RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers

Read RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers below, students should study RD Sharma class 6 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 6 Mathematics have been prepared by teacher of Grade 6. These RD Sharma class 6 Solutions have been designed as per the latest NCERT syllabus for class 6 and if practiced thoroughly can help you to score good marks in standard 6 Mathematics class tests and examinations

Exercise 3.1

 

Question 1:  Write down the smallest natural number. 

Solution 1:   The Smallest natural number is ‘1’.

 

Question 2:  Write down the smallest whole number. 

Solution 2:   The Smallest whole number is ‘0’.

 

Question 3:  Write down, if possible, the largest natural number. 

Solution 3:   As we know that every Natural number has a successor Number. So, there is not the largest Natural Number. 

 

Question 4:  Write down, if possible, the largest whole number. 

Solution 4:   As we know that every whole number has a successor Number. So, there is not the largest Whole Number. 

 

Question 5:  Are all natural numbers also whole numbers? 

Solution 5:   Yes, all the natural numbers also the whole numbers.

 

Question 6:  Are all whole numbers also natural numbers? 

Solution 6:   No, all the whole numbers are not the natural numbers because ‘0’ is not a Natural number but it is a Whole Number 

 

Question 7:  Give successor of each of the whole numbers?

(i) 1000909
(ii) 2340900
(iii) 7039999 

Solution 7: 

(i)       1000909 = 1000909 + 1 = 1000910

Successor of 1000909 is 1000910.

(ii)      2340900 = 2340900 + 1 = 2340901

Successor of 2340900 is 2340901.

(iii)  7039999 = 7039999 + 1 = 7040000

Successor of 7039999 is 7040000.

 

Question 8:  Write down the predecessor of each of the following whole numbers:

(i) 10000
(ii) 807000
(iii) 7005000 

Solution 8: 

(i)       10000 = 10000 - 1 = 9999

Predecessor of 10000 is 9999.

(ii)      807000 = 807000 - 1 = 806999

Predecessor r of 807000 is 806999.

(iii)  7005000 = 7005000 - 1 = 7004999

Predecessor of 7005000 is 7004999.

 

Question 9:  Represent the following numbers on the number line:

2,0,3,5,7,11,15 

Solution 9: 


RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers

Numbers are as shown in the above line.


Question 10:  How many whole numbers are there between 21 and 61? 

Solution 10:   There are 39 whole number between 21 and 61.

 

Question 11:  Fill in the blanks with the appropriate symbol < or >:

(i) 25______205
(ii) 170______107
(iii) 415______514
(iv) 10001______100001
(v) 2300014______2300041 

Solution 11: 

(i) 25 < 205

(ii) 170 > 107

(iii) 415 > 514

(iv) 10001 < 100001

(v) 2300014 < 2300041.

 

Question 12:  Arrange the following numbers is descending order:

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

Solution 12: 

Descending Series:

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

 

Question 13:  Write the largest number of 6 digits and the smallest number of 7 digits. Which one of these two is larger and by how much? 

Solution 13: 

Largest 6-digit number is: 999999

Smallest 7-digit number is: 1000000

The larger Number is 1000000 by 1

 

Question 14:  Write down three consecutive whole numbers just preceding 8510001. 

Solution 14: 

First Number: 8510001- 1 = 8510000,

Second Number: 8510000 – 1 = 8509999,

Third Number: 8509999- 1 = 8509998.

 

Preceding Numbers:

8510000, 8509999, 8509998.

 

Question 15:  Write down the next three consecutive whole numbers starting from 4009998. 

Solution 15: 

First Number: 4009998 + 1 = 4009999,

Second Number: 4009999 + 1 = 4010000,

Third Number: 4010000 + 1 = 4010001.

 

Successive Numbers:

4009999, 4010000, 4010001.

 

Question 16:  Give arguments in support of the statement that there does not exist the largest natural number. 

Solution 16:   Every natural number has its successes. Therefore, the largest natural number does not exist.

 

Question 17:   Which of the following statements are true and which are false? 

