CTAI Teacher Handbook Class 5

Computational
Thinking

Class 5

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. NCF-SE 23 carries this recommendation further implementation.

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

The book challenges learners with problems involving multi-step deduction, hidden
information, condition-based reasoning and transformations of numbers, shapes and
arrangements. Students engage with tasks that require interpreting constraints,
breaking complex situations into manageable parts and applying sequential logic to
arrive at justified solutions. The document also provides pedagogy, learning
resources, assessment support and classroom implementation guidelines to support
structured, 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 British Institute (Business Management and
Advertising), Frameworks Mumbai (Master’s in Computer Science), and Workstation (3D
Animation)), Field Expert

Material Production Group

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

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

Page 3

TABLE OF CONTENTS

SR. NO. CHAPTER PAGE
NO.

Introduction 5
1.
How to Use This Book? 8
2.
We the Travellers – I 9
3.
Fractions 15
4.
Angles as Turns 21
5.
We the Travellers – II 28
6.
Far and Near 35
7.
The Dairy Farm 41
8.
Shapes and Patterns 48
9.
Weight and Capacity 56
10.
Coconut Farm 68
11.
Symmetrical Designs 74
12.
Grandmother’s Quilt 82
13.
Racing Seconds 90
14.
Animal Jumps 95
15.
Maps and Locations 101
16.
Data Through Pictures 108
17.

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

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

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

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

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

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1. Relevance: Importance of introducing Computational Thinking (CT) and Artificial
Intelligence (AI)
Introducing CT and AI in Grade 5 is critical for building a foundation for the technology-enabled society
envisioned by national policies.
• Foundation for AI: CT is the intellectual backbone and cognitive framework required to understand
how intelligent systems operate. Skills like breaking problems into parts and spotting patterns are the
same reasoning processes that power AI and Machine Learning.
• AI-Readiness: The curriculum aims to develop AI-ready learners by building capacities in logical
thinking and problem-solving, helping them understand the role of AI in daily life.
• Holistic Development: Beyond technical skills, CT contributes to creative problem-solving, critical
thinking and ethical decision-making, which are essential for individual flourishing and responsible
digital citizenship.
• Interdisciplinary Connection: It helps students see that knowledge is not compartmentalised by
connecting disciplines like Mathematics, Science, and Humanities

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

3. Learning Outcomes:

ABSTRACT THINKING -
Students will be able to solve complex problems with multi-layered hidden cues, using:
• Different viewpoints of 3D objects.
• Changes in shapes after flips, turns, cuts/folds, or rotations, and changes in order and
directions (clockwise or counterclockwise).
• Hidden or missing parts in incomplete shapes or patterns.
• Mirror/Water images and identical halves based on symmetry.

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

DECOMPOSITION -
Students will be able to break down higher-order problems involving interconnected
clues, using information from:
• Number clues (place values, sum/difference/product)
• 3D objects and their parts (faces, edges, corners)

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• Step-by-step exchanges or transfers (money, objects, digits, quantities)
• Tables or charts with multiple pieces of information
• Conditions for counting/grouping/sorting items
• Pictures or visuals that represent certain numerical values

ALGORITHMIC THINKING
Students will be able to follow multi-layered rules to solve advanced problems involving:
• Number sequences formed using simple operations
• Movements on grids or direction-based paths
• Values that increase or decrease across steps
• Multi-step instructions involving moves, changes, transfers, swaps
• People/Events arranged in an order using attributes or chronological clues
• Simple counting instructions

4. Mapped with NEP and NCF 2023:
The curriculum is built upon the vision of the National Education Policy (NEP) 2020 and is directly aligned
with the National Curriculum Framework for School Education (NCF-SE) 2023.
• NEP 2020 Vision: It fulfils the goal of making India a global leader in emerging domains like AI and
machine learning by integrating them into school education.
• NCF-SE 2023 Alignment: The learning standards (Goals, Competencies, Outcomes) are derived
from the framework suggested in the National Curriculum Framework for School Education 2023
• Global Leadership: It fulfills the NEP goal of positioning India as a global leader in emerging domains
like AI and Machine Learning by integrating these topics early in school education.

5. Time Allocation
• Annual Hours: A total of 50 hours annually is suggested for the Preparatory Stage (Classes 3–5)
• Integrated Model: To ensure balance without overburdening students, this time it is not added as an
extra subject but is integrated into Mathematics and "The World Around Us" (TWAU) periods.
6. Approach / Pedagogy
The pedagogical approach for Grade 5 is designed to be experiential and activity-based:
• Hands-on Learning: Use of games, puzzles and interactive worksheets to teach systematic problem-
solving.
• Collaborative Tasks: Learning involves peer discussions and group work, allowing students to solve
problems collectively using structured resources.
• Systematic Decomposition: Teachers guide students to break larger, complex problems into smaller,
manageable parts.

7. Assessment:
Assessment at this stage shifts from rote memorisation to formative and competency-based
evaluation:
• Varied Methods: Evaluation tools include:
• Written tests specifically involving CT-focused puzzles
• Interactive group
• Qualitative tracking through a Teacher Observation Journal
• Creative Focus: The primary goal is to assess a student's ability to apply knowledge and
demonstrate creativity in finding solutions

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

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

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

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

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

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Chapter 1: We the Travellers - I
1. Which of the following CAN be implied if the given statement is true?
Statement: A 5-digit number is formed using single-digit whole numbers such that the units digit
is even and all the remaining digits are odd.
a) The number definitely lies between 11111 and 99999
b) The number formed CAN be a multiple of 100
c) The sum of the digits of the largest such number is 44
d) The sum of the digits is NOT divisible by 2
Answer: c
The statement says that a 5-digit number has an even digit in its units place and odd digits in all the other
places.
So, it’s basically an even number.
Now, let’s check the possibility of each of the options being true.
Option a: The number definitely lies between 11111 and 99999.
We know that our 5-digit number has odd digits in all other remaining places other than the units place.
11110 also satisfies this condition.
However, 11110 < 11111 and this number does not lie between 11111 and 99999.
So, the statement given in option a is False.
Option b: The number CAN be a multiple of 100.
5-digit multiples of 100 must have two 0s in the end, i.e. in units and tens place.
As our number has an even digit only in the units place and other places have odd digits, the statement
given in option b is False.
Option c: The sum of the digits of the largest such number is 44.
The largest number formed with the given condition will be 99998.
9 + 9 + 9 + 9 + 8 = 44
Hence, option c can be implied.
Option d: The sum of the digits is NOT divisible by 2.
As we know odd + odd = even, we have 4 odd digits which result in an even sum.
And when the sum is added to the even digit, the resulting sum is an even number.
So, the statement given in option d is False.
Hence, option c is the correct answer.
2. Rohan revealed his 5-digit phone password in a code word to Rajesh where all the digits are
NATURAL NUMBERS. Which of the following statements are SUFFICIENT to crack the code?
Statement 1: One of these digits is 1, and 2 appears thrice in the password
Statement 2: The first digit is smaller than all other digits and the 2nd digit is smaller than the 3rd
Statement 3: The largest digit in the password is 3 and it appears only once
a) Statements 1 and 2
b) Statements 2 and 3
c) All the statements are required
d) Cannot be determined even after using all the statements
Answer: b
From the information provided in statement 3, we deduce that the largest digit in the password is 3, and
it occurs only once. Statement 2 indicates that the first digit is smaller than all others, implying that the

Page 9

first digit is 1. Further, it states that the second digit is smaller than the third, making the second digit 2,
which is less than 3.
Consequently, the first three digits of the password are 123. Since the highest digit is 3 and appears only
once, the remaining two digits must be either 1 or 2.
However, statement 2 emphasizes that the first digit is smaller than all others, eliminating the possibility
of the remaining digits being 1. Therefore, the password is determined as 12322. Hence, only statements
2 and 3 are necessary to arrive at the correct password. Thus, the correct answer is option b.
3. What will come in place of "?"

a) b)

c) d)
Answer: c
Here, the terms on the left of the arrow are 5-digit numbers. On the right, the digits are arranged in
ascending order. The digits that remain in the same position with respect to the original number are
placed in the black circles and the remaining digits are placed in the white squares.
Based on the same rule, for the last term (8 2 1 5 4), arranging the digits in ascending order gives
1 2 4 5 8. Comparing positions with the original number, 2 and 5 remain in the same positions, so they
are placed in black circles, and the rest in white squares, which matches option c. Hence, the correct
answer is option c.
4. A 5-digit number is formed such that all the digits are different, and the sum of all the digits is 10.
What is the highest digit this number can have?
a) 3 b) 4 c) 5 d) 6
Answer: b
To address this question, we need to form a 5-digit number with unique digits and the sum of all its digits
is 10. If we use the highest single-digit numbers like 9, 8, and 7, the sum of all five digits will exceed 10.
By taking 6, we can have a sum of 10 as 6 + 1 + 1 + 1 + 1. However, we will not have unique digits. (The
same issue exists with 5 as well).
Therefore, we must have smaller digits.
Let’s consider 0, 1, 2, 3, and 4.
By selecting these distinct digits, we can create a 5-digit number where the sum of its digits equals 10.
(4 + 3 + 2 + 1 + 0 = 10)
In this context, since we are searching for the highest digit, the maximum value we can use is 4.
Therefore, the highest digit for the 5-digit number is '4'. Hence, the correct answer is option b.

