NCERT Solutions For Class 6 Maths Chapter 2 : Whole Numbers

NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

NCERT Solutions of Class 6 Maths Chapter 2 want to offer more knowledge on the numbers, which is an extension to the first chapter for grade 6 students. After southern knowledge about numbers in the first chapter, students will empower the Whole Numbers. NCERT Solutions for Class 6 Maths Chapter 2 offered by Vedantu provides a well-discipline, verified, structured material for the student's Victor several solved problems. These materials are used for competitive exams also. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English,Hindi will become easy to study if you have access to NCERT Solution for Class 6 Science , Maths solutions and solutions of other subjects.

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Access NCERT solutions for Maths Chapter 2 – Whole Numbers part-1

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.

So $0$ 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 $32=\left( 53-32 \right)-1$

                                               $=20$

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) $94$

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 $\left( >,< \right)$between them.

(a) $530,503$

Ans:

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

Here the smaller number is $503$.

So $503$ 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 $307$.

So $307$ 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 $56789$.

So $56789$ lies on the left of $98765$.

And $98765>56789$

Hence, $56789$ lies on the left of $98765$on 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 $9830415$.

So $9830415$ 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 goes on till infinity.

So the smallest natural number is $1$and 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 $9$.

Thus the predecessor of 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) $1$is the Smallest Whole Number.

Ans:

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

So the smallest whole number is $0$.

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 goes on till infinity.

So the smallest natural number is $1$and 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 goes 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.

$837+208+363=\left( 837+363 \right)+208$

$=1200+208$

$=1408$

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.

$1962+453+1538+647=\left( 1962+1538 \right)+\left( 453+647 \right)$

$=3500+1100$

$=4600$

So the sum of the numbers $1962+453+1538+647$ by suitable arrangement is $4600$.


2. Find the Product by Suitable Arrangement:

(a) $2\times 1768\times 50$

Ans:

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

And multiply them again to get the final answer.

$2\times 1768\times 50=\left( 2\times 50 \right)\times 1768$

$=\left( 100\times 1768 \right)$

$=176800$

So the product of $2\times 1768\times 50$ by suitable arrangement is $176800$.

(b) $4\times 166\times 25$

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.

$4\times 166\times 25=\left( 4\times 25 \right)\times 166$

$=\left( 100\times 166 \right)$

$=16600$

So the product of $4\times 166\times 25$ by suitable arrangement is $16600$.

(c) $8\times 291\times 125$

Ans:

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

And multiply them again to get the final answer.

$8\times 291\times 125=\left( 125\times 8 \right)\times 291$

$=\left( 1000\times 291 \right)$

$=291000$

So the product of $8\times 291\times 125$ by suitable arrangement is $291000$.

(d) $625\times 279\times 16$

Ans:

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

And multiply them again to get the final answer.

$625\times 279\times 16=\left( 625\times 16 \right)\times 279$

$=\left( 10000\times 279 \right)$

$=2790000$

So the product of $625\times 279\times 16$ by suitable arrangement is $2790000$.

(e) $285\times 5\times 60$

Ans:

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

 And multiply them again to get the final answer.

$285\times 5\times 60=285\times \left( 5\times 60 \right)$

$=\left( 285\times 300 \right)$

$=85500$

So the product of $285\times 5\times 60$ by suitable arrangement is $85500$.

(f) $125\times 40\times 8\times 25$

Ans:

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

And multiply them again to get the final answer.

$125\times 40\times 8\times 25=\left( 125\times 8 \right)\times \left( 40\times 25 \right)$

$=\left( 1000\times 1000 \right)$

$=1000000$

So the product of $125\times 40\times 8\times 25$ by suitable arrangement is $1000000$.


3. Find the Value of the Following:

(a) $297\times 17+297\times 3$

Ans:

It is in the form of $ab+ac$.

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$297\times 17+297\times 3=297\left( 17+3 \right)$

$=297\times 20$

$=5940$

So the value of $297\times 17+297\times 3$ is $5940$.

(b) $54279\times 92+8\times 54279$

Ans:

It is in the form of $ab+ac$.

