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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