CTAI Teacher Handbook Class 3

Computational
Thinking

Class 3

Teacher Handbook

CENTRAL BOARD OF SECONDARY EDUCATION

First Edition: March, 2026
Country of Publication: India
Published by: Central Board of Secondary Education, Integrated Office, Sector 23, Dwarka, New
Delhi-110077

Page 1

PREFACE
The National Education Policy (NEP) aims to position India as a leader in emerging
knowledge fields by integrating technologies like AI, Machine Learning, Big Data, and
Computational Thinking into school education. It promotes technology-enabled,
interactive, and gamified learning using tools such as Augmented Reality (AR), Virtual
Reality (VR), and virtual labs to foster creativity, problem-solving, and interdisciplinary
exploration. NCFSE 23 carries this recommendation further for implementation.

While Artificial Intelligence (AI) is an important requirement, Computational Thinking
(CT) should be a broader skill, developing a foundation for learning AI. It can cover
various aspects like Cybersecurity, basic network, etc. Hence, CBSE approaches this
by integrating Computational Thinking with AI and other technological advancements,
without dependence on any platform.

The book introduces foundational Computational Thinking skills through simple,
visual, and observation-based problems involving numbers, shapes, pictures, and
everyday situations. It focuses on recognising basic patterns, understanding simple
sequences, identifying similarities and differences, and following step-by-step
instructions. The document also provides age-appropriate pedagogical approaches,
learning resources, assessment support, and classroom implementation guidelines to
ensure joyful, inclusive, and skill-based learning, in alignment with NEP 2020.

Team CBSE

Page 2

ACKNOWLEDGEMENTS
Expert Committee

1. Dr. Karthik Raman, Core Leadership, IIT Madras Bodhan AI Foundation; Professor,
Department of Data Science and AI, Wadhwani School of Data Science and AI, IIT
Madras
2. Dr. Rajesh Kumar, Professor, Department of Electrical Engineering, MNIT, Jaipur
3. Dr. Seema Verma, Professor (ECE), Additional Project Director (Siemens CoE), NITTTR,
Bhopal
4. Dr. S. Neethi Core Leadership, IIT Madras Bodhan AI Foundation, Professor of Practice,
Department of Data Science and AI, Wadhwani School of Data Science and AI, IIT Madras
5. Dr. Arun L. Naik, Associate Professor, Mathematics Education, Azim Premji University,
Bengaluru
6. Dr. Aanchal Chomal, Associate Professor, School of Continuing Education and
University Resource Centre, Azim Premji University
7. Dr. Ankit Vijayvargiya, Assistant Professor, School of Technology, Dhirubhai Ambani
University, Gandhinagar
8. Sh. R P Singh, Associate Prof &Additional Director, CBSE
9. Mr. Mikin Lala, (IIT Roorkee and IIM Calcutta Alumnus), Field Expert
10. Ms. Rekha Malhotra, (Alumna of British Institute (Business Management and
Advertising), Frameworks Mumbai (Master’s in Computer Science), and Workstation (3D
Animation)), Field Expert

Material Production Group

1. Mr. Jay Thakkar, Senior Technical Officer, Center for Creative Learning, IIT Gandhinagar
2. Mr. Chris John, Project Scientist, Centre for Creative Learning, IIT Gandhinagar
3. Mr. Mikin Lala, (IIT Roorkee and IIM Calcutta Alumnus), Field Expert
4. Ms. Amatullah Mustafa Neemuchwala, Field Expert
5. Ms. Telidevara Sree Lasya, Field Expert
6. Mr. Parth Oza, Field Expert
7. Mr. Deep Mayekar, Field Expert
8. Mr. Mayank Patil, Field Expert
9. Mr. Shivraj Ugale, Field Expert
10. Mr. Raj Dhorade, Field Expert
11. Mr. Nilesh Vijay Rajput, Field Expert

The efforts of Prof. Manish Jain and his whole team from Centre for Creative Learning,
IIT Gandhinagar are also acknowledged and deeply appreciated.

Page 3

TABLE OF CONTENTS

SR. NO. CHAPTER PAGE NO.

Introduction 5
1.
How to use this Book? 8
2.
What’s in a Name? 9
3.
Toy Joy 21
4.
Double Century 27
5.
Vacation with My Nani Maa 32
6.
Fun with Shapes 37
7.
House of Hundreds – I 44
8.
Raksha Bandhan 49
9.
Fair Share 54
10.
House of Hundreds – II 59
11.
Fun at Class Party! 64
12.
Filling and Lifting 71
13.
Give and Take 76
14.
Time Goes On 82
15.
The Surajkund Fair 87
16.

Page 4

Introduction
Computational Thinking (CT) is a problem-solving approach that comprises Decomposition, Pattern
Recognition, Abstraction, Algorithm Design, Data Analysis and Troubleshooting. Computational
Thinking Skills involve solving complex problems that promote thinking skills such as critical &
creative thinking, abstraction and pattern recognition, as well as algorithmic thinking. Problem
identification and problem solving necessitate application of multidisciplinary understanding for
creating effective solutions.

Artificial intelligence (AI) is a cutting-edge technology that empowers machines and computers to
perform tasks that usually require mimicking human intelligence. These machines can perform
complex thinking processes such as data analysis, pattern recognition, prediction of trends, solving
problems and decision making. Thus, AI involves simulating cognitive processes associated with
human intelligence and is widely applicable in various sectors such as banking, healthcare, defense,
education, entertainment, agriculture and others for processing information, solving intricate
problems and for planning.

The National Education Policy (NEP) aims for India to emerge as a global leader in new emerging
knowledge domains such as artificial intelligence, machine learning, data analytics, 3 -D machining
etc. To realise this goal, the policy suggests teaching students Mathematics and Computational
Thinking, along with new subjects like Artificial Intelligence, Machine Learning, and Data Science
during their school education. The policy also focuses on technology -enabled learning and
classrooms by using tools like artificial intelligence, machine learning, and adaptive testing to create
knowledge.

The National Curriculum for School Education draws from this policy aspiration and emphasizes the
need to introduce these emerging domains of study and technologies in the school curriculum. It
recommends inclusion of subjects such as design thinking, augmented reality, virtual reality, artificial
intelligence, and computational thinking. Additionally, it promotes the use of gamified content,
interactive content, and immersive experiences (such as AR, VR, or virtual labs) to enhance student
learning. In a variety of subjects, including design, music, art, and sciences, these resources support
students in knowledge creation and exploration, and development of capacities such as problem-
solving, critical and creative thinking.

CBSE under the aegis of the Department of School Education and Literacy, Ministry of Education,
Govt. of India, is implementing a Curriculum on Computational Thinking and Artificial Intelligence (CT
&AI) to inculcate AI-readiness in school students. This curriculum will be implemented from classes
3rd to 8th, in the session 2026-27, and aims to develop AI-Ready learners, by focusing on
Computational Thinking Skills. The AI-readiness, so inculcated through CT Skills, will help develop
the capacities of learners to use computational thinking, such as logical thinking, problem solving,
pattern recognition, and so on, and understand the role and use of Artificial Intelligence in daily life.
The Curriculum aims to build strong foundations in computational thinking, digital literacy, and
responsible use of technology, along with nurturing innovation, critical thinking, and ethical decision-
making capacities.

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1. Relevance: Importance of Introducing Computational Thinking (CT)
Introducing Computational Thinking at the Grade 3 level is essential as it serves as the intellectual
backbone for future digital literacy and AI readiness.
• Foundation for AI: CT develops the reasoning processes—such as breaking problems into parts
and spotting patterns—that power modern technology and Artificial Intelligence
• Cognitive Development: It fosters essential human capacities, including logical thinking,
problem-solving, pattern recognition, and ethical decision-making
• Real-World Application: It equips students to approach daily-life challenges systematically,
transforming them into informed and creative digital citizens

2. Objectives (Curricular Goals)
For the Preparatory Stage (Classes 3–5), the curriculum defines three primary curricular goals:
• CG-1: Develop basic problem-solving skills with procedural fluency to solve daily-life problems
as a step toward formal computational thinking
• CG-2: Develop basic capacities of analytical thinking, verbal, and visual reasoning.
• CG-3: Demonstrate understanding of basic concepts of computers and knowledge of hardware
and software.

