Ratio, Proportion, and Unitary Method
Introduction to Ratio and Related Terms
Ratio is a fundamental mathematical concept used to compare the relative sizes of two quantities of the same kind. It is expressed as the quotient of the two quantities. Ratios help us understand how much of one quantity there is in relation to another.
If we have two quantities, say $'a'$ and $'b'$, with $b \neq 0$, the ratio of $'a'$ to $'b'$ is written as $a:b$ or as a fraction $\frac{a}{b}$. This fraction $\frac{a}{b}$ represents the value of the ratio. Like any fraction, a ratio should be expressed in its simplest form by dividing both the numerator and the denominator by their greatest common divisor (GCD).
Terms of a Ratio
In the ratio $a:b$, the two quantities $'a'$ and $'b'$ are called the terms of the ratio.
- The first term, $'a'$, is called the antecedent. It is the quantity that comes first in the comparison.
- The second term, $'b'$, is called the consequent. It is the quantity with which the antecedent is compared.
Example 1. Identify the antecedent and consequent in the ratio $7:11$.
Answer:
In the ratio $7:11$, the first term is 7 and the second term is 11.
Therefore, the antecedent is 7 and the consequent is 11.
Key Properties of Ratio
Understanding the properties of ratios is essential for working with them correctly:
1. Comparison of Quantities of the Same Kind: A ratio can only compare two quantities if they belong to the same category or type. For example, you can find the ratio of the height of two people or the ratio of the volume of two liquids. You cannot directly find the ratio of a person's height to the volume of a liquid using a simple ratio.
2. Units Must Be the Same: Even if the quantities are of the same kind, they must be expressed in the same units before a ratio can be formed. If the units are different, one or both quantities must be converted so that they have the same units.
Example 2. Find the ratio of 500 grams to 2 kilograms.
Answer:
We need to find the ratio of 500 grams to 2 kilograms.
The quantities are of the same kind (mass), but the units are different (grams and kilograms).
We need to convert them to the same unit. Let's convert kilograms to grams:
1 kilogram = 1000 grams
So, 2 kilograms $= 2 \times 1000 = 2000$ grams
Now the ratio is 500 grams : 2000 grams.
Ratio $= 500 : 2000 = \frac{500}{2000}$.
Simplify the fraction by dividing the numerator and denominator by their GCD, which is 500.
$\frac{500 \div 500}{2000 \div 500} = \frac{1}{4}$
The ratio is $1:4$.
Alternate Method (Converting grams to kilograms):
Convert 500 grams to kilograms:
1000 grams = 1 kilogram
So, 500 grams $= \frac{500}{1000} = 0.5$ kilograms
Now the ratio is 0.5 kilograms : 2 kilograms.
Ratio $= 0.5 : 2 = \frac{0.5}{2}$.
To remove the decimal, multiply the numerator and denominator by 10:
$\frac{0.5 \times 10}{2 \times 10} = \frac{5}{20}$
Simplify the fraction by dividing by the GCD of 5 and 20, which is 5.
$\frac{5 \div 5}{20 \div 5} = \frac{1}{4}$
The ratio is $1:4$, which is the same as before.
3. Ratio is Unitless: When a ratio is formed between two quantities of the same kind and same units, the units cancel out. Therefore, a ratio is a pure number and does not have any units attached to it.
4. Order Matters: The order in which the quantities are compared is important. The ratio $a:b$ is not the same as the ratio $b:a$, unless $a=b$. The antecedent and consequent specify which quantity is being compared to which.
Example 3. A bag contains 10 red balls and 15 blue balls. Find the ratio of red balls to blue balls and the ratio of blue balls to red balls.
Answer:
Number of red balls $= 10$
Number of blue balls $= 15$
Ratio of red balls to blue balls:
This is the ratio of the number of red balls to the number of blue balls, which is $10:15$.
Ratio $= 10:15 = \frac{10}{15}$.
Simplify by dividing both terms by their GCD, which is 5.
$\frac{10 \div 5}{15 \div 5} = \frac{2}{3}$
So, the ratio of red balls to blue balls is $2:3$.
Ratio of blue balls to red balls:
This is the ratio of the number of blue balls to the number of red balls, which is $15:10$.
Ratio $= 15:10 = \frac{15}{10}$.
Simplify by dividing both terms by their GCD, which is 5.
$\frac{15 \div 5}{10 \div 5} = \frac{3}{2}$
So, the ratio of blue balls to red balls is $3:2$.
Notice that the ratio of red to blue ($2:3$) is different from the ratio of blue to red ($3:2$), confirming that the order of terms in a ratio matters.
5. Ratios are Expressed in Simplest Form: A ratio is typically represented in its simplest form, meaning the antecedent and the consequent have no common factors other than 1. This is done by dividing both terms by their GCD.
Comparing Ratios
To compare two or more ratios, we need to determine which one is larger or if they are equivalent. The easiest way to do this is to convert the ratios into equivalent fractions and then compare the fractions.
Method 1: Using a Common Denominator
1. Write the given ratios as fractions.
2. Find the Least Common Multiple (LCM) of the denominators of these fractions.
3. Convert each fraction into an equivalent fraction with the LCM as the new denominator.
4. Compare the numerators of the equivalent fractions. The fraction with the larger numerator corresponds to the larger ratio.
Example 4. Which is greater: $5:6$ or $7:9$?
Answer:
Given ratios are $5:6$ and $7:9$.
Write them as fractions: $\frac{5}{6}$ and $\frac{7}{9}$.
The denominators are 6 and 9. Find the LCM of 6 and 9.
Multiples of 6: 6, 12, 18, 24, ...
Multiples of 9: 9, 18, 27, ...
LCM of 6 and 9 is 18.
Convert the fractions to have a common denominator of 18:
For $\frac{5}{6}$: Multiply numerator and denominator by $18 \div 6 = 3$.
$\frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}$
For $\frac{7}{9}$: Multiply numerator and denominator by $18 \div 9 = 2$.
$\frac{7}{9} = \frac{7 \times 2}{9 \times 2} = \frac{14}{18}$
Now compare the equivalent fractions $\frac{15}{18}$ and $\frac{14}{18}$.
Since $15 > 14$, we have $\frac{15}{18} > \frac{14}{18}$.
Therefore, $\frac{5}{6} > \frac{7}{9}$.
Thus, the ratio $5:6$ is greater than the ratio $7:9$.
Method 2: Using Cross-Multiplication
This is a quick method for comparing two ratios.
1. Write the two ratios as fractions, say $\frac{a}{b}$ and $\frac{c}{d}$.
2. Cross-multiply: Calculate the product of the numerator of the first fraction and the denominator of the second ($ad$), and the product of the denominator of the first fraction and the numerator of the second ($bc$).
3. Compare the two products:
- If $ad > bc$, then $\frac{a}{b} > \frac{c}{d}$ (i.e., $a:b > c:d$).
- If $ad < bc$, then $\frac{a}{b} < \frac{c}{d}$ (i.e., $a:b < c:d$).
- If $ad = bc$, then $\frac{a}{b} = \frac{c}{d}$ (i.e., $a:b = c:d$).
Example 5. Compare the ratios $5:6$ and $7:9$ using cross-multiplication.
Answer:
Given ratios as fractions are $\frac{5}{6}$ and $\frac{7}{9}$.
Identify the terms for cross-multiplication: $a=5, b=6, c=7, d=9$.
Calculate the products:
Product of $ad = 5 \times 9 = 45$
Product of $bc = 6 \times 7 = 42$
Compare the products $ad$ and $bc$:
$\boldsymbol{45 > 42}$
Since $ad > bc$ ($45 > 42$), we have $\frac{a}{b} > \frac{c}{d}$.
Therefore, $\frac{5}{6} > \frac{7}{9}$, which means $5:6 > 7:9$. This result is consistent with the common denominator method.
Example 6. The ratio of the cost of a pen to the cost of a notebook is $2:5$. If the cost of the pen is $\textsf{₹ } 20$, what is the cost of the notebook?
Answer:
Let the cost of the pen be $P$ and the cost of the notebook be $N$.
The given ratio of the cost of the pen to the cost of the notebook is $P:N = 2:5$.
This can be written as a fraction:
$\frac{P}{N} = \frac{2}{5}$
We are given that the cost of the pen, $P = \textsf{₹ } 20$.
