NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers

NCERT Solutions of Class 6 Maths Chapter 2 wants to offer more knowledge on numbers, which is an extension to the first chapter for grade 6 students. NCERT Solutions for Class 6 Maths Chapter 2 offered by Study Studio provides well-disciplined, verified, structured material for the student Victor to solve several solved problems. These materials are used for competitive exams also. Every NCERT Solution is provided to make the study simple and interesting on Study Studio. Subjects like Science, Maths, English, and Hindi will become easy to study if you have access to NCERT Solution for Class 6 Science, Maths solutions, and solutions of other subjects.

Important Topics Covered in This Chapter

The following list has been provided so that students can get an idea of the topics that are covered in this chapter before diving into the solution to the NCERT textbook questions.

Introduction
Whole Numbers
Number Line
Properties of Whole Numbers
Patterns in Whole Numbers

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Access NCERT Solutions for Maths Chapter 2 – Whole Numbers

Exercise (2.1)

1. Write the Next Three Natural Numbers After

10999.

Ans:

Natural numbers are those numbers which start from positive integers and go on till infinity.

To find the next three natural numbers, just add

1 to every preceding integer.

So,

10,999+1=11,000

11,000+1=11,001

11,001+1=11,002

Thus the three natural numbers after

10,999 are

11000,11001,11002 .

2. Write the Three Whole Numbers Occurring Just Before

10001 .

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

To find the three whole numbers occurring before the number, just subtract

1 from every preceding integer.

So,

10,001−1=10,000

10,000−1=9,999

9,999−1=9,998

Thus the three whole numbers occurring before

10,001 are

10000,9999,9998 .

3. Which is the Smallest Whole Number?

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

Since the whole numbers start with zero, the smallest whole number is zero.

0 is the smallest number.

4. How Many Whole Numbers are There Between

32 and 53 ?

Ans:

To find the number of whole numbers between two numbers, we have to list out the numbers between

32 and 53 .

The numbers are

33,34,35,…,52 .

Numbers between

53 and

32=(53−32)−1

=20

So there are

20 whole numbers between

32 and 53 .

5. Write the Successor of

(a)

2440701

Ans:

The successor is the number that comes after the given number.

It can be found by adding

1 to the given number.

So,

2440701+1=2440702

So the successor of

2440701 is

2440702 .

(b)

100199.

Ans:

The successor is the number that comes after the given number.

It can be found by adding

1 to the given number.

So,

100199+1=100200

So the successor of

100199 is

100200 .

(c)

1099999.

Ans:

The successor is the number that comes after the given number.

It can be found by adding

1 to the given number.

So,

1099999+1=1100000

So the successor of

1099999 is

1100000.

(d)

2345670.

Ans;

The successor is the number that comes after the given number.

It can be found by adding

1 to the given number.

So,

2345670+1=2345671

So the successor of

2345670 is

2345671.

6. Write the Predecessor of

(a)

Ans:

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

94−1=93

So the predecessor of

94 is

93.

(b)

10000

Ans:

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

10,000−1=9,999

So the predecessor of

10,000 is

9,999.

(c)

208090

Ans:

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

2,08,090−1=2,08,089

So the predecessor of

2,08,090 is

2,08,089.

(d)

7654321

Ans:

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

76,54,321−1=76,54,320

So the predecessor of

76,54,321 is

76,54,320.

7. In Each of the Following Pairs of Numbers, State Which Whole Number is on the Left of the Other Number on the Number Line. Also,Write Them with the Appropriate Sign

(>,<)

(>,<)between them.

(a)

530,503

Ans:

Numbers in the number line always increase from left to right.

Here the smaller number is

503.

503 lies on the left of

530.

And

530>503

Hence,

503 lies on the left of

530 on the number line and

530>503.

(b)

370,307

Ans:

Numbers in the number line always increase from left to right.

Here the smaller number is

307.

307 lies on the left of

370.

And

370>307

Hence,

307 lies on the left of

370 on the number line and

370>307.

(c)

98765,56789

Ans:

Numbers in the number line always increase from left to right.

Here the smaller number is

56789.

56789 lies on the left of

98765.

And

98765>56789

Hence,

56789 lies on the left of

98765on the number line and

98765>56789.

(d)

9830415,10023001

Ans:

Numbers in the number line always increase from left to right.

Here the smaller number is

9830415.

