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NCERT Exemplar Class 9 Maths solutions chapter 1 provide detailed answers for questions related to Number System. These problems and their solutions are devised for a better knowledge of the chapter 1 of NCERT Class 9 Maths Book. These solutions are extremely useful in understanding the concepts and know-how of real numbers. Experts have created these NCERT exemplar solutions for Class 9 Maths chapter 1 with a thorough analysis.
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The NCERT exemplar Class 9 Maths chapter 1 solutions provided for exemplar problems are detailed and expressive. These are helpful to pinpoint the critical understanding of concepts. The NCERT exemplar Class 9 Maths solutions chapter 1 are in accord with the CBSE syllabus for Class 9.
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Question:1
Every rational number is:
(A)A natural number
(B)An integer
(C)A real number
(D) A whole number
Answer:
Answer: [C]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples:
(A) All the positive integers from 1 to infinity are natural numbers. These cannot be fractions.
Examples: 1, 2, 3, 4 and so on
(B) An integer is a number which can be written without fractional components or decimal representation. They can be positive, negative or zero.
Examples:
(C) All numbers that we generally use are real numbers. They can be positive, negative, decimal, whole, natural, integer etc.
Examples:
(D) All the positive integers from 0 to infinity are whole numbers. These cannot be fractions.
Examples: 0, 1, 2, 3, 4 and so on
From the above definitions we can easily see that every rational number is a real number. Therefore option (C) is correct.
Question:2
(A)There is no rational number
(B)there is exactly one rational number
(C)there are infinitely many rational numbers
(D)no irrational number
Answer:
Answer: [C]
Solution.
Firstly let us define a rational number.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples:
So we can say that it is a type of real number.
Similarly we can see that irrational numbers are real numbers which cannot be represented as simple fractions.
Examples:
So we can say that between two rational numbers, there are infinitely many rational numbers.
For Example: between rational numbers 5 and 4, there are rational numbers like ………and so on.
Therefore option (C) is correct.
Question:3
Decimal representation of a rational number cannot be :
(A)Terminating
(B)non-terminating
(C)non-terminating repeating
(D)non-terminating non-repeating
Answer:
Answer: [D]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples:
Terminating decimals have a finite number of digits after decimal point,
Examples:
Non terminating decimals are the ones which keep on continuing after decimal point.
Examples:
Recurring decimals are those non terminating decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point. They are also called repeating decimals.
Examples:
Non-Recurring decimals are those non terminating decimals which do not have a particular pattern/sequence after the decimal point. They are also called non repeating decimals.
Examples:
So by the above definitions we can see that the decimal representation of a rational number cannot be non-terminating non repeating because decimal expansion of rational number is either terminating or non-terminating recurring (repeating).
Therefore option (D) is correct.
Question:4
The product of any two irrational number is
(A)always an irrational number
(B)always a rational number
(C)always an integer
(D)sometimes rational, sometimes irrational
Answer:
Answer: [D]
Solution.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples:
The product of two irrational numbers can be rational or irrational depending on the two numbers.
For example, is 2 which is a rational number
whereas is which is an irrational number.
Therefore option (D) is correct.
Question:5
The decimal expansion of the number is.
(A) A finite decimal
(B)
(C) non-terminating recurring
(D) non-terminating non-recurring
Answer:
Answer: [D]
Solution.
Terminating decimals have a finite number of digits after decimal point,
Examples:
Non terminating decimals are the ones which keep on continuing after decimal point.
Examples:
Recurring decimals are those non terminating decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point. They are also called repeating decimals.
Examples:
Non-Recurring decimals are those non terminating decimals which do not have a particular pattern/sequence after the decimal point. They are also called non repeating decimals.
Examples:
So, the decimal expansion of the number is non-terminating non-recurring. It is an irrational number which is a non-terminating non-recurring decimal expansion.
Therefore option (D) is correct.
Question:6
Which of the following is irrational?