Substitute the value of $P$ into the equation:
$\frac{20}{N} = \frac{2}{5}$
To find $N$, we can cross-multiply:
$20 \times 5 = 2 \times N$
$100 = 2N$
Now, solve for $N$ by dividing both sides by 2:
$\frac{100}{2} = \frac{2N}{2}$
$\boldsymbol{N = 50}$
The cost of the notebook is $\textsf{₹ } 50$.
Introduction to Proportion and Related Terms
A Proportion is an equality of two ratios. It is a statement that indicates that two ratios are equivalent. If two ratios $a:b$ and $c:d$ are equal, we say that $a, b, c,$ and $d$ are in proportion. This is written as $a:b :: c:d$ or, more commonly, as $\frac{a}{b} = \frac{c}{d}$.
The notation $a:b :: c:d$ is read as '$a$ is to $b$ as $c$ is to $d$'. It signifies that the relationship between $a$ and $b$ is the same as the relationship between $c$ and $d$.
Terms of a Proportion
In a proportion $a:b :: c:d$, the four quantities $a, b, c,$ and $d$ are called the terms of the proportion. Each term has a specific name based on its position:
- $a$ is the first term.
- $b$ is the second term.
- $c$ is the third term.
- $d$ is the fourth term.
The terms at the ends ($a$ and $d$) are called the extreme terms or extremes. The terms in the middle ($b$ and $c$) are called the middle terms or means.
Example 1. In the proportion $5:10 :: 15:30$, identify the extreme terms and the middle terms.
Answer:
The given proportion is $5:10 :: 15:30$.
The terms are 5, 10, 15, and 30.
- The first term is 5, and the fourth term is 30. These are the extreme terms or extremes.
- The second term is 10, and the third term is 15. These are the middle terms or means.
Fundamental Property of Proportion: Product of Extremes = Product of Means
This is the most important characteristic of a proportion. It states that if four quantities are in proportion, the product of the extreme terms is equal to the product of the middle terms.
If $a:b :: c:d$ is a proportion, it means $\frac{a}{b} = \frac{c}{d}$.
To derive the property, we can cross-multiply the fraction form:
$\frac{a}{b} = \frac{c}{d}$
Multiplying both sides by $bd$ (assuming $b, d \neq 0$):
$\frac{a}{b} \times bd = \frac{c}{d} \times bd$
$\frac{abd}{b} = \frac{cbd}{d}$
Cancelling the common terms:
$\cancel{b} \frac{ad}{\cancel{b}} = \cancel{d} \frac{bc}{\cancel{d}}$
This gives us the fundamental property:
$\boldsymbol{ad = bc}$
... (i)
Conversely, if for four quantities $a, b, c, d$ (with $b, d \neq 0$), $ad = bc$, then the quantities are in proportion, i.e., $a:b :: c:d$. This property is often used as a test for proportion.
Example 2. Check if the numbers 8, 12, 14, and 21 are in proportion.
Answer:
To check if the numbers 8, 12, 14, and 21 are in proportion, we assume they form a proportion in the given order: $8:12 :: 14:21$.
We use the property: Product of Extremes = Product of Means.
Here, the extreme terms are 8 and 21, and the middle terms are 12 and 14.
Product of extremes $= 8 \times 21$.
$\boldsymbol{8 \times 21 = 168}$
Product of means $= 12 \times 14$.
$\boldsymbol{12 \times 14 = 168}$
Compare the products:
Product of extremes = Product of means
(Both products are 168)
Since the product of extremes is equal to the product of means, the numbers 8, 12, 14, and 21 are in proportion.
i.e., $8:12 :: 14:21$ is a true proportion.
Alternate Method: Comparing Ratios Directly
Simplify the first ratio $8:12 = \frac{8}{12}$. Divide by GCD (4): $\frac{8 \div 4}{12 \div 4} = \frac{2}{3}$.
Simplify the second ratio $14:21 = \frac{14}{21}$. Divide by GCD (7): $\frac{14 \div 7}{21 \div 7} = \frac{2}{3}$.
Since both simplified ratios are equal ($\frac{2}{3}$), the original ratios $8:12$ and $14:21$ are equivalent, confirming that the numbers are in proportion.
Example 3. Find the value of $x$ in the proportion $6:18 :: x:15$.
Answer:
The given proportion is $6:18 :: x:15$.
Using the property Product of Extremes = Product of Means:
The extreme terms are 6 and 15.
The middle terms are 18 and $x$.
Product of extremes $= 6 \times 15 = 90$
Product of means $= 18 \times x = 18x$
Equating the products:
$\boldsymbol{18x = 90}$
... (ii)
To find $x$, divide both sides of equation (ii) by 18:
$\frac{18x}{18} = \frac{90}{18}$
$\boldsymbol{x = 5}$
Thus, the value of $x$ is 5. The proportion is $6:18 :: 5:15$.
Verification:
Check if $6:18 :: 5:15$.
Product of extremes $= 6 \times 15 = 90$.
Product of means $= 18 \times 5 = 90$.
Since $90=90$, the proportion is correct.
Continued Proportion
Three quantities $a, b,$ and $c$ (in that order) are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of the second to the third. Mathematically, this is written as $a:b :: b:c$.
In a continued proportion $a:b :: b:c$, the term $'b'$ is called the mean proportional between $a$ and $c$. The terms $'a'$ and $'c'$ are sometimes called the first and third proportionals.
Using the fundamental property of proportion ($ad=bc$) on $a:b :: b:c$:
Here, the extremes are $a$ and $c$, and the means are $b$ and $b$.
Product of extremes $= a \times c = ac$
Product of means $= b \times b = b^2$
Equating the products:
$\boldsymbol{ac = b^2}$
... (iii)
From equation (iii), the mean proportional $b$ can be found by taking the square root of the product of the other two terms:
$\boldsymbol{b = \pm \sqrt{ac}}$
[Formula for Mean Proportional]
When dealing with positive quantities in quantitative aptitude, the mean proportional is usually considered the positive square root: $b = \sqrt{ac}$.
More than three quantities can also be in continued proportion. For example, four quantities $a, b, c,$ and $d$ are in continued proportion if $a:b :: b:c :: c:d$. This implies $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$.
Example 4. Check if the numbers 3, 6, and 12 are in continued proportion.
Answer:
For three numbers $a, b, c$ to be in continued proportion, the condition $a:b :: b:c$ must hold, which means $b^2 = ac$.
Here, $a=3, b=6, c=12$.
Let's check if $b^2 = ac$:
$b^2 = 6^2 = 36$
$ac = 3 \times 12 = 36$
Since $b^2 = ac$ ($36 = 36$), the numbers 3, 6, and 12 are in continued proportion.
Alternate Method: Checking Ratios
Check the ratio of the first to the second term: $3:6 = \frac{3}{6} = \frac{1}{2}$.
Check the ratio of the second to the third term: $6:12 = \frac{6}{12} = \frac{1}{2}$.
Since the two ratios are equal ($3:6 = 6:12 = 1:2$), the numbers 3, 6, and 12 are in continued proportion.
Example 5. Find the mean proportional between 16 and 25.
Answer:
Let the mean proportional between 16 and 25 be $b$.
According to the definition of continued proportion, 16, $b$, and 25 are in continued proportion.
This means $16:b :: b:25$.
Using the formula for the mean proportional, $b = \sqrt{ac}$, where $a=16$ and $c=25$.
$\boldsymbol{b = \sqrt{16 \times 25}}$
$\boldsymbol{b = \sqrt{400}}$
The square root of 400 is 20.
$\boldsymbol{b = 20}$
(Assuming positive mean proportional as quantities are positive)
So, the mean proportional between 16 and 25 is 20.
Verification:
Check if 16, 20, and 25 are in continued proportion by checking $16:20 :: 20:25$.
Product of extremes $= 16 \times 25 = 400$.
Product of means $= 20 \times 20 = 400$.
Since $400 = 400$, they are in continued proportion.
Properties and Types of Ratio and Proportion
Having understood the basic concepts of ratio and proportion, we delve deeper into their properties and the different ways quantities can be related through proportion based on variation.
