CTAI Teacher Handbook Class 6

Computational
Thinking and
Artificial Intelligence

Class 6

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

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. Ms. 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 2

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.

The book engages learners with problems involving constraints, dependencies, logical
conditions, grids, data interpretation, and optimisation across numerical, spatial, and
real-life contexts. It introduces foundational Artificial Intelligence concepts such as
classification, pattern identification, data driven decision making, and ethical
awareness, enabling students to understand how logical rules and data influence
intelligent systems. The document further provides pedagogical guidance, learning
resources, assessment support, and classroom implementation guidelines to facilitate
competency-based learning in alignment with NEP 2020.

TEAM CBSE

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. Patterns in Mathematics 11

4. Lines and Angles 18

5. Number Play 28

6. Data Handling and Presentation 37

7. Prime Time 46

8. Perimeter and Area 57

9. Fractions 66

10. Playing with Constructions 72

11. Symmetry 82

12. The Other Side of Zero 90

PART-2 ARTIFICIAL INTELLIGENCE
SR CHAPTER TITLE ETHICAL PAGE
NO AWARE NO
NESS
1 Introduction to Artificial Intelligence and Everyday Responsi 99
Examples. ble usage
Learning Focus:
Meaning of AI, AI in daily life, AI and automation, Human vs
Machine Intelligence, Types of Learning in AI
2 Basic Data Concepts Data 101
Learning Focus: handling
Understanding data, Types of data, Collecting data, Organising care
data, Representing data
3 Simple Pattern Recognition and Decision Making Logical 104
Learning Focus: thinking
Understanding patterns, Identifying patterns, Observations and
conclusions,
Decision making
4 Ethics and Digital Responsibility Online 106
Learning Focus: behaviour
Responsible use of technology, Online safety, Privacy,
Password safety, Digital footprints

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 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 CT and AI

Introducing these concepts at the Grade 6 level is vital for several reasons:

• Future Readiness: It prepares students for the modern world of work where using data
effectively and applying AI ethically are essential capabilities.

• Holistic Development: It fosters core cognitive capacities such as reasoning, logical thinking,
and ethical decision-making, contributing to individual flourishing and responsible digital
citizenship

• Interdisciplinary Connection: Integrating CT and AI across subjects like Mathematics and
Science helps students see knowledge as interconnected rather than compartmentalized

• Innovation: It encourages an entrepreneurial mindset by teaching students to devise innovative
solutions to real-world challenges

2. Objectives (Curricular Goals & Competencies)

• 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: Develops 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 interpret and solve multi-step problems with layered and abstract clues,
using:

• Advanced viewpoints and cross-sections of 3D objects
• Combined transformation of shapes (multiple flips, rotations, reflections, cuts/folds)
• Changes in orientation, position, order, and direction (clockwise, anticlockwise, diagonal)
• Identifying hidden, overlapping, or implied parts in complex visual patterns
• Symmetry across multiple axes and composite mirror/water image reasoning
• Visual reasoning involving scale, proportion, and spatial relationships

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PATTERN RECOGNITION
Students will be able to identify, extend, and justify complex patterns involving multiple simultaneous
changes, formed using:

• Numbers with mixed operations and logical rules
• Shapes/images with changing attributes (size, position, count, orientation)
• Letters and symbols with positional or alphabetical logic
• Patterns involving alternation, skipping, grouping, or cyclic behaviour
• Mixed patterns combining numbers, shapes, and letters with dependency rules
DECOMPOSITION
Students will be able to break down higher-order problems involving interdependent clues and
constraints, using information from:

• Numerical clues involving place value, operations, factors, multiples, and comparisons
• Properties of 2D and 3D shapes (faces, edges, vertices, diagonals, angles
• Multi-step transfers or exchanges (money, quantities, digits, objects) with conditions
• Tables, grids, or charts requiring cross-referencing of multiple data points
• Conditional rules for counting, grouping, sorting, or eliminating possibilities
• Visual representations that encode numerical or logical values

ALGORITHMIC THINKING
Students will be able to follow, analyse, and apply multi-layered rules and procedures to solve
complex problems involving:

• Number sequences formed using combined operations and logical conditions
• Movement on grids involving direction, distance, turns, and path constraints
• Stepwise changes where values increase/decrease based on rules
• Multi-step instructions involving swaps, shifts, transfers, and rearrangements
• Ordering people, objects, or events using multiple attributes or clues
• Logical flow of steps, identifying necessary vs redundant information

Artificial Intelligence (AI) Learning Outcomes
Learners will be able to:

• Summarise the basic ideas and concepts of AI and its application
• Describe key differences between machine intelligence and human intelligence
• Explain the difference between automation and AI using practical, real -world cases.
• Differentiate the three fundamental AI methodologies, namely supervised, unsupervised,
and reinforcement learning
• Develop the skill of organizing and representing data and its various forms, including text,
numbers, images, and sounds

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• Recognize simple patterns in data and make decisions based on observations
• Demonstrate an understanding of ethics and digital responsibility in the use of AI, including
digital footprints, privacy, and responsible technology behaviour.
• Practice essential internet safety protocols, such as creating secure passwords,
maintaining safe online behaviour, and applying basic privacy measures while using
digital and AI tools.
• Apply conceptual knowledge of AI to everyday activities by recognising human-centred
design and ethical principles in how AI works and interacts with people

4. Mapped with NEP and NCF 2023
The Grade 6 curriculum is directly derived from the Aims of School Education outlined in the
National Curriculum Framework for School Education (NCF-SE) 2023. It fulfills the NEP 2020
mandate to integrate Machine Learning and Computational Thinking into the school journey to
foster creativity and interdisciplinary exploration.

5. Time Allocation
The Middle Stage (Classes 6–8) requires 100 hours annually. For Grade 6, this time is specifically
divided as follows:
• Advanced CT Skills: 40 hours
• Introductory AI Concepts: 20 hours
• Interdisciplinary Projects: 40 hours (20 hours each for two projects)
6. Approach / Pedagogy
The pedagogical approach for Grade 6 is activity-based and inquiry-driven:
• Experiential Learning: Students engage with complex puzzles, riddles, and hands-on real-world
problems
• Collaborative Work: The curriculum emphasizes group discussions, debates, and collaborative
projects to solve multidisciplinary challenges
• Project-Based Learning: Students use AI tools and data analysis to create solutions for
community or fictional city issues
7. Assessment
Assessment shifts from rote memorization to continuous, formative, and competency-based
evaluation. Methods include:
• Performance-Based Tools: Project presentations, assignments, and reflective journals
• Practical Evaluation: Written tests with CT puzzles, practical examinations, and interactive
activities
• Qualitative Feedback: Teachers use clear rubrics and Observation Journals to ensure
consistency in tracking student development

Page 8

How to Use This Book?
PART-1 Computational Thinking
Part 1 of 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 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: Patterns in Mathematics
1. The following series follows a fixed pattern:
2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, …
If the pattern continues, determine how many times the number 9 appears in the NEXT 20 terms.
a) 3 b) 4 c) 7 d) 8
Answer: a
The series is made up of consecutive natural numbers, starting from 2.
Each number follows this fixed rule:
A number n appears exactly (n−1) times
For example:
Number 2 appears (2 - 1) 1 time.
Number 3 appears (3 - 1) 2 times.
From the given series, one 6 has already appeared.
So, in the next 20 terms:
● 6 appears 4 more times
● 7 appears 6 times
● 8 appears 7 times
Total terms before 9 appears:
4 + 6 + 7 = 17
Out of the next 20 terms, 17 terms are not 9, leaving:
20 − 17 = 3
So, 9 appears 3 times in the next 20 terms.
The sequence is:

Hence, the correct answer is option a.
_____________________________________________________________________________________
2. The following number series is based on a pattern. One term is incorrect. Identify the incorrect
term.
2, 3, 6, 11, 18, 25, 38, 51, 66, 83
a) 11 b) 25 c) 51 d) 66
Answer: b
When analysing a number series, the first step is to check the difference between consecutive terms,
as many series are formed using a pattern of differences.

Step 1: Find the differences
● 3−2=1
● 6−3=3
● 11 − 6 = 5
● 18 − 11 = 7
● 25 − 18 = 7
● 38 − 25 = 13
● 51 − 38 = 13

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● 66 − 51 = 15
● 83 − 66 = 17

Step 2: Identify the pattern
The differences are expected to follow consecutive odd numbers:
1, 3, 5, 7, 9, 11, 13, 15, 17, …
Step 3: Verify the terms
Check the 4th and the 5th term, after the difference of 7, the next difference should be 9, which gives:
18 + 9 = 27
However, 25 is given instead.
Conclusion
Since 25 does not fit the pattern, it is the wrong term in the series.
The correct sequence is: 2, 3, 6, 11, 18, 27, 38, 51, 66, 83
Option b is the correct answer.
_____________________________________________________________________________________
3. What will come in place of "?" in the given series?
1@3, #5#, 7@9, #11#, 13@15, ?, 19@21
a) @17@ b) #15# c) 16#18 d) #17#
Answer: d
Each term is actually a group of three consecutive numbers (1-2-3, 4-5-6, and so on). However, only odd
numbers can be seen and even numbers in each group are replaced by symbols:
- The even numbers from the odd-positioned terms are replaced by “@”
- The even numbers from the even-positioned terms are replaced by “#”
We have to find the 6th term of the series (even-positioned term).
As it is an even-positioned term, the symbol that replaces the even numbers in that term is “#”.
As the previous term was 13@15 (which is 13-14-15), the next term will have 16-17-18.
When the even numbers in this term are replaced by “#”, we get: #17#.
Therefore, the 6th term of the series is: #17#, which is option d.
Hence, option d is the correct answer.
_____________________________________________________________________________________
4. The first four terms of a block series are shown below.
If the pattern continues in the same manner, which term of the series will be the first to contain
more than 20 blocks?

a) 5th term b) 6th term c) 7th term d) 8th term
Answer: b
In each next term, a new row is added at the bottom, with one extra block than the row above it
(continuing the same step pattern).
So, the 5th term will have an extra row at the bottom, and that row contains one more block (4 + 1 = 5)
than the row above it. Hence, the 5th term will have a total of 15 blocks.

