NCERT Solutions Class 8 Maths Chapter 16
NCERT Solutions for Class 8 Mathematics Chapter 16 – Playing with Numbers
Mathematics is a subject that requires consistency, hard work, and dedication. You can be a topper in this subject by practicing many questions, learning many concepts, and making your calculations faster. The knowledge of calculations covered in this subject will also be useful in other subjects.
Class 8 Mathematics Chapter 16 is related to Numbers. In this chapter, we will cover topics like general forms of numbers. We will learn to play with numbers such as reversing the digit – two-digit numbers & three-digit numbers and later learn about the divisibility test for the numbers.
To master Mathematics, students will need great study resources, which will help them build a strong understanding of each chapter. Extramarks has been providing high-quality and comprehensive study materials for many years.
Our study material is designed in simple language for a Class 8 student to easily understand all topics and sub-topics covered in the NCERT textbook. Experts design Extramarks NCERT Solutions for Class 8 Mathematics Chapter 16 with many years of mathematics experience. We provide notes, revision notes, important questions and solutions, and more. Students looking for the right study resources can register on our website and get full access to our solutions.
Key Topics Covered In NCERT Solutions for Class 8 Mathematics Chapter 16
In this chapter, Playing with Numbers of Class 8 Mathematics, we will learn about numbers. We will also cover topics like Numbers in general forms, playing with numbers which will cover reversing the two and three digits, and test of divisibility for the digits. This chapter will give students a lot of knowledge regarding numbers which play a great role in Mathematics.
NCERT Class 8 Mathematics Chapter 16 requires students to use their critical thinking ability and apply a wide range of formulas they have learned.
Introduction
In earlier classes, we learned about numbers, like natural numbers, whole numbers, integers, and rational numbers and how to find out the factors, multiples and relationships. This chapter is an extension of earlier chapters and discusses 2-digit, and 3-digit numbers and reversing them, puzzles involving letters of the digits, divisibility rules, etc.
Numbers in a general form
In this chapter, we will learn how to convert numbers into general form.
- For example, if there is a number 52, then its general form will be
52 = 50 + 2 = 5*10+2
Game with numbers
In this section, we will learn about some topics which are given below,
- Reversing the digit – two digits
- Reversing the digit – three digits
The topics given above are available in detail on our Extramarks website. To learn more about these topics, please visit our website and refer to NCERT Solutions for Class 8 Mathematics Chapter 16.
Letters of digits
Here we will solve the puzzle in which the letter takes the place of the digit in the problem of addition and multiplication.
We follow two rules in such puzzles:
- Each letter represents just one digit and stands for just one digit.
- The first digit of the number can’t be zero.
To view more examples of the application of these two rules, students can refer to our NCERT Solutions for Class 8 Mathematics Chapter 16.
Tests of Divisibility
This section discusses tests of divisibility. The test of divisibility for different digits is listed below:
- Divisibility BY 10
- To divide by ten, the last digits of the number must be zero.
- Divisibility BY 5
- To divide by five, the last digit of the number must be 5 or 0.
- Divisibility BY 2
- To divide by two, the last digit of the number must be 0,2,4,6,8.
- Divisibility BY 9 & 3
- To be divisible by three, if the sum of all digits of the number will divide by 3
- To be divisible by nine if the sum of all digits of the number will divide by 9.
Students need to memorize these divisibility rules as they will be very useful for doing quick calculations in future Mathematics classes. We have given additional tips for remembering these rules in our NCERT Solutions for Class 8 Mathematics Chapter 16. Students can register on the Extramarks website and get full access to our NCERT Solutions.
NCERT Solutions for Class 8 Mathematics Chapter 16: Exercise & Answer Solutions
Students can find answer solutions to all the exercises given in Class 8 Mathematics Chapter 16 in the NCERT Solutions for Class 8 Mathematics Chapter 16. It is designed in a way that students can easily understand all concepts and will be able to solve all types of questions related to the chapter. Students can get well-structured answer formats to help them get all solutions compiled in one place.
On our Extramarks website, you can search for the solutions related to Chapter 16. The solutions are completely accurate and error-free as the subject matter expertise makes them. As a result, students have access to correct and trustworthy study material. You can get all NCERT textbooks and related resources on the Extramarks website.
Click on the links below to view exercise-specific questions and solutions for NCERT Solutions for Class 8 Mathematics Chapter 16:
Chapter 16 Mathematics Class 8 NCERT Solutions – Exercise and Answer Solutions
Chapter 16 Playing with Numbers All Exercises | |
Exercise 16.1 | Question and Solutions |
Exercise 16.2 | Questions and Solutions |
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NCERT Exemplar Class 8 Mathematics
Students must practice a lot to excel in their competitive examinations. This can only be achieved if they have a collection of question banks related to their syllabus. Keeping this in mind, NCERT Exemplar Class 8 Mathematics was designed by experienced teachers. It has all NCERT-related questions that help students better understand the concepts they have learned.
