CTAI Teacher Handbook Class 7

Computational
Thinking and
Artificial Intelligence

Class 7

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.

The book engages learners with problems involving multi-layered constraints,
conditional dependencies, optimisation strategies, data interpretation and structured
decision-making across numerical, spatial and real-life contexts. It deepens Artificial
Intelligence understanding through concepts such as rule-based classification, data
representation, bias and fairness and decision-making systems, enabling students to
analyse how data, rules and assumptions influence intelligent behaviour. The
document also provides pedagogical guidance, learning resources, assessment
support and classroom implementation guidelines to facilitate competency-based
learning in alignment with NEP 2020.

TEAM CBSE

Page 2

ACKNOWLEDGEMENTS
Expert Committee

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

3. Large Numbers Around Us 12

4. Arithmetic Expressions 20

5. A Peek Beyond the Point 28

6. Expressions using Letter-Numbers 40

7. Parallel and Intersecting Lines 47

8. Number Play 56

9. A Tale of Three Intersecting Lines 73

10. Working with Fractions 80

PART-2 ARTIFICIAL INTELLIGENCE
SR NO CHAPTER TITLE ETHICAL PAGE
AWARENESS NO

1 AI Domains and Applications Understanding 89
Learning Focus: responsible AI use
AI domains, Predictive techniques, Small
datasets, Computer Vision, Natural
Language Processing, Data Science
2 AI in Industries Benefits and 91
Learning Focus: responsible
AI in healthcare, Education, Transport, application of AI
Communication, Benefits of AI
3 Data Visualization and Analysis Accurate and honest 93
Learning Focus: data representation
Collecting and organising data, Data
visualization, Interpreting data
4 Ethics and AI Bias Awareness Fair and responsible 96
Learning Focus: use of AI
Ethics in AI, Bias in AI, Responsible and fair
use, Digital citizenship

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 focussing on
Computational Thinking Skills. The AI-readiness, so inculcated through CT Skills, will help develop
the capacities of learners to use computational thinking, such as logical thinking, problem solving,
pattern recognition, and so on, and understand the role and use of Artificial Intelligence in daily life.
The Curriculum aims to build strong foundations in computational thinking, digital literacy, and
responsible use of technology, along with nurturing innovation, critical thinking, and ethical decision-
making capacities.

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1. Relevance: Importance of introducing Computational Thinking (CT) 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, ethical decision-making abilities, leading
to individual flourishing. It leads to creation of responsible digital citizens in society.
• Interdisciplinary Relevance: Embedding CT and AI concepts in the school curriculum helps
students to develop an integrated view of the world by connecting various disciplines such as
Mathematics, Science, Humanities etc.
• Innovation and Entrepreneurship: At the core CT and AI is about solving problems, devising
innovative solutions and recreating human thinking. This leads to an entrepreneurial and
innovative mindset in the learners.
• Ethical Awareness: Study of CT & AI will sensitise 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: 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 complex, multi-layered problems by:
• Visualising and analysing 3D objects and their transformations, including rotations, reflections,
cross-sections, and nets.
• Understanding compound transformations involving multiple flips, turns, folds, and
rearrangements.
• Identifying hidden relationships and constraints within incomplete figures, patterns, or logical
setups.
• Analysing symmetry, congruence, and proportional reasoning across different representations.
• Interpreting relative positions, orientations, and viewpoints of objects in advanced visual
scenarios.

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PATTERN RECOGNITION
Students will be able to recognise, extend, and predict complex patterns involving:
• Multi-rule numerical sequences, including alternating, nested, and dependent patterns
• Algebraic patterns using variables, expressions, and functional relationships
• Visual and geometric patterns formed through transformations or growth rules
• Letter and symbol-based patterns involving positional and logical dependencies
• Integrated patterns combining numbers, shapes, symbols, and logical conditions
DECOMPOSITION
Students will be able to break down real-world and abstract problems by:
• Separating interconnected conditions and constraints into manageable components
• Analysing number properties (factors, multiples, ratios, percentages, powers) within layered clues
• Deconstructing problems involving spatial reasoning, measurements, and geometry
• Interpreting tables, grids, charts, and flow-based information with multiple dependencies
• Breaking multi-step logical situations (movement, exchanges, comparisons, scheduling) into
ordered steps
• Translating visual or verbal information into structured data for systematic analysis
ALGORITHMIC THINKING
Students will be able to design and follow logical procedures to solve advanced problems
involving:
• Rule-based sequences and algorithms with conditional branching
• Grid-based navigation and pathfinding with constraints and decision points
• Step-wise transformations involving calculations, swaps, transfers, or positional changes
• Ordering and arranging elements (people, objects, events) using multiple attributes and logical
clues
• Solving problems using if–then reasoning, elimination strategies, and logical consistency checks
• Creating or analysing procedural steps to reach an optimal or valid solution

Artificial Intelligence (AI) Learning Outcomes
Learners will be able to:
• Distinguish key predictive techniques such as:
- Regression: The method of predicting a number based on patterns in past data
- Classification: The process by which a machine arranges things in a group based on what it
has learned
- Clustering: The process by which a system automatically puts similar items together
• Explain about the key domains of AI, namely:
- Data Science: learn to manage and extract insights from data
- Computer Vision: learn the basics of how machines understand and respond to visual
information
- Natural Language Processing: understand the basics and limitations of how computers
process and handle natural language inputs
• Explain what bias in AI means, and identify situations where AI can give unfair results
• Demonstrate courteous, safe, and responsible use of technology as part of good digital citizenship

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• Use safe practices for maintaining data privacy, including giving informed consent before personal
data is collected, used, shared, archived, or deleted
• Collect and organise simple structured data, interpreting patterns and trends, and create bar
charts, line graphs and pie charts
• Apply basic predictive approaches/techniques to a small dataset
• Explain uses of AI in healthcare, education, transport, and communication

4. Mapped with NEP and NCF 2023
• 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
• NCFSE 23 carries this recommendation further for implementation
• Learning standards are derived using the approach suggested with and are aligned to NCF SE-
2023
• Curricular Goals are derived from the Aims of Education, along with other relevant considerations

5. Time Allocation
The Middle stage (Class 6–8) suggests 100 hours annually, including the time for specific topics on
basics of AI and interdisciplinary projects. For Grade 7, the breakdown is:
• Advanced CT skills: 40 hrs. per academic year
• Introductory concepts of AI: 20 hrs. in an academic year
• Interdisciplinary projects: 40 hrs. total (20 hrs. allocated to each of the two required projects)

6. Approach / Pedagogy
• Use hands-on and real-world problems, collaborative and group work to solidify and apply
multidisciplinary foundational knowledge on Coding, Data Analysis and Artificial Intelligence tools
• Use complex puzzles, riddles and games to build on the computational thinking abilities taught
in the previous stage
• Deliver fundamental concepts of AI through explanations, demonstrations and hands -on
experience
• Organise group discussions, design collaborative projects that integrate CT & AI, and offer
guidance to students to carry out these projects
• Independent activities for students such as data collection, organisation and analysis created
using digital tools or manually
• Discussions, debates and case studies on ethical use of AI

7. Assessment
Assessment approaches move away from traditional summative assessment to continuous,
formative, and competency-based assessment. Methods for Class 7 include:
• Written Tests and Practical Examinations
• Project presentations, assignments, and reflective journals

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• Interactive Group activities like treasure hunts
• Teacher Observation Journal
• Thematic Projects and Reflections/Group Discussions

CT and AI Transition
• Computational Thinking (CT) forms the basis of learning AI.
• Skills like breaking problems into parts, spotting patterns, filtering essential information, and
designing step-wise procedures are the same reasoning processes that power AI and ML
systems.
• The curriculum begins with the introduction of CT and deepens it as we move across the stages;
AI is introduced later, once pre-requisite knowledge of CT is built for understanding AI.

Page 9

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 10

PART-1

COMPUTATIONAL THINKING

Page 11

Chapter 1: Large Numbers Around Us
1. ‘X’ is a 6-digit number formed using exactly three different digits. One of the digits appears once,
another appears twice, and the third appears three times. When 2 lakhs are added to ‘X’, the
resulting number ‘Y’ is still a 6-digit number. What is the highest possible value of ‘Y’?
a) 977889 b) 999999 c) 999988 d) 999887
Answer: c
Solution:
Given that X is a 6-digit number, where one digit occurs once, one digit occurs twice, and the other digit
occurs thrice.
Now, when 200000 is added to X, we get Y, which is also a 6-digit number.
To maximise the value of Y, we have to maximise the value of X.
Let X = __ __ __ __ __ __.
The first blank cannot have 8 or more than 8, as adding 200000 + 800000 gives 1000000, which is the
smallest 7-digit number. So, X must start with 7.
X = 7 __ __ __ __ __.
Also, we know that X is made of 3 different digits, where 1 digit occurs once, one digit occurs twice, and
the other digit occurs thrice.
Again, to maximise its value, the largest digit 9 must occur thrice, and the next largest digit 8 must occur
twice.
Hence, the HIGHEST possible value of X = 799988.
Therefore, the HIGHEST possible value of Y is 799988 + 200000 = 999988.
Option c is correct.
_____________________________________________________________________________________
2. What is the product of all the numbers on a telephone dial pad?
a) 12345 b) 32451 c) 362880 d) None of these
Answer: d
Solution:
A telephone dial pad contains the numbers 0 to 9.
When finding the product, all numbers are multiplied together.
0 is included in the dial pad.
We know that any natural number multiplied by 0 always results in 0.
So, whatever the product of the other digits be, the total product of the numbers on the dial -pad becomes
0.
As 0 is not present in the options, option d, ‘None of these’ is correct.
_____________________________________________________________________________________
3. Sam forms two different 6-digit numbers, X and Y, using digits from 0 to 9 (without repetition
within a number). Exactly two digits are common between X and Y.
● The digit 9 appears in the same position in both numbers
● Y ends with 0, has only one even digit, and its digits are arranged in descending order
● 6 is present in X, and the number of even digits on its left is equal to the number of even digits
on its right
If X is the largest possible number, what is the difference between X and Y?
a) 9310 b) 9210 c) 8420 d) 8310
Answer: a