3. Learning Outcomes
ABSTRACT THINKING
Students will be able to solve problems with hidden or unseen ideas, using:
• Different viewpoints of 3D objects
• Changes in shapes after flips, turns, folds, or rotations
• Hidden or missing parts in incomplete shapes or patterns

PATTERN RECOGNITION
Students will be able to identify simple patterns involving 1 or 2 changes in consecutive terms,
formed using:
• Numbers
• Shapes or images
• Letters
• Or a mix of the above

DECOMPOSITION
Students will be able to break down problems involving 2–3 clues, using information from:
• Number clues or number names
• 3D objects and their parts (faces, edges, corners)
• Step-by-step exchanges or transfers (money, objects, digits)
• Tables or charts with multiple pieces of information

ALGORITHMIC THINKING
Students will be able to follow clear step-by-step rules to solve problems involving:
• Number sequences formed using simple operations
• Movements on grids or direction-based paths
• Events arranged using before/after/in-between clues

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• Values that increase or decrease across steps
• Multi-step instructions involving moves, changes, or transfers
4. Mapped with NEP and NCF 2023
The curriculum is directly aligned with the vision of the National Education Policy (NEP) 2020 and
the National Curriculum Framework for School Education (NCF-SE) 2023.
• Integrated Learning: Following NCF-SE guidelines, CT is not treated as an isolated subject but
is integrated into Mathematics and "The World Around Us" (TWAU) to promote a multidisciplinary
understanding
• Competency-Based: The learning standards are derived from the Aims of School Education
defined in the NCF, moving from general capacities to specific classroom learning outcomes

5. Time Allocation
• Annual Hours: A total of 50 hours annually is suggested for the Preparatory Stage (Classes 3–
5)
• Integration: This time is embedded within the existing periods for Mathematics and TWAU to
ensure students are not overburdened with additional content load

6. Approach / Pedagogy
The pedagogical approach for Grade 3 is designed to be playful and experiential:
• Activity-Based: Learning is driven by fun math games, puzzles, and hands-on exercises rather
than rote memorization
• Resource Integration: Teachers use specialized CT workbooks and resource books that add
CT-focused questions to existing chapters in the standard textbooks
• Collaborative Learning: The curriculum encourages peer discussions and group tasks to solve
problems organized through worksheets

7. Assessment
Assessment for Grade 3 shifts away from traditional testing toward continuous and formative
methods:
• Specific Tools: Assessment methods include written tests involving puzzles, interactive group
activities, and the use of a Teacher Observation Journal to track student progress
• Focus: The goal is to assess the student's ability to apply knowledge and think creatively rather
than their ability to recall facts

Page 7

How to Use This Book?
This book is designed as a companion to the Mathematics textbook and is intended to be used
alongside regular classroom teaching. Since it follows the same chapter sequence, the
Mathematics teacher can seamlessly integrate it into daily instruction. As concepts are introduced
in class, the corresponding questions from this book can be used to deepen understanding and
encourage application.

Before beginning a chapter, the teacher is encouraged to go through the content of this book,
identify the underlying concepts required for each question, and plan how to align them with
classroom teaching. As these concepts are taught, the teacher can introduce the related thinking
questions to students.

It is important to note that the questions in this book are thinking-based and designed to promote
analysis, reasoning, and problem-solving. Teachers should adopt a facilitative approach, guiding
students through prompts and discussions rather than directly providing solutions. Students
should be given time to think and attempt independently, followed by classroom discussions
where different approaches are shared and explored.

Some chapters also include activities that build intuition and engagement. These should be
conducted before attempting the questions, as they help students approach the problems with
better understanding.

Teachers should approach this book with the mindset that the process of thinking is more
important than arriving at the correct answer. Creating a safe and encouraging environment where
students feel comfortable making mistakes, exploring multiple strategies, and expressing their
reasoning is essential. The goal is to nurture confident, independent thinkers rather than focus
solely on correctness.

Page 8

Chapter 1: What’s in a Name?
Activity Time
Activity: Caesar Cipher
Introduction
The need for secrecy when sending messages over the Internet is important for everyone. With this
activity, students are introduced to the concept of basic cryptography/encryption, and to simple
techniques for writing encrypted messages. Here, we will understand the Caesar cipher, use it to encrypt
and decrypt a message. This activity is a very simple form of encryption used by Julius Caesar in the 1st
century BCE.

Activity Time Description

Launch 5 min Start by asking students if they have ever shared a secret
message with a friend. Ask questions like: How did you
make sure others could not understand it? or Did you ever
use a code or secret language? The teacher can even write
a sentence on the board, which is encrypted, and a student
has to guess the message.
Tell students how long ago people also needed to send
secret messages. Introduce the idea that even today,
computers hide information using special codes to keep
messages safe on the Internet.
Explain that in this activity they will learn a simple way of
hiding messages using a tool called the Caesar Cipher,
and they will even make their own Caesar Cipher to send
secret messages to their friends.
Template: Caesar Cipher.pdf

Making & 20 Give students the alphabet shifting table question.
exploration of min Making of Caesar Cipher Model and Student’s Try out the
Caesar Cipher by activity with each other following worksheet questions.
Students
Student Worksheet:

Page 9

https://docs.google.com/document/d/1IRtpBWDWp6yOUi
RzpIGsbrlHZtGWFUkhOiPwRpv7-Co/edit?usp=sharing

Discussions & 10 Discussing the worksheets.
Conclusion min
Students learned how to hide messages using a simple
rule of shifting letters. They also discovered that while this
method works, it is not very strong because there are only
a few possible shifts. This shows why stronger encryption
methods are needed to keep information safe today.

CT Components
Algorithmic Thinking
We follow step-by-step process to encrypt and decrypt messages—choose a shift, match letters on the
cipher wheel, and write the encoded letters.

Decomposition
We break the task of sending a secret message into smaller steps: choosing a message, deciding the
shift, encoding each letter, and decoding it later.

Pattern Recognition
We can observe how letters consistently move forward in the alphabet by the same number of
positions. By noticing this repeating pattern, they can quickly understand how the cipher works and
even guess the shift used.

Generalisation
By trying different shift values (1, 2, 3, etc.), we can notice that the same rule works for any message or
shift.

Evaluation
We can think about how secure the Caesar Cipher really is by exploring that there are only 25 possible
shifts. This helps us to understand that some methods work but may not always be strong enough for
real security.

Page 10

Activity: Caesar Cipher
People have used secret messages for centuries to protect important information. Even today, much of
our communication online needs to be encrypted to keep our private data safe. For example, when your
parents withdraw money from an ATM or make a UPI payment, they enter a PIN. This number needs to
be hidden so that no one else can take their money, or when your parents buy something online and
enter their bank card details, the information needs to be protected so others cannot see or steal it.

A secret is information shared only with someone you completely trust through a message. Encryption is
the process of changing a message so that its original meaning is hidden from anyone except the
intended receiver. Though today we use sophisticated methods to encrypt information, here we will
understand encryption using a simple activity.

One simple way to do this is through ciphers. Cipher scrambles up the message using a key. For you to
unscramble and access the message, you must have the key. This way the cipher protects your message
from those who do not have the key.

A well-known example of cipher is the Caesar Cipher, named after the Roman leader Julius Caesar. He
used this method to send important messages securely. In the Caesar Cipher, the letters of the alphabet
are shifted by a certain number of places to be replaced by a new alphabet.

For example, if the alphabet is shifted one place, A becomes B, B becomes C, and so on, while Z comes
around to become A. This simple shifting method allows messages to be hidden from anyone who does
not know the key, the shift number.

Page 11

1. Complete the table below by shifting the alphabet.
Original
A B C D E F G H I J K L M N O P Q R S T U VWX Y Z
Placement
Shift 1 place
Shift 2 place
Shift 20 place
Answer:
Original
A B C D E F G H I J K L M N O P Q R S T U VWX Y Z
Placement
Shift 1 place B C D E F G H I J K L M N O P Q R S T U V W X Y Z A
Shift 2 place C D E F G H I J K L M N O P Q R S T U V W X Y Z A B
Shift 20 place U V W X Y Z A B C D E F G H I J K L M N O P Q R S T

CT Competency:
Algorithmic Thinking
Pattern Recognition

Making of Caesar Cipher
Print out the template file on A4 Paper for each student.
(Guide Students to Make Caesar Cipher)
Cut out both the two circles, and carefully cut out the arrow on the smaller circle; it will help as an
indicator. Put a toothpick or a thin stick through the centre of the two circles, with the smaller one on top
of the larger one. Make sure that the small circle is free to rotate over the bigger circle.

2. If the message “CAESAR CIPHER” is encoded with a shift of 3 places, which of the following is
the correct encoded message?
a) ECGUCT EKRJGT b) FDHVDU FLSKHU c) GEIWEV GMTLIV d) DBFTBS DJQIFS
Answer: b
Each letter in the message is shifted in the alphabet by a fixed number of positions.
Here, the shift is 3: C - F, A - D, E - H, S - V, A - D, R - U
So, CAESAR - FDHVDU, Similarly, CIPHER - FLSKHU

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CT Competency:
Algorithmic Thinking
Decomposition
Generalisation
Pattern Recognition
Activity: Encrypting messages through Caesar Cipher
Make groups of two students. Once the students make the Caesar Cipher, tell them to write a message,
encrypt it, and give it to their friends.

