Rs Aggarwal (2021, 2022) Solutions for Class 9 Maths Chapter 1 Number System are provided here with simple step-by-step explanations. These solutions for Number System are extremely popular among class 9 students for Maths Number System Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal (2021, 2022) Book of class 9 Maths Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal (2021, 2022) Solutions. All Rs Aggarwal (2021, 2022) Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 9:
Question 1:
Answer:
Yes, 0 is a rational number.
0 can be expressed in the form of the fraction , where and q can be any integer except 0.
Page No 9:
Question 2:
Yes, 0 is a rational number.
0 can be expressed in the form of the fraction , where and q can be any integer except 0.
Answer:
(i)
(ii)
(iii)
(iv) 1.3
(v) – 2.4
Page No 9:
Question 3:
(i)
(ii)
(iii)
(iv) 1.3
(v) – 2.4
Answer:
(i)
Let:
x = and y =
Rational number lying between x and y:
=
(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:
=
(iii)
Let:
x = 1 and y =
Rational number lying between x and y:
=
(iv)
Let:
x = and y =
Rational number lying between x and y:
=
(v)
A rational number lying between will be
Page No 9:
Question 4:
(i)
Let:
x = and y =
Rational number lying between x and y:
=
(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:
=
(iii)
Let:
x = 1 and y =
Rational number lying between x and y:
=
(iv)
Let:
x = and y =
Rational number lying between x and y:
=
(v)
A rational number lying between will be
Answer:
n = 3
Rational numbers between will be
There are infinitely many rational numbers between two given rational numbers.
â
â
Page No 9:
Question 5:
n = 3
Rational numbers between will be
There are infinitely many rational numbers between two given rational numbers.
â
â
Answer:
n = 4
n + 1 = 4 + 1 = 5
Thus, rational numbers between are .
Page No 9:
Question 6:
n = 4
n + 1 = 4 + 1 = 5
Thus, rational numbers between are .
Answer:
x = 2, y = 3 and n = 6
Thus, the required numbers are
Page No 9:
Question 7:
x = 2, y = 3 and n = 6
Thus, the required numbers are
Answer:
n = 5
n + 1 = 6
Thus, rational numbers between will be
Page No 10:
Question 8:
n = 5
n + 1 = 6
Thus, rational numbers between will be
Answer:
Let:
x = 2.1, y = 2.2 and n = 16
We know:
d = = 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18
Page No 10:
Question 9:
Let:
x = 2.1, y = 2.2 and n = 16
We know:
d = = 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18
Answer:
(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number
(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0.
(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.
(iv) Every integer is a rational number.
True, as rational numbers are of the form . All integers can be represented in the form .
(v) Every rational number is an integer.
False, as rational numbers are of the form . Integers are negative and positive numbers which are not in form.
For example, is a rational number but not an integer.
(vi) Every rational number is a whole number.
False, as rational numbers are of the form . Whole numbers are natural numbers together with a zero.
For example, is a rational number but not a whole number.
Page No 18:
Question 1:
(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number
(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0.
(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.
(iv) Every integer is a rational number.
True, as rational numbers are of the form . All integers can be represented in the form .
(v) Every rational number is an integer.
False, as rational numbers are of the form . Integers are negative and positive numbers which are not in form.
For example, is a rational number but not an integer.
(vi) Every rational number is a whole number.
False, as rational numbers are of the form . Whole numbers are natural numbers together with a zero.
For example, is a rational number but not a whole number.
Answer:
(i)
Denominator of is 80.
And,
80 = 245
Therefore, 80 has no other factors than 2 and 5.
Thus, is a terminating decimal.
(ii)
Denominator of is 24.
And,
24 = 233
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, is not a terminating decimal.
(iii)
Denominator of is 12.
And,
12 = 223
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, is not a terminating decimal.
(iv)
Denominator of is 375.
So, the prime factors of 375 are 5 and 3.
Thus, is not a terminating decimal.
(v)
Denominator of is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, is a terminating decimal.
Page No 19:
Question 2:
(i)
Denominator of is 80.
And,
80 = 245
Therefore, 80 has no other factors than 2 and 5.
Thus, is a terminating decimal.
(ii)
Denominator of is 24.
And,
24 = 233
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, is not a terminating decimal.
(iii)
Denominator of is 12.
And,
12 = 223
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, is not a terminating decimal.
(iv)
Denominator of is 375.
So, the prime factors of 375 are 5 and 3.
Thus, is not a terminating decimal.
(v)
Denominator of is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, is a terminating decimal.
Answer:
(i) = 0.625
By actual division, we have:
It is a terminating decimal expansion.
(ii)
= 0.28
By actual division, we have:
It is a terminating decimal expansion.
(iii) =
It is a non-terminating recurring decimal.
(iv) =
It is a non-terminating recurring decimal.
(v)
=
By actual division, we have:
It is nonterminating recurring decimal expansion.
(vi)
It is a terminating decimal expansion.
(vii)
It is a terminating decimal expansion.
(viii)
2 = =
By actual division, we have:
It is non-terminating decimal expansion.
Page No 19:
Question 3:
(i) = 0.625
By actual division, we have:
It is a terminating decimal expansion.
(ii)
= 0.28
By actual division, we have:
It is a terminating decimal expansion.
(iii) =
It is a non-terminating recurring decimal.
(iv) =
It is a non-terminating recurring decimal.
(v)
=
By actual division, we have:
It is nonterminating recurring decimal expansion.
(vi)
It is a terminating decimal expansion.
(vii)
It is a terminating decimal expansion.
(viii)
2 = =
By actual division, we have:
It is non-terminating decimal expansion.
Answer:
(i)
Let x = 0.222... .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222... .....(ii)
Subtracting (i) from (ii) we get
(ii)
Let x = 0.5353... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353... .....(ii)
Subtracting (i) from (ii) we get
(iii)
Let x = 2.9393... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393... .....(ii)
Subtracting (i) from (ii) we get
(iv)
Let x = 18.4848... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848... .....(ii)
Subtracting (i) from (ii) we get
(v)
Let x = 0.235235... .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235... .....(ii)
Subtracting (i) from (ii) we get
(vi)
Let x = 0.003232... .....(i)
we multiply x by 100.
