# NCERT Solutions Class 8 Maths Chapter 16 Exercise 16.1

All the information regarding numbers and how they can be divided by other numbers is covered in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 based on “Playing with Numbers.” Learning arithmetic involves a lot of numbers. They serve as the fundamental building blocks for practically all mathematical concepts and computations. Consequently, developing a thorough understanding of numbers and their characteristics is important. Students will become familiar with numbers, their properties, and divisibility tests by practising NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1.

For higher-level Mathematics courses, it is essential to have in-depth knowledge of this subject from the junior classes themselves. These NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 provide systematic instruction in numbers and mathematical concepts. Students who regularly practise these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 will develop the ability to attempt Number Divisibility tests. The set of solutions in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are appropriate for fostering students’ capacity for logical thought and problem-solving. Students can download the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 from the links on the Extramarks website and mobile application to study and practise with these solutions.

The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are an excellent tool for students’ exam preparation. Through numerical games and visuals provided on the Extramarks website, the complete nature of these solutions encourages active learning. The well-organised structure of NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 makes it easy to develop a thorough understanding of numbers in general form and how to perform divisibility tests on them.

The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are effective in enhancing students’ foundational numerical abilities.Using these well-organized resources to visualise number divisibility tests is quite effective.. The solutions’ progressively arranged question format allows for the best possible understanding of the entire subject. Students can click on the links provided on the Extramarks website and mobile application to prepare and practise the exercise-wise problems with these NCERT Solutions Class 8 Maths Chapter 16 Exercise 16.1 based on Playing with Numbers.

The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 have problems involving calculations based on locating digits in arithmetic operations, which are included in the chapter Playing with Numbers. These NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 consist of solutions to ten questions that offer a step-by-step comprehension of this chapter. Short answer questions in NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 provides a thorough study of numbers through the use of fundamental arithmetic procedures.

Finding digits for letters in number puzzles is an engaging method to learn the basics of Mathematics. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are a thorough manual that explains this information with appropriate examples. With the help of NCERT Solutions Class 8 Maths Chapter 16 Exercise 16.1, which are available on the Extramarks website and mobile application, students can learn and practise the different types of problems. Numbers in general form, their characteristics, and divisibility tests are all covered in detail in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1. These solutions cover a variety of themes but mostly introduce Numbers, Number forms, Games, Letters for Numbers, and Tests of Number Divisibility.

The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 offered by Extramarks will increase knowledge of number subsets’ fundamental structure. Students will develop the fundamental problem-solving abilities for higher-level arithmetic subjects by practising with the aid of these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1.

It is necessary to have a fundamental understanding of place value, general form, and numerical arithmetic in order to answer all of the questions in these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 offered by Extramarks. To master this chapter before tests, students can use these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 as a trustworthy resource. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are a great resource for efficiently retaining facts and ideas. Playing with Numbers has 14 questions that are mostly sums of short answers. The majority of these puzzles involve finding digits for letters and conducting divisibility checks. Clear ideas of number types and their characteristics are necessary to comprehend number forms and divisibility tests. With the aid of different interactive games, riddles, and visuals provided by Extramarks, students will be able to quickly understand these crucial ideas in NCERT Solutions Class 8 Maths Chapter 16 Exercise 16.1. The following is a list of some of the key ideas relating to numbers and the tests for their divisibility that are covered in these answers:

- Any number is capable of being represented in its general form. For instance, the generic form of the two-digit integer ab is ab = 10a + b.
- Any number that may be divided by two and has a divisible last digit 2, 4, 6, or 8 can be divided evenly by 2.
- The rule of divisibility of the digit 3 asserts that a number is divisible by the digit 3 if the sum of its digits is also divisible by 3.
- Any number whose last digit is 0 or 5 is always divisible by 5, according to the rule of divisibility by 5.
- Divisibility by 9: According to the rule of divisibility by 9, a number is divisible by the digit 9 if the sum of its digits is also divisible by 9.
- Any number that has a 0 as the last digit is divisible by 10.

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers (Ex 16.1) Exercise 16.1**

In NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1, students will learn about the various types of numbers in detail.These NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 will also cover the concept of divisibility by numbers. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 provided by Extramarks pertain to Chapter 16, which contains 2 exercises that cover each of the chapter’s themes. The following list includes the themes incorporated in various exercises. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 provide solutions to Exercise 16.1.