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5. Using the instructions given below, form two numbers.
Instruction 1: Form the largest 5-digit number using the digits in the 5-sided figures, without
repeating any digit.
Instruction 2: Form the smallest 3-digit number using the digits in the circles, without repeating
any digit.
What is the sum of the two numbers formed?

a) 99394 b) 93394 c) 98743 d) 98848
Answer: d
Numbers in five-sided shapes are:

The largest five-digit number formed using these digits: 98743.
Numbers in circles are:

The smallest three-digit number formed using these digits: 105.
Thus, the sum of both the numbers formed is: 98743 + 105 = 98848.
Hence, the correct answer is option d.
6. Using the digits at the ends of the vertical lines, form numbers by reading the digits from left to
right without rearranging them. Which option represents the greatest even number that can be
formed?

a) b) c) d)
Answer: d

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As shown in the image given below, option d makes the largest even number by reading the digits from
left to right without rearranging them, which is 78,946. Hence, the correct answer is option d.

7. A string of digits is given below. At MINIMUM, how many pairs of digits (may or may not be
adjacent to each other) must be interchanged such that the second largest 5-digit number is
formed from left to right?
Note: A digit can change its position any number of times

a) 2 b) 3 c) 4 d) 5
Answer: b
The digits given are 0, 7, 3, 6, and 5.
The largest 5-digit number that can be formed using these digits is 76530 (arranging the digits in
descending order).
To get the second largest 5-digit number, the digits in the higher place values must remain the
same. Any change in these places would make the number much smaller.
Therefore, we only interchange the last two digits (tens and ones place).
Thus, the second largest 5-digit number formed is 76503.
Now, we find the minimum number of digit swaps required to change 07365 - 76503.

Here, we can see that interchanging 3 and 5 will make both these digits have their target positions. As
one interchange sorts two place values, let’s swap 3 and 5.
Interchange 1: 3 and 5

Now, as all three remaining digits are in wrong positions, interchanging any two of them will sort only one
place value in any possible case (interchanging 0-6, 0-7, or 6-7 will give only one correct position in each
pair). For example, let’s interchange 6 and 7.
Interchange 2: 6 and 7

Then, only 6 will get its target position. (The same will be observed among other pairs too.)
Finally, interchanging the other two digits will give us the target value.
Interchange 3: 0 and 7

Hence, the minimum number of interchanges required is 3. Thus, the correct answer is option b.

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8. There is a 7-digit number X in which all the digits are different. Using the digits of X, two 5-digit
numbers Y and Z are formed, such that no digit is repeated within Y or within Z. What is the
MINIMUM number of digits that Y and Z must have in common?
a) 1 b) 2 c) 3 d) 4
Answer: c
Let's assume the 7-digit number X is ABCDEFG.
Since we need the least (MINIMUM) number of digits to be common, we will minimize the use of repeated
digits such that Y could be ABCDE and Z could be CDEFG.
As we can see, C, D, and E are there in both numbers Y and Z.
Thus, Y and Z will have at least three same digits.
Hence, the correct answer is option c.
9. A 5-digit number has the property that the number formed by its first three digits is the same as
the number formed by its last three digits (without changing the order of digits). What is the
minimum number of times any digit can appear in this number?
a) 3 b) 1 c) 2 d) 5
Answer: c
The number has 5 digits, and the first three digits must be exactly the same as the last three
digits.
However, a 5-digit number cannot contain six separate digits. Therefore, the middle digit (3rd digit)
becomes common to both groups of three digits. In other words, this digit is counted in both the first
three and the last three digits.
Because of this overlap, the shortest repeating pattern that satisfies the condition is ABABA, where
A and B are different digits.
In this pattern:
• A appears 3 times
• B appears 2 times
Thus, the minimum number of times a digit appears is 2.
The number cannot contain more than two different digits, otherwise the first three and last three
digits would not remain identical.
If the number used only one digit (for example, AAAAA), then a digit would appear 5 times, which is
not the minimum possible case.
Therefore, the valid pattern is ABABA, where the least number of times a digit appears is 2.
Hence, the correct answer is option c.
10. Which digit appears in the tens place of the second largest 5-digit number formed using 1, 5, and
7, where each digit is used at least once?
a) 5 b) 7 c) 1 d) Either a or c
Answer: c
The largest 5-digit number formed using 1, 5, and 7, where each digit is used at least once is 77751.
The second largest 5-digit number formed will be 77715.
So, 1 appears in the tens place. Hence, the correct answer is option c.

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P, Q, R, and S each has a different shape - Circle, Pentagon, Square, and Triangle
(not necessarily in the same order). They arranged the shapes as shown below,
where the Circle is placed at the bottom-most position.
• S placed his shape immediately above P's shape
• There are exactly 2 shapes placed above R's shape
Which of the following is Q's shape?

(a) Circle (b) Square (c) Pentagon (d) Triangle
Answer: A
Step 1: Identify the order in which the shapes are placed.
As mentioned in the question, the Circle is at the bottom-most position.
Clearly, the Pentagon is the topmost shape, as no other shape is overlapping it.
Between the Square and Triangle, the Triangle is on top of the Square, as the upper
corner of the Triangle is clearly overlapping the Square.
Hence, the order of the shapes from top to bottom is:
Pentagon - Triangle - Square - Circle.

Step 2: Based on the information provided, assign each shape to the respective
person.
S placed his shape immediately above P's shape.
Pentagon cannot be P’s shape, as there would be no scope for the shape of S above it.
So, P’s shape can either be Circle, Square or Triangle.
There are exactly 2 shapes placed above R's shape.
As per our order, the Square has two shapes above it.
Hence, R’s shape = Square.

So, P’s shape cannot be the Square too.
Also, if P’s shape is Circle, then the Square must be the shape of S, which is not
possible.

Thus, P’s shape = Triangle and the Pentagon (above the triangle) must be the shape
of S.

Therefore, Circle belongs to Q. Option a is the correct answer.

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Chapter 2: Fractions
1. Sam has some chocolates. He gave 1 more than half the chocolates to A and 1 less than half the
chocolates (of the original number of chocolates) to B. What is the difference between the number
of chocolates given to A and B?
a) 0 b) 1
c) 2 d) Cannot be determined
Answer: c
Let’s assume that Sam has a ‘2x’ number of chocolates.
So, half of the number of chocolates with Sam will be (2x)/2 = x.
Sam has given 1 more than half the chocolates to A.
So, he has given ‘x + 1’ chocolates to A ------(1)
Sam has given 1 less than half the chocolates to B.
So, he has given ‘x – 1’ chocolates to B --------(2)
Now, we need to find out the difference between the number of chocolates given to A and B.
From (1) and (2), the difference would be:
(x + 1) – (x – 1)
=x+1–x+1
=2
Hence, option c is the correct answer.
2. Three groups of people - A, B, and C - each have the same number of members. They take part
in three activities: 1, 2, and 3.
- Each person within a group can only be part of one activity
- Half of Group A does Activity 1, which is also done by all members of Group B
- Half of Group C does Activity 3, which is also done by the remaining members of Group A
- All remaining people do Activity 2
Which activity is done by the least number of people?
a) Activity 1 b) Activity 2
c) Activity 3 d) Both options a and b
Answer: b
Let activity 1 be named as A1, activity 2 be named as A2, and activity 3 be named as A3.
1. Half of group A does activity 1 which is also done by all the people in group B. Therefore,
Group A: (1/2) A1 + ?
Group B: A1
Group C: ?
2. Half of group C does activity 3 which is also done by the remaining people of group A.
Group A: (1/2) A1 + (1/2) A3
Group B: A1
Group C: (1/2) A3 + ?
3. Remaining people are doing activity 2
Group A: (1/2) A1 + (1/2) A3
Group B: A1
Group C: (1/2) A3 + (1/2) A2
Hence, we can see that Activity 2 is done by the least number of people. Therefore, the correct answer
is option b.

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3. If $ = 12, @ = half of 12, and # = quarter of 12, then find the value of:
$x@–#
a) 68 b) 69 c) 59 d) 36
Answer: b
It is given that $ = 12,
@ = half of 12 which is 12/2 = 6.
# = quarter of 12 which is 12/4 = 3.
On substituting the values:
$x@–#
= 12 x 6 – 3
= 72 – 3
= 69
Hence, the correct answer is option b.
4. What fraction of the total area of all the circles is shaded grey?

a) 10/24 b) 5/6 c) 5/24 d) 14/24
Answer: a
In the question image, we can see that the smallest portion of a shaded circle is a quarter of a circle.
So, it is easy for us to consider that each circle has four quarters and there are six circles. So, 6 x 4 = 24
quarters are present in the image, out of which 10 quarters are shaded.
Hence, 10 out of 24 quarters are shaded. 10/24 is the answer.
Option a is correct.

5. A piece of land is distributed between Raj and Rahul. Raj receives 1/2 of the total land, and Rahul
receives land equal to 1/2 of what Raj receives. What fraction of the land remains undistributed?
a) 1/2 b) 1/8
c) 1/4 d) The land is distributed completely
Answer: c
Raj receives 1/2 of the total land, so the remaining land is also 1/2.
Rahul receives land equal to 1/2 of what Raj receives, which is equivalent to 1/2 of 1/2 = 1/4.
So, Rahul receives 1/4 of the total land.
Since 1/4 is half of 1/2, the distributed land from the remaining portion becomes 1/4, and the other 1/4
is left.
Therefore, 1/4 of the total land remains undistributed, and the correct answer is option c.

