* **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