- 16.1 Introduction
- 16.2 Numbers In General form
- 16.3 Games With Numbers
- 16.4 Letters for Digits
- 16.5 Tests Of Divisibility
- Divisibility by 10
- Divisibility by 5
- Divisibility by 2
- Divisibility by 9 and 3

Extramarks provides NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 for students to download.

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers (Ex 16.1) Exercise 16.1**

Different types of numbers, such as Natural Numbers, Whole Numbers, Integers, and Rational Numbers, have all been studied by the students. They might have also researched different intriguing characteristics of these numbers, finding factors, multiples, and relationships between them. Students can also thoroughly investigate large numbers with the help of these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1. The solutions provided in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 can assist students in justifying divisibility trials. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are arranged in an easy-to-understand format for the convenience of students.

Here are a few important facts to remember while reading the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 by Extramarks:

- What is the general form of a number?

A two-digit number can be represented in general form as pq=10p+q.

- Adding and Reversing Two-Digit Numbers

A two-digit number is divisible by 11 when it is added to another number after being inverted. Additionally, the quotient is equal to the digits added together.

- The 3-digit numbers are then reversed and subtracted.

The result of reversing a three-digit number and subtracting the smaller number from the larger number is easily divisible by 99. Additionally, the quotient is equal to the variation between the chosen number’s first and third digits. A three-digit number is perfectly divisible by 111 if all possible combinations of the three digits are tallied together.

- Divisibility by 10

When the last digit of a number is 0, it is divisible by 10.

- Divisibility by 5

A number is divisible by 5 if its last digit is either 5 or 0.

- Divisibility by 2

Any number that ends with 0, 2, 4, 6, or 8 is divisible by 2.

- Divisibility by 9

A number is divisible by 9 only if the sum of its digits is divisible by 9.

The concepts embedded in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are clear and simple enough for students to understand. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 lay the proper conceptual basis for students to enable them to master arithmetic at higher levels.

**Exercise 16.1**

There are answers to all of the exercise questions in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 based on Playing with Numbers. Students frequently encounter certain doubts when answering the exercise questions, but with the assistance of the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1, students can positively understand all the concepts in the exercise. The NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 are offered on the Extramarks website and mobile application to assist students in quickly dispelling all of their academic concerns. Students can reinforce their fundamentals by downloading the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 provided by mentors at Extramarks.

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers (Ex 16.1) Exercise 16.1**

The ideas of Numbers in General form, Letters for Digits, and various divisibility tests are the main topics of the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1.

- It is possible to write numbers in a general form. The formula for the two-digit number ab will be ab = 10a + b.
- The formula for the three-digit number abc is abc = 100a + 10b + c.

Games with Numbers is another fascinating subject covered in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1.

- When solving puzzles or playing number games, the general form of numbers is useful.
- In games, players must reverse two- and three-digit numbers to create matching two-digit numbers.

Some number-related games and puzzles are included to enhance the content’s appeal and interest in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1. The concepts in the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 will be clarified while engaging with these activities. In some number games, letters and numbers are switched to create a code, and vice versa. These intriguing trick- number games are built around the idea of the general form of numbers and can be learned through the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1.

In these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1, students will study the divisibility of numbers by 11, as well as how a number that is divisible by m will also be divisible by each of m’s factors. The purpose of specific tests that are used to determine the divisibility will be highlighted in these NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1. The final section of the NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 is a summary with a focus on the key ideas.

**NCERT Solutions for Class 8 **

For Class 8 students, the NCERT solutions for Class 8 Mathematics are a dependable source of study materials. Holistic and credible Class 8 solutions are offered by Extramarks. All sixteen chapters of the NCERT Class 8 Mathematics textbook are covered in the Class 8 Mathematics Solutions. Each chapter has been thoroughly solved and is organised. The staged nature of the video tutorials facilitates students’ insight. The NCERT solutions provided are made available in PDF versions and video instructional formats for the convenience of students.

Anywher and at any time, students can study with the aid of online video lectures. The PDF versions of NCERT solutions are available for download if students want to study independently. The links provided on the Extramarks website and mobile application can be checked out by students. The NCERT Mathematics Class 8 Solutions are up-to-date with the most recent CBSE curriculum and cover all of the questions in the NCERT Class 8 Mathematics book.

The Class 8 Mathematics NCERT Solutions are made to make Mathematics easier for children to learn. Every question contains a thorough explanation to make the concepts used easy to comprehend. Students can begin using the chapter-by-chapter NCERT Mathematics solutions for Class 8 by clicking on the links.