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Similarly, the 6th term will have one extra row compared to the 5th term, and that row contains one
more block (5 + 1 = 6) than the row above it. Hence, the 6th term will have a total of 21 blocks, which is
the first term to have more than 20 blocks.
Hence, the correct answer is option b.

_____________________________________________________________________________________
5. What will come in place of "?" in the given series?

a) b) c) d)
Answer: d
First, observe how the numbers change from one term to the next.
Each term is a 2 × 2 grid containing four numbers.
When we compare consecutive terms, we see that each number in the grid increases by 1 as we
move to the next term.
For example:
● Top-left number: 1 - 2 - 3 - 4
● Top-right number: 2 - 3 - 4 - 5
● Bottom-left number: 2 - 3 - 4 - 5
● Bottom-right number: 3 - 4 - 5 - 6
Since this same pattern continues, the next term will have all four numbers increased by 1 compared
to the previous term.
So, the next grid will be:
● Top-left: 5
● Top-right: 6
● Bottom-left: 6
● Bottom-right: 7
This matches option d.
Thus, option d is the correct answer.

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6. What will come in place of "?" in the given series?

a) b) c) d)
Answer: c
First, observe the pattern in each position separately: top circle, middle square, and bottom circle.
Top circles:
The numbers increase by 3 each time:
3, 6, 9, 12, 15, 18
Bottom circles:
The differences increase by 1 each time:
4 to 5 (+1), 5 to 7 (+2), 7 to 10 (+3), 10 to 14 (+4)
So, the next increase is +5:
14 + 5 = 19
Middle squares: Each middle number is the sum of the numbers in the top and bottom circles:
3 + 4 = 7, 6 + 5 = 11, 9 + 7 = 16, 12 + 10 = 22, 15 + 14 = 29
So, the next middle number is:
18 + 19 = 37
Thus, the missing figure contains 18 (top), 37 (middle), and 19 (bottom).
Hence, the correct answer is option c.
_____________________________________________________________________________________
7. The first five terms of a series formed using grey hexagons and white diamonds are given below.
If the same pattern continues, how many diamonds will be present in the term where the number
of hexagons is 144?

a) 100 b) 135 c) 121 d) 169
Answer: c
From the pattern, we observe:
Term 1: 1 hexagon, 0 diamonds
Term 2: 4 hexagons, 1 diamond
Term 3: 9 hexagons, 4 diamonds
Term 4: 16 hexagons, 9 diamonds
Term 5: 25 hexagons, 16 diamonds

Here, the hexagons follow the pattern 1, 4, 9, 16, 25 and so on, while the diamonds follow the pattern 0,
1, 4, 9, 16 and so on. (where all are square numbers)
We notice that, in each term:

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Number of hexagons = n2
Number of diamonds = (n − 1)2
Now, we have to find the number of diamonds in a term that has 144 hexagons.
144 hexagons = 122
So, the number of diamonds = (12 − 1)2 = 112 = 121
So, there will be 121 diamonds when there are 144 hexagons.
Hence, the correct answer is option c.
_____________________________________________________________________________________
8. Given below are two sets of numbers, P and Q. Which number from Set P can be interchanged
with a number from Set Q such that both new sets follow a particular series or pattern?
Set P: (18, 22, 24, 27, 30)
Set Q: (21, 24, 27, 31, 36)
a) 18 b) 27 c) 22 d) 30
Answer: c
How to Approach Pattern Questions
● Look at the numbers and check if most of them already follow a clear pattern
● Identify the most natural pattern, such as:
■ Adding the same number each time
■ Increasing differences
■ Multiples, skips, or repeated operations
● Find the number that breaks this pattern
● Change or interchange only that number
● Verify the pattern again to ensure all numbers now follow the same rule
Step 1: Examine Set P
Set P: 18, 22, 24, 27, 30
At first glance, the differences are not consistent.
However, the last three numbers: 24, 27, 30 (clearly increase by 3).
This suggests that Set P may be intended to follow a +3 pattern.
If that is the case, the sequence would look like:
18, 21, 24, 27, 30
Comparing this with the given set, we see that 22 does not fit this pattern.
So, 22 is a possible incorrect term in Set P and would need to be replaced by 21 to make the pattern
consistent.
Step 2: Check if the required replacement exists in Set Q
Set Q: 21, 24, 27, 31, 36
Since 21 is already present in Set Q, an interchange is possible.
Now, before finalising, we must check whether replacing 21 with 22 in Set Q creates a valid pattern
there.
Step 3: Examine Set Q after the interchange
After replacing 21 with 22, Set Q becomes:
22, 24, 27, 31, 36
The differences are now:
+2, +3, +4, +5
This forms a clear and consistent increasing-difference pattern.
Step 4: Final Verification
After interchanging 22 and 21:
• Set P: 18, 21, 24, 27, 30 (constant increase of 3)

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• Set Q: 22, 24, 27, 31, 36 (differences increasing by 1)
Both sets now follow simple and logical patterns.
Thus, the number 22 from Set P is interchanged with the number 21 from Set Q.
Hence, option c is the correct answer.
_____________________________________________________________________________________
9. The first three terms of a series of circles are shown below. If the pattern continues in the same
manner, how many circles will be there in term 91?

a) 184 b) 180 c) 194 d) 204
Answer: a
In the first term, there are 4 circles arranged in 2 columns, with 2 circles in each column. In the second
term, one more column of circles is added, making a total of 3 columns.
Similarly, in term 3, the number of columns is 4.
Term 1 has 2 columns (1 + 1 = 2)
Term 2 has 3 columns (2 + 1 = 3)
Term 3 has 4 columns (3 + 1 = 4)
Hence, we can say that, the number of columns of circles in each term is 1 more than the position of that
term.
Based on the same rule, the number of columns in term 91 will be 91 + 1 = 92.
As each column contains two circles (one above and one below), the total number of circles in term 91
is 92 x 2 = 184.
Hence, there will be 184 circles in term 91.
The correct answer is option a.
_____________________________________________________________________________________
10. A pyramid has to be formed by combining cubes. Every level will have two fewer cubes than the
level below it. If a pyramid is formed using at most 30 cubes, what is the maximum number of
levels it can have?
a) 3 b) 4 c) 5 d) 6
Answer: c
To get the maximum number of levels, the top level must have the smallest possible number of cubes,
and the number of cubes should increase as we move to the lower levels.
Therefore, the number of cubes in the levels will be 1, 3, 5, 7, and 9 from top to bottom.
Total cubes in these 5 levels will be 25. After that, to add a level, we will need 11 cubes, which we don't
have.
Now, if we take 2 as the least number of cubes in the top level and keep adding as we go down the levels,
the number of cubes in the levels will be 2, 4, 6, 8, and 10 from top to bottom. Thus, in both arrangements,
the pyramid will have 5 levels.
Hence, option c is the correct answer.

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You must shoot exactly one item from each box, to eliminate it from its box. When
an item is shot, the same item in the adjacent box is also eliminated. What is the
MAXIMUM number of items that can be eliminated, after all 3 shots?

(a) 4 (b) 5 (c) 6 (d) 7
Answer: d
When we consider boxes A and B, we can see that two different shapes (rhombus and
triangle) are common between them.
So, if the triangle is shot in box A, the same shape will be eliminated from box B, leading
to two eliminations.
But, if you observe carefully, the triangle is the only shape that is common across all
three boxes.
This means that if you shoot the triangle from box B, the triangles from the boxes on
either side (A and C) will also be eliminated, making three eliminations in one go.
So, to maximise the number of items eliminated, it is ideal to shoot the triangle from box
B.

Then, the rhombus from box A will be shot and the same shape will be eliminated from
box B.

Finally, the arrow from box C will be shot and the same shape will be eliminated from box
B.

Hence, the maximum items that can be eliminated = 7 (3 triangles, 2 rhombuses, and 2
arrows). Option d is correct.

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Chapter 2: Lines and Angles
1. Eight friends (A, B, C, D, E, F, G, H) are sitting at equidistant positions around a circular table,
each facing towards the centre, as shown below.
- A is facing North-East
- B and D are facing perpendicular directions
- B is to the immediate left of A; while C is exactly between E and F
- B and F are facing opposite directions
If G is sitting exactly between B and D, what is the angle between the directions that A and C are
facing?

a) 45 degrees b) 90 degrees c) 135 degrees d) 180 degrees
Answer: b
B is to the immediate left of A. Also, B and F are facing opposite directions.