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Solution:
9 appears in the same position in both numbers.
Number Y ends with 0 and has only one even digit. So, the other digits are odd (as 0 is even) and
arranged in descending order.
The odd digits in descending order are: 9, 7, 5, 3, 1.
So, Y = 975310.
In number X, the digit 9 will also be at the first position from the left (same as in Y).
So, X = 9 _ _ _ _ _.
In number X, the digit 6 has an equal number of even digits on both sides, and X must be the largest
possible number.
The even digits are: 0, 2, 4, 6, 8.
If 6 has only one even digit on each side, then number X will contain three even digits, and the
remaining digits will be odd.
Since number Y contains all the odd digits and zero, three odd digits would be common between X and
Y. However, only two common digits are allowed.
Therefore, 6 cannot have only one even digit on each side; it must have more than one.
If 6 has two even digits on both sides, then it must be at the hundreds position.
So, X = 9 _ _ 6 _ _.
The remaining even digits are: 8, 4, 2, 0. Arranging them to get the largest possible value of X, we get
984620.
Now, find the difference:
984620 − 975310 = 9310
Therefore, the difference between the two numbers is 9310. Hence, the correct answer is option a.
_____________________________________________________________________________________
4. If each of the rows follows the same theme, what will come in place of “?”

a) b) c) d)
Answer: c
Solution:
In each term, two six-digit numbers are formed by replacing the grey circle with a digit. The maximum
possible value of the grey circle is given on the right. The comparison remains valid as long as the digit
in the grey circle is less than or equal to the digit given on the right. For example, in the first term, the
comparison holds when the grey circle is replaced with a digit less than or equal to 7.
178727 > 177827.
The condition is not valid for any number greater than 7.
In the question term, the numbers are: _ _ 506 _ < _ 5 _ 60 _.
Here, the comparison is valid as long as the grey circle is replaced with a digit less than or equal to 5.
555065 < 555605.
The condition is invalid for any other digit greater than 5. (For example, if the grey circle is
replaced with 6, then 665066 < 656606 is invalid)
Hence, the correct answer is option c.

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5. What will come in place of “?”

a) 0 b) 00 c) 000 d) 0000
Answer: c
Solution:
Before the arrow, the spelling of a number is written, and a few letters from that spelling are shown
after “-”.
Rule:
● The number of letters written after “-” tells how many zeros are to be removed from that number.
● The remaining zeros are shown on the right side of the arrow.
Understanding the example
Crore = 1,00,00,000 : has 7 zeros
Letters after “-” are “re”: 2 letters
Remaining zeros:
7 - 2 = 5: 00000
Applying the rule to the question
Lakh = 1,00,000 : has 5 zeros
Letters after “-” are “kh”: 2 letters
Remaining zeros:
5 - 2 = 3: 000
Hence, option c is the correct answer.
_____________________________________________________________________________________
6. Six number tokens are given. Rearrange them to form a 6-digit number such that:
● The difference between the first and last digits is as small as possible
● The hundreds digit is double the thousands digit
● No consecutive digits appear next to each other
● All given tokens must be used exactly once
How many different 6-digit numbers between 2 lakhs and 8 lakhs can be formed under
these conditions?

a) 1 b) 2 c) 3 d) More than 3
Answer: b
Solution:
As per the second condition, the hundreds digit is double the thousands digit.
So, there are only three possible cases, as per the available tokens.
Case A: __ __ 1 2 __ __
Case B: __ __ 2 4 __ __

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Case C: __ __ 4 8 __ __
However, the third condition states that no consecutive digits of the number series appear next to each
other. So, case A is invalid, as 1 and 2 (consecutive digits) are next to each other. Remaining cases:
Case B: __ __ 2 4 __ __
Case C: __ __ 4 8 __ __
Now, as per the first condition, the first and the last digits of the number have the least possible difference
between them.
The least possible difference between any two tokens is 1.
From the above tokens, the pairs that have a difference of 1 are: (1, 2), (4, 5), and (7, 8).
As all tokens must appear once in a number, let’s fill in cases B and C with the appropriate first and last
digits, where the digits are not repeated.
Case B: As the middle digits of case B are 2 and 4, the only possible pair for the first and last digits is (7,
8)
a) 7 __ 2 4 __ 8
b) 8 __ 2 4 __ 7
Case C: As the middle digits of case C are 4 and 8, the only possible pair for the first and last digits is (1,
2)
a) 1 __ 4 8 __ 2
b) 2 __ 4 8 __ 1
Fill in the remaining blanks in such a way that no two consecutive digits of the number series appear next
to each other.
Case B:
a) 7 5 2 4 1 8
b) 8 5 2 4 1 7
Case C:
a) 1 7 4 8 5 2
b) 2 7 4 8 5 1
Among all the numbers, the numbers that lie between 2 lakhs and 8 lakhs are 752418 and 274851.
Hence, option b is correct.
_____________________________________________________________________________________
7. Two five-digit numbers are formed using different digits from 0 to 9. Both numbers have some
digits missing, as shown below:
Number 1: _ 1 _ 9 _
Number 2: _ 4 _ 6 _
No two consecutive digits of the number series are present in the same number.
What is the maximum possible difference of the numbers formed?
a) 47525 b) 52663 c) 52429 d) 46733
Answer: b
Solution:
The digits not used in either number are 0, 2, 3, 5, 7, and 8.
No two consecutive digits are present in the same number.
Number 1 already contains 1 and 9. So, it can use 3, 5, and 7 only. (as 0, 2, and 8 are not allowed)
Number 2 already contains 4 and 6. So, it can use 0, 2, and 8 only. (as 3, 5, and 7 are not allowed)
To obtain the maximum difference, one number must be maximized and the other minimized.
We have to logically decide which number has a greater scope to be maximized and which can be
minimized.

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Clearly, we can see that the largest single digit ‘9’ is already present in the tens place of number 1. So,
the largest digit doesn’t play any role in our current task.
Let’s go to the next largest digit ‘8’.
We know that 8 belongs to number 2 and the largest place value of this number (the ten thousands) is
also empty. So, placing 8 in the ten thousands place will significantly increase the overall value of
number 2. (Hence, we get the maximum difference by having number 2 as large as possible and
number 1 as small as possible).
Number 1: _ 1 _ 9 _
Number 2: 8 4 _ 6 _
Let’s place the other digits (0 and 2) in number 2, accordingly.
Number 1: _ 1 _ 9 _
Number 2: 8 4 2 6 0
Similarly, to minimize number 1, we have to fill in the missing digits (3, 5, and 7) in ascending order.
Number 1: 3 1 5 9 7
Number 2: 8 4 2 6 0
The difference between both the numbers is: 84260 - 31597 = 52663.
Hence, the correct answer is option b.
_____________________________________________________________________________________
8. A, B, C, D, and E each have a different number card among one hundred, one thousand, ten
thousand, one lakh, and ten lakhs.
- C has an even number of zeros in his number
- A’s number has 2 zeros less than C’s number
- D’s number has 4 zeros more than A’s number
- The number of zeros in B is more than that of A but less than C
Which number card does E have?
a) One thousand b) Ten thousand c) One lakh d) Ten lakh
Answer: c
Solution:
Since A has 2 fewer zeros than C, and D has 4 more zeros than A, C must lie between A and D, and
must have an even number of zeros.
Therefore, C has ten thousand, A has one hundred, and D has ten lakh.
The number of zeros in B is more than that of A but less than C.
B must fall between A and C. So, B has a thousand.
The only card left for E is one lakh. So, the correct answer is option c.
_____________________________________________________________________________________
9. 4 friends, Alex, Bob, Calvin and David have a different profession – Doctor, Lawyer, Cricketer,
and Engineer and earn a different salary – ₹40,000, ₹50,000, ₹60,000 and ₹70,000 per year. (not
necessarily in the same order)
- The cricketer earns the highest
- Alex earns more than Bob, while the doctor earns more than David, who is the engineer
- Bob is the lawyer, and neither Bob nor David has an exact salary of ₹50,000
Which of these options shows the correct profession and salary of Calvin?
a) Cricketer - ₹70,000 b) Doctor - ₹60,000 c) Doctor - ₹50,000 d) Doctor - ₹40,000
Answer: c
Solution:
The 4 friends are Alex, Bob (lawyer), Calvin, and David (engineer). This leaves us with two professions,
doctor and cricketer (for Calvin and Alex), which have not been fixed. It is mentioned that the Cricketer
earns the highest.

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Since the lawyer (Bob) and the engineer (David) do not earn ₹50,000 (as given in the statement, neither
Bob nor David has an exact salary of ₹50,000), it must be the doctor who earns ₹50,000.

As “the doctor earns more than David, the engineer”, David must earn ₹40,000, as the doctor earns
₹50,000. The only salary remaining for the lawyer is ₹60,000.

Now, since Alex earns more than Bob, Alex must be the cricketer (₹70,000), and Calvin must be the
doctor (₹50,000).
Therefore, Calvin’s profession is a Doctor, with a ₹50,000 salary.

Option c is correct.
_____________________________________________________________________________________
10. The following are the clues to the 6-digit password of Nisha’s laptop:
35 tens
31 thousands
25 ones
423 hundreds
14 ten thousands
What is Nisha’s password?
a) 213625 b) 203675 c) 212375 d) 213675
Answer: d
Solution:
To find Nisha’s laptop password, add the values represented by each clue.
14 ten thousands = 14 × 10,000 = 1,40,000

Page 17

31 thousands = 31 × 1,000 = 31,000
423 hundreds = 423 × 100 = 42,300
35 tens = 35 × 10 = 350
25 ones = 25
Now add all the values:
1,40,000 + 31,000 + 42,300 + 350 + 25 = 2,13,675
Therefore, Nisha’s laptop password is 213675.
Hence, the correct answer is option d.