(A)
(B)
(C)
(D)
Answer:
Answer: [C]
Solution.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples:
(rational)
(rational)
(rational)
but is an irrational number.
Therefore option (C) is correct.
Question:7
Which of the following is irrational?
(A)
(B)
(C)
(D)
Answer:
Answer: [D]
Solution.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples:
Irrational cannot be expressed in form of p/q form and its decimal expansion is non- terminating non-recurring decimal expansions.
Non-Recurring non terminating decimals do not have a particular pattern/sequence after the decimal point. They are also called non repeating decimals.
Examples:
In the given options, only option D is non-terminating non-recurring decimal as it satisfies the above definition.
Therefore option (D) is correct.
Question:8
A rational number between and is –
(A)
(B)
(C)
(D)
Answer:
Answer: [C]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Now, we have and
(A) cannot be represented as simple fraction hence it is irrational.
(B) cannot be represented as simple fraction hence it is irrational.
(C) 1.5 is a rational number between and
Q
(D) 1.8 is a rational number but does not lie between and .
Q
Therefore option (C) is correct.
Question:9
The value of 1.999 ___in the form where p and q are integers and q 0, is
(A)
(B)
(C)2
(D)
Answer:
Answer. [C]
Solution.
Let x = 1.999……
Since, one digit is repeating, we multiply x by 10
we get, 10x = 19.999……..
so, 10x = 18 + 1.999………
10x = 18 + x
Therefore, 10x – x = 18, i.e., 9x = 18
i.e.,
Hence option C is correct answer.
Question:10
is equal to –
(A).
(B) 6
(C)
(D)
Answer:
Answer. [C]
Solution.
We can see that both the numbers and are irrational numbers.
So their decimal representation will be non terminating, non repeating. So we
cannot find the exact answer by adding their decimal representations.
Now can be written as
Taking common
Hence the correct answer is
Question:11
is equal to :
(A)
(B)
(C)
(D)
Answer:
Answer. [B]
Solution.
We know that
So we have,
This can be written as
Therefore option (B) is correct.
Question:12
The number obtained on rationalizing the denominator of is :
(A)
(B)
(C)
(D)
Answer:
Answer. [A]
Solution.
We have,
We have to rationalize it
[Multiplying numerator and denominator by ]
= [ (a – b) (a + b) = a^{2} – b^{2}]
Hence option A is correct.
Question:13
is equal to :
(A)
(B)
(C)
(D)
Answer:
Answer.[D]
Solution.
We have,
We have to rationalize it
[Multiplying and dividing by ]
= [ (a – b) (a + b) = a^{2} – b^{2}]
=
Hence option D is correct.
Question:14
After rationalizing the denominator of we get the denominator as:
(A)13
(B)19
(C)5
(D)35
Answer:
Answer. [B]
Solution.
We have,
We have to rationalize it
[Multiplying numerator and denominator by ]
= [ (a – b) (a + b) = a^{2} – b^{2}]
=
Therefore we get the denominator as 19.
Hence (B) is the correct option.
Question:15
The value of is equal to :
(A)
(B) 2
(C)4
(D) 8
Answer:
Answer. [B]
Solution.
Hence (B) is the correrct option.
Question:16
If = 1.4142 then is equal to :
(A)2.4142
(B)5.8282
(C)0.4142
(D)0.1718
Answer:
Answer. [C]
Solution
We have,
We have to rationalize it
[Multiplying numerator and denominator by ]
= [ (a – b) (a + b) = a^{2} – b^{2}]
=
=
=
=0.4142
Hence option C is correct.
Question:17
equals :-
(A)
(B)2^{-6 (C) (D) }
Answer:
Answer. [C]
Solution.
We have, =
So we get,
Hence option C is correct.
Question:18
The product equals:
(A)
(B)2
(C)
(D)
Answer:
Answer. [B]
Solution.
We have,
We know that
=
Now,
So,
Hence option B is correct.
Question:19
The value of is :
(A)
(B)
(C)9
(D)
Answer:
Answer. [A]
Solution.
