Unit 4 - Notes

PEA307 7 min read

Unit 4: Advanced ratio and proportion, Components and blending

Part 1: Advanced Ratio and Proportion

1. Concept of Ratio and Proportion

Ratio
A ratio is a mathematical relationship between two quantities of the same unit, indicating how many times the first quantity contains the second. It is denoted by $a : b$ or $\frac{a}{b}$ .

Antecedent and Consequent: In the ratio $a : b$ , $a$ is called the antecedent and $b$ is called the consequent.
Simplest Form: A ratio is in its simplest form when the antecedent and consequent have no common factors other than 1.

Key Properties of Ratios:

Duplicate Ratio: The duplicate ratio of $a : b$ is $a^2 : b^2$ .
Sub-duplicate Ratio: The sub-duplicate ratio of $a : b$ is $\sqrt{a} : \sqrt{b}$ .
Triplicate Ratio: The triplicate ratio of $a : b$ is $a^3 : b^3$ .
Sub-triplicate Ratio: The sub-triplicate ratio of $a : b$ is $\sqrt[3]{a} : \sqrt[3]{b}$ .
Compound Ratio: The compound ratio of $a : b$ and $c : d$ is $ac : bd$ .

Proportion
When two ratios are equal, they are said to be in proportion. It is represented as $a : b :: c : d$ or $\frac{a}{b} = \frac{c}{d}$ .

Terms: $a$ and $d$ are called the extremes, while $b$ and $c$ are called the means.
Fundamental Rule: Product of Extremes = Product of Means ( $a \times d = b \times c$ ).

Types of Proportions:

Fourth Proportional: If $a : b = c : d$ , then $d$ is the fourth proportional to $a, b,$ and $c$ . $d = \frac{bc}{a}$ .
Third Proportional: If $a : b = b : c$ , then $c$ is the third proportional to $a$ and $b$ . $c = \frac{b^2}{a}$ .
Mean Proportional: The mean proportional between $a$ and $b$ is $\sqrt{ab}$ .

Advanced Theorem: Componendo and Dividendo
If $\frac{a}{b} = \frac{c}{d}$ , then:

Componendo: $\frac{a+b}{b} = \frac{c+d}{d}$
Dividendo: $\frac{a-b}{b} = \frac{c-d}{d}$
Componendo and Dividendo: $\frac{a+b}{a-b} = \frac{c+d}{c-d}$

2. Problems Based on Ages

Age problems are direct applications of linear equations and ratios. They involve comparing the ages of two or more individuals at different points in time (past, present, and future).

Core Concepts and Rules:

Time Frames: Let the present age be .
- Age $n$ years ago = $x - n$
- Age $n$ years from now (hence) = $x + n$
Age Differences: The difference in ages between two people remains constant throughout their lives. If A is 5 years older than B today, A will always be 5 years older than B.
Ratio Method (Cross-Multiplication Method):
If the ratio of ages of A and B at present is $a : b$ , and after $n$ years the ratio becomes $c : d$ , the present ages can be found using the equations:
$\frac{ax + n}{bx + n} = \frac{c}{d}$

Detailed Example:
Problem: 10 years ago, the ratio of the ages of a father and his son was 3:1. After 10 years, the ratio of their ages will be 2:1. Find their present ages.
Solution:

Let the ages 10 years ago be $3x$ (father) and $x$ (son).
Present ages: Father = $3x + 10$ , Son = $x + 10$ .
Ages 10 years from now: Father = $3x + 10 + 10 = 3x + 20$ , Son = $x + 10 + 10 = x + 20$ .
According to the problem, the ratio in 10 years is 2:1.
$\frac{3x + 20}{x + 20} = \frac{2}{1}$
$3x + 20 = 2(x + 20)$
$3x + 20 = 2x + 40$
$x = 20$
Present Ages:
Father = $3(20) + 10 = 70$ years.
Son = $20 + 10 = 30$ years.

3. Problems Based on Partnership

Partnership problems deal with the distribution of profit or loss among partners who have invested capital in a business.

Types of Partnership:

Simple Partnership: All partners invest their capital for the same period of time.
- Rule: Profit is distributed in the ratio of their invested capitals.
- If A, B, and C invest $C_1, C_2,$ and $C_3$ , then Profit Ratio = $C_1 : C_2 : C_3$ .
Compound Partnership: Partners invest their capital for different periods of time.
- Rule: Profit is distributed in the ratio of the product of their capital and time.
- Profit Ratio = $(C_1 \times T_1) : (C_2 \times T_2) : (C_3 \times T_3)$

Working and Sleeping Partners:

Working (Active) Partner: A partner who manages the business. They often receive a specific percentage of the profit as a salary before the remaining profit is divided.
Sleeping Partner: A partner who only invests money and does not take part in management.

