CTAI Teacher Handbook Class 8

Computational
Thinking and
Artificial Intelligence

Class 8

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 networking, etc. Hence, CBSE approaches
this by integrating Computational Thinking with AI and other technological
advancements, without dependence on any platform.

Learners engage with problems involving powers and number systems, proportional
reasoning, geometric configurations and structured distributions, requiring
decomposition of multi-variable scenarios, identification of complex patterns and
design of stepwise algorithms under constraints. The Artificial Intelligence component
deepens understanding of the AI project lifecycle, data-driven decision-making and
ethical considerations such as bias and fairness, enabling students to critically analyse
how data and models influence outcomes. The document also provides pedagogical
guidance, resources, and assessment support aligned with NEP 2020 for effective
classroom implementation.

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 the British Institute (Business Management and
Advertising), Frameworks Mumbai (Advanced Certification in Computer Science), and
Workstation (3D Animation)), Field Expert

Material Production Group

1. 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
2. Dr. Ankit Vijayvargiya, Assistant Professor, School of Technology, Dhirubhai Ambani
University, Gandhinagar.
3. Mr. Jay Thakkar, Senior Technical Officer, Centre for Creative Learning, IIT Gandhinagar
4. Mr. Chris John, Project Scientist, Centre for Creative Learning, IIT Gandhinagar
5. Ms Rekha Malhotra {Alumna of the British Institute (Business Management and
Advertising), Frameworks Mumbai (Advanced Certification in Computer Science), and
Workstation (3D Animation)}, Field Expert
6. Mr. Mikin Lala, (IIT Roorkee and IIM Calcutta Alumnus), Field Expert
7. Ms. Amatullah Mustafa Neemuchwala, Field Expert
8. Ms. Telidevara Sree Lasya, Field Expert
9. Mr. Parth Oza, Field Expert
10. Mr. Deep Mayekar, Field Expert
11. Mr. Mayank Patil, Field Expert
12. Mr. Shivraj Ugale, Field Expert
13. Mr. Raj Dhorade, Field Expert
14. 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
PART-1 COMPUTATIONAL THINKING
SR. NO. CHAPTER PAGE
NO.

1. Introduction 5

2. How to Use this Book? 9

3. A Square and a Cube 11

4. Power Play 21

5. A Story of Numbers 29

6. Quadrilaterals 45

7. Number Play 52

8. We Distribute Yet Things Multiply 59

9. Proportional Reasoning 68

PART-2 ARTIFICIAL INTELLIGENCE

ETHICAL PAGE
SR. NO. CHAPTER TITLE
AWARENESS NO

Responsible
AI Project Lifecycle
1 problem solving 79

Understanding
Artificial Intelligence and Its
2 impact of AI 82
Applications
systems
Promoting
Data and Fairness in AI
3 fairness and 84
inclusivity
Ethical decision-
Ethics and Responsible AI
4 making in 90
technology

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 the 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, defence,
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 emphasises 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 computati onal thinking, digital literacy and
responsible use of technology, along with nurturing innovation, critical thinking, and ethical decision-
making capacities.

Page 5

1. Relevance: Importance of introducing Computational Thinking (CT) and Artificial
Intelligence (AI)
• Preparing for the future: To contribute to the world of work in modern societies, individuals
need capabilities such as problem solving, using data effectively, identifying patterns and
applying AI ethically for various purposes in life.
• Holistic Development: Study of CT and AI contributes to development of reasoning, logical
thinking, creative problem-solving skills, critical thinking, and ethical decision-making abilities,
leading to individual flourishing and the creation of responsible digital citizens.
• Interdisciplinary Relevance: Embedding CT and AI concepts helps students develop an
integrated view of the world by connecting various disciplines such as Mathematics, Science,
and Humanities, showing that knowledge is not compartmentalized.
• Innovation and Entrepreneurship: At its core, CT and AI are about solving problems and
devising innovative solutions, which leads to an entrepreneurial and innovative mindset.
• Ethical Awareness: Study of CT & AI will sensitize learners about the misuse and bias, fairness,
and inclusivity in AI systems.

2. Objectives: (Curricular Goals)
• CG-1: Develops skills and capacities of computational thinking, namely- decomposition, pattern
recognition, data representation, generalisation, abstraction, and algorithms to solve problems
where such techniques of computational thinking are effective.
• CG-2: Develop spatial and visual reasoning.
• CG-3: Gain foundational knowledge of AI, its types, and domains.
• CG-4: Understand key ethical terms such as bias and fairness in relation to AI.
• CG-5: Demonstrates proficiency to use Computer & other devices, computer applications for
learning and practical purposes such as data analysis, preparation of visual representations and
communication of ideas.

3. Learning Outcomes:
Computational Thinking (CT) Learning Outcomes
ABSTRACT THINKING
Students will be able to solve advanced, multi-layered problems involving abstract relationships
and hidden structures, using:
• properties and relationships of numbers (powers, factors, remainders, divisibility)
• generalization across different number systems (decimal, binary, ternary, Roman, Chinese
numerals)
• spatial visualization of 2D and 3D figures, including overlaps, intersections, and
transformations
• logical interpretation of symbols, codes and operations representing numerical or algebraic
ideas
• identification of essential information by ignoring irrelevant or misleading data

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PATTERN RECOGNITION
Students will be able to identify, compare and extend complex patterns involving multiple
simultaneous changes, formed using:
• Powers, exponents and numerical structures
• Relationships across different representations of the same number
• Geometric configurations and shape-based sequences
• Conditional patterns based on rules, constraints or dependencies
• Mixed patterns involving numbers, symbols, shapes, and movement

DECOMPOSITION
Students will be able to break down high-order logical problems into manageable components
by:
• Separating given conditions, constraints, and goals
• Analyzing multi-step processes such as distribution, transfers and exchanges
• Breaking numerical expressions into simpler equivalent forms
• Interpreting tables, grids, networks and diagrams with multiple dependencies
• Structuring problems involving multiple variables, positions or cases

ALGORITHMIC THINKING
Students will be able to design, follow and evaluate multi-step logical procedures to solve
problems involving:
• Rule-based transformations of numbers or symbols
• Stepwise movement on grids, tracks or paths with constraints
• Conditional instructions (if–then, either–or, must/must not)
• Sequential decision-making under given limitations
• Optimisation problems involving maximum or minimum outcomes

Artificial Intelligence (AI) Learning Outcomes
By the end of Grade 8, learners will be able to:
• Describe the stages of the AI project cycle as a stepwise structure (Define Problem, Collect
Data, Test AI Tools, Reflect and Improve)
• Apply no-code tools to tackle real-world problems and reflect on their utility/effectiveness
• Explain how AI uses data, find and research sources of bias in datasets, and apply basic
strategies to ensure fairness and inclusivity
• Recognize how bias in AI leads to unfair conclusions and realize the importance of
accountability, privacy, and serving human interests
• Explain the uses of AI in daily life and understand AI as a specific type of algorithm that uses
datasets, learning and prediction
• Analyse contributions of AI to fields like healthcare, automation, and education,
understanding both benefits and risks
• Describe AI ethics as the values and guidelines that ensure AI is created and used
responsibly

4. Mapped with NEP and NCF 2023:
• The National Education Policy (NEP) 2020 aims to position India as a leader in emerging fields
by integrating AI, Machine Learning and CT into school education

Page 7

• The National Curriculum Framework for School Education (NCF-SE), 2023 serves as the
foundation for implementation, drawing Curricular Goals from the Aims of Education.
• Learning standards are designed as foundational capacities that are progressive, age-
appropriate, and aligned to NCF-SE 2023.

5. Time Allocation:
The Middle Stage suggests 100 hours annually, allocated as follows for Grade 8:
• Advanced CT Skills: 40 hours per academic year
• Introductory Concepts of AI: 20 hours per academic year
• Interdisciplinary Projects: 40 hours total (20 hours for each of the two required projects)

6. Approach / Pedagogy:
• Activity-Based: Use of complex puzzles, riddles, and games to build on previous CT abilities
• Experiential Learning: Delivering fundamental AI concepts through explanations,
demonstrations, and hands-on experience
• Collaborative Work: Organising group discussions, debates and collaborative projects that
integrate CT & AI
• Inquiry-Based: Independent student activities such as data collection, organisation, analysis,
and creation of diagrams/flow charts using digital tools or manually
• Ethical Reflection: Case studies and debates on the social impact and ethical use of AI.

7. Assessment:
Assessment is continuous, formative and competency-based, focusing on the ability to apply
knowledge rather than rote memorization. Methods include:
• Written Tests and Practical Examinations
• Interactive Group Activities
• Thematic Projects and Reflective Journals
• Teacher Observation Journals and Group Discussions

8. CT and AI Transition:
Computational Thinking forms the intellectual backbone and foundation for learning AI. The
curriculum follows a phased approach where CT skills—like breaking problems into parts and
spotting patterns—build the cognitive structures necessary for students to eventually understand and
create AI-driven solutions.

Page 8

How to Use This Book?
PART-1 Computational Thinking
Part 1 of this handbook 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 read and 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.

PART-2 Artificial Intelligence

Part 2 of the handbook provides a structured introduction to Artificial Intelligence (AI) as a
technology that enables machines to learn from data, recognise patterns, and make decisions. The
concepts of AI are presented using simple explanations and real-life examples from areas such as
healthcare, education, transport, and communication.
Each chapter includes:
 Foundational understanding of AI concepts
 Real-life examples and applications of AI
 Introduction to key AI domains such as Data Science, Computer Vision, and Natural Language
Processing
 Activities and data-based tasks
 Reflection on ethical use of AI
The AI content progresses from introduction to application, including introductory predictive
techniques such as regression, classification, and clustering. The book emphasises ethical and
responsible use of AI, including introduction to bias, fairness, privacy, and safe use of technology,
enabling informed and thoughtful engagement with AI systems.

