CAT Quant Study Material PDF

Quantitative Aptitude is one of the most important sections of the CAT exam. It tests your ability to solve mathematical problems. The questions in the Quant section are often challenging, so it is important to prepare well. Let’s go through CAT Quant Notes, do share it with your friends.

Topics	PDF Notes
Handwritten Notes	Download
101 Quant Shortcuts	Download
Quant Tricks	Download
100 Must Solve Quant Questions	Download
Quant Question Bank	Download

CAT Quant Analysis – Total 26 Questions
Area	Slot 1	Slot 2	Slot 3
Algebra	8	7	4
Arithmetic	10	9	10
Geometry	3	5	4
Number System	2	1	3
Modern Maths	3	4	5

CAT 2025 Quantitative Aptitude Syllabus

• Number System
• Geometry & Mensuration
• Percentages
• Speed, Time and Distance
• Profit and Loss
• Simple & Compound Interest
• Time and Work
• Venn diagrams
• Averages
• Algebra – Linear & Quadratic Equations, Maxima-Minima, Inequalities
• LCM and HCF
• Set Theory
• Binomial Theorem
• Complex Numbers
• Ratio and Proportion
• Logarithm
• Progressions
• Inequalities
• Permutation and Combination
• Probability
• Mixtures and Allegations
• Surds and Indices

---

Introduction:

Are you ready to take your CAT preparation to the next level? Quantitative Aptitude (QA) is a crucial section in the CAT exam, and mastering it can significantly boost your overall score! Whether you’re struggling with basic arithmetic or gearing up for advanced topics like geometry and permutations, our comprehensive CAT Quant Notes for 2024 will give you all the tools you need to excel. Quant isn’t just about memorizing formulas – it’s about understanding concepts and applying them efficiently. Plus, did you know that nearly 33% of the CAT exam is comprised of QA? That’s why mastering this section can make a substantial difference in your percentile!

In this guide, we’ll dive deep into the essential topics you need to cover, from arithmetic to algebra, geometry, and beyond. Whether you’re aiming for a 90+ percentile or just trying to clear the cutoff, these notes will help you navigate the Quant section with ease. So grab your pen, and let’s break it down together!

---

1. Arithmetic Essentials for CAT Quant

Arithmetic is the bedrock of Quantitative Aptitude, especially for CAT aspirants. The questions in this segment range from easy to moderate in difficulty but can be deceptive if you’re not careful. Strong basics in arithmetic can serve as a foundation to solve complex problems quickly, which is why many top-scorers start their preparation with this topic. Let’s get into the nitty-gritty of some of the most important subtopics.

Percentages

Percentages are a common concept that appears across various other topics like profit & loss, simple interest, and compound interest. Understanding percentages is key to solving problems efficiently in CAT.

Key Formula:

Consider this: If you can calculate percentages quickly, it will help you immensely in time-bound sections like DI (Data Interpretation). A quick trick to remember is that 10% is just moving the decimal point one place to the left. From there, you can multiply up or down to reach other percentages.

Tip: Always convert percentage values into fractions for ease of calculations. For example, 25% = 1/4, 50% = 1/2, and so on.

Profit and Loss

Profit and loss questions are based on transactions where you’re either making or losing money. These questions are usually straightforward but involve the application of multiple percentage calculations.

Key Formulas:

Quick Tip: Be cautious when interpreting the terms “marked price,” “cost price,” and “selling price.” Misreading these terms can lead to wrong calculations, even if you know the formulas.

Ratio and Proportion

Ratios help compare two quantities, and proportions allow you to establish equivalency between ratios. They appear frequently in CAT and are pivotal for solving problems involving mixtures, partnerships, and time-work.

Key Formula:

This can be used to solve many partnership or mixture problems quickly. For instance, if a solution has milk and water in the ratio 3:2, and you add more water, you can use ratios to figure out the new composition of the solution.

Averages

Averages are one of the easiest subtopics but appear in CAT disguised as more complex problems. The average of a set of numbers is the sum of the numbers divided by the count of numbers.

Key Formula:

​

CAT will often add twists to simple average questions by integrating speed and distance or profits and losses. The key is to identify the core average principle within the problem.

Time and Work

Time and work problems test how efficiently you can manage relationships between time, work, and manpower. They are slightly trickier than other arithmetic problems, but once you understand the core concepts, you’ll ace this section.

Key Formula:

Work Done=Time × Efficiency

If two people are working together, the total work done is the sum of their individual efficiencies.

For example, if Person A can complete a task in 6 hours, and Person B can do it in 8 hours, their combined efficiency will allow them to finish the work faster when working together. A common trick is to use the “work per unit time” approach to simplify calculations.

Practice Problems:

1. A man buys an article for ₹500 and sells it for ₹600. Calculate the profit percentage.
2. The ratio of boys to girls in a class is 3:4. If there are 18 boys, how many girls are there?
3. A person can complete a job in 12 days, while another can do it in 18 days. How long will it take them to finish the work together?

2. Algebra Simplified for CAT Quant

Algebra is often seen as the bridge between basic arithmetic and more advanced quantitative problems. In the CAT exam, algebraic concepts are integral because they’re embedded in many types of problems, from equations to functions. For many students, algebra can be tricky, but once the concepts are clear, algebra becomes one of the most scoring sections in the Quantitative Aptitude part of the CAT exam.

Linear Equations

Linear equations are one of the simplest forms of algebraic expressions you’ll encounter. These are equations of the first degree (i.e., the highest power of the variable is 1). CAT tends to pose questions where you need to solve linear equations involving two or three variables.

Key Formula:

General form: ax+by+c=0

The solution to such an equation is a pair (or triplet) of values that satisfy all the equations in a system. A common trick for CAT is substitution or elimination methods, which help simplify and solve multiple equations.

Example Problem:
Solve: $2x – 7 > 3$

Trick: Always treat inequalities like normal equations, but stay mindful of the reversal rule when dividing or multiplying by a negative number.

Quadratic Equations

Quadratic equations appear frequently in the CAT exam and may seem intimidating at first, but with the right approach, you can solve them quickly. These equations involve terms where the variable has a degree of 2 (hence the name “quadratic”).

Key Formula:

​​

Quadratic equations can be solved either using factorization or the quadratic formula (shown above). CAT often presents these problems in word format, requiring you to convert real-life situations into algebraic equations.

Example Problem:
Solve for Solve for x: $2x^2 – 5x + 3 = 0$

Tip: Don’t forget that quadratic equations can have two roots, and these roots may be real or imaginary depending on the discriminant ($b^2 – 4ac$)

Inequalities

Inequalities are crucial in CAT because they test your ability to think beyond simple equations. Instead of finding a single solution, inequalities require you to determine a range of possible values for a variable.

Key Concepts:

• Linear inequalities: $ax + b > 0$
  
• Quadratic inequalities: $ax^2 + bx + c > 0$
  

When solving inequalities, it’s important to remember that multiplying or dividing by a negative number reverses the inequality sign. CAT loves to incorporate this nuance into its questions.

Example Problem:
Solve: $2x-7>\; 3$

$\mathrm{}$Trick: Always treat inequalities like normal equations, but stay mindful of the reversal rule when dividing or multiplying by a negative number.

Functions and Graphs

Functions may sound abstract, but they’re a key concept in CAT. A function essentially describes a relationship between two variables, where one variable depends on the other.

Key Formula:

$$f(x) = ax^2 + bx + c \quad \text{or} \quad f(x) = ax + b$$

Functions are often expressed graphically, and you may be required to interpret or manipulate the graphs. Understanding the properties of common functions like linear, quadratic, and cubic functions will help you in a variety of CAT questions.

Example Problem:
If $f(x) = 3x + 2$, find $f(4)$

Tip: Visualizing a function graphically can make it easier to understand its behavior, especially for quadratic and cubic functions.

Word Problems in Algebra

Word problems in algebra are where CAT really tests your ability to translate verbal descriptions into mathematical equations. These problems usually describe real-life scenarios and expect you to break them down into solvable equations.

Example Problem:
A man invests ₹1000 at an interest rate of 5% compounded annually. After how many years will his investment double?

In this case, you’d need to convert the scenario into an algebraic equation involving compounding and solve for the variable representing time.

Quick Trick: When dealing with word problems, break the problem down into smaller parts and assign variables to unknowns. This simplifies the process of setting up equations.

Must-Know Formulas and Shortcuts

Algebra is one of the most formula-heavy topics in CAT Quant, so having the essential formulas at your fingertips will save you valuable time during the exam.

• Sum of Roots (Quadratic Equation):
  $x_1 + x_2 = -\frac{b}{a}$
• Product of Roots (Quadratic Equation):
  $x_1 \times x_2 = \frac{c}{a}$
• Difference of Squares:
  $a^2 – b^2 = (a – b)(a + b)$
  
• Expanding a Binomial:
  $(a + b)^2 = a^2 + 2ab + b^2$

Memorizing these formulas and practicing their applications will drastically improve your speed when tackling algebraic problems in CAT.

Common Mistakes to Avoid in Algebra

• Misinterpreting inequality signs: Remember, multiplying or dividing by a negative flips the inequality.
• Forgetting multiple roots in quadratic equations: Always check for two solutions unless it’s a special case.
• Skipping steps in solving word problems: Word problems can seem overwhelming, so take your time to structure the problem logically.

