# NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers (Ex 2.3) Exercise 2.3

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers (Ex 2.3) Exercise 2.3

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Exercise – 2.3

1. Which of the following will not represent zero:

(a) $\mathbf{1}+\mathbf{0}$

(b) $\mathbf{0}\times \mathbf{0}$

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

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

Ans.

Note that, product of two zero is zero, so

$0\times 0=0$.

Zero divided by any nonzero number is zero, so

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

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

$1+0=1$.

Thus, option (a) is the correct answer.

2. If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

Ans.

Yes, the statement described in the question is appropriate, because if any nonzero number is multiplied by zero, then the resultant product is zero. For example:

$1\times 0=0,$ $2\times 0=0$, $3\times 0=0$, $10\times 0=0$, etc.

Again, if zero is multiplied by zero, then the resultant product is also zero, that is, $0\times 0=0$.

3. If the product of two whole numbers is $\mathbf{1}$, can we say that one or both of them will be $\mathbf{1}$ ? Justify through examples.

Ans.

No, the statement described in the question is not correct, because if any whole number other than $1$ is multiplied with $1$, then the resultant product cannot be $1$. For example:

$0\times 1=0$, $2\times 1=2$, $3\times 1=3$, $5\times 1=5$.

Now, if $1$ is multiplied by $1$, then the resultant is also $1$, that is $1\times 1=1$.

Thus, if the resultant of the product of two whole numbers is $1$, then both the numbers are needed to be $1$.

4. Find using distributive property :

(i) $\mathbf{728}\times \mathbf{101}$

Ans.

By the distributive property, if $a,b,c$ are three whole numbers, then

$a\times \left( b + c \right)=a\times b+a\times c$.

The given product is

$728\times 101$

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

$=728\times 100+728\times 1$, using the distributive property.

$=72800+728$, multiplying the numbers.

$=73528$.

(ii) $\mathbf{5437}\times \mathbf{1001}$

Ans.

The given product is

$5437\times 1001$

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

$=5437\times 1000+5437\times 1$, using the distributive property.

$=5437000+5437$, multiplying the numbers.

$=5442437$.

(iii) $\mathbf{824}\times \mathbf{25}$

Ans.

The given product is

$824\times 25$

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

$=824\times 20+824\times 5$, using the distributive property.

$=16480+4120$, multiplying the numbers.

$=20600$.

(iv) $\mathbf{4275}\times \mathbf{125}$

Ans.

The given product is

$4275\times 125$

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

$=4275\times 100+4275\times 20+4275\times 5$, using the distributive property.

$=427500+85500+21375$, multiplying the numbers.

$=534375$.

(v) $\mathbf{504}\times \mathbf{35}$

Ans.

The given product is

$504\times 35$

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

$=500\times 35+4\times 35$, applying the distributive property.

$=17500+140$, multiplying the numbers.

$=17640$.

5. Study the pattern :

$\mathbf{1\times 8+1=9}$;                      $\mathbf{12}\times \mathbf{8}+\mathbf{2}=\mathbf{98}$;                 $\mathbf{123}\times \mathbf{8}+\mathbf{3}=\mathbf{987}$;

$\mathbf{1234}\times \mathbf{8}+\mathbf{4}=\mathbf{9876}$;          $\mathbf{12345\times 8+5=98765}$

Write the next two steps. Can you say how the pattern works? (Hint: $\mathbf{12345=11111+1111+111+11+1}$).

Ans.

The next two steps will be

$123456\times 8+6=987654$;

$1234567\times 8+7=9876543$.

The pattern described in the question will be like the following way:

$\left( 1 \right)\times 8+1=9$

$\left( 12 \right)\times 8+2=\left( 11+1 \right)\times 8+2=98$

$\left( 123 \right)\times 8+3=\left( 111+11+1 \right)\times 8+3=987$

$\left( 1234 \right)\times 8+4=\left( 1111+111+11+1 \right)\times 8+4=9876$

$\left( 12345 \right)\times 8+5=\left( 11111+1111+111+11+1 \right)\times 8+5=98765$

$\left( 123456 \right)\times 8+6=\left( 111111+11111+1111+111+11+1 \right)\times 8+6=987654$

$\left(1234567\right)\times8+7=\left(1111111+111111+11111+1111+111+11+1 \right)\times 8+7=9876543$

## Class 6 Maths Exercise 2.3

Vedantu Class 6 NCERT solutions are ideal for students who are looking for question wise solutions for Integers. Our solutions are ideal for last-minute revisions before examinations to ensure a detailed knowledge of the chapter. The ease of usage and download has proved us to be a better student-related guide. With our app easily available from play store, many students now prefer our solutions as their study partner.

### Maths Chapter 2 Whole Numbers:

The numbers like 1,2,3 etc are known as natural numbers. If zero is included with these, they become whole numbers.

So now the whole numbers are natural numbers along with zero.

If you add 1 to a whole number, the solution becomes a successor. Similarly, if you remove 1 from a whole number, you get a predecessor.

Example 16 + 1= 17 is successor

16 - 1= 15 is a predecessor

Earlier in natural numbers that starts from 1, the number 1 had no predecessor, so including zero as the predecessor of 1 we get whole numbers.

Similarly, if you include negative and positive numbers along with zero are known as integers.

### Representation of Whole Numbers on a Number Line

The numbers are represented on a number line with zero at the start. The distance between numbers zero and 1 is called unit distance.

### Order of Numbers

The numbers towards the right are greater numbers. Whereas the numbers towards the left are smaller numbers.

8>4.

8 is towards right and 4 is placed towards left on the number line.

### Addition on the Number Line

For addition of numbers on the number line, we need to move the indicated steps towards right.

Example if we have to add 3 to 7, we shift 3 places towards right from 7.

### Subtraction on the Number Line

Similarly, for subtraction of numbers on the number line, we need to move the indicated steps towards the left.

Example if we have to subtract 3 from 7, we shift 3 places towards left from 7.

### Multiplication on the Number Line

Now for multiplication we need to move the same number of steps, for the indicated number of times starting from zero.

Example 4 ☓ 3

Starting from zero we shift 3 units at the same time 4 times. Now we will reach 12.

This is how multiplication can be done on the number line.

Exercise 2.1 has questions based on the above explanation. For the solutions, refer to our pdf for Whole numbers.

### Properties of Whole Numbers

While numbers are closed under addition as well as under multiplication.

That is the product of addition or multiplication of a whole number is also a whole number.

In other words, there is a commutative property.

### Patterns in Whole Numbers

The numbers can be arranged in a particular sequence that forms a square or rectangular shape.

Exercise 2.2 deals with such questions. Solutions to these questions are available in our NCERT solutions for Class 6.

### The Role of Zero

Zero when added to any whole numbers, the result is the same number.

Similarly, zero when multiplied by any number results in zero. This is the special property of zero.

Exercise 2.3 is based on the properties discussed above. The solutions of all the 5 questions inscribed in the NCERT textbook of Class 6 can be available in our pdf solutions.

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