Rational Numbers, a Linear Equation in One Variable, Understanding Quadrilaterals, Practical Geometry, Data Handling, Squares and Square Roots, Cubes and Cube Roots, Comparing Quantities, Algebraic Expressions and Identities, Visualizing Solid Shapes, Exponents and Powers, Direct and Inverse Proportions, Factorization, Introduction to Graphs, and Playing with Numbers are just a few of the chapters covered in the NCERT solutions for Class 8.

To better comprehend the ideas and achieve high marks in the school exams, students should practise with the assistance of NCERT solutions.

**Q.1 **Find the values of the letters in each of the following and give reasons for the steps involved.

$\begin{array}{l}\text{1.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A}\\ \underset{\xaf}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}5}}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}2}\text{\hspace{0.33em}}\\ \\ \text{2.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A}\\ \underset{\xaf}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}9\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}8}\\ \underset{\xaf}{\text{\hspace{0.33em}}\mathrm{C}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}3}\text{\hspace{0.33em}}\\ \\ \text{3.}\\ 1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}A}\\ \underset{\xaf}{\times \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}}\\ \underset{\xaf}{\text{\hspace{0.33em}}9\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}}\\ \\ \text{4.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\\ \underset{\xaf}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}7}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}}\\ \\ \text{5.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\times \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}3\text{\hspace{0.33em}}}\\ \underset{\xaf}{\mathrm{C}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}}\\ \\ \text{6.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\times \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}5\text{\hspace{0.33em}}}\\ \underset{\xaf}{\mathrm{C}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}}\\ \\ \text{7.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\times \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6\text{\hspace{0.33em}}}\\ \underset{\xaf}{\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}}\\ \\ \text{8.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1\\ \underset{\xaf}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}0}\\ \\ \text{9.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\\ \underset{\xaf}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1}\text{\hspace{0.33em}\hspace{0.33em}}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}8}\\ \\ \text{10.}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\\ \underset{\xaf}{\text{+\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}6\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{B}}\text{\hspace{0.33em}\hspace{0.33em}}\\ \underset{\xaf}{\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{A}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}9}\end{array}$

**Ans**

\begin{array}{l}\text{1}\text{. 3 A}\\ \underset{\xaf}{\text{+ 2 5}}\\ \underset{\xaf}{\text{B 2}}\\ \\ \text{The addition of A and 5 is 2 i}.\text{e}.,\text{a number whose}\\ \text{one\u2019s digit is 2}.\text{This is possible only when A = 7}.\text{}\\ \\ \text{Hence},\text{the addition of A and 5 will give 12 and}\\ \text{then 1 will be the carryover for the next step}.\\ \text{In the next step},\text{1}+\text{3}+\text{2}=\text{6}\text{.}\\ \text{So,B = 6}\text{.Therefore},\text{addition will be:}\\ \\ \text{3 7}\\ \underset{\xaf}{\text{+ 2 5}}\\ \underset{\xaf}{\text{6 2}}\\ \\ \text{Hence,}\text{\hspace{0.17em}}\text{A=7 and B=6}\text{.}\end{array}

\begin{array}{l}\text{2}\text{. 4 A}\\ \underset{\xaf}{\text{+ 9 8}}\\ \underset{\xaf}{\text{C B 3}}\\ \text{}\\ \\ \text{The addition of A and 8 gives us 3}\text{.}\\ \text{This is possible only when digit A is 5 since the addition of 5}\\ \text{and 8 will gives us 13 and thus},\text{1 will be the carry for the}\\ \text{next step}.\\ \\ \text{Now,}1+4+9=14.\\ \text{So B will be equal to 4 and 1 will be carry over to the next step}\text{.}\\ \text{Since there are no numbers in the next step which are}\\ \text{to be added,therefore C = 1}\\ \text{Hence, the addition will be as follows:}\\ \text{4 5}\\ \underset{\xaf}{\text{+ 9 8}}\\ \underset{\xaf}{\text{1 4 3}}\\ \\ \therefore \text{A}=5,\text{B= 4 and C=1}\end{array} \begin{array}{l}\text{}\end{array}

$\begin{array}{l}\begin{array}{l}\text{3. 1 A}\\ \underset{\xaf}{\text{\xd7 A}}\\ \underset{\xaf}{\text{9 A}}\end{array}\end{array}$

The multiplication of A with A itself gives a number whose ones digit is A again. This happens only when A = 1, 5, or 6.