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$54279\times 92+8\times 54279=54279\left( 92+8 \right)$

$=54279\times 100$

$=5427900$

So the value of $54279\times 92+8\times 54279$ is $5427900$.

(c) $81265\times 169-81265\times 69$

Ans:

It is in the form of $ab-ac$.

So we can use distributive property over subtraction, that is, $a\left( b-c \right)=ab-ac$

$81265\times 169-81265\times 69=81265\left( 169-69 \right)$

                                         $=81265\times 100$

                                         $=8126500$

So the value of $81265\times 169-81265\times 69$ is $8126500$.

(d) $3845\times 5\times 782+769\times 25\times 218$

Ans:

The equation can be reduced as,

$3845\times 5\times 782+769\times 25\times 218=3845\times 5\times 782+769\times 5\times 5\times 218$

$=3845\times 5\times 782+3845\times 5\times 218$

It is in the form of $ab+ac$.

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$3845\times 5\times 782+769\times 25\times 218=3845\times 5\left( 782+218 \right)$

$=3845\times 5\times 1000$

$=19225000$

So the value of $3845\times 5\times 782+769\times 25\times 218$ is $19225000$.


4. Find the Product Using Suitable Properties:

(a) $738\times 103$

Ans:

Write the given number $103$ as $\left( 100+3 \right)$

$738\times 103=738\times \left( 100+3 \right)$

Use the distributive law over addition, that is, $a\left( b+c \right)=ab+ac$

$738\times 103=\left( 738\times 100 \right)+\left( 738\times 3 \right)$

$=73800+2214$

$=76014$

The product of $738\times 103$ by using proper identity is $76014$.

(b) $854\times 102$

Ans:

Write the given number $102$ as $\left( 100+2 \right)$

$854\times 102=854\times \left( 100+2 \right)$

Use the distributive law over addition, that is, $a\left( b+c \right)=ab+ac$

$854\times 102=\left( 854\times 100 \right)+\left( 854\times 2 \right)$

$=85400+1708$

$=87108$

The product of $854\times 102$ by using proper identity is $87108$.

(c) $258\times 1008$

Ans:

Write the given number $1008$ as $\left( 1000+8 \right)$

$258\times 1008=258\times \left( 1000+8 \right)$

Use the distributive law over addition, that is, $a\left( b+c \right)=ab+ac$

$258\times 1008=\left( 258\times 1000 \right)+\left( 258\times 8 \right)$

 $=258000+2064$

 $=260064$

The product of $258\times 1008$ by using proper identity is $260064$.

(d) $1005\times 168$

Ans:

Write the given number $1005$ as $\left( 1000+5 \right)$

$1005\times 168=\left( 1000+5 \right)\times 168$

Use the distributive law over addition, that is, $a\left( b+c \right)=ab+ac$

$1005\times 168=\left( 1000\times 168 \right)+\left( 5\times 168 \right)$

$=168000+840$

$=168840$

The product of $1005\times 168$ by using proper identity is $168840$.


5. A taxi Driver Filled his Car's Petrol Tank with $40$ Liters of Petrol on Monday. The Next day, He Filled the Tank with $50$ Liters of Petrol. If the Petrol Costs $Rs.44$ Per Liter, How Much did he Spend on Petrol?

Ans:

Amount of Petrol filled on Monday$=40$ liters

Amount of petrol filled on next day$=50$ liters

Total amount of petrol filled$=40+50$

$=90$ liters

Cost of one liter petrol$=\text{Rs}\text{.44}$

Cost of $90$ liter petrol$=44\times 90$

$=\text{Rs}\text{.3960}$

Thus the taxi driver spent $\text{Rs}\text{.3960}$ on petrol.


6. A Vendor Supplies $32$ liters of milk to a hotel in the morning and $68$ liters of milk in the evening. If the milk costs $\text{Rs}\text{.15}$ per liter, how much money is due to the vendor per day?

