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

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 :**

**∛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 :**

**∛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 ?**

**∛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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