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Columns A, B, and C contain 3 numbers each, such that:
- The number of times each number appears in all 3 columns combined, is equal
to its value (e.g., 2 appears twice, 3 appears thrice, etc.)
- Each column contains AT LEAST TWO different numbers
If column A is the only column with THREE different numbers, then which
column has the highest sum?

(a) Column A (b) Column B
(c) Column C (d) Both columns B and C
Answer: d
Solution:
From the question, we can say that 4 appears four times, 2 appears twice, and 3
appears thrice.
It is stated that Column A is the only column with three different numbers,
meaning Column A contains 2, 3, and 4.
So, the empty circles of column A will have 2 and 3.

We already have two 2’s and two 3’s placed.
So, the remaining four circles must be filled with:
• One 3
• Three 4’s
Now, since Column B already contains a 2, we cannot place a 3 and 4 in it - doing so
would make three different digits in Column B, which violates the condition that only
Column A can have three different numbers.
Therefore, Column B must have two 4’s.
The remaining 3 and 4 will then go into Column C.

Sum of the digits of column A = 2 + 3 + 4 = 9
Sum of the digits of column B = 2 + 4 + 4 = 10
Sum of the digits of column C = 4 + 3 + 3 = 10
Hence, both columns B and C have the highest sum.
Hence, option d is the correct answer.

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Chapter 2: Arithmetic Expressions
1. Identify the odd one out from the following expressions, after INTERCHANGING their operators.
a) 204 + 3 ÷ 29 b) 67 - 2 × 37 c) 48 ÷ 6 × 3 d) 195 × 49 - 2
Answer: c
Solution:
We have to identify the odd one out after interchanging the operators present in the expressions.
Let us solve each equation one by one:
Option a: 204 + 3 ÷ 29
After interchanging the operators, the resultant expression is: 204 ÷ 3 + 29
= 68 + 29
= 97
Option b: 67 - 2 × 37
After interchanging the operators, the resultant expression is: 67 × 2 - 37
= 134 - 37
= 97
Option c: 48 ÷ 6 × 3
After interchanging the operators, the resultant expression is: 48 × 6 ÷ 3
= 288 ÷ 3
= 96
Option d: 195 × 49 - 2
After interchanging the operators, the resultant expression is: 195 - 49 × 2
= 195 - 98
= 97
In every option, the answer is 97 except in option c, where the resultant value is 96.
Hence, the correct answer is option c.
_____________________________________________________________________________________
2. There is a seminar from 11:00 AM to 2:00 PM, where students must spend AT LEAST 1 hour in
the seminar. The table displays the number of students entering and exiting the seminar at
different times. What is the MAXIMUM possible number of students who spent at least 2 hours in
the seminar?

a) 60 b) 80 c) 100 d) 120
Answer: c
Solution:
To maximize the number of students spending at least 2 hours in the seminar, we should aim to retain
as many students as possible for that duration.
1. Initial Entry at 11:00 AM:
• 120 students entered the seminar
• 60 students exited at 12:00 PM, meaning these 60 students cannot complete 2 hours
• At 12:00 PM, the remaining 60 students from the 11:00 AM batch are still in the hall

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2. At 12:00 PM:
• 40 new students entered the hall
• The total number of students in the seminar is now 100 (60 from the 11:00 AM batch + 40 new
students)
• The 60 students from the 11:00 AM batch have now completed 1 hour

3. At 1:00 PM:
• 80 more students entered
• 40 students left the hall, and to maximize the number staying for 2 hours, we assume these 40
students are from the 11:00 AM batch, meaning they have now completed 2 hours
• The remaining 20 students from the 11:00 AM batch stay in the hall
• The total number of students in the hall is now 140 (20 from 11:00 AM + 40 from 12:00 PM + 80
from 1:00 PM)

4. At 2:00 PM:
• No new students enter, and all 140 students exit
• By this time, the 40 students from the 12:00 PM batch have completed 2 hours

Final Calculation:
• 60 students from the 11:00 AM batch stayed until at least 1:00 PM, and 40 of them left exactly at
1:00 PM after completing 2 hours
• 20 students from the 11:00 AM batch remained until 2:00 PM, so they also completed 2 hours
• 40 students from the 12:00 PM batch stayed until 2:00 PM, completing 2 hours
Thus, the maximum possible number of students who spent at least 2 hours in the seminar is 100.
Option c is correct.
_____________________________________________________________________________________
3. In the given expression '8 @ 7 - 15 < 4 × 2 × 5', which operator cannot be used in place of '@',
such that the condition remains valid?
a) - b) ÷ c) + d) ×
Answer: d
Solution:
8 @ 7 - 15 < 4 × 2 × 5
The left side of the condition has 8 @ 7 - 15 and the right side has 4 × 2 × 5, which is 40.
Let’s solve it logically.
The LHS has -15 in it. So, whatever the value of 8 @ 7 be, its value further decreases due to the
subtraction of 15 from it.
So, Subtraction (-) and division (÷) reduce the value, and there is already a (-15) in the expression,
which decreases the value further.
So, the result will definitely be less than 40, and these operators can be used definitely.
Addition (+) increases the value slightly. Let’s check with “+”.
8 + 7 - 15 = 0, which is still less than 40. So, addition can be used here.

Multiplication (×) increases the value.
8 × 7 - 15 = 41, which is not less than 40. So, multiplication cannot be used.
Therefore, the operator that cannot be used is multiplication (×). Option d is correct.

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4. A bus has 40 seats, and some seats are already occupied, with one passenger in each seat.
● At the first stop, 5 passengers get off and 8 passengers get in
● At the second stop, 2 passengers get off and 4 passengers get in
● At the third stop, exactly half of the passengers get off
After this, half of the bus seats are occupied.
Which of the following expressions represents the number of passengers on the bus at the
beginning?
a) 5 × 5 + 5 b) 6 × 6 - 4 c) 5 × 5 + 10 d) 6 × 6 - 5
Answer: c
Solution:
This can be solved by moving backwards from the last stop. (reverse deduction)
Half of the bus seats are occupied at the end.
Since the bus has 40 seats, 20 passengers remain. (40/2 = 20)
At the third stop, half of the passengers got down.
This means before the third stop, there were 40 passengers, because half of 40 is 20.
At the second stop, 2 passengers got down and 4 got in.
If there are ‘X’ number of passengers and 2 got down and 4 got in, the expression for the number of
people at that time is:
X-2+4
=X+2
This means the number of passengers increased by 2 (overall).
So, before the second stop there were 38 passengers only, which later became 40.
At the first stop, 5 passengers got down and 8 got in.
If there are ‘Y’ number of passengers and 5 passengers got down and 8 got in, the expression for the
number of people at that time is:
Y-5+8
=Y+3
This means the number of passengers increased by 3 (overall), so at the beginning, there were 35
passengers. (35 + 3 = 38)
Among the given expressions,
a) 5 × 5 + 5 = 25 + 5 = 30
b) 6 × 6 - 4 = 36 - 4 = 32
c) 5 × 5 + 10 = 25 + 10 = 35
d) 6 × 6 - 5 = 36 - 5 = 31
Therefore, the correct answer is option c.
_____________________________________________________________________________________
5. The following expression has a whole number ‘n’. If a blank can either take ‘×4’ or ‘–12’, what is
the least possible value of ‘n’?
Expression: (n ____ ____ ) ____ = 96
a) 9 b) 18 c) 4 d) 72
Answer: a
Solution:
Think logically.
Multiplying by 4 makes the number grow bigger and subtracting 12 makes it smaller.

Page 22

If we want the smallest starting number ‘n’ and turn it into 96, we should multiply by 4 more times,
because even a small number can become big after multiplying. As subtraction decreases a number,
let’s try to avoid it as much as possible.
Now, the expression must make n into 96.
Let’s assume that all three operations performed on n are of ‘×4’ only.
So, n × 4 × 4 × 4 = 96
n × 64 = 96.
But, there is no such whole number which when multiplied by 64 gives 96 as the result.
Thus, we can say that not all the three operations are ‘×4’. At least one ‘−12’ must be used.
Now let’s try using two ‘×4’ operations and one ‘−12’ operation, since we want to subtract as little as
possible. There are different possible arrangements. Let’s test them carefully:
Case 1: Subtract first, then multiply twice
(n − 12 × 4) × 4 = 96
Divide by 4 on both sides.
(n − 12 × 4) = 24
n – 48 = 24
Add 48 on both sides.
n = 24 + 48
n = 72
This is a whole number.
Case 2: Multiply once, subtract, then multiply
(n × 4 − 12) × 4 = 96
(4n − 12) × 4 = 96
16n − 48 = 96
Add 48 on both sides.
16n = 144
n=9
This is also a whole number.
Case 3: Multiply twice, then subtract
(n × 4 × 4) − 12 = 96
16n − 12 = 96
Add 12 on both sides.
16n = 108
n = 108/16 = 6.75
n = 6.75
It is not a whole number.
Now, compare the valid whole numbers:
Case 1: n = 72
Case 2: n = 9
The smallest possible whole number is: 9.
Hence, option a is correct.