Choose a short message, like your name, your school name or the food item in your lunchbox that you
would like to give to your friend, and decide on a number that will represent the shift. To use the cipher,
first align the ‘A’ on the smaller circle with ‘A’ on the bigger circle. Now, turn the smaller circle such that
the arrow lies on the chosen shift number.

Now, to encode (encrypt) your message, look for each letter of your message on the smaller circle, and
write down the matching letter from the bigger circle. Do this one letter at a time until the entire message
is encoded.

Once you have finished encoding your message, give your friend the encoded message and the shift
letter (key). To decode (decrypt) the message, your friend should rotate the smaller circle so that the
arrow matches the given key. Then your friend finds each letter of the encrypted message from the bigger
circle and writes the corresponding letter on the smaller circle to reveal the original message.
3. If the message “THIS MESSAGE IS HIDDEN” is encoded as “ESTD XPDDLRP TD STOOPY”, by
how many positions has the alphabet been shifted?
a) 5 b) 11 c) 17 d) 7
Answer: b
To find the shift, compare a letter from the original message to the corresponding letter in the encoded
message. T - E, Moving 11 places from T in the alphabet comes around to E.
Similarly, H - S, I - T, S - D. Each letter is shifted by 11 positions, confirming that the shift used here is
11.
CT Competency:
Algorithmic Thinking
Decomposition
Generalisation
Pattern Recognition

Page 13

4. How many possible meaningful ways are there to shift the alphabet in a Caesar cipher?
a) 26 b) 52 c) 25 d) 50
Answer: c
We can shift letters by a certain number of positions, determined by the number of letters in the alphabet.
The English alphabet has 26 letters. But a shift of 26 would bring A to itself, leaving the message
unchanged. Therefore, only shifts from 1 to 25 produce encoded messages. So, there are 25 possible
meaningful shifts.

CT Competency:
Generalisation
Evaluation

Think about how secure the Caesar Cipher really is by exploring that there are only a few possible shifts.
This helps to understand that some methods work, but may not always be strong enough for real security.
This may work for your secret messages to your friends, but not for your parents’ bank transactions.

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Questions
1. Find the largest number name that is hidden in the series given below.
Note: The letters in the number name should appear together as continuous letters

a) 60 b) 6 c) 20 d) 70
Answer: c
If we arrange the given number names in descending order, we get: 70 > 60 > 20 > 6.
Among these, the numbers 70 and 60 are not present in the given series, while 20 and 6 are present.
Since the question asks for the largest number, and 20 is greater than 6, option c is the correct
answer.

______________________________________________________________________________ _______
2. Roy’s favourite number has:
• 3 letters in its name
• The number of vowels is double the number of consonants in its name
What is Roy’s favourite number?
a) 2 b) 6 c) 10 d) 1
Answer: d
The options given are TWO, SIX, TEN, and ONE.
Option d satisfies both the given conditions.
The word ONE has 3 letters in its spelling.
It contains 2 vowels (O, E) and 1 consonant (N), meaning the number of vowels is double the number
of consonants.
Hence, option d is the correct answer.
3. If each of the given boxes follows the same theme, what will come in place of “?”

a) NINE3 b) THREE9 c) THREE3 d) 9THREE
Answer: b
Each term has a number written in a black circle and a white circle. Below the circles, these two
numbers are presented in a specific format:
The number in the black circle is written first, followed by the number in the white circle.
The number in the black circle is represented in word form, while the number in the white circle is
presented in numeral form.
Therefore, following this logic, option b is the correct answer.

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4. If each of the following terms follows the same theme, what will come in place of "?"

a) b)

c) d)
Answer: c
Each term has two numbers written before the arrow: one in number name format and the other as a
digit. The series starts with the number given in number name format, and the length of the series is
determined by the value of the digit.
For the last term, "Four-3," the series starts at 4 and contains 3 digits. So, the answer is 4 5 6.
Hence, the correct answer is option c.
5. If each of the following terms follows the same theme, what will come in place of “?”

a) b) c) d)
Answer: a
In each term, the shape that appears the most times in the box (before the arrow) is shown after the
arrow. Following this logic, the shape in option a appears the most frequently in the question term.
Therefore, option a is the correct answer.
6. Find the odd one out.

a) b)

c) d)
Answer: b
Options a, c, and d represent road transport, while option b represents air transport. Hence, the correct
answer is option b.

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7. Which of the following number names given in the options will come second if arranged in
dictionary order of their names?
a) 5 b) 6 c) 8 d) 9
Answer: a
a) 5: FIVE b) 6: SIX c) 8: EIGHT d) 9: NINE
Arranging in dictionary order: EIGHT, FIVE, NINE, SIX
So, 5 will come second. Hence, the correct answer is option a.
8. Each shape has a letter in it. Spelling (or Name) of which of the following numbers, given in the
options can be formed by combining the letters in the circles only?

a) 1 b) 5 c) 9 d) 6
Answer: c
Option c - NINE can be formed.

Hence, the correct answer is option c.
9. Spelling of which of the following numbers given in the options is present in the group of letters
given below?
Note: The letters of the name should appear in the same order as in the word
TWLVETOWELEVENSVENTHREEIFEIHGT
a) 2 b) 7 c) 3 d) 8
Answer: c
In the given options only the spelling of number 3 is present in the group of letters given below:

Hence, the correct answer is option c.

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10. Which of the following numbers CANNOT be spelt using the letters present in the image given
below?
Note: If a letter appears in the image, it can be repeatedly used any number of times to spell the
number name

a) 50 b) 30 c) 85 d) 11
Answer: d
Letters present in the image: G, Y, F, T, L, R, V, H, I, E

Check each option (by spelling the number name):

● 50 - F I F T Y - all letters present

● 30 - T H I R T Y - all letters present

● 85 - E I G H T Y F I V E - all letters present

● 11 - E L E V E N - N is required but N is not present in the question image

So, eleven (11) cannot be formed. Thus, option d is the correct answer.
11. What will come in place of "?" in the given series?

a) b) c) d)
Answer: d
There are three repeating rules in the series:
The rectangles follow a consecutive number pattern (1, 2, 3, 4…) where numbers and number names
alternate.

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The diamond shows the sum of the two values in the rectangles, written alternately as a number and as a
number name.
The diamond’s position alternates. It appears below the rectangles in one term and above them in the
next.

Applying these rules to the next (fifth) term:

● The rectangles continue the sequence: after 7 and EIGHT, the next pair is 9 and TEN
● Their sum is 9 + 10 = 19. Since the previous diamond showed a number name
(FIFTEEN), this one should show the number 19
● The diamond should now appear below the rectangles

Hence, the final layout is 9 and TEN with 19 in the diamond below.

Thus, option d is the correct answer.
12. If you were to write the numbers given below in words, spelling of which of them would have
exactly one vowel in it?
a) 1 b) 2 c) 5 d) 9
Answer: b
When we write the numbers in words:

a) 1 (ONE)
Vowels: O, E - 2 vowels

b) 2 (TWO)
Vowels: O - 1 vowel

c) 5 (FIVE)
Vowels: I, E - 2 vowels

d) 9 (NINE)
Vowels: I, E - 2 vowels

Only TWO has exactly one vowel. So, the correct answer is option b.
13. X is a single digit number. If X has exactly five letters in its spelling, how many different values
can X possibly have?
Note: If a letter is repeated in the spelling of X, it has to be counted as many times as it appears in the
spelling. For example, the word "NINE" has 4 letters and not 3 letters
a) 1 b) 2 c) 3 d) 4
Answer: c
There are 10 single digit numbers from 0 to 9. Out of these numbers, only 3 (THREE), 7 (SEVEN), and 8
(EIGHT) have exactly five letters in their spelling. Hence, the correct answer is option c.

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Raj, Aryan, and Rahul went to a movie on different days.
1. Raj did not go to the movie on Monday, Wednesday, or Thursday.
2. Aryan went to the movie on Friday, and Rahul went to the movie on Sunday.
Which of the following days could Raj have gone to the movie?
(a) Thursday
(b) Tuesday
(c) Saturday
(d) Either Tuesday or Saturday
Answer: d
Raj did not go to the movie on Monday, Wednesday, or Thursday.
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Raj
Aryan
Rahul

Aryan went to the movie on Friday, and Rahul went to the movie on Sunday.
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Raj
Aryan
Rahul

The remaining days for Raj to go to the movie are Tuesday and Saturday.
Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Raj
Aryan
Rahul

Hence, the correct answer is option d.

