# Rd Sharma Solutions Class 6 Chapter 3

## Rd Sharma Solutions Class 6 Chapter 3 Whole Numbers

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Rd Sharma Class 6 Mathematics, Chapter 3, Whole Numbers. Here students can easily find exercise wise solution for chapter 3, Whole Numbers. Students will find proper solutions for Exercise 3.1. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.

Whole Numbers Exercise 3.1 Solution

Question no – (1)

Solution :

The smallest natural number is 1.

Question no – (2)

Solution :

The smallest whole number is 0

Question no – (3)

Solution :

The largest natural number does not exist.

Question no – (4)

Solution :

The largest whole number does not exist.

Question no – (5)

Solution :

Yes, all natural numbers are also whole numbers.

Question no – (6)

Solution :

No, all whole numbers are not natural numbers.

Question no – (7)

Solution :

(i) Given number, 1000909

Successor of the number is,

= 1000909 + 1

= 1000910

Therefore, 1000910 is successor.

(ii) Given number, 2340900

Successor of the number is,

= 2340900 + 1

= 2340901

Therefore, 2340901 is the successor.

(iii) Given number, 7039999

Successor of the number is,

= 7039999 + 1

= 7040000

Therefore, 7040000 is the successor.

Question no – (8)

Solution :

(i) Given number, 10000

The predecessor of the number is,

= (10000 – 1)

= 9999

Therefore, 9999 is the predecessor.

(ii) Given number, 807000

The predecessor of the number is,

= (807000 – 1)

= 806999

Therefore, 806999 is the predecessor.

(iii) Given number, 7005000

The predecessor of the number is,

= (7005000 – 1)

= 7004999

Therefore, 7004999 is the predecessor.

Question no – (9)

Solution :

Given numbers are,

= 2, 0, 3, 5, 7, 11, 15

∴ Now in the number line,

Question no – (10)

Solution :

The whole numbers between 21 and 61 are 39.

Question no – (11)

Solution :

(i) 25 < 205

(ii) 170 > 107

(iii) 415 < 514

(iv) 10001 < 100001

(v) 2300014 < 2300041

Question no – (12)

Solution :

Given numbers,

925, 786, 1100, 141, 325, 886, 0, 270

In descending order,

1100 > 925 > 886 > 786 > 325 > 270 > 141 > 0

Question no – (13)

Solution :

We know,

The largest number of 6 digit is 999999

The smallest no of 7 digits is 1000000.

And by,

(1000000 – 999999)

= 1

Therefore, 1000000 is the larger one by 1.

Question no – (14)

Solution :

The three consecutive whole numbers just proceeding 851001 are 8510000, 8509999, 850998.

Question no – (15)

Solution :

(i) 4009998 + 1

= 4009999,

(ii) 4009999 + 1

= 401000,

(iii) 401000 + 1

= 401001

So, next there consecutive whole numbers starting from 4009998 are 4009999, 401000, 401001.

Question no – (16)

Solution :

Every natural number has its successor.

Question no – (17)

Solution :

(i) Given statement is True.

(ii) Given statement is False.

(iii) Given statement is False.

(iv) Given statement is False.

Reason : Because 0 is the smallest whole number.

(v) Given statement is True.

(vi) Given statement is False.

(vii) Given statement is True.

(viii) Given statement is True.

(ix) Given statement is True.

(x) Given statement is False.

(xi) Given statement is False.

(xii) Given statement is True.

(xiii) Given statement is True.

(xiv) Given statement is True.

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Updated: June 7, 2023 — 10:24 am