B and D are facing perpendicular directions. Two possible arrangements are formed:

C is exactly between E and F.

However, as G is sitting exactly between B and D, case 2 is invalid.
As per case 1, the final arrangement can be seen:

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Now, we have to find the angle between the directions of A and C.
As the angle around a circle is 360°, and there are 8 equidistant positions, the angle between two adjacent
people = 360/8 = 45°.
Hence, the angle between A and C will be 90°, as highlighted below.

The correct answer is option b.
_____________________________________________________________________________________
2. Avi and Sam attend dance sessions in the afternoon.
- Avi’s session starts when the angle between the hour hand and the minute hand of the clock
is 60 degrees
- Sam’s session starts at 1:55 PM
What is the least possible difference between the starting times of both sessions?
a) 2 minutes b) 5 minutes c) 10 minutes d) 15 minutes
Answer: b
Given that Avi’s session started when the angle between the hour hand and the minute hand of the clock
is 60 degrees.
We know that the complete angle around a clock is 360 degrees.
So, 60 minutes = 360 degrees
Hence, 10 minutes = 60 degrees
So, 60 degrees exist between the hands of a clock, only when they are 10 minutes apart.
This means that when Avi’s session started, the hands of the clock were positioned so that they were 10
minutes apart.
Sam’s session starts at 1:55 PM and we need to find Avi’s session time.
To minimize the difference between the starting times of both the sessions, we have to consider a time
that gives 60-degree angle between the hands and it must be as close to 1:55 as possible.
The time that gives 60-degree angle immediately after 1:55 PM is 2:00 PM (One hand at 12 and another
hand at 2, which make 60 degrees)
Hence, the least possible difference between the starting times of both sessions is:
2:00 PM - 1:55 PM = 5 minutes.

Hence, option b is correct.

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3. An equilateral triangle (all the angles are equal) is given below. At minimum, by how many degrees
should the given triangle be rotated anticlockwise so that it looks exactly like the original triangle?

a) 30 degrees b) 60 degrees c) 120 degrees d) 90 degrees
Answer: c
Given that all three angles are 60° (equal angles), and all three sides of the triangle are equal (equilateral
triangle)

Now, we need to rotate the triangle in anti-clockwise direction, in such a way that side AC lies horizontally
(just like the current side AB)
We know that the angle made by a straight line is 180°.
Also, angle CAB is 60°.
To make the next side lie on the same straight line again, the triangle must rotate through the remaining
angle:
180° − 60° = 120°

Therefore, the triangle must be rotated 120° in the anticlockwise direction to look exactly the same
again.
As shown, rotating it by 60° does not give the same position (the triangle looks inverted), but rotating it
by 120° brings the triangle back to an identical appearance.

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Hence, option c is correct.
_____________________________________________________________________________________
4. What will come in place of “?”

a) b) c) d)
Answer: c
The number inside the circle decides how many equal parts the circle is divided into.
A full circle is 360 degrees.
So, we divide 360 by the number given inside the circle to find the number of parts.
● 360 ÷ 90 = 4 parts
● 360 ÷ 120 = 3 parts
● 360 ÷ 45 = 8 parts
Now for 60:
360 ÷ 60 = 6
So, the circle should be divided into 6 equal parts.
Hence, option c is the correct answer.
_____________________________________________________________________________________
5. A and B visit a park in the evening.
● When A visited the park, the minute hand pointed at 12, and the other hand was 150° clockwise
from it
● When B visited the park, the hour hand pointed at 5, and the other hand was 210° clockwise
from it
Based on the given information, which of the following statements is true?
a) A visited the park before B b) B visited the park before A
c) A and B visited exactly at the same time d) None of the above
Answer: c
On a clock:
• A full circle = 360°

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• There are 12 numbers, so the angle between two adjacent numbers is
360° ÷ 12 = 30°
When A visited the park:
• The minute hand was pointing at 12
• The other hand (hour hand) was 150° clockwise from the minute hand
To find how many numbers apart this is:
150° ÷ 30°= 5
So, the hour hand is 5 numbers ahead of 12, i.e., at 5.
The time when A visited the park was 5:00.

When B visited the park:
• The hour hand was pointing at 5
• The other hand (minute hand) was 210° clockwise from it.
To find how many numbers apart this is:
210° ÷ 30° = 7
Moving 7 numbers clockwise from 5 lands on 12.
The time when B visited the park was also 5:00.
As both of them visited in the evening, the time was 5:00 pm and it can be definitely implied that A and
B visited exactly at the same time. Therefore, the correct answer is option c.
_____________________________________________________________________________________
6. What will come in place of “?”

a) 40° b) 60° c) 90° d) 30°
Answer: b
A clock is a circle of 360° and has 12 numbers, so the gap between two numbers is 30°.
In each set, we look at both clocks and find the common angle covered by the hands.
● First set: 60°
The common part between the two clocks covers 2 number gaps (9 to 11),
2 × 30° = 60°
So, the centre shows 60°.
● Second set: 120°
The common part covers 4 number gaps (6 to 10),
4 × 30° = 120°
So, the centre shows 120°.
● Third set: 90°
The common part covers 3 number gaps (1 to 4),
3 × 30° = 90°
So, the centre shows 90°.

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● Question set:
The common part between the two clocks covers 2 number gaps (2 to 4),
2 × 30° = 60°
Thus, the missing value is 60°. Hence, option b is the correct answer.
_____________________________________________________________________________________
7. How many times in a day, on a 12 - hour format clock, do the minute and hour hands form a
straight line (i.e., an angle of 180°)?
a) 11 b) 12 c) 22 d) 24
Answer: c
In a clock, the minute hand moves faster than the hour hand. Because of this difference in speed, the
minute hand repeatedly moves ahead of the hour hand and at certain moments the two hands become
exactly opposite to each other, forming a straight line (180°).
Let us examine the clock hour by hour from 12:00 a.m. to 12:00 p.m.
These 12 hours can be divided into 12 intervals:
• 12:00 – 1:00
• 1:00 – 2:00
• 2:00 – 3:00
• …
• 11:00 – 12:00
In most of these hourly intervals, the hands form a straight line exactly once.
However, there is a special situation around 6 o’clock.
At 6:00, the hour hand is at 6 and the minute hand is at 12, so the two hands are already exactly opposite
each other, forming a straight line.
Now notice what happens when we count hour by hour:
• When considering the interval 5:00 to 6:00, we count the straight line that occurs at 6:00
• When considering the interval 6:00 to 7:00, we again count the same straight line at 6:00
So, the same occurrence at 6:00 gets counted twice, even though it happens only once in reality.
Therefore, from the 12 possible hourly cases, we must subtract 1 to remove this double counting.
Thus, in 12 hours, the hands form a straight line: 12 – 1 = 11 times
Since a day has 24 hours, this happens twice in a day: 11 x 2 = 22
Hence, the hands of a clock are in a straight line 22 times in a day. Option c is the correct answer.
_____________________________________________________________________________________
8. In the given figure, Triangle 2 is a right-angled triangle. Both triangles are arranged such that the
red circles completely overlap, meaning both the triangles meet at point P. The triangles cannot
be rotated or flipped. If an angle of 30° is formed between Triangle 1 and Triangle 2 at point P, find
the value of ∠APD?

a) 110° b) 120° c) 130° d) 150°
Answer: b

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Let’s solve it step by step:
When triangles 1 and 2 are arranged such that the red circles overlap each other, the arrangement looks
like this:

It is mentioned that an angle of 30° is formed between Triangle 1 and Triangle 2 at point P.
In the image, we can see the acute angle ∠APC, with vertex as P.
Thus, we can say that ∠APC = 30°.
As mentioned in question, Triangle 2 is a right-angled triangle
Thus, we can say that ∠CPD = 90°.
∠APD = ∠APC + ∠CPD
We can see that ∠APD = 30° + 90° = 120°.

Therefore, the correct answer is option b.
_____________________________________________________________________________________
9. Sam draws three line segments AB, BC, and BD (with a common point B) on a sheet of paper
∠ABC = 108° and ∠CBD = 162°, and these angles lie on either side of the line segment BC.
Later, he draws a pair of parallel line segments EF and GH, both perpendicularly intersecting BD.
How many pairs of parallel line segments appear on the sheet, finally?
a) 1 b) 2 c) 3 d) 4
Answer: c
It is given that ∠ABC = 108° and ∠CBD = 162°, and these angles lie on either side of BC.

So,
∠ABD = 360° − (108° + 162°)
∠ABD = 360° − 270°
∠ABD = 90°
This means AB is perpendicular to BD.
It is also given that EF and GH are drawn perpendicular to BD.

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Whatever the case, as AB is perpendicular to BD and both EF & GH are also perpendicular to the same
line segment BD, all of them are parallel to each other.