Solution 17: 

              Every whole number has its successor.                                                                                          True

              Every whole number has its predecessor.                                                                                       False

              0 is the smallest natural number.                                                                                                   False

              1 is the smallest whole number.                                                                                                     False

              0 is less than every natural number.                                                                                               True

              Between any two whole numbers there is a whole number.                                                              False

              Between any two non-consecutive whole numbers there is a whole number.                                     True

              The smallest 5-digit number is the successor of the largest 4 digit number                                       True

              Of the given two natural numbers, the one having more digits is greater.                                        True

              The predecessor of a two digit number cannot be a single digit number.                                          False

              If a and b are natural numbers and a < b, than there is a natural number c such that a<b<c.           False

              If a and b are whole numbers and a<b, then a+1< b+1.                                                                True

              The whole number 1 has 0 as predecessor.                                                                                    True

              The natural number 1 has no predecessor.                                                                                     True

 

 

Objective Type Questions


Mark the correct alternative in each of the following:

 

Question 1:  The smallest natural number is

(a) 0

(b) 1

(c) -1

(d) None of these 

Solution 1: (b)  1 is the smallest natural number.

 

Question 2:  The smallest whole number is

(a) 1

(b) 0

(c) -1

(d) None of these 

Solution 2: (b)  0 is the smallest whole number.

 

Question 3:  The predecessor of 1 in natural numbers is

(a) 0

(b) 2

(c) -1

(d) None of these 

Solution 3: (d)  1 is the smallest natural number and it does not have any predecessor.

 

Question 4:  The predecessor of 1 in whole numbers is

(a) 0

(b) -1

(c) 2

(d) None of these 

Solution 4: (a)

Predecessor of 1 is

= 1 – 1

= 0.

 

Question 5:  The predecessor of 1 million is

(a) 9999

(b) 99999

(c) 999999

(d) 1000001 

Solution 5: (c)  

1 million = 10, 00,000

= 10, 00,000 – 1

= 9, 99,999

 

Question 6:  The successor of 1 million is

(a) 10001

(b) 100001

(c) 1000001

(d) 10000001 

Solution 6: (c)  

1 million = 10, 00,000

= 10, 00,000 + 1

= 10, 00,001

 

Question 7:  The product of the successor and predecessor of 99 is

(a) 9800

(b) 9900

(c) 1099

(d) 9700 

Solution 7: (a) 

The successor of 99

= 99 + 1

= 100 

Predecessor of 99

= 99 − 1

= 98 

Product of successor and predecessor = 100 × 98 = 9800

 

Question 8:  The product of a whole number (other than zero) and its successor is

(a)  an even number

(b) an odd number

(c) divisible by 4

(d) divisible by 3 

Solution 8: (a) 

Let the whole number is = 3

The successor of 3 = 3 + 1 = 4

Product between them = 3 × 4 = 12

Hence, 12 is an even number.

 

Question 9: The product of the predecessor and successor of an odd natural number is always divisible by

(a) 2

(b) 4

(c) 6

(d) 8 

Solution 9: (b)

 

Question 10:  The product of the predecessor and successor of an even natural number is

(a) divisible by 2

(b) divisible by 3

(c) divisible by 4

(d) an odd number 

Solution 10: (d)

Let an even natural number = 4

The predecessor of 4 = 4 − 1 = 3

The successor of 4 = 4 + 1 = 5

Product between them = 3 × 5 = 15

Thus, the product is an odd number.

 

Question 11:  The successor of the smallest prime number is

(a) 1

(b) 2

(c) 3

(d) 4 

Solution 11: (c)  

The smallest prime number is 2

The Successor of 2 = 2 + 1

= 3

 

Question 12:  If x and y are co-primes, then their LCM is

(a) 1

(b) x/y

(c) xy

(d) None of these 

Solution 12: (c) 

The LCM of x and y is

 

Question 13: The HCF of two co-primes is

(a) the smaller number

(b) the larger number

(c) product of the numbers

(d) 1 

Solution 13: (d)  

The HCF of two co-prime numbers is 1.

 

Question 14:  The smallest number which is neither prime nor composite is

(a) 0

(b) 1

(c) 2

(d) 3 

Solution 14: (b) 

1 is the only smallest natural number which is neither prime nor composite.

 

Question 15:  The product of any natural number and the smallest prime is

(a) an even number

(b) an odd number

(c) a prime number

(d) None of these 

Solution 15: (a) 

The smallest prime number is 2.

