Class 9 Maths Chapter 1 Practice Question Answers - Number System

Ans: (b)
We know that π is irrational number and Irrational numbers have decimal expansions that neither terminate nor become periodic.
So, correct answer is option B.

Ans: (c)
According to definition of irrational number, If written in decimal notation, an irrational number would have an infinite number of digits to the right of the decimal point, without recurring digits.
Hence, a number having non terminating and non repeating decimal representation is an irrational number.
So, option C is correct.

Ans: (c)
Recall that to find a rational number between r and s, you can add r and s and divide the sum by 2, that is (r+s)/2 lies between r and s.
For example, 5/2 is a number between 2 and 3.
We can proceed in this manner to find many more rational numbers between 2 and 3.
Hence, we can conclude that there are infinitely many rational numbers between any two given rational numbers.    

Ans: (c)
Step-1: Explain property of rational numbers
A real number is a number which can be expressed in the form p/q , where q ≠ 0.
Step-2: Proving that every rational number is a real number
Real numbers are numbers that include bothe rational and irrational numbers.
Hence, every rational number is a real number.
Final Answer: Every rational number is a real number. The correct option is (C).

Ans: (a)
By definition, an irrational number in decimal form goes on forever without repeating (a non-repeating, non-terminating decimal). By definition, a rational number in decimal form either terminates or repeats.
By multiplying a non repeating non terminating number to repeating or terminating/repeating number, the result will always be a non terminating non repeating number.
So, option A is correct.

Ans: (d)
⇒ A rational number is a type of real numbers which can be expressed in the form of p/q , where q ≠ 0.
⇒ All the numbers are rational as they are in the form of p/q , where q ≠ 0.

Ans: (a)
The rationalizing factor of a+ √b  is a-√b  as the product of these two expressions give a rational number.

Ans: (c)
6 + √27 −(3 + √3) + (1 − 2√3) = 6+3√3 − 3 − √3 +1−2√3
= 4
4 is a positive rational number
Hence, correct answer is option C.

Ans:

Here, both 2 and 5 are integer.
Therefore, is a rational number.

Ans: Irrational, as decimal expansion is non-terminating non-recurring.

Ans: We need to rationalise

Now comparing this with a−6√3 , we get
a = 11

Ans: (c)
All numbers that can be written in the form of p/q , where p and q are integers are rational numbers.
Option A:

= 2/3
Hence, it is rational.
Option B:

Hence, it is a rational number.
Option C:

This cannot be simplified further.
This is an irrational number.
Option D:

This is a rational number.
Hence, option C is correct.

Ans: Rational, as decimal expansion is non-terminating recurring.

Ans: To get the rational numbers between -2/5 and 1/2
Take an LCM of these two numbers: -4/10 and 5/10
Multiply numerator and denominator by 2:
-8/20 and 10/20
All the numbers between -8/20 and 10/20 from the answer
Some of these numbers are -7/20, -6/20, -5/20  .... 0, 1/20

Ans: (d)
Let x=1.999...
Since, one digit is repeating, we multiply x by 10, we get 10x=19.99...
So, 10x = 18+1.999...=18+x
Therefore, 10x−x=18, i.e., 9x = 18

Hence, option D is correct answer.

Ans: (b)
Since the denominator of both rational numbers are same. So, for getting the rational numbers between the given rational numbers, we only have to consider the numerators of the rational numbers.
Two numbers between 1 & 4 are 2 and 3.
So, two rational numbers between the given rational numbers will be 2/5 and 3/5
So, correct answer is option B.

Ans: (c)
The value of π is equal to 3.14159265358… which is a non-terminating and non-repeating decimal hence, π is an irrational number.

Ans: We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.
Let us apply this concept to a binomial surd (√a + √b).
When we multiply this with the difference of the same two terms, that is, with (√a − √b), the product is:
(√a + √b)(√a − √b) = (√a)²− (√b)² = a − b (∵ a² − b² = (a + b)(a − b))
Since a−b is a rational number.
Hence, (√a  − √b) is the conjugate of (√a  + √b).

Ans: Since in 0.111222333444.... each number appears 3 times in a row is a non- terminating and non- recurring decimal expansion.
Hence, 0.111222333444.... is irrational.

Ans: We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.
Let us apply this concept to a binomial surd (10√2 + 3√5).
When we multiply this with the difference of the same two terms, that is, with (10√2 - 3√5), the product is:
(10√2 + 3√5)(10√2 − 3√5) = (10√2)² − (3√5)²
= (10 × 2) − (3 × 5) = 20 − 15 = 5( ∵ a² − b² = (a + b)(a − b))
Since 5 is a rational number.
Hence, (10√2  −3√5) is the conjugate of (10√2  + 3√5).

Ans: We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.
Let us apply this concept to a binomial surd (x + 3√y)
When we multiply this with the difference of the same two terms, that is, with (x − 3√y), the product is:
(x + 3√y)(x − 3√y) = (x)² − (3√y)²=x² − 9y (∵a² − b² = (a + b)(a − b))
Since x² − 9y is a rational number.
Hence, (x − 3√y) is the conjugate of (x+3√y).

Ans: We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.
Let us apply this concept to a binomial surd (√8  −5).
When we multiply this with the sum of the same two terms, that is, with (√8 + 5), the product is:
(√8 − 5)(√8 + 5) = (√8)²−(5)² = 8 − 25 = −17 (∵ a² − b² = (a + b)(a − b))
Since −17 is a rational number.
Hence, (8 + 5) is the conjugate of (8 - 5).

Ans: We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.
Let us apply this concept to a binomial surd (√x − 2√y).
When we multiply this with the sum of the same two terms, that is, with (√x  +2√y), the product is:
(√x − 2√y)(√x + 2√y) = (√x)² − (2y)² = x − 4y   (∵ a² − b² = (a + b)(a − b))
Since x − 4y is a rational number.
Hence, (√x  + 2√y) is the conjugate of (√x  − 2√y).

Ans:

Ans: We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.
Let us apply this concept to a binomial surd
When we multiply this with the difference of the same two terms, that is, with  the product is:

(∵ a² − b² = (a + b)(a − b))
Since  is a rational number.
Hence,  is the conjugate of