Thus, the pairs of parallel line segments are:
1) AB - EF
2) AB - GH
3) EF - GH
Hence, there are three such pairs of parallel line segments. Thus, option c is the correct answer.
_____________________________________________________________________________________
10. A line segment AB is shown below. Five points: C, D, E, F, and G lie on the line segment AB such
that
- All the points (including A and B) are at equal intervals
- The length of segment CB is equal to the length of segment GD
- Point E is immediately to the right of point C
- Point F cannot be next to E or D
Which segment among the following options is the longest?

a) GC b) FE c) GE d) FD
Answer: d
We cannot place C immediately next to B, as E is on the immediate right to C.
Also, we cannot place C immediately next to A, as the length of segment CB is equal to the length of
segment GD, and it is not possible to have both points G and D between C and B.

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There are three possibilities for the positions of C and E:

And the placement for G and D can be as follows:

As F cannot be next to E or D, the first and third cases are eliminated.
Hence, the final arrangement is:

Clearly, among the given options, FD is the longest line segment.
Thus, option d is the correct answer.

Page 26

There are four ropes: A, B, C, and D. Each rope has a different length and a
different colour chosen from Blue, Red, Green, and Yellow. Rope B is longer than
only the Green rope. The Blue rope is longer than B but shorter than C. Rope A is
not Blue. If the Yellow rope is the longest, which rope is Red?
(a) A (b) B (c) C (d) D
Answer: b
Ropes A, B, C, and D have different length and colour.
Rope B is longer than only the green rope.

The blue rope is longer than B but shorter than C.

A is not blue.

As per the question, the longest rope is yellow in colour.
This means that C is yellow.
Therefore, B is the red rope.

Hence, the correct answer is option b.

Page 27

Chapter 3: Number Play
1. Sachin plays a game using a standard die numbered 1 to 6. At each roll, he scores points equal
to the number shown. He may roll the die any number of times.
The game ends when the number 6 appears three times (the three 6’s need not be consecutive).
If the total score at the end of the game is 29, what is the minimum number of rolls Sachin could
have made?
a) 4 b) 5 c) 6 d) 7
Answer: c
It is given that the game ends when Sachin gets 6 on three turns.
So, he must have scored 6 three times, which gives:
3 × 6 = 18 points
Since his total score is 29 points, the remaining score must be:
29 − 18 = 11 points
To minimise the number of turns, we should assume the highest possible score in each remaining
turn.
If the remaining 11 points were scored as 6 + 5, the sequence of rolls would be:
6, 5, 6, 6, 6 (five turns)
However, this is not possible, because once Sachin gets three 6s, the game ends immediately. He
would not get another chance to roll the die.
So, the remaining 11 points must be scored without using another 6.
Since 6 cannot be used, let us consider the other possibilities to score 11.
• 5 + 5 + 1 = 11 points
• 5 + 4 + 2 = 11 points
• 4 + 4 + 3 = 11 points
However, in any case Sachin has to roll the dice at least 3 more times after 3 sixes.
Therefore, Sachin must have rolled the die at least six times, scoring:
1, 5, 5, 6, 6, 6 or 3, 4, 4, 6, 6, 6 or 2, 4, 5, 6, 6, 6 (in any order) to obtain a total of 29 points.
Hence, option c is the correct answer.
_____________________________________________________________________________________
2. In the given grid, each white square contains 1, 2, 3, or 4 hidden coins.
Each black square shows the maximum number of coins present in any of its adjacent white
squares.
If every row has the same total number of coins, what is the MAXIMUM possible number of white
squares that contain exactly one coin?
Note: Two squares are adjacent only if they share a common side. Squares that share a common corner
alone, are NOT considered as adjacent

a) 5 b) 6 c) 7 d) 8
Answer: c

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Let’s solve this in a logical way.
It is mentioned that the number on the black square represents the highest number of coins that its
adjacent square has. So, all the squares adjacent to 1 must have only one coin.

Now, the only option left for the black square with 3 is to have 3 coins in the square above it.

Similarly, the black square having 4 must have 4 coins in any of its adjacent squares, but they cannot be
adjacent to the black square having 2. So, 4 has only one possibility.

We know that every row has the same number of coins.
Observe Row A.
As it already has 4 coins in a square, the total number of coins in a row is definitely 6 or more than 6. (as
the other 2 squares of row 1 will have at least one coin each: 4 + 1 + 1 = 6).

However, if the total number of coins in a row is more than 6, then the last empty square of row D must
have more than 4 coins in it to get the total (1 + 1 + 5), which is invalid. (a square can have only 1 to 4
coins)
Therefore, the total number of coins in each row is definitely 6. Thus, rows A and D can be filled as:

Let’s move to Row B:
It has 3 coins already and we need to fill 3 more coins in two squares. The only possible combination of
3 here is 1 + 2 (in any order of the squares)

Similarly, row C has 1 coin already and we need to add 5 more coins in this row.
Now think, 5 coins can be made as a combination of 2 + 3 (or) 4 + 1.

Page 29

But, as we are trying to detect the highest possible number of squares that can have only one coin in
them, the combination that we choose must be 1 + 4.

Therefore, the number of squares that can have only one coin in them are 7.
Option c is correct.

_____________________________________________________________________________________
3. If each of the given terms follows the same theme, what will be the value of P + Q + R + S?

a) 134 b) 136 c) 119 d) 142
Answer: b
The following theme can be seen in each term:
In every term, the number in the black cell is split into 4 equal portions, as there are four white cells
adjoining it (the ones connected directly along its sides), where each portion is written in each cell. (for
example, in the first term, 96 ÷ 4 = 24)

Similarly, the numbers adjacent to the black cell (the ones connected to the black cell along its edges)
are split into three equal portions as each of them has three adjacent white cells. Each portion is written
in each outer cell. (for example, in the first term, 24 ÷ 3 = 8)

Based on the same theme, in the question term, we have 204 in the black cell.
As 204 ÷ 4 = 51, we must have 51 in each of its adjacent cells. Q = R = 51

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Now, each 51 is split into three equal portions.
As 51 ÷ 3 = 17, all the outer numbers will be 17. P = S = 17

P + Q + R + S = 17 + 51 + 51 + 17 = 136
Option b is the correct answer.
_____________________________________________________________________________________
4. An arrow between any two squares, always points towards the square having a larger number. A
and B are two DIFFERENT numbers. If the largest 5-digit number is formed using all five squares,
what will be the difference between the final number and 10000?

a) 44332 b) 54545 c) 54432 d) 44432
Answer: d
Let’s look at B first.
B has an arrow pointing towards it from 2. So, B must be greater than 2.
Also, B has an arrow pointing away from it towards 4. So, B must be less than 4.
Thus, B must be 3.
Now, let’s look at A.
A has an arrow pointing towards it from 2. So, A must be greater than 2.
Also, A has an arrow pointing away from it towards 5. So, A must be less than 5.
The only possible numbers are 3 and 4.
But we already know that B is 3.
So, A must be 4.
Thus, the values of all five squares are: 5, 4, 2, 3, and 4.
The largest 5-digit number formed using all five squares is: 54432.
Now, we have to determine the difference between this number and 10000.
54432 - 10000 = 44432.
Hence, option d is the correct answer.

Page 31

5. How many different triangles can be formed by connecting the black dots in the image below,
where the sum of the numbers at their corners is 12?

a) 7 b) 6 c) 8 d) 5
Answer: c
We need three numbers whose sum is 12.
The given image has only 3 unique numbers: 3, 4, and 5.
Let’s assume that all three corner numbers of a triangle are equal.
As 12/3 = 4, we need three 4s to connect a triangle. But, the above image has only two 4s.
So, we have to try combinations of different numbers from the above image.
If the triangle includes 3 and 4 as its corner numbers, then 3 + 4 = 7. We need a 5 at the third corner to
make a sum of 12. (3 + 4 + 5 = 12)
Similarly, if the triangle has 3 and 5 at its corners, then 3 + 5 = 8. We need a 4 to make a sum of 12. The
same is true if 4 and 5 are the corners of a triangle. (as 4 + 5 = 9, we need a 3 to make a sum of 12)
So, the only possible combination is 3, 4, and 5.
Since there are two 3’s, two 4’s, and two 5’s placed in different positions, we can form 8 different triangles
as shown below in the image.

Hence, the answer is option c.
_____________________________________________________________________________________
6. Amat and Ankit each picked three numbers from the set {1, 2, 3, 4, 5, 6}.
- 3 different numbers picked by Amat add up to give the highest possible sum
- 3 different numbers picked by Ankit add up to the second highest possible sum
Which pair of numbers were picked by both Amat and Ankit?
a) 3 and 4 b) 4 and 5 c) 5 and 6 d) 2 and 5
Answer: c
To attain the highest possible sum, select the three largest numbers from the set, which are 6, 5, and 4.