We have,
We know that
So,
Hence option A is correct
Question:20
Value of (256)^{0.16} × (256)^{0.09} is :
(A) 4
(B) 16
(C) 64
(D) 256.25
Answer:
Answer. [A]
Solution. We have,
(256)^{0.16} × (256)^{0.09} [ a^{m} × a^{n} = a^{m+n}]
= (256)^{0.16 + 0.09}
= (256)^{0.25}
=
=
Now 256 = 2^{8} = (2^{2})^{4} = 4^{4}
= (4^{4})^{1/4} [ (a^{m})^{n} = a^{mn}]
= 4
Hence option A is correct.
Question:21
Which of the following is equal to x?
(A)
(B)
(C)
(D)
Answer:
Answer. [C]
Solution.
(A) We have,
(B) We have,
(C) We have,
= x
(D) We have,
Hence option C is correct.
Answer:
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Given that x and y be rational and irrational number respectively.
So, is necessarily an irrational number.
For example, let
Then,
If possible, let us assume be a rational number.
Consider
On squaring both sides, we get
(using identity )
But we have assumed a is rational
is rational
is rational which is not true.
Hence our assumption was incorrect, so is irrational.
Hence proved
Question:2
Answer:
Answer: [xy is not necessarily an irrational number.]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Given that x and y be rational and irrational number respectively.
Let x = 0 (a rational number) and be an irrational number. then,
, which is not an irrational number.
Hence, xy is not necessarily an irrational number.
Question:3
Answer:
(i)
Answer: False
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples:
Irrational numbers are real numbers which cannot be represented as simple fractions.
Examples:
The given number is .
Here is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational it always gives an irrational number.
Hence the given number is irrational.
Therefore the given statement is False.
(ii)
Answer: False
Solution.
An integer is a number which can be written without fractional components or decimal representation. They can be positive, negative or zero.
Examples:
The given statement is “There are infinitely many integers between any two integers.”
This is false, because between two integers (like 1 and 9), there does not exist infinite integers.
Also, if we consider two consecutive integers (like 8 and 9), there does not exist any integer between them.
Therefore the given statement is False.
(iii)
Answer: False
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Examples:
So, it is a type of real number.
The given statement is: Number of rational numbers between 15 and 18 is finite.
If we see the definition of rational numbers as mentioned above, the given statement is false, because between any two rational numbers there exist infinitely many rational numbers.
Here we have rational numbers between 15 and 18 as:
and infinitely more.
Therefore the given statement is False.
(iv) Answer: True,
Solution:
There are infinitely many numbers which cannot be written in the form ,, p and q both are integers. These numbers are called irrational numbers
(v)
Answer: False
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number.
Irrational numbers are real numbers which cannot be represented as simple fractions.
The given statement is: The square of an irrational number is always rational.
This is False, e.g., let us consider irrational numbers and
which is a rational number.
which is an irrational number.
Hence, square of an irrational number is not always a rational number.
Therefore the given statement is False.
(vi)
Answer: False
Solution:
which is a rational number.
(vii)
Answer: False
Solution:
which is an irrational number.
Question:4
Answer:
(i)Answer. [Rational]
Solution.
We have,
= 14 = which follows rule of rational number.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
So, is a rational number.
(ii)Answer. [Irrational]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
So, it can be written in the form of as
But we know that is irrational
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence, is an irrational number
(iii)Answer. [Irrational]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
So this can be written in the form of as but we can see that (denominator) is irrational.
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence is irrational
(iv)Answer. [Rational]
Solution.
We have,
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Hence is a rational number.
(v)Answer. [Irrational]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We have,
So, it can be written in the form of as
But we know that both are irrational
(Irrational numbers are real numbers which cannot be represented as simple fractions.)
Hence, is an irrational number
(vi)Answer. [Rational]
Solution.
We have,
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.
So, is a rational number.
(vii)Answer. [Rational]
Solution.
We have,
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Also we can see that 0.5918 is a terminating decimal number hence it must be rational.
So, 0.5918 is a rational number.