100x = 0.3232... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232... .....(iii)
Subtracting (ii) from (iii) we get
(vii)
Let x = 1.32323... .....(i)
we multiply x by 10.
10x = 13.2323... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323... .....(iii)
Subtracting (ii) from (iii) we get
(viii)
Let x = 0.3178178... .....(i)
we multiply x by 10.
10x = 3.178178... .....(ii)
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178... .....(iii)
Subtracting (ii) from (iii) we get
(ix)
Let x = 32.123535... .....(i)
we multiply x by 100.
100x = 3212.3535... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35... .....(iii)
Subtracting (ii) from (iii) we get
(x)
Let x = 0.40777... .....(i)
we multiply x by 100.
100x = 40.7777... .....(ii)
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777... .....(iii)
Subtracting (ii) from (iii) we get
Page No 19:
Question 4:
(i)
Let x = 0.222... .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222... .....(ii)
Subtracting (i) from (ii) we get
(ii)
Let x = 0.5353... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353... .....(ii)
Subtracting (i) from (ii) we get
(iii)
Let x = 2.9393... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393... .....(ii)
Subtracting (i) from (ii) we get
(iv)
Let x = 18.4848... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848... .....(ii)
Subtracting (i) from (ii) we get
(v)
Let x = 0.235235... .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235... .....(ii)
Subtracting (i) from (ii) we get
(vi)
Let x = 0.003232... .....(i)
we multiply x by 100.
100x = 0.3232... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232... .....(iii)
Subtracting (ii) from (iii) we get
(vii)
Let x = 1.32323... .....(i)
we multiply x by 10.
10x = 13.2323... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323... .....(iii)
Subtracting (ii) from (iii) we get
(viii)
Let x = 0.3178178... .....(i)
we multiply x by 10.
10x = 3.178178... .....(ii)
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178... .....(iii)
Subtracting (ii) from (iii) we get
(ix)
Let x = 32.123535... .....(i)
we multiply x by 100.
100x = 3212.3535... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35... .....(iii)
Subtracting (ii) from (iii) we get
(x)
Let x = 0.40777... .....(i)
we multiply x by 100.
100x = 40.7777... .....(ii)
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777... .....(iii)
Subtracting (ii) from (iii) we get
Answer:
Given:
Let
First we take x and convert it into
100x = 236.3636... ...(iii)
Subtracting (i) from (iii) we get
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves.
...(iv)
Subtracting (ii) from (iv) we get
Adding x and y we get
=
Page No 19:
Question 5:
Given:
Let
First we take x and convert it into
100x = 236.3636... ...(iii)
Subtracting (i) from (iii) we get
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves.
...(iv)
Subtracting (ii) from (iv) we get
Adding x and y we get
=
Answer:
x = 0.3838... ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838... ...(ii)
Subtracting (i) from (ii) we get
99x = 38
Similarly, we take
y = 1.2727... ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727... ...(iv)
Subtract (iii) from (iv) we get
99y = 126
Page No 23:
Question 1:
x = 0.3838... ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838... ...(ii)
Subtracting (i) from (ii) we get
99x = 38
Similarly, we take
y = 1.2727... ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727... ...(iv)
Subtract (iii) from (iv) we get
99y = 126
Answer:
A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...
Page No 23:
Question 2:
A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...
Answer:
(i)
It is an irrational number.
(ii) = 19
So, it is rational.
(iii)
(iv) = 1.2
So, it is rational.
(v)
It is an irrational number
(vi) 4.1276
It is a terminating decimal. Hence, it is rational.
(vii)
(viii) 1.232332333...
(xi) 6.834834... is a rational number because it is repeating.
Page No 23:
Question 3:
(i)
It is an irrational number.
(ii) = 19
So, it is rational.
(iii)
(iv) = 1.2
So, it is rational.
(v)
It is an irrational number
(vi) 4.1276
It is a terminating decimal. Hence, it is rational.
(vii)
(viii) 1.232332333...
(xi) 6.834834... is a rational number because it is repeating.
Answer:
x be a rational number and y be an irrational number then x + y necessarily will be an irrational number.
Example: 5 is a rational number but is irrational.
So, 5 + will be an irrational number.
Page No 23:
Question 4:
x be a rational number and y be an irrational number then x + y necessarily will be an irrational number.
Example: 5 is a rational number but is irrational.
So, 5 + will be an irrational number.
Answer:
a be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but is irrational. And 6 is also an irrational number.
Page No 23:
Question 5:
a be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but is irrational. And 6 is also an irrational number.
Answer:
Product of two irrational numbers is not always an irrational number.
Example: is irrational number. And is a rational number. But the product of another two irrational numbers is which is also an irrational numbers.
Page No 23:
Question 6:
Product of two irrational numbers is not always an irrational number.
Example: is irrational number. And is a rational number. But the product of another two irrational numbers is which is also an irrational numbers.
Answer:
(i) 2 irrational numbers with difference an irrational number will be .
(ii) 2 irrational numbers with difference is a rational number will be
(iii) 2 irrational numbers with sum an irrational number
(iv) 2 irrational numbers with sum a rational number is
(v) 2 irrational numbers with product an irrational number will be
(vi) 2 irrational numbers with product a rational number will be
(vii) 2 irrational numbers with quotient an irrational number will be
(viii) 2 irrational numbers with quotient a rational number will be .
Page No 23:
Question 7:
(i) 2 irrational numbers with difference an irrational number will be .
(ii) 2 irrational numbers with difference is a rational number will be
(iii) 2 irrational numbers with sum an irrational number
(iv) 2 irrational numbers with sum a rational number is
(v) 2 irrational numbers with product an irrational number will be
(vi) 2 irrational numbers with product a rational number will be
(vii) 2 irrational numbers with quotient an irrational number will be
(viii) 2 irrational numbers with quotient a rational number will be .
