NCERT Solutions Class 9 Maths Chapter 1 Number System: Exercise 1.3

Exercise 1.3 in Class 9 Maths Chapter 1 helps you learn to find square roots and use square roots for problem solving. In this exercise, you will not only learn how to find the square roots of the real numbers but also how to find the value of square roots of real numbers without a calculator. Mastering this will benefit you greatly as you learn  maths in topics later on, like algebra, geometry etc. 

Our NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.3 provide well thought, straightforward step by step answers to the questions in your textbook. These solutions follow the latest CBSE syllabus framework and follow a converse approach taken to explain the strategies of solving by rewriting all explanations as simply and clearly as possible; so you can easily understand the process and all methods as easily as reviewing your textbook or on-line wiki journals. So even if you are completing your homework assignment or preparing for an exam, lessons learned using our NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.3 will provide invaluable and meaningful information to help you learn easier and answer questions with confidence.

1.0Download NCERT Solutions of Class 9 Maths Chapter 1 Number System Exercise 1.3 : Free PDF

NCERT Solutions Class 9 Maths Chapter 1 covers various topics related to square roots and its uses in different problems. Download the free pdf for the NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.3 here.

2.0NCERT Exercise Solutions Class 9 Chapter 1- Number Systems: All Exercises 

3.0NCERT Class 9 Maths Chapter 1 Exercise 1.3 : Detailed Solutions

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

(i) 36/100

(ii) 1/11

(iii) 1/8 (Note: The original text had 4 1/8 which is likely a typo for 1/8 or 33/8 for 4 1/8, assuming 4 1/8 is 4 + 1/8 = 33/8, not 48 1/8, so I'll interpret as 1/8 or 33/8 based on the solution)

(iv) 3/13

(v) 2/11

(vi) 329/400

Sol. (i) 36/100 = 0.36 (Terminating)

(ii) 1/11 = 0.090909..... = 0.09 (Non-Terminating Repeating)

[Long division of 1 by 11]

(iii) If it is 1/8: 1/8 = 0.125 (Terminating decimal)

If it is 33/8 (from 4 1/8 as a mixed fraction): 33/8 = 4.125 (Terminating decimal)

(iv) 3/13 = 0.230769230769...... = 0.230769 (Non-Terminating Repeating)

(v) 2/11 = 0.1818....... = 0.18 (Non-Terminating Repeating)

(vi) 329/400 = 0.8225 (Terminating)

[Long division of 329 by 400]

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?

Sol. Yes, we can predict decimal expansions without actually doing long division method as:

2/7 = 2 × (1/7) = 2 × 0.142857 = 0.285714

3/7 = 3 × (1/7) = 3 × 0.142857 = 0.428571

4/7 = 4 × (1/7) = 4 × 0.142857 = 0.571428

5/7 = 5 × (1/7) = 5 × 0.142857 = 0.714285

6/7 = 6 × (1/7) = 6 × 0.142857 = 0.857142

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

(i) 0.6

(ii) 0.47

(iii) 0.001

Sol. (i) Let x=0.6666.... (1)

Multiplying both the sides by 10:

10x=6.6666.... (2)

Subtract (1) from (2):

10x−x = 6.6666.... − 0.6666....

9x = 6

x = 6/9 = 2/3

(ii) Let x=0.4777.... (1)

Multiply both sides by 10:

10x=4.777.... (2)

Multiply both sides by 100 (from original x, or 10 from 10x):

100x=47.777.... (3)

Subtract (2) from (3):

100x−10x = 47.777.... − 4.777....

90x = 43

x = 43/90

(iii) Let x=0.001001001.... (1)

Multiply both sides by 1000:

1000x=1.001001.... (2)

Subtract (1) from (2):

1000x−x = 1.001001.... − 0.001001....

999x = 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.

Sol. Let x=0.99999.... (1)

Multiply both sides by 10:

10x=9.99999.... (2)

Subtract (1) from (2):

10x−x = 9.99999.... − 0.99999....

9x = 9

x = 9/9 = 1.

So, 0.99999... = 1 = 1/1.

This answer is surprising because it seems that 0.999... is strictly less than 1. However, mathematically, they are equivalent. This makes sense because the difference between 1 and 0.999... is infinitesimally small, approaching zero.

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.

Sol. For a rational number p/q where q is a prime number (other than 2 or 5), the maximum number of digits in the repeating block of its decimal expansion is q-1.

Here, q=17, so the maximum number of digits in the repeating block of 1/17 can be 17-1 = 16.

Performing the division:

1/17 = 0.0588235294117647... (repeating block has 16 digits)

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?

Sol. For a rational number of the form p/q (q≠0), where p and q are integers with no common factors other than 1, to have a terminating decimal representation, the prime factorization of q must have only powers of 2 or powers of 5 or both. That is, q must be of the form 2ᵐ×5ⁿ, where m and n are non-negative integers (whole numbers).

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

Sol. Non-terminating non-recurring decimal expansions are irrational numbers.

Examples:

0.01001000100001...

0.202002000200002...

0.003000300003...

(Other examples include π, √2, √3 etc.)

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

Sol. First, convert the given rational numbers to their decimal expansions:

5/7 = 0.714285... (repeating)

9/11 = 0.818181... (repeating)

We need to find three irrational numbers between 0.714285... and 0.818181...

Irrational numbers are non-terminating and non-repeating.

Three different irrational numbers can be:

1. 0.750750075000750000...
2. 0.780780078000780000...
3. 0.808008000800008...

9. Classify the following numbers as rational or irrational :

(i) √23

(ii) √225

(iii) 7√7 / 2√7

(iv) 1/√2

(v) 2π

Sol. (i) √23 = Irrational number (23 is not a perfect square)

(ii) √225 = 15 = Rational number (225 is a perfect square, 15 can be written as 15/1)

(iii) 7√7 / 2√7 = 7/2 = Rational number (√7 cancels out)

(iv) 1/√2: Since 1 is a rational number and √2 is an irrational number, their quotient (1/√2) is an irrational number.

(v) 2π: Since 2 is a rational number and π is an irrational number, their product (2π) is an irrational number.

4.0Key Features and Benefits Class 9 Maths Chapter 1 Number System: Exercise 1.3

• These exercises are in accordance with the CBSE syllabus, which takes both students to some elementary and advanced number operations in a clear manner.
• These problems improve logical and critical thinking which is an essential skill for solving problems in various competitive exams.
• Regularly practicing these NCERT Solutions helps students to develop analytical ability.
• Constant use of these solutions will develop students' confidence in maths and develop their ability of problem solving which is crucial for Olympiad.