RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

RD Sharma Solutions for Class 6 Mathematics

Created by: Abhishek Kapoor

Class 6 : RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

The document RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
All you need of Class 6 at this link: Class 6

Q1. Without drawing a diagram, find:

Solution: (i) 10th square number:

A square number can easily be remembered by the following rule

Nth square number = n x n

10th square number = 10 x 10 = 100

(ii) 6th triangular number:

A triangular number can easily be remembered by the following rule

Nth triangular number = n x ( n + 1 )2

Therefore, 6th triangular number = 6 x ( 6 + 1 )2 = 21

 

Q2. (i) Can a rectangle number also be a square number ?

(ii) Can a triangular number also be a square number?

Solution: (i) Yes, a rectangular number can also be a square number; for example, 16 is a square number also a rectangular number.

RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

(ii) Yes, there exists only one triangular number that is both a triangular number and a square number, and that number is 1.

 

Q3. Write the first four products of two numbers with difference 4 starting from in the following order:

1 , 2 , 3 , 4 , 5 , 6 , ………..

Identify the pattern in the products and write the next three products.

Solution: 1 x 5 = 5 (5 - 1 = 4)

2 x 6 = 12 (6 – 2 =4)

3 x 7 = 21 (7 – 3 = 4)

4 x 8 = 32 (8 – 4 = 4)

 

Q4. Observe the pattern in the following and fill in the blanks:

Solution: 9 x 9 + 7 =88

98 x 9 + 6 = 888

987 x 9 + 5 = 8888

9876 x 9 + 4 = 88888

98765 x 9 + 3 = 888888

987654 x 9 + 2 = 8888888

9876543 x 9 + 1 = 88888888

 

Q5. Observe the following pattern and extend it to three more steps:

Solution: 6 x 2 – 5 = 7

7 x 3 – 12 = 9

8 x 4 – 21 = 11

9 x 5 – 32 = 13

10 x 6 – 45 = 15

11 x 7 – 60 = 17

12 x 8 – 77 = 19

 

Q6. Study the following pattern:

1 + 3 = 2 x 2

1 + 3 + 5 = 3 x 3

1 + 3 + 5 + 7 = 4 x 4

1 + 3 + 5 + 7 + 9 = 5 x 5

By observing the above pattern, find:

Solution: (i) 1 + 3 + 5 + 7 + 9 + 11

= 6 x 6

= 36

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

= 8 x 8

= 64

(iii) 21 + 23 + 25 + ... + 51

= (21+23+25+...+51) can also be written as

(1+3+5+7+...+49+51) -(1+3+5+...+17+ 19)

(1+3 +5 +7+...+ 49 + 51)= 26 x 26 = 676

and, (1+3+5+...+17+ 19 ) = 10 x 10 = 100

Now,

( 21 + 23 +25 +...+ 51 )= 676 -100 = 576

 

Q7. Study the following pattern:

1x1 + 2x2 =RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

1x1 + 2x2 + 3x3 = RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

1x1 + 2 x 2 + 3 x 3 + 4 x 4 = RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

By observing the above pattern, write next two steps.

Solution: The next two steps are as follows:

1 x 1+2 x2+3 x 3+4 x4+5 x 5

=5x6x116

= 55

1 x 1+2 x2+3 x 3+4 x4+5 x 5+6 x6

=6x7x136

=91

 

Q8. Study the following pattern:

1 = RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

1 + 2 = RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

1 + 2 + 3 = RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev

1 + 2 + 3 + 4 = RD Sharma Solutions -Ex-4.5, Operations On Whole Numbers, Class 6, Maths Class 6 Notes | EduRev<

By observing the above pattern, find:

Solution: (i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= 10 x 112

= 55

(ii) 50 + 51 + 52 + ...+ 100

This can also be written as

(1 + 2 + 3 + ...+ 99 + 100) - (1 + 2 + 3 + 4 + ...+ 47 + 49)

Now,

(1+ 2 + 3 + ...+ 99 + 100 ) = 100 x 1012

and, (1 + 2 + 3 + 4 +...+ 47 + 49 ) = 49 x 502

So, (50 + 51 + 52 + ...+ 100 ) = 100 x 1012 – 49 x 502

= 5050 - 1225

= 3825

(iii) 2 + 4 + 6 + 8 + 10 +...+ 100

This can also be written as 2 x (1 + 2 + 3 + 4 + ...+ 49 + 50)

Now,

(1 + 2 + 3 + 4 + ...+ 49 + 50 ) = 50 x 512

= 1275

Therefore, (2 + 4 + 6 + 8 + 10 + ...+ 100) = 2 x 1275 = 2550

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