Answer:
(i) Let us assume, to the contrary, that is rational.
Then, , where p and q are coprime and .
Since, p and q are are integers.
is rational.
So, is also rational.
But this contradicts the fact that is irrational.
This contradiction has arisen because of our incorrect assumption that is rational.
Hence, is irrational.
(ii) Let us assume, to the contrary, that is rational.
Then, , where p and q are coprime and .
Since, p and q are are integers.
is rational.
So, is also rational.
But this contradicts the fact that is irrational.
This contradiction has arisen because of our incorrect assumption that is rational.
Hence, is irrational.
(iii) As,
Hence, is rational.
(iv) As,
Hence, is rational.
(v) As,
Hence, is rational.
(vi) As,
Hence, is rational.
Page No 23:
Question 8:
(i) Let us assume, to the contrary, that is rational.
Then, , where p and q are coprime and .
Since, p and q are are integers.
is rational.
So, is also rational.
But this contradicts the fact that is irrational.
This contradiction has arisen because of our incorrect assumption that is rational.
Hence, is irrational.
(ii) Let us assume, to the contrary, that is rational.
Then, , where p and q are coprime and .
Since, p and q are are integers.
is rational.
So, is also rational.
But this contradicts the fact that is irrational.
This contradiction has arisen because of our incorrect assumption that is rational.
Hence, is irrational.
(iii) As,
Hence, is rational.
(iv) As,
Hence, is rational.
(v) As,
Hence, is rational.
(vi) As,
Hence, is rational.
Answer:
As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ...
And,
Since,
So, irrational number between 2 ans 2.5 are:
Hence, a rational and an irrational number can be 2.1 and , respectively.
Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.
Page No 23:
Question 9:
As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ...
And,
Since,
So, irrational number between 2 ans 2.5 are:
Hence, a rational and an irrational number can be 2.1 and , respectively.
Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.
Answer:
There are infinite number of irrational numbers lying between .
As,
So, the three irrational numbers lying between are:
1.420420042000..., 1.505005000... and 1.616116111...
Page No 23:
Question 10:
There are infinite number of irrational numbers lying between .
As,
So, the three irrational numbers lying between are:
1.420420042000..., 1.505005000... and 1.616116111...
Answer:
The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52
The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...
Disclaimer: There are infinite number of rational and irrational numbers between 0.5 and 0.55.
Page No 23:
Question 11:
The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52
The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...
Disclaimer: There are infinite number of rational and irrational numbers between 0.5 and 0.55.
Answer:
So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
Page No 24:
Question 12:
So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
Answer:
The rational numbers between the numbers 0.2121121112... and 0.2020020002... are:
Disclaimer: There are an infinite number of rational numbers between two irrational numbers.
Page No 24:
Question 13:
The rational numbers between the numbers 0.2121121112... and 0.2020020002... are:
Disclaimer: There are an infinite number of rational numbers between two irrational numbers.
Answer:
The two irrational numbers between 0.16 and 0.17 are 0.161161116... and 0.1606006000...
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
Page No 24:
Question 14:
The two irrational numbers between 0.16 and 0.17 are 0.161161116... and 0.1606006000...
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
Answer:
(i) True
(ii) False
Example:
(iii) True
(iv) False
(v) True
(vi) False
Example:
(vii) False
Real numbers can be divided into rational and irrational numbers.
(viii) True
(ix) True
Page No 27:
Question 1:
(i) True
(ii) False
Example:
(iii) True
(iv) False
(v) True
(vi) False
Example:
(vii) False
Real numbers can be divided into rational and irrational numbers.
(viii) True
(ix) True
Answer:
Page No 28:
Question 2:
Answer:
Page No 28:
Question 3:
Answer:
Page No 28:
Question 4:
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page No 28:
Question 5:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
Page No 28:
Question 6:
Answer:
(i)
Hence, is rational.
(ii)
Since, the sum and product of rational numbers and an irrational number is always an irrational.
is irrational.
Hence, â is irrational.
(iii)
Hence, is rational.
(iv)
Since, the product of a rational number and an irrational number is always an irrational.
Hence, is rational.
Page No 28:
Question 7:
(i)
Hence, is rational.
(ii)
Since, the sum and product of rational numbers and an irrational number is always an irrational.
is irrational.
Hence, â is irrational.
(iii)
Hence, is rational.
(iv)
Since, the product of a rational number and an irrational number is always an irrational.
Hence, is rational.
Answer:
(i) As,
Hence, the number of chocolates distributed by Reema is 14.
(ii) The moral values depicted here by Reema is helpfulness and caring.
Disclaimer: The moral values may vary from person to person.
Page No 28:
Question 8:
(i) As,
Hence, the number of chocolates distributed by Reema is 14.
(ii) The moral values depicted here by Reema is helpfulness and caring.
Disclaimer: The moral values may vary from person to person.
Answer:
(i)
(ii)
(iii)
Page No 35:
Question 1:
(i)
(ii)
(iii)
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 2 as O and P, respectively.
(ii) At point A, draw AB OA such that AB = 1 units.
(iii) Join OB.
(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P.
Thus, point represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Page No 35:
Question 2:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 2 as O and P, respectively.
(ii) At point A, draw AB OA such that AB = 1 units.
(iii) Join OB.
(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P.
Thus, point represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 1 as O and A, respectively.
(ii) At point A, draw AB OA such that AB = 1 units.
(iii) Join OB.
(iv) At point B, draw DB OA such that DB = 1 units.
(v) Join OD.
(vi) With O as centre and radius OD, draw an arc intersecting the number line at point Q.
Thus, point Q represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Again, in right ODB,
Using Pythagoras theorem,
Page No 35:
Question 3:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 1 as O and A, respectively.
(ii) At point A, draw AB OA such that AB = 1 units.
(iii) Join OB.
