Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  NCERT Solutions: Number System (Exercise 1.3, 1.4 and 1.6)

NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)

Exercise 1.3

Q1. Write the following in decimal form and mention the kind of decimal expansion each has. 
(i) 36/100
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
= 0.36 (Terminating) 

(ii) 1/11
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)


(iii) NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
= 4.125 (Terminating)


(iv) 3/13 
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)


(v) 2/11 
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)


(vi) 329/400 
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
= 0.8225 (Terminating)

Q2. You know that 1/7 = NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6). Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, and 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:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)


Q3. Express the following in the form p/q, where p and q are integers and q ≠ 0.
(i) NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6) 
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
Assume that  x = 0.666…
Then,10x = 6.666…
10x = 6 + x
9x = 6
x = 2/3


(ii)NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
= (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) NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
Ans:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
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 very close to 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:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
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 a terminating one.
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 and non-recurring.
Ans: We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating and 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:
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)
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 and 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 the 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 and 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 and 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?
Ans: There is no 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-unit 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 the radius and B as the 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.
NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)


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 NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6) is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 1 - Number System (Exercise 1.3, 1.4 and 1.6)

1. What are the key concepts covered in NCERT Number System Exercise 1.3, 1.4, and 1.6?
Ans. NCERT Number System Exercises 1.3, 1.4, and 1.6 primarily cover concepts related to real numbers, rational and irrational numbers, and their properties. Exercise 1.3 focuses on identifying rational numbers, Exercise 1.4 deals with operations on rational numbers, and Exercise 1.6 introduces the concept of irrational numbers and their representation on the number line.
2. How do I represent irrational numbers on a number line as per NCERT guidelines?
Ans. To represent irrational numbers on a number line, first identify their approximate decimal values. For instance, the square root of 2 (√2) is approximately 1.414. Plot this value between 1 and 2 on the number line. You can also use methods like constructing a right triangle to accurately locate the position of an irrational number geometrically.
3. What strategies can I use to solve problems in Exercise 1.4 related to operations on rational numbers?
Ans. To solve problems in Exercise 1.4 involving operations on rational numbers, start by ensuring that all numbers are expressed in fraction form. Use common denominators for addition and subtraction, and for multiplication and division, simply multiply across the numerators and denominators. Always simplify your final answers to their lowest terms.
4. Are there any specific properties of rational and irrational numbers that I should memorize for my exam?
Ans. Yes, some key properties include: 1. Rational numbers can be expressed as a fraction a/b where a and b are integers and b ≠ 0. 2. The sum or product of two rational numbers is always rational. 3. An irrational number cannot be expressed as a simple fraction, and its decimal representation is non-repeating and non-terminating. 4. The sum or product of a rational and an irrational number is always irrational.
5. How can I effectively prepare for questions related to the number system in my exams?
Ans. To prepare effectively for number system questions, practice solving various types of problems from the NCERT exercises. Make sure to understand key concepts like rational vs. irrational numbers, properties of operations, and how to manipulate fractions. Additionally, review previous year question papers and take mock tests to improve your speed and accuracy in solving these problems.
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