Teachers should approach the 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 9

PART-1

COMPUTATIONAL THINKING

Page 10

Chapter 1: A Square and a Cube
1. In the following grid, all the circles follow the same theme. What will be the value of A + B?

a) 12 b) 14 c) 13 d) 15
Answer: c
In the given grid, the number in each circle is the sum of the squares of the digits from the grey cells
connected to it. For example:
42 + 32 = 16 + 9 = 25
22 + 32 = 4 + 9 = 13
52 + 92 = 25 + 81 = 106
Similarly,
92 + B2 = 117
81 + B2 = 117
B2 = 117 - 81
B2 = 36
B=6
Also,
B2 + A2 = 85
62 + A2 = 85
A2 = 85 - 36
A2 = 49
A=7
A = 7 and B = 6. So, A + B = 13
Hence, option c is correct.
_____________________________________________________________________________________
2. Sam writes a list of natural numbers. The list has three perfect cubes and three perfect squares.
If no number in the list has more than two digits, what is the MINIMUM number of distinct
numbers he must have written?
a) 3 b) 4 c) 5 d) 6
Answer: b
Sam writes a list of natural numbers of three perfect cubes and three perfect squares, where no
number has more than two digits.
To minimize the number of distinct elements in the list, Sam should use numbers that are both perfect
squares and perfect cubes.
Let us see the perfect cubes and perfect squares:
Perfect squares = 1, 4, 9, 16, 25, 36, 49, 64, 81
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Perfect cubes = 1, 8, 27, 64
Since 1 and 64 are both perfect squares and perfect cubes, Sam needs 2 more numbers in the list.
Therefore, a minimum of 4 numbers can be written.
Hence, option b is the correct answer.
_____________________________________________________________________________________
3. If AB is a two-digit number whose cube is in the form of a 4-digit number “__ __ __ C” such that
A < C < B, how many different values can C have?
a) 2 b) 3 c) 4 d) More than 4
Answer: a
We know that 10 × 10 × 10 = 1000 is the smallest 4-digit cube number. So, AB can be 10 or more than
10.
Similarly, 21 × 21 × 21 = 9261, is the largest 4-digit cube number. So, AB can be any number from 10
to 21.
Also, given that A<C<B
The units digit of the number is greater than the units digit of its cube.
Regardless of how many digits a number has, the units digit of its cube always depends on the units
digit of the original number. Let’s look at the units digits of a number and its cube number:

Example Digit in the units place of the Digit in the units place of its cube
original number

10 × 10 × 10 = 1000 0 0

11 × 11 × 11 = 1331 1 1

12 × 12 × 12 = 1728 2 8

13 × 13 × 13 = 2197 3 7

14 × 14 × 14 = 2744 4 4

15 × 15 × 15 = 3375 5 5

16 × 16 × 16 = 4096 6 6

17 × 17 × 17 = 4913 7 3

18 × 18 × 18 = 5832 8 2

19 × 19 × 19 = 6859 9 9

Hence, from the table, we can say that only 7 and 8 are possible units place digits of the number,
where the units place digit of the cube is smaller.
When B = 7 or 8, C can be 2 or 3.
Hence, C can take only two values, either 2 or 3.
Hence, option a is correct.

Page 12

4. 56 × k is a perfect cube where k is a natural number. What could be the smallest possible value
of k?
a) 36 b) 49 c) 56 d) 72
Answer: b
We are given that 56 × k is a perfect cube.
This means that the prime factors of 56 × k appear thrice or a number of times which is a multiple of 3.
We are asked to find the minimum possible value of k.
So, the smallest possible prime factors that are required to make the product of ‘56 × k’ a perfect cube
will be the value of k.
Now, prime factorization of 56 = 2 × 2 × 2 × 7
Here, 2 appears thrice but we only have one 7 which means k needs to be 7 × 7 so that the product 56
× k = 2 × 2 × 2 × 7 × 7 × 7 becomes a perfect cube.
Hence, the smallest possible value of k is 7 × 7 = 49.
Therefore, option b is correct.
_____________________________________________________________________________________
5. Each geometrical shape denotes a certain operation. What will come in place of “?”

a) 2888 b) 1914 c) 1924 d) 340
Answer: c
The perfect square of the number written inside a square is given on the right side of the arrow.
For example: 162 = 256 and 192 = 361
Similarly, the cube number of the number written inside a cube is given on the right side of the arrow.
153 = 3375 and 113 = 1331
So, let's first replace each figure with its correct value.
14 is present in a square and 12 is present in a cube.
So, the expression becomes:
142 + 123
= 196 + 1728
= 1924
Hence, the correct answer is option c.
_____________________________________________________________________________________
6. XYZ is a 3-digit number such that it is the square of a multiple of 5.
What will be the HIGHEST possible remainder of (XYZ)/100?
a) 10 b) 15 c) 20 d) 25
Answer: d
The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, ... and so on.
Among these, we cannot consider 5 and numbers above 30 (like 35, 40, .. and so on), as their squares
are not 3-digit numbers.

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The multiples of 5 either end with 0 or 5.
Squares of multiples ending with 0, will end with ‘00’ (10 2 = 100, 202 = 400)
Hence, dividing them by 100 results in zero as the remainder.
Squares of 15 and 25 will be: 152 = 225 and 252 = 625. So, the squares of numbers ending with 5 will
end with 25.
Dividing them by 100 will result in a remainder of 25. (Example: 625/100 gives 6 as quotient and 25 as
remainder)
In both cases, the remainder is either 0 or 25.
Since we want the highest possible remainder, the correct answer is 25, which is option d.
_____________________________________________________________________________________
7. A class teacher wrote the following pattern on the board, where an expression is written after
the “=” sign in each row. She then asked the class to find the expression for 133 - 123. Which
DIGIT appears the HIGHEST number of times in the expression that denotes 133 - 123?

a) 1 b) 2 c) 3 d) 4
Answer: a
Let’s solve this logically.
We can see that there exists a pattern in the expressions, in each row.
In each row, we can find the pattern n3 - (n - 1)3 = 1 + n × (n - 1) × 3, where n = 1, 2, 3, and so on. (for
example: 53 - 43 will be 1 + 5 × 4 × 3)
Hence, 133 - 123 = 1 + 13 × 12 × 3
So, in the resultant expression, there are three 1s, one 2, and two 3s.
1 occurs the highest number of times.
Therefore, option a is correct.
_____________________________________________________________________________________
8. Fill in the grid with the squares and cubes of digits 2 to 5 such that:
● Each square/cube number appears twice
● Both square and cube of the same number cannot appear in the same row/column
● Same numbers do not appear diagonally
What is the maximum possible sum that can be obtained by the cells present in the
configuration of Shape A (without rotation)?

a) 197 b) 161 c) 193 d) 216
Answer: a
Page 14

We need to fill the grid such that no number repeats diagonally; both square and cube of the same
number do not appear in the same row/column; and each number appears twice.
Solving for squares and cubes of 2:
• We need to place 4 and 8 once more.
• The existing 4 and 8 already occupy some rows and columns, so those rows and columns
cannot be used again.
• After eliminating those, only Row 2 and Row 4 are left where both numbers can go.
• Similarly, only Column 2 and Column 3 are left where both numbers can be placed.
Hence, we place 4 and 8 in Row 2 or Row 4 and Column 2 or Column 3:

4 cannot come beside 125, as it will be diagonal to 4. Hence, 8 will be next to 125 and 4 will be next to
16 as shown below:

Solving for squares and cubes of 3:
We need to place 9 and 27 once more.
• The existing 9 and 27 already block some rows and columns.
• After removing those, only Row 1 and Row 4 remain possible.
• Also, only Column 2 and Column 4 remain available.
Hence, we place 9 and 27 in Row 1 or Row 4 and Column 2 or Column 4.

27 cannot come above 125 as it will be diagonal to 27 and 9 cannot come next to 16 as it will be
diagonal to 9. Hence, 9 will be above 125 and 27 will be next to 16, as shown below:

Solving for squares and cubes of 4:
We need to place 16 and 64 once more.
• The existing 16 and 64 block certain rows and columns.
• The remaining possible rows are Row 2 and Row 3.
• The remaining possible columns are Column 2 and Column 4.

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Hence, we place 16 and 64 in Row 2 or Row 3 and Column 2 or Column 4.

16 cannot come below 125 as it will be diagonal to 16. Hence, 64 will be below 125 and 16 will be
below 4, as shown below:

Solving for squares and cubes of 5:
The square and cube of 5 are 25 and 125. We need to place 25 and 125 in the remaining cells. 125
cannot be placed in row 1 as it would be diagonal to 125. Hence, 125 should be in row 4 and 25 should
be in row 1 as shown below:

Consider Shape A such that we get the maximum sum from the numbers in it.
To get the maximum sum, we need to have the largest number in the grid to be present in Shape A.
Thus, 125 should be present in Shape A. 64 (the next largest) is also around 125, so we can also have
64, as shown below:

Among both the Shape A configurations, the one with 8 in it will give a larger sum.
Thus, the maximum possible sum of numbers of shape A is: 8 + 64 + 125 = 197.
Hence, option a is the correct answer.
_____________________________________________________________________________________
9. The image below shows a logic machine, where numbers move from Column to Column (starting
from column 1 as input and reach column 4) through the tunnel, where they change in a different
way, each time. What would be the sum of A, B, and C?

a) 2259 b) 1795 c) 180 d) 1788
Answer: d
Page 16

The following things are happening here:
In the first tunnel, the number in column 1 is squared, and 1 is deducted from it. For example:
42 = 16
16 - 1 =15.
Similarly, 222 = 484
484 - 1 = 483
Hence for 7: 72 = 49
49 - 1= 48.
A = 48

In the second tunnel, the digits of column two are added. For example:
15: 1 + 5 = 6.
483: 4 + 8 + 3 = 15.
Hence for 48, 48: 4 + 8 = 12.
B = 12

Finally, the number in the fourth column is the cube of the number in the third column. For example:
63 = 216
153 = 3375
Hence for 12: 123 = 1728.
C = 1728

Hence the sum of A, B and C is: 48 + 12 + 1728 = 1788.
Hence, option d is the correct answer.