(iv) At point B, draw DB OA such that DB = 1 units.
(v) Join OD.
(vi) With O as centre and radius OD, draw an arc intersecting the number line at point Q.
Thus, point Q represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Again, in right ODB,
Using Pythagoras theorem,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 3 as O and B, respectively.
(ii) At point A, draw AB OA such that AB = 1 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Page No 35:
Question 4:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 3 as O and B, respectively.
(ii) At point A, draw AB OA such that AB = 1 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 2 as O and B, respectively.
(ii) At point B, draw AB OA such that AB = 2 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Page No 35:
Question 5:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark points 0 and 2 as O and B, respectively.
(ii) At point B, draw AB OA such that AB = 2 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents on the number line.
Justification:
In right OAB,
Using Pythagoras theorem,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 4.7 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 4.7 2.85 = 1.85 units
In a right angled triangle OBD,
Page No 35:
Question 6:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 4.7 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 4.7 2.85 = 1.85 units
In a right angled triangle OBD,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 10.5 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 10.5 5.75 = 4.75 units
In a right angled triangle OBD,
Page No 35:
Question 7:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 10.5 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 10.5 5.75 = 4.75 units
In a right angled triangle OBD,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 7.28 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 7.28 4.14 = 3.14 units
In a right angled triangle OBD,
Page No 35:
Question 8:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 7.28 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 7.28 4.14 = 3.14 units
In a right angled triangle OBD,
Answer:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 9.5 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
(vii) From E, mark a point F on the same given line such that EF = 1 unit.
Thus, let us treat the given line as the number line, with B as 0, C as 1, E as and so on, then point F represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 9.5 5.25 = 4.25 units
In a right angled triangle OBD,
Page No 35:
Question 9:
![](/img/site_content/ask-answer/loader.gif)
To represent on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 9.5 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
(vii) From E, mark a point F on the same given line such that EF = 1 unit.
Thus, let us treat the given line as the number line, with B as 0, C as 1, E as and so on, then point F represents .
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB AO = 9.5 5.25 = 4.25 units
In a right angled triangle OBD,
Answer:
3 < 3.765 < 4
Divide the gap between 3 and 4 on the number line into 10 equal parts.
Now, 3.7 < 3.765 < 3.8
In order to locate the point 3.765 on the number line, divide the gap between 3.7 and 3.8 into 10 equal parts.
Further, 3.76 < 3.765 < 3.77
So, to locate the point 3.765 on the number line, again divide the gap between 3.76 and 3.77 into 10 equal parts.
Now, the number 3.765 can be located on the number line. This can be shown as follows:
![](/img/site_content/ask-answer/loader.gif)
Here, the marked point represents the point 3.765 on the number line.
Page No 35:
Question 10:
3 < 3.765 < 4
Divide the gap between 3 and 4 on the number line into 10 equal parts.
Now, 3.7 < 3.765 < 3.8
In order to locate the point 3.765 on the number line, divide the gap between 3.7 and 3.8 into 10 equal parts.
Further, 3.76 < 3.765 < 3.77
So, to locate the point 3.765 on the number line, again divide the gap between 3.76 and 3.77 into 10 equal parts.
Now, the number 3.765 can be located on the number line. This can be shown as follows:
![](/img/site_content/ask-answer/loader.gif)
Here, the marked point represents the point 3.765 on the number line.
Answer:
(Upto 4 decimal places)
4 < 4.6767 < 5
Divide the gap between 4 and 5 on the number line into 10 equal parts.
Now, 4.6 < 4.6767 < 4.7
In order to locate the point 4.6767 on the number line, divide the gap between 4.6 and 4.7 into 10 equal parts.
Further, 4.67 < 4.6767 < 4.68
To locate the point 4.6767 on the number line, again divide the gap between 4.67 and 4.68 into 10 equal parts.
Again, 4.676 < 4.6767 < 4.677
To locate the point 4.6767 on the number line, again divide the gap between 4.676 and 4.677 into 10 equal parts.
Now, the number 4.6767 can be located on the number line. This can be shown as follows:
![](/img/site_content/ask-answer/loader.gif)
Here, the marked point represents the point on the number line up to 4 decimal places.
Page No 43:
Question 1:
(Upto 4 decimal places)
4 < 4.6767 < 5
Divide the gap between 4 and 5 on the number line into 10 equal parts.
Now, 4.6 < 4.6767 < 4.7
In order to locate the point 4.6767 on the number line, divide the gap between 4.6 and 4.7 into 10 equal parts.
Further, 4.67 < 4.6767 < 4.68
To locate the point 4.6767 on the number line, again divide the gap between 4.67 and 4.68 into 10 equal parts.
Again, 4.676 < 4.6767 < 4.677
To locate the point 4.6767 on the number line, again divide the gap between 4.676 and 4.677 into 10 equal parts.
Now, the number 4.6767 can be located on the number line. This can be shown as follows:
![](/img/site_content/ask-answer/loader.gif)
Here, the marked point represents the point on the number line up to 4 decimal places.
Answer:
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in is .
Page No 43:
Question 2:
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in is .
