## CBSE Revision Notes Class 8 Maths Chapter 16: Playing with Numbers

CBSE Revision Notes Class 8 Maths Chapter 16 Playing with Numbers are provided to help the students understand and revise the concepts right from the beginning. The concepts taught in Class 8 are important to be understood as these concepts are continued in classes 9 and 10. To score good marks in Class 8 mathematics examination, it is advised to solve questions provided in the Revision Notes Class 8 Maths Chapter 16. These revision notes for Class 8 Maths help the students to revise all the concepts in a better way.

Swiflearn provides Revision Notes and keynotes chapter wise for the CBSE board exam in an easy-to-understand, free downloadable PDF format so students can use it for their studies and score better in their board exams. The CBSE Class 8 Revision Notes are made for the main subjects of Science and Maths. These core subjects are very critical as they are the stepping stones and plays a crucial role in student’s future. They might be tricky for students. The CBSE Class 8 Revision Notes for each chapter will enable them to have an expert studying pattern with which they can enjoy learning the subject and perform better in the exams.

CBSE Class 8 Maths Revision Notes are designed keeping in my mind the exam pattern and syllabus of NCERT 2020-21. Students can download the PDF for free and practice the questions to score well in the coming exams.

## CBSE Revision Notes Class 8 Maths Chapter 16: Playing with Numbers

**Number in general form:-**

Let us take the number 53 and write it as

53 = 50 + 3 = 10 × 5 + 3

Similarly, the number 42 can be written as

42 = 10 × 4 + 2

In general, any two digit number ab made of digits a and b can be written as

ab = 10 × a + b = 10a + b

ba = 10 × b + a = 10b + a

Let us now take number 369. This is a three digit number. It can also be written as

369 = 300 + 60 + 9 = 100 × 3 + 10 × 6 + 1 × 9

Also 497 = 100 × 4 + 10 × 9 + 1 × 7

Similarly, In general, a 3-digit number abc made up of digits a, b and c is written as abc = 100 × a + 10 × b + 1 × c = 100a + 10b + c

In the same way,

cab = 100c + 10a + b

bca = 100b + 10c + a