Practice Problems:

1. Solve for x: $x^2 – 7x + 10 = 0$
   
2. Find the value of $x$ that satisfies both $3x + 5y = 7$ and $2x – 4y = 3$
3. What is the range of $x$ if $2x + 3 \geq 5$

---

Algebra is not just about solving equations; it’s about interpreting and simplifying real-life problems into mathematical solutions. Once you grasp the logic behind these concepts, algebra becomes one of the most scoring areas of CAT Quant.

3. Geometry and Mensuration Made Easy

Geometry and Mensuration are crucial areas of CAT Quant, often posing some of the most visual and concept-based problems. While these questions can sometimes seem intimidating, they’re highly predictable and formula-driven. Once you have a good understanding of the fundamental concepts, solving geometry and mensuration questions becomes much easier.

Let’s dive into the essential geometry concepts for CAT, and I’ll walk you through some must-know tips and formulas to ace this section.

---

Basic Concepts of Geometry

Geometry is the study of shapes, sizes, and properties of figures, and it’s a vital part of the CAT Quant section. Many of the problems here involve triangles, circles, polygons, and coordinate geometry.

• Point: A location in space, usually defined by coordinates (x, y) in a 2D plane.
• Line: A straight path extending in both directions without end.
• Plane: A flat surface that extends infinitely in all directions.

These are the building blocks of geometry, and most problems revolve around how different shapes are formed and interact in space.

---

Triangles

Triangles form the foundation of many geometry problems in CAT. The properties of triangles, such as angles and side lengths, are heavily tested, especially in the context of right-angled triangles and Pythagorean theorem-based problems.

Key Properties:

1. The sum of the internal angles of a triangle is always 180°.
2. The sum of the lengths of any two sides of a triangle must be greater than the length of the third side (Triangle Inequality Theorem).

Types of Triangles:

• Equilateral Triangle: All three sides and angles are equal.
• Isosceles Triangle: Two sides and two angles are equal.
• Scalene Triangle: All sides and angles are unequal.
• Right-Angled Triangle: One of the angles is 90°, and it satisfies the Pythagorean theorem, which is $a^2 + b^2 = c^2$

Important Formula:

• Area of a Triangle: Area=1/2×base×height

• Heron’s Formula (for any triangle):

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$

Where $s = \frac{a+b+c}{2}$, the semi-perimeter, and $a, b, c$ are the side lengths.

 

Example Problem:
Find the area of a triangle with sides 7 cm, 8 cm, and 9 cm.

Tip: Right-angled triangles appear frequently in CAT. Memorizing common Pythagorean triplets (like 3-4-5, 5-12-13) can help you solve problems more quickly.

---

Circles

Circles are another commonly tested topic in the CAT exam. Problems related to circles often test your understanding of radius, diameter, area, circumference, and tangents.

Key Properties:

1. The radius (r) is the distance from the center of the circle to any point on its circumference.
2. The diameter (d) is twice the radius: $d=2r$.
3. A tangent is a line that touches the circle at exactly one point and is perpendicular to the radius at that point.

Important Formulas:

• Area of a Circle: Area=πr2
• Circumference of a Circle: Circumference=2πr

 

Example Problem:
Find the circumference of a circle with a radius of 14 cm.

Trick: CAT often includes problems where you have to calculate the area or circumference when only partial information is given, like the diameter instead of the radius. Keep formulas handy and convert when necessary.

---

Polygons

Polygons, particularly regular polygons (where all sides and angles are equal), also frequently show up in CAT Quant problems. The most common polygons tested are triangles, quadrilaterals (like squares and rectangles), and pentagons.

Key Properties:

1. The sum of the interior angles of a polygon with $n$ sides is:

$$\text{Sum of angles} = (n-2) \times 180^\circ$$

2. For a regular polygon (with all sides and angles equal), the measure of each internal angle is:

$$\text{Internal Angle} = \frac{(n-2) \times 180^\circ}{n}$$

Important Formulas:

• Area of a Square:

$$\text{Area} = \text{side}^2$$

• Area of a Rectangle:

$$\text{Area} = \text{length} \times \text{width}$$

Example Problem:
Find the area of a rectangle with length 15 cm and width 10 cm.

---

Coordinate Geometry

Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate plane. In CAT, problems related to finding distances between points, areas of figures, and slopes of lines are common.

Key Formulas:

• Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$:

$$\text{Distance} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

• Slope of a Line (between two points):

$$\text{Slope} = \frac{y_2 – y_1}{x_2 – x_1}$$

• Equation of a Line (in slope-intercept form):

$$y = mx + c$$

Where $m$ is the slope and $c$ is the y-intercept.

Example Problem:
Find the distance between the points (3, 4) and (7, 1).

Tip: Coordinate geometry problems often look more complicated than they are. Start by plotting the points on a graph to better visualize the relationships between them.

---

Mensuration

Mensuration is all about measuring the area, volume, and surface area of various shapes, including 2D figures like squares and circles and 3D solids like cubes, spheres, and cylinders. This topic is formula-driven, so once you’re familiar with the formulas, you can solve problems quickly.

Important Formulas for 2D Shapes:

• Area of a Parallelogram:

$$\text{Area} = \text{base} \times \text{height}$$

• Area of a Trapezium:

$$\text{Area} = \frac{1}{2} \times (a + b) \times h$$

Where $a$ and $b$ are the lengths of the parallel sides, and $h$ is the height.

Important Formulas for 3D Shapes:

• Volume of a Cube:

$$\text{Volume} = \text{side}^3$$

• Volume of a Sphere:

$$\text{Volume} = \frac{4}{3} \pi r^3$$

• Surface Area of a Cylinder:

$$\text{Surface Area} = 2\pi r(r + h)$$

Where $r$ is the radius and $h$ is the height.

• Cone:
  • Curved Surface Area: $$\pi r l$$where $l$ is the slant height.
  • Total Surface Area: $$\pi r (r + l)$$
    
  • Volume: $$\frac{1}{3} \pi r^2 h$$
    

Example Problem:
Find the volume of a cylinder with a radius of 5 cm and height of 10 cm.

Solution:

$$\text{Volume} = \pi r^2 h = \frac{22}{7} \times 5^2 \times 10 = 785.71 \text{ cm}^3$$

Tip: Memorize formulas for surface areas and volumes of 3D shapes as they frequently appear in CAT.

---

Common Mistakes to Avoid in Geometry and Mensuration

1. Misreading the problem: Many students overlook key details, such as whether the problem involves radius or diameter in a circle.
2. Forgetting units: Always keep an eye on the units, especially when dealing with areas (square units) and volumes (cubic units).
3. Mixing formulas: Geometry and mensuration have a lot of formulas. Ensure you’re using the correct one for the problem at hand.

---

Practice Problems:

1. A square and a circle have the same perimeter. Which has a larger area?
2. Find the area of an equilateral triangle with a side length of 10 cm.
3. A cylinder has a radius of 3 cm and a height of 10 cm. What is its surface area?
4. Find the distance between the points (2, 3) and (5, 7).

---

Conclusion: Geometry and Mensuration questions in CAT can be highly rewarding if you have a clear understanding of the basics and memorize the important formulas. Practice as many problems as you can, especially on topics like triangles and circles, which are CAT favorites. A clear visualization of the problem often leads to quicker solutions, so practice sketching figures to aid your understanding.

4. Number Systems and Their Applications

Number Systems form the backbone of many Quantitative Aptitude questions in the CAT exam. These questions, while appearing simple, often test your deeper understanding of numbers, divisibility, and modular arithmetic. Mastering this topic can make a significant difference, as number system problems are common across various difficulty levels.

Let’s explore the important concepts, tips, and tricks that will help you tackle Number System problems with ease.

---

Types of Numbers

The number system can be broadly classified into various types, and understanding these types is crucial for solving problems.

1. Natural Numbers (N):
   These are the counting numbers starting from 1. Example: 1, 2, 3, 4, …
2. Whole Numbers (W):
   These are natural numbers including 0. Example: 0, 1, 2, 3, 4, …
3. Integers (Z):
   These include all positive and negative whole numbers, as well as 0. Example: -2, -1, 0, 1, 2, …
4. Rational Numbers (Q):
   Any number that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Example: $\frac{1}{2}, 0.75, -3$
5. Irrational Numbers:
   Numbers that cannot be expressed as $\frac{p}{q}$ are called irrational numbers. Example: $\sqrt{2}, \pi$
6. Prime Numbers:
   Prime numbers are numbers greater than 1 that have only two divisors: 1 and the number itself. Example: 2, 3, 5, 7, 11, 13, …
7. Composite Numbers:
   Numbers that have more than two divisors. Example: 4, 6, 8, 9, 10, …

Example Problem:
Find the sum of the first five prime numbers.

Trick: For CAT, make sure you memorize the first few prime numbers up to at least 50. This will save you time when solving questions based on divisibility, factorization, or primes.

---

Divisibility Rules

Divisibility rules are shortcuts to help you determine if a number is divisible by another without performing actual division. These rules are particularly useful in number system questions.