(viii)Answer. [Rational]
Solution.
We have,
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
So, is a rational number.
(ix)Answer. [Rational]
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
All non-terminating recurring decimal numbers are rational numbers.
Non terminating Recurring decimals are those decimals which have a particular pattern/sequence that keeps on repeating
itself after the decimal point. They are also called repeating decimals.
Examples: 1/3 = 0.33333…, 4/11 = 0.363636…
Now, 10.124124 ………. is a decimal expansion which is a non-terminating recurring.
So, it is a rational number.
(x)Answer. [Irrational]
Solution.
Non terminating Recurring decimals are those decimals which have a particular pattern/sequence that keeps on repeating itself after the decimal point.
All non-terminating recurring decimal numbers are rational numbers.
Non terminating Non Recurring decimals are those decimals which do not have a particular pattern/sequence after the decimal point and it does not end.
All non-terminating non-recurring decimal numbers are irrational numbers.
1.010010001 ………. is non-terminating non-recurring decimal number, therefore it cannot be written in the form ,with p,q both as integers.
Thus, 1.010010001 ……….. is an irrational number.
Question:1
Answer:
(i)Answer. [Irrational]
Solution.
Given that
x^{2} = 5
On taking square root on both sides, we get
Irrational numbers are real numbers which cannot be represented as simple fractions.
So, x is an irrational number
(ii)Answer. [Rational]
Solution.
We have
y^{2} = 9
On taking square root on both sides, we get
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.
So, y can be written as where 1, 3, -3 are rational numbers.
Hence y is rational.
(iii)Answer. [Rational]
Solution.
We have
z^{2} = 0.04 =
On taking square root on both sides, we get
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q
should be rational when the fraction is expressed in the simplest form.
So, z can be written as where 1, 5, -5 are rational numbers.
Hence z is rational.
(iv)Answer. [Irrational]
Solution.
Given that
u^{2} =
On taking square root on both sides, we get
Any number which can be represented in the form of p/q where q is not equal to zero
is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
So, u can be written as where is irrational.
Hence u is irrational.
Question:2
Answer:
(i)Answer.
, and
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Between -1 and -2, many rational number can be written as:
(ii) Answer: ,,
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Between 0.1 and 0.11, many rational number can be written as:
0.103 =
0.104 =
0.105 =
(iii)Answer. and
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
We can write as and as
So, three rational number between are and
(iv)Answer.
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
L.C.M. of 4 and 5 is 20.
We can write as
and
So, three rational number between are
Question:3
(i) Answer. Rational number:
Irrational number: 2.040040004 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 2 and 3
Rational number: 2.5 =
and irrational number : 2.040040004
(ii) Answer. Rational number:
Irrational number 0.0105000500005 ……..
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0 and 0.1:
0.1 can be written as 0.10
Rational number: 0.019 =
and irrational number 0.0105000500005
(iii)Answer. Rational number
Irrational number : 0.414114111 ……
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions. They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
LCM of 3 and 2 is 6.
We can write as
and as
Also, = 0.333333….
And
So, rational number between and is
and irrational number : 0.414114111 ……
(iv)Answer. Rational number: 0
Irrational number: 0.151551555 …….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
and
Rational number between -0.4 and 0.5 is 0
And irrational number: 0.151551555 …….
(v) Answer. Rational number:
Irrational number: 0.151551555 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0.15 and 0.16
Rational number : 0.151 =
and irrational number 0.151551555
(vi) Answer. Rational number:
Irrational number: 1.585585558 ………
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions. They are non-terminating non-recurring in nature.
Between
Rational number: 1.5 =
and irrational number: 1.585585558
(vii) Answer. Rational number: 3
Irrational number: 3.101101110………
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 2.357 and 3.121
Rational number: 3
Irrational number: 3.101101110………
(viii) Answer. Rational number:
Irrational number: 0.000113133133 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0.0001 and 0.001
Rational number: 0.0002 =
Irrational number: 0.000113133133
(ix) Answer. Rational number: 1
Irrational number: 1.909009000 ……
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 3.623623 and 0.484848
A rational number between 3.623623 and 0.484848 is 1.
