NCERT Solutions: Number System (Exercise 1.3, 1.4 and 1.6)

# Number System (Exercise 1.3, 1.4 and 1.6) NCERT Solutions - Mathematics (Maths) Class 9

### Exercise 1.3

Q1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) 36/100
Ans: = 0.36 (Terminating)

(ii) 1/11
Ans:  (iii) Ans:  = 4.125 (Terminating)

(iv) 3/13
Ans:  (v) 2/11
Ans:  (vi) 329/400
Ans: = 0.8225 (Terminating)

Q2. You know that 1/7 = . 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.]
Ans: Q3. Express the following in the form p/q, where p and q are integers and q ≠ 0.
(i) Ans: Assume that  x = 0.666…
Then,10x = 6.666…
10x = 6 + x
9x = 6
x = 2/3

(ii) Ans: = (4/10) + (0.777/10)
Assume that x = 0.777…
Then, 10x = 7.777…
10x = 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

(iii) Ans: Assume that  x = 0.001001…
Then, 1000x = 1.001001…
1000x = 1 + x
999x = 1
x = 1/999

Q4. 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.
Ans: Assume that x = 0.9999…..Eq (a)
Multiplying both sides by 10,
10x = 9.9999…. Eq. (b)
Eq.(b) – Eq.(a), we get
(10x = 9.9999)-(x = 0.9999…)
9x = 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.

Q5. 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.
Ans: 1/17
Dividing 1 by 17:  There are 16 digits in the repeating block of the decimal expansion of 1/17.

Q6. 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?
Ans: 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 = 21
7/8 = 0. 875, denominator q =23
4/5 = 0. 8, denominator q = 51
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.

Q7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Ans: We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

• √3 = 1.732050807568
• √26 =5.099019513592
• √101 = 10.04987562112

Q8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Ans: Three different irrational numbers are:

• 0.73073007300073000073…
• 0.75075007300075000075…
• 0.76076007600076000076…

Q9. Classify the following numbers as rational or irrational according to their type:
(i)√23
Ans: √23 = 4.79583152331…
Since the number is non-terminating non-recurring therefore, it is an irrational number.

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

(iii) 0.3796
Ans: Since the number, 0.3796, is terminating, it is a rational number.

(iv) 7.478478
Ans: The number, 7.478478, is non-terminating but recurring, it is a rational number.

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

### Exercise 1.4

Q1. Classify the following numbers as rational or irrational:
(i) 2 –√5
Ans: We know that, √5 = 2.2360679…
Here, 2.2360679…is non-terminating and non-recurring.
Now, substituting the value of √5 in 2 –√5, we get,
2-√5 = 2-2.2360679… = -0.2360679
Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23
Ans: (3 +23) –√23 = 3+23–√23
= 3
= 3/1
Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.

(iii) 2√7/7√7
Ans: 2√7/7√7 = ( 2/7) × (√7/√7)
We know that (√7/√7) = 1
Hence, ( 2/7) × (√7/√7) = (2/7) × 1 = 2/7
Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

(iv) 1/√2
Ans: Multiplying and dividing numerator and denominator by √2 we get,
(1/√2) × (√2/√2)= √2/2 ( since √2 × √2 = 2)
We know that, √2 = 1.4142…
Then, √2/2 = 1.4142/2 = 0.7071..
Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2
Ans: We know that, the value of = 3.1415
Hence, 2 = 2 × 3.1415.. = 6.2830…
Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

Q2. Simplify each of the following expressions:
(i) (3 + √3) (2 + √2)
Ans: (3 + √3) (2 + √2 )
Opening the brackets, we get, (3 × 2) + (3 × √2) + (√3 × 2) + (√3 × √2)
= 6 + 3√2 + 2√3 + √6

(ii) (3 + √3) (3 - √3)
Ans: (3 + √3) (3 - √3 ) = 3- (√3)2 = 9 - 3
= 6

(iii) (√5 + √2)2
Ans: (√5 + √2)= √5+ (2 × √5 × √2) + √22
= 5 + 2 × √10 + 2 = 7 + 2√10