__Case 1: If A = 1__

The multiplication will be 11 × 1 = 11 but the tens digit given here is 9. Therefore, A = 1 is not possible.

__Case 2: If A = 5__

The multiplication will be 15 × 5 = 75. Thus, A = 5 is also not possible since the tens digit given is 9.

__Case 3: If A = 6__

If we take A = 6, then 16 × 6 = 96. Since the tens and one’s digit are same as the given question, therefore, A should be 6.

The multiplication will be as follows:

$\begin{array}{l}\text{1 6}\\ \underset{\xaf}{\text{\xd7 6}}\\ \underset{\xaf}{\text{9 6}}\end{array}$

\begin{array}{l}\text{4}\text{. A B}\\ \underset{\xaf}{\text{+ 3 7}}\\ \underset{\xaf}{\text{6 A}}\end{array}

We can observe that the addition of A and 3 is giving 6. There can be two cases.

__Case 1: First step is not producing a carryover.__

If first step is not producing a carryover then A comes to be 3 as 3 + 3 = 6.

Consider the first step in which the addition of B and 7 is giving A (i.e., 3), then B should be a number such that the unit’s digit of this addition is 3.This is only possible when B = 6.

Then A = 6 + 7 = 13. However, A is a single digit number. Hence, it is not possible.

__Case 2: First step is producing a carry__

If we are getting a carryover on adding B and 7, then A comes out to be 2 as 1 + 2 + 3 = 6.

Consider the first step in which the addition of B and 7 is giving A (i.e., 2).That means B should be a number such that the units digit of this addition is 2. It is only possible when B = 5 since 5 + 7 = 12.

The addition will be as follows:

\begin{array}{l}\text{2 5}\\ \underset{\xaf}{\text{+ 3 7}}\\ \underset{\xaf}{\text{6 2}}\end{array}

Hence, the values of A and B are 2 and 5 respectively.

\begin{array}{l}\text{5}\text{. A B}\\ \underset{\xaf}{\text{}\times \text{3}}\\ \underset{\xaf}{\text{C A B}}\end{array}

The multiplication of 3 and B gives a number whose ones digit is B again. This is possible only when B is 0 or 5.

__Case 1: If B = 5__

Multiplication of first step = 3 × 5 = 15

1 will be a carry for the next step.

We have, 3 × A + 1 = CA

This is not possible for any value of A.

__Case 2: If B = 0__

If B = 0, then there will be no carry for the next step.

We should obtain, 3 × A = CA

That is, the one’s digit of 3 × A should be A.

This is possible when A = 5 or 0.

Here, A cannot be 0 as AB is a two-digit number.

Therefore, A must be 5 only. The multiplication is as follows.

\begin{array}{l}\text{5 0}\\ \underset{\xaf}{\text{}\times \text{3}}\\ \underset{\xaf}{\text{1 5 0}}\end{array}

Hence, the values of A, B, and C are 5, 0, and 1 respectively.

\begin{array}{l}\text{6}\text{. A B}\\ \underset{\xaf}{\text{}\times \text{5}}\\ \underset{\xaf}{\text{C A B}}\end{array}

The multiplication of B and 5 is giving a number whose ones digit is B again. This is possible when B = 5 or B = 0 only.

__Case 1: If B = 5__

Then, B × 5 = 5 × 5 = 25

Thus, 2 will be a carry for the next step.

We have, 5 × A + 2 = CA, which is possible for A = 2 or 7

Again we have two cases.

The multiplication is as follows.

\begin{array}{l}\text{2 5}\\ \underset{\xaf}{\text{}\times \text{5}}\\ \underset{\xaf}{\text{1 2 5}}\end{array}

\begin{array}{l}\text{7 5}\\ \underset{\xaf}{\text{}\times \text{5}}\\ \underset{\xaf}{\text{3 7 5}}\end{array}

__Case 2: If B = 0__

B × 5 = B ⇒ 0 × 5 = 0

There will not be any carry in this step.

In the next step, 5 × A = CA

It can happen only when A = 5 or A = 0

However, A cannot be 0 as AB is a two-digit number.

Hence, A can be 5 only. The multiplication is as follows.

\begin{array}{l}\text{5 0}\\ \underset{\xaf}{\text{}\times \text{5}}\\ \underset{\xaf}{\text{2 5 0}}\end{array}

Hence, there are 3 possible values of A, B, and C.