Page 20

Chapter 2: Toy Joy
1. Which of the following options represents the top view of the object given below?

a) b) c) d)
Answer: d
To solve the question without errors it is important to logically approach the question. When viewed
from the top, a grey rectangle is visible. This rules out options a and c. Additionally, a white and a
black area will be visible. This leads to the conclusion that option d is the top view of the object given.

2. Count the number of surfaces (faces) in the given solid.

a) 6 b) 7 c) 8 d) 9
Answer: c
There are 5 faces visible from the front. The 6th face is on the left-side of the solid, the 7th face is on
the bottom of the solid and the 8th face is on the back-side of the solid. Hence, there are 8 faces and
option c is correct.

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3. The given shapes are to be placed in the buckets. Shapes with a number of edges greater than 5
are kept in a yellow bucket while the others are kept in the white bucket. Out of the shapes in the
yellow bucket, how many have EXACTLY 6 edges?

a) 2 b) 3 c) 4 d) 5
Answer: b
There are a total of 6 shapes in the yellow bucket, out of which only 3 shapes have exactly 6 edges.

Hence, the correct answer is option b.
4. Given below is a large box consisting of four smaller transparent boxes with some sides
coloured. Which of the boxes shown in the options is not a part of the larger box in the question
image?
Note: Rotation of option images is not allowed

a) b) c) d)
Answer: c
As shown in the image below, options a, b, and d are part of the larger box. Option c is not, as
there is no box with the top, back, and left sides coloured grey.

Hence, the correct answer is option c.

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5. If there is a dot on each corner of the cubes below, then how many dots are hidden?

a) 3 b) 4 c) 5 d) 6
Answer: b
Looking at the question logically, considering that each corner has a dot, and since there are 2 cubes
(with a total of 8 corners each), there should be a total of 16 dots. However, only 12 dots are visible,
indicating that there are 4 dots hidden. Hence, the correct answer is option b.
6. What will come in place of 'A' and 'B'?

a) b)

c) d)
Answer: c
The sequence is a repeating sequence of 3 differently shaped objects: Cone, Cube, Sphere. Thus, the
conical object will replace A and spherical object will replace B.
Thus, option c is the correct answer.
7. Find the odd one out.

a) b) c) d)
Answer: b
In all the boxes except the box in option b, the 2D figure shows the top view of the 3D figure in the same
box. In box b, the 2D figure (triangle) is not the top view of the 3D figure (cone).
Hence, option b is the odd one out.

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8. If the given solid is viewed from the direction of the arrow, what will it look like?
Note: You cannot rotate the question or option images

a) b) c) d)
Answer: c

From the side view, four blocks are visible. In this view, three blocks are stacked vertically, and the fourth
block extends outward towards left, from the middle block. Hence, the correct side view is option c.
9. How many edges of the solid shown below have the same-coloured dots on both of their
corners?

a) 2 b) 3 c) 4 d) 5
Answer: c
As shown below, there are 4 such edges of the solid that have the same-coloured dots on both of their
corners.

Thus, option c is the correct answer.

Page 24

10. Which of the following options can be formed by combining the shapes given below?

a) b) c) d)
Answer: d
When the 1st and 2nd shapes are combined, the resulting shape looks as shown below. This combined
shape is found in options b and d.
As highlighted below, in option a, the middle block is missing,

While in option c, the third block in the first row is missing.

By adding the 3rd figure, the resulting image looks as shown below, which matches option d.

Hence, the correct answer is option d.

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3. Two friends A and B play a game of tennis with 5 sets. No set can end in a
draw. Anyone who wins 3 or more sets will win the game. A won set 1, B
won sets 2 and 3. If B did not lose any EVEN NUMBERED set, who won the
tennis game?
(a) A
(b) B
(c) Game ended in a tie
(d) Cannot be Determined
Answer: b
B won sets 2 and 3.
B also won set 4 because B won all even numbered sets
(B did not lose any even numbered set and no set was drawn)
Therefore, B won 3 out of 5 sets. Therefore, B won the game.
Thus, option b is the correct answer.

Page 26

Chapter 3: Double Century
1. At minimum, how many more beads are required to show number 496 on the given abacus?

a) 487 b) 253 c) 10 d) 11
Answer: c
The given abacus shows:

● Hundreds place: 2 beads - 200

● Tens place: 4 beads - 40

● Ones place: 3 beads - 3

So, the number shown is 243.

We need to make 496.
Now, let’s find the difference between the two numbers:
496 − 243 = 253

This means we need 2 more hundreds, 5 more tens, and 3 more ones, i.e.,
2 + 5 + 3 = 10 more beads in total.

Hence, 10 more beads are required to show the number 496.

Hence, the correct answer is option c.
2. If each of the following terms follows the same theme, what will come in place of "?"

a) b)

c) d)
Answer: d

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In each term, a consecutive three-number series is followed. Before the arrow, one number from the
series is given, and after the arrow, the missing numbers are filled in according to their correct positions.
For example, in the first term, we have "__ 455 __". So, after the arrow, the missing numbers "454" and
"456" are placed.
Similarly, in the fourth term, we have "__ __ 439". So, after the arrow, the missing numbers "437" and
"438" will be placed. Hence, the correct answer is option d.
3. Each shape has a digit in it. Which of the following numbers in the options can be formed by
using the digits in the CIRCLES only?

a) 615 b) 142 c) 467 d) 146
Answer: d
The digits inside the circles are 6, 1, and 4. Using these digits, the number 146 can be formed, which
matches option d.

Thus, option d is the correct answer.
4. How many numbers will remain after Sam cancels all numbers less than 432 and all numbers
greater than 567 in the set given below?
451, 430, 568, 411, 392, 517, 533, 578, 420, 525, 566, 616, 571, 436
a) 5 b) 6 c) 7 d) 8
Answer: b
A systematic approach to avoid mistakes:
Step 1: Remove all the numbers less than 432
451, 568, 517, 533, 578, 525, 566, 616, 571, 436
Step 2: Remove all the numbers greater than 567
451, 517, 533, 525, 566, 436
Remaining numbers are:
451, 517, 533, 525, 566, 436
Hence, there are 6 numbers remaining.
Thus, option b is the correct answer.

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5. What will come in place of "?"

a) b) c) d)
Answer: b
In each figure, the digits shown on the left are used to form three 2-digit numbers. The digit pointed to by
the arrow becomes the tens digit in all three numbers. The remaining digits become the ones digits.
The three numbers formed are then arranged in ascending order.
In the 4th term, the arrow points to 8. Using 8 as the tens digit, we create 86, 89, and 85.
These are then arranged as 85, 86, and 89.
Thus, option b is the correct answer.
6. If you remove all the numbers greater than 5 from row 1, all the numbers less than 4 from row 2,
and all the numbers greater than 7 from row 3, then which row will have the largest count of
numbers remaining?

a) Row 1 b) Row 2
c) Row 3 d) Both rows 1 and 3 have equal numbers
Answer: c
If the numbers are removed from the rows as indicated in the question, then Row 3 will have the
highest count of remaining numbers. (Refer to the image given below).

Hence, the correct answer is option c.
7. Given below is a game board of snakes and ladders. Which ladder is most beneficial?

a) Black b) Blue c) Red d) Green
Answer: b

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The Black ladder will give you the benefit of 4 steps.
The Blue ladder will give you the benefit of 12 steps.
The Red ladder will give you the benefit of 6 steps.
The Green ladder will give you the benefit of 2 steps.
Thus, the Blue ladder is the most beneficial ladder.
Hence, the correct answer is option b.
8. In the image given below, what is the maximum number of dots that can be connected by
drawing a single straight line?

a) 3 b) 4 c) 5 d) 6
Answer: c
As shown in the image below, the greatest number of dots that can be connected with a single straight
line is 5. Hence, the correct answer is option c.

9. The rabbit has to reach the end of the path by stepping on EACH block WITHOUT skipping any
blocks in between. It collects an item from every 5th block it steps on. Which of the following
objects did it NOT collect?
Note: The rabbit can step on a block exactly once

a) b) c) d)
Answer: c
The rabbit collects the item from every 5th block it steps on. So, it will collect the objects that are
highlighted below. From the options given, only option c has the object which will not be collected by
the rabbit. Thus, option c is the correct answer.

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10. Altogether, 25 coins are to be arranged in different stacks in such a way that each stack has
one more coin than the previous stack. If the last stack has 7 coins, how many such
stacks will be required so that no coins are left?
a) 4 b) 5 c) 6 d) 7
Answer: b
It is given that each stack has one more coin than the previous stack, which means each stack has
one less coin than the next stack.
Since it is given that the last stack has 7 coins, we work backwards and count the total number of coins.
So, the earlier stack has 6 coins, 5 coins, and so on. We see that when we reach the stack with 3 coins,
we reach a total of 25 coins (7 + 6 + 5 + 4 + 3). Hence, 25 coins are arranged in 5 stacks.
Thus, option b is the correct answer.