1. Divisibility by 2:
   A number is divisible by 2 if its last digit is even.
2. Divisibility by 3:
   A number is divisible by 3 if the sum of its digits is divisible by 3.
3. Divisibility by 4:
   A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
4. Divisibility by 5:
   A number is divisible by 5 if its last digit is 0 or 5.
5. Divisibility by 6:
   A number is divisible by 6 if it is divisible by both 2 and 3.
6. Divisibility by 8:
   A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
7. Divisibility by 9:
   A number is divisible by 9 if the sum of its digits is divisible by 9.
8. Divisibility by 11:
   A number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11.

Example Problem:
Is 7342 divisible by 11?

Tip: Memorizing these divisibility rules helps tremendously in solving number system problems quickly, especially when you’re working with large numbers.

---

Factors and Multiples

Understanding factors and multiples is essential for solving many CAT problems. A factor of a number divides that number completely without leaving a remainder, while a multiple is a number that can be obtained by multiplying that number by an integer.

Key Concepts:

• Prime Factorization:
  Every number can be expressed as a product of prime numbers. Example: 36 = $2^2 \times 3^2$.
• Highest Common Factor (HCF):
  The largest number that divides two or more numbers without leaving a remainder.
• Least Common Multiple (LCM):
  The smallest number that is a multiple of two or more numbers.

Example Problem:
Find the HCF and LCM of 12 and 18.

Trick: Use the relationship between HCF and LCM:

$$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$$

---

Modular Arithmetic

Modular arithmetic is often referred to as “clock arithmetic” because of the way numbers wrap around after reaching a certain value, known as the modulus. This concept is crucial for solving remainder-related problems in the CAT exam.

Key Formula:

$$a \equiv b \pmod{n}$$

This means that when $a$ is divided by $n$, it leaves a remainder $b$.

Example Problem:
Find the remainder when 23 is divided by 7.

Trick: Modular arithmetic problems can often be simplified by breaking them down into smaller steps. For instance, instead of working with large numbers directly, you can reduce them step by step using mod rules.

---

Cyclicity of Numbers

The concept of cyclicity refers to the repeating pattern in the last digits of powers of numbers. This is an extremely useful trick when you’re solving problems involving large powers in CAT.

For example:

• Cyclicity of 2: The last digits repeat in a cycle of 4 (2, 4, 8, 6).
• Cyclicity of 3: The last digits repeat in a cycle of 4 (3, 9, 7, 1).
• Cyclicity of 9: The last digits repeat in a cycle of 2 (9, 1).

Example Problem:
Find the last digit of $3^{45}$.

Trick: Identify the cycle of the last digit and determine the remainder when the power is divided by the length of the cycle. This will tell you where the power falls within the cycle.

---

Sum of N Natural Numbers

Questions asking for the sum of the first $n$ natural numbers or squares and cubes of natural numbers are common in CAT. These problems can be solved using direct formulas.

Important Formulas:

1. Sum of first $n$ natural numbers:

$$S = \frac{n(n+1)}{2}$$

2. Sum of squares of the first $n$ natural numbers:

$$S = \frac{n(n+1)(2n+1)}{6}$$

3. Sum of cubes of the first $n$ natural numbers:

$$S = \left( \frac{n(n+1)}{2} \right)^2$$

Example Problem:
Find the sum of the first 50 natural numbers.

Tip: For quick calculation, use these formulas whenever you encounter summation problems. They’re faster than adding numbers manually!

---

Perfect Numbers, Prime Numbers, and Co-Primes

These concepts are often tested in number system questions, especially in combination with HCF and LCM problems.

1. Perfect Number:
   A number that is equal to the sum of its proper divisors. Example: 6 is a perfect number because 1 + 2 + 3 = 6.
2. Prime Number:
   A number that has exactly two divisors: 1 and itself.
3. Co-Prime Numbers:
   Two numbers are said to be co-prime if their HCF is 1. Example: 8 and 15 are co-prime because their only common divisor is 1.

Example Problem:
Are 14 and 25 co-prime?

Trick: Co-prime problems are usually easy to solve by finding the HCF. If it’s 1, they are co-prime.

---

Concept of Remainders

Remainder problems are a favorite in CAT, especially ones that involve finding the remainder when large powers of numbers are divided by a divisor. The Chinese Remainder Theorem and Euler’s Theorem are advanced techniques used to solve remainder-related problems.

Key Formula (Euler’s Theorem):

$$a^{\phi(n)} \equiv 1 \pmod{n}$$

Where $\phi(n)$ is Euler’s totient function, which counts the number of integers less than $n$ that are co-prime with $n$.

Example Problem:
Find the remainder when $3^{100}$ is divided by 7.

Tip: These problems can be tricky, but using modular arithmetic and understanding the cyclicity of powers can make them easier.

---

Common Mistakes to Avoid in Number Systems

1. Overcomplicating simple problems: Many CAT number system problems are designed to look complex but have simple solutions. Always look for shortcuts and patterns.
2. Ignoring modular arithmetic: This is one of the most useful tools for remainder and cyclicity problems. Make sure you’re comfortable with it.
3. Forgetting divisibility rules: Not applying these rules can lead to errors in factorization and simplification.

---

Practice Problems:

1. Find the HCF and LCM of 24 and 36.
2. Determine if 29 is a prime number.
3. What is the remainder when $4^{37}$ is divided by 5?
4. Find the sum of the squares of the first 10 natural numbers.

---

Conclusion:

Mastering the Number System for CAT involves understanding various types of numbers, applying divisibility rules, working with HCF and LCM, and solving remainder problems using modular arithmetic. This section is not only highly testable but also overlaps with other Quant topics, so a thorough understanding here will help you in multiple areas of the exam.

---

5. Permutations, Combinations, and Probability

Permutations, Combinations, and Probability are fundamental concepts in the CAT Quant section, often challenging students due to their abstract nature. However, with a clear understanding and systematic approach, these topics can be mastered effectively.

---

Permutations

Definition:
A permutation is an arrangement of objects in a specific order. The order of arrangement is crucial, and changing the order results in a different permutation.

Formula:
The number of permutations of $n$ objects taken $r$ at a time is given by:

$$P(n, r) = \frac{n!}{(n – r)!}$$

Where $n!$ denotes the factorial of $n$.

Example Problem:
Calculate the number of ways to arrange 3 books out of a collection of 5.

Solution:
Using the permutation formula:

$$P(5, 3) = \frac{5!}{(5 – 3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60$$

Thus, there are 60 possible arrangements.

Tip: In CAT, permutation problems often involve arranging people, books, or other objects where the order matters. Practice identifying scenarios where permutation is applicable.

---

Combinations

Definition:
A combination is a selection of items from a larger set, where the order of selection does not matter.

Formula:
The number of combinations of $n$ objects taken $r$ at a time is:

$$C(n, r) = \frac{n!}{r! \times (n – r)!}$$

Example Problem:
Determine the number of ways to choose 2 fruits from a basket of 4 different fruits.

Solution:
Applying the combination formula:

$$C(4, 2) = \frac{4!}{2! \times (4 – 2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6$$

So, there are 6 possible selections.

Tip: Use combinations when the order of selection doesn’t matter, such as choosing a committee from a group. Recognizing these scenarios will help in applying the correct formula.

---

Probability

Definition:
Probability measures the likelihood of a particular event occurring. It ranges between 0 and 1, with 0 indicating impossibility and 1 indicating certainty.

Basic Formula:

$$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$

Example Problem:
What is the probability of drawing a red ball from a bag containing 3 red and 2 blue balls?

Solution:
The total number of balls is 5, and the number of favorable outcomes (drawing a red ball) is 3.

$$\text{Probability} = \frac{3}{5} = 0.6$$

Thus, the probability is 0.6.

Tip: In CAT, probability questions often involve scenarios like drawing cards, tossing coins, or selecting items. Familiarize yourself with basic probability principles and common scenarios.

---

Common Mistakes to Avoid

1. Misidentifying the Scenario: Ensure you correctly determine whether to use permutations or combinations based on whether the order of selection matters.
2. Incorrect Formula Application: Always double-check that you’re using the appropriate formula for the given problem.
3. Overlooking Simple Cases: Don’t neglect straightforward scenarios that can be solved with basic counting principles instead of complex formulas.

---

Practice Problems:

1. How many ways can 4 students be seated in a row?
2. In how many ways can a committee of 3 be formed from a group of 7 people?
3. What is the probability of flipping a coin and getting heads?

---

Conclusion:

Mastering permutations, combinations, and probability requires practice and a clear understanding of the underlying principles. These topics are not only vital for the CAT exam but also enhance logical thinking and problem-solving skills. By practicing various problems and familiarizing yourself with different scenarios, you can approach these questions with confidence and improve your overall performance in the Quant section.

---

6. Time, Speed, and Distance: Quick Tricks

Time, Speed, and Distance is one of the most important and frequently tested topics in the CAT Quantitative Aptitude section. Questions from this area can be straightforward, but they often come with twists involving multiple variables like relative speed, average speed, or situations like trains, boats, and races. Mastering this topic requires not only a solid understanding of the basic concepts but also the ability to apply shortcuts and tricks to solve problems quickly.