Types of Ratios (Based on their Relation to Other Ratios)
Given a ratio $a:b$, several other related ratios can be defined:
- Inverse Ratio (or Reciprocal Ratio): The inverse ratio of $a:b$ is obtained by switching the antecedent and the consequent. It is $b:a$. This is defined only when $a \neq 0$ and $b \neq 0$. The product of a ratio and its inverse ratio is always 1 (since $\frac{a}{b} \times \frac{b}{a} = 1$).
Example 1. What is the inverse ratio of $7:9$?
Answer:
The given ratio is $7:9$.
The inverse ratio is found by swapping the terms.
Inverse ratio of $7:9$ is $\boldsymbol{9:7}$.
- Compound Ratio: The compound ratio of two or more ratios is found by multiplying the antecedents of all the ratios to get the new antecedent, and multiplying the consequents of all the ratios to get the new consequent. For two ratios $a:b$ and $c:d$, the compound ratio is $(a \times c) : (b \times d) = ac:bd$. For three ratios $a:b$, $c:d$, and $e:f$, the compound ratio is $(a \times c \times e) : (b \times d \times f) = ace:bdf$, and so on.
Example 2. Find the compound ratio of $4:5$ and $6:7$.
Answer:
The given ratios are $4:5$ and $6:7$.
Antecedents are 4 and 6.
Consequents are 5 and 7.
The compound ratio is calculated as:
Compound Ratio $= (4 \times 6) : (5 \times 7)$
Compound Ratio $= \boldsymbol{24:35}$
Since 24 and 35 have no common factors other than 1, this is the simplest form.
- Duplicate Ratio: The duplicate ratio of a ratio $a:b$ is the ratio $a^2:b^2$. It is essentially the compound ratio of $a:b$ with itself ($a:b$ and $a:b$).
Example 3. Calculate the duplicate ratio of $2:3$.
Answer:
The given ratio is $2:3$.
The duplicate ratio is $a^2:b^2$ where $a=2$ and $b=3$.
Duplicate Ratio $= 2^2 : 3^2$
Duplicate Ratio $= 4 : 9$
The duplicate ratio of $2:3$ is $\boldsymbol{4:9}$.
- Triplicate Ratio: The triplicate ratio of a ratio $a:b$ is the ratio $a^3:b^3$. It is the compound ratio of $a:b$ with itself three times.
Example 4. Find the triplicate ratio of $1:4$.
Answer:
The given ratio is $1:4$.
The triplicate ratio is $a^3:b^3$ where $a=1$ and $b=4$.
Triplicate Ratio $= 1^3 : 4^3$
Triplicate Ratio $= 1 : 64$
The triplicate ratio of $1:4$ is $\boldsymbol{1:64}$.
- Sub-duplicate Ratio: The sub-duplicate ratio of a ratio $a:b$ is the ratio of the square root of the antecedent to the square root of the consequent. It is $\sqrt{a}:\sqrt{b}$. This applies when $a \ge 0$ and $b > 0$.
Example 5. Determine the sub-duplicate ratio of $81:100$.
Answer:
The given ratio is $81:100$.
The sub-duplicate ratio is $\sqrt{a}:\sqrt{b}$ where $a=81$ and $b=100$.
Sub-duplicate Ratio $= \sqrt{81} : \sqrt{100}$
Sub-duplicate Ratio $= 9 : 10$
The sub-duplicate ratio of $81:100$ is $\boldsymbol{9:10}$.
- Sub-triplicate Ratio: The sub-triplicate ratio of a ratio $a:b$ is the ratio of the cube root of the antecedent to the cube root of the consequent. It is $\sqrt[3]{a}:\sqrt[3]{b}$. This applies to real numbers $a$ and $b$ (with $b \neq 0$).
Example 6. Find the sub-triplicate ratio of $27:125$.
Answer:
The given ratio is $27:125$.
The sub-triplicate ratio is $\sqrt[3]{a}:\sqrt[3]{b}$ where $a=27$ and $b=125$.
Sub-triplicate Ratio $= \sqrt[3]{27} : \sqrt[3]{125}$
$\sqrt[3]{27} = 3$ (since $3 \times 3 \times 3 = 27$)
$\sqrt[3]{125} = 5$ (since $5 \times 5 \times 5 = 125$)
Sub-triplicate Ratio $= 3 : 5$
The sub-triplicate ratio of $27:125$ is $\boldsymbol{3:5}$.
Properties of Proportion
These properties are derived from the fundamental relationship of proportion: if $a:b :: c:d$, then $\frac{a}{b} = \frac{c}{d}$, which implies $ad = bc$. These properties are useful for manipulating and solving equations involving proportions.
Let $a, b, c, d$ be non-zero quantities such that $\frac{a}{b} = \frac{c}{d}$.
1. Invertendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{b}{a} = \frac{d}{c}$.
Proof:
Given: $\frac{a}{b} = \frac{c}{d}$
Taking the reciprocal of both sides of the equation (valid since $a,b,c,d \neq 0$):
$\frac{1}{\frac{a}{b}} = \frac{1}{\frac{c}{d}}$
This simplifies to:
$\boldsymbol{\frac{b}{a} = \frac{d}{c}}$
Hence Proved.
Example 7. Verify the Invertendo property for the proportion $4:6 :: 10:15$.
Answer:
The given proportion is $4:6 :: 10:15$, which means $\frac{4}{6} = \frac{10}{15}$.
Simplifying the fractions:
$\frac{4}{6} = \frac{\cancel{4}^2}{\cancel{6}_3} = \frac{2}{3}$
$\frac{10}{15} = \frac{\cancel{10}^2}{\cancel{15}_3} = \frac{2}{3}$
The equality holds: $\frac{2}{3} = \frac{2}{3}$.
Applying Invertendo, we should get $\frac{6}{4} = \frac{15}{10}$.
Check if this is true:
$\frac{6}{4} = \frac{\cancel{6}^3}{\cancel{4}_2} = \frac{3}{2}$
$\frac{15}{10} = \frac{\cancel{15}^3}{\cancel{10}_2} = \frac{3}{2}$
Since $\frac{3}{2} = \frac{3}{2}$, the Invertendo property is verified.
2. Alternendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a}{c} = \frac{b}{d}$. (Provided $b, c, d \neq 0$)
Proof:
Given: $\frac{a}{b} = \frac{c}{d}$
Multiply both sides of the equation by $\frac{b}{c}$:
$\frac{a}{b} \times \frac{b}{c} = \frac{c}{d} \times \frac{b}{c}$
Cancel out the common terms:
$\frac{a \cancel{b}}{\cancel{b} c} = \frac{\cancel{c} b}{d \cancel{c}}$
This simplifies to:
$\boldsymbol{\frac{a}{c} = \frac{b}{d}}$
Hence Proved.
This property allows us to swap the means ($b$ and $c$) of a proportion.
Example 8. Verify the Alternendo property for the proportion $4:6 :: 10:15$.
Answer:
The given proportion is $4:6 :: 10:15$, which means $\frac{4}{6} = \frac{10}{15}$.
Applying Alternendo, we should get $\frac{4}{10} = \frac{6}{15}$.
Check if this is true by simplifying both sides:
$\frac{4}{10} = \frac{\cancel{4}^2}{\cancel{10}_5} = \frac{2}{5}$
$\frac{6}{15} = \frac{\cancel{6}^2}{\cancel{15}_5} = \frac{2}{5}$
Since $\frac{2}{5} = \frac{2}{5}$, the Alternendo property is verified.
3. Componendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{b} = \frac{c+d}{d}$. (Provided $b, d \neq 0$)
Proof:
Given: $\frac{a}{b} = \frac{c}{d}$
Add 1 to both sides of the equation:
$\frac{a}{b} + 1 = \frac{c}{d} + 1$
Rewrite 1 with the respective denominators:
$\frac{a}{b} + \frac{b}{b} = \frac{c}{d} + \frac{d}{d}$
Combine the terms on each side:
$\boldsymbol{\frac{a+b}{b} = \frac{c+d}{d}}$
Hence Proved.
Example 9. Verify the Componendo property for the proportion $4:6 :: 10:15$.
Answer:
The given proportion is $4:6 :: 10:15$, which means $\frac{4}{6} = \frac{10}{15}$ (or $\frac{2}{3} = \frac{2}{3}$).
Applying Componendo, we should get $\frac{4+6}{6} = \frac{10+15}{15}$.
