Ratios, Percentages, and Proportions

Overview

The concepts of ratio, proportion, and percentage form the bedrock of quantitative aptitude. These are not merely discrete mathematical topics but are, in fact, the fundamental language through which we compare quantities, express relationships, and analyze change. A firm command of these principles is indispensable for any rigorous analytical endeavor, providing the essential tools to deconstruct and interpret numerical information. In this chapter, we shall systematically develop a robust understanding of these concepts, moving from foundational definitions to their application in complex problem-solving scenarios.

For the GATE examination, proficiency in this chapter is of paramount importance. While direct questions on ratios or percentages are common, their true significance lies in their pervasive application across a multitude of other topics. Concepts from this chapter are intrinsically woven into the fabric of problems related to Data Interpretation, Profit and Loss, Simple and Compound Interest, and Mixtures and Alligations. Therefore, mastering the techniques presented herein will not only equip you to solve specific questions but will also enhance your speed and accuracy in a significant portion of the entire Quantitative Aptitude section. We will focus on both conceptual clarity and the development of efficient computational strategies tailored for a competitive examination environment.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Ratio and Proportion | Comparing quantities and establishing equivalence between ratios. |
| 2 | Percentages | Representing a number as a fraction of 100. |

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Learning Objectives

❗ By the End of This Chapter

After completing this chapter, you will be able to:

Analyze and solve problems involving direct and inverse proportions.

Calculate percentage changes, successive percentages, and their practical applications.

Apply the principles of ratio and proportion to solve problems involving mixtures, partnerships, and ages.

Formulate and solve equations based on proportional and percentage-based relationships in quantitative problems.

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We now turn our attention to Ratio and Proportion...

Part 1: Ratio and Proportion

Introduction

The concepts of ratio and proportion form a foundational pillar of quantitative aptitude, providing a systematic framework for comparing quantities and understanding their relative relationships. A ratio is a mathematical expression that quantifies the relationship between two numbers, indicating how many times the first number contains the second. A proportion, in turn, is a statement of equality between two ratios. Mastery of these concepts is indispensable for solving a wide array of problems in the GATE examination, particularly those involving mixtures, partnerships, time and work, and speed and distance.

In this chapter, we will undertake a rigorous examination of the principles governing ratios and proportions. We will begin by establishing the formal definitions and fundamental properties. Subsequently, we will explore advanced applications, including problems where the constituent parts of a ratio are altered, and scenarios where multiple ratios are interlinked through specific conditions. The methodologies presented herein are designed to foster both conceptual clarity and computational efficiency, which are critical under the time constraints of a competitive examination.

📖 Ratio

A ratio is a comparison of two quantities of the same kind, expressed as a quotient. The ratio of a quantity $a$ to a quantity $b$ is denoted as $a:b$ and is equivalent to the fraction $\frac{a}{b}$ . Here, $a$ is called the antecedent and $b$ is the consequent.

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Key Concepts

1. The Proportionality Constant

A ratio, such as $a:b$ , represents a relative comparison, not the absolute values of the quantities. To work with these quantities algebraically, we introduce a common multiplier, known as the proportionality constant, typically denoted by $k$ or $x$ . If two numbers are in the ratio $a:b$ , we can represent their actual values as $ak$ and $bk$ for some non-zero constant $k$ .

This conversion is the first and most crucial step in solving a majority of ratio-based problems. A direct consequence of this representation is a fundamental property concerning the sum of the quantities.

❗ The Sum of Ratio Terms

If a quantity $Q$ is divided into two parts in the ratio $a:b$ , then the parts can be represented as $ak$ and $bk$ . The total quantity is $Q = ak + bk = (a+b)k$ . It follows that the total quantity $Q$ must be an integer multiple of the sum of the ratio terms, $(a+b)$ . This property is exceptionally useful for eliminating incorrect options in multiple-choice questions.

Worked Example:

Problem: The number of engineering students to management students in a university is in the ratio $5:4$ . If the total number of students is 342, find the number of engineering students.

Solution:

Step 1: Represent the number of students using a proportionality constant.
Let the number of engineering students be $5k$ and the number of management students be $4k$ .

Step 2: Formulate an equation based on the total number of students.
The total number of students is the sum of the two groups.

5k + 4k = 342

Step 3: Solve for the proportionality constant $k$ .

9k = 342

k = \frac{342}{9}

k = 38

Step 4: Calculate the required quantity.
The number of engineering students is $5k$ .

5k = 5 \times 38 = 190

Answer: \boxed{190}

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2. Proportion and its Properties

When two ratios are equal, they are said to be in proportion. If $a:b$ is equal to $c:d$ , we write $a:b :: c:d$ or $\frac{a}{b} = \frac{c}{d}$ . In this proportion, $a$ and $d$ are called the extremes, while $b$ and $c$ are called the means. A key property is that the product of the extremes equals the product of the means, i.e., $ad = bc$ .

Several algebraic manipulations can be performed on proportions, which often simplify complex calculations. The most powerful of these is Componendo and Dividendo.

📐 Componendo and Dividendo

If $\frac{a}{b} = \frac{c}{d}$ , then by Componendo and Dividendo, we have:

\frac{a+b}{a-b} = \frac{c+d}{c-d}

Variables:

$a, b, c, d$ are the four terms of the proportion.

When to use: This rule is extremely effective in solving equations where the variable appears in fractional forms like

\frac{f(x)+g(x)}{f(x)-g(x)}

It simplifies the algebraic structure of the problem significantly.

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3. Problems Involving Change in Quantities

A common archetype of problems in GATE involves a scenario where an initial ratio is given, the quantities are altered by a certain amount, and a new ratio is formed. The key to solving such problems is to establish an algebraic equation based on this transformation.

Let the initial quantities be in the ratio $a:b$ . We represent them as $ak$ and $bk$ . If a value $c$ is added to both quantities, the new quantities become $ak+c$ and $bk+c$ . If the new ratio is $p:q$ , we can form the proportion:

\frac{ak+c}{bk+c} = \frac{p}{q}

Solving this equation for $k$ allows us to find the original quantities.

Worked Example:

Problem: The monthly incomes of two persons, A and B, are in the ratio $4:5$ . Their monthly expenditures are in the ratio $7:9$ . If each saves ₹2000 per month, find the monthly income of A.