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10. Every column follows a certain rule. What number should come in place of “?”

a) 17 b) 18 c) 19 d) 21
Answer: b
Step 1: Observe the pattern
Look at each column and check how the bottom number is formed using the top and middle numbers.

Step 2: Test the pattern with known columns
• Column 1:
Top = 16, Middle = 9
16 × 9 = 144
√144 = 12
• Column 2:
Top = 28, Middle = 7
28 × 7 = 196
√196 = 14
• Column 3:
Top = 25, Middle = 16
25 × 16 = 400
√400 = 20

Step 3: Identify the rule
Bottom number = √ (Top × Middle numbers)

Step 4: Apply the rule to Column 4
Top = 12, Middle = 27
12 × 27 = 324
√324 = 18

Therefore, option b is correct.

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Boxes are stacked in four columns A, B, C, and D, such that:
• Each box is labelled with a number from 1, 2, and 3, with labels on some of the boxes already
shown in the image given below
• No two adjacent boxes in the same column are labelled with the same number
Boxes are stacked in four columns A, B, C and D, such that:
• No two adjacent columns have the same number on the topmost box
• Each box is labelled with a number from 1, 2 and 3, with labels on some of the boxes
• For every column, the sum of the numbers labelled on the topmost and bottommost block is
already shown in the image given below
equal
• No two adjacent boxes in the same column are labelled with the same number
In the shaded boxes, which of these numbers will occur the HIGHEST number of times?
• No two adjacent columns have the same number on the topmost box
• For every column, the sum of the numbers labelled on the topmost and bottom most
block is equal
In the shaded boxes, which of these numbers will occur the HIGHEST number of
times?
a) 1 b) 2 c) 3 d) All of them occur equally
a)
Answer: b1 b) 2
c) 3
Solution: d) All of them occur equally
In column Answer: b
C, the shaded block cannot have 2 again, as no two adjacent blocks of a column have the
same number.
Also, if this block has 1, then the sum of the topmost and the bottommost block of this column will
be 2 + 1 = 3.
But this cannot be a possible sum because, in that case, all the columns must have the same sum
and column D already has 3 on the top and one more number in this column will add up to a sum
greater than 3.
So, the shaded block of column C cannot have 1 or 2. Hence, it definitely has 3.

By this, we can say that the sum of the topmost and bottommost blocks of each stack is 5.
Thus, column D will have 2 in the bottommost block (as 3 + 2 = 5).
Also, the middle block can have only 1, as 2 and 3 are adjacent to this block already.

Now, column B cannot have 1 as the topmost block, as the bottommost block cannot have 4 to
satisfy the sum.
Also, 2 cannot be present as well, as the topmost block of its adjacent column is already 2 (Column
C).
Hence, we can have only 3 here and the bottommost block will automatically have 2.

Now, the empty shaded block of column B is adjacent to both 3 and 1.
Hence, this block will have 2.

Page 19

Hence, this block will have 2.

Now, column A will have 2 as the topmost block and 3 as the bottommost block (based on the
topmost and bottommost blocks of column B).
After filling in the rest of the numbers, the final arrangement looks like the one given below.
Hence, 2 occurs the highest number of times (3 times) in the shaded blocks.

Option b is the answer.

Page 20

Chapter 2: Power Play
1. Each term is written in exponential form. If the terms are rearranged in ascending order of their
numerical values from left to right, how many terms remain in the same position as in the original
arrangement?

a) 0 b) 1 c) 2 d) 3
Answer: c
The given values are: 64, 16, 27, 256, 25, 729
The ascending order is: 16, 25, 27, 64, 256, 729
Hence, when they’re arranged in ascending order, only two terms appear in the same position in the
new arrangement, as shown in the image below:

Option c is the correct answer.
_____________________________________________________________________________________
2. What will come in place of “?” in the given series?
6, 26, 126, 626, ?
a) 3125 b) 3126 c) 926 d) 916
Answer: b
Each term is formed by taking a consecutive power of 5 (starting from 51) and adding 1.
51 + 1 = 6,
52 + 1 = 26,
53 + 1 = 126,
54 + 1 = 626.
Similarly, 55 + 1 = 3126.
Hence, option b is the correct answer.
_____________________________________________________________________________________
3. A team is to be formed from a group of 7 students: A, B, C, D, E, F, and G where:
• A and B cannot be in the same team, but at least one of them must be included in the team
• C and D must either both be included or both be excluded from the team
• E and F must either both be included or both be excluded from the team
If the team consists of 4 students, in how many different ways can it be formed?
a) 2 b) 4 c) 3 d) 6
Answer: b
Either A or B has to be in the team. Therefore, the combinations are:

Page 21

A___
B___
Now, either C and D, or E and F can be placed in a team. Thus, 2 more members of the team are fixed
which are either C and D, or E and F. Therefore, the combinations are:
ACD_
AEF_
BCD_
BEF_
The fourth person in the four combinations above will be G.
Therefore, the team can be formed in 4 different ways. Option b is correct.
_____________________________________________________________________________________
4. Let X, Y, and Z be single-digit whole numbers. The number XY is a two-digit number formed
using digits X and Y.
If 4000 < (XY)Z < 5000, what is the minimum possible value of XY?
a) 10 b) 16 c) 17 d) 15
Answer: b
To make a number lie between 4000 and 5000, the base and power must balance each other.
If the base is larger, the power must be smaller and vice versa.
Since we want the minimum possible value of XY, we should try to keep the power as large as possible.
So, let’s start with the largest possible power.
1) Z = 4
The smallest possible two-digit number is 10.
104 = 10000
This is greater than 5000, so it does not satisfy the condition.
Therefore, power 4 is not possible.
2) Try, Z = 3
Now check the cubes of two-digit numbers.
153 = 3375(less than 4000)
163 = 4096
Since 4096 lies between 4000 and 5000, this satisfies the condition.
Thus, the minimum possible value of XY is 16. Option b is correct.
_____________________________________________________________________________________
5. Raj must select three numbers from the grid from three DISTINCT shapes such that:
1. Each number is chosen from a different row
2. The number selected from Row R1 is the nth power of 3
3. No digit is repeated among the bases and exponents of the chosen numbers
Based on these rules, what is the sum of the three selected numbers?

a) 1881 b) 7968 c) 1482 d) 1100
Answer: c

Page 22

Each number has to be chosen from a different row and a different shape.
So, we have to take the numbers from a pentagon, a square, and a circle.
The number chosen from R1 must be a power of 3
It can either be 34 or 36.
Either circle or square is chosen.
Hence, the pentagon cannot be taken from R1.
As we have a pentagon again in R3, the number chosen from R3 is definitely from the pentagon.
As it is mentioned that no digit can be common among the bases and powers of the chosen numbers
and we are already taking a power of 3 from R1, our number from R3 must be from a pentagon, but
cannot have 3 again.
Hence, the number chosen from R3 is 27.

Now, as we have 27 from R3, we cannot again select 27 from R2.
We can either have 45 or 54 from R2

If we choose 45 from R2, we must take 36 from R1 (as digits cannot repeat).
But, this will give us numbers from two squares and a pentagon only.

So, we have to select 54 (from the circle) in R2 and then proceed to the square from R1.

Thus, the total value of the numbers selected is: 36 + 54 + 27
= (3 × 3 × 3 × 3 × 3 × 3) + (5 × 5 × 5 × 5) + (2 × 2 × 2 × 2 × 2 × 2 × 2)
= 729 + 625 + 128
= 1482
Option c is the correct answer.

Page 23

6. A man has a bag with a maximum capacity of 2 4 units. He earns certain points for the books he
carries in that bag and the load of each book is shown in the table below:
● For every Management book added to the bag, 2 Fiction books must be added
● For every Mathematics book added to the bag, 2 Physics books must be added
If he has to carry at least one book of each genre, what is the maximum number of points he can
earn without exceeding the bag’s capacity?

a) 116 b) 120 c) 112 d) 104
Answer: b
First, let’s simplify the power numbers and find the load and points of each book:
Management: load 4, points 32
Mathematics: load 2, points 16
Physics: load 1, points 8
Fiction: load 1, points 4
Conditions:
- At least 1 of each genre must be carried.
- For every 1 Management, he must carry 2 Fiction books, and for every 1 Mathematics, he must carry
2 Physics books.
- Total load must be less than or equal to 16. (since the capacity of the bag = 2 4 = 16)
So, the minimum compulsory load is 1 Mathematics and 1 Management, and with these, he automatically
carries 2 Physics and 2 Fiction.
Minimum load = 1 (4) + 1 (2) + 2 (1) + 2 (1) = 4 + 2 + 2 + 2 = 10
So, the minimum points earned = 32 + 16 + 2(8) + 2(4) = 32 + 16 + 16 + 8 = 72

As 10 out of 16 units are already filled in the bag, the remaining load = 6 units.
Use the remaining capacity optimally. Compare efficiency (points per unit load):
Physics: 8 per unit load (Good)
Fiction: 4 per unit load (Not preferable - Low value)
Management: 8 per unit load, but adds 2 units for fiction books, with only 8 extra points
Mathematics: 8 per unit load (Good)

Case A: Add 1 more Mathematics book (for which 2 more Physics books are required)
Total load added = 2 + 2 = 4 (still 2 more units remaining).
We can fill them with 2 more Physics books (we cannot add another Mathematics book - it needs 2 extra
Physics books, which exceeds the load)
Total points earned = Initial minimum points + Points added further (1 math + 4 physics)
= 72 + 16 + 8 + 8 + 8 + 8
= 120

Case B: Add only Physics books further.
To fill in the remaining capacity of 6 units, we can have 6 physics books.
Total points earned = Initial minimum points + Points added further (6 physics)
= 72 + 8 + 8 + 8 + 8 + 8 + 8
= 120
In any case, the maximum number of points that can be reached = 120.
Hence, option b is the right answer.
Page 24