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6. If each row follows the same theme, what will come in place of A and B?

a) A = 8, B = 12 b) A = 2, B = 3 c) A = 3, B = 2 d) A = 1, B = 6
Answer: b
In every row, the term written in the black box is the simplest form of the fraction that comes after the first
three terms of the row.
For example, in the first row, the fraction that comes after 1/6, 2/6 and 3/6 is 4/6.
The simplest form of 4/6 is 2/3.
Similarly, in the last row, the fraction that comes after 5/12, 6/12 and 7/12 is 8/12.
The simplest form of 8/12 is 2/3.
Hence, A/B = 2/3.
A = 2 and B = 3
Hence, the correct answer is option b.
7. Which of these options can we conclude from the following statement?
Statement: Three-quarters of total coins that Kiran has are of denomination five rupees.
a) All coins that Kiran has are of more than 2 rupees
b) All coins that Kiran has are of less than 10 rupees
c) Only one-fourth of the coins with Kiran are not of five rupees
d) Kiran has 30 coins of denomination five rupees
Answer: c
a) All coins that Kiran has are of more than 2 rupees
This tells us the value of the coins but does not tell us how many are five-rupee coins.
So, we cannot conclude anything about three-quarters being five-rupee coins.
Therefore, this option cannot be concluded from the given statement.
b) All coins that Kiran has are of less than 10 rupees
Again, this tells us only the range of coin values.
It does not say how many five-rupee coins Kiran has.
Therefore, this option cannot be concluded from the given statement.
c) Only one-fourth of the coins with Kiran are not of five rupees
If one-fourth are not five-rupee coins, the remaining three-fourths MUST be five-rupee coins.
For example, if Kiran has 40 coins:
● One-fourth (10 coins) are not of 5 rupees
● The remaining three-fourths (30 coins) are of 5 rupees
So, only option c directly tells us that three-quarters of Kiran’s coins are of five rupees.
Therefore, this option can be concluded from the given statement.
d) Kiran has 30 coins of denomination five rupees
We know only the number of five-rupee coins, but we do not know the total number of coins Kiran has.
Without the total, we cannot check whether 30 coins are three-fourths or not.

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Therefore, this option cannot be concluded from the given statement.
Thus, option c is the correct answer.
8. Sam makes three cuts on the same apple, as shown by the dotted lines in the image below.
Altogether, how many pieces of apple will he have?

a) 6 b) 7 c) 8 d) 9
Answer: c
Cut 1: When the apple is cut along the vertical dotted line by cut 1, we get 2 pieces as shown below:

Cut 2: As the apple has already been cut into 2 pieces by cut 1, when the apple is further cut along the
horizontal dotted line, we get 4 pieces altogether.

Cut 3: Then, cut 3 will result in double the existing number of pieces, where the upper half has 4 pieces,
and the lower half has 4 pieces.

In this way, the apple gets divided into 8 pieces in 3 cuts.
Hence, option c is correct.
9. If you take 6 times of me, you will get a whole and a half. If you take me out of a whole, you are
left with three times of me. Who am I?
a) Half b) Quarter c) Whole d) Zero
Answer: b
A whole and a half mean 1 and ½.
As we have 1 and ½, and we know that a whole has 2 halves, then we have three halves altogether
(whole and a half = ½ + ½ + ½)

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Now imagine dividing these three halves into 6 equal small pieces.
So, if you break 1½ into 6 equal parts, each small part will be one-fourth (¼). (As a quarter is nothing
but half of half).
Let’s check the answer by comparing it with the next statement, too.
The second clue says that when this number is taken out from a whole, the remaining part equals three
times the same number.
This means a whole is made up of four equal parts, because removing one part leaves three equal parts
behind. Therefore, the number is one-fourth (¼).
Option b is correct.
10. In a chocolate shop, four buyers come in one by one to buy chocolates. Each buyer purchases
half of the chocolates that are currently in the shop. For example, if there are X chocolates initially,
the first buyer buys X/2 chocolates. Then, from the remaining chocolates, the second buyer buys
half. This process continues until the fourth buyer, who buys only 1 chocolate. How many
chocolates were there in the shop initially?
a) 8 b) 32 c) 16 d) 20
Answer: c
If there are X chocolates in the shop initially, the buying process proceeds as follows:
The 1st buyer buys X/2 chocolates.
The 2nd buyer buys half of the remaining chocolates, which is X/4.
The 3rd buyer buys half of the chocolates left after the second buyer, which is X/8.
The 4th buyer buys half of the chocolates remaining after the third buyer, which is X/16.
It is given that the 4th buyer buys only 1 chocolate. Therefore, we can set up the equation:
X/16 = 1
X/16 = 1/1 = 16/16
X = 16
Hence, the correct answer is option c.

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Given below is a set of 8 cards, each having a WHOLE NUMBER on it. Cards A,
B, C, and D represent numbers that are either one more or one less than the
number on the card directly above them. If all 8 numbers on these cards are
different, what is the sum of B and C?

(a) 12 (b) 10
(c) 11 (d) Cannot be determined
Answer: A
It is mentioned that all 8 cards have different digits and 1, 4, 6, and 3 are already
present in the first row.
Hence, A, B, C, and D can take unique values from among 0, 2, 5, 7, 8, and 9 only.

Now, card A has 1 above it. As it is mentioned that each letter represents a value
either 1 more than or 1 less than the number above it, A can either be 0 or 2.

However, if A is 2, then D, which is below 3 can represent only 4. This is not possible
as 4 is already present.

Hence, A represents 0 and D represents 2.

Now, B, which is below 4 must represent only 5, as 3 is already present.
As a result, the value of C will be 7, as C has 6 above it and can represent only 5 or 7
as per the condition.

Clearly, B = 5 and C = 7.
Hence, B + C = 5 + 7 = 12, and option a is the correct answer.

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Chapter 3: Angles as Turns
1. P and Q are standing in a playground. P took a right-angle turn in the clockwise direction and Q
took a right-angle turn in the anti-clockwise direction, and now both of them face the same
direction. If P is currently facing North, which direction was Q initially facing?
a) East b) West
c) South d) Cannot be determined
Answer: a
P turned by a right angle in clockwise direction, and Q turned by a right angle in anti-clockwise direction
to face the same direction.
As P is currently facing North, Q must also be facing North.

So, Q turned by a right angle in an anti-clockwise direction to face the North.
We know that 1 right angle turn in anti-clockwise direction from East gives us North.

So, Q must be initially facing towards the East.
Option a is the correct choice.
2. Two arrows must pass through five blocks to reach the opposite side, making a right-angle
clockwise turn each time they enter a block. Which of the following options shows the correct
positions of these arrows when they are present in the shaded block?

a) b) c) d)
Answer: a
Each arrow rotates by a right angle in the clockwise direction each time it enters a block.
The positions of these arrows as they travel through the blocks are shown in the image below:

As we can see, when the arrows are in a grey block, they would look like the ones shown in option a.
Hence, option a is the correct answer.

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3. Prachi is facing North. She takes a right-angle turn in the clockwise direction. She then takes a
right-angle turn again in the clockwise direction. What direction is she facing now?
a) North b) East c) West d) South
Answer: d
First, Prachi took a right-angle turn in the clockwise direction, which means she turned right from the
north.

Then, Prachi took a right-angle turn in the clockwise direction from the east.

In the end, Prachi took two right turns and ended up facing south.
Hence, the correct answer is option d.
4. The given question image could be rotated clockwise or anticlockwise multiple times. Which of
the following options will you NOT get on rotating the given question image?

a) b) c) d)
Answer: b
Image in option b cannot be obtained on rotating the question image.
Hence, the correct answer is option b.

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5. Sam rotates Image 2 by a right-angle clockwise turn. If Tom places the rotated Image 2 on Image
1, which of the following statements is NOT true?

a) A and E overlap with each other b) 9 and 8 are opposite to each other
c) F and B are diagonal to each other d) 9 and 1 overlap with each other
Answer: d
Let’s first rotate image 2 by a right angle, clockwise.
The rotated version can be seen below:

Now, check each option one by one.
Compare the resultant positions of the letters/digits when rotated image 2 is placed on top of image 1.
Option a:
As we can see, A and E overlap with each other, as they’re in the same positions on both the grids.

Thus, option a is TRUE.
Option b:
As we can see, 9 is on the top edge of image 1 and 8 is on the bottom edge of image 2 (after rotating
image 2).
So, 9 and 8 will be opposite to each other, after placing image 2 on top of image 1.

Thus, option b is TRUE.
Option c:

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As we can see, B is at the top-right corner of image 1 and F is at the bottom-left corner of image 2.
Hence, F and B will be diagonal to each other, in the final arrangement.

Thus, option c is TRUE.
Option d:
As we can see, 9 and 1 are at different positions in their respective images.
So, they DO NOT overlap with each other, finally.

Thus, option d is NOT TRUE.
Hence, option d is the correct answer.
6. Pratik is facing north. He first makes a 1/4 clockwise turn, followed by a 1/2 anti-clockwise turn.
In which direction was he facing at the end?
a) North b) East c) West d) South
Answer: c
A full turn means turning all the way around, which is 4 right angles (360°).
So, we divide the full turn into 4 equal parts.
• 1/4 turn = one right-angle turn
• 1/2 turn = two right-angle turns
Now apply this to the question:
• 1/4 clockwise turn means one right-angle turn.
• 1/2 anticlockwise turn means two left-angle turns.
Now look at the turns together:
• 1 right-angle turn + 2 left-angle turns
One right-angle turn and one left-angle turn cancel each other, so we are left with one left-angle
turn. Starting from North, one left-angle turn will make Pratik face West. Hence, option c is the correct
answer.