Solution:

Step 1: Represent the incomes and expenditures using different proportionality constants.
Let the incomes of A and B be $4x$ and $5x$ , respectively.
Let their expenditures be $7y$ and $9y$ , respectively.

Step 2: Formulate equations based on the relationship: Income - Expenditure = Savings.
For person A:

4x - 7y = 2000 \quad \cdots(1)

For person B:

5x - 9y = 2000 \quad \cdots(2)

Step 3: Solve the system of linear equations.
Multiply equation (1) by 9 and equation (2) by 7 to eliminate $y$ .

36x - 63y = 18000 \quad \cdots(3)

35x - 63y = 14000 \quad \cdots(4)

Subtracting equation (4) from equation (3):

\begin{aligned}(36x - 63y) - (35x - 63y) & = 18000 - 14000 \\ x & = 4000\end{aligned}

Step 4: Calculate the required quantity.
The monthly income of A is $4x$ .

4x = 4 \times 4000 = 16000

Answer: \boxed{\text{₹}16,000}

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#
## 4. Ratios in Applications of Speed, Distance, and Time

The principles of ratio and proportion are directly applicable to problems involving motion. The fundamental relationship is $\text{Speed} = D/T$ . From this, we can derive two critical proportionalities:

For a constant time interval: Speed is directly proportional to distance.

S \propto D \implies \frac{S_1}{S_2} = \frac{D_1}{D_2}

For a constant distance: Speed is inversely proportional to time.

S \propto \frac{1}{T} \implies \frac{S_1}{S_2} = \frac{T_2}{T_1}

These relationships allow us to solve complex race and motion problems without necessarily calculating the absolute values of speed or time.

Worked Example:

Problem: Two runners, X and Y, have speeds in the ratio $5:6$ . They run a race. If X is given a head start of 50 meters, the race ends in a dead heat (they finish at the same time). What is the total length of the race course?

Solution:

Step 1: Analyze the condition of the race.
A "dead heat" implies that both runners take the same amount of time to finish the race.

Step 2: Use the proportionality for constant time.
Since time is the same for both runners, the ratio of their speeds is equal to the ratio of the distances they cover.

\frac{S_X}{S_Y} = \frac{D_X}{D_Y}

\frac{5}{6} = \frac{D_X}{D_Y}

Step 3: Relate the distances covered.
Let the length of the race course be $L$ meters.
Runner Y covers the full distance, $D_Y = L$ .
Runner X has a 50m head start, so X only needs to cover $D_X = L - 50$ .

Step 4: Formulate and solve the equation for $L$ .
Substitute the expressions for distance into the proportion.

\frac{5}{6} = \frac{L - 50}{L}

5L = 6(L - 50)

5L = 6L - 300

L = 300

Answer: The total length of the race course is 300 meters.

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Problem-Solving Strategies

💡 GATE Strategy: Option Elimination

In problems asking for a total quantity which is a sum of parts in a ratio $a:b:c$ , the total must be divisible by $(a+b+c)$ . Before beginning any calculation, check which of the given options satisfies this divisibility rule. Often, this can lead to the correct answer instantly or significantly narrow down the choices.

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Common Mistakes

⚠️ Avoid These Errors

❌ Confusing Ratio with Value: Assuming that a ratio of $7:3$ means the quantities are 7 and 3.

✅ Correct Approach: Always introduce a proportionality constant,

k

. The quantities are

7k

and

3k

. The value of

k

must be determined from other information in the problem.

❌ Incorrectly Modifying Ratios: When a quantity $c$ is added to two numbers $N_1$ and $N_2$ in ratio $a:b$ , writing the new ratio as $\frac{a}{b} + c$ .

✅ Correct Approach: The new numbers are

N_1+c

and

N_2+c

. The new ratio is

\frac{N_1+c}{N_2+c}

. If

N_1=ak

and

N_2=bk

, the new ratio is

\frac{ak+c}{bk+c}

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Practice Questions

:::question type="MCQ" question="The cost of a diamond varies directly as the square of its weight. A diamond weighing 20 grams costs ₹32,000. What will be the cost (in ₹) of a diamond weighing 50 grams?" options=["₹1,00,000","₹1,50,000","₹2,00,000","₹80,000"] answer="₹2,00,000" hint="Set up a direct proportion relationship: Cost ∝ (Weight)^2. Use the given information to find the constant of proportionality." solution="
Step 1: Establish the proportionality relationship.
Let $C$ be the cost and $W$ be the weight.
Given that $C \propto W^2$ , we can write $C = kW^2$ , where $k$ is the constant of proportionality.

Step 2: Use the given data to find the value of $k$ .
We are given that when $W = 20$ g, $C = ₹32,000$ .

32000 = k \times (20)^2

32000 = k \times 400

k = \frac{32000}{400} = 80

Step 3: Use the value of $k$ to find the cost for the new weight.
We need to find the cost when $W = 50$ g.

C = 80 \times (50)^2

C = 80 \times 2500

C = 200000

Result:
Answer: \boxed{₹2,00,000}
"
:::

:::question type="NAT" question="The ratio of zinc to copper in an alloy is $5:3$ . If 24 kg of zinc is added to the alloy, the ratio of zinc to copper becomes $7:3$ . What is the initial weight of copper in the alloy (in kg)?" answer="36" hint="Let the initial quantities be 5k and 3k. Form an equation based on the change in the quantity of zinc." solution="
Step 1: Represent the initial quantities of zinc and copper.
Let the initial weight of zinc be $5k$ kg.
Let the initial weight of copper be $3k$ kg.

Step 2: Formulate the new quantities after adding zinc.
24 kg of zinc is added. The amount of copper remains unchanged.
New weight of zinc = $5k + 24$ kg.
Weight of copper = $3k$ kg.

Step 3: Set up a proportion using the new ratio.
The new ratio of zinc to copper is $7:3$ .

\frac{5k + 24}{3k} = \frac{7}{3}

Step 4: Solve the equation for $k$ .
We can cancel the denominator 3 from both sides.

5k + 24 = 7k

2k = 24

k = 12

Step 5: Calculate the initial weight of copper.
The initial weight of copper is $3k$ .