7. How many possible combinations of 1 owl and 1 tree can be made from 3 owls and 5 trees?
a) 9 b) 10 c) 15 d) 16
Answer: c
We need to make a combination of 1 owl and 1 tree.
There are 3 owls and 5 trees.
● The 1st owl can be paired with all 5 trees = 5 combinations
● The 2nd owl can also be paired with all 5 trees = 5 combinations
● The 3rd owl can also be paired with all 5 trees = 5 combinations
Total combinations = 5 + 5 + 5 = 15
So, 15 different pairs of 1 owl and 1 tree can be formed.
Hence, option c is the correct answer.
_____________________________________________________________________________________
8. Sam has Rs. 4000 in such a way that he has Re. 1 coins in power of 10, Rs. 2 coins in power of
5, Rs. 5 coins in power of 2, and the remaining coins of Rs. 10. The amount of money of each
individual denomination of Re. 1, Rs. 2, and Rs. 5 is greater than or equal to thousand. At
MAXIMUM, how many coins of Rs. 10 can he have?
a) 53 b) 47 c) 38 d) 45
Answer: b
Step 1: Represent the amount from each type of coin
Sam has:
Re. 1 coins in powers of 10: amount = 1 × 10ⁿ
Rs. 2 coins in powers of 5: amount = 2 × 5ᵐ
Rs. 5 coins in powers of 2: amount = 5 × 2ᵏ
Rs. 10 coins = t (we need to find t)
The total amount is Rs. 4000.
So, 10ⁿ + (2 × 5ᵐ) + (5 × 2ᵏ) + 10t = 4000
Step 2: Find combinations where the amount is greater than or equal to Rs. 1000 for each coin type
The amount of money from each denomination of Re. 1, Rs. 2, and Rs. 5 must be greater than or equal
to Rs. 1000. To maximize the number of Rs.10 coins, we choose the smallest possible amount for each
of these denominations, that is at least Rs. 1000.
That means:
Amount in Re. 1 coins = 1000 or more
Amount in Rs. 2 coins = 1000 or more
Amount in Rs. 5 coins = 1000 or more
Step 3: Determine the LEAST possible number of coins of each type, to MAXIMISE the number of Rs.
10 coins.
(i) Re. 1 coins in powers of 10:
10³ = 1000
So, 10ⁿ = 1000
Hence, the least possible number of Re. 1 coins = 1000 [Amount = Rs. 1000]
(ii) Rs. 2 coins in powers of 5:
2 × 53 = 2 × 125 = 250 (less than 1000)
2 × 5⁴ = 2 × 625 = 1250 (greater than 1000)
So, the least possible number of Rs. 2 coins = 625 [Amount = Rs. 1250]
(iii) Rs. 5 coins in powers of 2:
5 × 27 = 5 × 128 = 640 (less than 1000)
5 × 2⁸ = 5 × 256 = 1280 (greater than 1000)
So, the least possible number of Rs. 5 coins = 256 [Amount = Rs. 1280]
Total so far = 1000 + 1250 + 1280 = 3530
Remaining amount for Rs. 10 coins = 4000 – 3530 = 470

Page 25

10t = 470
t = 47
Hence, Sam can have 47 coins of Rs. 10 at maximum. Option b is the correct answer.
_____________________________________________________________________________________
9. What will come in place of “?”

a) 243 b) 125 c) 343 d) 2187
Answer: c
In the given question, if the polygon’s border is black, the value after the arrow is equal to the number
inside the polygon raised to the power of the number of sides of the polygon.
For example, 4 is in a black square, whose sides are 4.

So, the value after the arrow is 44 = 4 x 4 x 4 x 4 = 256
However, if the polygon’s border is in red, then the number of sides is taken as the base and the
number inside it is the exponent.

It will be 35 = 243. (not 53, as in black shape)
Similarly, in the question term, we have a red shape with 7 sides.

It will be 73 = 343.
Therefore, option c is correct.
_____________________________________________________________________________________
10. Sam and Tim are playing a number-maximizing game. Their numbers are shown in the table below
(where A and B are undefined). They must modify their numbers using ONLY ONE of the following
rules:
● If the base is smaller than the exponent, the base n is replaced with 1/n
● If the base is greater than the exponent, the exponent n is replaced with 1/n
Sam and Tim choose A and B and apply the rule exactly once in a way that maximizes their own
numbers. What is the difference between the final numbers of Sam and Tim, where A and B are
single-digit natural numbers?

a) 2 b) 3 c) 1 d) 0
Answer: c
Page 26

Sam’s number is 4 A.
To maximize Sam’s number, we use the rule:
If the base is greater than the power, the exponent n is replaced with 1/n. (as 4 is already greater and
choosing a power greater than 4 will change 4 to 1/4 unnecessarily)
Here, the base 4 must be greater than the power A.
To keep the value as large as possible, we take the smallest natural value for the exponent, A = 1.
Applying the rule:
41 becomes 4(1/1)
4(1/1) = 4
So, the maximum value of Sam’s number is 4.

Now consider Tim’s number = B 2.
To maximize Tim’s number, we choose the largest possible value of the base (as the power is 2, which
is already a smaller number).
Since B is a single-digit natural number, the largest possible value is 9.
So, we take: 92
Here, base (9) > power (2), so the exponent is replaced with 1/2.
92 becomes 9(½)
9(½) = √9 = 3
So, the maximum value of Tim’s number is 3.

Difference between the numbers:
4−3=1
Hence, option c is the correct answer.

Page 27

Enter all the letters and the numbers of the Set in the empty squares of the grid given below,
such that:
- Every vowel must have an even number in at least one of its adjacent squares
- The letter 'H' is not adjacent to 'A' or 6
- Two consecutive numbers cannot be in any adjacent squares
What will come in place of “?”
Note: Squares are considered to be adjacent only if they share a common side. Squares sharing a
common corner are not considered adjacent

a) 6 b) Y c) E d) A
Answer: c
Among A, E, Y, 2, 4, 6, none of the digits can be placed in the top left corner cell, as no two
consecutive digits can be adjacent. Also, every vowel must have at least one even digit adjacent to
it. So, A and E also cannot be entered. Hence, the top left corner cell can have only ‘Y’.

Similarly, only 2 can be placed in the top right corner cell, as A and E cannot be placed here, as
vowels need an even digit next to them. Also, 4 and 6 are consecutive to 5 and cannot come in this
cell.

Now, H cannot have 6 or A adjacent to it.
So, 6 and A cannot come in place of the “?”.
Also, 4 cannot be placed here, as it is consecutive to both 3 and 5.
Hence, only E can come in place of the “?”.

Hence, option c is the correct answer.
The final arrangement is as follows:

Page 28

Chapter 3: A Story of Numbers
Activity Time
Measure your Height
Introduction
We will explore the concept of the positional number system through hands -on activities. We will
develop an intuition for various base representations of numbers by measuring students' heights,
recognizing that decimal numbers that we learn in school are just one way to represent numbers.

Activity Time Description

Launch 5 min The teacher poses the question to the students and invite a
couple of students to try their ideas.

Supporting Links:
Activity Video Link: https://youtu.be/bT_hDTvGAt8
Reference:

Problem- 25 Student’s Tryout the activity with each other.
solving and mins
algorithm Student Worksheet:
https://docs.google.com/document/d/1cycoY6rEfGqczQgfOrlneL
R_jWZEs9Iryhfzjp23QqA/edit?tab=t.0

As the teacher progresses through the activities, the students
progress along with the worksheet.

Discussions 5 min The discussion on the various algorithms, how a slight change in
and approach leads to a faster solution, and how computers apply
Explorations such algorithms

Page 29

CT Connection
Pattern recognition: Two see that the same patterns of representation appear in each of the base
systems and the symbols used in them.
Algorithmic Thinking: We use a step-by-step approach to measure the height using the decimal,
binary, and ternary pipes.
Decomposition: Breaking the conversion of the number to binary and ternary into one pipe at a time.
Abstraction: Two use the pattern seen in decimal and apply it to powers of two and powers of three.
Generalisation: Recognising a doubling pattern of 1, 2, 4, and so on, and adding different powers thus
generated yields the binary representation; we conclude that any number can be represented in terms
of powers of 2. Similarly, we see how to measure using ternary numbers. This method can thus be
generalised for any natural number n.
Logic: To understand why the digits start from zero and end before the base number n (0 to n - 1)

Activity: Measuring your Height
1: Measuring your height with Decimal Pipes
We generally measure our height in the decimal number system, i.e., using base -10 numbers. We use
powers of ten 100, 101, 102, 103, … (1, 10, 100, 1000, …) and digits from 0 to 9 (ten distinct symbols), to
represent any number that exists.
Imagine now that you are measuring heights using PVC pipes of fixed lengths:
1 inch, 10 inches, 100 inches, and so on.
Each pipe size is a power of 10. You also have 9 pipes of each length.

1. How many of the decimal pipes will it take to measure the height of a 66-inch-tall person?
a) 21 pipes: 5 pipes of 10 inches height and 16 pipes of 1 inch height
b) 7 pipes: 7 pipes of 10 inches height
c) 9 pipes: 6 pipes of 10 inches and 3 pipes of 2 inches height
d) 12 pipes: 6 pipes of 1 inch and 6 pipes of 10 inches
Answer: d
66 in decimal notation is represented as 6 x 10 + 6 x 1, that is, 6 of 10s and 6 of 1s, for a total of 12 pipes.
Competencies: Algorithmic thinking, logic, decomposition
_____________________________________________________________________________________
2: Measuring your height with binary pipes
We keep hearing that the language of computers is binary, the language of 0 and 1. But what are these
binary numbers? How do we understand these in light of what we already know?
Instead of using decimal pipes, which are powers of 10, what happens if we use pipes of length 1, 2, 4,
8, 16, 32, 64, and so on? What is special about these numbers?
They are powers of 2. We have exactly one copy of each pipe length. We will call these pipes Binary
pipes.