Answer:
(i)
On multiplying the numerator and denominator of the given number by , we get:
(ii)
On multiplying the numerator and denominator of the given number by , we get:
(iii)
On multiplying the numerator and denominator of the given number by , we get:
(iv)
On multiplying the numerator and denominator of the given number by , we get:
(v)
On multiplying the numerator and denominator of the given number by , we get:
(vi)
Multiplying the numerator and denominator by , we get
(vii)
Multiplying the numerator and denominator by , we get
(viii)
Multiplying the numerator and denominator by , we get
(ix)
Multiplying the numerator and denominator by , we get
Page No 43:
Question 3:
(i)
On multiplying the numerator and denominator of the given number by , we get:
(ii)
On multiplying the numerator and denominator of the given number by , we get:
(iii)
On multiplying the numerator and denominator of the given number by , we get:
(iv)
On multiplying the numerator and denominator of the given number by , we get:
(v)
On multiplying the numerator and denominator of the given number by , we get:
(vi)
Multiplying the numerator and denominator by , we get
(vii)
Multiplying the numerator and denominator by , we get
(viii)
Multiplying the numerator and denominator by , we get
(ix)
Multiplying the numerator and denominator by , we get
Answer:
(i)
(ii)
(iii)
Page No 43:
Question 4:
(i)
(ii)
(iii)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 43:
Question 5:
(i)
(ii)
(iii)
(iv)
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page No 44:
Question 6:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i)
(ii)
Page No 44:
Question 7:
(i)
(ii)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 44:
Question 8:
(i)
(ii)
(iii)
(iv)
Answer:
(i)
(ii)
Page No 44:
Question 9:
(i)
(ii)
Answer:
Comparing with the given expression, we get
a = 0 and b = 1
Thus, the values of a and b are 0 and 1, respectively.
Page No 44:
Question 10:
Comparing with the given expression, we get
a = 0 and b = 1
Thus, the values of a and b are 0 and 1, respectively.
Answer:
Page No 44:
Question 11:
Answer:
Adding (1) and (2), we get
, which is a rational number
Thus, is rational.
Page No 44:
Question 12:
Adding (1) and (2), we get
, which is a rational number
Thus, is rational.
Answer:
Subtracting (2) from (1), we get
Thus, the value of is .
Page No 44:
Question 13:
Subtracting (2) from (1), we get
Thus, the value of is .
Answer:
Adding (1) and (2), we get
Squaring on both sides, we get
Thus, the value of is 322.
Page No 44:
Question 14:
Adding (1) and (2), we get
Squaring on both sides, we get
Thus, the value of is 322.
Answer:
Adding (1) and (2), we get
Thus, the value of is 5.
Page No 44:
Question 15:
Adding (1) and (2), we get
Thus, the value of is 5.
Answer:
Subtracting (2) from (1), we get
Thus, the value of is .
Page No 45:
Question 16:
Subtracting (2) from (1), we get
Thus, the value of is .
Answer:
Subtracting (2) from (1), we get
Thus, the value of is .
Page No 45:
Question 17:
Subtracting (2) from (1), we get
Thus, the value of is .
Answer:
Adding (1) and (2), we get
Cubing both sides, we get
[Using (3)]
Thus, the value of is 52.
Page No 45:
Question 18:
Adding (1) and (2), we get
Cubing both sides, we get
[Using (3)]
Thus, the value of is 52.
Answer:
Disclaimer: The question is incorrect.
The question is incorrect. Kindly check the question.
The question should have been to show that .
Page No 45:
Question 19:
Disclaimer: The question is incorrect.
The question is incorrect. Kindly check the question.
The question should have been to show that .
Answer:
According to question,
Now,
Hence, .
Page No 45:
Question 20:
According to question,
Now,
Hence, .
Answer:
According to question,
Now,
Hence, the value of a2 + b2 – 5ab is 93.
Page No 45:
Question 21:
According to question,
Now,
Hence, the value of a2 + b2 – 5ab is 93.
Answer:
According to question,
Now,
Hence, the value of p2 + q2 is 47.
Page No 45:
Question 22:
According to question,
Now,
Hence, the value of p2 + q2 is 47.
Answer:
Hence, the rationalised form is .
Hence, the rationalised form is .
Hence, the rationalised form is .
Page No 45:
Question 23:
Hence, the rationalised form is .
Hence, the rationalised form is .
Hence, the rationalised form is .
Answer:
Hence, the value of correct to 3 places of decimal is −1.465.
Page No 45:
Question 24:
Hence, the value of correct to 3 places of decimal is −1.465.
Answer:
Now,
Hence, the value of x3 – 2x2 – 7x + 5 is 3.
Page No 45:
Question 25:
Now,
Hence, the value of x3 – 2x2 – 7x + 5 is 3.
Answer:
Hence, = 5.398 .
Page No 53:
Question 1:
Hence, = 5.398 .
Answer:
(i)
(ii)
(iii)
(iv)
Page No 53:
Question 2:
(i)
(ii)
(iii)
(iv)
Answer:
Page No 53:
Question 3:
Answer:
Page No 53:
Question 4:
Answer:
Page No 53:
Question 5:
Answer:
Page No 53:
Question 6:
Answer:
(i) (ab + ba)–1
(ii) (aa + bb)–1
Page No 53:
Question 7:
(i) (ab + ba)–1
(ii) (aa + bb)–1
Answer:
(i)
(ii) (14641)0.25
(iii)
(iv)
Page No 54:
Question 8:
(i)
(ii) (14641)0.25
(iii)
(iv)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 54:
Question 9:
(i)
(ii)
(iii)
(iv)
Answer:
(i)
(ii)
(iii)
(iv)
Page No 54:
Question 10:
(i)
(ii)
(iii)
(iv)
Answer:
(i)
(ii)
(iii)
Page No 54:
Question 11:
(i)
(ii)
(iii)
Answer:
Hence, the result in the exponential form is .
Page No 54:
Question 12:
Hence, the result in the exponential form is .
Answer:
Page No 54:
Question 13:
Answer:
(i)
(ii)
(iii)
Page No 54:
Question 14:
(i)
(ii)
(iii)
Answer:
Hence, the value of x is 6.
Hence, the value of x is 22.
Hence, the value of x is 5.
Hence, the value of x is 5.
Hence, the value of x is .
Page No 55:
Question 15:
Hence, the value of x is 6.
Hence, the value of x is 22.
Hence, the value of x is 5.
Hence, the value of x is 5.
Hence, the value of x is .
Answer:
(i)
Hence, .
(ii)
Hence, .
(iii)
Hence, .
(iv)
Hence, .
Page No 55:
Question 16:
(i)
Hence, .
(ii)
Hence, .
(iii)
Hence, .
(iv)
Hence, .
Answer:
Page No 55:
Question 17:
Answer:
Hence, m – n = 1.