3k = 3 \times 12 = 36

Result:
Answer: \boxed{36 \text{ kg}}
"
:::

:::question type="MCQ" question="The ratio of the ages of a father and his son 10 years ago was $3:1$ . The ratio of their ages 10 years hence will be $5:3$ . What is the ratio of their present ages?" options=["2:1","5:2","3:2","7:3"] answer="2:1" hint="Let their ages 10 years ago be 3x and x. Express their present ages and future ages in terms of x and form a proportion." solution="
Step 1: Define variables based on the past ages.
Let the ages of the father and son 10 years ago be $3x$ and $x$ , respectively.

Step 2: Express their present ages.
Present age of father = $3x + 10$ .
Present age of son = $x + 10$ .

Step 3: Express their future ages.
Age of father 10 years hence = $(3x + 10) + 10 = 3x + 20$ .
Age of son 10 years hence = $(x + 10) + 10 = x + 20$ .

Step 4: Form a proportion using the future ages.
The ratio of their ages 10 years hence will be $5:3$ .

\frac{3x + 20}{x + 20} = \frac{5}{3}

Step 5: Solve the equation for $x$ .

3(3x + 20) = 5(x + 20)

9x + 60 = 5x + 100

4x = 40

x = 10

Step 6: Calculate their present ages and find the ratio.
Present age of father = $3x + 10 = 3(10) + 10 = 40$ years.
Present age of son = $x + 10 = 10 + 10 = 20$ years.
Ratio of present ages = $40 : 20$ , which simplifies to $2:1$ .

Result:
Answer: \boxed{2:1}
"
:::

:::question type="MSQ" question="The monthly salaries of P and Q are in the ratio $3:4$ . The monthly salaries of Q and R are in the ratio $2:3$ . If the total monthly salary of P, Q, and R is ₹130,000, which of the following statements is/are correct?" options=["The monthly salary of Q is ₹40,000","The monthly salary of R is ₹60,000","The difference between the salaries of R and P is ₹30,000","The salary of P is half the salary of R"] answer="The monthly salary of Q is ₹40,000,The monthly salary of R is ₹60,000,The difference between the salaries of R and P is ₹30,000,The salary of P is half the salary of R" hint="First, find the combined ratio P:Q:R. Then use the total salary to find the value of each person's salary." solution="
Step 1: Find the combined ratio $P:Q:R$ .
We are given $P:Q = 3:4$ and $Q:R = 2:3$ .
To combine them, we need to make the term for Q the same in both ratios. The LCM of 4 and 2 is 4.
The first ratio is already $P:Q = 3:4$ .
Multiply the second ratio by 2: $Q:R = (2\times2) : (3\times2) = 4:6$ .
Now we can combine them: $P:Q:R = 3:4:6$ .

Step 2: Use the combined ratio and the total salary to find the proportionality constant.
Let the salaries of P, Q, and R be $3k, 4k$ , and $6k$ .
Total salary = $3k + 4k + 6k = 13k$ .

13k = 130000

k = 10000

Step 3: Calculate the individual salaries.
Salary of P = $3k = 3 \times 10000 = ₹30,000$ .
Salary of Q = $4k = 4 \times 10000 = ₹40,000$ .
Salary of R = $6k = 6 \times 10000 = ₹60,000$ .

Step 4: Evaluate each option.

Option A: The monthly salary of Q is ₹40,000. This is correct.

Option B: The monthly salary of R is ₹60,000. This is correct.

Option C: The difference between the salaries of R and P is $₹60,000 - ₹30,000 = ₹30,000$ . This is correct.

Option D: The salary of P (₹30,000) is half the salary of R (₹60,000). This is correct.

Result:
Answer: \boxed{\text{All four statements are correct.}}
"
:::

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Summary

❗ Key Takeaways for GATE

Proportionality Constant: Always convert a ratio $a:b$ into algebraic terms $ak, bk$ before performing calculations. This is the most fundamental step.

Sum of Terms Divisibility: The total of quantities in a ratio $a:b:c...$ must be divisible by $(a+b+c+...)$ . This is a powerful shortcut for eliminating options in MCQs.

Handling Changes: For problems where quantities are altered, set up an equation by representing the initial state with a constant $k$ , applying the changes, and equating the result to the new ratio.

Speed and Distance: Remember that for a fixed time, the ratio of speeds equals the ratio of distances covered ( $\frac{S_1}{S_2} = \frac{D_1}{D_2}$ ).

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What's Next?

💡 Continue Learning

This topic connects to:

Percentages: Ratios can be expressed as percentages and vice versa. Problems often combine these concepts, for example, by stating that one quantity is 20% more than another, which implies a ratio of $120:100$ or $6:5$ .

Mixtures and Alligations: The core of mixture problems involves finding the ratio in which two or more ingredients are to be mixed to produce a mixture of a desired characteristic.

Profit and Loss: Ratios are used to compare cost price, selling price, and profit, such as the ratio of cost price to selling price.

Master these connections for comprehensive GATE preparation!

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💡 Moving Forward

Now that you understand Ratio and Proportion, let's explore Percentages which builds on these concepts.

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Part 2: Percentages

Introduction

The concept of percentages is a cornerstone of quantitative aptitude and numerical computation, forming the bedrock upon which many complex problems in the GATE examination are built. A thorough command of percentages is not merely about performing calculations; it is about understanding relationships, comparisons, and changes in quantities in a standardized manner. From analyzing profit and loss in business scenarios to interpreting statistical data and calculating error margins, percentages provide a universal language for relative measurement.

In this chapter, we shall systematically explore the fundamental principles of percentages. We will begin with the core definitions and basic calculations, progressively advancing to more intricate applications such as successive percentage changes, percentage error, and the commercially vital concepts of profit, loss, and discount. Our focus will remain steadfastly on the problem-solving methodologies and conceptual clarity required to confidently address any percentage-based question that may appear in the GATE examination.

📖 Percentage

A percentage is a number or ratio expressed as a fraction of 100. It is a dimensionless number denoted using the percent sign, $\%$ . If we wish to express a part-to-whole relationship, the percentage value is obtained by multiplying the ratio $\frac{\text{Part}}{\text{Whole}}$ by 100.