An irrational number between 3.623623 and 0.484848 is 1.909009000 ……
(x) Answer. A rational number is
An irrational number is 6.375414114111……..
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 6.375289 and 6.375738:
A rational number is 6.3753 =
An irrational number is 6.375414114111……..
Question:4
Represent the following numbers on the number line. 7, 7.2,
Answer:
Solution.
Firstly we draw a number line whose mid-point is O. Mark positive numbers on right hand side of O and negative numbers on left hand side of O.
(i) Number 7 is a positive number. So we mark a number 7 on the right
hand side of O, which is at a 7 units distance form zero.
(ii) Number 7.2 is a positive number. So, we mark a number 7.2 on the right hand side of O, which is 7.2 units distance from zero.
(iii) Number or – 1.5 is a negative number so, we mark a number 1.5 on the left hand side of zero, which is at 1.5 units distance from zero.
(iv) Number or –2.4 is a negative number. So, we mark a number 2.4 on the left hand side of zero, which is at 2.4 units distance from zero.
Question:5
Answer:
Hint.
Solution.
Step I- Draw number line shown in the figure.
Let the point O represent 0 (zero) and point A represent 2 units from O.
Step II- Draw perpendicular AX from A on the number line and cut off arc AB = 1 unit
We have OA = 2 units and AB = 1 unit
Using Pythagoras theorem, we have.
OB^{2} = OA^{2} + AB^{2}
OB^{2} = (2)^{2} + (1)^{2} = 5
OB =
Taking O as the centre and OB = as radius draw an arc cutting the line at C.
Clearly, OC = OB = .
Hence, C represents on the number line.
Similarly for and we can plot the points as follows:
Question:6
Answer:
(i) Solution. AB = 4.5 units, BC = 1 unit
OC = OD = = 2.75 units
OD^{2} = OB^{2} + BD^{2}
So the length of BD will be the required one so mark an arc of length BD on number line, this will result in the required length.
(ii) Solution. Presentation of on number line.
Mark the distance 5.6 units from a fixed point A on a given line to obtain a point B such that AB = 5.6 units. From B mark a distance of 1 unit and mark a new point C. Find the mid point of AC and mark that point as O. Draw a semicircle with center O and radius OC. Draw a line
perpendicular to AC passing through B and intersecting the semicircle at O. Then BD =
(iii) Solution
Mark the distance 8.1 units from a fixed point A on a given line to obtain a point B such that AB = 8.1 units. From B mark a distance of 1 unit and mark the new point AB. Find the mid point of AC and mark a point as O. Draw a semi circle with point O and radius OC. Draw a line perpendicular to AC passing through B and intersecting
the semicircle at D. Then BD -
(iv) Solution
Mark the distance 2.3 unit from a fixed point A on a given line. To obtain a point B such that AB = 2.3 units. From B mark a distance of 1 unit and mark a new point as C. Find the mid point of AC and mark the point asO. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then BD =
Question:7
Answer:
(i) Answer.
Solution. We know that
0.2 can be written as
Now,
Hence the answer is
(ii) Answer.
Solution. Let x = 0.888….. .…(i)
Multiply RHS and LHS by 10
10 x = 8.88……. …(ii)
Subtracting equation (i) from (ii)
We get
10x – x = 8.8 – 0.8
Hence answer is
(iii) Answer.
Solution. Let x = …eq. (1)
Multiply by 10 on both sides
10x = …eq (2)
Subtracting equation (1) from (2)
We get
10x – x = –
9x = 47
x =
Hence the answer is
(iv) Answer.
Solution. Let x = …. Eq. (1)
Multiply by 1000 on both sides
1000 x = …eq.(2)
Subtracting eq. (1) from (2)
We get
1000 x – x = –
999x = 1
x =
Hence the answer is
(v) Answer.
