# Ncert Solutions For Class 6 Maths Ex 2.3

## Ncert Solutions For Class 6 Maths Chapter 2 Ex 2.3

Question 1:

Which out of the following questions will represent zero;

a) 2+0

b) 1$$\times$$0

c) $$\frac{0}{2}$$

d) $$\frac{20-20}{2}$$

a) 2 + 0 = 2

It does not represent zero, since the answer is 2.

b) 1*0 = 0

It represents zero, since number multiplied with anything is zero.

c) $$\frac{0}{2}$$ = 0

It represents zero, since number divided with anything is zero.

d) $$\frac{20-20}{2}$$ = 0

It represents zero.

Question 2:

The product of two numbers is zero; can we say that one or both of them will be zero or non-zero? State reasons.

First case

The product of 2 whole numbers will be zero only when one of the number is multiplied with zero

Ex)

5$$\times$$0 = 0

18$$\times$$0 = 0

Second case

The product of 2 whole numbers will be zero when both the numbers are multiplied are zero

0$$\times$$0 = 0

Third case

The product of 2 numbers will not be zero, if both are multiplied with a non-zero number

4$$\times$$3 = 12

5$$\times$$4 = 20

Question 3:

The product of two whole numbers is 1, can we say that one or both of them will be 1? State reasons.

First case

The product of 2 whole numbers will be one, if both the numbers are multiplied with 1.

Ex)

1$$\times$$1 =1

Second case

The product of 2 whole numbers will not be 1, if it is multiplied by numbers other than 1.

6$$\times$$5 = 30

So, it is clear that the product of two numbers will be 1 only when the numbers are multiplied with 1.

Question 4:

Find the solution using distributive property.

a) 725$$\times$$102

By distributive law

A$$\times$$ (B+C) = A$$\times$$B + A$$\times$$C

= 728$$\times$$ (100+2)

= 728$$\times$$100 + 728$$\times$$2

= 72800 + 1456

= 74,256

b) 5450$$\times$$1004

By distributive law,

A$$\times$$ (B+C) = A$$\times$$B + A$$\times$$C

= 5450$$\times$$ (1000+4)

= (5450$$\times$$1000) + (5450$$\times$$4)

= 5450000 + 21800

= 5471800

c) 724$$\times$$25

By distributive law,

A$$\times$$ (B+C) = A$$\times$$B + A$$\times$$C

= (700+24) $$\times$$25

= (700 + 25 – 1) $$\times$$25

= 700$$\times$$25 + 25$$\times$$25 – (1$$\times$$25)

= 17500 + 625 – 25

= 18100

d) 4225$$\times$$125

By distributive law,

A$$\times$$ (B+C) = A$$\times$$B + A$$\times$$C

= (4000 + 200 +100 -75) $$\times$$125

= (4000$$\times$$125) + (200$$\times$$125) + (100$$\times$$125) – (75$$\times$$125)

= (500000) + (25000) + (12500) – (9375)

= 528125

e) 508$$\times$$35

By distributive law,

A$$\times$$ (B+C) = A$$\times$$B + A$$\times$$C

= (500 + 8) $$\times$$35

= (500$$\times$$35) + (8$$\times$$35)

= 17500 + 280

= 17780

Question 5:

Analyze the pattern

(1$$\times$$8) + 1 = 91234$$\times$$8 + 4 = 9876

(12$$\times$$8) +2 = 981234$$\times$$ 8 + 5 = 98765

(123$$\times$$8) + 3 = 987

Follow the next two steps and explain the working of the pattern

(Note: 12345 = 11111 + 1111 + 111 + 11 + 1)

123456$$\times$$8 + 6 = 987648 + 6 = 987654

1234567$$\times$$8 + 7 = 9876536 + 7 =9876543

The answer is yes and the pattern works

W.K.T

123456 = (111111 + 11111 + 1111 + 111 + 11 + 1)

123456$$\times$$8 = (111111 + 11111 + 1111 + 111 + 11 +1) $$\times$$8

= (111111$$\times$$8) + (11111$$\times$$8) + (1111$$\times$$8) + (111$$\times$$8) + (11$$\times$$8) + (1$$\times$$8)

= 888888 + 88888 + 8888 + 888 + 88 + 8

= 987648

So,(123456$$\times$$8) + 6 = 987648 + 6 = 987654