Rational and Irrational Numbers Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Rational and Irrational Numbers Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn how to solve Multiple Choice Questions on Rational and Irrational Numbers very easily. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Rational and Irrational Numbers Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Rational and Irrational Numbers Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Board ICSE
Publications Goyal brothers Prakshan
Subject Maths
Class 9th
Chapter-1 Rational and Irrational Numbers
Writer RS Aggrawal
Book Name Foundation
Topics Solution of MCQs
Academic Session 2024-2025

Multiple Choice Questions 

Rational and Irrational Numbers Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Foundation Maths Solutions

Page- 22,23

Que-1: Which of the following is a rational number?

(a) π   (b) √2   (iii) 3.4    (iv) 1.010010001……..

Solution- (c) 3.4

Reason: a type of real number, which is in the form of p/q where q is not equal to zero.  Some of the examples of rational numbers are 1/2, 1/5, 3/4, and so on.

Que-2: Which of the following is an irrational number?

(a) 2.7   (b) 2.72    (c) √11    (d) 2/7

Solution- (c) √11

Reason:  a real number that cannot be expressed as a ratio of integers; for example, √2 is an irrational number.

Que-3: Which of the following is a prime number ?

(a) 51   (b) 57  (c) 71 (d) 81

Solution- (c) 71

Reason: A prime number is the one which has exactly two factors, which means, it can be divided by only “1” and itself.

Que-4: When 8.32 is expressed as a vulgar fraction, then it becomes :

(a) 208/824  (b) 824/99  (c) 800/99  (d) 416/45

Solution- (b) 824/99

Reason: 8.32 = 824/99 in vulgar fraction

Que-5: 0.3 when expressed as a ratio of two integers, becomes :

(a) 103/330    (b) 52/165   (c) 103/111    (d) 104/333

Solution- (a) 103/330

Reason: 0.3 = 103/330

Que-6: Only by inspecting the prime factors of the denominator, state which of the following fractions will be a terminating decimal?

(a) 7/12   (b) 2/15   (c) 3/16    (d) 4/21 

Solution- (c) 3/16

Reason:  The prime factorization of 16 is 2^4.
Since the denominator contains only the prime factor 2, 3/16​ will be a terminating decimal.

Que-7: Only by inspecting the prime factors of the denominator, state which of the following fractions will be a recurring decimal?
(a) 7/16   (b) 8/51   (c) 3/25    (d) 11/20

Solution- (b) 8/51

Reason: The prime factorization of 51 is 3×17.
Since the denominator contains the prime factors 3 and 17, which are neither 2 nor 5, 8/51​ will be a recurring decimal.

Que-8: The number which is to be subtracted from √72 to get √32 is :

(a) 2√10    (b) 4√2    (c) 3√2    (iv) 2√2

Solution- (d) 2√2

Reason: Let the number be x.
√72​−x = √32​
Solving for x, we get:
x =√72​−√32​
Now, we simplify the square roots:
√72 = √(36×2) = √36×√2 = 6√2​
√32 = √(16×2) = √16×√2 = 4√2
Substitute these values back into the equation:
x = 6√2​−4√2​
Combine like terms:​
x = (6−4)√2​
x = 2√2.

Que-9: If x = 3+2√2, then x+1/x =

(a) 4√2   (b) 6√2    (c) 6    (d) 4

Solution- (c) 6

Reason:x = 3+2√2
Then, 1/x = 1/(3+2√2) = [1×(3−2√2)]/(3+2√2)(3−2√2)
= (3−2√2)/(3)²−(2√2)² = (3−2√2)/(9−8) = (3−2√2)/1
∴x+1/x = (3+2√2)+(3−2√2) = 6 Ans.

Que-10: If x = 2-√2, then x-1/x =

(a) 4  (b) -4  (c) 2√2  (d) -2√2

Solution- (d) -2√2

Reason:  x = 2 – √2
To get 1/x, we can rationalize the denominator:
1/x = 1 / (2 – √2)
= (2 + √2) / (2 – √2)(2 + √2)
= (2 + √2) / 2
Now, we can substitute the values of x and 1/x into the expression x – 1/x:
x – 1/x = (2 – √2) – (2 + √2)/2
x -1/x = [4-2√2-2+√2]/2
x-1/x = 2(-2√2)/2
x-1/x = -2√2.

Que-11: If x = 5+2√6, then x²+1/x² =

(a) 98   (b) 142   (c) 49   (d) 138

Solution- (a) 98

Reason: X = 5-2√6
∴ x² = (5-2√6)²
=25-2.5.2√6+24 = 49-20√6
1/x² = 1/(5-2√6)²
= 1/(49-20√6)
= (49+20√6)/(49-20√6)(49+20√6)
= (49+20√6)/(2401-2400)
= 49+20√6
∴ x²+1/x²
= 49-20√6+49+20√6 = 98 Ans.

Que-12: Two rational numbers between -3/7 and -1/7 are :

(a) 4/14, 3/14   (b) -4/14, 3/14    (c) 4/14, -3/14    (d) -4/14, -3/14

Solution- (d) -4/14, -3/14

Reason:  First, let’s convert −37−73​ and −17−71​ to fractions with a common denominator of 14:
−3/7 = (−3×2)/(7×2) = −6/14​
−1/7 = (−1×2)/(7×2) = −2/14
We are looking for rational numbers between −6/14​ and −2/14.
−4/14, −3/14 are both between −6/14 and −2/14​.

Que-13: The correct ascending order of √3, ∛6, 4√7 is :

(a) ∛6, 4√7, √3   (b) 4√7, √3, ∛6    (c) √3, 4√7, ∛6    (d) ∛6, √3, 4√7

Solution- (b) 4√7, √3, ∛6

Reason: √3​:
√3 ≈ 1.732
∛6 (the cube root of 6):
∛6 ≈ 1.817
4√7​ (the fourth root of 7):
4√7 ≈ 1.626
Arranging these values in ascending order, we get:
4√7 < √3 < ∛6.

Que-14: The mixed surd for ∛432 is :

(a) 2∛6    (b) 6∛2    (c) 3∛6    (d) 6∛3

Solution- (b) 6∛2

Reason:  First, we factorize 432:
432 = 2^4 × 3^3
Next, we take the cube root of each factor:
3√432 = 3√(2^4×3^3)
We can separate the cube root into:
3√(2^4×3^3) = 3√(2^4) × 3√(3^3)
Since 3√(3^3) = 3, we have:
3√432 = 3√(2^4) × 3
We simplify 3√(2^4):
3√(2^4) = 2 × 3√2
Thus:
3√432 = 3 × 2 × 3√2 = 6∛2.

Que-15: What is the pure surd for 5∛2 ?

(a) ∛125    (b) ∛150    (c) ∛250    (d) ∛1000

Solution- (c) ∛250

Reason: Here, 5∛2​ has 5 outside the cube root. We convert 5 to its cube and move it inside the cube root:
5∛2 = ∛(5³.2)
Calculate 5³ :
5³ = 125
Thus:
5∛2 = ∛(125.2) = ∛250.

– : End of Rational and Irrational Numbers Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions : —

Return to :- RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)

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