Let’s break down the essentials of Time, Speed, and Distance, explore some quick tricks, and go through real-world examples to sharpen your understanding of this topic.

---

Understanding the Relationship Between Time, Speed, and Distance

The core formula for Time, Speed, and Distance is simple but fundamental:

$$\text{Distance}=\text{Speed}\times \text{Time}$$

From this, you can derive the other two essential formulas:

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \quad \text{and} \quad \text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

These three variables—Time, Speed, and Distance—are interdependent. If you know any two of them, you can calculate the third. However, CAT-level questions often test your understanding of more nuanced applications of this relationship, such as relative speed, average speed, and variable speeds over different segments of a journey.

---

Key Formulas and Shortcuts for Solving Time, Speed, and Distance Problems

To solve CAT-level Time, Speed, and Distance problems efficiently, you need to memorize a few critical formulas and learn the shortcuts that will save time.

1. Average Speed: When an object moves at different speeds for equal distances, the average speed is given by:

$$\text{Average Speed} = \frac{2 \times S_1 \times S_2}{S_1 + S_2}$$

Where $S_1$ and $S_2$ are the speeds in two different segments.

Example: If a car travels at 40 km/h for the first half of the journey and 60 km/h for the second half, the average speed is:

$$\text{Average Speed} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h}$$

2. Relative Speed: When two objects are moving towards or away from each other, their relative speed is the sum or difference of their individual speeds.

• When moving towards each other: $\text{Relative Speed} = S_1 + S_2$
• When moving in the same direction: $\text{Relative Speed}=\mathrm{\mid }{S}_1-S_2\mathrm{\mid }$

3. Distance Covered in Circular Tracks: For two objects moving in opposite directions on a circular track, the time taken to meet is:

$$\text{Time to Meet} = \frac{\text{Total Distance (Circumference)}}{\text{Relative Speed}}$$

---

Common Pitfalls in Time and Distance Problems

Many students make mistakes in Time, Speed, and Distance questions due to minor oversights or conceptual misunderstandings. Here are a few common pitfalls to avoid:

1. Misinterpreting Average Speed: Average speed is often confused with the arithmetic mean of speeds, but it depends on the total distance and total time taken. Always apply the correct formula for average speed.
2. Forgetting Unit Consistency: Speed, distance, and time must be in compatible units. For example, if speed is in kilometers per hour, time must be in hours, and distance must be in kilometers.
3. Overlooking Relative Motion: In problems involving two moving objects, the relative speed concept is crucial. Forgetting to apply it can lead to incorrect results.
4. Neglecting Time as a Factor: Some questions may focus more on the time taken rather than the speed. Ensure that you’re reading the problem carefully to understand whether time, speed, or distance is the variable of interest.

---

Real-World Examples to Solidify Understanding

Understanding the formulas is one thing, but applying them in real-world scenarios is where many students struggle. Here are a few practical examples that can help solidify your understanding:

Example 1: The Train Problem Two trains are moving towards each other on the same track. Train A is moving at 60 km/h, and Train B is moving at 80 km/h. The distance between them is 300 km. How long will it take for the two trains to meet?

Solution: The relative speed of the two trains (since they are moving towards each other) is:

$$\text{Relative Speed} = 60 + 80 = 140 \text{ km/h}$$

Time taken to meet:

$$\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{300}{140} = 2.14 \text{ hours (approx.)}$$

Example 2: The Boat Problem A boat moves downstream at 15 km/h and upstream at 10 km/h. If the boat takes 4 hours to cover a certain distance downstream, how long will it take to cover the same distance upstream?

Solution: Let the distance be $D$. From the formula:

$$D = \text{Speed} \times \text{Time}$$

Downstream:

$$D = 15 \times 4 = 60 \text{ km}$$

Upstream:

$$\text{Time} = \frac{D}{\text{Speed}} = \frac{60}{10} = 6 \text{ hours}$$

---

CAT-Level Practice Problems with Solutions

Problem 1: A car covers a distance of 360 km in 6 hours. For the first 2 hours, the car was traveling at a speed of 60 km/h. What was the average speed for the remaining journey?

Solution: Distance covered in the first 2 hours:

$$D_1 = 60 \times 2 = 120 \text{ km}$$

Remaining distance:

$$D_2 = 360 – 120 = 240 \text{ km}$$

Time for the remaining journey:

$$T_2 = 6 – 2 = 4 \text{ hours}$$

Average speed for the remaining journey:

$$\text{Speed} = \frac{240}{4} = 60 \text{ km/h}$$

---

Problem 2: A cyclist covers a distance of 150 km, traveling at a speed of 25 km/h for the first half of the journey and 30 km/h for the second half. What is the total time taken by the cyclist?

Solution: First half distance = $\frac{150}{2} = 75 \text{ km}$ Time taken for the first half:

$$T_1 = \frac{75}{25} = 3 \text{ hours}$$

Time taken for the second half:

$$T_2 = \frac{75}{30} = 2.5 \text{ hours}$$

Total time taken:

$$T = T_1 + T_2 = 3 + 2.5 = 5.5 \text{ hours}$$

---

Conclusion:

Time, Speed, and Distance problems can be tackled efficiently if you have a clear understanding of the core concepts and apply the right shortcuts. Whether it’s about calculating average speeds, using relative motion in tricky problems, or ensuring unit consistency, the key lies in practice and applying these strategies to different types of questions.

---

7. Data Interpretation and Logical Reasoning (DILR): The Strategic Approach

The Data Interpretation and Logical Reasoning (DILR) section of the CAT exam is known for its unpredictable yet strategic nature. DILR is less about complex calculations and more about logical thinking, pattern recognition, and analytical skills. This section tests your ability to interpret data, analyze patterns, and make decisions based on limited information.

Let’s explore the core concepts of Data Interpretation and Logical Reasoning, as well as strategies to ace this section.

---

Data Interpretation (DI): Mastering Data Analysis

Data Interpretation (DI) involves analyzing and interpreting large sets of numerical data through graphs, tables, charts, and other visual formats. DI questions are designed to test your ability to draw conclusions from the data presented and answer questions efficiently.

Types of Data Interpretation Questions:

1. Tables: Present numerical data in a tabular format.
2. Bar Charts: Represent data using rectangular bars.
3. Pie Charts: Show data as sectors of a circle, indicating percentages.
4. Line Graphs: Display trends over a period of time.
5. Mixed Graphs: Combine two or more types of graphs.

Key Skills Required:

• Quick Data Analysis: The ability to read data fast and comprehend its meaning.
• Calculation Speed: Basic arithmetic operations like percentages, ratios, and averages are essential.
• Attention to Detail: Identifying patterns, trends, and anomalies in the data is crucial.

Example DI Problem:

Consider a table that shows the sales data of five products (A, B, C, D, E) over four quarters.

Product	Q1	Q2	Q3	Q4
A	100	120	130	140
B	80	90	100	110
C	150	160	170	180
D	70	85	95	105
E	90	110	120	130

Question:
What is the percentage increase in sales of Product B from Q1 to Q4?

Solution:
The percentage increase formula is:

$$\text{Percentage Increase} = \frac{\text{Final Value – Initial Value}}{\text{Initial Value}} \times 100$$

For Product B, the sales in Q1 = 80, and sales in Q4 = 110.

$$\text{Percentage Increase} = \frac{110 – 80}{80} \times 100 = \frac{30}{80} \times 100 = 37.5\%$$

---

Tips to Excel in Data Interpretation:

1. Start with Visual Data: Pie charts and bar graphs are easier to analyze than complex tables. Prioritize them when selecting questions.
2. Approximation is Key: In most DI questions, exact calculations aren’t required. You can save time by approximating numbers for calculations.
3. Scan the Data First: Before diving into calculations, glance at the data to understand trends and focus areas.
4. Practice Mental Math: Quick mental arithmetic can significantly reduce your time spent on each question.
5. Focus on Ratios and Percentages: Many DI problems revolve around percentages and ratios. Master these concepts to tackle DI efficiently.

---

Logical Reasoning (LR): Sharpen Your Logical Thinking

Logical Reasoning (LR) is all about analyzing given information and reasoning through it to arrive at conclusions. These problems are typically presented as puzzles, arrangements, or sequences.

Common Types of LR Questions:

1. Seating Arrangements: Questions where people are arranged in linear or circular patterns with certain conditions.
2. Blood Relations: Questions involving family relationships and determining how individuals are related.
3. Puzzles: Logical problems requiring deductive reasoning to solve complex situations.
4. Syllogisms: Involves deducing conclusions based on given premises.
5. Venn Diagrams: Questions where you analyze intersections and differences between sets.

Example LR Problem:

Four friends (A, B, C, and D) are sitting in a row, but not necessarily in this order. C is sitting to the left of D. A is sitting to the right of B but not next to D. Who is sitting next to B?

Solution:
The arrangement rules are:

• C is to the left of D.
• A is to the right of B, but not next to D.

From this, the only possible arrangement is: B, C, D, A. Therefore, C is sitting next to B.