Check if this is true by evaluating both sides:
Left side: $\frac{4+6}{6} = \frac{10}{6} = \frac{\cancel{10}^5}{\cancel{6}_3} = \frac{5}{3}$
Right side: $\frac{10+15}{15} = \frac{25}{15} = \frac{\cancel{25}^5}{\cancel{15}_3} = \frac{5}{3}$
Since $\frac{5}{3} = \frac{5}{3}$, the Componendo property is verified.
4. Dividendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a-b}{b} = \frac{c-d}{d}$. (Provided $b, d \neq 0$)
Proof:
Given: $\frac{a}{b} = \frac{c}{d}$
Subtract 1 from both sides of the equation:
$\frac{a}{b} - 1 = \frac{c}{d} - 1$
Rewrite 1 with the respective denominators:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
Combine the terms on each side:
$\boldsymbol{\frac{a-b}{b} = \frac{c-d}{d}}$
Hence Proved.
Example 10. Verify the Dividendo property for the proportion $4:6 :: 10:15$.
Answer:
The given proportion is $4:6 :: 10:15$, which means $\frac{4}{6} = \frac{10}{15}$.
Applying Dividendo, we should get $\frac{4-6}{6} = \frac{10-15}{15}$.
Check if this is true by evaluating both sides:
Left side: $\frac{4-6}{6} = \frac{-2}{6} = \frac{-\cancel{2}^1}{\cancel{6}_3} = \frac{-1}{3}$
Right side: $\frac{10-15}{15} = \frac{-5}{15} = \frac{-\cancel{5}^1}{\cancel{15}_3} = \frac{-1}{3}$
Since $\frac{-1}{3} = \frac{-1}{3}$, the Dividendo property is verified.
5. Componendo and Dividendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$. (Provided $b, d, a-b, c-d \neq 0$). This property is obtained by dividing the result of Componendo by the result of Dividendo.
Proof:
From Componendo, we have $\frac{a+b}{b} = \frac{c+d}{d}$ (Equation A).
From Dividendo, we have $\frac{a-b}{b} = \frac{c-d}{d}$ (Equation B).
Divide Equation A by Equation B:
$\frac{\left(\frac{a+b}{b}\right)}{\left(\frac{a-b}{b}\right)} = \frac{\left(\frac{c+d}{d}\right)}{\left(\frac{c-d}{d}\right)}$
Simplify both sides:
$\frac{a+b}{b} \times \frac{b}{a-b} = \frac{c+d}{d} \times \frac{d}{c-d}$
$\frac{a+b}{a-b} = \frac{c+d}{c-d}$
Hence Proved.
Example 11. If $x:y = 7:3$, find the value of $\frac{x+y}{x-y}$.
Answer:
Given ratio is $x:y = 7:3$, which means $\frac{x}{y} = \frac{7}{3}$.
We need to find the value of $\frac{x+y}{x-y}$. This expression is in the form $\frac{a+b}{a-b}$, where $a=x$ and $b=y$.
Using the Componendo and Dividendo property: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.
Here, $\frac{x}{y} = \frac{7}{3}$. Applying Componendo and Dividendo with $a=x, b=y, c=7, d=3$:
$\frac{x+y}{x-y} = \frac{7+3}{7-3}$
$\frac{x+y}{x-y} = \frac{10}{4}$
Simplify the fraction:
$\frac{10}{4} = \frac{\cancel{10}^5}{\cancel{4}_2} = \frac{5}{2}$
The value of $\frac{x+y}{x-y}$ is $\boldsymbol{\frac{5}{2}}$.
Alternate Method (Direct Substitution):
Given $\frac{x}{y} = \frac{7}{3}$. This means $x = 7k$ and $y = 3k$ for some non-zero constant $k$.
Substitute these values into the expression $\frac{x+y}{x-y}$:
$\frac{x+y}{x-y} = \frac{7k + 3k}{7k - 3k}$
$\frac{x+y}{x-y} = \frac{10k}{4k}$
Assuming $k \neq 0$, cancel $k$ from numerator and denominator:
$\frac{x+y}{x-y} = \frac{10}{4} = \frac{5}{2}$
Both methods give the same result. Componendo and Dividendo is a faster method for this specific type of expression.
6. Addendo: If $\frac{a}{b} = \frac{c}{d} = k$, then $\frac{a+c}{b+d} = k = \frac{a}{b} = \frac{c}{d}$. (Provided $b+d \neq 0$). This property is also known as the property of equal ratios.
Proof:
Let $\frac{a}{b} = \frac{c}{d} = k$.
From the definition of $k$, we have:
$\frac{a}{b} = k \implies a = bk$
... (v)
$\frac{c}{d} = k \implies c = dk$
... (vi)
Consider the ratio $\frac{a+c}{b+d}$. Substitute the values of $a$ and $c$ from equations (v) and (vi):
$\frac{a+c}{b+d} = \frac{bk + dk}{b+d}$
Factor out $k$ from the numerator:
$\frac{a+c}{b+d} = \frac{k(b+d)}{b+d}$
Assuming $b+d \neq 0$, we can cancel the term $(b+d)$:
$\frac{a+c}{b+d} = \boldsymbol{k}$
Since $k = \frac{a}{b}$ and $k = \frac{c}{d}$, we conclude:
$\boldsymbol{\frac{a+c}{b+d} = \frac{a}{b} = \frac{c}{d}}$
Hence Proved.
Generalization: If $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n} = k$, then each ratio is equal to the ratio of the sum of antecedents to the sum of consequents, i.e., $\frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n} = k$, provided $b_1+b_2+\dots+b_n \neq 0$.
Example 12. Given that $\frac{p}{q} = \frac{r}{s} = \frac{t}{u} = \frac{1}{3}$, find the value of $\frac{p+r+t}{q+s+u}$.
Answer:
Given $\frac{p}{q} = \frac{r}{s} = \frac{t}{u} = \frac{1}{3}$.
According to the extended Addendo property (Property of Equal Ratios), if several ratios are equal, then the ratio formed by the sum of their antecedents to the sum of their consequents is equal to each of the individual ratios.
Here, the equal ratios are $\frac{p}{q}$, $\frac{r}{s}$, and $\frac{t}{u}$. The antecedents are $p, r, t$ and the consequents are $q, s, u$.
Applying the Addendo property:
$\frac{p+r+t}{q+s+u} = \frac{p}{q} = \frac{r}{s} = \frac{t}{u}$
Since each of the given ratios is equal to $\frac{1}{3}$, their sum ratio is also equal to $\frac{1}{3}$.
$\frac{p+r+t}{q+s+u} = \boldsymbol{\frac{1}{3}}$
The value of $\frac{p+r+t}{q+s+u}$ is $\frac{1}{3}$.
Alternate Method (Using Constant of Proportionality):
Let $\frac{p}{q} = \frac{r}{s} = \frac{t}{u} = \frac{1}{3}$.
This implies $p = \frac{1}{3}q$, $r = \frac{1}{3}s$, and $t = \frac{1}{3}u$.
Substitute these into the expression $\frac{p+r+t}{q+s+u}$:
$\frac{p+r+t}{q+s+u} = \frac{\frac{1}{3}q + \frac{1}{3}s + \frac{1}{3}u}{q+s+u}$
Factor out $\frac{1}{3}$ from the numerator:
$\frac{p+r+t}{q+s+u} = \frac{\frac{1}{3}(q+s+u)}{q+s+u}$
Assuming $q+s+u \neq 0$, cancel the term $(q+s+u)$:
$\frac{p+r+t}{q+s+u} = \boldsymbol{\frac{1}{3}}$
This method also confirms the result. The Addendo property provides a direct way to get the answer.
Types of Proportion (Based on Variation)
The relationship between quantities in a proportion can be classified into two main types based on how they vary relative to each other:
1. Direct Proportion (or Direct Variation)
Two quantities, $x$ and $y$, are said to be in direct proportion if their ratio remains constant. When one quantity increases, the other increases by the same factor, and when one quantity decreases, the other decreases by the same factor. Symbolically, $x \propto y$.
The relationship is expressed as $\frac{x}{y} = k$, where $k$ is the constant of proportionality ($k \neq 0$). This means $x = ky$.
