 # RD Sharma Solutions for Class 6 Maths Chapter 2: Playing with Numbers Exercise 2.4

## RD Sharma Solutions for Class 6 Maths Exercise 2.4 PDF

Mathematics is one of the challenging subjects and scoring well requires a lot of practice. Students can now practise the RD Sharma Solutions for Class 6 which are provided here. The solutions provided are based on the current CBSE syllabus and guidelines. The primary aim of creating chapter-wise and exercise-wise Solutions is to make the students familiar with the concepts. Each solution is provided with illustrations to help students understand the concepts in a shorter duration. RD Sharma Solutions for Class 6 Maths Chapter 2 Playing with Numbers Exercise 2.4 are provided here.

## Access answers to Maths RD Sharma Solutions for Class 6 Chapter 2: Playing with Numbers Exercise 2.4

1. In which of the following expressions, prime factorization has been done?

(i) 24 = 2 × 3 × 4

(ii) 56 = 1 × 7 × 2 × 2 × 2

(iii) 70 = 2 × 5 × 7

(iv) 54 = 2 × 3 × 9

Solution:

(i) We know that

24 = 2 × 3 × 4 is not a prime factorization since 4 is not a prime number.

(ii) We know that

56 = 1 × 7 × 2 × 2 × 2 is not a prime factorization since 1 is not a prime number.

(iii) We know that

70 = 2 × 5 × 7 is a prime factorization since 2, 5 and 7 are prime numbers.

(iv) We know that

54 = 2 × 3 × 9 is not a prime factorization since 9 is not a prime number.

2. Determine prime factorization of each of the following numbers:

(i) 216

(ii) 420

(iii) 468

(iv) 945

(v) 7325

(vi) 13915

Solution:

(i) 216

We know that Hence, the prime factorization of 216 is 2 × 2 × 2 × 3 × 3 × 3.

(ii) 420

We know that Hence, the prime factorization of 420 is 2 × 2 × 3 × 5 × 7.

(iii) 468

We know that Hence, the prime factorization of 468 is 2 × 2 × 3 × 3 × 13.

(iv) 945

We know that Hence, the prime factorization of 945 is 3 × 3 × 3 × 5 × 7.

(v) 7325

We know that Hence, the prime factorization of 7325 is 5 × 5 × 293.

(vi) 13915

We know that Hence, the prime factorization of 13915 is 5 × 11 × 11 × 23.

3. Write the smallest 4-digit number and express it as a product of primes.

Solution:

1000 is the smallest 4-digit number.

We know that

1000 = 2 × 500

= 2 × 2 × 250

So we get

1000 = 2 × 2 × 2 × 125

It can be further expanded as

1000 = 2 × 2 × 2 × 5 × 25

We get

1000 = 2 × 2 × 2 × 5 × 5 × 5

Therefore, the smallest 4-digit number is 1000 and can be expressed as 2 × 2 × 2 × 5 × 5 × 5.

4. Write the largest 4-digit number and give its prime factorization.

Solution:

9999 is the largest 4-digit number

We know that Therefore, 9999 is the largest 4-digit number and can be expressed as 3 × 3 × 11 × 101.

5. Find the prime factors of 1729. Arrange the factors in ascending order, and find the relation between two consecutive prime factors.

Solution:

We know that Hence, 1729 can be expressed as 7 × 13 × 19

We know that the consecutive prime factors of 1729 are 7, 13 and 19

13 – 7 = 6

19 – 13 = 6

Therefore, in the two consecutive prime factors each factor is 6 more than the previous factor.

6. Which factors are not included in the prime factorization of a composite number?

Solution:

1 and the number itself are not included in the prime factorization of a composite number.

For example, 4 is a composite number having prime factorization 2 × 2.

7. Here are two different factor trees for 60. Write the missing numbers:

(i) (ii) Solution:

(i) We know that

6 = 2 × 3

10 = 5 × 2 (ii) We know that

60 = 30 × 2

30 = 10 × 3

10 = 2 × 5 ### RD Sharma Solutions for Class 6 Maths Chapter 2 – Playing with Numbers Exercise 2.4

RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers Exercise 2.4 consists of answers for the fourth exercise along with illustrative examples, which help in understanding the concept of prime factorization.

### Key Features of RD Sharma Solutions for Class 6 Maths Chapter 2: Playing with Numbers Exercise 2.4

• The solutions created by subject experts help students to practise well and revise for the exams.
• By solving ample number of problems in each exercise, students can self assess their conceptual knowledge and areas of weakness.
• RD Sharma Solutions with explanations significantly help students in solving difficult problems more efficiently.
• The students will be familiar with the type of questions which might appear in the exam and the methods which can be used to solve them.