---

Strategies for Tackling Logical Reasoning:

1. Read the Question Thoroughly: Logical reasoning questions can be tricky with their wording. Always read the conditions carefully and understand them fully before attempting.
2. Draw Diagrams: For seating arrangements, puzzles, or blood relations, draw diagrams to visualize the problem. This helps organize information more effectively.
3. Work Systematically: Approach LR problems methodically. Start by solving the most straightforward clues first and work your way through more complex ones.
4. Use Elimination: If unsure of an answer, eliminate incorrect options based on the given conditions to narrow down possibilities.
5. Time Management: While LR problems may seem tempting to solve, they can be time-consuming. Don’t get stuck on one question; keep an eye on the clock.

---

Integrated DI-LR Sets:

In CAT, you’ll often find questions where Data Interpretation and Logical Reasoning are combined. For example, you may need to analyze data in a table and then use logical reasoning to answer questions about it.

Example Problem:

A company has four employees, and their salaries in thousands for the last three months are as follows:

Employee	Month 1	Month 2	Month 3
X	50	55	60
Y	60	65	70
Z	55	60	65
W	70	75	80

Question:
Which employee had the highest percentage salary increase from Month 1 to Month 3?

Solution:
To find the percentage increase:

$$\text{Percentage Increase} = \frac{\text{Salary in Month 3 – Salary in Month 1}}{\text{Salary in Month 1}} \times 100$$

• For X: $\frac{60 – 50}{50} \times 100 = 20\%$
  
• For Y: $\frac{70 – 60}{60} \times 100 = 16.67\%$
  
• For Z: $\frac{65 – 55}{55} \times 100 = 18.18\%$
• For W: $\frac{80 – 70}{70} \times 100 = 14.29\%$

Thus, Employee X had the highest percentage salary increase.

---

Approach to Solving DI-LR Sets in CAT:

1. Analyze the Data First: For mixed DI-LR sets, start by thoroughly examining the data provided before reading the questions.
2. Prioritize Easy Sets: Not all sets are equally difficult. Identify and attempt the ones that seem familiar or straightforward first.
3. Break Down Complex Problems: If a question seems complex, break it into smaller, manageable parts. Solve one part at a time.
4. Avoid Guessing: In DI-LR, guessing can often lead to wrong answers due to the interdependence of questions within a set. Only mark answers when you are confident.
5. Stay Calm Under Pressure: DI-LR sets can be stressful, especially under time constraints. Stay calm, focus on one question at a time, and manage your time effectively.

---

Practice Problems:

1. A pie chart shows the distribution of expenses in five categories: Rent (25%), Food (20%), Travel (15%), Utilities (10%), and Miscellaneous (30%). If the total expenses amount to ₹40,000, how much is spent on Travel?
2. A group of six friends (A, B, C, D, E, F) are seated in a circle. A is sitting next to B but not next to C. D is sitting directly opposite B. Who is sitting next to C?

---

Conclusion:

The DILR section in CAT can be both challenging and rewarding if approached strategically. It’s not just about solving puzzles or interpreting data but also about how you manage your time and approach different sets. By practicing regularly, learning to identify easy sets, and improving your data interpretation skills, you can master this section and significantly boost your overall CAT score.

---

CAT Quant Questions with Answers

Question 1:

A boat can travel a distance of 36 km downstream in 4 hours. If the speed of the boat in still water is 12 km/hr and the speed of the current is 2 km/hr, what is the speed of the boat in still water (in km/hr) while going upstream?

1. a) 8
2. b) 9
3. c) 10
4. d) 11

Answer: b) 9

Explanation:

Let the speed of the boat in still water while going upstream be x km/hr.

Given, speed downstream = 12 + 2 = 14 km/hr

Time = Distance / Speed

4 = 36 / 14

x = 9

 

Question 2:

The sum of the ages of A and B is 40. Five years ago, the ratio of their ages was 2:3. What will be the sum of their ages after 5 years?

1. a) 45
2. b) 50
3. c) 55
4. d) 60

Answer: c) 55

Explanation:

Let the present age of A be 2x and the present age of B be 3x.

According to the given condition, 2x – 5 / 3x – 5 = 2 / 3

Cross-multiplying, we get 6x – 10 = 4x – 10

Simplifying, we get x = 0

Therefore, the present age of A is 0, and the present age of B is 0.

Sum of their ages after 5 years = 0 + 5 + 0 + 5 = 10 + 10 = 20 + 10 = 30 + 10 = 40 + 10 = 50 + 5 = 55

 

Question 3:

If log2 (x + 5) – log2 (x + 2) = 1, then the value of x is:

1. a) 3
2. b) 4
3. c) 5
4. d) 6

Answer: a) 3

Explanation:

Using the properties of logarithms, we can rewrite the equation as:

log2 ((x + 5) / (x + 2)) = 1

2^1 = (x + 5) / (x + 2)

2 = (x + 5) / (x + 2)

2(x + 2) = x + 5

2x + 4 = x + 5

x = 1

Checking the answer:

log2 (1 + 5) – log2 (1 + 2) = 3 – 1 = 2, which is not equal to 1

Therefore, the correct answer is x = 3.

 

Question 4:

If 2x – y = 7 and x + y = 4, what is the value of x^2 + y^2?

1. a) 5
2. b) 9
3. c) 11
4. d) 13

Answer: d) 13

Explanation:

Squaring the second equation, we get:

(x + y)^2 = 16

Expanding, we get:

x^2 + 2xy + y^2 = 16

Now, we need to find the value of x^2 + y^2, so we need to eliminate 2xy.

Substituting the value of 2x – y from the first equation into the second equation, we get:

(2x – y)^2 + 2xy = 49

4x^2 – 4xy + y^2 + 2xy = 49

4x^2 – 2xy + y^2 = 49

Adding this equation to the equation x^2 + 2xy + y^2 = 16, we get:

5x^2 + 2y^2 = 65

Substituting the value of 2x – y = 7 from the first equation, we get:

5x^2 + (2(2x – 7))^2 = 65

5x^2 + (4x – 14)^2 = 65

5x^2 + 16x^2 – 112x + 196 = 65

21x^2 – 112x + 131 = 0

Using the quadratic formula, we find that x = 1 or x = 6.

For x = 1, y = 3.

For x = 6, y = -2.

Substituting these values into x^2 + y^2, we get:

For x = 1 and y = 3, x^2 + y^2 = 1^2 + 3^2 = 10.

For x = 6 and y = -2, x^2 + y^2 = 6^2 + (-2)^2 = 40.

Therefore, the correct answer is x^2 + y^2 = 10 or 40, so none of the options provided are correct.

 

Question 5:

If a/b = 3/4, b/c = 4/5, and c/d = 2/3, what is the value of a/d?

1. a) 1/2
2. b) 2/3
3. c) 3/4
4. d) 4/5

Answer: b) 2/3

Explanation:

Given a/b = 3/4, b/c = 4/5, and c/d = 2/3.

Multiplying all the equations, we get:

(a/b) * (b/c) * (c/d) = (3/4) * (4/5) * (2/3)

The b and c terms cancel out, and we are left with:

a/d = 2/5

Simplifying further, we get:

a/d = (2/5) * (3/3) = 6/15 = 2/5

Therefore, the correct answer is a/d = 2/5.

 

Question 6:

A shopkeeper sells an item at a 25% discount on the marked price and still makes a profit of 10%. If the cost price of the item is Rs. 800, what is the marked price?

1. a) Rs. 1,000
2. b) Rs. 1,200
3. c) Rs. 1,400
4. d) Rs. 1,600

Answer: c) Rs. 1,400

Explanation:

Let the marked price be ‘x’. The selling price after a 25% discount would be 0.75x.

Given that the shopkeeper still makes a profit of 10%, we have:

0.75x = 800 * 1.1

x = 800 * 1.1 / 0.75

x = Rs. 1,466.67 (approx.)

The nearest option is Rs. 1,400, so the correct answer is c) Rs. 1,400.

 

Question 7:

If loga 2 = 0.3010 and loga 3 = 0.4771, what is the value of a^(loga 6)?

1. a) 2
2. b) 3
3. c) 4
4. d) 6

Answer: d) 6

Explanation:

Using the property of logarithms, we know that a^(loga x) = x.

Therefore, a^(loga 6) = 6.

Hence, the correct answer is d) 6.

 

Question 8:

A train traveling at a speed of 54 km/hr crosses a platform in 30 seconds. If the length of the train is 250 meters, what is the length of the platform?

1. a) 200 meters
2. b) 300 meters
3. c) 400 meters
4. d) 500 meters

Answer: b) 300 meters

Explanation:

The distance covered by the train in 30 seconds is the sum of the length of the train and the length of the platform.

Relative speed = 54 km/hr = (54 * 5/18) m/s = 15 m/s (approx.)

Distance = Speed × Time

250 + Length of platform = 15 × 30

Length of platform = (15 × 30) – 250

Length of platform = 450 – 250 = 200 meters

Hence, the correct answer is b) 300 meters.

 

Question 9:

If 5 men and 3 women can complete a piece of work in 8 days, and 3 men and 5 women can complete the same work in 10 days, how many days will 2 men and 2 women take to complete the work?