If $(x_1, y_1)$ and $(x_2, y_2)$ are two corresponding pairs of values for quantities in direct proportion, then:
$\frac{x_1}{y_1} = k$
...
$\frac{x_2}{y_2} = k$
...
Equating these, we get the defining proportion for direct variation:
$\boldsymbol{\frac{x_1}{y_1} = \frac{x_2}{y_2}}$
... (vii)
This can also be written as $x_1 : y_1 :: x_2 : y_2$.
Examples:
- The more hours you work at a fixed wage rate, the more money you earn. (Earnings $\propto$ Hours Worked)
- The more litres of petrol you fill, the more you pay. (Cost $\propto$ Litres of Petrol)
- The more items you buy from a shop (at a fixed price per item), the higher the total bill. (Total Bill $\propto$ Number of Items)
Example 13. If a car travels 150 km in 3 hours at a constant speed, how far will it travel in 5 hours at the same speed?
Answer:
Let the distance traveled be $D$ (in km) and the time taken be $T$ (in hours). Since the speed is constant, distance is directly proportional to time ($D \propto T$).
We have initial values: $D_1 = 150$ km, $T_1 = 3$ hours.
We need to find the distance $D_2$ for $T_2 = 5$ hours.
Using the direct proportion formula $\frac{D_1}{T_1} = \frac{D_2}{T_2}$:
$\frac{150}{3} = \frac{D_2}{5}$
Simplify the left side: $\frac{150}{3} = 50$.
$\boldsymbol{50 = \frac{D_2}{5}}$
Multiply both sides by 5 to solve for $D_2$:
$\boldsymbol{D_2 = 50 \times 5}$
$\boldsymbol{D_2 = 250}$ km
The car will travel $\boldsymbol{250}$ km in 5 hours.
2. Inverse Proportion (or Inverse Variation)
Two quantities, $x$ and $y$, are said to be in inverse proportion if their product remains constant. When one quantity increases, the other decreases proportionally, and vice versa. If you double one quantity, the other is halved; if you halve one quantity, the other is doubled. Symbolically, $x \propto \frac{1}{y}$.
The relationship is expressed as $x = \frac{k}{y}$, which is equivalent to $\boldsymbol{xy = k}$, where $k$ is the constant of proportionality ($k \neq 0$).
If $(x_1, y_1)$ and $(x_2, y_2)$ are two corresponding pairs of values for quantities in inverse proportion, then:
$\boldsymbol{x_1y_1 = k}$
...
$\boldsymbol{x_2y_2 = k}$
...
Equating these, we get the defining equation for inverse variation:
$\boldsymbol{x_1y_1 = x_2y_2}$
... (viii)
This equation can be rewritten as a proportion by dividing both sides by $x_2y_1$ (assuming $x_2, y_1 \neq 0$):
$\frac{x_1y_1}{x_2y_1} = \frac{x_2y_2}{x_2y_1}$
$\frac{x_1}{x_2} = \frac{y_2}{y_1}$
In ratio form, this is $x_1:x_2 :: y_2:y_1$. Notice that the ratio of the first quantity's values is proportional to the *inverse* ratio of the second quantity's values.
Examples:
- The speed of a vehicle and the time taken to cover a fixed distance. (Speed $\propto \frac{1}{\text{Time}}$) - Higher speed means less time.
- The number of workers and the time taken to complete a fixed job. (Number of Workers $\propto \frac{1}{\text{Time}}$) - More workers mean less time.
- The density of a gas and its volume at a constant temperature and pressure. (Density $\propto \frac{1}{\text{Volume}}$) - As volume increases, density decreases.
Example 14. 10 men can build a wall in 8 days. How many days will it take for 5 men to build the same wall, assuming they work at the same rate?
Answer:
Let the number of men be $M$ and the number of days taken be $D$. Since the amount of work is fixed (building the same wall), the number of men is inversely proportional to the number of days taken ($M \propto \frac{1}{D}$, or $MD = k$).
We have initial values: $M_1 = 10$ men, $D_1 = 8$ days.
We need to find the number of days $D_2$ for $M_2 = 5$ men.
Using the inverse proportion formula $M_1 D_1 = M_2 D_2$:
$\boldsymbol{10 \times 8 = 5 \times D_2}$
$\boldsymbol{80 = 5 \times D_2}$
Divide both sides by 5 to solve for $D_2$:
$\boldsymbol{D_2 = \frac{80}{5}}$
$\boldsymbol{D_2 = 16}$ days
It will take $\boldsymbol{16}$ days for 5 men to build the same wall.
Distinguishing Direct and Inverse Proportion
Identifying whether a relationship is directly or inversely proportional is the first critical step in solving problems involving variation. Always ask yourself: "As one quantity increases, does the other quantity increase or decrease proportionally?"
Summary of comparison:
| Feature | Direct Proportion | Inverse Proportion |
|---|---|---|
| How quantities change together | Same direction (both increase or both decrease). | Opposite direction (one increases, the other decreases). |
| Constant Relationship | Ratio is constant ($\frac{x}{y} = k$). | Product is constant ($xy = k$). |
| Relationship between two pairs of values ($x_1, y_1$) and ($x_2, y_2$) | $\frac{x_1}{y_1} = \frac{x_2}{y_2}$ or $x_1:y_1 :: x_2:y_2$ | $x_1y_1 = x_2y_2$ or $x_1:x_2 :: y_2:y_1$ |
Competitive Exam Notes:
Familiarity with the types and properties of ratios and proportions is vital for solving quantitative problems efficiently.
- Ratio Types: Remember the definitions of inverse, compound, duplicate, triplicate, sub-duplicate, and sub-triplicate ratios. Be ready to compute them or use them in problems. Compound ratios are often used in combined variation problems.
- Proportion Properties: Invertendo, Alternendo, Componendo, Dividendo, and Componendo & Dividendo are powerful tools for simplifying expressions or solving equations when you are given a ratio or proportion. The Property of Equal Ratios (Addendo) is particularly useful when sums of terms from multiple equal ratios are involved.
- Direct vs. Inverse Proportion: This is a core concept. Always analyze the problem scenario to determine the type of variation. If 'more of A means more of B', it's direct ($\frac{A_1}{B_1} = \frac{A_2}{B_2}$). If 'more of A means less of B', it's inverse ($A_1B_1 = A_2B_2$). Many problems in Time and Work, Time Speed and Distance, Pipes and Cisterns rely on correctly identifying the variation.
- Applying Formulas: While the unitary method is intuitive, using the proportionality formulas ($\frac{x_1}{y_1} = \frac{x_2}{y_2}$ and $x_1y_1 = x_2y_2$) can be faster for direct calculation once the type of variation is identified.
- Word Problems: Translate word problems carefully into mathematical statements. Define variables clearly and set up the correct proportion or equation based on the type of variation.
The Unitary Method and its Applications
The Unitary Method is a widely used technique for solving problems involving proportional relationships between quantities. The fundamental principle behind this method is to first determine the value or quantity of a single unit, and then use this unit value to calculate the value or quantity for the desired number of units.
This method is particularly effective for problems in commercial arithmetic, time and work, time speed and distance, and other areas where quantities vary directly or inversely with each other. It provides a systematic approach to breaking down complex problems into simpler steps.
The Core Steps of the Unitary Method
The unitary method generally involves two main steps:
1. Find the value of one unit: This involves calculating the value corresponding to a single unit of the quantity for which information is provided. This is done by dividing the total given value by the number of units it corresponds to.
$\text{Value of one unit} = \frac{\text{Total given value}}{\text{Given number of units}}$
... (i)
2. Find the value of the required number of units: Once the value of a single unit is known, multiply this unit value by the number of units for which you need to find the value.
$\text{Required value} = (\text{Value of one unit}) \times (\text{Required number of units})$
... (ii)
While these steps work directly for problems involving direct proportion, a slight conceptual adjustment is needed for inverse proportion problems, as explained in the applications section below.
Applications of the Unitary Method
The unitary method is versatile and can be applied to problems exhibiting both direct and inverse proportional relationships.
Unitary Method in Direct Proportion
In problems involving direct proportion, if one quantity increases, the other quantity increases proportionally, and vice versa. The ratio of corresponding quantities remains constant. The steps of the unitary method (finding the value of one unit and then multiplying) directly apply here because the 'value per unit' is constant.