Page 30

We have only one copy of each pipe for the following reason. Two 2-inch pipes make a 4-inch pipe, but
we already have a 4-inch pipe. Two pipes of 4 inches make an 8 inches pipe, but we already have an 8-
inch pipe; likewise, we can do this for pipes of all lengths.
Let’s measure a height in binary. For example, 55 = 32 + 16 + 4 + 2 + 1.
We will see that if we keep a 64-inch pipe, it's more than the height we are measuring, so we don’t use a
64-inch pipe. We use the next lower power of 2, 32. We then keep 16, the next lower power. We then
keep the next lower power of 2, which is 8, but this is more than the height we are measuring. So, we
don’t use it and move on to the next power of 2. This process continues.
We can tabulate the information below. Using this, we can say that the binary representation of 55 is
(110111)2.

55 in binary 25=32 24=16 23=8 22=4 21=2 20=1

(110111)2 1 1 0 1 1 1
_____________________________________________________________________________________
1. Which all binary pipes will we use to measure the height of a 66-inch-tall person?
a) 63, 3
b) 32, 16, 8, 4, 2, 1, 2, 1
c) 64, 2
d) 32, 32, 2

Page 31

Answer: c
Let’s understand what is happening. We first place a 64-inch-tall block. Now, when we try to keep the
other heights over 64 inches, like 32, 16, 8, 4, etc., we will see that these heights will exceed the height
we have to measure. So, we pick the next lower height until we reach the one that fits. In the number 66,
we have 1 copy of 64, 0 copies of 32, 0 copies of 16, 0 copies of 8, 0 copies of 4, 1 copy of 2, and 0
copies of 1. If we write this information in the same way we write place values for decimals, we get:

27=128 26=64 25=32 24=16 23=8 22=4 21=2 20=1

0 1 0 0 0 0 1 0

Thus, 66 in decimal in binary representation is (1000010) 2. We write a (- - - -)2, subscript 2 to tell us that
this is a base 2 representation of a number.
Competencies: algorithmic thinking, decomposition, abstraction, generalisation, pattern recognition
_____________________________________________________________________________________
2. Binary representation of 10 is:
a) 1010 b) 1111 c) 1000 d) 1100
Answer: a
We can draw a table similar to the one shown above and list which powers of 2 can be added to get 10.

23 22 21 20

1 0 1 0

Thus, 10 = 1 x 8 + 1 x 2 = (1010)2.
Competencies: algorithmic thinking, abstraction, generalisation, pattern recognition
_____________________________________________________________________________________
3. Why can't we use the digit “2” in the representation of a number in binary?
a) It is illegal, as only 0 and 1 are given to us in the rules
b) We can write any number using 0 and 1; there is no need for 2
c) Computers dislike it
d) It is too big
Answer: b

Page 32

If we used the digit “2” in the binary system in the ones place, it would denote 2 x 1 inch = 2
inches. But this is equivalent to using one 2-inch block, which is already present among the height
blocks. This means redundancy.
That is why digits must be less than the base. Similarly, two blocks of 4 inches make 8 inches,
but we already have an 8-inch block; likewise, two 8 inches is 16 inches, and there is already an
16 inches block
Competencies: abstraction, generalisation, pattern recognition, logic
_____________________________________________________________________________________
3: Measuring your height with ternary pipes
Instead of base 10 (decimal) and base 2 (binary), what if we set the base to something else, for example,
3?
Then we will have pipes of lengths 1, 3, 9, 27, 81…, all powers of 3, and 2 pipes of each length.

Let’s see an example of measuring 55 inches in height in ternary.

We will see that if we keep the 81-inch pipe, it's taller than the height we are measuring, so we don’t use
it. We use the next lower power of 3, 27. We keep two of these, which is the maximum number of pipes
of each type allowed.
We then keep 2 pipes of 9, the next lower power, but this overshoots the height, so we remove one pipe.
Even with one 9-inch pipe, we are still over the height we are measuring.
So, remove this pipe and keep two 3-inch pipes. But this overshoots the height, so we remove one pipe.
Even with one 3-inch pipe, we are still over the height we are measuring.
So, we remove that and keep two pipes of the next lower power of 3, which is 1, but this is more than the
height we are measuring. So, we remove one pipe and see that we get the correct height measurement.
55 inches is 54 + 1 = 2 x 27 + 1 x 1.

55 33 32 31 30

(2001)3 2 0 0 1

1. When we measure 66 inches with ternary pipes, how will its representation be?
a) 2011 b) 2101 c) 2110 d) 2210
Answer: c
To measure 66 inches, we use two 27-inch pipes, one 9-inch pipe, one 3-inch pipe, and zero 1-
inch pipes.

Page 33

66 = 2 × 27 + 1 × 9 + 1 × 3 + 0 × 1 = (2110)3

66 33 32 31 30

(2110)3 2 1 1 0

Competencies: algorithmic thinking, decomposition, abstraction, generalisation, pattern
recognition
_____________________________________________________________________________________
2. What is the largest digit allowed in Base 3?
a) 3 b) 2 c) 1 d) 9
Answer: b
As we saw in other representations, the digits allowed in a base number system are numbers
less than the base. In case 3, we have three numbers: 0, 1, and 2.
Competencies: abstraction, generalisation, pattern recognition, logic
_____________________________________________________________________________________
3. How will we write 100 in base 3 (ternary) notation?
a) (3201)3 b) (2333)3 c) (11000)3 d) (10201)3
Answer: d
100 = 1 × 81 + 0 × 27 + 2 × 9 + 0 × 3 + 1 × 1 = (10201)3

100 34 33 32 31 30

(10201)3 1 0 2 0 1

Competencies: algorithmic thinking, decomposition, abstraction, generalisation, pattern recognition
_____________________________________________________________________________________

Conclusion
We see that a number can be represented in many forms and symbols. Decimal notation is one of the
many possible base number systems (Positional value system).
The fundamental idea is that we arrange the powers of the base number n in increasing order from right
to left.
The face value at each place tells us how many times the corresponding power of the base is used to
represent the number.
We also see that the base n system requires n distinct digits. These digits are 0 till n - 1.

Page 34

Questions
1. Some terms are missing in between the sequence. Find the missing terms in the order from left
to right. VII, VI, VI, VII, V, VIII , ______ , _______ , III , X
a) VI, IX b) V, VIII c) IV, IX d) IV, V
Answer: c
The sequence is written in Roman Numerals. The sequence is formed by two alternate series. VII, VI,
V, ____, III and VI, VII, VIII, ____, X.
So, the missing terms will be IV and IX.

Thus, option c is correct.
_____________________________________________________________________________________
2. Find the odd one out from the following:

a) b) c) d)
Answer: c
Option a: The Roman number is 1 and the shape has three sides.
Option b: The Roman number is 2 and the shape has four sides.
Option c: The Roman number is 3 and the shape has six sides.
Option d: The Roman number is 3 and the shape has five sides.
The value of the Roman number in every shape is two less than the number of sides in that shape,
except in option c.
Hence, option c is the correct answer.
_____________________________________________________________________________________
3. In each of the given options, the weight of one bag is written in the Roman numeral form and
that of the other bag is in Hindu - Arabic form. Identify the option that displays the tilt of the
balance correctly.
Note: The balance tilts towards the larger weight

a) b)

c) d)

Page 35

Answer: b
We first convert the Roman numeral to Arabic form and then compare the numbers:
The following table shows the relation between Hindu - Arabic numerals and Roman numerals:

Also, a Roman numeral of smaller value when written on the right of a Roman numeral of greater value
is added to the Roman numeral of greater value.
Example: VI = 5 + 1 = 6, where I (1) written to the right of V (5) is added to it.
And a Roman numeral of smaller value when written on the left of a Roman numeral of greater value is
subtracted from the Roman numeral of greater value.
Example: IV = 5 − 1 = 4, where I (1) written to the left of V (5) is subtracted from it.

Therefore, DCCLIX can be expressed as 500 + 100 + 100 + 50 + 9 = 759 and
DCXXXVII can be expressed as 500 + 100 + 10 + 10 + 10 + 7 = 637.

Now, let’s check each of the given options:
Option a: 763 < DCCLIX
763 < 759 (Incorrect)

Option b: 763 > DCCLIX
763 > 759 (Correct)

Option c: 631 = DCXXXVII
631 = 637 (Incorrect)

Option d: 631 > DCXXXVII
631 > 637 (Incorrect)
Hence, option b is correct.
_____________________________________________________________________________________
4. On the abacus, select two poles, A and B, where pole B has fewer beads than pole A. Exactly
one bead from pole A must be moved to pole B.
If more than one pole has fewer beads than A, move the bead to the pole that has the maximum
number of beads among those poles.
After making this single move, what is the highest possible 6-digit number that can be formed?

a) 3,62,046 b) 4,72,035 c) 4,72,046 d) 4,62,045
Answer: b
The current number is 3,72,045.
We must move EXACTLY one bead from pole A to pole B, which has fewer beads than A.

Page 36

If there are multiple such poles, we must place it on the pole that has the highest number of beads
among them.
To make the largest number, we should try to move a bead to the pole of Lakhs value.
For this, if we choose a bead from the ten thousands’ pole, we have to move it to the ones pole, which
gives a result of 3,62,046.

However, if we choose the tens pole (which has 4 beads), we can move the bead to the Lakhs pole
(which has 3 beads, which is the highest count among the poles having fewer beads than the tens
pole).

Thus, the final answer is 4,72,035.
Option b is the correct answer.
_____________________________________________________________________________________
5. What will come in place of “?”
Note: All the numerals on the left of each term are from the same number system

a) 18 b) 90 c) 36 d) 180
Answer: d
Here, the Egyptian numerals are considered

Page 37

The individual numbers are added first and then sent as input.
‘c’ is obtained by multiplying the inputs.
For example, in the first term, we have input ‘a’ as 3 + 3 = 6
Input ‘b’ = 10
So, c is obtained by multiplying a and b, 6 x 10 = 60.
Likewise, in the question term, input a = 3 + 3 = 6
Input b = 10 + 10 + 10 = 30
c = a x b = 6 x 30 = 180.
Option d is the correct answer.
_____________________________________________________________________________________
6. A 3 - digit code is formed using Chinese numerals (Zongs) and Roman numerals. Each row
represents a guess, along with a statement describing how correct the guess is. Using these
clues, determine the correct code.

a) b) c) d)
Answer: b
From the third hint, we know 3 (in Roman numeral), 8 (in Roman numeral), and 2 (in Chinese numeral)
are not part of the code.