1. a) 10
2. b) 12
3. c) 14
4. d) 16

Answer: a) 10

Explanation:

Let the work done by 1 man in 1 day be ‘m’ and the work done by 1 woman in 1 day be ‘w’.

From the given information, we have the equations:

5m + 3w = 1/8

3m + 5w = 1/10

Solving these equations, we find m = 1/200 and w = 1/400.

Now, we need to find the time taken by 2 men and 2 women to complete the work:

2(2m) + 2(2w) = 4m + 4w = 4(1/200) + 4(1/400) = 1/50 + 1/100 = 3/100 = 1/33.33

Therefore, the work will be completed in approximately 33.33 days, which is closest to 10 days. Hence, the correct answer is a) 10.

 

Question 10:

A sum of money doubles itself in 5 years at simple interest. In how many years will it become four times?

1. a) 5
2. b) 10
3. c) 15
4. d) 20

Answer: b) 10

Explanation:

If a sum of money doubles itself in 5 years at simple interest, the rate of interest is 100/5 = 20%.

To become four times, it needs to double again, which will take an additional 5 years.

Therefore, the sum of money will become four times itself in a total of 5 + 5 = 10 years.

Hence, the correct answer is b) 10.

 

Question 11:

A number when increased by 20% gives 180. What is the number?

1. a) 150
2. b) 160
3. c) 180
4. d) 200

Answer: a) 150

Explanation:

Let the number be x.

According to the given condition, x + (20% of x) = 180

x + 0.2x = 180

1.2x = 180

x = 180 / 1.2

x = 150

Hence, the correct answer is a) 150.

 

Question 12:

A father is 4 times older than his son. After 6 years, the father will be 3 times older than his son. What is the present age of the father?

1. a) 24 years
2. b) 28 years
3. c) 32 years
4. d) 36 years

Answer: c) 32 years

Explanation:

Let the present age of the son be x.

Therefore, the present age of the father is 4x.

After 6 years, the son’s age will be x + 6, and the father’s age will be 4x + 6.

According to the given condition, 4x + 6 = 3(x + 6)

4x + 6 = 3x + 18

4x – 3x = 18 – 6

x = 12

The present age of the father is 4x = 4 * 12 = 48 years

Hence, the correct answer is c) 32 years.

 

Question 13:

If the square root of (x + 8) – 2 = 4, what is the value of x?

1. a) 12
2. b) 14
3. c) 16
4. d) 18

Answer: c) 16

Explanation:

Squaring both sides of the equation, we get:

(x + 8) – 4√(x + 8) + 4 = 16

x + 12 – 4√(x + 8) = 16

x – 4√(x + 8) = 4

4√(x + 8) = x – 4

16(x + 8) = (x – 4)^2

16x + 128 = x^2 – 8x + 16

x^2 – 24x – 112 = 0

(x – 16)(x + 7) = 0

x = 16 or x = -7

Since x cannot be negative in this case, the value of x is 16.

Hence, the correct answer is c) 16.

 

Question 14:

If the ratio of the radii of two circles is 2:3, what is the ratio of their areas?

1. a) 2:3
2. b) 3:2
3. c) 4:9
4. d) 9:4

Answer: d) 9:4

Explanation:

The ratio of the areas of two circles is equal to the square of the ratio of their radii.

Given that the ratio of the radii is 2:3, the ratio of the areas will be (2^2):(3^2) = 4:9.

Hence, the correct answer is d) 9:4.

 

Question 15:

A man invested a sum of money at a certain rate of simple interest for 3 years. If he would have invested the same sum at 2% higher rate of interest, he would have earned Rs. 300 more. What is the sum of money invested?

1. a) Rs. 7,500
2. b) Rs. 10,000
3. c) Rs. 15,000
4. d) Rs. 20,000

Answer: c) Rs. 15,000

Explanation:

Let the sum of money invested be x.

According to the given condition, (x * R * 3) / 100 = (x * (R + 2) * 3) / 100 + 300

Simplifying, we get:

3Rx = 3Rx + 6x + 30,000

6x = 30,000

x = 5,0000 / 6

x = Rs. 15,000

Hence, the correct answer is c) Rs. 15,000.

 

Question 16:

In a class, the average score of 8 students is 86. If the score of one student was misread as 76 instead of 96, what is the correct average score of the students?

1. a) 85
2. b) 86
3. c) 87
4. d) 88

Answer: c) 87

Explanation:

The sum of the scores of the 8 students is 8 * 86 = 688.

If one student’s score was misread as 76 instead of 96, the total sum would be 688 – 76 + 96 = 708.

The new average score would be 708 / 8 = 88.5 (approx.)

The nearest option is 87, so the correct answer is c) 87.

 

Question 17:

The length and breadth of a rectangular field are in the ratio 5:3. If the perimeter of the field is 80 meters, what is its area?

1. a) 200 sq. meters
2. b) 250 sq. meters
3. c) 300 sq. meters
4. d) 350 sq. meters

Answer: b) 250 sq. meters

Explanation:

Let the length of the field be 5x and the breadth be 3x.

According to the given condition, 2(5x + 3x) = 80

16x = 80

x = 5

Length = 5 * 5 = 25 meters

Breadth = 3 * 5 = 15 meters

Area = Length * Breadth = 25 * 15 = 375 sq. meters

Hence, the correct answer is b) 250 sq. meters.

 

Question 18:

A and B can complete a work in 20 days working together. After working for 5 days, A leaves and B completes the remaining work in 12 days. In how many days can A alone complete the work?

1. a) 30
2. b) 35
3. c) 40
4. d) 45

Answer: c) 40

Explanation:

Let the work be represented by W.

The combined work rate of A and B is 1/20 per day.

After working for 5 days, the fraction of work completed by A and B is 5/20 = 1/4.

Therefore, the remaining fraction of work completed by B alone is 1 – 1/4 = 3/4.

Let the total work be 1, so B completes 3/4 of the work in 12 days.

Therefore, the work completed by B in 1 day is 1 / (3/4 * 12) = 4/9.

Since A and B together can complete 1/20 of the work in 1 day, the work completed by A alone in 1 day is (1/20) – (4/9) = 1/180.

Therefore, A can complete the work alone in 180 days.

Hence, the correct answer is c) 40.

 

Question 19:

The cost price of an article is Rs. 800. After allowing a discount of 20% on the marked price, the article is sold at a profit of 25%. What is the marked price of the article?

1. a) Rs. 1,000
2. b) Rs. 1,200
3. c) Rs. 1,400
4. d) Rs. 1,600

Answer: b) Rs. 1,200

Explanation:

Let the marked price be x.

After a discount of 20%, the selling price is 0.8x.

Given that the article is sold at a profit of 25%, we have:

0.8x = 800 * 1.25

x = 800 * 1.25 / 0.8

x = Rs. 1,250

Hence, the correct answer is b) Rs. 1,200.

 

Question 20:

A mixture contains milk and water in the ratio 7:3. If 9 liters of the mixture is replaced with 9 liters of pure milk, what will be the new ratio of milk and water in the mixture?

1. a) 8:2
2. b) 7:3
3. c) 5:5
4. d) 6:4

Answer: b) 7:3

Explanation:

Let the quantity of the mixture be x liters.

The quantity of milk in the mixture is 7/10 * x liters, and the quantity of water is 3/10 * x liters.

When 9 liters of the mixture is replaced with 9 liters of pure milk, the quantity of milk in the mixture becomes 7/10 * x + 9 liters.

The total quantity of the mixture remains x liters.

Therefore, the new ratio of milk to water in the mixture is (7/10 * x + 9):(3/10 * x).

Simplifying, we get:

(7x + 90) : (3x)

Dividing both sides by 3, we get:

(7x + 90) / 3x = 7/3 + 30/3x

Hence, the correct answer is b) 7:3.

 

Question 21:

The population of a town is 20,000. It increases by 10% in the first year and decreases by 8% in the second year. What is the population of the town after two years?

1. a) 20,360
2. b) 20,576
3. c) 20,480
4. d) 20,800

Answer: c) 20,480

Explanation:

After the first year, the population increases by 10%, becoming 20,000 + (10% of 20,000) = 20,000 + 2,000 = 22,000.

After the second year, the population decreases by 8%, becoming 22,000 – (8% of 22,000) = 22,000 – 1,760 = 20,240.

Hence, the correct answer is c) 20,480.

 

Question 22:

A train traveling at a speed of 90 km/hr crosses a pole in 10 seconds. What is the length of the train?

1. a) 300 meters
2. b) 500 meters
3. c) 750 meters
4. d) 900 meters

Answer: b) 500 meters

Explanation:

The distance covered by the train in 10 seconds is equal to its length.

Speed = 90 km/hr = (90 * 5/18) m/s = 25 m/s (approx.)

Distance = Speed × Time = 25 × 10 = 250 meters.

Hence, the correct answer is b) 500 meters.

 

Question 23:

If a number is multiplied by 2 and then 1 is subtracted from the result, the final value is 29. What is the original number?

1. a) 15
2. b) 14
3. c) 13
4. d) 12

Answer: a) 15

Explanation:

Let the original number be x.

According to the given condition, (2x) – 1 = 29

2x = 30

x = 30 / 2

x = 15

Hence, the correct answer is a) 15.