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Foundational Concepts

1. Conversions: Fractions, Decimals, and Percentages

The ability to fluidly transition between fractions, decimals, and percentages is fundamental. This interchangeability is the key to simplifying complex calculations.

Percentage to Fraction: To convert $x \%$ to a fraction, we write it as $\frac{x}{100}$ . For example, $25\% = \frac{25}{100} = \frac{1}{4}$ .
Fraction to Percentage: To convert a fraction $\frac{a}{b}$ to a percentage, we compute $(\frac{a}{b} \times 100) \%$ . For instance, $\frac{3}{5} = (\frac{3}{5} \times 100)\% = 60\%$ .
Percentage to Decimal: To convert $x \%$ to a decimal, we divide by 100, which is equivalent to shifting the decimal point two places to the left. For instance, $72.5\% = 0.725$ .
Decimal to Percentage: To convert a decimal to a percentage, we multiply by 100, or shift the decimal point two places to the right. For instance, $0.42 = 42\%$ .

2. Percentage Change

One of the most frequent applications of percentages in competitive examinations involves the calculation of change over time or between two values. This could be a change in price, population, or any other measurable quantity.

📐 Percentage Change

\text{Percentage Change} = \frac{\text{Final Value} - \text{Initial Value}}{\text{Initial Value}} \times 100

Variables:

Final Value: The value after the change has occurred.

Initial Value: The original value before the change.

When to use: This formula is used to quantify the relative increase or decrease of a quantity. A positive result signifies a percentage increase, while a negative result indicates a percentage decrease.

Worked Example:

Problem: The monthly salary of an employee was ₹40,000. After an appraisal, it was increased to ₹46,000. Calculate the percentage increase in the salary.

Solution:

Step 1: Identify the initial and final values.

\text{Initial Value} = 40,000

\text{Final Value} = 46,000

Step 2: Apply the percentage change formula.

\text{Percentage Change} = \frac{46,000 - 40,000}{40,000} \times 100

Step 3: Simplify the expression.

\text{Percentage Change} = \frac{6,000}{40,000} \times 100

\text{Percentage Change} = \frac{6}{40} \times 100 = \frac{3}{20} \times 100

Step 4: Compute the final answer.

\text{Percentage Change} = 15\%

Answer: The percentage increase in the salary is $15\%$ .

3. Successive Percentage Change

It is a common scenario for a quantity to undergo multiple percentage changes in succession. A frequent error is to simply add or subtract the percentages; this is incorrect because the base value upon which the subsequent percentage is calculated changes after the first modification.

Consider a quantity that is first changed by $A \%$ and then by $B \%$ . The net effective percentage change is not simply $(A+B) \%$ . We must account for the change in the base.

📐 Successive Percentage Change

\text{Net Change} \% = \left( A + B + \frac{AB}{100} \right) \%

Variables:

$A$ : The first percentage change.

$B$ : The second percentage change.

When to use: Use this formula when a quantity undergoes two consecutive percentage changes. Note that increases should be taken as positive values (e.g.,

+20

for a

20 \%

increase) and decreases as negative values (e.g.,

-10

for a

10 \%

decrease).

Worked Example:

Problem: The price of a product is first increased by $20 \%$ and then decreased by $10 \%$ . Find the net percentage change in the final price.

Solution:

Step 1: Identify the values of $A$ and $B$ .

The first change is a $20 \%$ increase, so $A = +20$ .
The second change is a $10 \%$ decrease, so $B = -10$ .

Step 2: Apply the successive percentage change formula.

\text{Net Change} \% = \left( 20 + (-10) + \frac{(20)(-10)}{100} \right) \%

Step 3: Simplify the expression.

\text{Net Change} \% = \left( 20 - 10 - \frac{200}{100} \right) \%

\text{Net Change} \% = (10 - 2) \%

Step 4: Compute the final answer.

\text{Net Change} \% = 8\%

Answer: The net change is an $8\%$ increase in the price.

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Applications in Profit, Loss, and Discount

The commercial application of percentages is a topic of great importance for GATE. Understanding the terminology and the relationships between different price points is crucial.

1. Core Terminology

Let us first define the fundamental terms used in profit and loss calculations.

Cost Price (CP): The price at which an item is purchased or manufactured.

Selling Price (SP): The price at which an item is sold.

Marked Price (MP) or List Price: The price printed on the item's label. This is the price at which the seller intends to sell the item before any discounts.

Profit (or Gain): Occurs when $SP > CP$ . Profit = $SP - CP$ .

Loss: Occurs when $SP < CP$ . Loss = $CP - SP$ .

Discount: A reduction offered on the Marked Price. Discount = $MP - SP$ .

The relationship can be visualized as follows:

Cost Price (CP)

+ Profit

Selling Price (SP)

- Discount

Marked Price (MP)

2. Key Formulas

The following formulas are essential for solving problems in this domain.

📐 Profit and Loss Percentage

\text{Profit} \% = \left( \frac{SP - CP}{CP} \right) \times 100

\text{Loss} \% = \left( \frac{CP - SP}{CP} \right) \times 100

Variables:

$CP$ = Cost Price

$SP$ = Selling Price

When to use: To calculate the percentage profit or loss. It is critical to remember that the base for these calculations is always the Cost Price, unless explicitly specified otherwise.

📐 Discount Percentage

\text{Discount} \% = \left( \frac{MP - SP}{MP} \right) \times 100

Variables:

$MP$ = Marked Price

$SP$ = Selling Price

When to use: To calculate the discount offered as a percentage of the marked price.

3. Special Case: Constant Selling Price

A classic problem type involves two articles sold at the same selling price, one at a gain of $x \%$ and the other at a loss of $x \%$ . In such a scenario, there is always an overall loss.

Let the selling price of each article be $S$ .
Let the profit and loss percentage be $x \%$ .