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6. Which operators should be interchanged to get the minimum value of the given expression?
8×4+2-1
a) × and + b) + and - c) - and × d) No change
Answer: c
Solution:
Let us solve the current expression: 8 × 4 + 2 - 1 = 32 + 2 - 1 = 33.
To find the minimum value by interchanging the operators, let's analyse the problem logically instead of
trying each pair of operators given in the options.
Since we need the least possible result, we should consider changing the multiplication operator
because it is placed between two of the highest numbers in the given expression.
By interchanging it with the subtraction operator, which is placed between two of the lowest numbers,
we can reduce the overall value.
Let’s apply this change to the expression: Original expression: 8 × 4 + 2 - 1
Interchange multiplication with subtraction: New expression: 8 - 4 + 2 × 1
Now, solve the new expression step by step:
2×1=2
8-4+2=6
Thus, the minimum value is 6.
Option c is the correct choice.
_____________________________________________________________________________________
7. In certain code, 1 is coded as 2, 3 is coded as 10, 5 is coded as 26.
In the same way, 4 is coded as A and 6 is coded as B.
If C is the sum of A and B, which of the following is correct regarding C?
a) C > 55 b) C < 50 c) C = 50 d) C < 55
Answer: d
Solution:
Here, the pattern for coding is to multiply the number by itself and add 1. For example:
1 = (1 × 1) + 1 = 2
3 = (3 × 3) + 1 = 10
So, 4 = (4 × 4) + 1 = 17
A = 17
Also, 6 = (6 × 6) + 1 = 37
B = 37
Now,
C=A+B
C = 17 + 37 = 54
We know that 54 < 55
So, C < 55
Therefore, the correct answer is option d, C < 55.
_____________________________________________________________________________________
8. In the following expression, each ball with a mathematical operator is swapped with the ball
immediately to its left, to form a new arithmetic expression. What will be the value of NEW
EXPRESSION - ORIGINAL EXPRESSION?

a) 63 b) 73 c) -63 d) -73
Answer: d

Page 24

Solution:
The value of the original expression is:
87 + 15 - 3
= 102 - 3
= 99
Let’s form the new expression, as per the given instructions.
Given that, each ball with an operator is swapped with the ball immediately to its left.

After swapping the balls having mathematical operators with the balls on their immediate left, the
resultant expression is:

= 8 + 71 - 53
= 79 - 53
= 26
New expression - Original expression
= 26 - 99
= - 73
Thus, the correct answer is option d.
_____________________________________________________________________________________
9. A stone with an initial value of 10 is dropped through one of the holes at the START and moves
in a straight path to reach the END. As it passes each numbered bar along the way, its value
changes according to the operation shown on that bar. What is the maximum value the stone
can have when it reaches the END?

a) 60 b) 88 c) 53 d) 68
Answer: b
Solution:
As we need the highest possible value, we should try to touch any of these multiplication bars, so that
there is a good increase in the stone’s value.
However, if we concentrate on bar ‘×3’, we would miss the bars ‘×2’ and ‘×4’, which would be a great
loss. So, let’s ignore bar ‘×3’.
We should avoid -32 and -10 as it would decrease the overall value.
Hence, we just take -12 over -32 and -10.

Page 25

The value we get from the above hole is:
Bar 1: (10 × 2) = 20
Bar 2: 20 + 5 = 25
Bar 3: 25 × 4 = 100
Bar 4: 100 - 12 = 88
Hence, option b is the correct answer.
_____________________________________________________________________________________
10. Each arrow has a mathematical operator. When placed in the centre box, its operator replaces
the letter A in the equation it points to. Which arrow satisfies an equation based on this rule?

a) b) c) d)
Answer: d
Solution:
Let us check each option one by one by replacing A with the operator shown on the arrow.
Option a: Operator = ‘–’
Top equation becomes:
5–4=1
Since 1 ≠ 20, this option is incorrect.
Option b: Operator = ‘÷’
Left equation becomes:
7 ÷ 6 = 1.16…
Since 1.16 ≠ 13, this option is incorrect.
Option c: Operator = ‘+’
Right equation becomes:
9 + 3 = 12
Since 12 ≠ 6, this option is incorrect.
Option d: Operator = ‘÷’
Bottom equation becomes:
14 ÷ 7 = 2
Since 2 = 2, this equation holds true.
Therefore, the correct answer is option d.

Page 26

Tom, Sam, Mariya, John, and Bob each selected a shape. They placed their shapes
in the grid below.
Tom’s shape is neither a circle nor a triangle
Sam’s shape is in column number 3
Mariya’s shape is not in the first three rows
John’s shape is neither square nor circle
Which of the following shapes is chosen by Bob?

(a) Square (b) Circle (c) Hexagon (d) Triangle
Answer: b
Solution:
• Sam: The shape in column 3 is a star.
• Mariya: Not in the first three rows, so it must be a hexagon because Sam has a star.
• Tom: Since it's not a circle or a triangle, it is a square because Sam has a star and
Mariya has a hexagon.
• John: Since it's neither a square nor a circle, it has to be a triangle, the only shape left.
• Bob has selected a circle, which is the last remaining shape.
Hence, option b is the correct answer.

Page 27

Chapter 3: A Peek Beyond the Point
Activity Time
Introduction
We know that numbers can be represented in powers of 10. Though this is not the only way to represent
the number. In this activity, we will understand the binary representation and powers of 2, using punched
cards. Consider a set of cards with the alphabet written on them, and we shuffle them. Using a stick and
holes on the top of the cards, we can magically arrange the cards alphabetically in a few seconds. With
this activity, we will discover the patterns in punching & the principle behind the trick.

Activity Time Description

Launch 5 min Teacher demonstrates the activity

Supporting Links:
Print file:
https://drive.google.com/file/d/15IRE5Pr1BrkjBab4oXqoi
k6rTDeFesbK/view?usp=drive_link
Activity Video Link:
https://www.youtube.com/watch?si=ggcG_e4GWMByA
MBh&v=F3DagixkqsQ&feature=youtu.be

Reference:
https://docs.google.com/document/d/1jJQmWEnU21zht
Mbp3ht_FX_E1HldpVs6iizucnafxEI/edit?tab=t.0

Trial by Students 15 min Students try out the activity.

Page 28

Student Worksheet:
https://docs.google.com/document/d/1Sr0B87i2x6a3tvT
kcNL0vaHDBz_OhYcDtXKAEYuR_eU/edit?usp=sharing

As the teacher progresses through the activities, the
students progress along with the worksheet.
Students can make their own punch cards using the
template after class hours.

Discussions and 10 min Attempting the worksheets and the discussion based on
Explorations the activity

CT Components
Algorithmic Thinking
We used a step-by-step approach to convert a decimal number to binary.
Decomposition
Breaking the conversion of the number to binary into one bit at a time.
Abstraction
Using different types of holes to represent the binary number system.
Logic
Every alphabet can correspond to a number, and every number can be represented in the binary system
with a punching pattern.
Generalisation
Recognising a doubling pattern of 1, 2, 4, and so on, and adding different powers thus generated gives
us the binary representation, and we conclude that any number can be represented in terms of powers
of 2.

Page 29

Activity Description: Punched Cards
Consider a set of cards with the alphabet written on them, and we shuffle them. The objective now is to
arrange them in alphabetical order. One method is to find the card A first, then B, and so on. This might
take some time.
Now, notice that the cards have been punched on the top. There are an O-shaped and a U-shaped
punch, and using these holes, we can sort the cards in an interesting way. Insert a stick in the set of
cards starting from the rightmost hole. Once you lift the stick, some will come along with it. Now, bring
them in front & repeat this for all sets of holes. Magically, the cards are sorted.
In our day-to-day life, we use the decimal number system, which uses the numbers 0, 1, 2, 3, 4, 5, 6, 7,
8, and 9. The decimal number system is based on bundles of ten (Powers of 10, which is 1,10,100,
1000,...). Similarly, the binary number system is based on bundles of two (Power of 2, which is 1,2,4,8,...).
So, if we compare the two numbering systems for place values, this is what it looks like….

Decimal (Bundles of Ten) Binary (Bundles of
Two)
Units Place 1 1
First Bundle 10 = 10 2=2
Second Bundle 10x10 = 100 2x2 = 4
Third Bundle 10x10x10 = 1000 2x2x2 = 8

The binary number system (popularly known as the language of computers) deals with only two numbers:
0 and 1. We can convert any decimal number to its binary equivalent and vice versa. You can convert
any decimal number into a binary representation by writing it with the power of 2.
for example, 23 = 16 + 4 + 2 + 1. Show this to the students.

23 in decimal 101 = 10 100 = 1

23 2 3

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

(10111)2 1 0 1 1 1

You can also do this by an algorithm, which is presented below (which you may present to the student):
i. Divide the decimal number by the base (2) of the binary number.
ii. Note the remainder from the first step till the end of the division process.
iii. The remainders obtained as 0 or 1 are written in order from the last step to the first step of division.
Let’s find out the binary equivalent of the decimal number 11.

Page 30

Divisor Decimal Number Corresponding Remainder
2 11 1
2 5 1
2 2 0
1

The highlighted cells represent the binary representation, written from bottom to top. So, 11 = (1011)2.
The binary number 1011 can be changed to its corresponding decimal number as:
A B C D
23 x 1 22 x 0 21 x 1 20 x 1

Decimal number = A+B+C+D = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20
= (1x8)+(0 x 4) + (1 x 2) + (1 x 1) =8+2+1 = 11
1. What will be the binary representation for the decimal number 14?
a) 1010 b) 1100 c) 1110 d) 1111
Explanation
c) 1110
Decimal 14 can be written in power of 2s as 8 + 4 + 2 = 1 x 23 + 1 x 22 + 1 x 21 + 0 x 20 = 1110
CT Competency
Algorithmic Thinking
Decomposition
Generalisation
The number of digits or bits (binary digits) in a binary number depends on the number. For example,
numbers up to 7 can be represented in a 3-bit binary number, numbers from 8 to 15 in a 4-bit binary
number, numbers from 16 to 31 similarly need an additional bit, hence a 5-bit binary number, and so
on. There are 26 letters in English; hence, 5 sets of holes on each card for sorting.
_____________________________________________________________________________________
2. You want to perform a similar punched card activity with Hindi alphabets. How many holes are
needed on each card?
a) 5 b) 6 c) 8 d) 10
Explanation
b) 6
There are 52 Hindi alphabets. Using 6 bits/holes, we can represent numbers from 0 up to 63.
CT Competency
Abstraction
Logic
Punching Pattern on the Card
Every alphabet can correspond to a number, and every number can be represented in the binary
system. The following table contains the corresponding conversion from decimal to binary.
Letter Corresponding Corresponding
Decimal Number Binary Number
A 1 00001