Page 24

7. A, B, C, D, E, and F are seated around a table, facing the centre, with two chairs left empty, as
shown below. Two new members join the group. E, C, and B move one seat clockwise, while D
and F move one seat anticlockwise. After these changes, the new members occupy the empty
chairs. Who is sitting between the two new members?

a) A b) B c) E d) F
Answer: a
E, C, and B move one seat clockwise. The changes in position are shown below:

D and F move one seat anticlockwise. The changes in position are shown below:

Finally, the empty seats would be available on either side of A, as shown in the image above.
Thus, A will be sitting between the new members. Hence, the correct answer is option a.
8. Rohan and Kamini are facing East. Rohan takes 2 quarter turns in clockwise direction and Kamini
takes 2 quarter turns in anti-clockwise direction and then Kamini takes 1 quarter turn in clockwise
direction. Which of the following will lead Kamini and Rohan to face opposite directions?
a) Rohan taking 1 quarter turn in a clockwise direction
b) Kamini taking 1 quarter turn in a clockwise direction
c) Rohan taking 2 quarter turns in an anti-clockwise direction
d) Both options a and b
Answer: b
Breaking it down step by step:
• Initially, Rohan and Kamini both face East.
• Rohan takes 2 quarter turns in clockwise direction, positioning himself facing West.
• Kamini, on the other hand, takes 2 quarter turns in anti-clockwise direction and 1 quarter turn in
clockwise direction. However, the anti-clockwise and clockwise turns cancel each other out, leaving
her with only 1 quarter turn in anti-clockwise direction, resulting in her facing North.
• For Kamini and Rohan to face in opposite directions, Kamini needs to take one more quarter turn in
the clockwise direction.
Hence, the correct answer is option b.

Page 25

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

a) b) c) d)
Answer: d
In the given series, each term is formed by rotating the previous term by a right angle in the clockwise
direction.

Following this logic, the answer will be option d, as shown in the image above.
10. If both the blue boxes shift to the next corner in clockwise direction, how will the image look after
the shift?

a) b) c) d)
Answer: d
As shown below, when both blue boxes move to the next corner in the clockwise direction, the figure will
look like option d. Thus, option d is the correct answer.

Page 26

In the grid given below, colour some of the empty blocks such that each bird has
exactly one coloured block adjacent to it. What is the MINIMUM number of blocks
that need to be coloured?
Note: For two blocks to be adjacent they MUST share one common side. Blocks
having only a common corner are not adjacent

(a) 2 (b) 3 (c) 4 (d) 5
Answer: A
We need to find the minimum number of empty blocks that need to be coloured so that
every bird has exactly one adjacent coloured block.

On colouring the above highlighted block, we will be able to satisfy the condition for
three birds, which is the highest possible number of birds, with just one block. Hence
this block will be coloured.

The birds in the first two rows each need one coloured block adjacent to them. They
share two common adjacent blocks, as shown. By colouring either of them, the
remaining birds will each have one adjacent coloured block. As shown below, only two
blocks need to be coloured for every bird to have exactly one coloured block next to it.
1

2

Hence, option a is correct.

Page 27

Chapter 4: We the Travellers - II
1. Which statement is DEFINITELY true about a five-digit number that consists of three odd digits
and two even digits?
a) The sum of all the 5 digits is an odd number
b) The sum of all the 5 digits is an even number
c) The sum of all the 5 digits is a prime number
d) Cannot be determined
Answer: a
Consider a five-digit number consisting of three odd digits and two even digits. Expressing this, we get:
Odd + Odd + Odd + Even + Even.
We know that the sum of two odd numbers is even. Therefore, the expression simplifies to:
Even + Odd + Even + Even.
Now, adding even numbers together always results in an even sum (even + even + even = even).
Therefore, the expression further simplifies to: Even + Odd = Odd.
This proves that the sum of the digits of a five-digit number that consists of three odd digits and two even
digits is an odd number.
To illustrate, take the example of the digits 1, 2, 3, 4, and 5. The sum of these digits (1 + 2 + 3 + 4 + 5) is
15, which is indeed an odd number.
Hence, the correct answer is option a.
2. A has 5 gold coins and B has 7 silver coins. B can exchange 2 silver coins for 1 gold coin with A.
If B exchanges as many coins as possible, what will be the total number of coins (both gold and
silver) with A at the end?
a) 8 b) 6 c) 5 d) 2
Answer: a
A has 5 gold coins, and B has 7 silver coins. B can exchange 2 silver coins for 1 gold coin with A. This
means:
• 1 gold coin = 2 silver coins
B wants to exchange as many silver coins as possible. Since B has 7 silver coins and can only exchange
them in multiples of two, the maximum he can exchange is 6 silver coins. For these 6 silver coins, B will
receive:
• 6 silver coins ÷ 2 = 3 gold coins
So, B gives 6 silver coins to A and receives 3 gold coins in return.
Final count of coins with A:
• Gold coins: A initially had 5 gold coins but gave 3 to B, so A now has:
5 - 3 = 2 gold coins
• Silver coins: A receives 6 silver coins from B, so A now has:
0 + 6 = 6 silver coins
Total coins with A:
• Total coins = 2 gold coins + 6 silver coins = 8 coins
Hence, the total number of coins with A at the end is 8. Thus, the correct answer is option a.

Page 28

3. In how many different ways can you fill in all the shaded cells such that the numbers in the column
are arranged in descending order and each cell has a unique number?

a) 2 b) 3 c) 4 d) 5
Answer: c
The numbers in the column should be arranged in descending order.
• The number at the top is 30, and the number at the bottom is 25
• The possible numbers that can fill the 3 empty cells are 26, 27, 28, and 29
The possible arrangements of these numbers in the shaded cells are shown below:

The number of possible arrangements is 4. Hence, the correct answer is option c.
4. John has Rs. 12 with him. He went to a shop and bought 3 balls and 2 balloons. After buying these
items, he still has Rs. 2 remaining, which is exactly the cost of 1 ball. What is the cost of 1 balloon?
a) Rs. 1 b) Rs. 2 c) Rs. 3 d) Rs. 4
Answer: b
Out of the Rs. 12, he has only Rs. 2 left with him, which means he spent Rs. 10.
The cost of 1 ball is Rs. 2. If 1 ball costs Rs. 2, then 3 balls would have cost Rs. 6.
So, the remaining Rs. 4, he spent on buying 2 balloons.
This means that each balloon costs Rs. 2.
Hence, the correct answer is option b.
5. If you add 1 to the digit in the ones place and subtract 2 from the digit in the tens place in the
numbers given below, how many of the new numbers will be even?
4434, 2655, 3561, 1636, 1253, 8094, 1371, 3038, 2335, 1045
a) 5 b) 6 c) 7 d) 8
Answer: b
We are told to add 1 to the digit in the ones place and subtract 2 from the digit in the tens place
for each number. However, to check whether the resulting number is even, we only need to look at the
final units digit, because a number is even if its units digit is even.
Since we add 1 to every units digit:

Page 29

● If the original units digit is odd, then odd + 1 = even
● If the original units digit is even, then even + 1 = odd
So, only numbers that originally end in an odd digit will become even after the change.
Step 1: Identify numbers ending in an odd digit
From the given list:
4434 - ends in 4 (even)
2655 - ends in 5 (odd)
3561 - ends in 1 (odd)
1636 - ends in 6 (even)
1253 - ends in 3 (odd)
8094 - ends in 4 (even)
1371 - ends in 1 (odd)
3038 - ends in 8 (even)
2335 - ends in 5 (odd)
1045 - ends in 5 (odd)
Step 2: Count them
Numbers ending in an odd digit:
2655
3561
1253
1371
2335
1045
Total = 6 numbers
So, the answer is option b.
6. You have to place 4 numbers from 1, 2, 3, 4, and 5 in the squares below to satisfy the given
comparisons. How many different arrangements of numbers are possible if box B contains
number 1? Note: You cannot use the same number twice in the same arrangement

a) 8 b) 9 c) 11 d) 12
Answer: d
We are told:
The numbers must satisfy: A > B < C < D
● B contains 1, so we only have 2, 3, 4, 5 left for A, C, and D
● We must place the remaining three numbers in A, C, and D such that:
1. A > 1
2. C > 1
3. D > C
Since every number (2, 3, 4, 5) is already greater than 1, the only real condition we must satisfy is: C <
D
So, we must count how many ordered pairs (C, D) can be made from {2, 3, 4, 5} where C < D.
Then the remaining number automatically goes to A.
Step 1: List all valid (C, D) pairs where C < D
Using the set {2, 3, 4, 5}:
● If C = 2, then D = 3, 4, 5 - 3 pairs