Page No 55:
Question 18:
Hence, m – n = 1.
Answer:
Page No 57:
Question 1:
Answer:
Since, the sum and product of a rational and an irrational is always irrational.
So, and are irrational numbers.
Also, π is an irrational number.
And, 0 is an integer.
So, 0 is a rational number.
Hence, the correct option is (d).
Page No 57:
Question 2:
Since, the sum and product of a rational and an irrational is always irrational.
So, and are irrational numbers.
Also, π is an irrational number.
And, 0 is an integer.
So, 0 is a rational number.
Hence, the correct option is (d).
Answer:
Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3
But 1.101100110001... is an irrational number
So, the rational number between –3 and 3 is 0.
Hence, the correct option is (a).
Page No 57:
Question 3:
Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3
But 1.101100110001... is an irrational number
So, the rational number between –3 and 3 is 0.
Hence, the correct option is (a).
Answer:
We have,
And,
Also,
Since,
So, the two rational numbers between are .
Hence, the correct opion is (c).
Page No 57:
Question 4:
We have,
And,
Also,
Since,
So, the two rational numbers between are .
Hence, the correct opion is (c).
Answer:
As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.
So, every point on a number line represents a unique number.
Hence, the correct option is (d).
Page No 57:
Question 5:
As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.
So, every point on a number line represents a unique number.
Hence, the correct option is (d).
Answer:
(c)
Because 225 is a square of 15, i.e., = 15, and it can be expressed in the form, it is a rational number.
Page No 57:
Question 6:
(c)
Because 225 is a square of 15, i.e., = 15, and it can be expressed in the form, it is a rational number.
Answer:
(d) a real number
Every rational number is a real number, as every rational number can be easily expressed on the real number line.
Page No 57:
Question 7:
(d) a real number
Every rational number is a real number, as every rational number can be easily expressed on the real number line.
Answer:
(c) are infinitely many rational numbers
Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.
Page No 57:
Question 8:
(c) are infinitely many rational numbers
Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.
Answer:
(b) either terminating or repeating
As per the definition of rational numbers, they are either repeating or terminating decimals.
Page No 58:
Question 9:
(b) either terminating or repeating
As per the definition of rational numbers, they are either repeating or terminating decimals.
Answer:
(d) neither terminating nor repeating
As per the definition of irrational numbers, these are neither terminating nor repeating decimals.
Page No 58:
Question 10:
(d) neither terminating nor repeating
As per the definition of irrational numbers, these are neither terminating nor repeating decimals.
Answer:
As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.
So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.
Hence, the correct option is (d).
Page No 58:
Question 11:
As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.
So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.
Hence, the correct option is (d).
Answer:
(d) 3.141141114...
Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.
Page No 58:
Question 12:
(d) 3.141141114...
Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.
Answer:
Since,
Hence, the correct option is (d).
Page No 58:
Question 13:
Since,
Hence, the correct option is (d).
Answer:
We have,
And,
Since,
So, the rational number which does not lie between is .
Hence, the correct option is (b).
Page No 58:
Question 14:
We have,
And,
Since,
So, the rational number which does not lie between is .
Hence, the correct option is (b).
Answer:
Since, π has a non-terminating non-recurring decimal expansion.
So, π is an irrational number.
Hence, the correct option is (c).
Page No 58:
Question 15:
Since, π has a non-terminating non-recurring decimal expansion.
So, π is an irrational number.
Hence, the correct option is (c).
Answer:
(c) a non-terminating and non-repeating decimal
Because is an irrational number, its decimal expansion is non-terminating and non-repeating.
Page No 58:
Question 16:
(c) a non-terminating and non-repeating decimal
Because is an irrational number, its decimal expansion is non-terminating and non-repeating.
Answer:
Since, = 15, which is an integer,
0.3799 is a number with terminating decimal expansion, and
is a number with non-terminating recurring decimal expansion
Also, 23 is a prime number.
So, is an irrational number.
Hence, the correct option is (a).
Page No 58:
Question 17:
Since, = 15, which is an integer,
0.3799 is a number with terminating decimal expansion, and
is a number with non-terminating recurring decimal expansion
Also, 23 is a prime number.
So, is an irrational number.
Hence, the correct option is (a).
Answer:
So, there are 6 digits in the repeating block of digits in the decimal expansion of .
Hence, the correct option is (b).
Page No 58:
Question 18:
So, there are 6 digits in the repeating block of digits in the decimal expansion of .
Hence, the correct option is (b).
Answer:
Since,
, which is a rational number,
, which is a rational number,
, which is an irrational number, and
, which is a rational number
Hence, the correct option is (c).
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Question 19:
Since,
, which is a rational number,
, which is a rational number,
, which is an irrational number, and
, which is a rational number
Hence, the correct option is (c).
Answer:
(d) sometimes rational and sometimes irrational
For example:
is an irrational number, when it is multiplied with itself it results into 2, which is a rational number.
when multiplied with , which is also an irrational number, results into , which is an irrational number.
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Question 20:
(d) sometimes rational and sometimes irrational
For example:
is an irrational number, when it is multiplied with itself it results into 2, which is a rational number.
when multiplied with , which is also an irrational number, results into , which is an irrational number.
Answer:
(d) Every real number is either rational or irrational.
Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.
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Question 21:
(d) Every real number is either rational or irrational.
Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.
Answer:
(d) is irrational and is rational.
Because the value of is neither repeating nor terminating, it is an irrational number. , on the other hand, is of the form , so it is a rational number.
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Question 22:
(d) is irrational and is rational.
Because the value of is neither repeating nor terminating, it is an irrational number. , on the other hand, is of the form , so it is a rational number.
Answer:
Since, and are irrational numbers,
And,
So, the rational number lying between is 1.6 .
Hence, the correct option is (c).
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Question 23:
Since, and are irrational numbers,
And,
So, the rational number lying between is 1.6 .
Hence, the correct option is (c).
Answer:
Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.