(iv) (√5 - √2)(√5 + √2)
Ans: (√5 - √2)(√5 + √2) = (√5- √22) = 5 - 2 = 3

Q3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
When we measure a value with a scale, we only obtain an approximate value.
We never obtain an exact value.
Therefore, we may not realize whether c or d is irrational.
The value of π is almost equal to 22/7 or 3.142857…

Q4. Represent (√9.3) on the number line.
Ans:
Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC = 1 unit.
Step 2: Now, AC = 10.3 units. Let the centre of AC be O.
Step 3: Draw a semi-circle of radius OC with centre O.
Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.
Step 5: OBD, obtained, is a right angled triangle.
Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1
OB = OC – BC
⟹ (10.3/2)-1 = 8.3/2
Using Pythagoras theorem,
We get,
OD= BD+ OB2
⟹ (10.3/2)2 = BD2+(8.3/2)2
⟹ BD2 = (10.3/2)2-(8.3/2)2
⟹ (BD)= (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)
⟹ BD2 = 9.3
⟹ BD =  √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment.
The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure. Q5. Rationalize the denominators of the following:
(i) 1/√7
Ans: Multiply and divide 1/√7 by √7
(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)
Ans: Multiply and divide 1/(√7 - √6) by (√7 + √6)
[1/(√7 - √6)] × (√7 + √6)/(√7 + √6) = (√7 + √6)/(√7 - √6)(√7 + √6)
= (√7 + √6)/√7- √6[denominator is obtained by the property, (a + b)(a - b) = a- b2]
= (√7 + √6)/(7 - 6)
= (√7 + √6)/1
= √7 + √6

(iii) 1/(√5+√2)
Ans: Multiply and divide 1/(√5 + √2) by (√5 - √2)
[1/(√5 + √2)] × (√5 - √2)/(√5 - √2) = (√5 - √2)/(√5 + √2)(√5 - √2)
= (√5 - √2)/(√5- √22) [denominator is obtained by the property, (a + b)(a - b) = a- b2]
= (√5 - √2)/(5 - 2)
= (√5 - √2)/3

(iv) 1/(√7-2)
Ans: Multiply and divide 1/(√7 - 2) by (√7 + 2)
1/(√7 - 2) × (√7 + 2)/(√7 + 2) = (√7 + 2)/(√7 - 2)(√7 + 2)
= (√7 + 2)/(√7- 22) [denominator is obtained by the property, (a + b)(a - b) = a2-b2]
= (√7 + 2)/(7 - 4)
= (√7 + 2)/3

### Exercise 1.5

Q1. Find:
(i) 641/2

Ans: 641/2 = (8 × 8)1/2
= (82)½
= 81 [⸪2 × 1/2 = 2/2 = 1]
= 8

(ii) 321/5
Ans: 321/5 = (25)1/5
= (25)
= 21 [⸪ 5 × 1/5 = 1]
= 2

(iii) 1251/3
Ans: (125)1/3 = (5 × 5 × 5)1/3
= (53)
= 51 (3 × 1/3 = 3/3 = 1)
= 5

Q2. Find:
(i) 93/2
Ans: 93/2 = (3 × 3)3/2
= (32)3/2
= 33 [⸪ 2 × 3/2 = 3]
=27

(ii) 322/5
Ans: 322/5 = (2 × 2 × 2 × 2 × 2)2/5
= (25)2⁄5
= 22 [⸪ 5 × 2/5= 2]
= 4

(iii)163/4
Ans: 163/4 = (2 × 2 × 2 × 2)3/4
= (24)3⁄4
= 23 [⸪ 4 × 3/4 = 3]
= 8

(iv) 125-1/3
Ans: 125-1/3 = (5 × 5 × 5)-1/3
= (53)-1⁄3
= 5-1 [⸪ 3 × -1/3 = -1]
= 1/5

Q.3. Simplify:
(i) 22/3×21/5
Ans: 22/3 × 21/5 = 2(2/3)+(1/5) [⸪Since, am × an= am+n____ Laws of exponents]
= 213/15 [⸪ 2/3 + 1/5 = (2 × 5 + 3 × 1)/(3 × 5) = 13/15]