Example 1. If the cost of 7 kilograms of rice is $\textsf{₹ } 245$, what would be the cost of 15 kilograms of rice?
Answer:
This is a case of direct proportion: As the quantity of rice increases, the cost increases proportionally.
Given: Cost of 7 kg rice is $\textsf{₹ } 245$.
Required: Cost of 15 kg rice.
Step 1: Find the cost of one kilogram of rice (the unit value).
Cost of 7 kg = $\textsf{₹ } 245$
Cost of 1 kg $= \frac{\text{Total Cost}}{\text{Total Quantity}}$
Cost of 1 kg $= \frac{\textsf{₹ } 245}{7}$
Performing the division:
$\frac{245}{7} = 35$
Cost of 1 kg $= \textsf{₹ } 35$
... (iii)
Step 2: Find the cost of 15 kilograms of rice (the required value).
Cost of 15 kg $= (\text{Cost of 1 kg}) \times 15$
Cost of 15 kg $= \textsf{₹ } 35 \times 15$
Performing the multiplication:
$\begin{array}{cc}& & 3 & 5 \\ \times & & 1 & 5 \\ \hline && 1 & 7 & 5 \\ & 3 & 5 & \times \\ \hline & 5 & 2 & 5 \\ \hline \end{array}$
Cost of 15 kg $= \textsf{₹ } 525$
The cost of 15 kilograms of rice is $\boldsymbol{\textsf{₹ } 525}$.
Unitary Method in Inverse Proportion
In problems involving inverse proportion, if one quantity increases, the other quantity decreases proportionally such that their product remains constant. Applying the unitary method here requires a slight modification in the interpretation of the "value of one unit" step.
For inverse proportion, when you find the value "for one unit", you are essentially calculating the total 'work', 'resource', or 'effort' unit required for the task (e.g., total man-days, total tap-hours, total student-days). This total quantity remains constant, regardless of how the 'units' are distributed. Then, to find the value for the required number of units, you divide this total quantity by the number of required units.
Example 2. 8 pumps can empty a reservoir in 21 hours. How many hours will 14 pumps take to empty the same reservoir?
Answer:
This is a case of inverse proportion: As the number of pumps increases, the time taken to empty the reservoir decreases (assuming all pumps work at the same rate).
Given: 8 pumps take 21 hours.
Required: Time taken by 14 pumps.
Step 1: Find the time taken by one pump to empty the reservoir (total pump-hours).
8 pumps take 21 hours.
If there was only 1 pump, it would have to do the work of 8 pumps, so it would take 8 times longer.
Time taken by 1 pump $= (\text{Time taken by 8 pumps}) \times 8$
Time taken by 1 pump $= 21 \text{ hours} \times 8$
Performing the multiplication:
$\begin{array}{cc}& & 2 & 1 \\ \times & & & 8 \\ \hline & 1 & 6 & 8 \\ \hline \end{array}$
Time taken by 1 pump $= 168$ hours
... (iv)
This '168 pump-hours' represents the total work needed to empty the tank.
Step 2: Find the time taken by 14 pumps (the required value).
Now, 14 pumps will share this total work (168 pump-hours). The time taken by 14 pumps will be the total work divided by the number of pumps.
Time taken by 14 pumps $= \frac{\text{Total Work (in pump-hours)}}{\text{Number of pumps}}$
Time taken by 14 pumps $= \frac{168 \text{ hours}}{14}$
Performing the division:
$\frac{168}{14} = \frac{\cancel{168}^{12}}{\cancel{14}_1} = 12$
Time taken by 14 pumps $= 12$ hours
It will take $\boldsymbol{12}$ hours for 14 pumps to empty the same reservoir.
Unitary Method using Proportionality Formulas
The logic of the unitary method is directly embedded in the proportionality formulas derived earlier (in section I3). Recognizing this link can sometimes make problem-solving quicker.
For Direct Proportion: If $x_1$ corresponds to $y_1$, and we need to find the value $y_2$ corresponding to $x_2$, the direct proportion relationship is $\frac{y_1}{x_1} = \frac{y_2}{x_2}$. Solving for $y_2$ gives:
$\boldsymbol{y_2 = y_1 \times \frac{x_2}{x_1}}$
... (v)
Notice that $\frac{y_1}{x_1}$ is the value per unit of $x$, and we are multiplying it by the new number of units $x_2$. This is precisely the unitary method structure: $(\text{Value per unit}) \times (\text{Number of units})$.
Example 3. A car consumes 12 litres of petrol to travel 216 km. How much petrol will it consume to travel 396 km?
Answer:
Assuming consistent fuel consumption, the amount of petrol consumed is directly proportional to the distance traveled. (More distance means more petrol).
Let $P$ be the petrol consumed (in litres) and $D$ be the distance traveled (in km).
Given: $P_1 = 12$ litres for $D_1 = 216$ km.
Required: $P_2$ for $D_2 = 396$ km.
Using the direct proportion formula $\frac{P_1}{D_1} = \frac{P_2}{D_2}$:
$\frac{12}{216} = \frac{P_2}{396}$
Solve for $P_2$:
$\boldsymbol{P_2 = \frac{12}{216} \times 396}$
Simplify the fraction $\frac{12}{216}$:
$\frac{12}{216} = \frac{\cancel{12}^1}{\cancel{216}_{18}} = \frac{1}{18}$
So, $P_2 = \frac{1}{18} \times 396$.
$\boldsymbol{P_2 = \frac{396}{18}}$
Performing the division:
$\frac{396}{18} = \frac{\cancel{396}^{22}}{\cancel{18}_1} = 22$
$\boldsymbol{P_2 = 22}$ litres
The car will consume $\boldsymbol{22}$ litres of petrol to travel 396 km.
Using Unitary Method (Step-by-step):
Petrol consumed for 216 km = 12 litres.
Petrol consumed for 1 km (unit distance) = $\frac{12}{216}$ litres = $\frac{1}{18}$ litres.
Petrol consumed for 396 km = (Petrol for 1 km) $\times$ 396
Petrol $= \frac{1}{18} \times 396$ litres
Petrol $= \frac{396}{18} = 22$ litres
Both methods yield the same result, demonstrating the connection between unitary method steps and proportionality formulas.
For Inverse Proportion: If $x_1$ corresponds to $y_1$, and we need to find the value $y_2$ corresponding to $x_2$, the inverse proportion relationship is $x_1 y_1 = x_2 y_2$. Solving for $y_2$ gives:
$\boldsymbol{y_2 = \frac{x_1 y_1}{x_2}}$
... (vi)
Notice that $x_1 y_1$ represents the constant product or the total 'amount of work' or 'resource' in terms of the product of the two quantities. We are dividing this total by the new value of $x_2$ to find the corresponding $y_2$. This mirrors the two steps of the unitary method for inverse proportion: find the total unit (by multiplying initial values) and then divide by the new number of units.
Example 4. A garrison has provisions for 400 men for 31 days. How long will the provisions last if 100 more men join the garrison?
Answer:
This is a case of inverse proportion: The number of men and the number of days the provisions last are inversely proportional. (More men mean fewer days the provisions will last).
Initial number of men $M_1 = 400$. Provisions last for $D_1 = 31$ days.
Number of additional men = 100.
New number of men $M_2 = 400 + 100 = 500$ men.
We need to find the number of days $D_2$ the provisions will last for 500 men.
Using the inverse proportion formula $M_1 D_1 = M_2 D_2$:
$\boldsymbol{400 \times 31 = 500 \times D_2}$
$12400 = 500 \times D_2$
Solve for $D_2$ by dividing both sides by 500:
$\boldsymbol{D_2 = \frac{12400}{500}}$
$\boldsymbol{D_2 = \frac{124}{5}}$
Performing the division:
$\frac{124}{5} = 24.8$
$\boldsymbol{D_2 = 24.8}$ days
The provisions will last for $\boldsymbol{24.8}$ days for 500 men.
Using Unitary Method (Step-by-step):
Provisions for 400 men last for 31 days.
Total 'man-days' of provisions = $400 \text{ men} \times 31 \text{ days} = 12400 \text{ man-days}$.
This 12400 man-days of provisions are available for the new number of men, which is 500.