So, 0.853853853... is a rational number.
Hence, the correct option is (d).
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Question 24:
Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.
So, 0.853853853... is a rational number.
Hence, the correct option is (d).
Answer:
Since, the product of a non-zero rational number with an irrational number is always an irrational number.
Hence, the correct option is (a).
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Question 25:
Since, the product of a non-zero rational number with an irrational number is always an irrational number.
Hence, the correct option is (a).
Answer:
Let
Multiplying both sides by 10, we get
Subtracting (1) from (2), we get
Hence, the correct answer is option (b).
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Question 26:
Let
Multiplying both sides by 10, we get
Subtracting (1) from (2), we get
Hence, the correct answer is option (b).
Answer:
(c)
Let x = 1.6666666... ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666... ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
x =
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Question 27:
(c)
Let x = 1.6666666... ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666... ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
x =
Answer:
(b)
Let x = 0.545454... ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545... ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
x = =
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Question 28:
(b)
Let x = 0.545454... ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545... ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
x = =
Answer:
(c)
Let x = 0.3222222222... ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222... ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222... ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
∴ x =
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Question 29:
(c)
Let x = 0.3222222222... ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222... ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222... ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
∴ x =
Answer:
(d) none of these
Let x = 0.12333333333... ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333... ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333... ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111
x =
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Question 30:
(d) none of these
Let x = 0.12333333333... ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333... ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333... ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111
x =
Answer:
(c)
An irrational number between a and b is given as .
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Question 31:
(c)
An irrational number between a and b is given as .
Answer:
(d) 61/4
An irrational number between
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Question 32:
(d) 61/4
An irrational number between
Answer:
(c)
An irrational number between a and b is given as .
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Question 33:
(c)
An irrational number between a and b is given as .
Answer:
Let
Multiplying both sides by 10, we get
Subtracting (1) from (2), we get
Let
Multiplying both sides by 10, we get
Subtracting (3) from (4), we get
Sum of and =
Hence, the correct answer is option (b).
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Question 34:
Let
Multiplying both sides by 10, we get
Subtracting (1) from (2), we get
Let
Multiplying both sides by 10, we get
Subtracting (3) from (4), we get
Sum of and =
Hence, the correct answer is option (b).
Answer:
Let
Multiplying both sides by 100, we get
Subtracting (1) from (2), we get
Let
Multiplying both sides by 100, we get
Subtracting (3) from (4), we get
So,
Hence, the correct answer is option (c).
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Question 35:
Let
Multiplying both sides by 100, we get
Subtracting (1) from (2), we get
Let
Multiplying both sides by 100, we get
Subtracting (3) from (4), we get
So,
Hence, the correct answer is option (c).
Answer:
Hence, the correct answer is option (b).
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Question 36:
Hence, the correct answer is option (b).
Answer:
Thus, the given expression when simplified is positive and rational.
Hence, the correct answer is option (b).
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Question 37:
Thus, the given expression when simplified is positive and rational.
Hence, the correct answer is option (b).
Answer:
Thus, the given expression when simplified is positive and rational.
Hence, the correct answer is option (b).
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Question 38:
Thus, the given expression when simplified is positive and rational.
Hence, the correct answer is option (b).
Answer:
Hence, the correct answer is option (c).
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Question 39:
Hence, the correct answer is option (c).
Answer:
Hence, the correct answer is option (a).
Page No 60:
Question 40:
Hence, the correct answer is option (a).
Answer:
Hence, the correct answer is option (b).
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Question 41:
Hence, the correct answer is option (b).
Answer:
Hence, the correct answer is option (ii).
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Question 42:
Hence, the correct answer is option (ii).
Answer:
Hence, the correct answer is option (b).
Page No 61:
Question 43:
Hence, the correct answer is option (b).
Answer:
Hence, the correct answer is option (c).
Page No 61:
Question 44:
Hence, the correct answer is option (c).
Answer:
Hence, the correct answer is option (b).
Page No 61:
Question 45:
Hence, the correct answer is option (b).
Answer:
Hence, the correct answer is option (d).
Page No 61:
Question 46:
Hence, the correct answer is option (d).
Answer:
The value of is .
Hence, the correct option is (a).
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Question 47:
The value of is .
Hence, the correct option is (a).
Answer:
The value of is 2.
Hence, the correct option is (b).
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Question 48:
The value of is 2.
Hence, the correct option is (b).
Answer:
The value of is 3.
Hence, the correct option is (a).
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Question 49:
The value of is 3.
Hence, the correct option is (a).
Answer:
Hence, the correct option is (c).
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Question 50:
Hence, the correct option is (c).
Answer:
Simplified value of is 1.
Hence, the correct option is (b).
Page No 61:
Question 51:
Simplified value of is 1.
Hence, the correct option is (b).
Answer:
The value of is .
Hence, the correct option is (c).
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Question 52:
The value of is .
Hence, the correct option is (c).
Answer:
Simplified value of is 5.
Hence, the correct option is (d).
Page No 61:
Question 53:
Simplified value of is 5.
Hence, the correct option is (d).
Answer:
The value of is 3.
Hence, the correct option is (a).
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Question 54:
The value of is 3.
Hence, the correct option is (a).
Answer:
x can be .
Hence, the correct option is (d).
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Question 55:
x can be .
Hence, the correct option is (d).
Answer:
Hence, the correct option is (b).
Page No 62:
Question 56:
Hence, the correct option is (b).
Answer:
Hence, the correct option is (b).
Page No 62:
Question 57:
Hence, the correct option is (b).
Answer:
Hence, the correct option is (b).
Page No 62:
Question 58:
Hence, the correct option is (b).
Answer:
Hence, the correct option is (c).
Page No 62:
Question 59:
Hence, the correct option is (c).
Answer:
Hence, the correct option is (a).
Page No 62:
Question 60:
Hence, the correct option is (a).
Answer:
Hence, the correct answer is option (d).
Page No 62:
Question 61:
Hence, the correct answer is option (d).