9830415 lies on the left of

10023001.

And

9830415<10023001

Hence,

9830415 lies on the left of

10023001 on the number line and

9830415<10023001.

8. Which of the Following Statements are True (T) and Which are False (F):

(a) Zero is the Smallest Natural Number.

Ans:

Natural numbers are those numbers which start from positive integers and go on till infinity.

So the smallest natural number is

1and not zero.

So the given statement “Zero is the smallest natural number” is false.

(b)

400 is the Predecessor of

399.

Ans:

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

399−1=398

So the predecessor of

399 is

398.

So the given statement “

400 is the predecessor of

399” is false.

(c) Zero is the Smallest Whole Number.

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

So the smallest whole number is zero.

So the given statement “Zero is the smallest whole number” is true.

(d)

600 is the successor of

599.

Ans:

The successor is the number that comes after the given number.

It can be found by adding

1 to the given number.

So,

599+1=600

So the successor of

599 is

600.

So the given statement “

600 is the successor of

599:” is true.

(e) All Natural Numbers are Whole Numbers.

Ans:

Natural numbers are those which start from positive integers to infinity.

Whole numbers are those numbers which start from zero and go on till infinity.

So natural numbers will also come under whole numbers.

So all natural numbers are whole numbers.

So the given statement “All natural numbers are whole numbers” is true.

(f) All Whole Numbers are Natural Numbers.

Ans:

Natural numbers are those which start from positive integers to infinity.

Whole numbers are those numbers which start from zero and go on till infinity.

So all natural numbers are whole numbers but not all whole numbers are natural numbers since natural will miss zero from whole numbers.

So the given statement “All whole numbers are natural numbers” is false.

(g) The Predecessor of a Two Digit Number is Never a Single Digit Number.

Ans:

Let us consider a two digit number.

Let it be

10.

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

10−1=9

So the predecessor of

10 is

Thus the predecessor of a two digit number is a single digit number in this case.

So the given statement “The predecessor of a two digit number is never a single digit number’ is false.

(h)

1is the Smallest Whole Number.

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

So the smallest whole number is

So the given statement “

1 is the smallest whole number” is false.

(i) The Natural number

1 has no predecessor.

Ans:

Natural numbers are those which start from positive integers and go on till infinity.

So the smallest natural number is

1and it has no predecessor.

So the given statement “The natural number

1 has no predecessor” is true.

(j) The Whole Number

1 has no Predecessor.

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

The predecessor is the number that comes before the number.

It can be found by subtracting

1 from the given number.

So,

1−1=0

So the predecessor of

1 is

0 which is a whole number.

Thus the given statement “The whole number

1 has no predecessor” is false.

(k) The Whole Number

13 Lies Between

11 and

12.

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

0,1,2,3,…,11,12,13,…

The whole number

13 lies after

11,

12 and not between them.

So the given statement “The whole number

13 lies between

11 and

12 “ is false.

(l) The whole Number

0 has no Predecessor.

Ans:

Whole numbers are those numbers which start from zero and go on till infinity.

So the smallest number in the whole number is zero and it has no predecessor.

So the given statement “The whole number

0 has no predecessor” is true.

(m) The Successor of Two Digit Number is Always a Two Digit Number”

Ans:

Let us consider a two digit number.

Let it be

99.

The successor is the number that comes after the given number.

It can be found by adding

1 to the given number.

So,

99+1=100

So the successor of

99 is

100.

Thus the successor of a two digit number is a three digit number in this case.

So the given statement “The successor of a two digit number is always a two digit number’ is false.

Exercise (2.2)

1. Find the Sum by Suitable Rearrangement:

(a)

837+208+363

Ans:

Add unit places of a two digit number and check if we get an easier number.

And then add those numbers and simplify them like this to get the final answer.

837+208+363=(837+363)+208

=1200+208

=1408

So the sum of the numbers

837+208+363 by suitable arrangement is

1408.

(b)

1962+453+1538+647

Ans:

Add unit places of a two digit number and check if we get an easier number.

And then add those numbers and simplify them like this to get the final answer.

1962+453+1538+647=(1962+1538)+(453+647)

=3500+1100

=4600

So the sum of the numbers

1962+453+1538+647 by suitable arrangement is

4600.

2. Find the Product by Suitable Arrangement:

(a)

2×1768×50

Ans:

Arrange the numbers in such a way that we get easier numbers while multiplying them.