Detailed Example:
Problem: A, B, and C started a business. A invested Rs. 10,000 for 12 months, B invested Rs. 15,000 for 8 months, and C invested Rs. 20,000 for 6 months. If the total profit at the end of the year is Rs. 10,200, find C's share.
Solution:

Calculate the Capital $\times$ Time for each:
A = $10,000 \times 12 = 120,000$
B = $15,000 \times 8 = 120,000$
C = $20,000 \times 6 = 120,000$
Find the profit sharing ratio:
$A : B : C = 120,000 : 120,000 : 120,000 = 1 : 1 : 1$
Calculate C's share:
C's share = $\frac{1}{1 + 1 + 1} \times 10,200 = \frac{1}{3} \times 10,200 =$ Rs. 3,400.

Part 2: Components and Blending

1. Concept of Alligation and Mixtures

Definitions:

Mixture: Mixing two or more ingredients together in a certain ratio.
Alligation: A rule that enables us to find the ratio in which two or more ingredients at a given price must be mixed to produce a mixture of a desired price.

Rule of Alligation:
When two ingredients of different prices (or strengths) are mixed to form a mixture of a mean price, the ratio of their quantities is inversely proportional to the differences in their prices from the mean price.

Let $C$ be the cost price of a cheaper ingredient.
Let $D$ be the cost price of a dearer (more expensive) ingredient.
Let $M$ be the mean price (cost price of the mixture).

The Alligation Cross:

TEXT

Cost of Cheaper (C)                 Cost of Dearer (D)
                      \             /
                    Mean Price (M)
                      /             \
          (D - M)                         (M - C)

Formula:

\frac{\text{Quantity of Cheaper}}{\text{Quantity of Dearer}} = \frac{D - M}{M - C}

Note: Alligation can be applied to prices, percentages (like alcohol/water mixtures), speeds, averages, etc. The unit of $C, D,$ and $M$ must always be strictly identical (e.g., all must be Cost Price, not Selling Price).

Detailed Example:
Problem: In what ratio must rice at Rs. 20 per kg be mixed with rice at Rs. 28 per kg so that the mixture is worth Rs. 25 per kg?
Solution:

Cheaper price ( $C$ ) = 20
Dearer price ( $D$ ) = 28
Mean price ( $M$ ) = 25
Apply Alligation Cross:
$(D - M) = 28 - 25 = 3$
$(M - C) = 25 - 20 = 5$
Required ratio = $(D - M) : (M - C) = 3 : 5$ .

2. Problems Based on Removal of Mixture (Replacement)

These problems involve a vessel full of a pure liquid (or a mixture). A certain quantity is removed and replaced by an equal quantity of another liquid (usually water). This process may be repeated multiple times.

The Successive Replacement Formula:
Let a vessel contain $V$ units of pure liquid (initial quantity).
Let $x$ units of this liquid be drawn out and replaced with water.
If this operation is performed $n$ times, the formula to find the amount of pure liquid left is:

$\text{Final Quantity of Pure Liquid} = V \times \left(1 - \frac{x}{V}\right)^n$

Derivation logic: After the first operation, the fraction of the original liquid left is $\left(1 - \frac{x}{V}\right)$ . Because the mixture becomes homogeneous, every subsequent removal of $x$ volume removes the pure liquid in that exact same proportion. Over $n$ iterations, the multiplier compounds.

Variations of the Formula:

Different Replacement Quantities: If different quantities $x, y, z$ are removed and replaced successively, the formula becomes:
$\text{Final Quantity} = V \times \left(1 - \frac{x}{V}\right) \times \left(1 - \frac{y}{V}\right) \times \left(1 - \frac{z}{V}\right)$
Finding the Ratio: The ratio of pure liquid left to the total volume (which remains $V$ ) is:
$\frac{\text{Final Quantity of Pure Liquid}}{\text{Total Volume}} = \left(1 - \frac{x}{V}\right)^n$

Detailed Example:
Problem: A cask contains 50 liters of pure wine. 5 liters of wine is drawn out and replaced with water. This process is repeated 2 more times (total 3 times). Find the amount of wine left in the cask.
Solution:

Initial volume ( $V$ ) = 50 liters.
Amount replaced each time ( $x$ ) = 5 liters.
Number of times operation is performed ( $n$ ) = 3.
Apply formula:
$\text{Wine left} = 50 \times \left(1 - \frac{5}{50}\right)^3$
$\text{Wine left} = 50 \times \left(1 - \frac{1}{10}\right)^3$
$\text{Wine left} = 50 \times \left(\frac{9}{10}\right)^3$
$\text{Wine left} = 50 \times \frac{729}{1000}$
$\text{Wine left} = \frac{5 \times 729}{100} = \frac{3645}{100} = 36.45$ liters.
Result: After 3 operations, 36.45 liters of pure wine is left in the mixture. (The remaining 13.55 liters is water).

Unit 3

Unit 5