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OP Malhotra Class 9 ICSE Maths Solutions Chapter 10 Rational and Irrational Numbers
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 1, Rational and Irrational Numbers. Here students can easily find step by step solutions of all the problems for Rational and Irrational Numbers, Exercise 1a, 1b and 1c Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Rational and Irrational Numbers Exercise 1(a) Solution :
Question no – (1)
Solution :
(i) Given, 1/2 and 3/4
∴ 1/2 + 3/4/2
= 2 + 3/4/2
= 4/5 × 1/2
= 5/8
Hence, the rational number between 1/2 and 3/4 will be 5/8
(ii) 1st rational number,
= 0.1 + 0.2/2
= 0.3
= 1.15
=- 15/100
= 3/20
2nd rational number,
= 3/20 + 2/10/
= 3 + 4/20/2
= 7/20 × 1/2
= 7/40
Thus, two rational numbers between 0.1 and 0.2 are 3/20 and 7/40
(iii) Infinite Rational numbers can find between two given rational numbers.
Question no – (2)
Solution :
(i) 4/5 and 7/13
1st number,
= 4/5 + 7/13/2
= 52 + 35/65/2
= 87/65 × 1/2
= 87/130.
Now, 2nd number,
= 87/130 + 7/10/2
= 87 + 70/130/2
= 157/130 × 1/2
= 157/260
Hence, two Rational number are 87/130 and 157/260
(ii) 3/4 and 1 1/5.
1st Rational no = 3/4 + 1 1/5/2
= 3/4 + 6/5/2
= 1/2 (15 + 24/20)
= 1/2 × 39/20
= 39/40
Now, the 2nd number will be,
= (39/40 + 6/5/2)
= 1/2 (39 + 48/40)
= 1/2 × 87/40
= 87/80
Therefore, two rational numbers are 39/40 and 87/80
Question no – (3)
Solution :
1st rational number,
= 0 + 0.2/2
= 0.2/2
= 0.1
2nd rational number,
= 0 + 0.1/2
= 0.1/2
= 0.05
3rd rational number,
= 0.1 + -0.2/2
= 0.3/2
= 0.15
Hence, three Rational numbers between 0 and 0.2 is 0.1, 0.05 and 0.15
Question no – (4)
Solution :
1st Rational number is,
= 3 + 4/2
= 7/2
2nd Rational number between 3 and 7/2 is
= 3 + 7/2/2
= 1/2 (6 + 7)/2
= 1/2 × 13/2
= 13/4
3rd Rational number between = 7/3 and 4
= 7/3 + 4/2
= 1/2 (7 + 8/2)
= 1/2 × 15/2
= 15/4
Hence, three rational numbers between 3 and 4 are 7/2, 13/4 and 15/4
Question no – (5)
Solution :
Given, 1 3/4 to 4 3/8
= 7/4 to 35/8
So, make same denominator x by 2
= 14/8 to 35/8
Now, (35 – 14) = 21
So, there are 21 terms between 14/8 to 35/8
Now, 1/7th of 21 terms
= 1/7 × 21
= 3rd terms
∴ 3rd terms is
= 17/8
= 2 1/8
Question no – (6)
Solution :
1st Rational numbers = 1 + (- 1/2)
= – 1 – 1/2
= 1/2 × (- 3/2)
= – 3/4
2nd no between – 1 and – 3/4
= 1/2 (- 1 + (- 3/4))
= 1/2 [- 1 – 3/4]
= 1/2 [- 4 – 3/4]
= 1/2 × (- 7/4)
= – 7/8
3rd Rational no between – 3/4 and – 1/2
= 1/2 [- 3/4 + (- 1/2)]
= 1/2 [ – 3/4 – 1/2]
= 1/2 [ – 3 – 2/4]
= 1/2 × – 5/4
= – 5/8
4th no is = – 5/8 + (- 1/2)/2
= 1/2 × [- 5 – 4/8]
= 1/2 × (- 9/8)
= – 9/16
Thus, the four rational numbers are -3/4, -7/8, -5/8 and -9/16
Question no – (7)
Solution :
Given, 12/125
= 0.096
= 125 ÷ 0.09
= 1388.8
Question no – (8)
Solution :
Given, 0.03 = 0.0333333
multiply by 10, 100
Let, x = 0.03333…
x × 10 = 0.33333
x × 100 = 3.3333
Now, (100x – 10x)
= 9 90x = (3.3333 ….. -0.33333….)