(i) 5, 0, and 2 (ii) 2, 5, and 1 (iii) 7, 5, and 3

\begin{array}{l}\text{7}\text{. A B}\\ \underset{\xaf}{\text{}\times \text{6}}\\ \underset{\xaf}{\text{B B B}}\end{array}

The multiplication of 6 and B gives a number whose one’s digit is B again.

It is possible only when B = 0, 2, 4, 6, or 8

__Case 1: If B = 0__

Then the product will be 0. Therefore, this value of B is not possible.

__Case 2: If B = 2__

If B = 2, then B × 6 = 12 and 1 will be a carry for the next step.

6A + 1 = BB = 22 ⇒ 6A = 21 which is not possible.

__Case 3: If B = 6__

If B = 6, then B × 6 = 36 and 3 will be a carry for the next step.

6A + 3 = BB = 66 ⇒ 6A = 63 .Therefore the value of A is not possible.

__Case 4: If B = 8__

Then B × 6 = 48 and 4 will be a carry for the next step.

6A + 4 = BB = 88 ⇒ 6A = 84 and hence, A = 14, but A is a single digit number. Therefore, this value of A is not possible.

__Case 5: If B = 4__

If B = 4, then B × 6 = 24 and 2 will be a carry for the next step.

6A + 2 = BB = 44 ⇒ 6A = 42 and hence, A = 7

The multiplication is as follows.

\begin{array}{l}\text{7 4}\\ \underset{\xaf}{\text{}\times \text{6}}\\ \underset{\xaf}{\text{4 4 4}}\end{array}

Hence, the values of A and B are 7 and 4 respectively.

\begin{array}{l}\text{8}\text{. A 1}\\ \underset{\xaf}{\text{+ 1 B}}\\ \underset{\xaf}{\text{B 0}}\end{array}

The addition of 1 and B is giving 0 i.e., a number whose ones digits is 0. This is possible only when digit B is 9. In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step,

1 + A + 1 = B

Clearly, A is 7 as 1 + 7 + 1 = 9 = B

Therefore, the addition is as follows.

\begin{array}{l}\text{7 1}\\ \underset{\xaf}{\text{+ 1 9}}\\ \underset{\xaf}{\text{9 0}}\end{array}

Hence, the values of A and B are 7 and 9 respectively.

\begin{array}{l}\text{9}\text{. 2 A B}\\ \underset{\xaf}{\text{+ A B 1}}\\ \underset{\xaf}{\text{B 1 8}}\end{array}

The addition of B and 1 is 8. This is possible only when digit B is 7. In the next step,

A + B = 1

Clearly, A is 4 since, 4 + 7 = 11 and 1 will be a carry for the next step.

In the next step,

1 + 2 + A = B

1 + 2 + 4 = 7

Therefore, the addition is as follows.

\begin{array}{l}\text{2 4 7}\\ \underset{\xaf}{\text{+ 4 7 1}}\\ \underset{\xaf}{\text{7 1 8}}\end{array}

Hence, the values of A and B are 4 and 7 respectively.

\begin{array}{l}\text{10}\text{. 1 2 A}\\ \underset{\xaf}{\text{+ 6 A B}}\\ \underset{\xaf}{\text{A 0 9}}\end{array}

Since both A and B are unknown to us in the first step, so let’s find the value of A from the next step.

In the next step, 2 + A = 0

It is possible only when A = 8

If we put the value of A = 8 in the first step, we get the value of B as 1.

In the second step, 2 + 8 = 10 and 1 will be the carry for the next step.

1 + 1 + 6 = A

Therefore, the addition is as follows.

\begin{array}{l}\text{1 2 8}\\ \underset{\xaf}{\text{+ 6 8 1}}\\ \underset{\xaf}{\text{8 0 9}}\end{array}

Hence, the values of A and B are 8 and 1 respectively.

##### FAQs (Frequently Asked Questions)

## 1. Where can students access the NCERT Class 8 Maths Chapter 16 Exercise 16.1?

Students can access the Class 8 Maths Exercise 16.1 Solution on the Extramarks website and Extramarks mobile application.

## 2. Is the Class 8 Maths Exercise 16.1 tough?

The difficulty level of the Class 8 Maths Chapter 16 Exercise 16.1 is average and students can easily manage to master the exercise with the help of the NCERT Solutions For Class 8 Maths Chapter 16 Exercise 16.1.