Cost price of the first article ( $CP_1$ ):

S = CP_1 \left(1 + \frac{x}{100}\right) \implies CP_1 = \frac{S}{1 + x/100}

Cost price of the second article ( $CP_2$ ):

S = CP_2 \left(1 - \frac{x}{100}\right) \implies CP_2 = \frac{S}{1 - x/100}

Total Selling Price = $2S$
Total Cost Price = $CP_1 + CP_2$

\begin{aligned}\text{Total Cost Price} & = S \left( \frac{1}{1+x/100} + \frac{1}{1-x/100} \right) \\ & = S \left( \frac{(1-x/100) + (1+x/100)}{(1+x/100)(1-x/100)} \right) \\ & = S \left( \frac{2}{1 - (x/100)^2} \right) \\ & = \frac{2S}{1 - (x/100)^2}\end{aligned}

Since $1 - (x/100)^2 < 1$ , we have Total CP > Total SP, indicating a loss. The overall loss percentage can be shown to be:

\text{Overall Loss} \% = \left( \frac{x^2}{100} \right) \%

💡 Exam Shortcut: Constant SP

If two items are sold at the same selling price, one at a profit of $x \%$ and the other at a loss of $x \%$ , the transaction always results in a net loss. The net loss percentage is given by:

\text{Net Loss} \% = \frac{x^2}{100} \%

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Advanced Percentage Applications

1. Percentage Error

Percentage error measures the discrepancy between an observed or calculated value and a true or accepted value. It is a common way to express the precision of a calculation.

📐 Percentage Error

\text{Percentage Error} = \frac{|\text{True Value} - \text{Measured Value}|}{\text{True Value}} \times 100

Variables:

True Value: The correct or expected value.

Measured Value: The observed or incorrect value.

When to use: To find the relative error of a measurement or calculation as a percentage of the correct value. The absolute value in the numerator ensures the error is always positive.

Worked Example:

Problem: A researcher was required to add a number $N$ to itself. Instead, the researcher multiplied the number $N$ by itself. If the resulting percentage error was $1200 \%$ , and $N > 0$ , find the value of $N$ .

Solution:

Step 1: Define the True Value and Measured Value.

The correct operation was addition:

\text{True Value} = N + N = 2N

The incorrect operation was multiplication:

\text{Measured Value} = N \times N = N^2

Step 2: Set up the percentage error equation.

\text{Percentage Error} = \frac{|\text{True Value} - \text{Measured Value}|}{\text{True Value}} \times 100

Substituting the values:

1200 = \frac{|2N - N^2|}{2N} \times 100

Step 3: Solve the equation for $N$ .

\begin{aligned}1200 & = \frac{|2N - N^2|}{2N} \times 100 \\12 & = \frac{|2N - N^2|}{2N} \\24N & = |2N - N^2|\end{aligned}

Since

N > 0

, we can write this as

24N = |N(2-N)|

. This gives two possibilities:

Case 1: $2N - N^2 \ge 0 \implies N(2-N) \ge 0 \implies 0 < N \le 2$ .
In this case, $24N = N(2-N)$ .
Since $N > 0$ , we can divide by $N$ :
$24 = 2-N$
$N = 2 - 24$
$N = -22$ . This is not possible as $N > 0$ and $N \le 2$ .

Case 2: $2N - N^2 < 0 \implies N(2-N) < 0 \implies N > 2$ (since $N > 0$ ).
In this case, $24N = -(2N - N^2) = N^2 - 2N$ .
Since $N > 0$ , we can divide by $N$ :
$24 = N - 2$
$N = 24 + 2$
$N = 26$ . This is consistent with $N > 2$ .

Answer: $N = \boxed{26}$ .

2. Population Subsets and Nested Percentages

Many problems in GATE involve calculating percentages of a subset of a population. For example, $X \%$ of students are from one branch, and of those students, $Y \%$ are female. The key is to correctly identify the base for each percentage calculation.

Worked Example:

Problem: In an organization, $60 \%$ of employees are engineers. Of the remaining employees, $50 \%$ are managers. If $75 \%$ of engineers and $80 \%$ of managers are postgraduates, what percentage of the total employees are postgraduates?

Solution:

Step 1: Assume a convenient total number of employees. Let the total number of employees be $100$ .

Step 2: Calculate the number of employees in each category.

Number of engineers = $60 \%$ of $100 = 60$ .
Remaining employees = $100 - 60 = 40$ .
Number of managers = $50 \%$ of remaining employees = $50 \%$ of $40 = 20$ .

Step 3: Calculate the number of postgraduates in each category.

Postgraduate engineers = $75 \%$ of engineers = $0.75 \times 60 = 45$ .
Postgraduate managers = $80 \%$ of managers = $0.80 \times 20 = 16$ .

Step 4: Calculate the total number of postgraduates and express it as a percentage of the total.

Total postgraduates = Postgraduate engineers + Postgraduate managers

\text{Total Postgraduates} = 45 + 16 = 61

Percentage of postgraduates in the organization:

\text{Percentage} = \frac{\text{Total Postgraduates}}{\text{Total Employees}} \times 100 = \frac{61}{100} \times 100 = 61\%

Answer: $\boxed{61\%}$ of the total employees are postgraduates.

---

Problem-Solving Strategies

💡 Assume a Convenient Base

In problems that deal exclusively with percentages and ratios without any absolute values, assuming a base value of $100$ can significantly simplify the arithmetic. As demonstrated in the population subset example, this converts all percentage calculations into simple operations on the number $100$ .

💡 The Base Value Method

The most critical step in any percentage problem is to correctly identify the base (the "whole" or the denominator) for the calculation. Phrases like "percent of the remaining" or "percent of those who qualified" signal that the base has changed from the overall total to a specific subset. Always ask: "percent of what?"

---

Common Mistakes

⚠️ Avoid These Errors

❌ Adding/Subtracting Percentages in Successive Change: A $20\%$ increase followed by a $20\%$ decrease does not result in a $0\%$ net change. The second percentage is calculated on a new, larger base.

✅ Correct Approach: Use the successive change formula:

A + B + \frac{AB}{100}

. For a

20\%

increase and

20\%

decrease, this is

20 - 20 + \frac{(20)(-20)}{100} = -4\%

, a net

4\%

loss.

❌ Using Selling Price as Base for Profit/Loss: Calculating profit percentage as $\frac{\text{Profit}}{SP} \times 100$ .

✅ Correct Approach: Profit and loss percentages are always calculated with the Cost Price (

CP

) as the base, unless the problem explicitly states otherwise.

❌ Confusing Discount and Markup: Calculating discount on the cost price or markup on the selling price.