From the second hint, we know that among 8 (in Chinese numerals), 1 (in Chinese numerals), and 6 (in
Chinese numerals), one of them is in the correct position, but it will appear in Roman numerals.
Since 8 is not present in the secret code, only 1 and 6 can possibly be in the code.

In the fifth hint, among 2, 3, and 4, only one digit is correct. Since 2 and 3 are not part of the code, 4 must
be part of the code.
However, it is in the incorrect position and also in the incorrect numeral system. Hence, 4 must be written
in Chinese numerals.

In the first and fourth hints, the common digits are 5 and 7.
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In Hint 1, 7 appears as a Chinese numeral in the units place, but the hint states that it is in the wrong
position and the wrong numeral. Therefore, if 7 were part of the code, it would have to be written in
Roman numerals.
In Hint 4, 7 appears as a Chinese numeral in the hundreds place, and the hint indicates that the numeral
type is correct. Since this contradicts the earlier conclusion about the numeral type of 7, it follows that 7
is not part of the code.
According to Hint 1, 5 should be in Chinese numerals and placed either in the hundreds or units position.
Hint 4 states that two digits are in the correct numeral but in the wrong positions. Since 7 is not part of
the code, the remaining digits must be 5 and 6, confirming that both are part of the code.
Combining Hints 1 and 4, 5 must be in the hundreds position and written in Chinese numerals.

Digit 6 is also part of the code, and the only remaining position for it is the units place.
Hint 4 indicates that the numeral type of 6 should be Roman numerals.
Hence, the final code will be:

Hence, the correct answer is option b.
_____________________________________________________________________________________
7. If certain numbers are coded as shown in the image, what would be the code for 208?

a) b)

c) d)
Answer: a
The symbols in the code add up to the original number, with 1 and 3 defined at the beginning.
From the first code, we know that 1 is coded as I and 3 is coded as a circle.
In the code for 12, the circle anyways represents 3, and 12 − 3 = 9.
Hence, the code for 9 is an upward triangle.
In the code for 34, we know that a circle represents 3 and I represents 1.
So, the value of two circles and one I is: 3 + 3 + 1 = 7.
Therefore, 34 − 7 = 27, which is the value represented by the square.
In the code for 87, two circles represent 6, so the remaining value is 81 (87 − 6 = 81), which is coded as
an inverted triangle.
So, the values of the shapes are: 1, 3, 9, 27, 81.

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The original number is coded into the above numbers, represented by the shapes. (these shapes add up
to the number)
The code appears to follow numerals with base 3, where:
30 = 1, 31 = 3, 32 = 9, 33 = 27, 34 = 81.
Hence, the code for 208 will be:
208 can also be written as: 81 + 81 + 27 + 9 + 9 + 1, which in code will be written as:

Hence, the correct answer is option a.
_____________________________________________________________________________________
8. Two 2-digit numbers are represented below by stacking balls on poles. If you can shift/move
EXACTLY ONE ball from a pole to any of the other three poles, what will be the MINIMUM possible
difference between the numbers, finally?

a) 1 b) 2 c) 11 d) 13
Answer: b
The numbers shown are 56 (5 tens, 6 ones) and 43 (4 tens, 3 ones). To get the smallest possible
difference, we should try to make their tens digits closer, because tens change the value the most.
There are two effective moves:
Take 1 ball from the tens pole of 56 and add it to the ones pole of 43.
New numbers: 46 and 44. So, Difference = 2.
Take 1 ball from the ones pole of 56 and add it to the tens pole of 43.
New numbers: 55 and 53. So, Difference = 2.
In both cases, the minimum possible difference is 2.
Hence, option b is correct.
_____________________________________________________________________________________
9. Sam wants to create a 4-digit password using the buttons (Chinese numerals - Zongs) from the
given screens. He must press the buttons in the same sequence as he wants the digits in the
password to appear.
● Every next button pressed must have a greater digit than the previous button
● No two adjacent digits in the password can be selected from the same screen and from
buttons of the same colour
How many different passwords can Sam form?

a) 1 b) 2 c) 3 d) 4
Answer: c
Digits in screen 1 and screen 2 have Chinese numerals, so let us first identify each digit:

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Screen 1: Digits in white: 1, 4, and 7 & Digits in black: 5 and 8
Screen 2: Digits in white: 2 and 9 & Digits in black: 1, 3, and 6
As every next digit selected is greater than the previous digit, Sam must start with smaller digits only.
As per the image, starting with a button 4 or more than 4, will not allow Sam to select digits in
ascending order as well as alternate colours and screens. (as 4 is there in screen 1, if we start with 4,
we can only have 4 from screen 1, 6 from screen 2, and 7 from screen 1. After this, the 4th digit of the
password cannot be selected)
The pattern of colour will be Black - White - Black – White, OR White - Black - White - Black.

CASE 1: Assume that Sam started with a black digit of Screen 1.
He must start with 5 only. As discussed above, the password cannot start with 4 or more than 4 (as 6
from screen 2 is also from a black button and cannot be chosen immediately after 5).
So, this case is invalid.

CASE 2: Assume that Sam started with a white digit of Screen 1.
He can start with 1 only. (as the password cannot start with 4)
If he starts with 1 from screen 1, the password could be 1 - 3 - 4 - 6

CASE 3: Assume that Sam started with a black digit of Screen 2.
He can either start with 1 or 3 only.
Case 3 A: If he starts with 1, the password could be 1 - 4 - 6 - 7
Case 3 B: If he starts with 3, the password could be 3 - 4 - 6 - 7

CASE 4: Assume that Sam started with a white digit of Screen 2.
He can start with 2 only.
After choosing 2 (white) from screen 2, he can choose only five (black) from screen 1. After this, there
isn’t any other suitable white button to be selected as the 3rd digit of the password (as 9 is the highest
available number and cannot come in the 3rd place of the 4-digit number).
So, this case in invalid.

Therefore, three different passwords can be formed: 1 - 3 - 4 - 6, 1 - 4 - 6 - 7, and 3 - 4 - 6 - 7
Option c is correct.
_____________________________________________________________________________________
10. Rearrange the numbers in Set A into ascending order, as shown in Set B. The blocks can be
rearranged only by swapping adjacent blocks. What is the MINIMUM number of swaps required
to do this?
Note: Blocks that have common sides are considered to be adjacent. Blocks that have a common corner
alone are not adjacent

a) 3 b) 4 c) 5 d) 6
Answer: b
The Set A has digits in Mesopotamian numeral system.
As we know:

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Thus, we substitute the values in Set A:

We need to rearrange Set A into ascending order as shown in Set B, using only swaps between
adjacent blocks (blocks that share a common side).
Set A is:
642
513
Ascending order should be arranged from least to highest as:
123
456
Now we arrange step by step using minimum adjacent swaps.
First, we need to place 1 in the top-left box.
Currently, 1 is in the bottom middle.
To move 1 correctly:
• Interchange 1 and 5. Now, 5 moves toward its correct lower row position.

• Then, interchange 1 and 6. Now, 1 reaches the top-left position, and 6 shifts toward its correct side.

Next, we arrange the remaining numbers with minimum swaps.
• Interchange 2 and 4. Now, 2 moves closer to its correct position in ascending order.

• Then, interchange 3 and 4. Now, both 3 and 4 reach their correct ascending positions.

After these four adjacent swaps, all numbers are arranged in proper ascending order.
Therefore, the minimum number of swaps required is 4.
Hence, option b is the correct answer.

Page 42

As shown below, A, B, and C are positioned at different starting points on a circular
track. B runs at half the speed of A, while C runs at twice the speed of A.
If all three start running at the same time and in the same direction, what could be the
positions of A and B by the time C completes one full round of the track?

(a) (b) (c) (d)
Answer: b
Assume that A, B and C are running in a clockwise direction.
Step 1: Identify the distance that could be covered by each of them, based on the
speed mentioned:
• C runs at twice the speed of A
If C completes one full round, A completes half the round.
• B runs at half the speed of A
If A completes half the round, B can finish only a quarter distance.
Step 2: Determine the final positions of A and B, assuming that C has finished one full
round:
By the time C completes one full round, we know that A could finish only half the round.
So, A appears exactly on the opposite side of his current position. (A’s final position would be
the same
as B’s current position).
Based on this, the options that show a different position for A, can be eliminated.

Options c and d represent A’s position at a point which is more than half the distance from
A’s current position.
Hence, options c and d are incorrect.
Also, we know that when A completes half the round, B should be able to cover only a
quarter distance.
But, in option a, we can see that, B reached A’s initial position, by the time A reached B’s
initial position.

As C finishes a round, A finishes half the round (and reaches B’s initial position) and B
Page 43
round (and reaches C’s initial position), as shown below:

So, option a clearly shows that both A and B covered the same distance, which is also a
contradiction.
Hence, option a is also incorrect.
As C finishes a round, A finishes half the round (and reaches B’s initial position), and B finishes
a quarter round (and reaches C’s initial position), as shown in option b.
Hence, option b is the correct answer.