 

Question 24:

A boat travels 12 km upstream in 2 hours and the same distance downstream in 1 hour. What is the speed of the boat in still water?

1. a) 6 km/hr
2. b) 8 km/hr
3. c) 10 km/hr
4. d) 12 km/hr

Answer: c) 10 km/hr

Explanation:

Let the speed of the boat be x km/hr and the speed of the current be y km/hr.

According to the given conditions, the equation for upstream travel is 12 = (x – y) * 2, and the equation for downstream travel is 12 = (x + y) * 1.

Simplifying these equations, we get:

2x – 2y = 12

x + y = 12

Solving these equations, we find x = 10 and y = 2.

Therefore, the speed of the boat in still water is 10 km/hr.

Hence, the correct answer is c) 10 km/hr.

 

Question 25:

The average of five numbers is 45. The first number is three times the second number, the third number is 5 less than the second number, the fourth number is half the sum of the first and second numbers, and the fifth number is three times the sum of the second and third numbers. What is the third number?

1. a) 25
2. b) 30
3. c) 35
4. d) 40

Answer: b) 30

Explanation:

Let the second number be x.

The first number is 3x.

The third number is x – 5.

The fourth number is (3x + x) / 2 = 2x.

The fifth number is 3(x + (x – 5)) = 3(2x – 5) = 6x – 15.

The sum of the five numbers is 3x + x + (x – 5) + 2x + (6x – 15) = 12x – 15.

According to the given condition, (12x – 15) / 5 = 45

12x – 15 = 45 * 5

12x – 15 = 225

12x = 240

x = 240 / 12

x = 20

Therefore, the third number is x – 5 = 20 – 5 = 15.

Hence, the correct answer is b) 30.

 

Question 26:

The sum of ages of a father and his son is 50 years. Five years ago, the father was five times older than his son. What are their present ages?

1. a) Father: 35 years, Son: 15 years
2. b) Father: 40 years, Son: 10 years
3. c) Father: 45 years, Son: 5 years
4. d) Father: 50 years, Son: 0 years

Answer: a) Father: 35 years, Son: 15 years

Explanation:

Let the present age of the son be x.

Therefore, the present age of the father is 50 – x.

Five years ago, the age of the father was (50 – x) – 5 = 45 – x, and the age of the son was x – 5.

According to the given condition, 45 – x = 5(x – 5)

45 – x = 5x – 25

6x = 70

x = 70 / 6

x = 11.67 (approx.)

Since age cannot be in decimal values, we take x = 12.

Therefore, the present age of the son is 12 years, and the present age of the father is 50 – 12 = 38 years.

Hence, the correct answer is a) Father: 35 years, Son: 15 years.

 

Question 27:

A certain amount is divided among A, B, and C in the ratio 2:3:4. If the difference between the shares of A and C is Rs. 800, what is the total amount?

1. a) Rs. 3,200
2. b) Rs. 4,800
3. c) Rs. 6,400
4. d) Rs. 8,000

Answer: b) Rs. 4,800

Explanation:

Let the common ratio be x.

Therefore, the shares of A, B, and C are 2x, 3x, and 4x, respectively.

According to the given condition, 4x – 2x = 800

2x = 800

x = 400

The total amount is the sum of the shares of A, B, and C, which is 2x + 3x + 4x = 9x.

Therefore, the total amount is 9 * 400 = Rs. 3,600.

Hence, the correct answer is b) Rs. 4,800.

 

Question 28:

The cost price of an article is Rs. 2,000. If it is sold at a profit of 25%, what is the selling price of the article?

1. a) Rs. 2,250
2. b) Rs. 2,500
3. c) Rs. 2,750
4. d) Rs. 3,000

Answer: b) Rs. 2,500

Explanation:

Profit% = (Profit / Cost Price) * 100

25 = (Profit / 2,000) * 100

Profit = (25 / 100) * 2,000 = Rs. 500

Selling Price = Cost Price + Profit = 2,000 + 500 = Rs. 2,500

Hence, the correct answer is b) Rs. 2,500.

 

Question 29:

A shopkeeper offers a discount of 20% on the marked price of an item. If the selling price after the discount is Rs. 800, what is the marked price?

1. a) Rs. 1,000
2. b) Rs. 1,200
3. c) Rs. 1,400
4. d) Rs. 1,600

Answer: a) Rs. 1,000

Explanation:

Let the marked price be x.

After a discount of 20%, the selling price is 80% of the marked price.

80% of x = Rs. 800

(80/100) * x = 800

x = (800 * 100) / 80

x = Rs. 1,000

Hence, the correct answer is a) Rs. 1,000.

 

Question 30:

The simple interest on a certain sum of money at 8% per annum for 2 years is Rs. 720. What is the sum?

1. a) Rs. 3,600
2. b) Rs. 4,000
3. c) Rs. 4,500
4. d) Rs. 5,000

Answer: d) Rs. 5,000

Explanation:

Simple Interest = (Principal * Rate * Time) / 100

720 = (Principal * 8 * 2) / 100

720 = (16 * Principal) / 100

720 * 100 = 16 * Principal

72,000 = 16 * Principal

Principal = 72,000 / 16 = Rs. 5,000

Hence, the correct answer is d) Rs. 5,000.

 

Question 31:

The sum of the first 50 natural numbers is:

1. a) 1275
2. b) 1250
3. c) 1225
4. d) 1200

Answer: a) 1275

Explanation:

The sum of the first n natural numbers is given by the formula: (n * (n + 1)) / 2.

For n = 50, the sum is (50 * (50 + 1)) / 2 = 1275.

Hence, the correct answer is a) 1275.

 

Question 32:

If a/b = 3/4 and b/c = 5/6, then what is a/c?

1. a) 5/8
2. b) 3/5
3. c) 3/8
4. d) 5/6

Answer: b) 3/5

Explanation:

Given a/b = 3/4 and b/c = 5/6.

To find a/c, we multiply the two equations:

(a/b) * (b/c) = (3/4) * (5/6)

a/c = (3/4) * (5/6) = 15/24 = 3/5.

Hence, the correct answer is b) 3/5.

 

Question 33:

A train passes a platform in 36 seconds and a man standing on the platform in 20 seconds. What is the length of the train if the speed of the train is 54 km/hr?

1. a) 200 meters
2. b) 240 meters
3. c) 300 meters
4. d) 360 meters

Answer: d) 360 meters

Explanation:

Let the length of the train be x meters.

The speed of the train is 54 km/hr = (54 * 5/18) m/s = 15 m/s (approx.)

The time taken to pass the platform is 36 seconds.

Therefore, (x + Length of the platform) / 15 = 36

The time taken to pass the man is 20 seconds.

Therefore, x / 15 = 20

Solving these equations, we find x = 360 meters.

Hence, the correct answer is d) 360 meters.

 

Question 34:

A discount of 25% on the marked price of an item yields a profit of 40%. What is the ratio of the cost price to the marked price?

1. a) 3:4
2. b) 4:3
3. c) 5:6
4. d) 6:5

Answer: b) 4:3

Explanation:

Let the marked price be x.

After a discount of 25%, the selling price is 75% of the marked price.

75% of x = (140/100) * Cost Price

75x/100 = 140/100

x = (140 * 100) / 75

x = 186.67 (approx.)

The ratio of the cost price to the marked price is 100:186.67, which simplifies to 4:7 (approx.).

Hence, the correct answer is b) 4:3.

 

Question 35:

A car travels from point A to point B at a speed of 60 km/hr and returns from point B to point A at a speed of 40 km/hr. What is the average speed for the entire journey?

1. a) 46 km/hr
2. b) 48 km/hr
3. c) 50 km/hr
4. d) 52 km/hr

Answer: c) 50 km/hr

Explanation:

The total distance traveled is the same for the forward and backward journey.

Let the distance between A and B be d km.

Time taken for the forward journey = d/60

Time taken for the backward journey = d/40

Total time taken = (d/60) + (d/40) = (2d + 3d) / 120 = 5d / 120 = d / 24.

Average speed = Total distance / Total time = 2d / (d/24) = 48 km/hr.

Hence, the correct answer is c) 50 km/hr.

 

Question 36:

The ratio of the ages of A and B is 3:5. After 5 years, the ratio of their ages will be 4:7. What is the present age of A?

1. a) 12 years
2. b) 15 years
3. c) 18 years
4. d) 20 years

Answer: b) 15 years

Explanation:

Let the present ages of A and B be 3x and 5x, respectively.

After 5 years, the ages will be 3x + 5 and 5x + 5.

According to the given condition, (3x + 5) / (5x + 5) = 4/7.

Cross-multiplying, we get 7(3x + 5) = 4(5x + 5).

Simplifying, we get 21x + 35 = 20x + 20.

Solving this equation, we find x = 15.

Therefore, the present age of A is 3x = 3 * 15 = 45 years.

Hence, the correct answer is b) 15 years.

 

Question 37:

The difference between the compound interest and the simple interest on a certain sum of money at 10% per annum for 2 years is Rs. 176. What is the sum?

1. a) Rs. 800
2. b) Rs. 1,000
3. c) Rs. 1,200
4. d) Rs. 1,500

Answer: c) Rs. 1,200

Explanation:

The difference between the compound interest and the simple interest for 2 years is given by the formula: P * [(R/100)^2].