Thus, multiplied by any natural number we get an even number.

 

Question 16:  Every counting number has an infinite number of

(a) factors

(b) multiples

(c) prime factors

(d) None of these 

Solution 16: (b)

Every counting number has an infinite number of multiples.

 

Question 17:  The product of two numbers is 1530 and their HCF is 15. The LCM of these numbers is

(a) 102

(b) 120

(c) 84

(d) 112

Solution 17: (a)

Product of two numbers = HCF × LCM

1530 = 15 × LCM

1530/15 = LCM

102 = LCM

Thus, the LCM of two numbers 102.

 

Question 18:  The least number divisible by each of the numbers 15, 20, 24 and 32 is

(a) 960

(b) 480

(c) 360

(d) 640 

Solution 18: (b) 

We know that the LCM of 15, 20, 24 and 32 is

Prime Factorization of 15 is = 3 × 5

Prime Factorization of 20 is = 2 × 2 × 5

Prime Factorization of 24 is = 2 × 2 × 2 × 3

Prime Factorization of 32 is = 2 × 2 × 2 × 2 × 2

Thus, the LCM = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 480

 

Question 19:  The greatest number which divides 134 and 167 leaving 2 as remainder in each case is

(a) 14

(b) 19

(c) 33

(d) 17 

Solution 19: (c) 

Subtract the remainder from given number

134 – 2 = 132

167 – 2 = 165

LCM of 132 and 165

Prime Factorization of 132 is = 2 × 2 × 3 × 11

Prime Factorization of 165 is = 3 × 5 × 11

HCF = 3 × 11 = 33

Thus, the greatest number which divides 134 and 167 leaving 2 as remainder is 33.

 

Question 20:  Which of the following numbers is a prime number?

(a) 91

(b) 81

(c) 87

(d) 97 

Solution 20: (d) 

Prime Factorization of 91 is = 1 × 7 × 13

Prime Factorization of 81 is = 1 × 3 × 3 × 3 × 3

Prime Factorization of 87 is = 1 × 3 × 29

Prime Factorization of 97 is = 1 × 97

Thus, the prime numbers is 97.

 

Question 21:  If two numbers are equal, then

(a) their LCM is equal to their HCF

(b) their LCM is less than their HCF

(c) their LCM is equal to two times their HCF

(d) None of these 

Solution 21: (a)

If two numbers are equal, then

HCF = LCM

 

Question 22:  a and b are two co-primes. Which of the following is/are true?

(a) LCM (a, b) = a × b

(b) HCF (a, b) = 1

(c) Both (a) and (b)

(d) Neither (a) nor (b) 

Solution 22: (c)

LCM and product of two co-prime numbers are equal.

Hence, the HCF of two co-prime numbers is 1.

 

RD Sharma Solutions Class 6 Maths
RD Sharma Solutions Class 6 Maths Chapter 1 Knowing our Numbers
RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers
RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers
RD Sharma Solutions Class 6 Maths Chapter 6 Fractions
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals
RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra
RD Sharma Solutions Class 6 Maths Chapter 9 Ratio Proportion and Unitary Method
RD Sharma Solutions Class 6 Maths Chapter 10 Basic Geomatrical Concepts
RD Sharma Solutions Class 6 Maths Chapter 11 Angles
RD Sharma Solutions Class 6 Maths Chapter 12 Triangle
RD Sharma Solutions Class 6 Maths Chapter 13 Quadrilaterals
RD Sharma Solutions Class 6 Maths Chapter 14 Circles
RD Sharma Solutions Class 6 Maths Chapter 15 Pair of Lines and Transversal
RD Sharma Solutions Class 6 Maths Chapter 16 Understanding Three Dimensional Shapes
RD Sharma Solutions Class 6 Maths Chapter 17 Symmetry
RD Sharma Solutions Class 6 Maths Chapter 18 Basic Geometrical Tools
RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions
RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration
RD Sharma Solutions Class 6 Maths Chapter 21 Data Handling Presentation of Data
RD Sharma Solutions Class 6 Maths Chapter 22 Data Handling Pictographs
RD Sharma Solutions Class 6 Maths Chapter 23 Data Handling Bar Graphs