4. My 4-digit phone passcode is made using only odd numbers, and no number
is used more than once. 3 is the smallest digit in the code, and is also the last
digit of the code. The largest digit and the smallest digit of the code have only
one digit in between them.
5. What is the hundred's digit (2nd digit from the left) of my phone pass code?
6. Note: None of the digits in the pass code are repeated. Each digit appears only
once
(a) 7
(b) 5
(c) 9
(d) Either option a or c
Answer: c
In the phone passcode, 3 is the smallest odd digit. So, 1 cannot be in the passcode.
Since none of the digits in the passcode are repeated, the remaining odd digits
must be 5, 7, and 9.
In the code, 3 is the last digit of the code (_ _ _ 3).
The largest and smallest digits in the code have exactly one digit between them.
Therefore, the largest digit must be in the hundreds place. Among the remaining
digits, 9 is the largest. So, 9 is placed in the hundreds place (_ 9 _ 3).
Thus, the correct answer is option c.

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Chapter 4: Vacation with My Nani Maa
1. Jars A, B, and C together have 9 balls, with each ball having a number written on it. Each number
appears EXACTLY 3 times. What is the sum of the numbers in Jar C?

a) 8 b) 9 c) 7 d) 6
Answer: a
There are a total of 9 balls, some of which are already placed in the jars.
Step 1: Determine the number of balls in Jar C
● Jar A has 4 balls, and Jar B has 2 balls
● Total balls in A and B = 4 + 2 = 6
● Since the total number of balls is 9, the number of balls in Jar C is:
● 9 - 6 = 3 balls.

Step 2: Identify the numbers in Jar C
Each ball number appears exactly 3 times across all jars. Let's analyse:
● Ball numbered 1 has already appeared three times (2 in Jar A, 1 in Jar B)
● Ball numbered 2 has appeared twice (once in Jar A, once in Jar B). So, the third
occurrence must be in Jar C
● Ball numbered 3 has appeared only once (once in Jar A). So, the remaining two
occurrences must be in Jar C

Thus, the balls in Jar C are:
● One ball numbered 2
● Two balls numbered 3

Step 3: Calculate the sum of numbers in Jar C
2+3+3=8
Final Answer: The sum of the numbers in Jar C is 8. Hence, the correct answer is option a.
2. For each shape, if we remove the smallest digit, the difference between the remaining digits of
which two shapes will be the LEAST?

a) b)

c) d)
Answer: d

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The smallest digit from each shape is removed.

Remaining digits of each shape:
Circle: 7
Pentagon: 5
Triangle: 6
Rectangle: 6
As the triangle and rectangle have the same digit 6, their difference is 0, which is the LEAST possible
difference between any two shapes.
Hence, option d is correct.
3. A, B, and C have some amount of money (in rupees) as shown below:
A = 50, B = 100, and C = 150
How much money should B and C together give to A such that each of them ends up with the
same amount of money?
a) 15 b) 25 c) 50 d) 75
Answer: c
Let's break down the situation. A, B, and C currently have 50, 100, and 150 rupees, respectively. The
total money among them is 300 rupees. Ideally, each person should receive an equal share, which
amounts to 300/3 = 100 rupees per person.
However, B and C already have 100 and 150 rupees, respectively. To ensure that everyone ends up
with the same amount, B and C need to give some money to A.
B, who has 100 rupees, doesn't need to give anything to A since B's amount is already equal to the
target amount.
C, with 150 rupees, needs to give A enough money to balance the amounts. The amount C needs to
give to A is 150 - 100 = 50 rupees.
Therefore, when B and C together contribute, they should give 50 rupees to A to ensure that each
person has an equal share. Hence, the correct answer is option c.
4. Starting from the left end of a row, there are three kites, followed by two umbrellas, then three
balls, and then three bottles in that order. Which item is located 7th from the right end of the
given row?

a) Umbrella b) Kite c) Bottle d) Ball
Answer: a
The easier approach to solve this is to count the items from the RIGHT END OF THE ROW:
3 bottles + 3 balls + 2 umbrellas + 3 kites (this sequence is in reverse order: from right to left)
After 3 bottles and 3 balls, the 7th item will be an umbrella only. Hence, the correct answer is option a.

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5. Tom (a boy) has 2 brothers. Each of his sisters has 2 sisters. How many children does Tom's
mother have?
a) 7 b) 6 c) 5 d) 4
Answer: b
Tom has 2 brothers, making it 3 boys including him. Additionally, each of his sisters has 2 sisters,
giving a total of 3 sisters. So, the total number of children Tom's mother has is 6. Hence, the correct
answer is option b.
6. The given boxes have some candies, and if EACH of the given boxes should have EXACTLY 3
candies, then how many MORE candies do we need in total?

a) 1 b) 2 c) 3 d) 4
Answer: b
The 1st box has 2 candies; therefore, we need to add 1 more candy to make it 3.
2+1=3
The 2nd box has 3 candies. So, no additional candies are needed.
The 3rd box also has 2 candies, which means we need to add 1 more candy to reach 3.
2+1=3
In total, we need 2 more candies, 1 for the 1st box and 1 for the 3rd box, so that each box has exactly
3 candies. Hence, the correct answer is option b.
7. A frog is standing on a number line at 0. It jumps on the number line such that in each jump, it
moves 4 steps forward but then slips back 2 steps.
In how many jumps will the frog reach number 18 for the first time?
a) 8 b) 9 c) 10 d) 11
Answer: a
In each jump, the frog moves 4 steps forward but then slips back 2 steps, effectively covering 2 steps
after each complete jump.
After 7 jumps, the frog will be at 14 (since 7 × 2 = 14).
On the 8th jump, it will move 4 steps forward from 14 and reach 18 before slipping back.
This is the first time the frog reaches 18.
Therefore, the frog reaches number 18 for the first time on the 8th jump.
Option a is the correct answer.
8. X, Y, and Z start a game with 100 points each. After each round, the winner takes 10 points from
each player. Z won only the fourth round, and they played 5 rounds in total. What is Z's final
score? Note: No round can end in a draw
a) 70 b) 80 c) 90 d) 100
Answer: b

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After each round, the winner receives 10 points from each player. They played 5 rounds, in which Z won
only 1 round (the fourth round). This means Z lost the remaining 4 rounds.
Initially, Z has 100 points. By losing 4 rounds, Z lost (4 x 10 points) = 40 points (these points were given
to the winners of each round).
On winning one round, Z received 10 points from both X and Y: (2 x 10 points) = 20 points.
Therefore, the final score of Z is calculated as follows:
Final score of Z = 100 - 40 + 20 = 80 points.
Hence, the correct answer is option b.
9. Which number will you NEVER get if you add a single digit number and a two-digit number?
a) 11 b) 102 c) 105 d) 109
Answer: d
The greatest single digit number is 9 and the greatest two-digit number is 99. Thus, the maximum sum
we can get is 108. Since 109 is greater than 108, we can never get it. Hence, the correct answer is
option d.
10. A and B are playing a game, where the winner of every round receives 3 points and 1 point will
be reduced from the loser's score. If they start with 10 points each, what could be the MAXIMUM
possible difference between their final scores, if they play 5 rounds altogether?
a) 20 b) 25 c) 15 d) 10
Answer: a
Step 1: Maximizing the Score Difference
To achieve the maximum difference, one player must win all 5 rounds, while the other loses all 5 rounds.
Winner’s final score:
● Gains 3 points per round: 3 × 5 = 15 points
● As the players start with 10 points, Total score = 10 + 15 = 25
Loser’s final score:
● Loses 1 point per round: 1 × 5 = 5 points lost
● As the players start with 10 points, Total score = 10 - 5 = 5
Step 2: Finding the Maximum Difference
Difference = 25 - 5 = 20
Thus, the maximum possible difference between their final scores is 20.
Thus, option a is the correct answer.

Page 35

There are 3 identical glasses A, B, and C which are filled equally. Now,
(A) Half of the water in glass A is poured into glass C
(B) Half of the water in glass C is poured into glass B
Which glass has the maximum amount of water?
(a) A
(b) B
(c) C
(d) Either A or C
Answer: b
Let us say that each glass has 4 litres of water.
A) Half of the water in glass A is poured into glass C
Thus, glass A will have 2 litres and glass C will have 6 litres.
B) Half of the water in glass C (3 litres) is poured into glass B
Thus, glass B will have 7 litres of water, C will have 3 litres of water. We know
from the first step that A has 2 litres of water.
Hence, B will have the maximum amount of water.
Thus, option b is the correct answer.