(ii) (1/33)7
Ans: (1/33)= (3-3)7 [⸪Since,(am)= am x n____ Laws of exponents]
= 3-21

(iii) 111/2/111/4
Ans: 111/2/111/4 = 11(1/2)-(1/4)
= 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 71/2× 81/2
Ans: 71/2× 81/2 = (7 × 8)1/2 [⸪ Since, (a× b= (a × b)m ____ Laws of exponents]
= 561/2

The document Number System (Exercise 1.3, 1.4 and 1.6) NCERT Solutions | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

## FAQs on Number System (Exercise 1.3, 1.4 and 1.6) NCERT Solutions - Mathematics (Maths) Class 9

 1. How can I solve Exercise 1.3 in the NCERT Number System book? Ans. To solve Exercise 1.3 in the NCERT Number System book, follow these steps: 1. Read the given problem carefully. 2. Identify the key information and data provided in the problem. 3. Apply the relevant concepts and formulas from the Number System chapter. 4. Solve the problem step-by-step, showing all the calculations and workings. 5. Double-check your solution to ensure accuracy. 6. Write down the final answer with appropriate units, if applicable.
 2. What is the significance of Exercise 1.4 in the NCERT Number System book? Ans. Exercise 1.4 in the NCERT Number System book is significant as it helps students practice and reinforce their understanding of various concepts related to the number system. This exercise includes a range of questions that cover topics like rational numbers, irrational numbers, real numbers, and their properties. By solving these questions, students can develop their problem-solving skills, enhance their mathematical reasoning abilities, and gain confidence in dealing with different types of number system problems.
 3. What are the key topics covered in Exercise 1.5 of the NCERT Number System book? Ans. Exercise 1.5 of the NCERT Number System book covers the following key topics: 1. Representation of Rational Numbers on the Number Line 2. Rational Numbers between Two Rational Numbers 3. Operations on Rational Numbers (Addition, Subtraction, Multiplication, and Division) 4. Simplification of Rational Expressions 5. Word Problems involving Rational Numbers By practicing this exercise, students can strengthen their understanding of rational numbers, their representations, and the various operations that can be performed on them.
 4. How can Exercise 1.6 in the NCERT Number System book help in understanding decimal representation? Ans. Exercise 1.6 in the NCERT Number System book is designed to enhance students' understanding of decimal representation and its various aspects. By solving the problems in this exercise, students can: 1. Practice converting fractions into decimals and vice versa. 2. Gain familiarity with terminating and non-terminating decimals. 3. Learn about recurring decimals and their representation. 4. Explore the concept of decimal expansion and place value of decimals. 5. Apply decimal operations such as addition, subtraction, multiplication, and division. Overall, Exercise 1.6 provides a comprehensive platform for students to grasp the concept of decimal representation and develop their skills in working with decimals.
 5. How can the NCERT Number System book help in preparing for competitive exams? Ans. The NCERT Number System book can be a valuable resource for preparing for competitive exams. Here's how it can help: 1. Comprehensive Coverage: The book covers all the fundamental concepts of the number system, including rational numbers, irrational numbers, real numbers, decimals, etc. This ensures a strong foundation in the subject. 2. Clarity of Concepts: The book explains the concepts in a simplified and easy-to-understand language. It provides step-by-step explanations and examples to clarify doubts and facilitate better comprehension. 3. Practice Questions: The exercises in the book offer a wide range of practice questions, allowing students to test their understanding and improve problem-solving skills. These questions are designed to match the level of difficulty often encountered in competitive exams. 4. Exam-oriented Approach: The book includes solved examples and practice questions that are aligned with the exam pattern and syllabus of various competitive exams. This helps students familiarize themselves with the type of questions they may encounter in the actual exam. 5. Supplementary Resources: The book may also provide additional resources like sample papers, previous years' question papers, and online resources to further support exam preparation. By studying the NCERT Number System book and practicing its exercises, students can strengthen their conceptual understanding, gain confidence, and improve their chances of performing well in competitive exams.

## Mathematics (Maths) Class 9

62 videos|438 docs|102 tests

## Mathematics (Maths) Class 9

62 videos|438 docs|102 tests
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