And multiply them again to get the final answer.

2×1768×50=(2×50)×1768

=(100×1768)

=176800

So the product of

2×1768×50 by suitable arrangement is

176800.

(b)

4×166×25

Ans:

Arrange the numbers in such a way that we get easier numbers while multiplying them.

And multiply them again to get the final answer.

4×166×25=(4×25)×166

=(100×166)

=16600

So the product of

4×166×25 by suitable arrangement is

16600.

(c)

8×291×125

Ans:

Arrange the numbers in such a way that we get easier numbers while multiplying them.

And multiply them again to get the final answer.

8×291×125=(125×8)×291

=(1000×291)

=291000

So the product of

8×291×125 by suitable arrangement is

291000.

(d)

625×279×16

Ans:

Arrange the numbers in such a way that we get easier numbers while multiplying them.

And multiply them again to get the final answer.

625×279×16=(625×16)×279

=(10000×279)

=2790000

So the product of

625×279×16 by suitable arrangement is

2790000.

(e)

285×5×60

Ans:

Arrange the numbers in such a way that we get easier numbers while multiplying them.

And multiply them again to get the final answer.

285×5×60=285×(5×60)

=(285×300)

=85500

So the product of

285×5×60 by suitable arrangement is

85500.

(f)

125×40×8×25

Ans:

Arrange the numbers in such a way that we get easier numbers while multiplying them.

And multiply them again to get the final answer.

125×40×8×25=(125×8)×(40×25)

=(1000×1000)

=1000000

So the product of

125×40×8×25 by suitable arrangement is

1000000.

3. Find the Value of the Following:

(a)

297×17+297×3

Ans:

It is in the form of

ab+ac

ab+ac.

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

297×17+297×3=297(17+3)

=297×20

=5940

So the value of

297×17+297×3 is

5940.

(b)

54279×92+8×54279

Ans:

It is in the form of

ab+ac.

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

54279×92+8×54279=54279(92+8)

=54279×100

=5427900

So the value of

54279×92+8×54279 is

5427900.

(c)

81265×169−81265×69

Ans:

It is in the form of

ab−ac.

So we can use distributive property over subtraction, that is,

a(b−c)=ab−ac

81265×169−81265×69=81265(169−69)

=81265×100

=8126500

So the value of

81265×169−81265×69 is

8126500.

(d)

3845×5×782+769×25×218

Ans:

The equation can be reduced as,

3845×5×782+769×25×218=3845×5×782+769×5×5×218

=3845×5×782+3845×5×218

It is in the form of

ab+ac.

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

3845×5×782+769×25×218=3845×5(782+218)

=3845×5×1000

=19225000

So the value of

3845×5×782+769×25×218 is

19225000.

4. Find the Product Using Suitable Properties:

(a)

738×103

Ans:

Write the given number

103 as

(100+3)

738×103=738×(100+3)

Use the distributive law over addition, that is,

a(b+c)=ab+ac

738×103=(738×100)+(738×3)

=73800+2214

=76014

The product of

738×103 by using proper identity is

76014.

(b)

854×102

Ans:

Write the given number

102 as

(100+2)

854×102=854×(100+2)

Use the distributive law over addition, that is,

a(b+c)=ab+ac

854×102=(854×100)+(854×2)

=85400+1708

=87108

The product of

854×102 by using proper identity is

87108.

(c)

258×1008

Ans:

Write the given number

1008 as

(1000+8)

258×1008=258×(1000+8)

Use the distributive law over addition, that is,

a(b+c)=ab+ac

258×1008=(258×1000)+(258×8)

=258000+2064

=260064

The product of

258×1008 by using proper identity is

260064.

(d)

1005×168

Ans:

Write the given number

1005 as

(1000+5)

1005×168=(1000+5)×168

Use the distributive law over addition, that is,

a(b+c)=ab+ac

1005×168=(1000×168)+(5×168)

=168000+840

=168840

The product of

1005×168 by using proper identity is

168840.

5. A taxi Driver Filled his Car’s Petrol Tank with

40 Litres of Petrol on Monday. The Next day, He Filled the Tank with

50 Litres of Petrol. If the Petrol Costs

Rs.44 Per Litre, How Much did he Spend on Petrol?