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● If C = 3, then D = 4, 5 - 2 pairs
● If C = 4, then D = 5 - 1 pair
Total valid (C, D) pairs = 3 + 2 + 1 = 6
These are:
1. (2, 3)
2. (2, 4)
3. (2, 5)
4. (3, 4)
5. (3, 5)
6. (4, 5)
Step 2: For each pair, A can be any of the remaining numbers
For each (C, D) pair, one number remains unused.
That number goes to A (since any number > 1 works for A).
For example:
● If (C, D) = (2, 3), leftover = 4 or 5 - 2 possibilities
● If (C, D) = (2, 4), leftover = 3 or 5 - 2 possibilities
● If (C, D) = (2, 5), leftover = 3 or 4 - 2 possibilities
● etc
Each of the 6 pairs gives 2 possible arrangements.
Total arrangements:
Total = 6 pairs × 2 choices for A
= 12 arrangements
Hence, the correct answer is option d.
7. Bob has 4 coins with different values: 2, 4, 6, and 8.
Coin C is worth more than Coin A.
The total value of Coins B and C is equal to the value of Coin D.
Which TWO coins can Bob use to buy a candy that costs Rs. 14?
a) Coins A and D b) Coins B and D c) Coins A and C d) Coins C and D
Answer: d
Let's assign values to the coins based on the given conditions:
1. The coins have values 2, 4, 6, and 8
2. Coin C is worth more than Coin A, so C > A
3. The sum of Coins B and C equals Coin D, meaning B + C = D
Now, let's determine the values of each coin:
Setting D = 6, we get:
B+C=6
2+4=6
If B and C are assigned the values 2 and 4, then A must be given the highest value, which is 8. However,
this would make A greater than C, contradicting the given condition.
• Suppose D = 8 (the highest value coin)
• For B + C = D, we need two coins that sum to 8
• The only possible pair is B = 2 and C = 6
• Since C (6) is greater than A, we set A = 4
Now, Bob needs two coins that sum to 14:
• The available coins are 2, 4, 6, and 8
• The only pair that sums to 14 is 6 and 8
Answer:

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Bob can use Coin C (6) and Coin D (8) to buy the candy worth Rs. 14.
Hence, the correct answer is option d.
8. How many different three-digit even numbers can you make using the digits given below?
Note: You can use any one, any two, or all the digits, and you may repeat the digits
509
a) 3 b) 4 c) 5 d) 6
Answer: d
To form a three-digit even number, the last digit must be even.
From the digits 5, 0, and 9, the only even digit is 0.
So, the units place must be 0.
This means every number will look like: _ _ 0
Now we can choose the hundreds and tens digits from 5, 0, and 9.
Digits can be repeated, so we list all possible combinations:
● 5 in hundreds place and 5 in tens place - 550
● 9 in hundreds place and 9 in tens place - 990
● 5 in hundreds place and 9 in tens place - 590
● 9 in hundreds place and 5 in tens place - 950
● 5 in hundreds place and 0 in tens place - 500
● 9 in hundreds place and 0 in tens place - 900
So, we can form 6 different three-digit even numbers. Hence, the correct answer is option d.
9. X is a 4-digit number which has only even numbers and only 2 unique digits. What could be the
minimum value of the sum of the digits of the number X?
a) 2 b) 4 c) 8 d) 12
Answer: a
Since we have to look for the minimum value, we should make this 4-digit number with the smallest digits
possible.
The two smallest even numbers are 0 and 2.
The number must be 4-digit and contain exactly two unique digits.
Now, the number must be 4-digit, so the first digit cannot be 0. Therefore, the first digit must be 2.
To keep the sum of the digits as minimum as possible, we should avoid repeating 2 more than necessary.
If we repeat 2 (for example: 2200 or 2220), the sum of the digits increases.
So, to get the minimum sum, we use 2 only once and fill the remaining places with 0.
Four-digit number using 0 and 2 = 2000. Thus, the sum of digits = 2 + 0 + 0 + 0 = 2
Hence, the correct answer is option a.

Page 32

10. A, B, and C each walk a certain distance.
- The person who covered the least distance walks 25 km, and another person walks 6 times this
distance
- C walks 12 km more than B, while A did not cover the least distance
If only two among A, B, and C are allowed to walk further, how many more kilometres must they
walk so that their COMBINED distance (from the start) equals the distance walked by the third
person?
a) 92 km b) 78 km c) 88 km d) 62 km
Answer: c
Given:
The least distance = 25 km and one person walks 6 × 25 = 150 km
C walks 12 km more than B.
So, the three distances must be:
25 km, 150 km, and one unknown value
Also, C = B + 12
Now, as it is given that the least distance is 25 km, C is not the person who walked the least, as he
walked 12 km more than B.
Hence, either A or B must have walked 25 km.
But it is mentioned that A did not cover the least distance.
Hence, B = 25
Then, C = 25 + 12 = 37
A = 150 (highest value)
Then, the distances are 25 km, 37 km, and 150 km.
Only two among A, B, and C can walk further and the total distance that these two people walk together
must be equal to the distance walked by the third person.
So, to satisfy the above condition, the person who walked the highest distance already, cannot walk
further.
So, B and C must be walking further.
As they have already walked 25 + 37 = 62 kms, the distance that they should walk further will be
150 – 62 = 88 km
Hence, option c is correct.

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Count the number of rectangles which contain the black circle.
Note: Please count the squares also as rectangles for the purpose of this question

(a) 3 (b) 4 (c) 5 (d) 6
Answer: c
There are 5 rectangles that contain the black circle in the given figure, as highlighted
below. Hence, the correct answer is option c.

Page 34

Chapter 5: Far and Near
1. Ram tied his cat to a small pole using a 10 feet long rope. At noon, Ram placed some milk 15 feet
away FROM THE PLACE WHERE THE CAT WAS STANDING at that time. Can the cat reach the
milk?
(Assume that the cat's body length is negligible and it is tied in middle of an open field)
a) Definitely yes
b) Definitely no
c) Sometimes yes, sometimes no; it depends on the exact location of the milk
d) None of these
Answer: c
As seen in the image, in Case 1 when the cat is between the milk and the pole, then the cat may not be
able to reach.

In Case 2 when the cat and milk are on the either side of the pole, there may be a case where the cat
can reach the milk.

Hence, the answer is option c.

Page 35

2. Given below are the results for Javelin and High Jump competition.
- Tom threw the javelin 12 m farther than Roy
- Kim jumped 1 inch higher than Roy
- Roy jumped 2 inches less than John
Which participant has secured the same position in both the sports?

a) Mary b) Tom c) Roy d) John
Answer: c
1. Tom threw the javelin 12 m farther than Roy. Since Roy threw till 40 m, Tom would throw till 52 m.
2. Roy jumped 2 inches less than John. Since John jumped 29 inches, Roy would have jumped 27 inches.
3. Kim jumped 1 inch higher than Roy. Since Roy jumped 27 inches, Kim jumped 28 inches.
Thus, with the updated results, we can see in the table below that Roy secured the same position in both
sports. Hence, the correct answer is option c.

3. Soham collected 10 sticks, each measuring 6 cm. He breaks each stick into two equal halves.
Using all the stick pieces, what is the maximum number of squares of side 3 cm he can form if
each piece can be used only once and no piece can be shared by two squares?
a) 3 b) 4 c) 5 d) 6
Answer: c
If 10 sticks of 6 cm each are broken into two equal halves, there will be a total 20 sticks of 3 cm each. 4
sticks will be needed to make 1 square.
Therefore, at maximum a total of 5 squares can be made. Hence, the correct answer is option c.
4. A snail moves forward by 3 cm in 1 minute, and then immediately after that moves backward by
2 cm in 1 minute. It follows the same rhythm in its movement. How much time will the snail take
to reach the food placed 10 cm away for the first time?
a) 20 minutes b) 10 minutes c) 4 minutes d) 15 minutes
Answer: d
If a snail moves 3 cm forward in a minute and then immediately after that moves 2 cm backward in one
minute, it is effectively moving 1 cm in 2 minutes.

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To reach 7 cm, it will take 14 minutes.
In the 15th minute, it will move 3 cm more and reach the food at a distance of 10 cm.
Hence, the correct answer is option d.
5. A and B are two friends. They are standing at two opposite ends of a football field, which is 60
metres long. They both start running towards each other. A runs with a speed which is double
the speed of B. There is a football at the centre of the field. When they meet each other, who is
farther from the football?
a) A
b) B
c) Need to know the speed of A or B to know the distance
d) None of these
Answer: d
The football is kept exactly at the centre of the 60-metre field.
So, it is 30 metres from A, and 30 metres from B initially.
A runs twice as fast as B, so A covers the distance faster.
But they meet at a point that divides the field in the ratio of their speeds.
Since A : B = 2 : 1,
A will cover 40 metres, and
B will cover 20 metres before they meet.
This meeting point is 40 metres from A’s starting point,
and since the ball is at 30 metres, the meeting point is 10 metres beyond the ball.
But the key point is:
Both A and B meet at the same point.
So, at the moment they meet, both are at exactly the same distance from the football.
Therefore, no one is farther. Hence, the correct answer is option d.
6. I have to cut a 50-metre piece of cloth into sections of 1 metre each. How many cuts will I have to
make at minimum? Note: I cannot double up the cloth to make 2 layers
a) 47 b) 48 c) 49 d) 50
Answer: c
To cut 50 metre piece of cloth in 1 metre each, we will cut it in 50 pieces. To cut 50 metres into 50 pieces,
49 cuts are required, as the 49th cut will automatically give you two pieces.
A student can also think about how many cuts will be required to cut a cloth into two pieces - ONE; to cut
into three pieces - TWO.
So, the number of cuts required is always one less than the number of pieces required.
Hence, the correct answer is option c.
7. Rishi visited Goa and made new friends. His new friends stay near the beach.
Parish lives 800 metres east of the beach.
Anil lives 1400 metres east of the beach.
Sunil’s house is exactly between the houses of Parish and Anil.
What distance must Rishi travel if he visits the houses of Sunil and Parish, starting from the beach
and returning to the beach?
a) 2500 m b) 3800 m c) 1900 m d) 2200 m
Answer: d

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Parish lives 800 m east of the beach.
Anil lives 1400 m east of the beach.
Sunil lives exactly between Parish and Anil.
So, the distance between Parish and Anil is:
1400 – 800 = 600 m
Half of this distance is:
600 ÷ 2 = 300 m
So, Sunil’s house is 300 m east of Parish’s house.
Therefore, the distance of Sunil's house from the beach is:
800 + 300 = 1100 m
Now Rishi has to:
1. Start from the beach
2. Visit Sunil’s house (1100 m away)
3. Visit Parish’s house (which lies on the way back, since it is at 800 m)
4. Return to the beach
The farthest point he needs to travel to is Sunil’s house: 1100 m.
So, total distance covered:
1100 m (going) + 1100 m (coming back) = 2200 m
Option d is correct.
8. A group of trees with different heights is shown below. How many trees have both a taller and a
shorter tree immediately next to them, compared to their own height?

a) 1 b) 2 c) 3 d) 4
Answer: b
The trees highlighted 1 and 2 have both a taller and a shorter tree immediately next to them. Hence, the
correct answer is 2, option b.