Page 44

Chapter 4: Quadrilaterals
1. Raj colored the following shapes (Trapezium, Parallelogram, Square, Kite) in Red, Yellow, Green,
and Blue, not necessarily in the same order.
● The shape colored in Red has no parallel sides
● The shape with each of its angles equal to 90o, is not Yellow
● The shape with two pairs of parallel sides is not Green
Which shape did he color in yellow?

a) Trapezium b) Parallelogram c) Square d) Kite
Answer: b
-The shape coloured in Red has no parallel sides
In the given shapes, only ‘Kite’ has no parallel sides.
Hence, Raj coloured the Kite in Red.
-The shape with two pairs of parallel sides is not Green
In the given shapes, the shapes with two pairs of parallel sides are the Square and Parallelogram. So,
they are not Green.
The Trapezium is green.
- The shape with each of its angles equal to 90o, is not Yellow
The square is not yellow
The Parallelogram is Yellow and the Square must be Blue
Hence, the Parallelogram is coloured in Yellow.
So, the correct answer is option b.
_____________________________________________________________________________________
2. How many quadrilaterals are there in the given figure?

a) 8 b) 9 c) 10 d) 11
Answer: d
There are 11 quadrilaterals as shown in the image:

Page 45

Option d is the correct answer.
_____________________________________________________________________________________
3. How many rectangles are there in the given figure?
Note: For the purpose of this question, please count all squares also as rectangles

a) 8 b) 9 c) 10 d) 11
Answer: c

Hence, the correct answer is option c.

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4. In the given tangram, how many quadrilaterals are made with exactly 3 colored tiles?

a) 4 b) 3 c) 2 d) 1
Answer: b
A quadrilateral is a closed figure with 4 sides.
We must look for quadrilaterals formed by combining exactly 3 differently colored pieces from the
tangram.

Altogether, we can spot 3 distinct quadrilaterals made with exactly 3 colors.
Therefore, option b is correct.
_____________________________________________________________________________________
5. Below are three families of shapes and a set of Property Cards.
- A property card should be placed into a family if the property applies to that family
- A single property may belong to more than one family
After placing all the property cards, which family will have the maximum number of properties?

a) Square Family b) Rectangle Family c) Parallelogram Family d) Both a and c
Answer: a
All sides are equal and all angles are 90°: Square
All angles are 90° and opposite sides are equal: Rectangle, Square
Opposite sides are parallel: Parallelogram, Rectangle, Square
Diagonals are equal and bisect each other: Rectangle, Square
Opposite sides are equal and parallel: Parallelogram, Rectangle, Square
So, the number of properties in each family is:
● Rectangle Family: 4 properties
● Square Family: 5 properties
● Parallelogram Family: 2 properties
Square Family will have the highest number of properties.
Hence, option a is the correct answer.

Page 47

6. In the grid, each letter stands for the initial of a shape: Square, Rhombus, Kite, Trapezium, and
Parallelogram. How many pairs of adjacent letters represent two shapes in which the diagonals
bisect each other?

a) 3 b) 4 c) 5 d) 6
Answer: c
The shapes whose diagonals bisect each other are Square, Rhombus, and Parallelogram.
So, we only look for the letters S, R, and P in the grid.
Now check the grid and count the pairs where two of these letters are placed next to each other.
We find 5 such adjacent pairs in the grid.

Hence, option c will be the correct answer.
_____________________________________________________________________________________
7. Find the option that does NOT belong to the elements in the given set.
Set = (Angle between diagonals of a Rhombus, Each angle of a rectangle, Each angle of a
square)
a) Angle inscribed in a semicircle
b) Half of sum of opposite angles of a Cyclic Quadrilateral
c) Each angle of an Equilateral Triangle
d) Angle opposite to Hypotenuse in a Right - Angled Triangle
Answer: c
First, find the common value of the given set.
Set:
● Angle between diagonals of a rhombus = 90°
● Each angle of a rectangle = 90°
● Each angle of a square = 90°
So, all elements in the set are 90°.
Now check the options:
a) Angle inscribed in a semicircle = 90°
b) Half of the sum of opposite angles of a cyclic quadrilateral
Opposite angles sum to 180°, half = 90°
c) Each angle of an equilateral triangle = 60°
d) Angle opposite the hypotenuse in a right-angled triangle = 90°
Option c does not belong to the set.
Hence, option c is the correct answer.
_____________________________________________________________________________________
8. Which shapes from the given options will NOT form a quadrilateral when any two identical
shapes are joined along any one of their sides?
Note: You can rotate the shapes but they cannot overlap each other
a) Triangle b) Rhombus c) Trapezium d) Pentagon
Answer: d
Page 48

As shown below, only the pentagon, when joined along a side, will never form a quadrilateral.

Option d is correct.
_____________________________________________________________________________________
9. Two transparent sheets are shown below. Sheet A is a rectangle and Sheet B is a right-angled
triangle. Rearrange the sheets so that the red points overlap exactly (without rotating or flipping
the sheets). How many quadrilaterals are formed in the final figure?

a) 3 b) 4 c) 5 d) 6
Answer: c
As shown in the figure below, when both the sheets overlap at the same red point, the final figure
contains 5 quadrilaterals.

Therefore, the correct answer is option c.
_____________________________________________________________________________________
10. Paul and Sam are at different positions and start walking in opposite directions but along parallel
paths. After covering the same distance, each of them turns toward the starting point of the other
person and walks straight (without taking any other turns) until they reach each other’s starting
points. What can we say for certain about the shape formed by BOTH their paths together?
a) The shape of the paths is a rectangle b) It has no lines of symmetry
c) Opposite sides are of same length d) Both options a and c
Answer: c
Paul and Sam walk on parallel paths starting from different positions, but in opposite directions.

Page 49

After covering the same, both of them turn towards the starting point of the other person and walk straight
until they reach the other's starting point.
There will be two possible cases for the starting positions. They can lie straight on the same vertical line
or not on the same vertical line (diagonally placed).
If the starting points are not on the same vertical line (A and B as shown in the image below), a
square/rectangular/parallelogram/rhombus is formed (based on the distance they walk):

In the second case, when the starting points lie on the same vertical line, the figure formed is a
parallelogram, where the opposite sides are parallel to each other, as shown in the image below:

In any of the cases, we cannot confirm the exact shape formed. However, we can see that the opposite
sides of the shapes are equal in any case.
Therefore, the correct answer is option c.

Page 50

Given below are two boards, X and Y. Clicking any circle on Board Y changes
its colour from black to white, or white to black.
What is the MINIMUM number of clicks required to transform Board Y into the
MIRROR IMAGE of Board X?

(a) 2 (b) 3 (c) 5 (d) 4
Answer: d
Visualize and detect what the mirror image of board X looks like:
As we need to transform board Y to the mirror image of board X, first, we need to
understand what the mirror image of board X looks like.
We know that, in a mirror image, the right side of the object appears on the left and
vice versa.
So, the mirror image of board X looks like the one shown below:

Compare board Y and the mirror image of board X:
• Check and determine which circles of board Y differ from those of the mirror
image of board X.

Clearly, we can see that there are two black circles in board Y that should actually
be white, and two white circles that should actually be black, to get the mirror
image of board X.
The other circles exactly match with the mirror image of board X. Hence, they need
not be changed.
Determining the minimum number of clicks:
As it is mentioned that clicking a circle changes its colour from black to white and
vice versa, we have to click each of the above highlighted circles of board Y once,
to get the required pattern.
Hence, the minimum number of clicks required = 4
Therefore, the correct answer is option d.

Page 51

Chapter 5: Number Play
1. A, B, C, D, and E are five distinct whole numbers, with 16 as the smallest number. When
arranged in ascending order, the difference between any two consecutive numbers is 8.
- It is given that D is the greatest number and A is the least
- Also, B is greater than E but less than C
Which of the following numbers is definitely divisible by 32?
a) A b) B c) C d) E
Answer: b
Since the numbers are arranged in ascending order with a constant difference of 8, and the smallest
number is 16, the five numbers must be: 16, 24, 32, 40, 48
Now, let us assign these values using the given conditions.
● A is the least. So, A = 16
● D is the greatest. So, D = 48
● We are told that, B is greater than E but less than C
This gives the order: D > C > B > E > A
Matching this order with the numbers:
● A = 16
● E = 24
● B = 32
● C = 40
● D = 48
Now, check which number is definitely divisible by 32.
Only 32 is divisible by 32, and this value corresponds to B.
Hence, option b is the correct answer.
_____________________________________________________________________________________
2. A control panel has three bulbs - A, B, and C. Each bulb is associated with one fixed divisor
(greater than 3). A bulb glows whenever the entered number is divisible by its own divisor. A
technician tested the panel with several numbers and noted which bulbs glowed, as shown in
the table below.
Using this information, if 56 is entered, which bulb(s) will glow?

a) A and C b) Only B c) B and C d) A and B
Answer: c
From the table:
Bulb C glows for 35 and 42. The common divisor greater than 3 is 7.
So, C glows whenever the input is divisible by 7.
Bulb B glows for 12 and 16. The common divisor greater than 3 is 4.
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Hence, B glows whenever the input is divisible by 4.
Bulb A glows for 12, 18, and 42. The common divisor greater than 3 is 6.
So, A glows whenever the input is divisible by 6.
Now check for 56:
56 ÷ 6 is not an integer. A will not glow
56 ÷ 4 = 14. B glows
56 ÷ 7 = 8. C glows
Therefore, bulbs B and C will glow.
Option c is correct.
_____________________________________________________________________________________
3. Using the digits shown in the Box, form the largest possible 6-digit number that is divisible by 9
(without repeating the digits). In the number formed, which place values contain digits that point
to a greater digit than themselves in the box?