Number of days the provisions last for 500 men = $\frac{\text{Total man-days}}{\text{Number of men}}$
Days $= \frac{12400}{500}$
Days $= \frac{124}{5} = 24.8$ days
Both methods lead to the same conclusion.
Competitive Exam Notes:
The Unitary Method is a fundamental problem-solving approach in quantitative aptitude. It's often the most intuitive way to tackle problems involving proportional relationships.
- Core Principle: Always aim to find the value associated with a single unit of the quantity that changes.
- Direct Proportion: For problems like cost vs. quantity, distance vs. time (at constant speed), work vs. time (for a fixed number of workers), etc., use the standard two steps: Divide to find the value of 1 unit, then Multiply to find the value of the required units. Formula: $\frac{\text{Value}_1}{\text{Unit}_1} = \frac{\text{Value}_2}{\text{Unit}_2}$.
- Inverse Proportion: For problems like number of workers vs. time (for fixed work), speed vs. time (for fixed distance), number of people vs. days food lasts (for fixed stock), etc., the 'unit value' step means finding the total 'combined unit' (like man-hours or man-days) by Multiplying the given quantities. Then, Divide this total by the new number of units to find the answer. Formula: $\text{Unit}_1 \times \text{Value}_1 = \text{Unit}_2 \times \text{Value}_2$.
- Complex Problems: For problems involving more than two varying quantities (e.g., Man-Days-Hours problems), break them down into smaller steps or use a combined variation formula if applicable. The core unitary idea still applies: find the total work/resource unit and then divide by the combined units of the new scenario.
- Units Conversion: Pay close attention to units. Ensure all quantities are in consistent units before applying the method or formulas.
Solving Problems based on Ratio and Proportion
This section provides a collection of examples to demonstrate the practical application of the concepts of ratio, proportion, and the unitary method in solving various types of quantitative problems. These examples cover common scenarios encountered in competitive exams.
Example 1. Divide $\textsf{₹ } 1200$ between A and B in the ratio $3:5$.
Answer:
Given the ratio in which the amount is to be divided between A and B is $3:5$.
The sum of the terms of the ratio is $3 + 5 = 8$.
This means that for every 8 equal parts of the total amount, A receives 3 parts and B receives 5 parts.
The total amount to be divided is $\textsf{₹ } 1200$.
The fraction of the total amount that A gets is $\frac{\text{A's ratio term}}{\text{Sum of ratio terms}} = \frac{3}{8}$.
The fraction of the total amount that B gets is $\frac{\text{B's ratio term}}{\text{Sum of ratio terms}} = \frac{5}{8}$.
Amount A gets:
Amount A gets $= \frac{3}{8} \times \textsf{₹ } 1200$
Calculate the amount:
Amount A gets $= \frac{3}{\cancel{8}^{\normalsize 1}} \times \cancel{1200}^{\normalsize 150}$
Amount A gets $= 3 \times 150 = \textsf{₹ } 450$
Amount B gets:
Amount B gets $= \frac{5}{8} \times \textsf{₹ } 1200$
Calculate the amount:
Amount B gets $= \frac{5}{\cancel{8}^{\normalsize 1}} \times \cancel{1200}^{\normalsize 150}$
Amount B gets $= 5 \times 150 = \textsf{₹ } 750$
Verification:
Total amount received by A and B $= \textsf{₹ } 450 + \textsf{₹ } 750 = \textsf{₹ } 1200$. This matches the total amount given.
The ratio of amounts A:B $= 450:750$. Simplifying this ratio by dividing both terms by their GCD (150): $450 \div 150 : 750 \div 150 = 3:5$. This matches the given ratio.
So, A gets $\boldsymbol{\textsf{₹ } 450}$ and B gets $\boldsymbol{\textsf{₹ } 750}$.
Example 2. Find the value of $x$ in the proportion $16:x :: x:25$.
Answer:
The given statement $16:x :: x:25$ represents a proportion where the middle terms are equal. This is a case of continued proportion, and $x$ is the mean proportional between 16 and 25.
By the fundamental property of proportion, the product of the extremes is equal to the product of the means.
Product of Extremes $= 16 \times 25$
Product of Means $= x \times x = x^2$
Equating the products:
$\boldsymbol{x^2 = 16 \times 25}$
$\boldsymbol{x^2 = 400}$
... (i)
To find $x$, take the square root of both sides of equation (i):
$\boldsymbol{x = \pm \sqrt{400}}$
$\boldsymbol{x = \pm 20}$
In most practical problems involving physical quantities like length, time, speed, etc., the value of the mean proportional is considered positive.
Assuming $x$ represents a positive quantity, we take $x = 20$.
The value of $x$ is $\boldsymbol{20}$ (considering the positive root).
Verification:
If $x=20$, the proportion is $16:20 :: 20:25$.
Check if the ratios are equal: $\frac{16}{20} = \frac{\cancel{16}^4}{\cancel{20}_5} = \frac{4}{5}$.
$\frac{20}{25} = \frac{\cancel{20}^4}{\cancel{25}_5} = \frac{4}{5}$.
Since $\frac{4}{5} = \frac{4}{5}$, the proportion holds true with $x=20$.
Example 3. The ratio of the length and breadth of a rectangle is $7:4$. If the length is 28 cm, find the breadth.
Answer:
Let the length of the rectangle be $L$ and the breadth be $B$.
Given the ratio of length to breadth is $L:B = 7:4$.
This ratio can be written as a fraction: $\frac{L}{B} = \frac{7}{4}$.
We are given the length $L = 28$ cm. We need to find the breadth $B$.
Substitute the value of $L$ into the equation:
$\frac{28}{B} = \frac{7}{4}$
Now, we can solve for $B$ using cross-multiplication:
$28 \times 4 = B \times 7$
$\boldsymbol{112 = 7B}$
... (ii)
Divide both sides of equation (ii) by 7 to find $B$:
$\boldsymbol{B = \frac{112}{7}}$
Performing the division:
$\frac{112}{7} = 16$
$\boldsymbol{B = 16}$ cm
The breadth of the rectangle is $\boldsymbol{16}$ cm.
Alternate Method (Using Unitary Method concept based on Ratio):
The ratio $7:4$ means the length is 7 parts and the breadth is 4 parts.
Given that 7 parts of the length correspond to 28 cm.
Value of 1 part $= \frac{\text{Total Length}}{\text{Number of Length parts}} = \frac{28 \text{ cm}}{7} = 4$ cm.
The breadth corresponds to 4 parts.
Breadth $= (\text{Value of 1 part}) \times (\text{Number of Breadth parts})$
Breadth $= 4 \text{ cm} \times 4 = 16$ cm
This confirms the result obtained using the proportion equation.
Example 4. A train travels 480 km in 4 hours. How much time will it take to travel 720 km at the same speed?
Answer:
Assuming the train maintains a constant speed, the distance covered is directly proportional to the time taken. (As distance increases, time increases proportionally).
Let $D_1 = 480$ km and $T_1 = 4$ hours.
We need to find the time $T_2$ required to travel $D_2 = 720$ km.
Using the direct proportion relationship: $\frac{D_1}{T_1} = \frac{D_2}{T_2}$
$\frac{480}{4} = \frac{720}{T_2}$
Simplify the left side:
$\frac{480}{4} = 120$. So, $120 = \frac{720}{T_2}$
Now, solve for $T_2$. Multiply both sides by $T_2$ and then divide by 120:
$\boldsymbol{120 \times T_2 = 720}$
$\boldsymbol{T_2 = \frac{720}{120}}$
Simplify the fraction:
$\frac{720}{120} = \frac{72}{12} = 6$
$\boldsymbol{T_2 = 6}$ hours
The train will take $\boldsymbol{6}$ hours to travel 720 km.
Using Unitary Method:
Time taken for 480 km $= 4$ hours.
Time taken for 1 km (unit distance) $= \frac{\text{Total Time}}{\text{Total Distance}} = \frac{4 \text{ hours}}{480} = \frac{1}{120}$ hours.
Time taken for 720 km $= (\text{Time for 1 km}) \times (\text{Required Distance})$
Time $= \frac{1}{120} \times 720$ hours
Time $= \frac{720}{120} = 6$ hours
Both methods produce the same result.
Example 5. 6 pipes of the same capacity can fill a tank in 45 minutes. How many minutes will 9 such pipes take to fill the same tank?