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9. Statement: The lamp is A unit(s) taller than the cup but 2 units shorter than the B . What
will come in place of A and B, in the given statement?

a) A - 2, B - lunch box b) A - 1, B - bottle
c) A - 1, B - lunch box d) A - 2, B - bottle
Answer: b
The heights of the objects are as follows:

Lamp = 4 units, Cup = 3 units, Bottle = 6 units, Lunch Box = 8 units.
Therefore, the lamp is 1 unit taller than the cup but 2 units shorter than the bottle.
Hence, the correct answer is option b.
10. Mahesh knew that the distance between A and B was more than 3 km but less than 7 km. Ravi
knew that the distance was more than 5 km but less than 10 km. If both of them are correct and
the distance can only be an EXACT (integer) number of kilometres, how much must they pay if
the taxi charges Rs. 8 per km?
a) Rs. 48 b) Rs. 56 c) Rs. 40 d) Rs. 64
Answer: a
Mahesh says the distance is more than 3 km but less than 7 km.
So, the possible integer distances are:
4 km, 5 km, 6 km.
Ravi says the distance is more than 5 km but less than 10 km.
So, the possible integer distances are:
6 km, 7 km, 8 km, 9 km.
To satisfy both of their statements, we take the common distance in both lists.
Common value = 6 km.
Now, the taxi fare is Rs. 8 per km.
Total fare = 6 × 8 = Rs. 48. Hence, the correct answer is option a.

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The locker has a secret code. Identify the code with the help of the given code
key chart.

(a) 541 (b) 585 (c) 582 (d) 581
Answer: d
The locker has a secret code made up of three digits, represented by blocks with a
"?" symbol. We need to identify the corresponding numbers in the Code Key Chart
by matching the patterns of dots on the blocks.
Step 1: Identifying the First Digit
• The first block has black dots on the left and white dots on the right
• Looking at the Code Key Chart, the matching box contains 5 (refer image 1)
• First digit = 5
Step 2: Identifying the Second Digit
• The second block has only a black dot in the bottom left corner
• In the Code Key Chart, the matching box contains 8 (refer image 2)
• Second digit = 8
Step 3: Identifying the Third Digit
• The third block has black dots on the right and white dots on the left
• In the Code Key Chart, the matching box contains 1 (refer image 3)
• Third digit = 1
Thus, the secret locker code is 581.
Hence, the correct answer is option d.

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Chapter 6: The Dairy Farm
1. A mason laid 100 bricks in a row. He grouped bricks together, where every group has 25 bricks.
If 8 bricks from each group are removed, then how many bricks are left in total?
a) 32 b) 64 c) 68 d) 92
Answer: c
Divide the bricks into groups:
The total number of bricks is 100. He divides them into groups of 25 bricks each.
So, the number of groups formed is:
100 ÷ 25 = 4 groups
Remove bricks from each group:
From each group, 8 bricks are removed.
Total bricks removed:
8 × 4 = 32 bricks
Calculate the remaining bricks:
100 − 32 = 68 bricks left
Hence, option c is the correct answer.
2. Two types of boxes, A and B, are being transported. Each carton can hold either 8 boxes of type
A or 10 boxes of type B. If a total of 76 boxes were transported, what is the minimum number of
cartons required, given that all the cartons are full?
a) 6 b) 7 c) 8 d) 9
Answer: c
Let’s solve it logically.
The question asks for the minimum number of cartons required.
To keep the count as low as possible, we have to fill in the cartons with more boxes.
The carton can hold 10 boxes of type B.
As the total number of boxes is 76, the maximum number of type B boxes that can be transported is 70.
(7 cartons x 10)
This leaves 6 boxes behind, but the next carton cannot be filled completely (as a carton can either hold
8 or 10 boxes).
So, let’s decrease the number of type B boxes to 60 (6 cartons x 10).
Now, 76 – 60 = 16 boxes will be left.
Now, we can accommodate 16 boxes in two cartons (8 in each).
This makes the total count 8 (6 cartons with 10 boxes each and 2 cartons with 8 boxes each)
Hence, the minimum number of cartons that can transport 76 boxes = 8.
Thus, the correct answer is option c.

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3. 8 teams are divided into two pools of 4 teams. The teams within a pool play with every other team
in the pool twice. After these matches, the top team from each pool is selected and they play a
final to decide the winner. How many matches will be played in total?
a) 24 b) 25 c) 28 d) 29
Answer: b
There are 8 teams divided into 2 pools of 4 teams each.
In one pool of 4 teams, each team plays with every other team, twice.
Let A, B, C, and D be the 4 teams in one pool.
6 different pairs can be made: AB, AC, AD, BC, BD, CD.
Since each team plays with every other team twice, that makes 12 matches in one pool.
With 2 pools, the total becomes 12 + 12 = 24 matches.
Then the top team from each pool plays 1 final match.
So, 24 + 1 = 25 matches in total.
Option b is the correct answer.
4. Raj took an entrance exam where he attempted 38 questions out of 40, out of which he got 3
questions wrong. If the marking scheme per question is shown below, what is the maximum
marks Raj could score?

a) 76 b) 79 c) 81 d) 83
Answer: b
Let’s solve this question using a logical approach.
Raj attempted 38 questions out of 40 and he got 3 questions wrong.
So, we can say that he did not attempt 2 questions.
As per the table, section A grants 0 and section B grants -1 for each un-attempted question.
To maximize Raj’s score, we’ll have to assume that Raj’s un-attempted questions are from section A.
The maximum marks that Raj can score from section B is 15 x 3 = 45 (assuming that all questions of this
section are answered correctly).
Now, among the 38 questions that he had answered, 15 are from section B.
So, the remaining 23 questions are from section A.
Also, we know that he got 3 questions wrong.
So, his status in section A would be 20 correct answers + 3 wrong answers.
Score of section A = (20 x 2) - 3(2) = 40 - 6 = 34.
Adding the scores of section A and section B, the total score is 45 + 34 = 79.
Hence, the maximum possible score would be 79. Thus, the correct answer is option b.

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5. 80 kg of material needs to be transported to a place that is 10 km away.
The material must be divided into only one fixed load size for the entire transport. The possible
load sizes are 10 kg, 20 kg, or 40 kg. Any number of couriers may be used.
Each courier:
- Charges ₹10 per hour
- Travels at 10 km/hr when empty
- Travels at 5 km/hr while carrying 10 kg
- Travels at 2 km/hr while carrying 20 kg
- Travels at 1 km/hr while carrying 40 kg
A courier can carry at most 40 kg at a time.
After delivering a load, the courier must return empty before the next trip. Even after the final
delivery, the courier returns empty, and all return journeys are charged.
What is the minimum total cost required to transport all 80 kg of material?
a) Rs. 220 b) Rs. 240 c) Rs. 260 d) Rs. 280
Answer: a
We must move 80 kg over 10 km, and we can choose only one load size: 10 kg, 20 kg, or 40 kg.
Couriers charge ₹10 per hour and must return empty every time (even after the last trip), so every delivery
includes a go trip (loaded) and a return trip (empty).
Case 1: If we choose 40 kg per trip, we need 2 deliveries (80 ÷ 40).
• Going with 40 kg: speed = 1 km/hr i.e. 10 km takes 10 hours
• Returning empty: speed = 10 km/hr i.e. 10 km takes 1 hour
So, one round trip takes 11 hours.
For 2 round trips: 11 × 2 = 22 hours
Cost = 22 × ₹10 = ₹220
Case 2: If we choose 20 kg per trip, we need 4 deliveries.
• Going with 20 kg: speed = 2 km/hr i.e. 10 km takes 5 hours
• Returning empty: 1 hour
One round trip = 6 hours
For 4 round trips: 6 × 4 = 24 hours
Cost = 24 × ₹10 = ₹240
Case 3: If we choose 10 kg per trip, we need 8 deliveries.
• Going with 10 kg: speed = 5 km/hr i.e. 10 km takes 2 hours
• Returning empty: 1 hour
One round trip = 3 hours
For 8 trips: 3 × 8 = 24 hours
Cost = 24 × ₹10 = ₹240
Comparing all options, the lowest cost is ₹220 when carrying 40 kg per trip.
Therefore, option a is the minimum possible and the correct answer.
6. In a theatre, seats are arranged in rows labelled A to F and columns labelled 1 to 5. If the entire
row C and the entire columns 3 and 5 are empty and the rest of the seats are sold out, then how
many tickets are sold out in total?
a) 13 b) 17 c) 15 d) 14
Answer: c
The theatre has 6 rows with 5 seats and 5 columns with 6 seats. Therefore, the total number of seats in
the theatre is 6 x 5 = 30.