a) Hundreds and Ones b) Ones and Tens c) Thousands and Tens d) Ones and Thousands
Answer: d
We need to form the largest 6-digit number divisible by 9 using the digits in the BOX. The
divisibility rule of 9 states that the sum of the digits must be divisible by 9.
The digits in the BOX are 1, 2, 3, 5, 7, 8, 9, and their sum is 1 + 2 + 3 + 5 + 7 + 8 + 9 = 35.
Since we must form a 6-digit number, we remove one digit.
If we remove 8, the sum becomes 35 − 8 = 27, which is divisible by 9.
So, the digits used are 1, 2, 3, 5, 7, 9, and the largest 6-digit number formed is 975321.
Now check the arrows in the BOX:
● 5 points to 7, so 5 qualifies. In 975321, 5 is in the thousands place.
● 1 points to 7, so 1 qualifies. In 975321, 1 is in the ones place.
Therefore, the place values that contain a digit that points to a greater number in the BOX are ones
and thousands.
Hence, option d is correct.
_____________________________________________________________________________________
4. Sam has two different two-digit numbers such that:
- Both numbers are divisible by 9
- None of the numbers is a multiple of 18
What is the LEAST possible sum of the two numbers?
a) 45 b) 72 c) 63 d) 54
Answer: b
Step 1: List the two-digit multiples of 9
18, 27, 36, 45, 54, 63, 72, 81, 90, 99
Remove multiples of 18:
18, 36, 54, 72, 90
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Remaining numbers:
27, 45, 63, 81, 99
Choose the smallest such pair:
27 and 45
27 + 45 = 72
Hence, option b is the correct answer.
_____________________________________________________________________________________
5. The numbers of the SET will be distributed to the Box 1 and Box 2, based on the conditions
given below:
● If the number is a multiple of 4 it goes to Box 1
● If the number is a multiple of 6 it goes to Box 2
● If the number is a multiple of both 6 and 4 it goes in both the boxes
Find the ratio of the number of BLACK CIRCLES in Box 1 to the number of WHITE CIRCLES in
Box 2, finally.

a) 2:1 b) 3:2 c) 5:2 d) 5:3
Answer: c
The numbers that will go into Box 1 are:

The numbers that will go into Box 2 are:

The number of black circles in Box 1 is 5.
The number of white circles in Box 2 is 2.
So, the ratio will be 5:2.
Option c is correct.

Page 54

6. Ashish and Devika together choose two different numbers from 2 to 24 (both inclusive). The
chosen numbers are multiplied, and the product is divided by 25. The remainder decides the
result:
● If the remainder is less than 10, Devika wins
● If the remainder is more than 14, Ashish wins
● In all other cases, the game is a draw
What is the minimum possible sum of the numbers chosen by Ashish and Devika such that
Ashish wins the game?
a) 7 b) 8 c) 9 d) 6
Answer: b
For Ashish to win, the remainder after dividing the product by 25 must be greater than 14.
So, Ashish wins if the remainder is 15 or more. (minimum possible remainder)
So, if a number leaves a remainder of 15, when divided by 25, it is possible that the number is 15 more
than the multiple of 25.
As we need to choose the smallest possible case, let’s assume that the multiple is 25 itself.
As 25 + 15 = 40, when 40 is divided by 25, it leaves a remainder of 15.
Ashish wins in this case.
The product of the numbers chosen = 40
The numbers chosen could be 5 and 8 (5 × 8 = 40)
Hence, the smallest possible sum in this case is 5 + 8 = 13.
However, let’s check if it is also possible for the product to be less than 40 and still the remainder is 15,
when divided by 25.
When a smaller number is divided by a larger number, the quotient is 0 and the remainder is the smaller
number itself (for example: 6 ÷ 25 will give us 0 as quotient and 6 as remainder).
So, the product of the two numbers chosen could be 15 too.
15: 15 ÷ 25 leaves remainder 15. So, Ashish wins
To have the product as 15, the numbers chosen must be (1, 15) or (3, 5).
1 + 15 = 16 and 3 + 5 = 8 (which is the minimum possible sum)
Therefore, the LEAST possible sum will be 8. Hence, option b is the correct answer.
_____________________________________________________________________________________
7. On the six faces of a die, even numbers from 2 to 12 are written.
● All multiples of 4 are written on adjacent faces such that no two of these faces are opposite
to each other
● The sum of the numbers on any two opposite faces is always greater than 10
Which of these pairs are written on opposite faces?
a) 8 and 10 b) 12 and 6
c) 2 and 12 d) Cannot be determined
Answer: c
- All multiples of 4 are written on adjacent faces such that no two of these faces are opposite to each
other.
4, 8, and 12 are written on three faces that have a common corner.
- The sum of the numbers on any two opposite faces is always greater than 10.
2 cannot be written opposite to 8 or 4.
Hence, it is opposite to 12.
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Similarly, 6 cannot be opposite to 4.
Hence, 6 is opposite to 8 and 10 is opposite to 4.
Thus, the pairs of opposite faces are:
2 and 12
6 and 8
10 and 4
Therefore, option c is the correct answer.
_____________________________________________________________________________________
8. In the year 2020, Simran’s age was a multiple of 6. In the year 2024, her age was a multiple of 11.
Which of the following could possibly be the age of Simran in the year 2024?
a) 11 years b) 22 years c) 33 years d) 44 years
Answer: b
Simran’s age in 2020 was a multiple of 6.
Her age in 2024 is 4 years more than her age in 2020 and is a multiple of 11.
We check the given options and see which option is a multiple of 6, 4 years ago.
● 11: 11 − 4 = 7 (not a multiple of 6)
● 22: 22 − 4 = 18 (a multiple of 6)
● 33: 33 − 4 = 29 (not a multiple of 6)
● 44: 44 − 4 = 40 (not a multiple of 6)
Only 22 satisfies both conditions.
So, in the year 2024, Simran’s age could be 22.
Hence, the correct answer is option b.
_____________________________________________________________________________________
9. The middle number of 5 consecutive even numbers is 5p. Exactly two numbers in the sequence
are divisible by 4, and only one is divisible by 5. Among the following values, which could be the
possible value of p.
a) 2 b) 3 c) 4 d) 5
Answer: a
The middle number of the sequence is given as 5p.
Since the numbers are five consecutive even numbers, the sequence must be:
5p − 4, 5p − 2, 5p, 5p + 2, 5p + 4
Use the condition about divisibility by 5
We are told that exactly one number in the sequence is divisible by 5.
● The middle number is 5p, which is always divisible by 5
● The other four numbers differ from 5p by 2 or 4, so they are not divisible by 5
So this condition is already satisfied as long as the middle number is the only multiple of 5, which is
fine for any value of p.
Divisibility by 4:
We are told that exactly two numbers in the sequence are divisible by 4.
For this to happen, the middle number must not be a multiple of 4.
Instead, it should be 2 more or 2 less than a multiple of 4.
This way, the numbers just before and just after the middle number become multiples of 4.
Trying the options, when p = 2: (as p = 1 will give us odd numbers)

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● Middle number = 10 (which is 2 more than 8)
● The sequence becomes: 6, 8, 10, 12, 14
Here:
● 8 and 12 are divisible by 4 (exactly two numbers)
● 10 is the only number divisible by 5
All conditions are satisfied. Hence, option a is the correct answer.
_____________________________________________________________________________________
10. A cryptarithmetic multiplication is shown below. Here, P, Q, M, and N are distinct single -digit
natural numbers. MN represents a two-digit number, and PQN represents a three-digit number.
What is the smallest possible value of P + M?

a) 3 b) 1 c) 4 d) 5
Answer: a
The product of 6 and N results in a number with N in the unit's place.
This is true for the numbers 0, 2, 4, 6, and 8:
0×6=0
2 × 6 = 12
4 × 6 = 24
6 × 6 = 36
8 × 6 = 48
Since, we have to find the smallest possible value of P+M, we start by considering the smallest
possible value of M, which is 2. (M cannot be 1, as the highest value of MN in that case will be 18,
where 18 × 6 = 108. This will have 1 in the hundreds place of the product. This will make M = P, which
is not possible).
So, let M = 2.
If M = 2, then the values possible for MN are: 20, 24, 26, and 28 (as 0, 4, 6, and 8 give the same units
digit after multiplying by 6)
20 × 6 = 120 (In this case, M = Q = 2, which is invalid.)
24 × 6 = 144 (In this case, Q = N = 4, which is invalid.)
26 × 6 = 156 (In this case, P = 1, M = 2, N = 6, and Q = 5. All distinct). Here, we get the smallest
possible values for P and M (1 and 2)

Similarly, 28 × 6 = 168 (In this case, P = 1, M = 2, N = 8, and Q = 6. All distinct)
Therefore, the smallest possible value of P + M = 1 + 2 = 3
Hence, option a is correct.

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There are 9 switches in a row on a switchboard. 3 of them belong to lights and
the remaining are of fans.
Every two consecutive light switches have exactly two fan switches between
them. The switch at the extreme right is NOT a light switch and the switch at
the extreme left is NOT a fan switch.
Which of them is definitely a light switch?
(a) 2nd switch from the left (b) 3rd switch from the left
(c) 6th switch from the right (d) 5th switch from the left
Answer: c
There are 9 switches altogether, out of which 3 of them are of lights and 6 are of fans.
Also, it is given that every two consecutive light switches (L) have exactly two fan
switches (F) between them.
So, there are three possible cases:
Case 1: FLFFLFFLF
Case 2: LFFLFFLFF
Case 3: FFLFFLFFL
However, the question mentions that the switch at the extreme right is not a light
switch.
So, case 3 can be eliminated.
Also, the switch at the extreme left in not a fan switch.
Based on this statement, case 1 is also eliminated.
Therefore, the correct sequence of switches is: LFFLFFLFF
As we can see, the 6th switch from the right is a light switch.
Thus, option c is the correct answer.

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Chapter 6: We Distribute Yet Things Multiply
1. What will come in place of “?”

a) b) c) d)
Answer: a
Observe the figures carefully. Each new figure adds one new circle, and the numbers follow a clear
multiplication pattern.
Step 1: Observe the numbers
Look at the numbers that appear in order:
● First figure: 1
● Second figure: 2
● Third figure: 6
● Fourth figure: 24
Now notice how each number is formed:
● 1×1=1
● 1×2=2
● 2×3=6
● 6 × 4 = 24
So, the rule is: Each new number is obtained by multiplying the previous number by the next
natural number. (starting from × 1)
Step 2: Find the next number
Following the same pattern:
● 24 × 5 = 120
So, the next number must be 120.
Hence, option a is the correct answer.

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