Page 36

Chapter 5: Fun with Shapes
1. Given below is an incomplete grid. Which option will complete the grid such that ALL the grey
boxes TOGETHER form a square?

a) b) c) d)
Answer: c
To see if it makes a square, count each small box as one unit. A square has all sides equal, so the
grey boxes together should make equal sides. As shown below, only option c makes a square because
it has the same number of grey boxes on every side.

Hence, option c is the correct answer.
2. Count the number of rectangles in the given figure.
Note: Please count all squares as rectangles for the purpose of this question

a) 3 b) 4 c) 5 d) 6
Answer: b

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As highlighted below, there are 4 rectangles in the given figure.

Thus, option b is the correct answer.
3. When the sheet is folded along the red dotted line, which two identical shapes will overlap?

a) Triangle b) Square c) Circle d) Diamond
Answer: a
When the paper is folded along the dotted lines, corner A will overlap with corner B, and corner C will
overlap with corner D.

Hence the folder paper will look like:

Here, the shapes with dotted edges represent the shapes from part A that overlap with part B of the
paper.
As we can see, the only shape that overlaps with its identical shape is the triangle.
Hence, option a is the correct answer.
4. Which of the following shapes CAN NEVER give a 4-sided figure, when folded only once into
two equal halves?

a) b) c) d)
Answer: d

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As shown below, option d will not form a four-sided figure when folded into two equal halves.

Thus, option d is the correct answer.
5. If the boxes keep getting added in the same pattern, then in which term, a complete square (made
up of boxes) will be formed?

a) 7 b) 8 c) 9 d) 11
Answer: c
Here, in every next term, the number of squares is increasing by 1.
In the 1st term, there is 1 square, in the 2nd term, there are two squares, the 3rd term has 3 squares,
the 4th term has 4 squares and so on.
Also, the new square is added each time, in a specific order.
The sequence begins by building a vertical stack (column) of three squares - bottom to top, then the
second column is built from top to bottom and the third column is built again from bottom to top.
Thus, if we follow the same rule, we will get 9 squares in the 9th term, which form a square shape with
three rows and three columns.
Hence, the correct answer is option c.

6. The given shapes must be arranged to form a square without removing any shape. Which of the
following actions are required to do so?

a) All the green shapes must take one left turn
b) All the green shapes must take one right turn
c) All the orange shapes must take one right turn
d) All the orange shapes must take one left turn
Answer: c

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As seen in the image, when the orange shapes take a right turn, the resulting shapes form a rectangle
which is a closed shape. Hence, the correct answer is option c.

7. Which of the following shapes is NOT HIDDEN in the given figure?

a) b) c) d)
Answer: c
As highlighted below, option c is not hidden in the given figure.

Thus, option c is the correct answer.
8. What will come in place of "?"

a) b) c) d)
Answer: b

Page 40

In the given analogy, the two shapes on the top combine to form a perfect square.
This shows that the pieces fit together such that all outer edges become straight and equal, completing
the square.
Similarly, in the second case, we must choose the shape that, when combined with the given shape,
also forms a complete square.
Options a and c can be eliminated because their slanted edges do not align properly with the given
shape. When joined, they would leave uneven or non-straight outer edges, which cannot form a square.
Option d can be eliminated because, despite having straight edges, its shape and size do not match the
missing portion. When combined, it would either overlap or leave a gap, so a perfect square cannot be
formed.
Option b fits correctly with the given shape. When the two are combined, all outer edges align to form
four straight, equal sides, resulting in a perfect square.
Therefore, option b is the correct answer.

9. Identify the image which will complete the square grid given below.
Note: You CANNOT rotate the question image or the option images to find the answer

a) b) c) d)
Answer: a
As shown below, option a will complete the given square.

Thus, option a is the correct answer.

Page 41

10. The figure below is formed of different blocks. Which block appears the MOST number of times
in the given figure?

a) b)

c) d) All of them occur equally
Answer: a
Let us analyse each of the options:
Option a: The block in option a appears three times (refer to Image A)
Option b: The block in option b appears twice (refer to Image B)
Option c: The block in option c appears twice (refer to Image C)
Option d: All of the blocks do not appear the same number of times
Hence, the block in option a appears the MOST number of times in the given figure.

Thus, option a is the correct answer.
11. How many times is Image X present in the Question Image?
Note: Rotation of Image X or the Question Image is not allowed

a) 1 b) 2 c) 4 d) 5
Answer: c

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As highlighted below, Image X is present 4 times in the Question Image.
Thus, option c is the correct answer.

A coach must make 2 teams of two players each from four friends A, B, C,
and D such that each player can be in only one team. He needs to follow the
following conditions:
a. A cannot team up with B
b. C cannot team up with B
Which of the following is a possible team?
(a) D and A
(b) C and D
(c) C and B
(d) D and B
Answer: d
According to the given conditions, A and C cannot team up with B. So, only
person who can team up with B is D.
Therefore, A and C are remaining for the second team.
Hence, the correct answer is option d.

Page 43

Chapter 6: House of Hundreds – I
1. How many 3-digit numbers can you make with only 4 and 5, if one of the digits is present
exactly twice?
a) 3 b) 5 c) 6 d) 9
Answer: c
3-digit numbers have to be formed using 4 and 5, where one of the digits is used exactly twice.
So, the numbers must be formed using (4, 4, 5) or (4, 5, 5).
The possible combinations are: 445, 454, 544, 455, 545, 554.
Therefore, 6 such numbers can be formed. Hence, the correct answer is option c.
2. If the hundreds digits of each of the numbers given below are shifted to the units place, then
how many numbers will be less than 150, finally?
314, 616, 511, 127, 814, 915, 118, 714
a) 3 b) 4 c) 5 d) 6
Answer: b
Step 1: Shift the hundreds digits: 143, 166, 115, 271, 148, 159, 181, 147
Step 2: Identify numbers less than 150: 143, 115, 148, 147
There are 4 numbers less than 150. Hence, the correct answer is option b.
3. What is the difference between the largest and the second largest three-digit numbers formed
using the digits 0, 3, and 1 such that no digit is repeated within a number?
a) 180 b) 99 c) 9 d) 1
Answer: c
The largest 3-digit number formed using 0, 3, and 1 is: 310
The second largest 3-digit number formed using 0, 3, and 1 is 301.
Hence, the difference is 310 - 301 = 9. Thus, the correct answer is option c.
4. Find the odd one out.
Note: It is not related to the number of letters, vowels, or consonants of the words
a) Thirty-Two tens and two ones b) Three hundreds and Twelve ones
c) Three hundred and twenty-two d) Three hundreds, two tens and two ones
Answer: b
In this question, we have to find the odd one out based on the values each option represents, rather
than the structure or letters of the words.
Here’s how each option translates to a numerical value:
Option a: Thirty-Two tens and two ones
Thirty-Two tens = 320

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Two ones = 2
So, thirty-two tens and two ones mean 320 + 2 = 322

Option b: Three hundreds and Twelve ones
Three hundred = 300
Twelve ones = 12
So, three hundreds and Twelve ones = 300 + 12 = 312

Option c: Three hundred and twenty-two = 322

Option d: Three hundreds, two tens and two ones
Three hundred = 300
Two tens = 20
Two ones = 2
So, three hundreds, two tens and two ones = 300 + 20 + 2 = 322
As all the options except option b represent 322.
Thus, option b is the correct answer.
5. In a 3-digit number, the digit in the unit’s place is twice the digit in the hundreds place. What
could be the lowest possible value of the sum of the digits?
a) 0 b) 3 c) 4 d) 5
Answer: b
We have to find the lowest possible value of the sum of the digits.
So, we must use the lowest possible digits.
Thus, the digit in the hundreds place = 1 (If we take 0, then it will not be a 3-digit number).
The digit in the unit’s place is twice the digit in the hundreds place = 2.
The number could be 102.
Sum of digits = 3.
Hence, the correct answer is option b.
6. I am given a number 435. If the first (hundreds) and third (units) digits of this number are
interchanged, then the new number will be ________ the original number.
a) smaller than b) same as c) greater than d) cannot say
Answer: c
Given number: 435
After interchanging, the number becomes: 534
Hence, the new number will be greater than the original number.
Therefore, the correct answer is option c.