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As Row C is empty, and since 1 row has 5 seats, we deduct 5 seats from the total, resulting in 30 - 5 =
25 seats.
Additionally, columns 3 and 5 are empty, and considering 1 column has 6 seats, we deduct 12 seats.
However, only 10 seats will be deducted, as 2 seats were already accounted for while deducting seats
for Row C (since 1 seat is always common between a row and a column).
After deducting 10 seats from the remaining 25 seats (25 – 10 = 15), the total number of tickets sold is
15. Hence, the correct answer is option c.

ALTERNATE APPROACH
One row and two columns are completely empty, while five rows and three columns are filled with seats.
To determine the number of tickets sold or seats occupied, multiply the number of occupied rows by the
number of occupied columns. This calculation results in 5 rows × 3 columns = 15 occupied seats.

7. If the number in each square is the product of the two numbers in the circles connected to it, what
is the value of B?

a) 9 b) 6
c) 4 d) Cannot be determined
Answer: b
It is mentioned in the question that the number in each square is the product of the two numbers in the
circles connected to it.
As can be seen in the image below,
A × B = 24, A × C = 36, and B × C = 54
So,
B must be a common factor of 24 and 54.
Common factors of 24 and 54 are 1, 2, 3, 6.
Check which value fits all conditions.
If B = 6, then A × 6 = 24, so A = 4
and 6 × C = 54, so C = 9.
Now verify: A × C = 4 × 9 = 36, which matches the given number.
Thus, B = 6.
Hence, the correct answer is option b.

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8. If we place the numbers 5, 6, and 7 in the boxes shown below and the product is 455, which
number will go in the red box?

a) 5 b) 6
c) 7 d) Either option a or option c
Answer: a
We want the product of the two numbers to be 455.
We are placing 5, 6, and 7 in the three, empty boxes:
Step 1: Look at the units digit
● The units digit of 455 is 5
● A multiplication ends in 5 only when one number ends in 5 and the other number is odd
So, yes, there are two ways the “5” can appear in the units place:
1. Top number ends in 5
2. Bottom number is 5
So, we need to check both.
Case 1: RED box = 5
Then the top number looks like:
_ 5
× _
_________
Now try the remaining digits 6 and 7:
● Try 65 × 7 = 455 i.e. Correct!
● Try 75 × 6 = 450 i.e. Not 455
So, one valid result exists here.
Case 2: Bottom number = 5
Then the multiplication becomes:
_ _
× 5
______
Check all possible two-digit numbers formed by 6 and 7:
● 67 × 5 = 335
● 76 × 5 = 380
Neither equals 455.
So, this case is not possible.
Conclusion:
Even though the 5 could be in the bottom number in theory, it does NOT produce 455 with the
remaining digits.

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The only way to get 455 is:
65
× 7
-----
455
So, the red box must contain 5. Hence, the correct answer is option a.
9. In a classroom, there are 5 desks in each row. The number of rows is not known. First, all the
desks in the front row are removed. Then, one desk is added to the leftmost side of each
remaining row. After these changes, the total number of desks in the classroom becomes one of
the following: 18, 20, 22, or 28.
Which of these could represent the number of rows remaining after the front row was removed at
the start?
a) 3 b) 4 c) 5 d) 7
Answer: a
Each row has 5 desks initially.
Now, a desk is added on the leftmost side of each row, so all the rows will now have 6 desks and hence
the number of desks will be the product of 6 and some number.
Among the given numbers (18, 20, 22, and 28), only 18 is a multiple of 6 and hence the number of desks
present now are 18.
Since each row has 6 desks, the number of rows will be 18/6 = 3.
Hence, the correct answer is option a.
10. The numbers in Row C have been changed to the numbers in Row D by using a specific rule.
Which of the following options shows that rule?

a) 8 × C = D b) 8 × D = C
c) C + 210 = D d) Both a and c are correct
Answer: a
Let us check each option one by one to find the correct rule that converts Row C into Row D.
Option a: 8 × C = D
30 × 8 = 240, 60 × 8 = 480, 10 × 8 = 80, 90 × 8 = 720, and 150 × 8 = 1200.
All the values match Row D, so this rule works correctly.
Option b: 8 × D = C
240 × 8 = 1920, which does not match 30.
So, this rule is incorrect.
Option c: C + 210 = D
30 + 210 = 240 (matches)
60 + 210 = 270, which does not match 480.
So, this rule does not work for all numbers.
Option d: Both a and c are correct
Since option c is incorrect, this option is also incorrect.
Hence, option a is the correct answer.

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Given below are 6 boxes, A, B, C, D, E, and F, each having a pair of shapes. They
need to be categorised into 4 groups, as follows:
- If both shapes in a box are of the same colour, they belong to Group 1. If not,
they belong to Group 2
- If both shapes in a box are the same (irrespective of their colour), they belong to
Group 3. If not, they belong to Group 4
- One box can be part of more than one group
How many boxes belong to BOTH groups 2 and 4?

(a) 3 (b) 4 (c) 2 (d) 1
Answer: A
The following boxes should be categorised into four groups and the same box can
belong to more than one group.

If both shapes in a box are of the same colour, they belong to Group 1. If not, they
belong to Group 2. Accordingly, only C and E belong to group 1 as they have shapes of
same colour, while the rest belong to group 2.

If both shapes in box are same (irrespective of their colour), they belong to Group 3. If
not, they belong to Group 4. Accordingly, B and E belong to group 3, as they have
same shapes, while the rest belong to group 4.

Clearly, three boxes A, D, and F belong to both groups 2 and 4.
Hence, option a is the correct answer.

Page 47

Chapter 7: Shapes and Patterns
1. On a wall, there are some triangular tiles with equal sides, of which some have been removed,
leaving a black coloured wall. How many such triangular tiles have been removed?

a) 4 b) 5 c) 6 d) 7
Answer: b
As highlighted in the image below, 5 tiles have been removed. Hence, the correct answer is option b.

2. Which of the following additional lines, shown in the options, should be drawn so that the square
given in the question image is divided into 8 EQUAL parts?
Note: You cannot rotate the question or option images

a) b) c) d)
Answer: d
Let’s check each option
Option a:

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Option a does not show the square divided in 8 equal parts.
Option b:

Option b does not show the square divided in 8 equal parts.
Option c:

Option c does not show the square divided in 8 equal parts.
Option d:

Option d divides the square in 8 equal parts.
Hence, option d is correct.
3. Sam divided a rectangle into 6 equal parts using lines such that each line further divided the
rectangle into more parts. Which of the below options could represent the number of lines used
by Sam?
a) The rectangle is divided using 3 horizontal and 3 vertical lines
b) The rectangle is divided using 3 horizontal and 2 vertical lines
c) The rectangle is divided using 2 horizontal and 2 vertical lines
d) The rectangle is divided using 2 horizontal and 1 vertical line
Answer: d
As shown in the image below, the rectangle divided by 2 horizontal and 1 vertical line will give 6 equal
parts. Hence, option d is the correct answer.

Page 49

4. A paper piece is in the form of a regular pentagon. At most how many times can it be folded in
half? Note: Once the paper is folded, it cannot be unfolded
a) 0 b) 1 c) 2 d) 3
Answer: b
A regular pentagon can be folded in half only along a line of symmetry. One such fold is possible.
After the first fold, the new shape cannot be divided into two equal halves. Since the paper cannot be
unfolded, no further fold in half is possible.
Therefore, the paper can be folded in half at most once. Hence, the correct answer is option b.

5. Given below is an incomplete arrangement of a 3 x 3 x 3 cube where all the smallest cubes are of
the same size. How many such smallest cubes are required to complete the cube?

a) 4 b) 5 c) 6 d) 7
Answer: c
Let’s see which cubes have been removed by looking at the given arrangement.
We can see that the cubes above 1, 2, and 3 are missing.
Also, the cubes in front of 4, 5, and 6 are missing.

Let’s place red cubes in the empty spaces next to 1, 2, 3, 4, 5, and 6.

As we can see, there are 6 small cubes that were missing from the given arrangement.

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Thus, the correct answer is option c.
6. In how many different ways can Set A (without rotating) be placed completely on the grid so that
no two squares overlap and no two adjacent squares have the same pattern?
Note: Squares that share common sides are considered to be adjacent. Squares that share a common
corner alone, are NOT adjacent

a) 4 b) 5 c) 6 d) 7
Answer: c
First, we mark all the positions on the grid where the L-shape from Set A can fit without going outside the
boundary.
Next, we remove the positions where Set A cannot be placed because one of its patterned squares would
be adjacent to a square on the grid that has the same pattern.
For example, at positions 2 and 6, the striped square of Set A becomes adjacent to another striped
square already in the grid. So, these placements are not allowed.

Now, we check the remaining positions.
In these positions, the L-shape fits completely, and none of its squares is adjacent to another square with
the same pattern.

Finally, after testing all the valid positions, we see that Set A can be placed in 6 different ways.
Hence, option c is the correct answer.