✅ Correct Approach: Discount is always calculated on the Marked Price (

MP

). Markup is the amount added to the Cost Price (

CP

) to arrive at the

MP

---

Practice Questions

:::question type="MCQ" question="The population of a town increases by $10\%$ in the first year and decreases by $10\%$ in the second year. If the population at the end of the second year is 49,500, what was the population at the beginning of the first year?" options=["49,000","50,000","50,500","51,000"] answer="50,000" hint="Use the successive percentage change concept or set up an algebraic equation. Let the initial population be P." solution="
Step 1: Let the initial population be $P$ .
After the first year ( $10\%$ increase), the population becomes $P \times (1 + \frac{10}{100}) = 1.1P$ .

Step 2: In the second year, the population decreases by $10\%$ . The new base is $1.1P$ .
The final population is $(1.1P) \times (1 - \frac{10}{100}) = (1.1P) \times 0.9$ .

Step 3: Simplify the expression for the final population.
Final Population = $0.99P$ .

Step 4: Equate this to the given final population and solve for $P$ .

0.99P = 49,500

P = \frac{49,500}{0.99} = \frac{4,950,000}{99}

P = 50,000

Result: The initial population was 50,000.
Answer: \boxed{50,000}
"
:::

:::question type="NAT" question="A number is incorrectly divided by 4 instead of being multiplied by 4. If the correct value was positive, what is the percentage error in the result?" answer="93.75" hint="Define the correct (True Value) and incorrect (Measured Value) operations. Apply the percentage error formula." solution="
Step 1: Let the number be $x$ .

Step 2: Define the True Value and Measured Value.
True Value (correct operation) =

x \times 4 = 4x

Measured Value (incorrect operation) =

x \div 4 = \frac{x}{4}

Step 3: Apply the percentage error formula.

\text{Percentage Error} = \frac{|\text{True Value} - \text{Measured Value}|}{\text{True Value}} \times 100

\text{Percentage Error} = \frac{|4x - \frac{x}{4}|}{4x} \times 100

Step 4: Simplify the expression.

\text{Percentage Error} = \frac{|\frac{16x - x}{4}|}{4x} \times 100

\text{Percentage Error} = \frac{\frac{15x}{4}}{4x} \times 100

\text{Percentage Error} = \frac{15x}{16x} \times 100 = \frac{15}{16} \times 100

Step 5: Calculate the final value.

\text{Percentage Error} = 0.9375 \times 100 = 93.75\%

Result: The percentage error is 93.75.
Answer: \boxed{93.75}
"
:::

:::question type="MCQ" question="A shopkeeper sells two laptops for ₹30,000 each. On one, he gains $25\%$ , and on the other, he loses $25\%$ . What is his net gain or loss in the entire transaction?" options=["No profit, no loss","6.25% profit","6.25% loss","12.5% loss"] answer="6.25% loss" hint="This is a classic case of two items sold at the same price with equal profit and loss percentages. A shortcut formula can be applied." solution="
Step 1: Identify the scenario. Two items are sold at the same selling price (₹30,000 each), with a profit of $x\%$ on one and a loss of $x\%$ on the other, where $x=25$ .

Step 2: Recall the shortcut formula for this specific case. The transaction always results in a net loss.

\text{Net Loss} \% = \frac{x^2}{100} \%

Step 3: Substitute the value of $x=25$ into the formula.

\text{Net Loss} \% = \frac{25^2}{100} \% = \frac{625}{100} \%

\text{Net Loss} \% = 6.25\%

Result: The shopkeeper has a net loss of $6.25\%$ in the transaction.
Answer: \boxed{6.25\% \text{ loss}}
"
:::

:::question type="MSQ" question="In a class of 150 students, $40\%$ are girls. $60\%$ of the boys and $80\%$ of the girls passed an examination. Which of the following statements is/are correct?" options=["The number of boys who passed is 54.","The total number of students who passed is 102.","The number of girls who failed is 12.","The pass percentage of the entire class is 70%."] answer="A,B,C" hint="Calculate the number of boys and girls first. Then, calculate the number of passed/failed students for each group and for the total class." solution="
Step 1: Calculate the number of boys and girls.
Total students = 150.
Number of girls = $40\%$ of $150 = 0.40 \times 150 = 60$ .
Number of boys = $150 - 60 = 90$ .

Step 2: Evaluate option A.
Number of boys who passed = $60\%$ of $90 = 0.60 \times 90 = 54$ .
Statement A is correct.

Step 3: Evaluate option B.
Number of girls who passed = $80\%$ of $60 = 0.80 \times 60 = 48$ .
Total students who passed = (Boys passed) + (Girls passed) = $54 + 48 = 102$ .
Statement B is correct.

Step 4: Evaluate option C.
Number of girls who failed = Total girls - Girls who passed = $60 - 48 = 12$ .
(Alternatively, girls who failed = $20\%$ of $60 = 0.20 \times 60 = 12$ ).
Statement C is correct.

Step 5: Evaluate option D.
Pass percentage of the entire class =

\frac{\text{Total passed}}{\text{Total students}} \times 100 = \frac{102}{150} \times 100

\frac{102}{150} \times 100 = \frac{1020}{15} = 68\%

The pass percentage is

68\%

, not

70\%

.
Statement D is incorrect.

Result: The correct statements are A, B, and C.
Answer: \boxed{\text{A, B, C}}
"
:::

---

Summary

❗ Key Takeaways for GATE

Base is King: The most critical aspect of any percentage problem is identifying the correct base value for the calculation. A change in wording (e.g., "of the remainder") often implies a change in the base.

Master the Formulas: Be fluent with the formulas for Percentage Change ( $\frac{\Delta V}{V_i} \times 100$ ), Successive Change ( $A+B+\frac{AB}{100}$ ), and Percentage Error ( $\frac{|\text{True} - \text{Measured}|}{\text{True}} \times 100$ ).

Profit/Loss/Discount Chain: Understand the relationship: Profit/Loss is based on Cost Price ( $CP$ ), while Discount is based on Marked Price ( $MP$ ). The Selling Price ( $SP$ ) connects these two domains.

Utilize Shortcuts Wisely: For specific, recurring patterns like the constant selling price problem (resulting in a $\frac{x^2}{100}\%$ loss), shortcuts can save valuable time. However, always be certain that the problem conditions perfectly match the shortcut's assumptions.