Solution. Let x = 0.2555 ….. …eq.(1)
Multiply by 10 on both sides
10x = 2.555… …eq.(2)
Multiply by 100 on both sides
100x = 25.55… …eq. (3)
Subtracting eq. (2) from (3),
We get
100x – 10x = 25.555… – 2.555…
90x = 23
x =
Hence the answer is
(vii) Answer.
Solution. Let x = 0.00323232….. …eq.(1)
Multiply with 100 on both sides.
We get
100x = 0.3232… …eq(2)
Multiply again with 100 on both sides,
10000x = 32.3232… …eq(3)
Equation (3) – (2)
we get,
10000 x – 100x = 32.3232… – 0.3232…
9900x = 32
x =
x =
Hence the answer is
(viii) Answer.
Solution. Let x = 0.404040……. …(1)
Multiplying by 100 on both sides
we get
100x = 40.40… …(2)
Subtracting equation (1) from (2)
we get
100x – x = 40.40 – 0.40
99x = 40
x =
Hence the answer is
Question:8
Show that 0.142857142857 ……. =
Answer:
Solution. Let, x = 0.142857142857… …(i)
Multiply (i) by 1000000, we get
1000000 x = 142857.142857… …(ii)
Subtracting equation (i) from (ii), we get
1000000 x – x = 142857.142857… - 0.142857…
999999 x = 142857
Hence, 0.142857 …….
Hence proved
Question:9
Answer:
(i) Answer.
Solution.
We know that,
45 =
20 =
So we get
Hence the answer is
(ii) Answer.
Solution. We have,
We know that,
So we get
Taking LCM (3,4) = 12
(iii) Answer.
Solution. We have
We know that
12 =
6 =
So we get,
=
=
Hence the number is .
(iv)Answer.
Solution. We have,
We know that
28 =
So we can write,
=
=
Hence the answer is
(v) Answer.
We know that
27 =
So, =
(Rationalising the denominator)
(Taking common)
Now LCM (1,1,3) = 3
Hence the answer is 19.63
(vi) Answer.
Solution. Given,
We know that (a + b)^{2} = a^{2} – 2ab + b^{2}
Comparing the given equation with the identity, we get:
= 3 + 2 –
Hence the answer is
(vii) Answer. 0
Solution. We have,
We know that
81 =
216 =
32 =
225 =
So,
= 3 – 8 × 6 + 15 × 2 + 15
= 3 – 48 + 30 + 15
= – 45 + 45
= 0
Hence the answer is 0
(viii) Answer.
Solution. We have,
We know that, 8 =
So,
Hence the answer is
(ix) Answer.
Solution. We have,
LCM (3,6) = 6
Hence the answer is .
Question:10
Answer:
(i) Answer.
Solution. We have,
Rationalising the denominator, we get:
Hence the answer is
(ii) Answer.
Solution. We have ,
We know that, 40 = (2) (2) (10)
Rationalising the denominator, we get:
Hence the answer is:
(iii) Answer.
Solution. We have
Rationalising the denominator, we get:
Hence the answer is
(iv) Answer.
Solution. We have
Rationalising the denominator, we get:
Using the identity (a – b) (a + b) = a^{2} – b^{2}
We get:
Hence the answer is
(v) Answer.
Solution. We have,
Rationalising the denominator, we get:
Using (a – b) (a + b) = a^{2} – b^{2}
and (a + b)^{2} = a^{2} + b^{2} + 2ab
Hence the answer is
(vi)Answer.
Solution. We have,
Rationalising the denominator, we get:
Using the identity (a – b) (a + b) = a^{2} – b^{2}
We get:
Hence the answer
(vii) Answer.
Solution. We have,
Rationalising the denominator, we get:
Using (a – b) (a + b) = a^{2} – b^{2}
and (a + b)^{2} = a^{2} + b^{2} + 2ab
Hence the answer is
(viii)
Answer:
Solution:
We have
Rationalize
(ix) Answer:
Solution:
We have
Rationalize
Question:11
Answer:
(i) Answer. a = 11
Solution. We have,
LHS =
Rationalising the denominator, we get:
{Using (a – b) (a + b) = a^{2} – b^{2}}
Now RHS
Hence a = 11 is the required answer
(ii)Answer.