Even if students have many questions to practice, they must practice questions they will be facing in their examinations. Hence, the book has problems from the exam point of view, thereby making the students’ mindsets exam-oriented. As a result, they find their real examinations easy.
You can get NCERT Exemplar from the Extramarks website. The book will surely change your preparation process and make you step ahead in whatever you do. Thus ensuring you stand out from the rest of the class. This will help build your overall personality and make a difference in your dealings in real life. Hence, it is a perfect resource to include in the study material.
Key Features of NCERT Solutions for Class 8 Mathematics Chapter 16
Good results come from accurate and correct resources. Extramarks study materials are prepared by the best Mathematics teachers who have spent years in the teaching profession. So our study materials are rated the best in the industry by students, parents, and teachers.
Below are a few key features of our NCERT Solutions for Class 8 Mathematics Chapter 16:
- The solution is prepared with easy-to-understand language using a lot of visual demonstrations to make learning easier for Class 8 students. Many students find learning more interesting when studying from Extramarks study materials.
- A lot of exercises are given in our solutions. So students learn to solve problems within a stipulated time and can complete their papers on time.
- After completing our NCERT Solutions for Class 8 Mathematics Chapter 16, students enjoy dealing with numbers and their applications.
Q.1 If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Ans-
If a number is a multiple of 9, then the sum of its digits will be divisible by 9.
Sum of digits of 21y5 = 2 + 1 + y + 5 = 8 + y
Hence, 8 + y should be a multiple of 9.
This is possible when 8 + y is any one of these numbers 0, 9, 18, 27, and so on …
Since y is a single digit number, therefore, y can be 1 only.
Q.2 If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Ans-
If a number is a multiple of 9, then the sum of its digits will be divisible by 9.
Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z
Hence, 9 + z should be a multiple of 9.
This is possible when 9 + z is any one of these numbers 0, 9, 18, 27, etc.
Since z is a single digit number, the sum can be either 9 or 18.
Therefore, z should be either 0 or 9.
Q.3 If 24x is a multiple of 3, where x is a digit, what is the value of x?
Ans-
If a number is a multiple of 3, then the sum of its digits should also be a multiple of 3.
Since 24x is a multiple of 3, the sum of its digits is a multiple of 3.
Sum of digits of 24x = 2 + 4 + x = 6 + x
Hence, 6 + x is a multiple of 3.
This is possible when 6 + x is any one of these numbers 0, 3, 6, 9, and so on.
Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15.
Thus, the value of x comes to 0 or 3 or 6 or 9 respectively.
Therefore, x can have any of the four different values 0, 3, 6, or 9.
Q.4 If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Ans-
If a number is a multiple of 3, then the sum of its digits should also be a multiple of 3.
Since 31z5 is a multiple of 3, the sum of its digits will be a multiple of 3.
That is, 3 + 1 + z + 5 = 9 + z is a multiple of 3.
This is possible when 9 + z is any one of 0, 3, 6, 9, 12, 15, 18, and so on.
Since z is a single digit number, 9 + z= 9 or 9 + z= 12 or 9 + z= 15 or 9 + z= 18
Therefore, the value of z = 0 or z = 3 or z = 6 or z = 9.
Q.5 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. Are NCERT Solutions for Class 8 Mathematics Chapter 16 reliable?
NCERT Solutions for Class 8 Mathematics Chapter 16 is written by experienced teachers in the field of Mathematics. Moreover, it is reviewed and re-checked by the academic team. It is designed keeping in mind the latest CBSE syllabus.
Even the students of well-known schools have passed their examinations with flying colours after referring to Extramarks NCERT solutions. A number of teachers and parents also appreciate its original content. Hence, the NCERT solutions are reliable, and students can ensure their success once they study from the NCERT solutions provided by Extramarks.
2. How can I improve my understanding of numbers in Mathematics?
You can improve your understanding of numbers in Mathematics by studying from NCERT Solutions for Class 8 Mathematics Chapter 16. The chapter briefly talks about the role of numbers in Mathematics. It has various sub-topics which help develop understanding and students’ interest in the study of numbers.
You will also find various activities and illustrations in which you will play with numbers and their operations and start loving the subject. Students relying on Extramarks solutions have increasingly improved their knowledge leading to improved exam scores.
3. How to test the divisibility of a number by 8?
If the last three digits of a number are divisible by eight or if the last three digits are zero, then the number is divisible by 8.