---

What's Next?

💡 Continue Learning

A strong foundation in percentages is a prerequisite for several other quantitative topics.

Ratios and Proportions: Percentages are simply a specific type of ratio where the denominator is 100. Many problems combine concepts from both topics.
Data Interpretation: Charts, graphs, and tables presented in GATE frequently use percentages to represent data. Your ability to quickly calculate and compare percentages is essential for this section.
Simple and Compound Interest: Interest rates are expressed as percentages. The entire framework of interest calculations is built upon the principles of percentage increase over time.

Master these connections to build a comprehensive and robust understanding of quantitative aptitude for the GATE examination.

---

Chapter Summary

📖 Ratios, Percentages, and Proportions - Key Takeaways

In this chapter, we have explored the fundamental concepts of ratios, percentages, and proportions, which form the bedrock of quantitative aptitude. For success in the GATE examination, it is imperative that the following key principles are thoroughly understood and committed to memory.

Ratio as a Fundamental Comparison: A ratio $a:b$ is an ordered comparison of two quantities. We have seen that operations such as finding the compounded ratio, the duplicate ratio ( $a^2:b^2$ ), and the triplicate ratio ( $a^3:b^3$ ) are essential tools for solving complex problems.

The Principle of Proportionality: A proportion is an equality of two ratios. The fundamental property, if $a:b::c:d$ , then $ad = bc$ (product of extremes equals product of means), is the basis for solving problems involving the fourth, third, and mean proportionals.

Percentage as a Standardized Fraction: A percentage is simply a ratio with a denominator of 100. Fluency in converting between percentages, fractions, and decimals is a critical skill that significantly enhances calculation speed. For instance, recognizing that $12.5\%$ is equivalent to $\frac{1}{8}$ is indispensable.

Percentage Change and Successive Changes: The formula for percentage change, $\frac{\text{Change}}{\text{Original Value}} \times 100$ , is universally applicable. It is crucial to remember that successive percentage changes of $a\%$ and $b\%$ are not additive; the net effective change is given by $(a + b + \frac{ab}{100})\%$ .

Direct and Inverse Variation: We established that many quantitative relationships can be modeled using direct proportion ( $y \propto x$ ) or inverse proportion ( $y \propto \frac{1}{x}$ ). The ability to correctly identify the relationship between variables is key to setting up and solving word problems.

Application in Mixtures and Alligations: Ratios and percentages are the core of mixture problems. The rule of alligation is a powerful and efficient technique derived from these principles to determine the ratio in which ingredients must be mixed to achieve a desired concentration or price.

---

Chapter Review Questions

:::question type="MCQ" question="The monthly salaries of two individuals, Aarav and Bhuvan, are in the ratio $5:8$ . If Aarav's salary is increased by 20% and Bhuvan's salary is decreased by 25%, the new sum of their salaries becomes ₹72,000. What was the original monthly salary of Aarav?" options=["₹30,000","₹48,000","₹36,000","₹40,000"] answer="A" hint="Represent the original salaries using a common variable, say $x$ . Apply the percentage changes to these expressions and form an equation based on the new sum." solution="
Let the original monthly salaries of Aarav and Bhuvan be $5x$ and $8x$ , respectively.

Aarav's salary is increased by 20%.
The new salary of Aarav is $5x + 20\%$ of $5x$ .

\text{New Salary (Aarav)} = 5x \times \left(1 + \frac{20}{100}\right) = 5x \times 1.20 = 6x

Bhuvan's salary is decreased by 25%.
The new salary of Bhuvan is $8x - 25\%$ of $8x$ .

\text{New Salary (Bhuvan)} = 8x \times \left(1 - \frac{25}{100}\right) = 8x \times 0.75 = 6x

The new sum of their salaries is given as ₹72,000.

\text{New Salary (Aarav)} + \text{New Salary (Bhuvan)} = 72,000

6x + 6x = 72,000

12x = 72,000

x = \frac{72,000}{12} = 6,000

The question asks for the original monthly salary of Aarav, which was $5x$ .

\text{Original Salary (Aarav)} = 5 \times 6,000 = 30,000

Thus, the original monthly salary of Aarav was ₹30,000.
Answer: $\boxed{₹30,000}$
"
:::

:::question type="NAT" question="A 60-litre vessel contains a mixture of milk and water with milk constituting 80%. From this vessel, 12 litres of the mixture is taken out and replaced with an equal amount of pure water. If this process is repeated one more time, what is the final percentage of milk in the vessel?" answer="51.2" hint="Use the formula for replacement: Final Quantity = Initial Quantity $\times (1 - \frac{\text{amount replaced}}{\text{total volume}})^n$ , where $n$ is the number of times the process is repeated." solution="
First, we determine the initial quantity of milk in the vessel.
Total volume of mixture = 60 litres.
Initial percentage of milk = 80%.

\text{Initial quantity of milk} = 80\% \text{ of } 60 = \frac{80}{100} \times 60 = 48 \text{ litres}

The process involves removing a quantity of the mixture and replacing it with pure water. This affects only the quantity of milk, as the total volume of the mixture is restored to 60 litres after each operation. We can calculate the final quantity of milk using the replacement formula.

Here:

Initial quantity of milk = 48 litres

Total volume of the vessel = 60 litres

Quantity replaced in each step = 12 litres

Number of times the process is repeated, $n = 2$

The formula for the final quantity of a component after

n

replacements is:

\text{Final Quantity} = \text{Initial Quantity} \times \left(1 - \frac{\text{Quantity Replaced}}{\text{Total Volume}}\right)^n

\text{Final Milk} = 48 \times \left(1 - \frac{12}{60}\right)^2

\text{Final Milk} = 48 \times \left(1 - \frac{1}{5}\right)^2

\text{Final Milk} = 48 \times \left(\frac{4}{5}\right)^2

\text{Final Milk} = 48 \times \frac{16}{25} = \frac{768}{25} = 30.72 \text{ litres}

To find the final percentage of milk, we compare the final quantity of milk to the total volume of the mixture.