Solution. Given that,
LHS =
Rationalising the denominator, we get:
LHS
{Using (a – b) (a + b) = a^{2} – b^{2}}
Now RHS
Comparing both , we get
Hence is the correct answer
(iii) Answer:
Solution:
Given:
LHS =
Rationalize
=
=
=
=
(iv) Answer. a = 0, b = 1
Solution. Given,
LHS
Using (a – b) (a + b) = a^{2} – b^{2}
(a + b)^{2} = a^{2} + b^{2} + 2ab
(a - b)^{2} = a^{2} + b^{2} - 2ab
RHS
Now LHS = RHS
a = 0, b = 1
Hence the answer is a = 0, b = 1
Question:12
If a = 2 + then find the value of
Answer:
Answer.
Solution. Given that a =
We have
Rationalising,
Using (a – b) (a + b) = a^{2} – b^{2}
Now,
Hence the answer is
Question:13
Answer:
(i) Answer. 2.3093
Solution. Given:
Rationalising,
(Given that )
= 2.3093
Hence the answer is 2.3093
(ii) Answer. 2.449
Solution. Given:
Rationalising,
Putting the given values,
We get :
Hence the answer is 2.449
(iii) Answer. 0.462852
Solution. Given that
This can be written as
Now putting the given values,
We get :
= 0.462852
Hence the answer is 0.462852
(iv) Answer. 0.414
Solution. Given:
Rationalising,
Using (a – b) (a + b) = a^{2} – b^{2 }
Putting the given value of
We get
= 1.414 – 1
= 0.414
Hence the answer is 0.414
(v) Answer. 0.318
Solution. Given that
Rationalising,
Using (a – b) (a + b) = a^{2} – b^{2 }
Putting the given values,
We get,
= 1.732 – 1.414
= 0.318
Hence the answer is 0.318
Question:14
(i) Answer. 6
Solution. (1^{3} + 2^{3} + 3^{3})^{1/2}
We know that
1^{3} = 1.1.1 = 1
2^{3} = 2.2.2 = 8
3^{3} = 3.3.3 = 27
Putting these values we get
(1^{3} + 2^{3} + 3^{3})^{1/2}
Hence the answer is 6
(ii) Answer.
Solution.
We know that
8 = 2.2.2 = 2^{3}
32 = 2.2.2.2.2 = 2^{5}
Hence the answer is
(iii) Answer. 9
Solution. Given
We know that
27 = 3.3.3 = 3^{3 }
= 3^{2} = 9
Hence the answer is 9
(iv) Answer. 5
Solution. Given
We know that
= 5^{1} = 5
Hence the answer is 5
(v) Answer.
Solution. We have
Now we know that
9 = 3.3 = 3^{2}
27 = 3.3.3 = 3^{3 }
Hence the answer is ^{}
(vi) Answer. – 3
Solution. We have ,
We know that 64 =4.4.4=4^{3 }
= – 3
Hence the answer is – 3
(vii) Answer. 16
Solution. Given ,
We know that
8 = 2.2.2 = 2^{3}
16 = 2.2.2.2 = 2^{4}
32 = 2.2.2.2.2 = 2^{5 and }
Hence the answer is 16.
Express 0.6 + in the form where p and q are integers and .
Answer:
Answer.
Solution. Let x = 0.6
Multiply by 10 on LHS and RHS
10x = 6
So, the from of 0.6 =
Let y =
Multiply by 10 on LHS and RHS
10y = 7.7777 …….
10y – y =
= 7.77777 ….. – 0.77777 ……
9y = 7
So the from of 0.7777 =
Let z = 0.47777…
Multiply by 10 on both side
10z = 4.7777 ….
10z – z =
9z = 4.3
So the from of 0.4777 …. =
Therefore, form of 0.6 + is,
Hence the answer is
Question:2
Answer:
Answer. 1
Solution.