= 3.00….
= x = 30/90
= 1/30
Therefore, the Vulgar fraction will be 1/30
Question no – (9)
Solution :
(i) 6.46
= 6.46 = 6.464646 …
Let, x = 6.464646….
Multiple by 100
100x = 646.4646
now, 100x – x = (646 – 6)
= 640
= 99x = 640
= x = 640/99
(ii) 0.136
= 0.136 = 0.1363636….
Let, x = 0.1363636….
∴ 10x = 1.363636….
∴ 100x = 136.363636…..
Now, Subtract,
100x – 10x = 136 – 1
= 990x = 135
= x = 135/990
= 27/198
= 3/22
(iii) 3.146
= 3.146 = 3.146146146…..
Let, x = 3.146146…….
∴ 100 × x = 3146.146
Now, Subtract,
= 1000x – x = 3146 – 3
= 999x = 3143
= x = 3143/999
Question no – (10)
Solution :
(i) 7/4
= 7/4 = 1.75
= 4 ÷ 7.00
= 1.75
(ii) 29/50
= 50 ÷ 2900
= 0.58
(iii) 17/32
= -0.53125
= 32 ÷ 170000
= 0.5312.5
Question no – (11)
Solution :
(i) 1/9
= 9 ÷ 1000 = 0.111
∴ 1/9 = 0.1
Fore more better understanding :
(ii) – 4/3
= – 1.3
= 3 ÷ 40000
= 1.3333
Fore more better understanding :
(iii) 1/6
= 1/6 = 1.16
= 6 ÷ 10000
= 0.1666
Fore more better understanding :
Rational and Irrational Numbers Exercise 1(b) Solution :
(1) Look at the following real numbers
Solution :
(i) which are rational?
= Here, Rational numbers are 0, 3/5, -5, -√9, 6.37, -2/7, 0.03, 4
(ii) which are irrational?
= Irrational numbers are
= √3, √8, √11
(iii) which are positive integers?
= Positive integer are, 4
(iv) which are negative integers?
= Negative integer are -√9, √-5
(v) which number is neither positive nor negative?
= The number which is neither positive nor negative is O
Question no – (2)
Solution :
(i) All rational numbers are real numbers
→ True
(ii) All real numbers are rational numbers
→ False
(iii) Some real numbers are rational numbers
→ True
(iv) All integers are rational numbers
→ True
(v) No rational number is also an irrational number
→ True
(vi) There exists a whole number that is not a natural number
→ True
Question no – (3)
Solution :
(i) 0.578
= Rational
(ii) 0.573 333…
= Rational
(iii) 0.688 434 445 4…
= Irrational
(iv) 0.727 374 75…
= Irrational
(v) 0.638 754 71…
= Rational
(vi) 0.471 7171…
= Rational
(vii) 283
= Rational number
(viii) 289.387 000…
= Rational
(ix) 5.93
= Rational number
(x) 2.30987
= Rational number
(xi) 0.585 885 888…
= Irrational number
Question no – (4)
Solution :
Three distinct irrational numbers are √2, √3, √5, √7
Question no – (5)
Solution :
To Prove, √3 is not a rational numbers
Let, √3 be a rational number
p/q = √3 = [p & q are integer that have no common factor q ≠ 0]
Squaring both side,
p3/q3 = 3
= p2 = 3q2
p2 is an earn number.