Ans:

Amount of milk supply in the morning$=32$ liters

Amount of milk supply in the evening$=68$ liters

Total amount of milk supply$=32+68$

$=100$liters

Cost of one liter milk$=\text{Rs}\text{.15}$

Cost of $100$ liter milk$=15\times 100$

$=\text{Rs}\text{.1500}$

Amount of money due to the vendor per day is $\text{Rs}\text{.1500}$


7. Match the Following:

(i) $425\times 136=425\times \left( 6+30+100 \right)$

Ans:
$425\times 136=425\times \left( 6+30+100 \right)$

This is in the form of $a\left( b+c \right)=ab+ac$ which is a distributivity multiplication under addition which is in the option (c).

(ii) $2\times 48\times 50=2\times 50\times 48$

Ans:
$2\times 48\times 50=2\times 50\times 48$

This is in the form of $a\times b\times c=a\times c\times b$ which is commutative under multiplication which is in the option (a).

(iii) $80+2005+20=\left( 80+20+2005 \right)$

Ans:
$425\times 136=425\times \left( 6+30+100 \right)$

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.

So $1+0$ does not represent zero.

(b) $0\times 0$

Ans:

The given numbers represent the multiplication of the numbers,

So, $0\times 0=0$

So $0\times 0$represents zero.

(c) $\dfrac{0}{2}$

Ans:

The given numbers represent the division of the numbers,

So, $\dfrac{0}{2}=0$

So $\dfrac{0}{2}$ represents zero.

(d) $\dfrac{10-10}{2}$

Ans:

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

So, $\dfrac{10-10}{2}=0$

So $\dfrac{10-10}{2}$ 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 $4,0$in which one of them is zero.

Product of these numbers is $4\times 0=0$

If we consider both the numbers to be zero, then$0\times 0=0$. 

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 $1$.

$1\times 1=1$

Consider either of the numbers to be any other whole number, let it be $3$.

$3\times 1=3$

So the product of both the whole number will not always be equal to $1$.

No, we cannot say that the product of one or both the numbers will be $1$.


4. Find Using Distributive Property:

(a) $728\times 101$

Ans:

We can $101$ as $101=\left( 100+1 \right)$

$728\times \left( 100+1 \right)$

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$728\times \left( 100+1 \right)=728\times 100+728\times 1$

$=72800+728$

$=73528$

So the value of $728\times 101$ is $73528$.

(b) $5437\times 1001$

Ans:

We can $1001$ as $1001=\left( 1000+1 \right)$

$5437\times \left( 1000+1 \right)$

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$5437\times \left( 1000+1 \right)=5437\times 1000+5437\times 1$

$=5437000+5437$

$=5442437$

So the value of $5437\times 1001$ is $5442437$.

(c) $824\times 25$

Ans:

We can $25$ as $25=\left( 20+5 \right)$

$824\times \left( 20+5 \right)$

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$824\times \left( 20+5 \right)=\left( 824\times 20 \right)+\left( 824\times 5 \right)$

$=16480+4120$

$=20600$

So the value of $824\times 25$ is $20600$.

(d) $4275\times 125$

Ans:

We can $125$ as $125=\left( 100+20+5 \right)$

$4275\times \left( 100+20+5 \right)$

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$4275\times \left( 100+20+5 \right)=\left( 4275\times 100 \right)+\left( 4275\times 20 \right)+\left( 4275\times 5 \right)$

$=427500+85500+21375$

$=534375$

So the value of $4275\times 125$ is $534375$.

(e) $504\times 35$

Ans:

We can $504$ as $504=\left( 500+4 \right)$

$\left( 500+4 \right)\times 35$

So we can use distributive property over addition, that is, $a\left( b+c \right)=ab+ac$

$\left( 500+4 \right)\times 35=500\times 35+4\times 35$

$=17500+140$

$=17640$

So the value of $504\times 35$ is $17640$.


