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NCERT Solutions for Class 8 Maths Chapter 16: Playing with Numbers - Exercise 16.2

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NCERT Solutions for Class 8 Maths Chapter 16 (EX 16.2)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.2 and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 16 - Playing with Numbers

Exercise:

Exercise - 16.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Access NCERT Solution for Class 8 Mathematics Chapter 16.2 - Playing with Numbers

Exercise 16.2

1. If $21y5$ is a multiple of \[{\mathbf{9}}\], where \[{\mathbf{y}}\]is a digit, what is the value of \[{\mathbf{y}}\]?

Ans: If a number is a multiple of \[9\], then the sum of its digits will be divisible by \[9\].

Sum of digits of $21y5$

\[ = {\text{2}} + 1 + y + 5\]

\[ = 8 + y\].

 Hence, \[8 + y\]should be a multiple of\[9\]. 

When \[8 + y\] is any of these numbers \[0,9,18,27,\] and so on, this is possible.

However, because \[y\] is a single digit value, the sum can only be \[9\].

As a result, \[y\] should only be $1$.


2. If $31z5$ is a multiple of \[{\mathbf{9}}\], where\[\;{\mathbf{z}}\] is a digit, what is the value of \[\;{\mathbf{z}}\]? 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 \[0,9,18,27,\]and so on. However, since $z$ is a single digit number, this sum can be either \[9\] or $18$. Therefore, $z$ should be either $0$ or \[9\].


3. If \[{\mathbf{24x}}\] is a multiple of \[{\mathbf{3}}\], where \[{\mathbf{x}}\] is a digit, what is the value of \[{\mathbf{x}}\]?  (Since \[{\mathbf{24x}}\] is a multiple of \[{\mathbf{3}}\], its sum of digits \[{\mathbf{6}} + {\mathbf{x}}\] is a multiple of \[{\mathbf{3}}\]; so, \[{\mathbf{6}} + {\mathbf{x}}\] is one of these numbers: \[\mathbf{0},\mathbf{3},\mathbf{6},\mathbf{9},\mathbf{12},\mathbf{15},\mathbf{18}\]. But since \[{\mathbf{x}}\] is a digit, it can only be that \[{\mathbf{6}} + {\mathbf{x}}\]\[ = {\mathbf{6}}\] or \[{\mathbf{9}}\] or \[{\mathbf{12}}\] or \[{\mathbf{15}}\]. Therefore, \[\mathbf{x}=\mathbf{0}\]  or \[{\mathbf{3}}\] or \[{\mathbf{6}}\] or \[{\mathbf{9}}\]. Thus, \[{\mathbf{x}}\] can have any of four different values).

Ans: 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 $15$ and thus, the value of x comes to 0 or \[3\] or $6$ or $9$ respectively. Thus, x can have its value as any of the four different values \[0,3,6,\]or $9$. 


4. If $31z5$ is a multiple of \[{\mathbf{3}}\], where \[\;{\mathbf{z}}\] is a digit, what might be the values of \[\;{\mathbf{z}}\]? 

Ans: Since $31z5$  is a multiple of \[3\], the sum of its digits will be a multiple of \[3\]. That is, \[3 + 1 + z + 5\]

\[ = 9 + z\]

Thus, \[9 + z\]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. Because $z$ is a single digit number, the value of \[9 + z\] can only be $9$ or $12$ or \[15\] or \[18\] , and so the value of $x$ is 0 or [3], [6], or $9$. As a result, $z$ can take on any of the four possible values \[0,3,6,\] or $9$.


NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers

Opting for the NCERT solutions for Ex 16.2 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 16.2 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Playing with Numbers textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 16 Exercise 16.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 16 Exercise 16.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 16 Exercise 16.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.