p is also an earn no (i)
Let, p = 3r, r (- N)
= p2 = 9r2 – (ii)
By substituting, 3q2 for p2 in
= 3q2 = 9r2
= q2 = 3r2
∴ a2 is multiple of 3
∴ q is a multiple of 3 |
∴ p and q both are multiples of 3 |
∴ 3 is a common factor of p and q
∴ √3 is not a rational number
(ii) √5
Let, √5 is a rational number
Let, √5 = p/q where p and q are integers and have common factors
Now, 5 = p2/q2
= p2 = 5q2 – (i)
5q2 is multiple of 5
p2 is multiple of 5 (ii)
p is multiple of 5
Let, p = 5m
= p2 = 25m2 (iii)
∴ 25x2 = p2
= 25m2 = 5q2
= q2 = 5m2
a2 is multiple of 5 (iv)
From (i) and (ii)
5 is a common factor of p and q
∴ √5 is not rational number.
Question no – (6)
Solution :
Reasons :
1st √100 + √36
= 10 + 6
= 16
2nd √100 + 36
= √136
= 11.68
Therefore, √100 + √36 and √ 100 + 36 are not same,
Rational and Irrational Numbers Exercise 1(c) Solution :
Question no – (1)
Solution :
(i) √1/3
= 1 × √3/√3 × √3
= √3/3…(Simplified)
(ii) √5/12
= √5 √12/√12 √12
= √60/√144
= √2 × 2 × 15/12
= 2 √15/2
= √15/6…(Simplified)
(iii) √1 46/75
= √1 46/75 = √121/75
= √11 × 11/5 × 5 × 3
= 11/5√3
= 11 × √3/5 3
= 11 × √3/15…(Simplified)
Question no – (2)
Solution :
Given, √112 – √63 + 224/√28
= √112 – √63 + 224/√28
= √2 × 2 × 2 × 2 × 7 – √3 × 3 × 7 + 224/2 × 2 × 7
= 4√7 – 3√7 + 224/2√7
= √7 + 112/√7
= √7 + 112√2/7
= √7 (1 + 16)
= 17√7…(Simplified)
Question no – (3)
Solution :
Given, 4 √18/ √12 – √63 + 224/ √28
= 4√18/√12 – 8√75/√32 + 9√2/√3
= 4√2 × 3 × 3/2 × 2 × 3 – 8√5 × 5 × 5/√2 × 2 × 2 × 2 × 2 + 9√2/√3
= 4 × 3√2/2√3 – 8 × 5 √3/4√2 + 9√2/√3
= 6√2 × √3/√3 × √3 – 10√3 × √2/√2 × √2 + 9√2 × √3/√3 × √3
= 6√6/3 – 10√6/2 + 9√6/3
= 2√6 – 5√6 + 3√6
= 0…(Simplified)
Question no – (4)
Solution :
(i) 1/4 – √3
= 1 × (4 + √3)/(4 – √3) (4 + √3)
= 4 + √3/(4)2 – (√3)2
= 4 + √3/16 – 3
= 4 + √3/13
(ii) 2/√5 + √3
= 2 × (√5 – √3)/(√5 + √3) (√5 – √3)
= 2 × (√5 – √3)/(5 – 3)
= 2 × (√5 – √3)/2
= (√5 – √3)
(iii) 1/2√5 – √3
= 1 × (2√5 + √3)/(2√5 – √3) (2 √5+ √3)
= 2√5 + √3/(2√5)2 – (√3)2
= 2√5 + √3/20 – 3
= 2√5 + √3/17
(iv) √3 + √2/√3 – √2