Answer:
Hence, the correct answer is option (d).
Page No 62:
Question 62:
Hence, the correct answer is option (d).
Answer:
Hence, the correct answer is option (b).
Page No 62:
Question 63:
Hence, the correct answer is option (b).
Answer:
So, the simplest rationalisation factor of is .
Hence, the correct answer is option (d).
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Question 64:
So, the simplest rationalisation factor of is .
Hence, the correct answer is option (d).
Answer:
Simplest rationalisation ractor of is .
Hence, the correct answer is option (b).
Page No 62:
Question 65:
Simplest rationalisation ractor of is .
Hence, the correct answer is option (b).
Answer:
Rationalisation factor of will be .
Hence, the correct answer is option (d).
Page No 62:
Question 66:
Rationalisation factor of will be .
Hence, the correct answer is option (d).
Answer:
Rationalisation of denominator gives
Hence, the correct answer is option (d).
Page No 62:
Question 67:
Rationalisation of denominator gives
Hence, the correct answer is option (d).
Answer:
Hence, the correct answer is option (c).
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Question 68:
Hence, the correct answer is option (c).
Answer:
Hence, the correct answer is option (c).
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Question 69:
Hence, the correct answer is option (c).
Answer:
Given:
Hence, the correct answer is option (b).
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Question 70:
Given:
Hence, the correct answer is option (b).
Answer:
Hence, the correct answer is option (c).
Page No 63:
Question 71:
Hence, the correct answer is option (c).
Answer:
Given that
Hence, the correct answer is option (b).
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Question 72:
Given that
Hence, the correct answer is option (b).
Answer:
This is of the form
Hence, the correct answer is option (d).
Page No 63:
Question 73:
This is of the form
Hence, the correct answer is option (d).
Answer:
This is in the form
So, we have
Thus,
Hence, the correct answer is option (c).
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Question 74:
This is in the form
So, we have
Thus,
Hence, the correct answer is option (c).
Answer:
Hence, the correct answer is option (c).
Page No 63:
Question 75:
Hence, the correct answer is option (c).
Answer:
Given:
Hence, the correct answer is option (a).
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Question 76:
Given:
Hence, the correct answer is option (a).
Answer:
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.
Page No 64:
Question 77:
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.
Answer:
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Page No 64:
Question 78:
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Answer:
(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
It is known that e and are irrational numbers, but Reason is not the correct explanation.
Page No 64:
Question 79:
(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
It is known that e and are irrational numbers, but Reason is not the correct explanation.
Answer:
(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
Page No 64:
Question 80:
(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
Answer:
(a) Because it is a non-terminating and repeating decimal, it is a rational number.
(b) is an irrational number.
(c) ...
Hence, its period is 6.
(d)
Page No 64:
Question 81:
(a) Because it is a non-terminating and repeating decimal, it is a rational number.
(b) is an irrational number.
(c) ...
Hence, its period is 6.
(d)
Answer:
(a)
(b)
(c)
(d)
Page No 65:
Question 1:
(a)
(b)
(c)
(d)
Answer:
Sum of a rational number and an irrational number is an irrational number.
Example: 4 + represents sum of rational and an irrational number where 4 is rational and is irrational.
Page No 65:
Question 2:
Sum of a rational number and an irrational number is an irrational number.
Example: 4 + represents sum of rational and an irrational number where 4 is rational and is irrational.
Answer:
Page No 65:
Question 3:
Answer:
So, will terminate after 3 decimal places.
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Question 4:
So, will terminate after 3 decimal places.
Answer:
Page No 66:
Question 5:
Answer:
Page No 66:
Question 6:
Answer:
A number which is non terminating and non recurring is known as irrational number.
There are infinitely many irrational numbers between 5 and 6.
One of the example is 5.40430045000460000....
Page No 66:
Question 7:
A number which is non terminating and non recurring is known as irrational number.
There are infinitely many irrational numbers between 5 and 6.
One of the example is 5.40430045000460000....
Answer:
Hence, the value of is .
Page No 66:
Question 8:
Hence, the value of is .
Answer:
Hence, the rationalised form is .
Page No 66:
Question 9:
Hence, the rationalised form is .
Answer:
Hence, .
Page No 66:
Question 10:
Hence, .
Answer:
Hence, = 11.
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Question 11:
Hence, = 11.
Answer:
Hence, .
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Question 12:
Hence, .
Answer:
Page No 66:
Question 13:
Answer:
For a = 1and b = 2,
Thus, the value of (ab + ba)–1 when a = 1 and b = 2 is .
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Question 14:
For a = 1and b = 2,
Thus, the value of (ab + ba)–1 when a = 1 and b = 2 is .
Answer:
Page No 66:
Question 15:
Answer:
Let the two irrational numbers be and .
Sum of these irrational numbers , which is rational
Product of these irrational numbers , which is rational
Page No 66:
Question 16:
Let the two irrational numbers be and .
Sum of these irrational numbers , which is rational
Product of these irrational numbers , which is rational
Answer:
Yes, the product of a rational and an irrational number is always an irrational number.
Example:
2 is a rational number and is an irrational number.
Now, , which is an irrational number.
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Question 17:
Yes, the product of a rational and an irrational number is always an irrational number.
Example:
2 is a rational number and is an irrational number.
Now, , which is an irrational number.
Answer:
The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let .
Now,
, which is an irrational number
Also,
, which is a rational number
Page No 66:
Question 18:
The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let .
Now,
, which is an irrational number
Also,
, which is a rational number
Answer:
The reciprocal of
Page No 66:
Question 19:
The reciprocal of
Answer:
The value of
Page No 66:
Question 20:
The value of
Answer:
Page No 66:
Question 21:
Answer:
We have,
Taking square root from both sides, we get
Multiplying both sides by 10, we get
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Question 22:
We have,
Taking square root from both sides, we get
Multiplying both sides by 10, we get
Answer:
Page No 66:
Question 23:
Answer:
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