Ans:

Amount of Petrol filled on Monday

=40 litres

Amount of petrol filled on next day

=50 litres

Total amount of petrol filled

=40+50

=90 litres

Cost of one litre petrol

=Rs.44

Cost of

90 litre petrol

=44×90

=Rs.3960

Thus the taxi driver spent

Rs.3960 on petrol.

6. A Vendor Supplies

32 litres of milk to a hotel in the morning and

68 litres of milk in the evening. If the milk costs

Rs.15 per litre, how much money is due to the vendor per day?

Ans:

Amount of milk supply in the morning

=32 litres

Amount of milk supply in the evening

=68 litres

Total amount of milk supply

=32+68

=100 litres

Cost of one litre milk

=Rs.15

Cost of

100 litre milk

=15×100

=Rs.1500

Amount of money due to the vendor per day is

Rs.1500

7. Match the Following:

(i)

425×136=425×(6+30+100)

Ans:

425×136=425×(6+30+100)

This is in the form of

a(b+c)=ab+ac which is a distributivity multiplication under addition which is in the option (c).

(ii)

2×48×50=2×50×48

Ans:

2×48×50=2×50×48

This is in the form of

a×b×c=a×c×b which is commutative under multiplication which is in the option (a).

(iii)

80+2005+20=(80+20+2005)

Ans:

425×136=425×(6+30+100)

This is in the form of

a+b+c=a+c+b which is commutative under addition which is in the option (b).

Exercise (2.3)

1. Which of the Following will not Represent Zero:

(a)

1+0

Ans:

The given numbers represent the sum the numbers,

So,

1+0=1

So it gives an answer which is a non-zero number.

1+0 does not represent zero.

(b)

0×0

Ans:

The given numbers represent the multiplication of the numbers,

So,

0×0=0

0×0represents zero.

(c)

Ans:

The given numbers represent the division of the numbers,

So,

02=0

02 represents zero.

(d)

10−10

10−102

Ans:

The given numbers represent the difference of the numbers followed by division.

So,

10−10

10−102=0

10−10

10−102 represents zero.

2. If the Product of Two Whole Numbers is Zero, Can We Say That One or Both of Them Will Be Zero? Justify Your Examples.

Ans:

Let us consider two numbers as

4,0in which one of them is zero.

Product of these numbers is

4×0=0

If we consider both the numbers to be zero, then

0×0=0.

Yes, we can say that one or both of the whole numbers will be zero.

3. If the Product of Two Whole Numbers is

1, Can We Say That One or Both of the Numbers will be

1? Justify Through Examples.

Ans:

Let us consider both the whole number to be

1×1=1

Consider either of the numbers to be any other whole number, let it be

3×1=3

So the product of both the whole number will not always be equal to

No, we cannot say that the product of one or both the numbers will be

4. Find Using Distributive Property:

(a)

728×101

Ans:

We can

101 as

101=(100+1)

728×(100+1)

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

728×(100+1)=728×100+728×1

=72800+728

=73528

So the value of

728×101 is

73528.

(b)

5437×1001

Ans:

We can

1001 as

1001=(1000+1)

5437×(1000+1)

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

5437×(1000+1)=5437×1000+5437×1

=5437000+5437

=5442437

So the value of

5437×1001 is

5442437.

(c)

824×25

Ans:

We can

25 as

25=(20+5)

824×(20+5)

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

824×(20+5)=(824×20)+(824×5)

=16480+4120

=20600

So the value of

824×25 is

20600.

(d)

4275×125

Ans:

We can

125 as

125=(100+20+5)

4275×(100+20+5)

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

4275×(100+20+5)=(4275×100)+(4275×20)+(4275×5)

=427500+85500+21375

=534375

So the value of

4275×125 is

534375.

(e)

504×35

Ans:

We can

504 as

504=(500+4)

(500+4)×35

So we can use distributive property over addition, that is,

a(b+c)=ab+ac

(500+4)×35=500×35+4×35

=17500+140

=17640

So the value of

504×35 is

17640.

5.Study the pattern:

1×8+1=9

12×8+2=98

123×8+3=987

1234×8+4=9876

12345×8+5=98765

Ans:

Since it was like

12…n×8+n=n…1

The next two terms are

123456×8+6=987654

1234567×8+7=9876543

And the pattern will be like:

1×8+1=9

12×8+2=98

123×8+3=987

1234×8+4=9876

12345×8+5=98765

The next two terms are

987654,9876543.

NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers Free PDF

NCERT Solutions of Maths book for Class 6 are available in PDF format for free to download. It helps the students to practise whenever they want. Additionally, the book can be stored for future purposes like academic exams or competitive exams or olympiads etc. Students also need not bother about the uninterrupted internet connection. The free PDF is available to practise more examples in their leisure Time and also can verify or recall the sums during the time of examination. It also benefits the students to work together so they can share their knowledge with them.

2.1 Introduction

At the beginning of the chapter, the Whole Numbers of Class 6 by the NCERT Solutions are recalling the numbers and explaining the concepts of successor and predecessor. Students can easily understand because these are similar to before and after numbers. As the level of the student increases, the terminology also changes at times. So the student should be aware of all the terms because several students may lose their marks without knowing the terms even though they knew the concept very well.

2.2 Whole Numbers

After attaining the knowledge of the successor and predecessor, the students may get a doubt, what is the predecessor of one. So, in order to clarify this, Aryabhatta discovered the numeral called zero. All the natural numbers, including zero, are nothing but Whole Numbers. NCERT Solutions of the Maths book of Chapter 2 has explained it in detail. So we can say that the national numbers are a subset of Whole Numbers.

2.3 Number Line

The NCERT solutions of Class 6 Maths Chapter 2 Whole Numbers PDF has introduced A New concept to the students, which is called the number line. Students can understand it as simple as the numbers because it is representing the numbers on your straight line. The number line can be used to represent additions, subtractions, and multiplications too. The experience of teachers of NCERT solutions has been verified thoroughly and explained the concept with several examples. Also, students can assess themselves by using the exercises and practice questions.

2.4 Properties of Whole Numbers

As of now, the students got knowledge of ‘what are the Whole Numbers?’. So the NCERT Solutions of Study Studio wants to give more on each topic. For that purpose, students are about to learn about the properties of Whole Numbers. The Whole Numbers may satisfy different properties. They are as follows:

Closure Property: The Whole Numbers can hold the closure property for both the addition and multiplication.
Associative Property: The Whole Numbers also satisfy the associated property for both the addition and multiplication.
Distributive of Multiplication Over Addition: It means that the Whole Numbers can satisfy the distributivity of multiplication only on additions but not on multiplications.

2.5 Patterns in Whole Numbers

NCERT Solutions of Chapter 2 Mathematics Class 6 taught the students in an understandable way about the patterns available in the Whole Numbers. The pattern means the way of representing or the appearance of that particular number. Students can learn it very enthusiastically. Because here the numbers can appear in different shapes like a straight line, square, rectangle, or a triangle or any other. So the students can enjoy more and more numbers to represent their favourite shapes.

Class 6 Maths Chapter 2 Exercises

Chapter 2 Whole Numbers All Exercises in PDF Format
Exercise 2.1	8 Question & Solutions
Exercise 2.2	7 Questions & Solutions
Exercise 2.3	5 Questions & Solutions

Key Takeaways of the NCERT Solutions of Class 6 Maths Chapter 2: Whole Numbers

Students can take many things from NCERT Solutions on the official website of Study Studio. It is like an assorted back which consists of all types of requirements. The benefits of the NCERT solutions are:

The explanation will be understandable and liberally according to the level of the student.

It provides a test purpose and unsolved questions more than enough to become strong in that particular subject for every chapter.

The free PDF benefits the students in several ways and there will not be any further issues of the affordability of the parents.

The doubts of the students can be clarified through the live session available on the official website.

Along with this, students can also view additional study materials provided by Vedantu, for Class 6 Maths Chapter 1 – Knowing Our Numbers.

NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers

Important Topics Covered in This Chapter

Access NCERT Solutions for Maths Chapter 2 – Whole Numbers

Exercise (2.1)

Exercise (2.2)

Exercise (2.3)

NCERT Solutions for Class 6 Maths Chapter 2 – Whole Numbers Free PDF

2.1 Introduction

2.2 Whole Numbers

2.3 Number Line

2.4 Properties of Whole Numbers

2.5 Patterns in Whole Numbers

Class 6 Maths Chapter 2 Exercises

Key Takeaways of the NCERT Solutions of Class 6 Maths Chapter 2: Whole Numbers

Related Questions

Conclusion

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