Page 31

B 2 00010
C 3 00011
D 4 00100
E 5 00101
.. .. ..
.. .. ..
Y 25 11001
Z 26 11010

These binary numbers indicate to us a unique way to punch. Where there is 0, punch an O-shaped
hole, and where there is 1, punch a U-shaped hole. For example: 01010 is the binary equivalent of J,
so the holes on card J from left to right will be O U O U O.
_____________________________________________________________________________________
3. Can you identify which card the binary pattern 10101 belongs to?
a) P b) S c) T d) U
Explanation
c) T
The alphabet T comes 21st in the order starting from A as 1. Therefore, it is represented by the decimal
number 21, and that can be written in binary as 10101.
CT Competency
Abstraction
Logic
_____________________________________________________________________________________
4. The alphabet R comes 18th in the order starting from A as 1. What will be the punching pattern
on Card R?
a) U0U0U b) UU0U0 c) U00U0 d) 0UU00
Explanation
c) UOOUO
The alphabet R comes 18th in the order starting from A as 1. Therefore, it is represented by the
decimal number 18, and that can be written in binary as 10010.
CT Competency
Abstraction
Logic
Punched cards, utilising binary choices is analogous to their role in powering early computing machines
such as IBM sorters and Turing's Bombe. This binary concept has its roots in study of poetry in ancient
India, which was well studied by Acharya Pingala's and documented in his work Chandaḥśāstra (~2
BCE).

Page 32

Questions
1. Two decimal numbers differ by 3.6. In one of the numbers, the decimal point is moved one place
to the right. After this change, both numbers become equal. Which was the smaller of the two
original numbers before shifting the decimal point?
a) 0.4 b) 0.36 c) 3.6 d) 4
Answer: a
Solution:
Let the smaller number be x.
Since the two numbers differ by 3.6, the larger number is: x + 3.6
When the decimal point of one number is moved one place to the right, that number becomes 10 times
its original value. After this change, both numbers become equal.
So, 10x = x + 3.6
Now solve:
10x − x = 3.6
9x = 3.6
x = 3.6/9 = 0.4
So, the smaller original number was 0.4
Hence, option a is the correct answer.
_____________________________________________________________________________________
2. What will come in place of “?”

a) 2000 b) 4000 c) 40000 d) 8000
Answer: b
Solution:
Before the arrow, each term has some shaded blocks and a number next to it, in KG.
The shaded blocks represent a fraction (out of 10).
This fraction is multiplied by the given number.
The result (KG) is converted into grams and written after the arrow.
For example, in the first pair, the shaded fraction is 3/10, which is 0.3
0.3 multiplied by 2 kg gives 0.6 kg.
Now, this 0.6 kg is converted into grams.
We know that 1 kg = 1000g. So, 0.6 kg = 0.6 x 1000 = 600 g.
Hence, 600 is given after the arrow.
Similarly, in the question term, 8 out of 10 blocks are shaded.

Page 33

So, the fraction is 0.8
Multiplying with 5 kg, we get: 0.8 × 5 = 4 kg
Converting it into grams: 4 kg = 4000 g
Hence, option b, 4000 is the correct choice.
_____________________________________________________________________________________
3. Find the next term in the given series below:
85.41, 88.71, 89.01, 92.31, 92.61, 95.91, 96.21, __
a) 96.51 b) 99.21 c) 99.51 d) None of these
Answer: c
Solution:
Each term in the given series is obtained by adding 3.3 and 0.3 to the previous term alternately.
85.41 + 3.3 = 88.71
88.71 + 0.3 = 89.01
89.01 + 3.3 = 92.31
92.31 + 0.3 = 92.61
As the last given term 96.21 is 0.3 more than its previous term, the next term is obtained by adding 3.3
to 96.21, which gives 99.51.
Hence, the correct answer is option c.
_____________________________________________________________________________________
4. There are three sensors - A, B, and C. Each sensor displays the measurement of the length of
the same item, but in different units: metres, centimetres, or kilometres. Only one sensor shows
the correct length. The other two show values that are 10 times greater and 10 times smaller
than the actual length (the units remain unchanged). Based on the readings shown, which
sensor displays the correct length?

a) Sensor A b) Sensor B
c) Sensor C d) Cannot be determined
Answer: c
Solution:
Let’s take one sensor to be correct each time. So, the other two sensors will be showing 10 times
greater and 10 times smaller than the actual length. (in any order)
Case 1: Let’s assume that sensor A is displaying the correct length.
As A = 0.04506 is in metres, let’s convert it into cm and km.
0.04506 m = 0.04506 x 100 = 4.506 cm.
However, Sensor B is showing 450.6, which is neither 10 times greater than nor 10 times smaller than
4.506
So, our assumption is incorrect. Sensor A is not displaying the correct value.
Case 2: Let’s assume that sensor B is displaying the correct length.
As B = 450.6 is in centimetres, let’s convert it into m and km.
450.6 cm = 450.6 ÷ 100 = 4.506 m
However, Sensor A is showing 0.04506, which is neither 10 times greater than nor 10 times smaller
than 4.506.

Page 34

So, our assumption is incorrect. Sensor B is not displaying the correct value.
Case 3: Let’s assume that sensor C is displaying the correct length.
As C = 0.0004506 is in kilometres, let’s convert it into m and cm.
0.0004506 km = 0.0004506 x 1000 = 0.4506 m
Sensor A is showing 0.04506, which is 10 times smaller than 0.4506
So, when sensor C is correct, sensor A is displaying a value which is 10 times smaller.
Now, let’s check if sensor B is showing a value which is 10 times larger.
Convert km to cm.
0.0004506 km = 0.0004506 x 100000 = 45.06
Sensor B is displaying 450.6, which is 10 times larger than 45.06
Hence, both conditions are satisfied.
Therefore, sensor C is displaying the correct value.
Option c is the correct answer.
_____________________________________________________________________________________
5. What will come in place of “?” in the given series?

a) b) c) d)
Answer: c
Solution:
The given series follows these rules:
- In each term, the decimal number at the bottom is the average of the top two numbers in grey
blocks.
- The grey blocks in every next term are a combination of the top-right number and the bottom
number of its previous term.
For example, in the second term, the numbers in the grey cells are 4 and 4.5
4 is the top-right number of the previous term (the first term)
4.5 is the bottom number of the first term (average of 5 and 4)
Likewise, the terms follow this rule of averages:
(4 + 4.5)/2 = 4.25
(4.5 + 4.25)/2 = 4.375
(4.25 + 4.375)/2 = 4.3125
(4.375 + 4.3125)/2 = 4.34375
Hence, in the next term, both grey blocks must have numbers: 4.3125 and 4.34375, and their average
must be written below.
(4.3125 + 4.34375)/2 = 4.328125
As the term in option c follows this rule, option c is the correct choice.

Page 35

6. Four friends A, B, C, and D each select a distinct point on the following number line (from 0 to 1)
such that:
- A’s point is 4th to the right of D’s point
- The value of C’s point is at least 0.4 and B’s point has the highest value among all friends
- No two friends choose adjacent points
What is the LEAST possible sum of the points chosen by all of them?

a) 2.0 b) 1.9 c) 1.8 d) 1.4
Answer: c
Solution:
Clearly, as there are 9 points between 0 and 1 on the number line, we can say that the value of each
point is 0.1 more than the previous value.
We have to minimise the sum of the points chosen by all 4 friends.
So, we must minimise the individual values.
Let’s check if the least value 0 can be assigned to any friend.
Given that B’s point has the highest value among all. So, B cannot have 0.
A is 4th to the right of D. So, A cannot have 0.
The value of C’s point is at least 0.4 and C also cannot have 0.
So, let’s take 0 for D. Thus, A and D can be placed as follows:

Now, B and C are remaining.
As the value of C’s point is at least 0.4, its value can be anything greater than 0.3
However, as A is already 0.4, we cannot assign any value till 0.5 for C (no two friends chose adjacent
points).
So, the least possible value for C is 0.6. Hence, B will further get 0.8 (as 0.7 cannot be assigned).

Hence, the values are: 0, 0.4, 0.6, and 0.8
Therefore, the least possible sum is 0 + 0.4 + 0.6 + 0.8 = 1.8
Hence, option c is the correct choice.
_____________________________________________________________________________________
7. Alex has to fill in the blank with a DECIMAL NUMBER formed using every digit among 0, 1, 2, 3,
and 4, each exactly once, where:
- The sum of the digits before the decimal point is equal to the sum of the digits after the decimal
point
- The digits before the decimal point are in descending order and the digits after the decimal
point are in ascending order
- The conditions of “<” are satisfied
How many different numbers can be placed in the blank?

a) 2 b) 3 c) 4 d) More than 4
Answer: b
Solution:

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The digits given are: 0, 1, 2, 3, and 4
The total sum of all digits is 10.
We have to form a decimal number where the sum of the digits before the decimal point is equal to the
sum of the digits after the decimal point.
So, the digits must be divided into two groups such that both groups have equal sum, that is, 5 and 5.
So, the groups of numbers before and after the decimal point can be: (1, 4, 0) and (2, 3), OR (1, 4) and
(0, 2, 3).
Following the descending order before the decimal point and ascending order after the decimal point,
the possible numbers are:
41.023, 32.014, 410.23, 320.14.
Also, it is given that 34.201 < the decimal number < 412.03
Based on this only three numbers are possible from the above list: 41.023, 320.14, and 410.23
Hence, 3 different numbers can be formed. Thus, the correct answer is option b.
_____________________________________________________________________________________
8. Raj measures the lengths of two objects, Object A and Object B, using a ruler marked in
centimetres.
● On Day 1, Object A measures 5.2 cm. Its length decreases by 1 line on Day 2, 2 lines on Day
3, 3 lines on Day 4, and so on, following the same pattern
● On Day 1, Object B measures 1.2 cm. Its length increases by 1 line on Day 2, 2 lines on Day 3,
3 lines on Day 4, and so on, following the same pattern
On which day will the difference between the lengths of the two objects be the least?
Note: The scale is in centimetres

a) Day 3 b) Day 4 c) Day 5 d) Day 6
Answer: c
Solution:
Let’s solve it logically.
The length of Object A on Day 1 = 5.2 cm
The length of Object B on Day 1 = 1.2 cm
Difference between their lengths = 5.2 − 1.2 = 4.0 cm
The pattern of increase or decrease in the lengths in each day is 1 line (0.2 cm), 2 lines (0.4 cm), 3
lines (0.6 cm), 4 lines (0.8 cm), and so on.
Hence, the sequence is 0.2, 0.4, 0.6, 0.8, and so on. (in both cases, either + or -)
The lengths of object A from Day 1 are: 5.2, 5, 4.6, 4, 3.2, 2.2 and so on.
The lengths of object B from Day 1 are: 1.2, 1.4, 1.8, 2.4, 3.2, 4.2 and so on.
Clearly, the 5th term of the above patterns is the same (3.2)
Hence, the least difference between the lengths of both objects can be seen on the 5th day.
Option c is correct.
_____________________________________________________________________________________
9. Some numbers are given in the boxes below. In each number, interchange the digits
immediately to the left and right of the decimal point. After making this change, how many
boxes will contain a number greater than 498.540?

a) 1 b) 2 c) 3 d) 4
Answer: b

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Solution:
The digits on the immediate sides of the decimal point interchange their positions.
Box A: 491.893 becomes 498.193, which is less than 498.540
Box B: 497.860 becomes 498.760, which is greater than 498.540
Box C: 491.855 becomes 498.155, which is less than 498.540
Box D: 490.993 becomes 499.093, which is greater than 498.540
Hence, only two boxes satisfy the condition.
Option b is the correct answer.
_____________________________________________________________________________________
10. Seven decimal numbers, A-G, are given.
1. Arrange the numbers in ascending order to form Arrangement 1
2. Modify the numbers as follows:
○ For numbers in odd positions, move the decimal point one place to the right
○ For numbers in even positions, move the decimal point one place to the left
3. Arrange the modified numbers again in ascending order to form Arrangement 2
In Arrangement 2, which pair(s) of numbers have exactly three numbers between them?

a) A and E b) G and D c) C and F d) E and D
Answer: d
Solution:
After arranging the numbers in ascending order, we get:

Now, for odd positions: decimal moves right (×10) and for even positions: decimal moves left (÷10)

Arranging the above numbers again in ascending order we get:

Among the options given, only E and D have exactly three numbers between them.
Hence, option d is correct.

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The numbers given below are to be placed in the grid in such a way that the sum
of two adjacent numbers in the grid should be greater than 9 but less than 17.
What will come in place of A and B?
Numbers: 2, 3, 5, 11, 15

(a) A = 5 and B = 2 (b) A = 3 and B = 5
(c) A = 2 and B = 3 (d) A = 5 and B = 3
Answer: d
Solution:
Let us approach the question logically and systematically.
The number after 14 can only be 2, resulting in 16, which is greater than 9 but less than
17.

The number before 12 can only be 3. When you add 3 to 12, you get 15, which is greater
than 9 but less than 17. Placing any other number would result in a sum equal to 17 or
greater than 17.

The number 11 can be placed after 1 or 2. When 1 or 2 is added to 11, the results are 12
and 13, both of which are greater than 9 and less than 17.
11
OR
11

Numbers left with us: 5 and 15
The number 15 can be placed only after 1. When you add 1 to 15, you get 16, which is
greater than 9 but less than 17. Placing it next to any other number would result in a sum
equal to 17 or greater than 17. So, we place 15 after 1. We also place 11 after 2.

The only place left for 5 is A. 5 + 11 = 16, and 5 + 8 = 13. Both conditions are satisfied.
Therefore, A = 5.

Therefore, A = 5 and B = 3 as highlighted above.
Hence, option d is the correct answer.

Page 39

Chapter 4: Expressions using Letter-Numbers
1. It is given that x means ‘greater than’, $ means ‘equal to’, < means ‘not less than’, A means ‘less
than’ and + means ‘not greater than’. Which of the following options has the same meaning as
the given expression?
axbAc
a) a $ c A b b) b < a x c c) c x b A a d) c + b < a
Answer: c
Solution:
Let's replace the symbols of the expression given in the question: a x b A c.
The resultant statement will be: a > b < c
a > b < c can be written as a > b, c > b
Option a: Replacing symbols in option a. The expression becomes:
a $ c A b: a = c < b.
We know c > b. Hence, c < b is false.
Option b: Replacing symbols in option b. The expression becomes:
b < a x c: b is not less than a > c.
We know a > b. Hence, b is not less than a is false.
Option c: Replacing symbols in option c. The expression becomes: c x b A a: c > b < a.
We know, a > b < c, which is also written in option c. Hence, option c is true.
Option d: Replacing symbols in option d. The expression becomes:
c + b < a: c is not greater than b is not less than a.
We know a > b. Hence, b is not less than a is false.
Only option c is true. Hence, the correct answer is option c.
_____________________________________________________________________________________
2. A factory ‘S’ sends goods to four cities: A, B, C, and D through pipelines (represented by arrows),
as shown below. The demand of each city is mentioned above that city (for example, the demand
at City A is 400 units). The demands are satisfied exactly, and no extra units are left in the pipelines.
If 300 units are sent through the pipeline from A to B, how many units are sent from D to B?

a) 200 b) 600 c) 800 d) 1000
Answer: c
Solution:
Assume that ‘X’ units are sent from D to B.
Now, the units that travel from A to B (300 units, as given in the question) and those that travel from D
to B together satisfy the demand of the cities at the receiving end.
So, 300 + X must satisfy the demand at city B, which is 400.
But, if you observe carefully, some units also travel from B to C, to satisfy the demand at city C. So, city
B must receive the units equal to its own demand as well as the demand of city C.

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Hence, (300 + X) units must travel through both pipelines to satisfy the demand at cities B and C.
Therefore, the equation is:
300 + X = 400 + 700
300 + X = 1100
X = 1100 - 300
X = 800.
Hence, the number of units sent from D to B is 800. Thus, option c is the correct choice.

_____________________________________________________________________________________
3. Each letter in Set 1 corresponds to exactly one different number from Set 2 (in any order).
Set 1: A, B, C, P, Q, R
Set 2: 1, 2, 4, 5, 7, 9
The following conditions are given:
● A and C are consecutive numbers, with A > C
● P + Q = R, with P > Q
● B is greater than both A and C
Using this information, find the value of: (4P+3B) − (2P+3C) + (2A−3B)
a) 9 b) 13 c) 15 d) 11
Answer: d
Solution:
A and C are consecutive numbers, and the possible pairs are (2, 1) and (5, 4).
Case 1: If A and C are (5, 4), then the remaining numbers are: 1, 2, 7, 9
Among these, we must have a possible set for P, Q and R, where P + Q = R.
The only possible case will be 7 + 2 = 9.
Hence, the last digit left is 1, which will be assigned to B.
However, it is given that B is greater than both A and C.
Hence, this case is invalid.
Case 2: If A and C are (2, 1), then the remaining numbers are: 4, 5, 7, 9
Among these, we must have a possible set for P, Q and R, where P + Q = R.
The only possible case will be 5 + 4 = 9, where P = 5, Q = 4 and R = 9.
Hence, B = 7 (B > A,C)
Now, we have to find the value of (4P + 3B) − (2P + 3C) + (2A − 3B)
Rather than substituting the values directly, let’s write the expression in its simplest form for easier
substitution.
(4P + 3B) − (2P + 3C) + (2A − 3B) = 4P + 3B − 2P − 3C + 2A − 3B
= 2P − 3C + 2A
Now, substitute the values: 2(5) − 3(1) + 2(2) = 10 − 3 + 4 = 11
Hence, the correct answer is option d.

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4. The sum of the ages of four people A, B, C, and D is 35. Each one of them is at least 8 years old.
What is the MAXIMUM possible age of the oldest person?
a) 9 years b) 10 years c) 11 years d) 12 years
Answer: c
Solution:
Let the ages of A, B, C, and D be a, b, c, and d respectively.
It is given that:
a + b + c + d = 35
Let the oldest person's age be X.
To get the maximum possible age for the oldest person, we need to minimize the ages of the rest of the
people. So, let the rest of the people be 8. (each of them is at least 8 years old)
So, 8 + 8 + 8 + X = 35
24 + X = 35
X = 35 - 24
X = 11
Hence, the MAXIMUM possible age of the oldest person is 11 years.
Hence, option c is the correct answer.
_____________________________________________________________________________________
5. 3 whole numbers A, B, and C, satisfy the given conditions. At maximum, how many different
possible values can C have?

a) 4 b) 5 c) 3 d) 2
Answer: a
Solution:
We are given:
• A, B, and C are whole numbers (0 and above)
• The conditions are: A > B > C and A + B = 10
To find how many different values C can take at maximum, we follow these steps:
Step 1: Understanding A and B
Since A + B = 10 and A > B,
we cannot take A = 5 and B = 5 (because A must be strictly greater than B).
Also, we cannot take A < 6, because then B would have to be greater than A to make their sum 10
(which breaks the A > B condition).
So, we start checking from the highest possible A value, working downward, ensuring:
•A>B
• A + B = 10
•Once A and B are fixed, C must be a whole number less than B.
Step 2: Try valid A and B pairs
• A = 9, B = 1: A + B = 10, A > B
C < 1: Only possible C = 0
• A = 8, B = 2: A + B = 10, A > B
C < 2: C can be 0 or 1
• A = 7, B = 3
C < 3: C can be 0, 1, or 2
• A = 6, B = 4
C < 4: C can be 0, 1, 2, or 3
Now, let’s list all possible distinct C values from these cases:
C = 0, 1, 2, 3