Page 32

This combination results in a sum of 6 + 5 + 4 = 15.
Now, to find the 2nd-highest possible sum, we must get a total just less than 15.
To keep the sum as large as possible, we should retain the two largest numbers, 6 and 5.
If we remove either 6 or 5, the total would decrease too much.
Since 6 + 5 + 4 gives 15 (the highest), we replace 4 with the next largest remaining number, which is 3.
This gives 6 + 5 + 3 = 14.
No other combination gives a sum greater than 14 but less than 15.
Thus, Amat picked {6, 5, 4} and Ankit picked {6, 5, 3}.
So, the pair of numbers picked by both Amat and Ankit is 5 and 6.
Hence, option c is correct.
_____________________________________________________________________________________
7. A box contains six numbered circles coloured black and white.
A 6-digit number must be formed using all six circles, and the colours must alternate throughout
the number. The first circle may be either black or white.
Among all such numbers that can be formed, take the second smallest number and the second
largest number. What is the difference between these two numbers?

a) 683233 b) 684324 c) 685314 d) 583297
Answer: b
Step 1: Identify the digits
● White digits: 9, 9, 6
● Black digits: 4, 3, 2
All six digits must be used, and colours must appear alternately.
Second Greatest Number
To form the greatest number, the first digit should be the largest possible.
● Largest white = 9
● Largest black = 4
Since 9 > 4, the number starts with White.
Pattern: W – B – W – B – W – B
Greatest number formed: 949362
To get the second greatest, keep the higher place values the same and make the smallest possible
change.
Changing white digits affects higher places, so interchange the last two black digits.
Second greatest number: 949263
Second Smallest Number
To form the smallest number, the first digit should be the smallest possible.
● Smallest black = 2
● Smallest white = 6
Since 2 < 6, the number starts with Black.
Pattern: B – W – B – W – B – W
Smallest number formed: 263949

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To get the second smallest, interchange the last two black digits. (since the last two white digits are
9, there wouldn’t be any change in the number’s value if we interchange them)
Second smallest number: 264939
Final Step
949263 − 264939 = 684324
Hence, option b is the correct answer.
_____________________________________________________________________________________
8. I am a 5-digit number made up of both even and odd digits. I read the same backwards and
forwards (palindrome). The greatest difference between at least two of my digits is 9. What is the
smallest possible sum of my digits?
a) 9 b) 10 c) 11 d) 13
Answer: c
The number is a 5-digit palindrome, so it has the form ABCBA.
It must:
● Contain both even and odd digits
● Have the greatest difference between at least two of my digits equal to 9
A difference of 9 is possible only if 0 and 9 are both present, so the number must include these digits.
To keep the sum of digits as small as possible:
● Use 9 only once (since repeating 9 would increase the sum)
● Place 0 in the repeated positions B and B, because repeated digits affect the sum more
The first digit A cannot be 0, so the smallest possible value for A is 1.
Thus, the smallest possible number is 10901.
Digit sum:
1 + 0 + 9 + 0 + 1 = 11
All conditions are satisfied, so the smallest possible sum of digits is 11.
Therefore, option c is correct.
_____________________________________________________________________________________
9. Count the number of blocks that are connected to at least one block containing a smaller number
and one block containing a larger number than the number in it.

a) 2 b) 3 c) 4 d) 5
Answer: c
As shown below, there are 4 numbers which are connected to at least one smaller and one larger number
than itself.

Therefore, option c is the correct answer.

Page 34

10. A 3 × 3 grid of digits is given below. Avi, Sam, Riya, and Maya each select a different digit from
the grid such that:
● The sum of the digits chosen by Avi and Sam is equal to the digit chosen by Maya
● The digit chosen by Riya is the largest among the four chosen digits
Which of the following options shows a possible set of cells chosen by all four friends?
Note: You cannot rotate the question or option images

a) b) c) d)
Answer: d
Riya picks the largest digit among all four friends. So, from the four selected digits, the remaining three
must satisfy: Avi’s digit + Sam’s digit = Maya’s digit
We should look for an option where, when the highest digit is allotted to Riya, the next highest digit
(Maya’s choice) is the sum of the remaining two digits.
Option a: There are two possible placements as shown below:

Case A: If 9 is taken by Riya, the next highest digit 7 is NOT the sum of 2 and 4.
Case B: If 6 is taken by Riya, the next highest digit 5 is NOT the sum of 3 and 4.

Option b: There are two possible placements as shown below:

Case A: If 7 is taken by Riya, the next highest digit 6 is NOT the sum of 4 and 5.
Case B: If 9 is taken by Riya, the next highest digit 4 is NOT the sum of 0 and 3.

Option c: There are two possible placements as shown below:

Case A: If 7 is taken by Riya, the next highest digit 4 is NOT the sum of 0 and 2.
Case B: If 9 is taken by Riya, the next highest digit 6 is NOT the sum of 4 and 5.

Page 35

Option d: There are two possible placements as shown below:

Case A: If 6 is taken by Riya, the next highest digit 5 is NOT the sum of 2 and 4.
Case B: If 9 is taken by Riya, the next highest digit 7 is THE SUM OF 3 and 4.

On checking the options, only option d satisfies both conditions. Hence, option d is correct.
_____________________________________________________________________________________

Four friends A, B, C, and D each selected a different shape from the shapes shown
below, such that:
• A selected a shape that is neither beside B's shape nor D's shape
• C and D did not select the triangle
Who among them chose the circle?

(a) A (b) B (c) C (d) D
Answer: d
A selected a shape which is neither beside B’s shape nor D’s shape.
A selected a shape which is next to only C’s shape
So, A can select a shape only at the extreme ends, because he can have only 1 shape (C’s
shape) next to his shape.
Hence, A can either choose the star or circle.
If A’s shape is a circle, then automatically, C’s shape is the triangle.
However, it is already given that C and D did not select the triangle.
Thus, A must have selected the star and C must have chosen the rhombus.
Hence, B’s shape is triangle (as D did not select the triangle) and D’s shape would be the
circle. Thus, the correct answer is option d.

Page 36

Chapter 4: Data Handling and Presentation
1. What will come in place of "?"

a) b) c) d)
Answer: d
Each group of blocks has letters positioned on top of them. When the blocks are considered in
descending order of their height, the letters follow the same order of size after the arrow.
On the right side of the arrow, the letters are positioned in a proper order of their size, where the letter
on the tallest block becomes the bigger outermost letter of the image (and the remaining letters follow
the same order inward)
In the 4th term, arranging the blocks in descending order results in the letter arrangement (in decreasing
size) U-V-S-T. So, the figure after the arrow must show U as the outermost letter, V inside it, then S, and
finally T placed further inside, which matches Option d.
Therefore, option d is the correct answer.
_____________________________________________________________________________________
2. What will come in place of "?"

a) b) c) d)
Answer: c

Page 37

In each pair, the number of lines on the right is equal to the number of letters in the word on the left.
For example, the word “ONE” has 3 letters. So, it is represented by 3 lines.
Following the same logic, “SEVEN” has 5 letters. So, it should be represented by 5 lines.
Among the given options, option c shows 5 lines. Therefore, option c is the correct answer.
_____________________________________________________________________________________
3. Sales of 4 milk stores are given below, in the form of a pictograph. If each symbol in the pictograph
represents ‘x’ litres, where x is a whole number and the sales made by any store is less than 100
litres, what is the MAXIMUM possible quantity of milk sold (in litres) by all the stores in total?

a) 255 L b) 272 L c) 306 L d) 283 L
Answer: b
From the given pictograph, we have sales made by each store as:
A: 3x
B: 4.5x
C: 6x
D: 3.5x
As the sales made by any store is less than 100 litres, and the highest sales shown in the pictograph is
6x (Store C), the largest possible value of 6x must also be less than 100.
The highest multiple of 6 that is less than 100 is 96 (as x is a whole number, 6x is also a whole number).
The maximum value of x can be 96/6 = 16
Hence, the maximum possible quantity of milk sold by all stores is 3x + 4.5x + 6x + 3.5x
= 17x
= 17 × 16 = 272 litres
Option b is the correct answer.
_____________________________________________________________________________________
4. The weights (in kg) of five friends, A, B, C, D, and E, are represented in the graph below (not
necessarily in the same order).
- E is lighter than C
- Both B and D are lighter than E, and the difference between the weights of B and E is the same
as the difference between the weights of D and E
- The difference between the weights of A and any other friend is 10 kg or less
What is the difference between the weights of B and C?

a) 10 kg b) 5 kg c) 20 kg d) 15 kg
Answer: c

Page 38

From the bar graph, the five weights are:
15, 15, 20, 25, 35
It is given that B and D are lighter than E and the difference between B and E is the same as the difference
between D and E.
This means that both B and D are of the same weight, and are less than E.
Hence, we can assign the bars of 15 kg to B and D, as only 15 kg are repeated.

Also, it is given that the difference between the weights of A and any other friend is 10 or less than 10
kg. A can either be 20 kg, 25 kg or 35 kg only.
35 - 20 gives a difference of 15 kg, which is more than 10 kg difference.
So, A cannot be either 35 kg or 20 kg. Therefore, A can only be 25 kg.