Page 45

7. Nishit remembers that his friend Shreya’s score is more than 160 but less than 190, whereas
their friend Nitigya remembers that Shreya’s score is more than 170 but less than 200. If Nishit
and Nitigya both remember correctly, and the unit’s place of Shreya’s score is 0, what is her
score?
a) 150 b) 160 c) 170 d) 180
Answer: d
The unit’s digit of Shreya’s score is ‘0’.
As per Nishit, Shreya’s score can only be 170 or 180.
As per Nitigya, Shreya’s score can only be 180 or 190.
Between the two, only 180 is common. Therefore, Shreya’s score is 180.
Hence, the correct answer is option d.
8. Lalit has 3 marbles and 3 boxes, where each box represents a place value, as labelled.
He places the marbles into the boxes to form a number.
For example, in the image shown below, the number 120 is formed.
How many different numbers can Lalit form if he uses all 3 marbles every time to make a number?

a) 6 b) 9 c) 10 d) 11
Answer: c
The numbers Lalit can form depend on how he places the 3 marbles in the 3 boxes.
1. All 3 marbles together:
• 3, 30, 300
2. 2 marbles together and 1 separate:
• 120, 102, 12, 210, 21, 201
3. All 3 marbles separate:
• 111
So, the total number of different numbers that can be formed = 3 + 6 + 1 = 10
Hence, Lalit can form 10 different numbers.
Thus, option c is the correct answer.
9. In a 3-digit number, the digit at the unit's place represents the total number of 0's this number
has. Which of the following must be the digit at the unit's place in this number?
a) 0 b) 1 c) 2 d) 3
Answer: b
Let's assume a 3-digit number is XYZ.
Z is the unit digit, and it represents the total number of 0's this number has.

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If the units digit is 0, it means the number should contain 0 zeroes. But since Z itself is 0, the number
already contains at least one zero. This contradicts the condition. So, Z cannot be 0.

Z can be 1. In that case, a possible number could be 101 (or any other number in which the unit digit is
1 and the tens digit is 0. The digit at the hundreds place could be any digit from 1 to 9).

Z cannot be 2 as well because in that case the values of X and Y will be 0, and then the number won't
be a 3-digit number. Instead, it will become a single-digit number.

Z cannot be 3 as well because we have only 2 digits remaining, excluding the unit digit.
Hence, the correct answer is option b.
10. If P and Q are two 3-digit numbers, then which of the following CANNOT be the difference of P
and Q?
a) 950 b) 150 c) 300 d) 750
Answer: a
To determine which number cannot be the difference of P and Q we need to find the maximum
possible difference. By subtracting the smallest three-digit number (100) from the largest three-digit
number (999), we obtain a maximum possible difference of 899. When comparing this to the given
options, we notice that the number 950 exceeds the maximum possible difference of 899. Since, 950
cannot be the difference of P and Q. Hence, the correct answer is option a.

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If every row follows the same theme, what will come in the place of "A" and "B"?

(a)

(b)

(c)

(d)
Answer: c
The total number of small blocks next to each number corresponds to its numerical
value.
• In the first row, 3 has 3 small blocks, 5 has 5 small blocks (3 on the left + 2 on the
right), and 2 has 2 small blocks
• Applying the same pattern to the last row:
• 2 has 2 blocks next to it
• 5 must have 5 blocks next to it. Since there are already 2 blocks on the right, B must
be 3 blocks
• A has one block on the left and 3 blocks on the right (B), making a total of 4 blocks.
So, A = 4
Thus, the correct answer is option c (A = 4, B = 3 blocks).

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Chapter 7: Raksha Bandhan
1. Every class that visits the National Museum is divided into groups of 6 students for each tour
guide. Among the given options, which of the classes would not be able to form groups of 6
students with none left over?
Note: The number in the option is the number of students in that class
a) 36 b) 42 c) 38 d) 48
Answer: c
Only those classes which have a student count which is a multiple of 6 will be able to form groups of 6
students with no one left over. Since the class with 38 students will only be able to form 6 groups of six
students and 2 students will be left over, the correct answer is option c.
2. What will come in place of "?" in the given series?

a) b) c) d)
Answer: a
The following things are happening in the given sequence:
- The series follows an even number sequence starting from 0, 2, 4, and so on
- The number of shaded circles around the shape matches the number inside the shape
- The shading begins at the left and right corners, and in each next term, the adjacent circles
are shaded in sequence from both sides
Following this pattern, the next even number in the series is 8, with 8 circles shaded.
Therefore, option a is the correct answer.
3. I have to distribute a few sweets. If I keep 2, 3, or 4 sweets in a pack, then I am left with one
sweet after all packs are made. But if I put 5 sweets in a pack, I am left with none. What is the
MINIMUM number of sweets I have?
Note: I have more than one pack
a) 25 b) 50 c) 54 d) 55
Answer: a
As no sweets are left when packed in groups of 5, the total number of sweets must occur in the 5 table.
(5, 10, 15, 20....). Since 1 sweet is left when they are packed in groups of 2, 3, or 4, the total number is
not divisible by 2, 3, or 4.
Now, we have to identify the MINIMUM possible number of sweets.

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If 5 is the minimum number of sweets, we would have only one pack.
If 10 is the minimum number of sweets, then the sweets can be distributed in packs of 2 as well.
Similarly, 15 and 20 also cannot be the minimum number of sweets as they can be distributed equally
in packs of 3 and 4 respectively.
Hence, 25 is the MINIMUM possible number. Thus, option a is the correct answer.
4. Which of the following statements is/are necessarily required to answer the question below?
Question: Which number am I?
Statement 1: I am not an even number.
Statement 2: If you multiply any number with me then the answer is the same as the number
which was multiplied with me.
a) Only statement 1 is required
b) Only statement 2 is required
c) Both statements 1 and 2 are required
d) Cannot be answered even if we use both the statements together
Answer: b
Using only statement 1, we can’t determine the value of the number. Thus, statement 1 alone is
insufficient to answer the question.
Any number multiplied by 1 will give you the same number. Hence, only from statement 2 we know that
the number is 1. Thus, option b is the correct answer.
5. PQ and RS are 2-digit numbers in the equation given below. How many different possible values
can “RS” have?

a) 12 b) 13 c) 14 d) 15
Answer: d
It is given that RS = 4 x PQ
RS is a 2-digit number that is the product of 4 and another 2-digit number PQ.
Now, as RS is a 2-digit number, it must be less than 100.
So, we have to find out the possible 2-digit values of “PQ”, which when multiplied by 4, give a result
less than 100.
We know that 10 is the smallest 2-digit number.
4 x 10 gives 40.
The least possible value of RS is 40.
Similarly, all numbers from 10 to 24 when multiplied by 4 give an answer less than 100.
25 or a greater number when multiplied by 4 gives at least 100.
So, there are 15 such possible values for RS.
Hence, the correct answer is option d.

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6. Reema put 10 x 7 plastic cups in the first stack, 10 x 8 plastic cups in the second stack, and
10 x 9 plastic cups in the third stack. If this pattern continues, how many plastic cups will
Reema put in the fifth stack?
a) 100 b) 108 c) 110 d) None of these
Answer: c
1st stack = 10 x 7 = 70 cups
2nd stack = 10 x 8 = 80 cups
3rd stack = 10 x 9 = 90 cups
Reema puts 10 more cups in each subsequent stack. So, the fourth stack should contain 100 cups and
similarly the fifth one should contain 110 cups. Hence, the correct answer is option c.
7. I have some toys. If I take 2 times the number of toys with me and divide them equally into
groups of 30, then 3 groups will be formed. What is the number of toys I have?
a) 5 b) 20 c) 45 d) 180
Answer: c
3 Groups of 30 toys will altogether have 30 + 30 + 30 = 90 toys.
If 90 is 2 times the number of toys with me, then I must have 45 toys.
Hence, the correct answer is option c.
8. If all the given triangles follow the same pattern, what will come in place of "?"

a) 15 b) 19 c) 32 d) 17
Answer: b
The number in the middle of the triangle is obtained by multiplying the two numbers at the top and then
subtracting the number written at the bottom of the triangle from the product.
So, 9 × 3 = 27, and 27 – 8 = 19.
Therefore, option b is the correct answer.

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9. What will come in place of "?"

a) 24 b) 30 c) 25 d) 15
Answer: b
The number in the outermost circle is the product of the sum of the numbers in the layer below
(hexagon) it and the number in the middle circle.
For example, 36 is the product of 9 + 3 = 12 and 3.
Similarly, 5 + 5 = 10 and 10 x 3 = 30 is the answer.
Hence, the correct answer is option b.
10. What will come in place of "?"

a) b) c) d)
Answer: c
The number on the right is the product of the number of lines of different colours on the left. The
number is written in the same colour as the horizontal lines.
For example, 1st pair has 2 vertical lines and 2 horizontal lines. So, their product will be 2 × 2 = 4
(written in purple colour)
The 2nd pair has 3 grey and 4 green lines; thus, their product will be 3 × 4 = 12. (written in green
colour)
Similarly, the last pair has 3 blue and 3 orange lines, so their product will be 3 × 3 = 9 (written in blue
colour).
Hence, the correct answer is option c.

