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) \(\frac{20-20}{2}\)

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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)

Answer:

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

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