Now, as E is lighter than C, we get the following graph:

Therefore, the difference between the weights of B and C is 35 - 15 = 20 kg.
Option c is the correct answer.
_____________________________________________________________________________________
5. The bar graph shows the number of marbles collected by Monica, Rachel, Ross, Chandler, and
Joey.
- Rachel has collected more marbles than Joey and less than that of Chandler
- Monica has collected as much as Rachel and Ross collected together
How many marbles did Chandler collect?

a) 18 b) 14 c) 20 d) Cannot be determined
Answer: c

Page 39

From given graph, we have the count of marbles as: 18, 12, 30, 14, 20.
From the second instruction we know, Monica has collected as much as Rachel and Ross collected
together.
We get only one pair that adds up to the other number in the graph i.e., 18 + 12 = 30.
So, Monica has collected 30 marbles.
Which means marbles collected by Ross and Rachel are 18 and 12.
From the first instruction we know, Rachel has collected more marbles than Joey and less than that of
Chandler.
So, Rachel must have a value such that one remaining number is smaller than Joey and one is greater
than Chandler.
Among the remaining numbers 14 and 20, this is only possible if Rachel = 18, because 14 is less than
18 and 20 is greater than 18.
So, Joey = 14 and Chandler = 20. Therefore, Chandler collected 20 marbles.
Hence, option c is the correct answer.
_____________________________________________________________________________________
6. The pictograph represents students’ fruit preferences. Let X denote the number of students who
like Watermelon and Y denote those who like Muskmelon.
● X is half the total number of students who like Strawberry and Banana
● Y equals the total number of students who like Pear and Mango
Which fruit has a number of students equal to the difference between X and Y?

a) Strawberry b) Banana c) Pear d) Mango
Answer: a
Each symbol in the pictograph represents 2 students.
Step 1: Find the number of students for each fruit
● Strawberry:
6 symbols: 6 × 2 = 12 students
● Banana:
4 symbols: 4 × 2 = 8 students
● Pear:
4 full symbols and 1/2 symbol
4 × 2 = 8 students
1/2 symbol = 1 student
Total = 9 students
● Mango:
6 full symbols and 1/2 symbol
6 × 2 = 12 students
1/2 symbol = 1 student
Total = 13 students

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Step 2: Find X (Watermelon)
X = (Strawberry + Banana) ÷ 2
X = (12 + 8) ÷ 2
X = 20 ÷ 2
X = 10 (represented by 5 symbols, as each symbol = 2 students)
Step 3: Find Y (Muskmelon)
Y = Pear + Mango
Y = 9 + 13
Y = 22 (represented by 11 symbols, as each symbol = 2 students)
Step 4: Find the difference between X and Y
Y − X = 22 − 10 = 12
Step 5: Match the difference with a fruit
12 students like Strawberry.
Hence, option a is the correct answer.
_____________________________________________________________________________________
7. A pictograph with some missing data is shown below. Extra symbols need to be added (but none
can be removed or shifted) to complete the pictograph, such that:
- The number of students increases by 300 for every gap of 4 years
- The number of students in each year is a multiple of 100
What is the MINIMUM number of symbols that must be added to satisfy all the above conditions?

a) 8.5 b) 8 c) 9 d) 7.5
Answer: c
Given:
● 1996 = 400
● 1998 = 550
● 2000 = 450
● 2002 = 600
● 2004 = 700
Each 4-year gap must be 300, and all final numbers must be multiples of 100. Only addition is allowed.
1996 is already a multiple of 100.
So, no new symbol is required here.
To make the difference between year 1996 and 2000 as 300,
We need to have the number of students in the year 2000 as 700. (since 400 + 300 = 700)
As we already have 450 in the year 2000, we need 250 more students. — (1)
So, 2000 will have 700 students.
● 1996 = 400
● 1998 = 550
● 2000 = 450 + 250
● 2002 = 600
● 2004 = 700

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1998 has 550 students. Here, as the number of students must be a multiple of 100, we need 50 more
students — (2)
So, 1998 will have 550 + 50 = 600 students.
● 1996 = 400
● 1998 = 550 + 50
● 2000 = 450 + 250
● 2002 = 600
● 2004 = 700
To have an increase of 300 from year 1998 to 2002, we need 600 + 300 = 900 students in 2002.
As 2002 already has 600 students, we need 900 - 600 = 300 more students in 2002
1998 becomes 600, and 2002 becomes 900.
● 1996 = 400
● 1998 = 550 + 50
● 2000 = 450 + 250
● 2002 = 600 + 300
● 2004 = 700
Finally, to make an increase of 300 from year 2000 to 2004, we need to have 300 more students in 2004
(since both have 700 students each)
● 1996 = 400
● 1998 = 550 + 50
● 2000 = 450 + 250
● 2002 = 600 + 300
● 2004 = 700 + 300
Students to be added:
50 + 250 + 300 + 300 = 900
Since 1 symbol = 100 students, 900 ÷ 100 = 9 symbols
Minimum symbols added = 9.
Hence, option c is the correct answer.
_____________________________________________________________________________________
8. What will come in place of "?"

a) b) c) d)
Answer: d
Here, the tile with the ball and its immediate neighbours (left and right) turn black; and every other tile
gets a bottle.
In the question term, the ball is on the 3rd tile, so tiles 2, 3, and 4 must be black and the rest must have
bottles. The diagram matching this pattern is option d, which is the correct answer.

Page 42

9. The bar graph shows the quiz scores of six students A, B, C, D, E, and F. The students are grouped
as follows:
● Group 1: Students whose scores are composite numbers
● Group 2: Students whose score forms a perfect square when added to any other student’s
score
● Group 3: Students whose score is greater than the average score of all six students
How many students belong to exactly two of these groups?

a) 2 b) 3 c) 4 d) 5
Answer: b
From the bar graph, the scores are:
9, 7, 12, 14, 10, 16
Group 1: Composite numbers
Composite numbers among the scores are: 9, 10, 12, 14, and 16
So, the students in group 1 are: A, C, D, E, and F.
Group 2: Scores that can form a perfect square when added to another score
● 9 + 7 = 16 (perfect square)
● 9 + 16 = 25 (perfect square)
The scores in this group are: 9, 7, and 16
So, the students in group 2 are: A, B, and F.
Group 3: Scores greater than the average
Average score = (9 + 7 + 12 + 14 + 10 + 16) ÷ 6
Average = 68 ÷ 6 ≈ 11.33
Scores greater than the average are: 12, 14, and 16
So, the students in group 3 are: C, D, and F.
Group membership count
● 9 (A) - Group 1 and Group 2 (2 groups)
● 7 (B) - Group 2 only (1 group)
● 12 (C) - Group 1 and Group 3 (2 groups)
● 14 (D) - Group 1 and Group 3 (2 groups)
● 10 (E) - Group 1 only (1 group)
● 16 (F) - Group 1, Group 2, and Group 3 (3 groups)
A, C, and D belong to exactly two groups. Hence, 3 students belong to exactly two groups.
Option b is the correct answer.

Page 43

10. Raj saved some amount of money every month from January to June.
● He saved a total of Rs. 30,000 in the first three months and Rs. 30,000 again in the last three
months as well
● He did not save any money in February
● The maximum total saved in any two consecutive months is Rs. 40,000
● While drawing the chart, Raj arranged the savings in increasing order without matching them
with the month names written below
What was the actual amount saved in June?

a) Rs. 22000 b) Rs. 9000
c) Rs. 3000 d) Cannot be determined
Answer: d
1. In the first 3 months, and the next 3 months, he saved Rs. 30,000 each time.
Which gives,
January + February + March = 30,000
April + May + June = 30,000
2. In February he didn’t save any money.
January + March = 30,000
Only two amounts adding up to 30,000 are 8000 and 22,000
3. The highest amount he saved altogether in two sequential months is 40,000.
The only amounts that add up to 40,000 are 18,000 and 22,000. Since 22,000 is in the first quarter, its
adjacent month must have 18,000.
January cannot have 22,000 because February’s savings are 0.
Therefore, Raj saved 22,000 in March and 18,000 in April.
From 2 and 3, we get, March = Rs. 22,000, April = Rs. 18,000, and January = Rs. 8000.
Hence, June can be either Rs. 3000 or Rs. 9000.
Therefore, option d is correct.

Page 44

The circles in the grid below can only contain numbers from 1 to 6.
- Every pair of connected circles contains consecutive numbers
- No two circles belonging to the same row or column can have the same number
- A number can be placed in more than one circle
What is the difference between A and B?

(a) 1 (b) 2 (c) 3 (d) 4
Answer: a
Given that the pairs of connected circles must have consecutive numbers, with no row
or column having the same number, twice.
A can be either 3 or 5, as it is connected to 4.
However, as 3 is already present in the same row, A will be 5.

Now, this empty circle can either have 2 or 4 in it, as it is connected to 3.

But, if it has 2, its next connected circle should either have 3 or 1.
However, the grid cannot be filled further, as the question condition cannot be satisfied
in any case, as shown.

Hence, the grid can be filled completely, only if we proceed with 4.
As shown below, A = 5 and B = 6, and the difference between A and B is 1.
Hence, option a is the correct answer.

Page 45

Chapter 5: Prime Time
Activity Time
Factors, Multiples, HCF and LCM
Introduction
Imagine using numbers and their relationships to create beautiful artworks. In this activity, we will take a
circle with 12 equally spaced points. By connecting these points using a fixed rule, we will create
beautiful shapes such as regular polygons and stars.

This activity helps students understand:
• Factors and multiples of a number
Activity Time Description

Launch 5 min Teacher demonstrate the activity

Supporting Links:
Activity Video Link:
https://youtu.be/L_PqVp8Ro2g?si=XRF5cFMvsXzSMYI
A&t=382

Trail by Students 15 min Students Tryout the activity with each other.

Student Worksheet:
https://docs.google.com/document/d/1TGjoEGiumTgstd
yNznQy8pUJ10izCgaAJg1hGtMwTRY/edit?usp=sharing

Discussions and 15 min Attempting the worksheets and the discussion based on
Explorations the activity

Page 46

CT Components
Algorithmic Thinking:
In this activity, students will follow a rule - Starting from a number on a circle with 12 equally spaced
points, add a step number (say n) to it and keep doing it until you reach back to the starting point. This
becomes the algorithm for every circle.

