# NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1

NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 (Ex. 16.1) Playing with Numbers for academic session 2020-2021 updated as per latest CBSE Syllabus. All the contents are in videos and PDF file format free to use online or download.

Hindi Medium and English Medium solutions are given separately so that students do not face any difficulty. Class 8 Maths exercise 16.1 is very interesting exercise to solve or play with fun.## Class 8 Maths Chapter 16 Exercise 16.1 Solution

Class: 8 | Mathematics |

Chapter: 16 | Playing with Numbers |

Exercise: 16.1 | PDF and Video Solution |

### CBSE NCERT Class 8 Maths Chapter 16 Exercise 16.1 Solution in Hindi and English Medium

### Class 8 Maths Chapter 16 Exercise 16.1 Solution in Videos

#### Number in Generalised Form

A number is said to be in a generalised form if it is expressed as the sum of the products of its digits with their respective place values.

##### Two-digit Numbers

Consider a 2-digit number having the digit a at its tens place and the digit b at its units place. This number is (10a + b), where a can be any whole number from 1 to 9 and b can be any whole number from 0 to 9.

Examples:

(i) 36 = 10 x 3 + 6

(ii) 74 = 10 x 7 + 4

(iii) 90 = 10 x 9 + 0

(iv) 99 = 10 x 9 + 9

##### Three-digit Numbers

Consider a 3-digit number having a, b, c as its hundreds digit, tens digit and unit digit respectively.

This number is (100a + 10b + c), where a can be any whole number from 1 to 9, b can be any whole number from 0 to 9 and c can be any whole number from 0 to 9.

Example:

(i) 137 = 100 x 1 + 10 x 3 +7

##### In a 2-digit number, the units digit is four times the tens digit and the sum of the digits is 10. Find the number.

Solution: Let the tens digit be x.

Then, the units digit = 4x.

x + 4x = 10

5x = 10, x = 2.

Tens digit = 2 and units digit = (4 x 2) = 8.

Hence, the required number is (10 x 2 + 8) = 28.

##### The sum of the digits of a two-digit number is 8. The number obtained by interchanging its digits is 18 more than the original number. Find the original number.

Let the tens and units digits of the required number be a and b respectively. Then,

a + b = 8 (i)

Original number = 10a + b

Number obtained by interchanging its digits = 10b + a

So, (10b + a) – (10a + b) = 18

Or, 9(b – a) = 18

(b – a) = 2

Adding (i) and (ii), we get: 2b = 10 b = 5

Putting b = 5 in (i)

We get: a + 5 = 8

Or a = 3.

So, the original number = (10a + b) = (10 x 3 + 5) = 35

##### How can we write a 3-digit number in generalized form?

We can write a three digits’ number as:

Let number is abc, then we can write

100a + 10b + c

##### In a 3-digit number, the hundreds digit is twice the tens digit while the units digit is thrice the tens digit. Also, the sum of its digits is 18. Find the number.

Let the tens digit be x. Then, the hundreds digit = 2x and the units digit = 3x.

So, 2x + x + 3x = 18 or, 6x = 18 or, x = 3.

So, hundreds digit = (2 x 3) = 6, tens digit = 3 and units digit = (3 x 3) = 9.

Hence, the required number = (100 x 6 + 10 x 3 + 9) = 639.

##### In a 2-digit number, the units digit is four times the tens digit and the sum of the digits is 10. Find the number.

Let the tens digit be x. Then, the units digit = 4x

So, x + 4x = 10

Or, 5x = 10 or, x = 2

As given, tens digit = 2 and units digit = (4 x 2) = 8

Hence, the required number is (10 x 2 + 8) = 28