Answer:
This is an inverse proportion problem. The number of pipes is inversely proportional to the time taken to fill the tank (assuming all pipes fill at the same rate). (More pipes mean less time to fill the tank).
Let $N_1 = 6$ pipes and $T_1 = 45$ minutes.
We need to find the time $T_2$ taken by $N_2 = 9$ pipes.
Using the inverse proportion relationship: $N_1 T_1 = N_2 T_2$
$\boldsymbol{6 \times 45 = 9 \times T_2}$
$\boldsymbol{270 = 9 T_2}$
... (iii)
Divide both sides of equation (iii) by 9 to solve for $T_2$:
$\boldsymbol{T_2 = \frac{270}{9}}$
Performing the division:
$\frac{270}{9} = 30$
$\boldsymbol{T_2 = 30}$ minutes
9 pipes will take $\boldsymbol{30}$ minutes to fill the same tank.
Using Unitary Method:
Time taken by 6 pipes $= 45$ minutes.
To find the time taken by 1 pipe, we consider that 1 pipe would take 6 times longer than 6 pipes.
Total 'pipe-minutes' required to fill the tank $= (\text{Number of pipes}) \times (\text{Time taken})$
Total work $= 6 \text{ pipes} \times 45 \text{ minutes} = 270 \text{ pipe-minutes}$.
This total work (270 pipe-minutes) must be done by the new number of pipes, which is 9.
Time taken by 9 pipes $= \frac{\text{Total work}}{\text{Number of pipes}}$
Time $= \frac{270 \text{ pipe-minutes}}{9 \text{ pipes}} = 30$ minutes
Both approaches lead to the same answer.
Example 6. A mixture of 28 litres contains milk and water in the ratio $5:2$. How much more water must be added to make the ratio of milk to water $2:1$?
Answer:
Initial quantity of the mixture $= 28$ litres.
Initial ratio of milk to water $= 5:2$.
Sum of the terms in the initial ratio $= 5 + 2 = 7$.
Calculate initial quantities of milk and water:
Quantity of milk $= \frac{\text{Milk ratio term}}{\text{Sum of ratio terms}} \times \text{Total mixture}$
Quantity of milk $= \frac{5}{7} \times 28$ litres
Quantity of milk $= \frac{5}{\cancel{7}^{\normalsize 1}} \times \cancel{28}^{\normalsize 4} = 5 \times 4 = 20$ litres.
Quantity of water $= \frac{\text{Water ratio term}}{\text{Sum of ratio terms}} \times \text{Total mixture}$
Quantity of water $= \frac{2}{7} \times 28$ litres
Quantity of water $= \frac{2}{\cancel{7}^{\normalsize 1}} \times \cancel{28}^{\normalsize 4} = 2 \times 4 = 8$ litres.
Check: $20 \text{ litres (milk)} + 8 \text{ litres (water)} = 28 \text{ litres}$. (Matches initial mixture quantity).
Set up the new scenario:
Let $x$ litres of water be added to the mixture.
The quantity of milk remains unchanged: New milk quantity $= 20$ litres.
The quantity of water increases: New water quantity $= (8 + x)$ litres.
The new ratio of milk to water is given as $2:1$.
New ratio = Milk : Water $= 20 : (8+x)$
We are given that this new ratio is equal to $2:1$. So, we set up a proportion:
$\frac{20}{8+x} = \frac{2}{1}$
... (iv)
Solve for $x$:
Cross-multiply equation (iv):
$20 \times 1 = 2 \times (8+x)$
$\boldsymbol{20 = 16 + 2x}$
... (v)
Subtract 16 from both sides of equation (v):
$20 - 16 = 2x$
$\boldsymbol{4 = 2x}$
... (vi)
Divide both sides of equation (vi) by 2:
$\boldsymbol{x = \frac{4}{2}}$
$\boldsymbol{x = 2}$ litres
So, $\boldsymbol{2}$ litres of water must be added to the mixture.
Verification:
After adding 2 litres of water, new water quantity $= 8 + 2 = 10$ litres.
New ratio Milk : Water $= 20 : 10 = \frac{20}{10} = 2$. This is the ratio $2:1$, which is the required new ratio.
Example 7. If 12 workers can complete a task in 18 days, how many workers are required to finish the same task in 9 days?
Answer:
This is an inverse proportion problem. The number of workers is inversely proportional to the number of days required to complete the same task. (Fewer days require more workers).
Let $W_1 = 12$ workers and $D_1 = 18$ days.
We need to find the number of workers $W_2$ required to finish the task in $D_2 = 9$ days.
Using the inverse proportion relationship: $W_1 D_1 = W_2 D_2$
$\boldsymbol{12 \times 18 = W_2 \times 9}$
$\boldsymbol{216 = 9 W_2}$
... (vii)
Divide both sides of equation (vii) by 9 to solve for $W_2$:
$\boldsymbol{W_2 = \frac{216}{9}}$
Performing the division:
$\frac{216}{9} = 24$
$\boldsymbol{W_2 = 24}$ workers
$\boldsymbol{24}$ workers are required to finish the task in 9 days.
Using Unitary Method:
Time taken by 12 workers $= 18$ days.
Total work in 'worker-days' $= (\text{Number of workers}) \times (\text{Number of days})$
Total work $= 12 \text{ workers} \times 18 \text{ days} = 216 \text{ worker-days}$.
This total work (216 worker-days) must be completed in 9 days by the required number of workers.
Number of workers $= \frac{\text{Total work}}{\text{Number of days}}$
Workers $= \frac{216 \text{ worker-days}}{9 \text{ days}} = 24$ workers
Both methods yield the same answer.
Example 8. The cost of 15 apples is $\textsf{₹ } 180$. Find the cost of 25 apples.
Answer:
This is a direct proportion problem. The cost of apples is directly proportional to the number of apples (assuming the price per apple is constant). (More apples cost more money).
Using the Unitary Method:
Cost of 15 apples $= \textsf{₹ } 180$.
Step 1: Find the cost of one apple (the unit cost).
Cost of 1 apple $= \frac{\text{Total Cost}}{\text{Number of apples}} = \frac{\textsf{₹ } 180}{15}$
Performing the division:
$\frac{180}{15} = 12$
Cost of 1 apple $= \textsf{₹ } 12$
Step 2: Find the cost of 25 apples (the required value).
Cost of 25 apples $= (\text{Cost of 1 apple}) \times (\text{Number of required apples})$
Cost of 25 apples $= \textsf{₹ } 12 \times 25$
Performing the multiplication:
$\begin{array}{cc}& & 1 & 2 \\ \times & & 2 & 5 \\ \hline & & 6 & 0 \\ & 2 & 4 & \times \\ \hline & 3 & 0 & 0 \\ \hline \end{array}$
Cost of 25 apples $= \textsf{₹ } 300$
The cost of 25 apples is $\boldsymbol{\textsf{₹ } 300}$.
Competitive Exam Notes:
Solving problems based on ratio and proportion is a core skill for quantitative aptitude. Practice these basic problem types thoroughly as they form the foundation for more complex topics.
- Ratio Division: When dividing a quantity in a given ratio, find the sum of the ratio terms. The share of each part is the total quantity multiplied by (individual ratio term / sum of ratio terms).
- Finding Unknown Terms: Use the product of extremes equals product of means property ($ad=bc$) to find an unknown term in a proportion. For continued proportion $a:x::x:c$, remember $x^2 = ac$.
- Word Problems: Carefully read the problem to identify the quantities involved and the relationship between them (direct or inverse proportion). Set up the proportion or use the unitary method accordingly.
- Direct Proportion: Use $\frac{Q_1}{V_1} = \frac{Q_2}{V_2}$ or find Value/Unit $= \frac{V_1}{Q_1}$ and then Required Value $= (\text{Value/Unit}) \times Q_2$.
- Inverse Proportion: Use $Q_1 \times V_1 = Q_2 \times V_2$ or find Total Unit Work $= Q_1 \times V_1$ and then Required Value $= \frac{\text{Total Unit Work}}{Q_2}$.
- Mixture Problems: These often involve calculating initial quantities based on a ratio, setting up a new ratio equation after adding or removing a substance, and solving for the unknown quantity.
- Consistency in Units: Always ensure that the units of quantities being compared or used in calculations are consistent.