\text{Final Percentage of Milk} = \frac{\text{Final Milk}}{\text{Total Volume}} \times 100

\text{Final Percentage of Milk} = \frac{30.72}{60} \times 100 = 0.512 \times 100 = 51.2\%

Therefore, the final percentage of milk in the vessel is 51.2.
Answer: $\boxed{51.2}$
"
:::

:::question type="MCQ" question="The cost of a diamond is directly proportional to the square of its weight. A diamond weighing 20 grams, with an original price of ₹50,000, breaks into two pieces whose weights are in the ratio $2:3$ . What is the total financial loss incurred due to this breakage?" options=["₹20,000","₹24,000","₹26,000","₹30,000"] answer="B" hint="First, establish the constant of proportionality from the original diamond's price and weight. Then, calculate the weights of the two smaller pieces and find their individual costs using the same proportionality rule." solution="
Let $C$ be the cost of the diamond and $W$ be its weight.
According to the problem, $C \propto W^2$ , which can be written as $C = kW^2$ , where $k$ is the constant of proportionality.

For the original diamond:
$W = 20$ grams and $C = ₹50,000$ .
We can find the value of $k$ :

50,000 = k \times (20)^2

50,000 = k \times 400

k = \frac{50,000}{400} = 125

So, the cost formula is

C = 125 W^2

The diamond breaks into two pieces with weights in the ratio $2:3$ . The total weight is still 20 grams.
Weight of the first piece, $W_1 = \frac{2}{2+3} \times 20 = \frac{2}{5} \times 20 = 8$ grams.
Weight of the second piece, $W_2 = \frac{3}{2+3} \times 20 = \frac{3}{5} \times 20 = 12$ grams.

Now, we calculate the cost of each piece using the formula $C = 125 W^2$ .
Cost of the first piece, $C_1 = 125 \times (8)^2 = 125 \times 64 = ₹8,000$ .
Cost of the second piece, $C_2 = 125 \times (12)^2 = 125 \times 144 = ₹18,000$ .

The total value of the two broken pieces is the sum of their individual costs:

\text{New Total Value} = C_1 + C_2 = 8,000 + 18,000 = ₹26,000

The financial loss is the difference between the original price and the new total value.

\text{Loss} = \text{Original Price} - \text{New Total Value}

\text{Loss} = 50,000 - 26,000 = ₹24,000

The total loss incurred is ₹24,000.
Answer: $\boxed{₹24,000}$
"
:::

:::question type="NAT" question="In an election between two candidates, 10% of the voters on the voters' list did not cast their votes, and 60 votes cast were found to be invalid. The successful candidate secured 47% of all the voters on the list and defeated his rival by 308 votes. What is the total number of voters on the list?" answer="6200" hint="Let the total number of voters be $V$ . Express the number of votes cast, valid votes, and votes for each candidate in terms of $V$ . Use the winning margin to form the final equation." solution="
Let the total number of voters on the voters' list be $V$ .

Step 1: Determine the number of votes cast.
10% of voters did not vote, so 90% of voters cast their votes.

\text{Votes Cast} = (100\% - 10\%) \text{ of } V = 90\% \text{ of } V = 0.90V

Step 2: Determine the number of valid votes.
Out of the votes cast, 60 were invalid.

\text{Valid Votes} = \text{Votes Cast} - 60 = 0.90V - 60

Step 3: Determine the votes for the winner and the loser.
The winner secured 47% of the total voters on the list ( $V$ ).

\text{Winner's Votes} = 47\% \text{ of } V = 0.47V

The loser's votes are the remaining valid votes.

\text{Loser's Votes} = \text{Valid Votes} - \text{Winner's Votes}

\text{Loser's Votes} = (0.90V - 60) - 0.47V = 0.43V - 60

Step 4: Use the winning margin to form an equation.
The winner defeated the rival by 308 votes.

\text{Winner's Votes} - \text{Loser's Votes} = 308

0.47V - (0.43V - 60) = 308

0.47V - 0.43V + 60 = 308

0.04V = 308 - 60

0.04V = 248

Step 5: Solve for $V$ .

V = \frac{248}{0.04} = \frac{24800}{4} = 6200

The total number of voters on the list is 6200.
Answer: $\boxed{6200}$
"
:::

---

What's Next?

💡 Continue Your GATE Journey

Having completed this chapter on Ratios, Percentages, and Proportions, you have established a firm foundation for many of the most important chapters in Quantitative Aptitude. The principles mastered here are not isolated concepts but rather the essential language through which a significant portion of quantitative reasoning is expressed.

This chapter builds upon your primary understanding of arithmetic and fractions, structuring them into powerful tools for comparison and analysis. As you move forward, you will see these tools applied repeatedly.

Key connections to upcoming chapters include:

Profit and Loss: The principles of percentage increase and decrease you have mastered here are the very language of profit, loss, markup, and discounts.

Simple and Compound Interest: The calculation of interest is a direct application of percentages, and compound interest, in particular, is a study in successive percentage changes over time.

Time & Work and Time, Speed & Distance: The concepts of direct and inverse proportion will be instrumental in solving problems where quantities like work, time, speed, and distance are interrelated.

Data Interpretation: Perhaps most significantly, a strong command of ratios and percentages is non-negotiable for success in Data Interpretation. Nearly all DI sets require the swift and accurate calculation of percentage changes, growth rates, and fractional contributions to a total.

By internalizing the concepts from this chapter, you are now well-equipped to tackle these more advanced applications.

Ratios, Percentages, and Proportions

Ratios, Percentages, and Proportions

Overview

Chapter Contents

Learning Objectives

Part 1: Ratio and Proportion

Introduction

Key Concepts

1. The Proportionality Constant

2. Proportion and its Properties

3. Problems Involving Change in Quantities

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

What's Next?

Part 2: Percentages

Introduction

Foundational Concepts

1. Conversions: Fractions, Decimals, and Percentages

2. Percentage Change

3. Successive Percentage Change

Applications in Profit, Loss, and Discount

1. Core Terminology

2. Key Formulas

3. Special Case: Constant Selling Price

Advanced Percentage Applications

1. Percentage Error

2. Population Subsets and Nested Percentages

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

What's Next?

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Quantitative Aptitude

Basic Geometry

Graphical Data Representation

Powers, Exponents, and Logarithms

Series and Progressions

More Resources

Study Notes

Short Notes

Test Series

Mock Tests

Previous Year Papers

Chapter-wise PYQs

Chapter Practice

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