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There are 3 rods A, B, and C placed at different levels as shown below.
• Rod A moves 1 level up
• Rod B moves 2 levels down
• Rod C moves 2 levels up
HOW MANY levels are COMMON between ALL three rods?

(a) 0
(b) 1
(c) 2
(d) 3
Answer: b
After the movement of all three rods as given in the question, only one level will
be common between all three rods. So, the answer is option b.

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Chapter 8: Fair Share
1. A piece of paper was folded in half to get a semicircle. What was the original shape before the
fold?

a) b) c) d)
Answer: d
A circle can be folded in half to get a semicircle. So, the original shape of the paper is a circle. Hence,
the correct answer is option d.

2. If the alphabets are divided into two equal halves - from A to M and N to Z, which letter in the
other half would be in the same position as the letter J?
a) Q b) V c) X d) W
Answer: d
We divide the alphabet into two equal halves:
• First half: A to M (13 letters)
• Second half: N to Z (13 letters)
Write the first half with positions:
A (1), B (2), C (3), D (4), E (5), F (6), G (7), H (8), I (9), J (10), K (11), L (12), M (13)
So, J is the 10th letter in the first half.
Now, write the second half with positions:
N (1), O (2), P (3), Q (4), R (5), S (6), T (7), U (8), V (9), W (10), X (11), Y (12), Z (13)
The 10th letter in the second half is W.
Hence, the correct answer is option d.
3. 3 people A, B, and C have to cover a distance of 12 m each. A has covered half the distance. B
has covered half the distance covered by A, and C has covered double the distance covered by
B. Who among the following has covered the least distance?
a) A
b) B
c) C
d) All three of them have covered equal distance
Answer: b

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A has covered half the total distance, which means A has covered 6 m.
B has covered half the distance covered by A, which means B has covered 3 m.
C has covered double the distance covered by B, which means C has covered 6 m.
Thus, A and C have each covered 6 m, while B has covered 3 m. Hence, B has covered the least
distance. Thus, option b is the correct answer.
4. Fill in the blank.
Half of 4 + Half of 8 = ___
a) 12 b) 10 c) 8 d) 6
Answer: d
Half of 4 + Half of 8
=2+4
=6
Hence, the correct answer is option d.
5. If ‘#’ means ‘half of 2’, ‘$’ means ‘double of 2’, and ‘&’ means ‘2’, then which of these options
results in 3?
a) 6 – $ – # b) 6 – # – & c) 6 + & + # d) 6 – # + &
Answer: b
Given:
‘#’ means half of 2
‘$’ means double of 2
‘&’ means 2
Step 1: Find the values of each symbol
Half of 2 = 2 ÷ 2 = 1, so # = 1
Double of 2 = 2 × 2 = 4, so $ = 4
‘&’ means 2
Step 2: Check each option
Option a: 6 – $ – #
=6–4–1
=1
Option b: 6 – # – &
=6–1–2
=3
Option c: 6 + & + #
=6+2+1
=9
Option d: 6 – # + &
=6–1+2
=7
Hence, the correct answer is option b.

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6. A circular piece of paper is cut into four identical quarters. Two or more quarters are joined
together. Which of these figures can possibly be made by joining them? Note: You cannot overlap
the quarters
a) Triangle b) Semicircle c) Trapezium d) Rectangle
Answer: b
If a circle is cut from a paper into 4 identical (i.e. equal) parts, each part would be a quarter of a circle,
if 2 quarters are joined together, they CAN make a semicircle. Hence, option b is the correct answer.

7. There are 3 different circular sheets: A, B, and C. The radius of sheet B is twice the radius of
sheet A and half the radius of sheet C. Which sheet requires the HIGHEST number of folds to
attain the shape of a quarter circle?
a) A
b) B
c) C
d) All of them require the same number of folds
Answer: d
You can convert the circle to a semi-circle by making one-fold along the centre line (diameter of the
circle).
Then you can make another fold along the centre of the semi-circle to make a quarter.
The same process is applicable to circles of any radius.
Therefore, any circle requires exactly 2 folds.
Hence, the correct answer is option d.
8. If the number on the bottom face of the cube is double the number on the front face, what is the
sum of the number on the top face and bottom face?

a) 8 b) 7 c) 6 d) 4
Answer: b
Given:
The number on the bottom face of the cube is double the number on the front face.
From the figure:
● Number on the front face = 2
● Number on the top face = 3

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Step 1:
Number on the bottom face = Double of number on the front face
=2×2=4
Step 2:
Sum of the numbers on the top face and bottom face
=3+4=7
Hence, the correct answer is option b.
9. Three friends are playing a game. In this game, a staircase with numbers written on it is given.
The students shown on the right side have to stand on any random step. If a student stands on a
step where the numbers in the column below that step add up to either half or double their age,
then he/she will win the game. Which of A, B, C, D, or E is the step on which no one will win?

a) Step A b) Both steps B and C
c) Step D d) Step E
Answer: c
The numbers in the column below the steps will add up as follows:
Step A = 8 (Double of Nia's age)
Step B = 2 + 2 = 4 (Half of Tina's age)
Step C = 7 + 4 + 5 = 16 (Double of Tina's age)
Step D = 1 + 4 + 5 + 3 = 13
Step E = 3 + 2 + 3 + 2 + 4 = 14 (Double of Ron's age)
Only Step D does not add up to half or double of any of the friends' ages.
Hence, the correct answer is option c.
10. What will come in place of "?"
3_4_6 → 4_8_3

2_2_4 → 3_4_2

2_6_8 → ?

a) 4_12_8 b) 4_12_4 c) 3_3_4 d) 3_12_4
Answer: d
Following changes are happening
- 1st number increases by 1
- 2nd number doubles
- 3rd number becomes half
Similarly, if we make the following changes in the 3rd row then we will get 3_12_4.
Thus, option d is the correct answer.

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Sam removed a pair of digits from SET A such that the sum of the removed
digits equals the sum of the remaining digits in SET A.
Which of the following is NOT a possible two-digit number that can be
formed using the remaining digits?

(a) 14
(b) 32
(c) 42
(d) Both options b and c
Answer: c
We need to determine which two numbers can be removed from the given SET A
= {1, 2, 3, 4} such that the sum of the removed numbers equals the sum of the
remaining numbers.
Step 1: Calculate the total sum: 1 + 2 + 3 + 4 = 10
Since we are removing two numbers whose sum is equal to the sum of the
remaining two numbers, the sum of the removed numbers should be: 10 ÷ 2 = 5
Step 2: Find valid pairs
The pairs of numbers that sum to 5 are:
• (4,1) - Remaining digits: {2,3}
• (3,2) - Remaining digits: {1,4}
Step 3: Form possible two-digit numbers
• If (4,1) is removed - Remaining digits: {2,3} - Possible numbers: 23, 32
• If (3,2) is removed - Remaining digits: {1,4} - Possible numbers: 14, 41
Step 4: Identify the incorrect option
The given options: 14, 32, 42
• 14 - Possible
• 32 - Possible
• 42 - Not possible (since 4 and 2 are never together as remaining digits)
Thus, the correct answer is option c.

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Chapter 9: House of Hundreds – II
1. In a 3-digit number, the digit in the tens place is twice the digit in hundreds place. What could be
the maximum possible sum of the digits?
a) 21 b) 12 c) 27 d) 18
Answer: a
The digit in the tens place is twice the digit in the hundreds place.
So, the tens place and hundreds place digit pair can be (2, 1), (4, 2), (6, 3), or (8, 4).
Since we have to find the maximum possible value of the sum of the digits, we will use the maximum
possible digits.
So, the tens place digit will be 8, the hundreds place digit will be 4, and the unit place digit will be 9
(highest possible digit). So, the 3-digit number will be 489 and the sum of the digits will be 4 + 8 + 9 = 21.
Hence, the correct answer is option a.
2. Product of two numbers is 100. Which of the given options is DEFINITELY true?
a) Sum of these numbers is 25
b) Sum of these numbers is 20
c) Sum of these numbers is 52
d) None of these
Answer: d
The product of two numbers can be 100 in a number of ways. 20 and 5, 50 and 2, 4 and 25, and so on.
So, the sum of the two numbers cannot be uniquely determined. Hence, the correct answer is option d.
3. With regards to the figure given below,
X = (The smallest 3-digit number made using different digits from the circles) + (any one digit
present ONLY in a rectangle)
What is the highest possible value of X?

a) 255 b) 253 c) 252 d) 247
Answer: b
The smallest 3-digit number made using different digits from the circles is 247.
To maximize the value of X, we have to add the largest one digit present ONLY in a rectangle, which is
6.
So, X = 247 + 6 = 253.
Note: 8 is present in the rectangle as well as the triangle. So, it cannot be considered
Hence, the correct answer is option b.

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