5.Study the pattern:

        $1\times 8+1=9$

      $12\times 8+2=98$

    $123\times 8+3=987$

  $1234\times 8+4=9876$

$12345\times 8+5=98765$

Ans:

Since it was like $12...n\times 8+n=n...1$

The next two terms are $123456\times 8+6=987654$

                                   $1234567\times 8+7=9876543$

And the pattern will be like:

        $1\times 8+1=9$

      $12\times 8+2=98$

    $123\times 8+3=987$

  $1234\times 8+4=9876$

$12345\times 8+5=98765$

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 practice whenever they want. Adding, 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 practice 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 the 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 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 Vedantu 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 and eagerly to know all the numbers and their shapes.


We Cover all Exercises in the Chapter Given Below:

Exercise - 2.1 - 8 Questions with Solutions.

Exercise - 2.2 - 7 Questions with Solutions.

Exercise - 2.3 - 5 Questions with Solutions.


Key Takeaways of the NCERT Solutions of Class 6 Maths Chapter 2: Whole Numbers 3

Students can take many things from NCERT Solutions on the official website of Vedantu. 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 issue of the affordability of the parents.

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

FAQs (Frequently Asked Questions)

1. Explain the Closure Property on Addition and Multiplication with an Example.

The closure property of Whole Numbers will satisfy both additions and multiplications.

For addition, a + b = b + a

3000 + 5000 = 8000

5000 + 3000 = 8000

For multiplication,  a*b = b*a

25 * 50 = 1250

50 * 25 = 1250

2. Explain the Associative Property on Addition and Multiplication with an Example.

The Whole Numbers can satisfy the associated property on both additions and multiplications similar to the closure property. Let's see an example of this.

For addition, a+(b+c) = (a+b)+c 

                        2+ (4+6) = (2+4)+6 = 12

For multiplication, a*(b*c )= (a*b) *c

                                 2*(4*6) = (2*4)*6 = 48

3. What are the Properties Available in Mathematics for Numbers?

Generally, we have six properties for the numbers. However, there will not be any compulsion that all members should satisfy all the properties. Some of the properties were satisfied and some may not. The basic properties available are -

  • Closure property

  • Associative property

  • Commutative property

  • Distributive property

  • Identity property

  • Inverse property etc.

4. Where can I access the solutions for Class 6 Maths Chapter 2?

Class 6 Chapter 2 is an easy chapter if practised regularly. To make it easier, you can easily avail the solutions on Vedantu. The solutions can easily be accessed free of cost via the link given. There are a variety of modules and example papers available on the Vedantu website and the Vedantu mobile app for those who are interested and want to do well in their exams. 

5. What is the whole number in Class 6 Maths Chapter 2?

Whole numbers are basic counting numbers in mathematics: 0, 1, 2, 3, 4,... These include 55, 88, 69856555 etc. Natural numbers beginning with 1 are included in the definition of whole numbers. Positive integers and 0 are included in whole numbers. For more information and guidance you can visit the Vedantu Site (vedantu.com) or the Vedantu app. The problems of this chapter can be tricky sometimes and hence it is advisable to practice them carefully even if you find them simple. Practice well for your exams!

6. Do I need to practice all the questions provided in Class 6 Maths Chapter 2 NCERT Solutions?

Indeed. It is very important that you practice and answer all questions since they cover a variety of subjects and concepts and will give you a good understanding of the kind of questions that might be set from those areas and the framework of the question paper. These questions also help you learn how different questions from the same topic may be set. Each exercise should be thoroughly revised. The Vedantu website and Vedantu mobile app both provide a variety of modules as well as example papers on these subjects if you're so inclined.

7. What is the number 0?

Zero is a number that can be classified as a whole number, a real number, and a non-negative integer. It is not classified as undercounting, odd, positive natural or negative whole numbers and neither a complex number. It is quite tricky to classify zero into different categories. It can be included in multiple equations involving complex numbers though. Practice all the problems related to these topics and other topics of this chapter as well in order to score well in them.

8. Can zero be classified as a natural number?

Zero is a number that is a whole number, a real number, and a non-negative integer. It is neither a counting, odd, positive natural, or negative whole number, nor is it a complex number. It is difficult to categorise zero in numerous ways. It can, however, be used in numerous equations involving complex values.  If you are interested in obtaining various modules and example papers relating to these areas, you can simply get them through the Vedantu website as well as the Vedantu mobile app.

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