Number System Class 9 Chapter 1 Exercise 1.3 NCERT Solutions

Exercise 1.3 of Chapter 1 Number System Class 9 explains how rational numbers appear in decimal form and how to identify whether a number is terminating, non-terminating, repeating, or non-terminating non-repeating. This understanding is very important because decimal identification helps classify numbers correctly.

These NCERT Solutions for Number System Class 9 Exercise 1.3 explain the rules of decimal expansions using simple logic and examples. With step-by-step answers, you can understand better and score full marks in exams.

Class 9 Chapter 1 Number System Exercise 1.3 Questions

Class 9 Chapter 1 Number System Exercise 1.3 Questions require you to apply the prime factorisation method and check what kind of decimal expansion results. Questions along with the answers are given here for clarity:

Class 9 Chapter 1 Number System Exercise 1.3 Questions (Page: 14)

1. Write the following in decimal form and say what kind of decimal expansion each has :

(i) 36/100 

Solution:

= 0.36 (Terminating) 

(ii)1/11 

Solution: 

 Solution: = 4.125 (Terminating) 

(iv) 3/13 

Solution:

(v) 2/11 

Solution: 

(vi) 329/400 

Solution:

  = 0.8225 (Terminating)

2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?

[Hint: Study the remainders while finding the value of 1/7 carefully.]

Solution:

3. Express the following in the form p/q, where p and q are integers and q 0.

(i)  

Solution: Assume that x = 0.666… Then,10 x = 6.666… 10 x = 6 + x 9 x = 6 x = 2/3

(ii)

Solution:

0.47―=0.4777..

= (4/10)+(0.777/10) Assume that x = 0.777… Then, 10 x = 7.777… 10 x = 7 + x x = 7/9 (4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 ) = (36/90)+(7/90) = 43/90

Solution:  

Assume that x = 0.001001… Then, 1000 x = 1.001001… 1000 x = 1 + x 999 x = 1 x = 1/999

4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Solution: Assume that x = 0.9999…..

Eq (a) Multiplying both sides by 10, 10 x = 9.9999…. 

Eq. (b) Eq.(b) – Eq.(a), we get 10 x = 9.9999 – x = -0.9999… _____________ 9 x = 9 x = 1 

The difference between 1 and 0.999999 is 0.000001 which is negligible. Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.

Solution: 1/17 Dividing 1 by 17: 

There are 16 digits in the repeating block of the decimal expansion of 1/17.

6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Solution: We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. 

For example: 1/2 = 0. 5, denominator q = 2 1 7/8 = 0. 875, denominator q =2 3 4/5 = 0. 8, denominator q = 5 1 We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Solution: We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

1. √3 = 1.732050807568

2. √26 =5.099019513592

3. √101 = 10.04987562112

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Solution:

Three different irrational numbers are:

1. 0.73073007300073000073…

2. 0.75075007300075000075…

3. 0.76076007600076000076…

9.  Classify the following numbers as rational or irrational according to their type:

(i)√23 

Solution: √23 = 4.79583152331… Since the number is non-terminating and non-recurring therefore, it is an irrational number. 

(ii)√225 

Solution: √225 = 15 = 15/1 Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Solution: Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

Solution: The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…

Solution: Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

NCERT Maths Class 9 Number System Exercise 1.3 PDF

NCERT Maths Class 9 Number System Exercise 1.3 PDF provides well-structured solutions. This makes it easier for you to revise decimal expansion concepts anytime. With the help of the solutions, you can clearly understand how the decimal form is obtained. 

For last-minute exam preparation and regular revision, you can use Number System Class 9 Chapter NCERT Solutions. It has easy-to-follow solutions with step-by-step explanations: 

NCERT Maths Class 9 Number System Exercise 1.3 PDF

Number System ONE SHOT Full Chapter Class 9th Maths Youtube Video