Pattern Recognition:
Once the different shapes have been formed for the circle with different step numbers, students will try
to recognise the different shapes that are formed. They will then try to see a pattern between the shape
formed, and the step numbers and points on the circle.

Generalisation:
After recognising the pattern of the shapes formed on the circle with different step numbers, students
will generalise the shapes formed for any circle with m equidistant points and n step numbers.
Activity
You are given a circle with 12 equidistant points.
You have to connect these points using a rule.
Rule:
1. Starting from 2, add 2 to it and connect it with the next number, i.e. 4.
2. Add 2 to 4, then connect it to the next.
3. Keep adding 2 until you reach the starting point.

1. How many steps did it take to reach the starting point? What shape do you get?
a) Three b) Four c) Five d) Six
Answer: d
We can see that the 6 steps formed equal chords on the circle, thus giving us a regular hexagon. Also, if
we add 2 to the last point again, we will just retrace the steps we took in the first sequence.

Competencies: Algorithmic Thinking

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Now starting from any point, add 3 to it and join to the next point. Keep adding 3 until you reach
the starting point.

2. How many steps would be needed to reach the starting point?
a) Two b) Three c) Four d) Five
Answer: c
Starting from 3 and adding 3 at every step, we proceed like this
3 → 6 → 9 → 12 → 3
After 4 steps, we return to 3.
Shape: Since the 4 steps formed equal chords on the circle, we get a square.
Note: If we add 3 to 3 again, we will just retrace the steps that we made in the first sequence.
Repeat the same process by
a) Starting from 4 and adding 4
b) Starting from 5 and adding 5
c) Starting from 6 and adding 6
Until you reach the starting point.

What shapes do you get? Record your answers in the table below.

Number added No. of steps till the starting Shape formed
point

2 6 Hexagon

3 4 Square

4

5

6

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Note: When the starting point is 5, adding 5 to it gives 10. Now, adding 5 to 10 gives 15, which is not on
the circle. But since the numbers lie on a circle, we can keep going around it. So 15 will go to 3 (12 + 3
= 15). Now add 5 to 3, and we get 8. Keep going in the same manner and we will get this sequence of
steps:
5→ 10→3→8→1→6→11→4→9→2→7→12→5
So, after 12 steps, we will reach 5 again.

Do you notice any patterns in the numbers we are adding, the number of points on the circle, and
the shape being formed in the circle?

Why did you get polygons on adding 2, 3 and 4, while a star on adding 5? Is there any relationship
between these numbers and 12, the number of points on the circle?

Competencies: Algorithmic Thinking, Pattern Recognition
Explanation: The numbers 2, 3, and 4 are factors of 12, while 5 is not. To reach 12 - 2, 3, and 4 take 6,4
and 3 steps respectively and reach the starting point. But when we start from 5, we have to keep on going
until we reach a number that is a common multiple of both 5 and 12, so that the starting point is the same
as the ending point. So 5 takes 12 steps, adding to 60, which is a common multiple of both 5 and 12 and
in the process, it covers all points, giving us a 12-pointed star.

Explorations
1. What shapes will you get when you add 7, 8, 9, 10 and 11 until you reach the starting point?
Try to find out the relationship between the numbers and the number of points on a circle.

Competencies: Algorithmic Thinking, Pattern Recognition, Generalisation

Page 49

Explanation - We will observe that 7, 8, 9, 10, and 11 will have the same shape as in the cases of 5, 4,
3, 2, and 1, respectively. And if you look at these numbers on the circle, they are the reflections of each
other. In case 2, if I mark my lines in order in different colours and then repeat the procedure for 10, we
will be able to see that the hexagons obtained are reflections of each other.

In the first image, we start from 2, the blue string connects to 4. Then 4 connects to 6 through a yellow
string, and so on.
In the second image, we start from 10, add 10 to get 20, which will go to 8 (12 + 8), so the first string is
blue from 10 → 8. Then 8 plus 10 is 18, which goes to 6 using a yellow string and so on.
So, the numbers that are reflections will have the same shapes.
We can also generalise - If there are m points on a circle, and we know the shapes formed by step
number n, then m-n will also have the same shape.
_____________________________________________________________________________________
2. Two circles, one with 16 points and the other with 20 points, are given. Which of the two will give
a square when added to which number?

a) The 16-point circle, on adding 4
b) The 16-point circle, on adding 12
c) The 20-point circle, on adding 5
d) The 20-point circle, on adding 15

Competencies: Pattern Recognition, Decomposition
Explanation: All 4 options are correct.

Page 50

For 16 point circle, to get a square, we need to take 4 steps. So, 16 divided by 4 is 4, hence adding 4
gives us a square. Also, 16-4 = 12 also gives a square.
We can similarly find for 20 point circle.

For further exploration on the teachers’ part: Try to explore the same exercise with a 20-point circle
and different step numbers. Try to see what happens when step numbers are not proper factors, but
have a common factor with the number of points on the circle.

Page 51

Questions
1. Find the ODD one out.

a) b) c) d)
Answer: b
In each term, the numbers written in white blocks are multiplied to give their product, which is written in
the black shaded block.
The only option which does not follow this logic is option b. Thus, option b is the correct answer.
_____________________________________________________________________________________
2. If "X" is a number formed by the addition of two or more DIFFERENT single-digit prime numbers,
which of the following options CANNOT be a value of the number "X"?
a) 8 b) 10 c) 11 d) 15
Answer: c
The single-digit prime numbers are:
2, 3, 5, and 7.
The number X must be formed by adding two or more different prime numbers from this list.
Let us check each option one by one:
Option a: 8 can be the sum: 3 + 5 = 8.
Option b: 10 can be the sum: 3 + 7 = 10
Option c: 11 cannot be the sum of any two or more single digit prime numbers, as 3 + 7 results in 10 (1
less than 11) and 5 + 7 results in 12 (1 more than 11). So, 11 cannot be the sum in any possible way.
Option d: 15 can be the sum: 3 + 5 + 7 = 15. Hence, the correct answer is option c.
_____________________________________________________________________________________
3. The sum of three consecutive natural numbers is equal to Y. Which of the following numbers will
Y always be divisible by?
a) 3 b) 2 c) 4 d) Cannot be determined
Answer: a
Since the numbers are three consecutive natural numbers, the result must hold true for any such set.
Check a few examples:
● 1 + 2 + 3 = 6, divisible by 3 (also by 2, but not by 4)
● 2 + 3 + 4 = 9, divisible by 3 (not by 2, or by 4)
● 6 + 7 + 8 = 21, divisible by 3 (not by 2, or by 4)
So, divisibility by 2 and 4 is not consistent, but divisibility by 3 appears every time. This happens
because among any three consecutive numbers, one number is always a multiple of 3.
The other two numbers are 1 more and 1 less than that multiple, and together they also add up to a
multiple of 3. Therefore, the sum of any three consecutive natural numbers is always divisible by 3.
Hence, option a is the correct answer.
_____________________________________________________________________________________
4. A two-digit number is 1 more than a multiple of 3. How many different possible values can it have?
a) 29 b) 30 c) 31 d) 32
Answer: b

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A two-digit number which is one more than a multiple of 3 can be shown as 3x + 1.
The largest two-digit multiple of 3 is 99, and adding 1 to it would result in 100, which is a 3-digit number,
which breaks the condition.
So, the maximum value 3x + 1 can be is 97:
3x + 1 = 97
3x = 97 - 1
3x = 96
x = 32.
Also, let’s check the least value x can have:
For x = 1:
3x + 1 = 3(1) + 1 = 4 (since it is a single digit, x cannot be 1)
For x = 2:
3x + 1 = 3(2) + 1 =7 (since it is a single digit, x cannot be 2)
Now, the values which x can have for 3x+1 to be a two-digit number could be anywhere from 3 to 32.
1 to 32 gives 32 values for x and as x cannot be 1 or 2, excluding two values, we get 32 - 2 = 30.
Hence, x has 30 values, and the correct answer is option b.
_____________________________________________________________________________________
5. The product of a number 'PQ' and 7 is 'RSQ', where P, Q, R, and S are distinct digits. Which of the
following CANNOT be the possible value of 'RSQ'?

a) 210 b) 315 c) 420 d) 105
Answer: d
As per the question, when you multiply a number by 7, the last digit of the product is the same as the last
digit of the original number (the problem says both end with Q).
Let’s check each of the options, one by one:
210 ends with 0. So, Q = 0. If the product is 210 then the original two-digit number must be 30. That gives
digits 3, 0, 2, and 1, which are all different. Thus, possible.
315 ends with 5. So, Q = 5. If the product is 315 the original two-digit number must be 45. That yields
digits 4, 5, 3, and 1, which are all different. Thus, possible.
420 ends with 0. So, Q = 0. Product 420 comes from 60: digits 6,0,4, and 2, which are all different. Thus,
possible.
105 ends with 5. So, Q = 5. Product 105 would come from 15, that means the original two-digit number
starts with 1 and the product’s hundreds digit is also 1. So, the hundreds digit (R) and the original tens
digit (P) would both be 1 - they’re not distinct, which breaks the problem condition.
105 cannot be the product because it forces two digits to be the same.
Hence, the correct answer is option d.
_____________________________________________________________________________________
6. Nausheen thinks of 5 consecutive numbers. 3 of these numbers are prime and 3 of these are even
numbers. What is the sum of these numbers?
a) 10 b) 15 c) 20 d) Cannot be determined
Answer: c
In the question, it is mentioned that Nausheen thinks of three prime numbers and 3 even numbers.
So, there are 3 + 3 = 6 numbers.