Rationalise the denominators:
Hence the answer is 1.
Question:3
Answer:
Answer. 2.0632
Solution. Given that :
,
Using (a – b) (a + b) = a^{2} – b^{2 }
Putting the given values,
= 2.0632
Hence the answer is 2.0632.
Question:4
Answer:
Answer. 7
Solution.Given that :-
On rationalizing the denominator, we get
Using (a – b) (a + b) = a^{2} – b^{2}
Also,
Substituting the values of a and
We get,
= (3)^{2} = 9
Hence the correct answer is 7.
Question:5
If and then find the value of x^{2} + y^{2}.
Answer:
Answer. 98
Solution.We use the identity
So,
Using (a – b) (a + b) = a^{2}– b^{2}= 98
Hence the answer is 98.
Question:6
Answer:
Answer. 2^{48}
Solution. Given,
We know that,
256 = 2.2.2.2.2.2.2.2 = 2^{8}
=
Hence the answer is 2^{48}
Question:7
Answer:
Answer. 214
Solution. We have,
We know that
216 = 6.6.6 = 6^{3}
256 = 4.4.4.4 = 4^{4}
243 = 3.3.3.3.3 = 3^{5}
So,
= 4 × 6^{2} + 4^{3} + 2 × 3
= 4 × 36 + 64 + 6
= 144 + 70
= 214
Hence the answer is 214
Topics covered in NCERT exemplar Class 9 Maths solutions chapter 1 deals with the understanding of:
◊ Real Numbers as Rational numbers and Irrational numbers.
◊ Problems to find a rational number between two given numbers or to find an irrational number between two given numbers are dealt with here.
◊ The concepts of the number line and locating rational number and irrational numbers on the number line are explained.
◊ The decimal expressions of real numbers and fractions are explained in categories like terminating or non-terminating as well as recurring or non-recurring.
◊ Questions based on operations performed on real numbers and exponents for real numbers are explained appropriately.
◊ Laws of exponents for real numbers such as dealing with multiplications of powers and addition of powers of real numbers are explained through NCERT exemplar Class 9 Maths solutions chapter 1.
◊ The concept of rationalising the denominator by multiplication of conjugate of an irrational number is explained through NCERT exemplar solutions for Class 9 Maths chapter 1.
These Class 9 Maths NCERT exemplar solutions chapter 1 will help the students to understand the concept and knowledge of the real numbers. Students of Class 9 can use these solutions as reference material for better study and practice sums of real numbers.
The NCERT exemplar Class 9 Maths solutions chapter 1 Number System will be adequate to solve problems of other reference books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths or RS Aggarwal Class 9 Maths etcetera.
NCERT exemplar Class 9 Maths solutions chapter 1 pdf download will be made available to resolve the doubts encountered while attempting the exemplar of chapter 1.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
Number system weighs approximately 8-10% (discretional to the paper setters; varies from school to school for Class 9 Maths) of the total marks of the paper, however, being one on the basic building blocks for Maths of higher Classes (Class 10, Class 11 and Class 12), the student should understand and practice NCERT exemplar Class 9 Maths chapter 1.
The NCERT exemplar Class 9 Maths solutions chapter 1 equip the student with a multidimensional approach to the problems and understanding the concept of Number System.
A clear understanding of Real Number can prepare a student in solving the problems based on fundamentals of maths which ranges from 2-5% marks of the whole paper.
NCERT exemplar Class 9 Maths solutions chapter 1 will provide the student multifaceted problems to hone the skills and prepare well for the competitive exams such as JEE Main.
Pi is an irrational number because it is a non-terminating non-recurring decimal number. Recently, a tech company employee has broken the Guinness World Record by calculating Pi till 31.4 trillion (1012) decimal places (See it is never-ending…..)
NCERT exemplar Class 9 Maths solutions chapter 1 explains that both the sets of rational and irrational numbers are infinite; hence, it cannot be deduced.
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