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Total = 4 values
Values beyond this won’t work, as:
• A = 5, B = 5: A is not greater than B
• A < 5: B becomes ≥ A or invalid
Final Answer: 4
C can take four different whole number values: 0, 1, 2, 3.
Option a is the correct answer.
_____________________________________________________________________________________
6. If the sum of the ages of Shubham and Shivam is 36. What would be the sum of their ages after
36 years?
a) 108 b) 136 c) 72 d) 96
Answer: a
Solution:
Let Shubham’s present age be: x
Let Shivam’s present age be: y
Given:
x + y = 36
After 36 years:
Shubham’s age will be: x + 36
Shivam’s age will be: y + 36
So, the sum of their ages after 36 years:
(x + 36) + (y + 36)
= x + y + 72
We know that the sum of the ages of Shubham and Shivam is 36:
x + y = 36
Thus, substituting that in the expression:
36 + 72 = 108
The correct answer is 108.
Hence after 36 years, both of them would age by 36 years. So, the sum would increase by 72 years.
Hence, the correct answer is option a.
_____________________________________________________________________________________
7. A, B, and Z are natural numbers such that A + B = Z and:
● The value of B is twice the value of A
● The value of B is greater than 4
● The value of Z is less than 10
What could be the value of A?
a) 1 b) 2 c) 3 d) 4
Answer: c
Solution:
Given:
A+B=Z
Also, B = 2A (Since B is twice the value of A)
B>4
Z < 10
We can substitute B in the first equation with 2A from the second condition:
A + 2A = Z
3A = Z
Now, using the third condition (B > 4), and since (B = 2A), we get:
2A > 4

Page 43

A>2
And using the fourth condition (Z < 10), and since (3A = Z), we get:
3A < 10
A < 10/3
Combining the two inequalities for A:
2 < A < 10/3
2 < A < 3.33
The only integer value for A that satisfies all conditions is:
A = 3.
Option c is the correct answer.
_____________________________________________________________________________________
8. In the given image, each number equals the sum of the two numbers immediately below it. If P,
Q, and R are unknown, what is the value of P × Q + R?

a) 42 b) 48 c) 38 d) 36
Answer: c
Solution:
Each number equals the sum of the two numbers directly below it. Thus,
Value of Q would be: Q = 1 + P
Value of R would be: R = P + 3
14 at top is the result of the sum of Q and R:
14 = Q + R
14 = (1 + P) + R (substituting the value of Q in the equation)
14 = 1 + P + (P + 3) (substituting the value of R in the equation)
14 = 4 + 2P
14 – 4 = 2P
10 = 2P
5=P
Finding the value of Q and R:
Q=1+P=1+5
Q=6
R=P+3=5+3
R=8
Thus, the values of P, Q and R are 5, 6, and 8 respectively.
So, the value of P × Q + R would be:
5×6+8
= 30 + 8
= 38.
Hence, the correct answer is option c.
_____________________________________________________________________________________
9. 5A, 5B, and C9 are 2-digit numbers, where A, B, and C represent one of the digits of the
respective numbers, and A + B + C = 15. Find the MAXIMUM possible value of 3B + 4C + 8, if:

a) 43 b) 39 c) 31 d) 36
Answer: b

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Solution:
As 5B > C9, C cannot be a value which is 5 or more than 5.
So, to maximize the value of 3B + 4C, we have to assign 4 to C.
We get: 5A > 5B > 49
As A + B + C = 15 and C = 4,
We now have A + B = 11.
If B = 9, (As B must be a single digit only) we have 5A > 59 > 49.
Then, we cannot assign an appropriate value for A.
So, B = 9 is invalid.
Similarly, if B is a value equal to 6 or more than 6, A cannot have an appropriate value that makes 5A >
5B.
So, the maximum possible value of B is 5.
So, 56 > 55 > 49.
Hence, the maximum possible value of 3B + 4C + 8 = 3(5) + 4(4) + 8
= 15 + 16 + 8
= 31 + 8
= 39.
Option b is correct.
_____________________________________________________________________________________
10. All the cells having ‘x’ in the given calendar represent dates that are consecutive multiples of a
particular number. The first ‘x’ represents the first Thursday of the month, as shown below. Which
of the following is the last day of the month?

a) Monday b) Sunday c) Saturday d) Friday
Answer: b
Solution:
Let the particular number be k.
The four ‘x’ marked dates are consecutive multiples of k:
k, 2k, 3k, 4k
It is given that each ‘x’ occurs exactly 6 days after the previous one (as per the image).
Hence,
2k−k = 6
k=6
Therefore, the dates marked ‘x’ are:
6, 12, 18, 24
(If we consider 12, 18, 24, 30, then 12 would be the first multiple. But it would not match the condition
that the first ‘x’ is the first Thursday of the month. Hence, k = 6 is correct.)

Page 45

From the calendar, 24th November is a Monday.
Now,
Last date of November = 30
30 − 24 = 6
So, the last day is 6 days after Monday:
Monday + 6 = Sunday
Hence, the last day of the month is Sunday. Thus, the correct answer is option b.
_____________________________________________________________________________________

Numbers from 1-9 are to be placed in the grid such that no two consecutive
numbers are placed adjacent to each other.
For example, the bottom-left circle has 1, so the circle above it and the circle on
the immediate right cannot have the number 2.
Each group of circles which are inter-connected by lines must contain either only
odd or only even numbers.
Find the values of A and B.

(a) A = 9 and B = 3 (b) A = 3 and B = 9 (c) A = 7 and B = 3 (d) A = 9 and B = 8
Answer: a
Solution:
A and B are connected to circles with odd numbers. Therefore, A and B cannot be an
even number. Thus, option d can be eliminated.
Now, the value of A cannot be 7 as it is consecutive to 6 which is in the adjacent circle.
So, option c can also be eliminated.
Now, if we keep A = 3, then for the numbers 2 and 4 we will have only one circle left that
is above 6. Because if A = 3, then the circle above A cannot have 2 and the circle below
6 also cannot have 2 (as 1 is adjacent to it). The same condition applies for the number
4 as well due to the numbers 3 and 5. So, option B is also eliminated.
Hence, the correct answer is option a.

Page 46

Chapter 5: Parallel and Intersecting Lines
1. Each side of a square is divided into three equal parts by marking points on it. Using these points
as endpoints, line segments are drawn inside the square, such that each line segment is parallel
to at least one side of the square. What is the maximum number of squares that can be found in
the resulting figure?
a) 6 b) 9 c) 14 d) 16
Answer: c
Solution:
As shown below, when each side of the square is divided into three equal parts and line segments are
drawn using these dots as endpoints, the resultant figure looks like:

The number of squares seen are 14, as highlighted below:

Option c is the correct answer.
_____________________________________________________________________________________
2. The following arrangement is formed using a cube and a cuboid. How many edges are parallel to
the edge highlighted by the red line?

a) 5 b) 6 c) 7 d) 8
Answer: c
Solution:
There are 4 edges parallel to the red edge in the cuboid as shown below:

There are 3 edges parallel to the red edge in the cube as shown:

Page 47

Altogether, 4 + 3 = 7 edges are parallel to the edge highlighted by the red line.
Hence, the correct answer is option c.
_____________________________________________________________________________________
3. In the following solid, every edge has at least one other edge parallel to it. How many different
pairs of PARALLEL FACES are present in the solid?

a) 8 b) 7 c) 6 d) 5
Answer: b
Solution:
The below image shows the three faces which are parallel to each other:

Hence, the pairs of parallel faces would be: (1,2) (1,3) (2,3)
The below images show the three faces that are parallel to each other:

Hence, the pairs of parallel faces would be: (4,5) (4,6) (5,6).
The below images show the two faces that are parallel to each other:

Hence, the pair of parallel faces would be: (7,8)
Since there are a total of 7 parallel pairs, the correct answer is option b.
_____________________________________________________________________________________
4. In a clock, the hour hand and the minute hand are perpendicular to each other, where the minute
hand is at 12. How many different possible positions the hands of the clock can have at this time?
a) 1 b) 2 c) 3 d) 4
Answer: b

Page 48

Solution:
If the minute hand is at 12, the hour hand can be at 3 or 9, so that the hands are perpendicular.
Hence, option b, 2 is the answer.
_____________________________________________________________________________________
5. If each of the given terms follows the same theme, what will come in place of “?”

a) 7 - 1 b) 2 - 8 c) 1 - 8 d) 5 - 11
Answer: b
Solution:
In each term, the hands of the clock form a straight line (180°). The pair of numbers written below the
clock represent the two positions the hands should point to, in order to form a line that is perpendicular
(90°) to the existing straight line.
Also, the numbers listed below are in the order: Smaller number - Larger number
For example, in the first term, when the hands point towards 3 and 9, the new line is perpendicular to
the existing line, as shown below:

Following this rule, in the question term, we must choose appropriate positions that form a
perpendicular line to the one formed by 5 and 11. As shown below, it is formed by 2 and 8.

Hence, the correct answer is option b.
_____________________________________________________________________________________
6. Figures A and B are formed using two transparent squares, each. If figure A is placed on top of
figure B (without rotation), such that it completely overlaps figure B, how many right-angled
triangles can be seen in the resultant image?

a) 10 b) 12 c) 13 d) 14
Answer: b

Page 49

Solution:
Both Figure A and Figure B are squares of the same dimensions, 4 cm × 4 cm.
Figure A is placed on Figure B without rotation.
As it is given that the figures are made using squares, all the angles of each square will be 90°. So, the
triangles formed at these corners are right angled triangles.
As shown in the given figure, a total of 12 right-angled triangles are formed when the two figures are
overlapped.
Therefore, the correct answer is option b.