Page 51

7. What is the minimum number of matchsticks needed to make a figure which is divided into 6
equal parts?
Note: You can't change the placement of the matchsticks already in the image

a) 4 b) 5 c) 6 d) 7
Answer: b
We cannot change the position of the matchsticks already given in the figure. We need to add the
minimum number of matchsticks so that the figure is divided into 6 equal parts.
Step 1: First, complete the outer square by adding matchstick 1 at the top side.
Step 2: Next, complete the inner structure of the figure by adding matchstick 2 vertically from the top to
the middle point and matchstick 3 horizontally on the right side.
Step 3: Now the figure is almost divided into equal sections. Since the figure must be divided into 6 equal
parts, add two more matchsticks: matchstick 4 and matchstick 5 vertically to complete the divisions.
As shown in the solution, 5 matchsticks are required to form a square, which is divided into 6 equal parts.
Hence, the correct answer is option b.

8. Given below is the net of an unfolded cube. Which of the following pairs of numbers would be on
the opposite faces of the cube when the net is folded?

a) 6 and 4 b) 1 and 3 c) 5 and 6 d) 1 and 5
Answer: d
Let's use the given information step by step.
• Starting with 1 as the top face, the numbers 4 and 2 are adjacent to 1. Upon folding, they become
opposite to each other, with 2 on the left face and 4 on the right face, while 1 remains on the top face (as
shown in Figure 1).
• After the first fold, 3 and 6 will become opposite to each other, resulting in 3 being on the back face and
6 on the front face (as shown in Figure 2).
• The last remaining number, 5, naturally occupies the bottom face of the cube and is opposite to 1 (as
shown in Figure 3).

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Among the given options, the only option that gives the numbers that end up being opposite after the
folding process is option d) 1 and 5
Hence, the correct answer is option d.

9. Which of the following set of pieces CANNOT be arranged in the grey space to complete the
rectangle?
Note: You cannot rotate or overlap the given pieces

a) b) c) d)

Answer: c
The approach to solving this question is to first consider the empty space shown in the question image.
For each option, start by taking the larger piece of the two and check if it can fit within the blank area in
the question image.
Once the larger piece is placed, check if the remaining space is sufficient to accommodate the smaller
piece, ensuring that both pieces together will fully cover the empty area.
Repeat this process for all the given options.

Only the set of pieces given in option c cannot be arranged in the grey space. Thus, option c is the correct
answer.
10. Two faces of a plain white cube are painted red. Now the cube is divided into 27 small cubes of
equal dimensions. How many such smaller cubes will have at least one face painted?
a) 6 b) 12
c) 15 d) Cannot be determined
Answer: d
The cube is divided into 27 smaller cubes, which means it is cut into 3 parts along each edge (3 × 3 × 3),
as shown in the image given below. Thus, each face of the cube will be divided into 9 smaller squares,

Page 53

and each face of the cube will have 9 smaller cubes on that face.

However, if the coloured faces are adjacent, there would be three cubes which would have two faces
coloured.
Thus, the number of smaller cubes with at least one face coloured will be 15.
However, if the coloured faces are opposite, there would be 18 such smaller cubes with at least one
coloured face. Hence, the exact number cannot be determined. Thus, the correct answer is option d.

Page 54

A, B, C, D, E, and F are 6 friends.
A, C, and F are girls, while the others are boys.
C and E are playing chess, while the others are playing Ludo.
How many boys are playing Ludo?
(a) 1 (b) 2 (c) 3 (d) 4
Answer: b
The correct approach to solve the given question is to organize all the
information in a table.
From the given information, we know that A, C, and F are girls.

That means B, D, and E are boys.

It is given that C and E are playing chess and others are playing Ludo.

So, the boys who are playing Ludo are B and D.

Hence, the answer is option b.

Page 55

Chapter 8: Weight and Capacity
Activity Time
Measuring Weight
Introduction
In this activity, we can use the understanding of a weighing scale to search for a particular weight. This
can be done in many ways, and some of these methods will be faster than others. This is similar to how
a computer also searches. In this activity, we will explore search algorithms using a weighing balance
and see how we can speed up the process using different approaches. We will separate a lighter fake
weight from 8 proper weights among 9 identical-looking weights. We convey to students that the same
reasoning can be applied to the analysis of computer algorithms.
Activity Time Description

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

Activity Video Link: Computational Thinking with तराज़ू
(Weighing Scale) | 15th August | Ep 6 - 3030 Eklavya

Problem-solving 25 mins Student’s Tryout the activity with each other.
and algorithm
Student Worksheet: Khotta Baant aur Tarazoo| CT
CBSE | Student

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

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

CT Components
This activity develops in a way that helps students discover important ideas, such as breaking a problem
into smaller problems (decomposition), identifying patterns, developing algorithms, and finding faster
ways to complete a task, thus analysing algorithms. These are fundamental ideas involved in
computational thinking.

Page 56

Activity: Searching with a Weighing Scale
Imagine we have 9 identical-looking weights out, of which one is Lighter, a fake weight. We have to
figure out which one is lighter using only a weighing scale (and nothing else).
Our goal is to find the fake in the least number of weighings.

How will we figure this out? What process will we use to determine the fewest weighings needed to
identify the lighter fake weight? Always assume that you have the worst luck and will get the fake weight
in the last possible step.

Make the students observe these things:

• A 2-pan weighing balance can be in 3 positions during weighing. Either the left or the right side could
be heavier, or both sides could be balanced.
• We weigh two weights, one on either side, and the weighing balance shows that they are equal. From
this, we can conclude that. The fake weight is in the remaining weights.
We know that there is only one fake weight that is lighter than the rest. This means all the other
weights weigh the same. So, if the weighing scale is balanced, the weights are not fake. This implies
that the lighter weight is in the rest of them.
• By observing how many weights can be discarded after each weighing, we can determine which
method is faster. The method that helps us remove more weight per round is faster than the rest of
the methods

A: Initial attempt
Let’s take two weights at a time and compare. Taking two weights at a time can be done in two different
ways:

• In the first method, we keep the weight on one side constant (let’s say the right) and choose a new
weight for the left side at each weighing.
• Second method, we chose two new weights for every weighing.

Demonstrate both methods partially to the students. Do not completely show the students the methods.
Or you can invite a couple of students whose answers seem like a good method to come and do it in
front of the class.

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1. Is one of the methods above faster (fewer weighings) than the other, and if yes which one?
a) Yes, method one is faster
b) Yes, method two is faster
c) No, both take the same number of weighings
Answer: B
Solution: In method one, we can remove only one weight per weighing, whereas in the second method,
we can remove two weights. Removing more weights in each weighing means fewer weights to compare
in future rounds.

Competency: Analysis, observations, deduction
Demonstrate both methods to the students so that they can verify their answers.
2. How many minimum weighings will it take to guarantee figuring out the fake (lighter) weight when
you compare two weights at a time?
(This means we are assuming you always have bad luck and get the fake one in the last possible step)
a) 5 weighings b) 8 weighings c) 4 weighings d) 7 weighings
Answer: c
We compare two weights; if they are equal, we know neither is fake, so we remove both. Similarly, we
have to compare two more pairs. Now, in the fourth attempt, we have 3 weights remaining. If the two
chosen weights are unequal, then the lighter one is the one we have found; if they are equal, the
remaining weight is the lighter one. This type of strategy is called Divide and Conquer.
Competency: Pattern recognition, logic
For the first method suggested above, you have to compare one fixed weight with the remaining 8 weights
to find the one that differs in weight. Using this method, we can figure out the lighter weight in 7 weighings
(Why not 8?). This is a Linear search.
Key observations after the above activity:
• If weights are equal when weighed, then they are both not the fake weight (Why?)
• We can remove a maximum of 2 weights in each round
• We need to remove the maximum possible weights in each round, and divide-and-conquer is more
efficient than linear/sequential search.
• If we have 3 weights remaining, and compare any two. Then, if the two chosen weights are unequal,
then the lighter one is the fake; if they are equal, the remaining weight is the fake weight.
B: A faster way (Binary search)

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To do it faster, we need to remove as many weights as possible in each round and use divide-and-
conquer better. We can divide the 9 weights into 2 equal parts of 4 each, leaving 1 weight.
In the first weighing, we keep 4 on the left and 4 on the right, and the remaining is set aside.

1. After one weighing, how many weights remain?
a) 1 b) 4 c) 5 d) 8
Answer: b
When we compare 4 weights with 4 other weights, the weighing scale is either balanced or not balanced.
In the worst-case situation, they are not balanced. This means the lighter side will be above the other 4.
So, we can remove the 4 weights on the heavier side, and we can also remove the one we didn’t weigh,
since it's not a fake weight either. 5 proper weights get removed from the search space.

Competency: Algorithmic thinking, decomposition, generalization
2. If the fake is among the 4 remaining weights, how many more weighings are needed (minimum)
to find the fake weight?
a) 1 b) 2 c) 3 d) 4
Answer: b
We divide the 4 weights into 2 groups of 2 and compare them. One side will be lighter than the other; it
will have the fake weight. So, we have 2 weights left. We further split these two weights, compare them,
and then find the fake weight.

Competency: algorithmic thinking, analysis, decomposition
We figured out the fake weight in 3 weighings. This method of dividing a problem into two parts and then
finding a solution is known as Binary search.
Can we find the fake weight in fewer weighings than this?
C: Fastest possible algorithm in a three-state system
Can we do this faster than the binary search? Let’s bring together the various points we observed above.
We saw that the weighing balance is a three-state system. We also saw that we can infer information
about a weight without actually weighing it. We also saw that with three weights, we can infer which one
is the fake in a single weighing.
Let’s try dividing into 3 equal parts, and now we will weigh two groups of three.

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