= (√3 + √2) (√3 + √2)/( √3 – √2) (√3 + √2)
= (√3 + √2)2/3 – 2
= 3 + 2 + 2√3. √2/1
= 5 + 2√6/1
= 5 + 2 √6
Question no – (5)
Solution :
(i) 4 + √5/4 – √5 + 4 – √5/4 + √5
= (4 + √5)² + (4 – √5)²/(4 + √5) (4 – √5)
= 16 + 5 + 2 × 4 √5 + 16 + 5 – 2 × 4 √5/(4)² – (√5)²
= 21 + 8√5 + 21 – 8√5/11
= 42/11
(ii) 3/5 – √3 + 2/5 + √3
= 3 (5 + √3) + 2 (5 – √3)/(5 – √3) (5 + √3)
= 15 + 3√3 + 10 – 2√3/(5)2 – (√3)2
= 25 + √3/25 – 3
= 25 + √3/22
Question no – (6)
Solution :
(i) 3 + √2/3 – √2 = a + b√2
= (3 + √2) (3 + √2)/(3 – √2) (3 + √2) = a + b√2
= (3 + √2)2/9 – 2 = a + b√2
= (9 + 2 × 3 × √2 + 2/7 = a + 2√2
= 11 + 6√2/7 = a + b√2
= 11/7 + 6√2/7 = a + b√2
Comparing both,
= a = 11/7 and b = 6/7
(ii) 5 + 2√3/7 + 4√3 a + b√3
∴ (5 + 2√3) (7 – 4√3)/(7 + 4√3) (7 – 4√3) = a + b√3
= 35 – 20√3 + 14√3 – 8 × 3/(7)2 – (4√3)2
= 35 – 209√3 + 14√3 – 24/49 – 48 = a + b√3
= 11 – 6√3 = a + b√3
Comparing, a = 11, b = – 6
(iii) √7- 1/√7 + 1 – √7 + 1/√7 = a + b√7
= (√7 – 1)2 – ) (√7 + 1)2/(√7 + 1) (√7 + 1) (√7 – 1) = a b√7
= (7 + 1 – 2√7) – (7 + 1 + 2√7)/7 – 1 a = + √7
= 8 – 2√7 – 8 – 2√7/6 = a + b√7
= 4√7/6 = a + b √7
Comparing we, get,
a = 0
b = – 4/0 = – 2/3
Hence, a = 0 and b = -2/3
Question no – (7)
Solution :
Given, 1/√2 + √3 + √5
= 1 × √2 + √3 – √5/(√2 + √3 + √5) (√2 + √3 – √5)
= √2 + √3 – √5/(√2 + √3)2 – (√5)2
= √2 + √3 – √5/2 + 3 + 2√6 – 5
= √2 + √3 – √5/2√6
Now, (√2 + √3 – √5) √6/2√6 × √6
= (√2 + √3 – √5) √6/2 × 6
= √12 + √18 – √30/12
= 2√3 + 3√2 – √30/12
= 2√3 + 3√2 – √30/12
Question no – (8)
Solution :
(i) 1/3 – √2
= 3 + √2/(3- √2) (3 + √2)
= 3 + √2/(3)² (√2)²
= 3 + √2/9 – 2
= 3 + √2/7
= 3 + 1.412/7
= 4.414/7
= 0.631
Thus, the value will be 0.631
(ii) 2/√3 – √2
= 2 (√3 + √2)/(√3 – √2) (√3 + √2)
= 2(√3 + √2)/3 – 2
= 2 √3 + 2
= 2 (1.732 + 1.414)
= 2 × 3.146
= 6.282
Thus, the value will be 6.282
Question no – (9)
Solution :
Given, 3 – 5√5/3+ 2√5
= (3 – 5√5) (3 – 2√5)/(3 + 2√5) (3 – 2√5)
= 9 – 6√5 15√5 + 50/(3)2 – (2√5)2
= 9 – 21√5 + 50/9 – 20
= 59 – 21√5/- 11
If is not express in from of (a√5 – b)
Question no – (10)