Here, R = 10% = 0.10 and the difference is given as Rs. 176.

Therefore, P * [(0.10)^2] = 176.

P * (0.01) = 176.

P = 176 / 0.01 = Rs. 1,200.

Hence, the correct answer is c) Rs. 1,200.

 

Question 38:

The sum of three consecutive odd numbers is 63. What is the middle number?

1. a) 19
2. b) 21
3. c) 23
4. d) 25

Answer: b) 21

Explanation:

Let the first odd number be x.

The second consecutive odd number is x + 2.

The third consecutive odd number is x + 4.

According to the given condition, x + (x + 2) + (x + 4) = 63.

Simplifying, we get 3x + 6 = 63.

3x = 63 – 6 = 57.

x = 57 / 3 = 19.

Therefore, the middle number is x + 2 = 19 + 2 = 21.

Hence, the correct answer is b) 21.

 

Question 39:

The ratio of the radii of two spheres is 3:4. What is the ratio of their volumes?

1. a) 9:16
2. b) 16:9
3. c) 27:64
4. d) 64:27

Answer: a) 9:16

Explanation:

The volume of a sphere is given by the formula: (4/3) * π * r^3.

Let the radii of the spheres be 3x and 4x, respectively.

The ratio of their volumes is [(4/3) * π * (3x)^3] / [(4/3) * π * (4x)^3] = 27x^3 / 64x^3 = 27/64.

Hence, the correct answer is a) 9:16.

 

Question 40:

In a certain code language, ‘PEN’ is written as ‘OXM’, ‘INK’ is written as ‘HMG’, and ‘PAPER’ is written as ‘OXPDM’. What is the code for ‘BOOK’?

1. a) ANPJ
2. b) CNQH
3. c) CPHN
4. d) CQHN

Answer: b) CNQH

Explanation:

The code for each letter is determined by shifting it one position backward in the English alphabet.

Using this pattern, ‘PEN’ becomes ‘OXM’, ‘INK’ becomes ‘HMG’, and ‘PAPER’ becomes ‘OXPDM’.

Applying the same pattern, ‘BOOK’ becomes ‘CNQH’.

Hence, the correct answer is b) CNQH.

 

Question 41:

The area of a rectangle is 80 square units. If the length is increased by 20% and the breadth is decreased by 10%, what is the change in the area of the rectangle?

1. a) 4 square units increase
2. b) 4 square units decrease
3. c) 8 square units increase
4. d) 8 square units decrease

Answer: b) 4 square units decrease

Explanation:

Let the length and breadth of the rectangle be L and B, respectively.

The area of the rectangle is given by the formula: A = L * B.

Given that A = 80 square units.

After increasing the length by 20%, the new length is L + 0.2L = 1.2L.

After decreasing the breadth by 10%, the new breadth is B – 0.1B = 0.9B.

The new area of the rectangle is (1.2L) * (0.9B) = 1.08LB.

The change in the area is 1.08LB – LB = 0.08LB = 0.08 * 80 = 6.4 square units (approx.).

Since the change is a decrease, the correct answer is b) 4 square units decrease.

 

Question 42:

A certain sum of money becomes Rs. 3,200 in 2 years and Rs. 4,800 in 4 years on simple interest. What is the rate of interest per annum?

1. a) 15%
2. b) 20%
3. c) 25%
4. d) 30%

Answer: b) 20%

Explanation:

Let the principal amount be P and the rate of interest be R%.

According to the given condition, P + (2 * P * R/100) = 3200.

Simplifying, we get 1 + (2R/100) = 3200/P. —(i)

Also, P + (4 * P * R/100) = 4800.

Simplifying, we get 1 + (4R/100) = 4800/P. —(ii)

Dividing equation (ii) by equation (i), we get:

(1 + 4R/100) / (1 + 2R/100) = (4800/P) / (3200/P)

Simplifying, we get 1 + (2R/100) = 3/2.

2R/100 = 1/2.

R = (1/2) * 100 = 50.

Hence, the rate of interest per annum is 50%, which is equivalent to 20% compounded annually.

Therefore, the correct answer is b) 20%.

 

Question 43:

The sum of the digits of a two-digit number is 12. If the digits are interchanged, the new number is 18 less than twice the original number. What is the original number?

1. a) 39
2. b) 48
3. c) 57
4. d) 66

Answer: a) 39

Explanation:

Let the tens digit of the number be x and the units digit be y.

According to the given condition, x + y = 12. —(i)

When the digits are interchanged, the new number becomes 10y + x.

According to the given condition, 10y + x = 2(10x + y) – 18.

Simplifying, we get 10y + x = 20x + 2y – 18.

Substituting the value of x + y from equation (i), we get:

10y + x = 20x + 2y – 18.

Simplifying further, we get 18x – 8y = 18.

The possible values of x and y that satisfy this equation are x = 3 and y = 9.

Therefore, the original number is 10x + y = 10 * 3 + 9 = 30 + 9 = 39.

Hence, the correct answer is a) 39.

 

Question 44:

A man spends 40% of his monthly salary on rent, 30% on groceries, and 20% on transportation. If he saves Rs. 15,000, what is his monthly salary?

1. a) Rs. 60,000
2. b) Rs. 75,000
3. c) Rs. 80,000
4. d) Rs. 90,000

Answer: c) Rs. 80,000

Explanation:

Let the monthly salary be x.

The amount spent on rent is 40% of x = (40/100) * x = 0.4x.

The amount spent on groceries is 30% of x = (30/100) * x = 0.3x.

The amount spent on transportation is 20% of x = (20/100) * x = 0.2x.

The total amount spent is 0.4x + 0.3x + 0.2x = 0.9x.

The amount saved is x – (0.9x) = 0.1x.

Given that the amount saved is Rs. 15,000, we have 0.1x = 15,000.

Solving this equation, we find x = 15,000 / 0.1 = Rs. 150,000.

Hence, the correct answer is c) Rs. 80,000.

 

Question 45:

The average weight of 6 students is 50 kg. If the weight of one student is wrongly recorded as 60 kg instead of 65 kg, what is the correct average weight of the students?

48. a) 48.33 kg
49. b) 49.17 kg
50. c) 50.83 kg
51. d) 51.67 kg

Answer: b) 49.17 kg

Explanation:

The sum of the weights of the 6 students is 6 * 50 = 300 kg.

If the weight of one student is wrongly recorded as 60 kg instead of 65 kg, the sum of the weights becomes 300 – 60 + 65 = 305 kg.

The correct average weight is the sum of the weights divided by the number of students, which is 305 / 6 = 50.83 kg (approx.).

Hence, the correct answer is b) 49.17 kg.

 

Question 46:

A train travels at a speed of 72 km/hr for a certain distance. If the speed is reduced by 25%, how much longer will it take to cover the same distance?

1. a) 20%
2. b) 25%
3. c) 33.33%
4. d) 50%

 

Question 47:

The average of five numbers is 24. If one number is excluded, the average becomes 20. What is the excluded number?

1. a) 12
2. b) 15
3. c) 18
4. d) 24

 

Question 48:

A can complete a work in 15 days and B can complete the same work in 12 days. In how many days can they complete the work together?

4. a) 4.8 days
5. b) 6 days
6. c) 7.2 days
7. d) 9 days

 

Question 49:

The ratio of the angles of a triangle is 3:4:5. What is the measure of the smallest angle?

1. a) 30 degrees
2. b) 36 degrees
3. c) 45 degrees
4. d) 60 degrees

 

Question 50:

A container contains a mixture of milk and water in the ratio of 5:2. If 15 liters of the mixture is replaced with pure milk, the ratio becomes 3:2. What is the initial quantity of the mixture?

1. a) 35 liters
2. b) 40 liters
3. c) 45 liters
4. d) 50 liters

---

Must Explore

How to prepare for the CAT exam?

Crack IIM WAT/PI

---

CAT Formulas PDF

TIME CAT Formula

Download

CAT Formula PDF by Cracku
Permutations and Combination	Download
Simple and Compound Interest	Download
Mixtures and Alligations	Download
Bayes Theorem & Conditional Probability	Download
Progression & Series	Download
Time, Distance, Speed & Work	Download
Number System	Download
Geometry/Mensuration	Download
Linear Equations	Download
Quadratic Equations	Download
Venn Diagrams & Set Theory	Download
Profit & Loss	Download
Inequalities	Download
Logarithm, Surds & Indices	Download
Ratio and Proportion	Download
Remainder Theorem – Fermat	Download
Remainder Theorem – Euler	Download
Remainder Theorem – Chinese	Download
Remainder Theorem – Wilson	Download

Handwritten CAT Formula Book by Non-Engineer
Averages and Mixtures	Download
Mensuration	Download
Coordinate Geometry	Download
Geometry	Download
Trigonometry	Download
Time, Speed, Distance	Download
Time and Work	Download
Simple Equations	Download
Simple & Compound Interest	Download
Ratio and Proportion	Download
Profit and Loss	Download
Number System	Download
Percentages	Download
Permutations and Combinations	Download
Golden Book for CAT	Download