. The LCM of $4$ and $6$ is $12$.
3. Scale $A : B = 3 : 4$ by multiplying by $3 \rightarrow 9 : 12$.
4. Scale $B : C = 6 : 5$ by multiplying by $2 \rightarrow 12 : 10$.
5. Combined Ratio $A : B : C = 9 : 12 : 10$.

Multi-Part Division of Quantities

To divide a total quantity $T$ among $k$ entities in the ratio $a_1 : a_2 : \dots : a_k$:

$$\text{Share of Part } i = \left( \frac{a_i}{\sum_{j=1}^k a_j} \right) \times T$$

Worked Example: Divide PKR $105,000$ among three research projects in the ratio $2 : 3 : 5$.

• Sum of ratio parts $= 2 + 3 + 5 = 10$.
• Share 1 $= \frac{2}{10} \times 105,000 = \text{PKR } 21,000$.
• Share 2 $= \frac{3}{10} \times 105,000 = \text{PKR } 31,500$.
• Share 3 $= \frac{5}{10} \times 105,000 = \text{PKR } 52,500$.

---

Direct, Inverse & Compound Proportion

Direct Proportion

Two variables $x$ and $y$ are directly proportional ($y \propto x$) if their ratio remains constant:

$$\frac{y_1}{x_1} = \frac{y_2}{x_2} = k$$

Inverse Proportion

Two variables $x$ and $y$ are inversely proportional ($y \propto \frac{1}{x}$) if their product remains constant:

$$x_1 \cdot y_1 = x_2 \cdot y_2 = k$$

Compound Proportion & Universal Work Formula

When work involves multiple varying factors—such as workers ($M$), working days ($D$), daily hours ($H$), and total work output ($W$)—the variables relate via the Universal Work Equality:

$$\frac{W_1}{M_1 \cdot D_1 \cdot H_1} = \frac{W_2}{M_2 \cdot D_2 \cdot H_2}$$

Worked Example: If $15$ laborers working $8$ hours a day can construct a $300$-meter wall in $12$ days, how many days will $20$ laborers take to construct a $500$-meter wall working $10$ hours a day?

• Given: $M_1 = 15, D_1 = 12, H_1 = 8, W_1 = 300$.
• Given: $M_2 = 20, D_2 = ?, H_2 = 10, W_2 = 500$.
• Substitute into formula: $\frac{300}{15 \times 12 \times 8} = \frac{500}{20 \times D_2 \times 10}$
• Simplify denominators: $15 \times 12 \times 8 = 1,440$, and $20 \times 10 = 200$.
• $\frac{300}{1,440} = \frac{500}{200 \cdot D_2} \Rightarrow \frac{5}{24} = \frac{5}{2 \cdot D_2} \Rightarrow 2 \cdot D_2 = 24 \Rightarrow D_2 = 12 \text{ days}$.

---

Mixture Concentration Balance & The Rule of Alligation

Mixture problems involve blending two or more components with different concentrations or unit prices to create a desired mean concentration.

Weighted Mass Balance Equation

When combining volume $V_1$ at concentration $C_1$ with volume $V_2$ at concentration $C_2$, the final mixture concentration $C_m$ is:

$$C_m = \frac{V_1 C_1 + V_2 C_2}{V_1 + V_2}$$

The Rule of Alligation

The Rule of Alligation provides a fast visual cross-subtraction matrix to determine the ratio in which two ingredients of price/concentration $c$ (cheaper) and $d$ (dearer) must be mixed to produce a mixture of mean price/concentration $m$.

Cheaper Rate (c)              Dearer Rate (d)
                       \    /
                         Mean (m)
                       /    \
(d - m)                        (m - c)

$$\frac{\text{Quantity of Cheaper}}{\text{Quantity of Dearer}} = \frac{d - m}{m - c}$$

Worked Example: In what ratio must a $20\%$ alcohol solution be mixed with a $50\%$ alcohol solution to produce a $30\%$ alcohol solution?

• Cheaper $c = 20\%$, Dearer $d = 50\%$, Target Mean $m = 30\%$.
• Difference 1 ($d - m$) $= 50 - 30 = 20$.
• Difference 2 ($m - c$) $= 30 - 20 = 10$.
• Ratio of Cheaper to Dearer $= \frac{20}{10} = \frac{2}{1}$ (or $2 : 1$).

---

Successive Dilution of Solutions

When a container holds a volume $V$ of pure liquid (e.g., milk or acid), and $x$ units are drawn out and replaced with water, repeated $n$ times, the final remaining volume of pure liquid $Q_n$ is:

$$Q_n = V \left( 1 - \frac{x}{V} \right)^n$$

Worked Example: A vessel initially contains $80$ liters of pure milk. $8$ liters of milk are withdrawn and replaced with water. This process is repeated once more ($n=2$). Calculate the final quantity of pure milk remaining.

• Initial Volume $V = 80$, Withdrawn Volume $x = 8$, Operations $n = 2$.
• $Q_2 = 80 \left( 1 - \frac{8}{80} \right)^2 = 80 \left( 1 - 0.10 \right)^2 = 80 \times (0.90)^2 = 80 \times 0.81 = 64.8 \text{ liters}$.

---

Weighted Averages & Measures of Central Tendency

Weighted Average

When individual data points $x_i$ possess varying weights $w_i$, the simple arithmetic mean is invalid. Instead, the Weighted Average $\bar{x}_w$ must be calculated:

$$\bar{x}_w = \frac{\sum_{i=1}^k w_i x_i}{\sum_{i=1}^k w_i}$$

Measures of Central Tendency & Data Spread

Worked Example: A student scores $70$ on Test 1 (weight $20\%$), $80$ on Test 2 (weight $30\%$), and $90$ on the Final Exam (weight $50\%$). Calculate the student's weighted average score.

• $\bar{x}_w = \frac{(70 \times 20) + (80 \times 30) + (90 \times 50)}{20 + 30 + 50} = \frac{1,400 + 2,400 + 4,500}{100} = \frac{8,300}{100} = 83.0$.