Solution :
Given, 1/√2 – 1 + 2√3 + 1 = √2 + √3
L.H.S = 1/√2 – 1 + 2√3 + 1
= √2 + 1/(√2 – 1) (√2 + 1) + 2√3 -1)/(√3 + 1) (√- 1)
= √2 + 1/2 – 1 + 2 (√3 – 1)/3 – 1
= √2 + 1 + 2(√3 – 1)/2
= √2 + 1 √3 – 1
= 2 + √3
= R.H.S
∴ L.H.S = R.F.S…(Proved)
Question no – (11)
Solution :
Given, 6√2/√3 + √6 – 4√3/√6 + √2 + 2√6/√2 + √3
= 6√2/√3 + √6 – 4√3/√6 + √2 + 2√6/√2 + √3
= 6√2 (√3 – √6)/3 – 6 – 4√3 (√6 – √2)/6 – 2 + 2√6 (√2 + √3)/2 – 3
= 6√2 (√3 – √6)/- 3 – 4√3 (6 – √2)/- 4 + 2√6 (√2 – √3)/- 1
= – 2√2 (√3 – √6) – √3 (√6 – √2) – 2√6 (√2 – √3)
= – 2√6 + 2√12 – √18 + √6 – 2√2 + 2√18
= – 2√6 + 4√3 – 3√2 + √6 – 4√3 + 6√2
= – √6 + 3√2
= 3√2 – √6
= √9 × 2 – √6
= √18 – √6…(Simplified)
Question no – (12)
Solution :
(i) 6/2√3 – √6 + √6/√3 + √2 – 4√3/√6 – √2
= 6(2√3 + √6)/(2√3 – √6) (2√3 + √6) + √6 (√3 – √2)/( √3 + √2) (√3 – √2) – 4√3 (√2 + √2)/ (√6 – √2) (√6 + √2)
= 6(2√3 + √6)/12 – 6 + √6 (√3 – √2)/3 – 2 – 4√3 (√6 + √2)/6 – 2
= 6(2√3 + √6)/6 + √6 (√3 – √2)/1 – 4√3 (√6 + √2)/4
= 2√3 + √6 + √18 – √12 √ 3 (6 + √2)
= 2√3 + √6 + 3√2 – 2√3 – √18 – √6
= 2√3 + √6 + 3√2 – 2√3
= 3√2 + √6
= 0…(Simplified)
(ii) 7√3/√10 + √3 – 2√5/√6 + √5 – 3√2/15 + 3√2
= 7√3 (√10 – √3)/10 =- 3 – 2√5 (√6 – √5)/6 – 5 – 3√2 (15 – 3√2)/15 – 18
= 7 √3 (√10 – √3)/7 – 2√5(√6 √5/31 – 3√2(√15 – 3√2/-3
= √30 – 3 – 2√30 + 10 + √30 – 6
= 1
Question no – (13)
Solution :
As per the question, x = 2 + √3
∴ 1/x = 1/2 + √3
= 2 – √3/(2 + √3) (2 – √3)
= 2 – √3/4 – 3
= 2 – √3
∴ x + /x = 2 √3 + 2 – √3 = 4
= (x + 1/X)² = 42
= x² + 1/x²
= 16 – 2
= 14
Therefore, the value will be 14
Question no – (14)
Solution :
Here, x = √2 + 1
= 1/x = √2 – 1
So, x + 1/x = √2 + 1 + √2 – 1
= 2√2
= (x + /x)² = (2√2)
= x2 + 1/x² (2√2)2 – 2
= 8 – 2
∴ x² + 1/x² = 6
Therefore, the value will be 6
Question no – (15)
Solution :
Given, x + 1/x
= x + 1/x = 5 – √21/2 + 5 + √21/2
= 5 – √21 + 5 + √21/2
= 10/2
= 5
Now, squaring both side,
= (x + 1/x)² = (5)²
= x2 + 1/x² + 2 = 25
= x2 + 1/x²
= 25 – 2
= 23
Therefore, the value will be 23
Next Chapter Solution :
👉 Chapter 2 👈