_____________________________________________________________________________________
7. In triangle ABC, BC is the base.
Point P lies on AB and point Q lies on BC such that PQ is parallel to AC.
Point R lies on AB and point S lies on AC such that RS is parallel to BC.
Lines PQ and RS intersect at a point O inside the triangle.
Which of the following quadrilaterals must be a parallelogram?
a) QPAC b) ORBQ c) QOSC d) SAPO
Answer: c
Solution:
Since RS ∥ BC (RS is parallel to BC; ∥ means parallel to), any segment on RS is parallel to BC.
So, OS ∥ BC
Now look at QC.
Q lies on BC, so QC is part of BC.
Therefore, OS ∥ QC
First pair of opposite sides parallel.

Since PQ ∥ AC, any segment on PQ is parallel to AC.
So, OQ ∥ AC
Now look at SC.
S lies on AC, so SC is part of AC.
Therefore, OQ ∥ SC
Second pair of opposite sides parallel.

A quadrilateral is a parallelogram if both pairs of opposite sides are parallel.

Page 50

In quadrilateral QOSC:
OS ∥ QC
OQ ∥ SC
So, both pairs of opposite sides are parallel.
Therefore, QOSC is a parallelogram. Hence, the correct answer is option c.

[Please note that the above triangle is taken as a reference. This case is applicable to any type of
triangle ABC].
_____________________________________________________________________________________
8. Two lines intersect at point K.
One angle is (ax − 10)° and its vertically opposite angle is (bx + 20)°.
Which of the following is DEFINITELY true?
a) a = b b) a ≠ b c) x = 10 d) a + b = 0
Answer: b
Solution:
Since the given angles are vertically opposite, they must be equal.
So, ax − 10 = bx + 20
a x − b x = 30
(a − b)x = 30
So, x = 30 / (a − b)
Now check each option.
Option a: a = b
If a = b, then (a − b) = 0.
But then (a − b)x = 0, which cannot equal 30.
So, this is impossible.
Option b: a ≠ b
For (a − b)x to equal 30, (a − b) must not be zero.
So, a and b must be different.
This must always be true.
Option c: x = 10
If x = 10, then:
30/(a − b) = 10
So, a − b = 3.
This is possible for some values of a and b. But x could also be 30, 15, 6, etc., depending on (a − b).
So, x = 10 is possible, but not necessary.
Option d: a + b = 0
Since a and b are positive integers, their sum cannot be zero.
So, this is impossible.
Hence, option b is the correct answer.

Page 51

9. A computer draws two intersecting lines. One angle formed at the intersection is 70°, but
originally, it was intended to be 40°. Which of the following options shows the respective
‘Expected and Resultant’ measures of one of the other angles?
a) 140 degrees and 110 degrees b) 110 degrees and 140 degrees
c) 140 degrees and 170 degrees d) 170 degrees and 140 degrees
Answer: a
Solution:
One angle formed at the intersection is 70°, but originally, it was intended to be 40°.
So, Expected angle = 40 degrees and Resultant angle = 70 degrees.
Now, if an angle of 40 degrees was expected, the other angle would be 180 - 40 = 140 degrees.
Hence, the other expected angle = 140 degrees.
However, as the original angle resulted in 70 degrees, this makes the other angle 180 - 70 = 110
degrees.
So, Expected vs Resultant = 140 degrees - 110 degrees
Option a is the correct choice.
_____________________________________________________________________________________
10. ABCD is a square of side length 4 m. Raj starts from point A and walks North to reach point D.
From point D, he further walks 4 m without taking any turns and reaches point P. At P, he turns
135° clockwise and walks a certain distance and stops only when he reaches a point on the
square. Which of the following statements is INCORRECT with respect to the boundary of the
square and the path taken?

a) PC ∥ DB b) ∠PCD = ∠DCA c) PC ≠ AC d) ∠DPC = ∠DCP
Answer: c
Solution:
Given that Raj walks from A to D and then walks 4 m from D to reach P.

At P, he takes 135° clockwise turn.
A 90° clockwise turn from D makes him face towards East. As he turned 135° (90° + 45°) in clockwise
direction, he starts walking towards C, as shown.

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Hence, he stops when he reaches point C.
From the image, we can say that ∠DPC is 45°. (∠DPC = 180° - (90° + 45°))

Also, triangle PDC is a right angled triangle.
All three angles of this triangle add up to 180°.
∠DPC + ∠PDC + ∠PCD = 180°.
45° + 90° + ∠PCD = 180°.
∠PCD = 180° - 45° - 90°
∠PCD = 45°.
Therefore, ∠PCD = ∠DPC = 45°
Option d is true.
Let’s have a line joining points B and D.

Both PC and DB are the hypotenuses of congruent right-angled triangles (triangle PDC and triangle
DAB of sides 4m.)
Here, PC is parallel to DB and is of the same length as DB.
Option a is true.
Also, in the above case, we have seen that PC is of the same length as DB, which is also the diagonal
of the square ABCD.
So, PC should also be equal to another diagonal of the square, which is AC.
PC ≠ AC is false
Option c is INCORRECT.

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As AC is the diagonal of the square ABCD, it bisects the angle of the square.
∠DCA = 90°/2 = 45°
We already know that ∠PCD = 45°.
Thus, ∠DCA = ∠PCD
Option b is true.
Hence, option c is incorrect.
Thus, option c is the right answer.

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A, B, and C are all different numbers having any value among 1, 2, and 5. What will
come in place of C so that no two consecutive numbers are placed next to each
other in any direction (vertically, horizontally, or diagonally)?

(a) 2 (b) 1 (c) 5 (d) None
Answer: c
Solution:
To solve the question, we start by finding the value of A:
A is adjacent to 3 and 4. Since no two consecutive numbers can be adjacent, A cannot
be 2 (because 2 and 3 are consecutive) and A cannot be 5 (because 5 and 4 are
consecutive).
Therefore, the only possible value for A is 1.

1

Now, the remaining numbers for B and C are 2 and 5.
Next, we check the value of C:
C is adjacent to 3, and since 2 and 3 are consecutive numbers, C cannot be 2.
Hence, C must be 5.

5
1

Finally, the remaining number for B will be 2.

5
2 1

Hence, the correct answer is option c.

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Chapter 6: Number Play
Activity Time
Introduction
Solving puzzles always seems complex. But have you wondered how computers solve puzzles? We will
see a simplified way to solve some puzzles by following simple instructions and using the binary computer
language. This will give us insight into how we can break complex problems into smaller pieces that can
be easily solved. By combining smaller problems, computers can solve a larger one.
Activity Time Description

Launch 5 min The teacher will demonstrate by solving a puzzle quickly
using the cards, then ask students how it worked.
Template:
https://drive.google.com/file/d/10sRoRMg0KRobPqwhkrU
asJX9Ekvwnsz3/view

Exploring Logic 25 Students spend time understanding the cards.
punch card by min Then they start solving basic logical statements using the
Students cards.
They will then solve some puzzles with multiple
statements using the cards.
Students can make their own punch cards using the
template after class hours.
Student Worksheet:
https://docs.google.com/document/d/1RxupyjbvHrgOy26a
48e8cUMihTm5Ri46BFfxaSTt9IA/edit?tab=t.0

Discussions & 10 Discussing the worksheets.
Conclusion min
The students will continue their discussion of binary
numbers from the previous activity. They discuss their
understanding of the logical statements and how to
deduce from given informations
They can also discuss how a complex problem is broken
down into individual logical steps and combined to solve a
larger problem.

CT Components
Algorithmic Thinking
We follow a step-by-step process to search for the required card using the statements given in the
puzzle.
Decomposition
We break the task of solving a logic puzzle into solving individual steps using the binary logic of true or
false.

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Pattern Recognition
They observe that the number of cards possible is related to the number of holes at the top. They see
how true and false, on and off, and U-shaped and O-shaped holes are identical.
Generalisation
From solving a puzzle in 3-holed cards, they see the extension to 5-holed cards.
Logic
These puzzles require solving each step logically.

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Activity: Logical puzzle punch cards
In this activity, you will use punched cards to solve logical puzzles. By stacking, aligning, and filtering the
cards, we will see how complex reasoning can be built from simple decisions - True or False, mirroring
how binary logic works inside a computer.
The scenario is that you have the following lights in the room: Yellow Y, Pink P, Red R, Green G, and
Blue B. By following the clues given in the Case Cards, you have to figure out which lights are ON in the
given Puzzle. To solve these puzzles, we have a deck of cards representing all possible combinations of
these lights being ON or OFF.

Understanding the cards
To solve these puzzles, we have a deck of cards representing all possible combinations of these lights
being ON or OFF. Each card has five positions for holes on the top, and each position corresponds to a
coloured light mentioned above in the given order.
Each position has two possible hole types: U-shaped or O-shaped. Each card represents one possible
combination. If a light is ON in a card, then the hole corresponding to it is U-shaped. If a light is OFF in
a card, then the hole corresponding to it is O-shaped.
Eg: Yellow Y is OFF, Pink P is ON, Red R is ON, Green G is OFF, Blue B is OFF is the card where
the holes are O U U O O.
Each hole, thus, has 2 possible states. If we have two holes, then we have 2 × 2 = 4 possible states.
Likewise, if we have 3 holes, we have 2 × 2 × 2 = 8 states. Therefore, if we have 5 holes, we will have
2 × 2 × 2 × 2 × 2 = 32 states. That is the number of distinct cards we have.

Figure: All the possible three-hole states and how they come from the two-hole states

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Figure: Some possible cards with 5 binary digits

Procedure
If the puzzle says a light is ON, we need to remove all cards where that light is OFF. For that, we put the
stick into the deck of cards through the hole corresponding to that light and pull out all the cards that get
stuck to it. We remove these cards because they got stuck due to O-shaped holes that indicate the light
is OFF.
If we know some light is OFF in the puzzle, we remove all the cards that do not come out with the stick;
these are the cards with U-shaped holes in positions corresponding to the given colour, which indicate
that the light is ON.
Ensuring the above helps us solve the logical puzzles through punch cards.
Let’s get familiar with the cards and the rules.

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