RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.5

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.5 is about Operations on Whole Numbers, an important topic in the CBSE Class 6 syllabus.

This exercise teaches students about different types of numbers like square numbers, triangular numbers, and rectangular numbers. It also helps students spot patterns and solve number problems easily.

The solutions are simple and easy to understand, helping students learn how to use formulas and solve questions step by step. This exercise is very useful for building strong math skills for exams.

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.5 Overview

Exercise 4.5 in Chapter 4, Operations on Whole Numbers, deals with special number patterns and properties of whole numbers, helping students deepen their understanding of how numbers relate and behave in sequences and formulas.

What the Exercise Covers:

1. Square Numbers and Triangular Numbers

• Students learn to find the nth square number using the formula  n × n. 

• They also learn to calculate the nth triangular number using the formula  n × (n + 1)/ 2.

• The exercise asks for specific numbers, like the 10th square number or the 6th triangular number, encouraging practice with these formulas

2. Relationship Between Different Types of Numbers

• Students explore whether a rectangular number can be a square number, and whether a triangular number can also be a square number.

• This helps students see connections between different mathematical concepts and understand that some numbers can belong to multiple categories.

3. Patterns in Products of Numbers

• The exercise provides a pattern involving multiplication of pairs of numbers with a fixed difference (like difference 4), then asks students to identify and continue the pattern.

• This develops students’ skills in observing numerical relationships and predicting the next terms in sequences.

4. Number Patterns with Addition and Multiplication

• Several problems show interesting patterns involving multiplication and addition of numbers, such as 9×9+7=88, and students fill in the missing numbers following the pattern.

• This section trains students to analyze patterns and apply logical reasoning.

5. Sum of Odd Numbers and Their Patterns

• Students study sums of sequences of odd numbers, noting how sums of the first n odd numbers relate to perfect squares.

RD Sharma Solutions for Class 6 Maths Operations on Whole Numbers Chapter 4 Exercise 4.5

Here are the detailed solutions for RD Sharma Class 6 Maths Chapter 4 Exercise 4.5 on Operations on Whole Numbers. 

1. Without drawing a diagram, find

(i) 10th square number

(ii) 6th triangular number

Solution:

(i) 10th square number

The square number can be remembered using the following rule

Nth square number = n × n

So the 10th square number = 10 × 10 = 100

(ii) 6th triangular number

The triangular number can be remembered using the following rule

Nth triangular number = n × (n + 1)/ 2

So the 6th triangular number = 6 × (6 + 1)/ 2 = 21

2. (i) Can a rectangular 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.

Example – 16 is a rectangular number which can also be a square number.

(ii) Yes. A triangular number can also be a square number.

Example – 1 is a triangular number which can also be a square number.

3. 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:

We know that

1 × 5 = 5

2 × 6 = 12

3 × 7 = 21

4 × 8 = 32

So the first four products of two numbers with difference 4

5 – 1 = 4

6 – 2 = 4

7 – 3 = 4

8 – 4 = 4

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

9 × 9 + 7 = 88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 × 9 + 4 = ………

98765 × 9 + 3 = ………

987654 × 9 + 2 = ……….

9876543 × 9 + 1 = ……….

Solution:

9 × 9 + 7 = 88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 × 9 + 4 = 88888

98765 × 9 + 3 = 888888

987654 × 9 + 2 = 8888888

9876543 × 9 + 1 = 88888888

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

6 × 2 – 5 = 7

7 × 3 – 12 = 9

8 × 4 – 21 = 11

9 × 5 – 32 = 13

….. × ……. – ….. = …….

….. × ……. – ….. = …….

….. × ……. – ….. = …….

Solution:

6 × 2 – 5 = 7

7 × 3 – 12 = 9

8 × 4 – 21 = 11

9 × 5 – 32 = 13

10 × 6 – 45 = 15

11 × 7 – 60 = 17

12 × 8 – 77 = 19

6. Study the following pattern:

1 + 3 = 2 × 2

1 + 3 + 5 = 3 × 3

1 + 3 + 5 + 7 = 4 × 4

1 + 3 + 5 + 7 + 9 = 5 × 5

By observing the above pattern, find

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

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

(iii) 21 + 23 + 25 + ….. + 51

Solution:

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

By using the pattern

1 + 3 + 5 + 7 + 9 + 11 = 6 × 6

= 36

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

By using the pattern

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8

= 64

(iii) 21 + 23 + 25 + ….. + 51

We know that

21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19)

By using the pattern

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

(1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100

So we get

21 + 23 + 25 + ….. + 51 = 676 – 100 = 576

7. Study the following pattern:

1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6

1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6

By observing the above pattern, write next two steps.

Solution:

By using the pattern

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5

On further calculation

= (5 × 6 × 11)/6

So we get

= 55

By using the pattern

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6

On further calculation

= (6 × 7 × 13)/6

So we get

= 91

8. Study the following pattern:

1 = (1 × 2)/ 2

1 + 2 = (2 × 3)/ 2

1 + 2 + 3 = (3 × 4)/ 2

1 + 2 + 3 + 4 = (4 × 5)/ 2

By observing the above pattern, find

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

(ii) 50 + 51 + 52 + ……. + 100

(iii) 2 + 4 + 6 + 8 + 10 + …….. + 100

Solution:

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

We get

= (10 × 11)/2

On further calculation

= 55

(ii) 50 + 51 + 52 + ……. + 100

We can write it as

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

So we get

(1 + 2 + 3 + …… + 99 + 100) = (100 × 101)/2

(1 + 2 + 3 + 4 + ….. + 47 + 49) = (49 × 50)/2

By substituting the values

50 + 51 + 52 + ……. + 100 = (100 × 101)/2 + (49 × 50)/2

On further calculation

= 5050 – 1225

We get

= 3825

(iii) 2 + 4 + 6 + 8 + 10 + …….. + 100

We can write it as

2 (1 + 2 + 3 + 4 + …… + 49 + 50)

So we get

= 2 × (50 × 51)/2

On further calculation

= 2 × (1275)

We get

= 2550

RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.5 PDF

Students can download the RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.5 PDF from the link below.

This PDF provides step-by-step solutions to all questions in the exercise, helping students understand important concepts like square numbers, triangular numbers, number patterns, and sums of sequences.

Using this PDF, students can practice effectively and improve their problem-solving skills for better exam preparation.

Benefits of Using RD Sharma Solutions for Class 6 Maths Chapter 4 Exercise 4.5 

• The solutions break down each problem into simple steps, making it easier for students to understand complex concepts like square numbers, triangular numbers, and number patterns.

• Regular practice with these solutions helps students develop strong analytical and reasoning abilities needed for solving mathematical problems confidently.

• The exercise covers important topics from the Class 6 Maths syllabus, ensuring students are well-prepared for their school exams.

• By practicing the types of questions frequently asked in exams, students gain confidence and reduce exam anxiety.

• These solutions are a reliable reference for completing homework accurately and revising before tests.