6.5 Ratios, Proportions, Mixtures & Averages

Ratios, Proportions, Mixtures & Averages

Quick Answer: Ratios and averages represent core quantitative techniques across arithmetic and Data Interpretation. Combine ratios like $A:B = 2:3$ and $B:C = 4:5$ by equating the common term $B$ to yield $A:B:C = 8:12:15$. Solve mixture problems using the Rule of Alligation: $\frac{Q_c}{Q_d} = \frac{d - m}{m - c}$. For repeated liquid replacement, apply $\text{Remaining} = a(1 - \frac{b}{a})^n$. When correcting an average with a recorded error, adjust the sum directly: $\text{New Average} = \text{Old Average} + \frac{\text{Correct} - \text{Incorrect}}{N}$.

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1. Ratios and Proportions: Theory and Operations

A ratio expresses the quantitative relation between two amounts of the same kind ($a : b = \frac{a}{b}$). Multiplying or dividing both antecedent ($a$) and consequent ($b$) by the same non-zero number leaves the ratio unchanged.

Types of Ratios

• Duplicate Ratio of $(a : b)$: $a^2 : b^2$
• Sub-Duplicate Ratio of $(a : b)$: $\sqrt{a} : \sqrt{b}$
• Triplicate Ratio of $(a : b)$: $a^3 : b^3$
• Sub-Triplicate Ratio of $(a : b)$: $\sqrt[3]{a} : \sqrt[3]{b}$
• Inverse / Reciprocal Ratio of $(a : b)$: $\frac{1}{a} : \frac{1}{b} = b : a$
• Compound Ratio: The compound ratio of $(a : b)$ and $(c : d)$ is $(a \times c) : (b \times d)$.

Bridging and Compounding Multi-Variable Ratios

When given individual pairwise ratios, merge them into a single continuous ratio by matching common terms:

$\text{Given: } A : B = 2 : 3 \quad \text{and} \quad B : C = 4 : 5$

• Multiply the first ratio by $4$ (the $B$-term of the second): $A : B = 8 : 12$
• Multiply the second ratio by $3$ (the $B$-term of the first): $B : C = 12 : 15$
• Combined Ratio: $A : B : C = 8 : 12 : 15$.

$\text{Given: } A : B = 1 : 2, \quad B : C = 3 : 4, \quad C : D = 5 : 6$

• $A : B : C : D = (1 \times 3 \times 5) : (2 \times 3 \times 5) : (2 \times 4 \times 5) : (2 \times 4 \times 6) = 15 : 30 : 40 : 48$.

Proportions and Proportional Terms

An equality of two ratios is a proportion ($a : b = c : d$, written as $a : b :: c : d$).

• Fundamental Rule: Product of Extremes $=$ Product of Means ($a \times d = b \times c$).
• Fourth Proportional to $a, b, c$: Value $x$ such that $a : b :: c : x \implies x = \frac{b \times c}{a}$.
• Third Proportional to $a, b$: Value $x$ such that $a : b :: b : x \implies x = \frac{b^2}{a}$.
• Mean Proportional between $a$ and $b$: Value $x$ such that $a : x :: x : b \implies x^2 = ab \implies x = \sqrt{ab}$.

2. Mixtures and the Rule of Alligation

Alligation is a modified weighted average method that determines the ratio in which two ingredients of different values must be mixed to produce a blend of a desired mean value.

The Alligation Formula

Let $c$ be the unit cost of the cheaper ingredient, $d$ be the unit cost of the dearer ingredient, and $m$ be the mean value of the resulting mixture ($c < m < d$):

$\frac{\text{Quantity of Cheaper } (Q_c)}{\text{Quantity of Dearer } (Q_d)} = \frac{d - m}{m - c}$

• Worked Example: In what ratio must wheat at ₹$42/\text{kg}$ be mixed with wheat at ₹$54/\text{kg}$ to produce a mixture worth ₹$47/\text{kg}$?
  • $c = 42, d = 54, m = 47$
  • $d - m = 54 - 47 = 7$
  • $m - c = 47 - 42 = 5$
  • Required Ratio ($Q_c : Q_d$): $7 : 5$.

Repeated Dilution / Replacement Formula

When a container initially holds $a$ units of a pure liquid, and $b$ units are drawn off and replaced with water, and this process is repeated $n$ times in total:

$\text{Quantity of Pure Liquid Remaining} = a \left(1 - \frac{b}{a}\right)^n$ $\frac{\text{Pure Liquid Remaining}}{\text{Total Mixture Volume}} = \left(1 - \frac{b}{a}\right)^n$

• Worked Example: A cask contains $80\text{ litres}$ of pure milk. $8\text{ litres}$ are drawn out and replaced with water. This process is repeated two more times ($n = 3$ operations total).
  • $a = 80, b = 8, n = 3$
  • $\text{Pure Milk Remaining} = 80 \left(1 - \frac{8}{80}\right)^3 = 80 \left(1 - \frac{1}{10}\right)^3 = 80 \times (0.9)^3 = 80 \times 0.729 = 58.32\text{ litres}$.
  • $\text{Water in cask} = 80 - 58.32 = 21.68\text{ litres}$.

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3. Mathematical Averages and Weighted Means

An average (arithmetic mean) represents the central value of a discrete set of observations.

Foundational Formulations

$\text{Average } (\bar{x}) = \frac{\text{Sum of all observations } (\sum x)}{\text{Total number of observations } (N)}$ $\text{Sum of Observations} = \text{Average} \times N$

Arithmetic Properties of Averages

1. If each observation in a dataset is increased by $k$, the new average becomes $\bar{x} + k$.
2. If each observation is decreased by $k$, the new average becomes $\bar{x} - k$.
3. If each observation is multiplied by $k$, the new average becomes $\bar{x} \times k$.
4. If each observation is divided by $k$ ($k \neq 0$), the new average becomes $\bar{x} / k$.

Weighted Average Formula

When combining distinct sub-groups with different averages $\bar{x}_1, \bar{x}_2, \dots, \bar{x}_k$ and sample weights $w_1, w_2, \dots, w_k$:

$\bar{x}_w = \frac{\sum w_i \bar{x}_i}{\sum w_i} = \frac{w_1 \bar{x}_1 + w_2 \bar{x}_2 + \dots + w_k \bar{x}_k}{w_1 + w_2 + \dots + w_k}$

Correcting Misrecorded Observations in Averages

When one or more data points were recorded incorrectly during data entry, compute the corrected mean without recalculating from scratch:

$\text{Corrected Sum} = \text{Original Sum} + (\text{Correct Value} - \text{Incorrect Value})$ $\text{Corrected Average} = \text{Original Average} + \frac{\text{Correct Value} - \text{Incorrect Value}}{N}$

• Worked Example: The average marks of $50$ students in a class was calculated as $68$. Later, it was discovered that a score of $48$ was misread as $84$.
  • $N = 50, \text{Old Average} = 68$
  • $\text{Difference} = \text{Correct (48)} - \text{Incorrect (84)} = -36$
  • $\text{Corrected Average} = 68 + \frac{-36}{50} = 68 - 0.72 = 67.28$.

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4. Age-Based Word Problems

Age problems apply linear algebraic relationships to individuals across temporal reference points (past, present, future).

Key Principles

• Invariance of Age Difference: The age difference between two individuals remains constant across all points in time ($A - B = \text{constant}$).
• If the present ages of $A$ and $B$ are in ratio $x : y$, their ages can be represented as $kx$ and $ky$.
• Their ages $n$ years ago were $(kx - n)$ and $(ky - n)$.
• Their ages $m$ years hence will be $(kx + m)$ and $(ky + m)$.