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But, Nausheen actually thinks of only 5 consecutive numbers and not 6.
So, one of the numbers that he thinks will be an even number, as well as a prime number.
We know that the only even prime number is 2.
So, to have 3 prime numbers, the sequence must include 2.
Also, in the set of any 5 consecutive numbers, there can be 3 even numbers only if the sequence starts
with an even number. So, the set starts with 2. (as it cannot start with 1, which is odd)
Therefore, the numbers are:
2, 3, 4, 5, 6
Check:
Even numbers: 2, 4, 6 (3)
Prime numbers: 2, 3, 5 (3)
Sum = 2 + 3 + 4 + 5 + 6 = 20
Hence, the correct answer is option c.
_____________________________________________________________________________________
7. In a certain language, if 3 is coded as ‘free’, 5 is coded as ‘dive’, 10 is coded as ‘hen’, then what
could be the possible code for the PRIME FACTOR of 169, in the same language?
a) heighten b) throwing c) routine d) titan
Answer: c
Observe the given codes carefully:
● 3: free
● 5: dive
● 10: hen
Notice that each code word rhymes with the number name:
● 3: three - free
● 5: five - dive
● 10: ten - hen
So, the pattern is:
The code is a word that rhymes with the number.
Now, find the prime factor of 169.
169 = 13 × 13
So, the prime factor is 13.
The number name is thirteen.
Among the options, the word that rhymes with thirteen is routine.
Hence, the correct answer is option c.
_____________________________________________________________________________________
8. If "X" is a number formed by the multiplication of 2 single-digit prime numbers, which of the
following options is ALWAYS TRUE about the number "X"?
a) The number "X" cannot be even b) The number "X" is a 2-digit number
c) The number "X" is a multiple of 4 d) The number "X" is less than 50
Answer: d
Option a: 2 x 3 = 6. Hence, option a is not always true.
Option b: 2 x 3 = 6 is a single digit number. Hence, option b is not always true.
Option c: 3 x 5 = 15 which is not a multiple of 4. Hence, option c is not always true.
Option d: If we take the two largest single digit prime numbers (5 and 7) and multiply them, we get 35
which is less than 50. Hence, option d is always TRUE.
Thus, the correct answer is option d.

Page 54

9. In the letter grid below, the spelling of which prime number is immediately followed by the spelling
of its succeeding number in the number series?
ONEFIIVESIXTWOTHREENINETENELEVNTWELVEONEFOURSIXSEVEN
a) 3 b) 5 c) 11 d) 2
Answer: d
Step 1: Identify the prime numbers in the grid:

Step 2: Among the prime numbers found, identify the number which is immediately followed by its
succeeding number (next number of the original number series).

As shown, only TWO is the prime number, which is immediately followed by the spelling of its
succeeding number (THREE) in the number series.

Hence, option d is the correct answer.
_____________________________________________________________________________________
10. Alex, Jim, and Sam were born on prime-numbered dates in the same month and year. Sam was
born on a Friday, and the month started on a Sunday.
- The difference between the dates of Alex’s and Jim’s birthdays is 20
- Jim was born 10 days after Sam
On which day of the week was Alex born?
a) Thursday b) Wednesday c) Tuesday d) Saturday
Answer: c
The prime-numbered dates in a month are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (and 31, if applicable).
The month started on a Sunday. Therefore, the first Friday falls 5 days later:
Sunday + 5 days = Friday
So, the date of the first Friday is:
1+5=6
Adding 7 to get the remaining Fridays, the Friday dates are:
6, 13, 20, 27
Among these, only 13 is a prime number.
Hence, Sam was born on the 13th of the month.
Jim was born 10 days after Sam, so Jim’s birth date is:
13 + 10 = 23rd
The difference between the birth dates of Alex and Jim is 20 days. So, Alex’s birth date is:
23 − 20 = 3rd
All these dates are prime numbers. So, all conditions are satisfied.
Since the month started on Sunday:
• 2nd falls on Monday
• 3rd falls on Tuesday
Therefore, Alex was born on a Tuesday. Hence, option c is the correct answer.

Page 55

The image given below contains ONLY white and grey squares. You can change the
colour of a square by clicking on it.
- If you click a white square, the colour of that square and all squares that share a
common corner (but do not share a common side) with it will change
- If you click a grey square, its colour and the colour of the squares that share a
common side with it will change
What is the minimum number of squares that must be clicked to make the number of
grey and white squares equal?

(a) 1 (b) 2 (c) 3 (d) 4
Answer: a
Let's approach the question logically to achieve an equal number of grey and white
squares in a grid with a total of 16 squares. Currently, there are 11 white squares and 5
grey squares.
To balance the count, we aim for 8 squares of each colour. Since we have more white
squares, we need to increase the number of grey squares by 3.
By observing the grid, we need to identify a white square to click, as clicking it will not only
change its colour but also the colours of squares that share a common corner (without
sharing a common side) with it.
To achieve the desired change of only 3 white squares to grey, we look for a white square
that shares a common corner with two other white squares and one grey square. In the
grid, such a square is identified and marked. By clicking this square, we can achieve an
equal number of white and grey squares.
Therefore, strategically clicking the one highlighted white square will help us reach
the goal of having 8 grey and 8 white squares in the grid.
Hence, the correct answer is option a.

Page 56

Chapter 6: Perimeter and Area
1. Figures A and B represent squares which are divided into different parts, as shown in the image
below. Which of the following options correctly represents the relation between the division?

a) b)

c) d)
Answer: c
Given that the side of each square = 4 cm
Therefore, Area of the square = 4 × 4 = 16 sq. cm
Area of one small square = 16/4 = 4 sq. cm
Red and Blue triangles = Half of the area of square
Thus, Red and Blue triangles = 1/2 of 16 = 8 sq. cm
Area of any smaller coloured triangle = 1/2 of 8 = 4 sq. cm
Area of bigger white triangle = 1/2 × 4 × 4 = 8 sq. cm
Option a:
Squares: 4 + 4 = 8
Triangle: 4
According to option a, 8 < 4.
8 is not less than 4. So, option a is incorrect.
Option b:
Squares: 4 + 4 + 4 = 12
Triangles: 8
12 < 8
According to option b, 12 < 8.
12 is not less than 8. So, option b is incorrect.
Option c:
Squares: 4 + 4 + 4 = 12
Triangles: 8 + 4 = 12
12 = 12
Areas match perfectly. So, option c is correct.

Page 57

Option d:
Square: 4
Triangles: 8
4≠8
According to option d, 4 = 8.
So, option d is incorrect.
Therefore, the correct answer is option c.
_____________________________________________________________________________________
2. The perimeter of a square X is 16 cm. The square is cut along its diagonal to form two identical
triangles.
Which of these statements is DEFINITELY FALSE based on the above information?
a) Two sides of the triangle meet each other perpendicularly
b) The length of at least one side of each triangle is 4 cm
c) The combined area of the triangles is 16 cm2
d) The combined perimeter of the triangles is 16 cm
Answer: d
Perimeter of a square = 16 cm = 4 x Side length.
So, side length = 16 cm / 4 = 4 cm.

When the square is cut into two triangles, then the two triangles formed are right angled triangles.
So, two sides of the triangle meet each other perpendicularly.
Option a is true.
Also, the length of at least one side of each triangle is 4 cm as shown in the image above.
Option b is true.
As the side of the square = 4cm, the area of the square = 4 x 4 = 16 cm 2.
The combined area of both the triangles is also 16 cm2.
Option c is true.
Now, the combined perimeter of the triangles will be more than 16 cm as now we have 1 new side in
each triangle, as shown below:

Therefore, option d is false.
The correct answer is option d.

Page 58

3. A larger block is formed using 4 small cubes (2 grey cubes and 2 white cubes), as shown below.
How many faces of the larger block have equal grey and white areas?

a) 4 b) 5 c) 3 d) 2
Answer: a
Given: A large block is made of 4 small cubes - 2 grey (bottom) and 2 white (top).
Let's analyse each face:

Front View (as shown in the image):
•Top = 2 white cubes
•Bottom = 2 grey cubes
Equal grey and white - Yes!

Back View
•Since all cubes are the same front-to-back, the back will look exactly like the front.
Equal grey and white - Yes!

Left Side View
•On the left face, we’ll see the left half of the structure.
•That will be: 1 white cube on top, 1 grey cube on bottom
Equal grey and white - Yes!

Right Side View
•Same logic - we see the right half
•1 white on top, 1 grey on bottom
Equal grey and white - Yes!

Top View
•Topmost cubes are both white
Only white visible - Not equal

Bottom View
•Bottommost cubes are both grey
Only grey visible - Not equal

Final Answer:
•Equal grey and white faces: Front, Back, Left, Right = 4 faces

Hence, the correct answer is option a.

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