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CBSE Questions for Class 9 Maths Number Systems Quiz 2 - MCQExams.com
CBSE
Class 9 Maths
Number Systems
Quiz 2
Two rational numbers between
2
3
and
5
3
are :
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0%
1
6
and
2
6
0%
1
2
and
2
1
0%
5
6
and
7
6
0%
2
3
and
4
3
Explanation
Changing the denominators of both numbers to 6, we get
2
3
=
4
6
&
5
3
=
10
6
Numbers between the given rational numbers from the options are
5
6
&
7
6
So, correct answer is option C.
State true or false:
There are numbers which cannot be written in the form
p
q
, where
q
≠
0
and
both
p, q are integers.
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0%
True
0%
False
Explanation
The statement is true as there are Irrational numbers which dont satisfy the condition of rational numbers i.e irrational number cannot be written in the form of
p
q
q
≠
0
,
w
h
e
r
e
p
,
q
are integers.
Example,
√
3
,
√
99
The value of
2
√
3
+
√
3
is equal to
Report Question
0%
2
√
6
0%
3
√
3
0%
4
√
6
0%
6
Explanation
2
√
3
+
√
3
=
(
2
+
1
)
√
3
=
3
√
3
Between any two rational numbers,
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0%
there is no rational number
0%
there is exactly one rational number
0%
there are infinitely many rational numbers
0%
there are only rational numbers and no irrational numbers
Explanation
Re
call that to find a rational number between
r
and
s
,
you can add
r
and
s
and divide the sum by
2
,
that is
r
+
s
2
lies between r and s.
For example,
5
2
is a number between
2
and
3.
We can proceed in this manner to find many more rational numbers between
2
and
3.
Hence, we can conclude that there are infinitely many rational numbers between any two given rational numbers.
Every rational number is
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0%
A natural number
0%
An integer
0%
A real number
0%
A whole number
Explanation
Step-1: Explain property of rational numbers
A real number is a number which can be expressed in the form
p
q
,
where
q
≠
0.
Step-2: Proving that every rational number is a real number
Real numbers are numbers that include bothe rational and irrational numbers.
Hence, every rational number is a real number.
Final Answer: Every rational number is a real number. The correct option is (C).
The product of a non - zero rational number with an irrational number is always :
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0%
Irrational number
0%
Rational number
0%
Whole number
0%
Natural number
Explanation
By definition, an irrational number in decimal form goes on forever without repeating (a non-repeating, non-terminating decimal). By definition, a rational number in decimal form either terminates or repeats.
By multiplying a non repeating non terminating number to repeating or terminating/repeating number, the result will always be a non terminating non repeating number.
So, option A is correct.
√
2
,
√
3
are
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0%
Whole numbers
0%
Rational numbers
0%
Irrational numbers
0%
Integers
Explanation
√
2
,
√
3
are irrational numbers.
Because these two values cannot be written in the form of p/q where p and q both are integers and q should not be equal to zero .
State whether the given statement is True or False.
After rationalising the denominator of
5
3
√
2
−
2
√
3
, we get its denominator as
7.
Report Question
0%
True
0%
False
Explanation
The denominator of
5
3
√
2
−
2
√
3
is
3
√
2
−
2
√
3
Rationalising the denominator, we get
(
3
√
2
−
2
√
3
)
(
3
√
2
+
2
√
3
)
=
(
3
√
2
)
2
−
(
2
√
3
)
2
=
18
−
12
=
6
So, the denominator is
6
.
Thus, answer is
false.
Find the product.
(
a
2
)
(
2
a
22
)
(
4
a
26
)
Report Question
0%
8
a
40
0%
8
a
50
0%
8
a
30
0%
8
20
a
Explanation
(
a
2
)
(
2
a
22
)
(
4
a
26
)
=
(
a
2
)
×
(
2
a
22
)
×
(
4
a
26
)
=
8
a
2
+
22
+
26
=
8
a
50
The value of
(
6
+
√
27
)
−
(
3
+
√
3
)
+
(
1
−
2
√
3
)
when simplified is :
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0%
positive and irrational
0%
negative and rational
0%
positive and rational
0%
negative and irrational
Explanation
6
+
√
27
−
(
3
+
√
3
)
+
(
1
−
2
√
3
)
=
6
+
3
√
3
−
3
−
√
3
+
1
−
2
√
3
=
4
4
is a positive rational number
Hence, correct answer is option C.
The value of
3
0
+
7
0
5
0
is:
Report Question
0%
2
0%
0
0%
9
5
0%
1
5
Explanation
⇒
3
0
+
7
0
5
0
⇒
1
+
1
1
[
∵
x
0
=
1
]
⇒
2
1
⇒
2
∴
3
0
+
7
0
5
0
=
2
√
5
is an irrational number.
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0%
True
0%
False
Explanation
An irrational number is any real number that cannot be expressed as a ratio
a
b
,
where
a
and
b
are integers and
b
is non-zero.
√
5
is irrational as it can never be expressed in the form
a
b
.
Find the five rational numbers between
1
2
and
3
2
Report Question
0%
0.5
<
0.6
<
0.7
<
0.8
.
.
.
<
1.1
<
.
.
.
<
1.15
<
1.50
0%
0.5
<
0.6
<
1.7
<
3.8
.
.
.
<
1.8
<
.
.
.
<
1.15
<
1.50
0%
0.5
<
0.6
<
0.7
<
2.8
.
.
.
<
1.1
<
.
.
.
<
1.15
<
1.50
0%
0.5
<
0.6
<
0.7
<
0.8
.
.
.
<
3.1
<
.
.
.
<
1.15
<
1.50
Which of the following numbers are rational ?
Report Question
0%
1
0%
−
6
0%
3
1
2
0%
All above are rational
Explanation
⇒
A rational number is a type of real numbers which
can be expressed in the form of
p
q
,
where
q
≠
0.
⇒
All the numbers are rational as they are in the form of
p
q
,
where
q
≠
0.
i.e,
1
1
,
−
6
1
,
7
2
The rationalizing factor of
(
a
+
√
b
)
is
Report Question
0%
a
−
√
b
0%
√
a
−
b
0%
√
a
−
√
b
0%
None of these
Explanation
The rationalizing factor of a+
√
b
is a-
√
b
as the product of these two expressions give a rational number.
Simplify:
3
√
3
+
10
√
3
Report Question
0%
13
√
3
0%
10
√
3
0%
12
√
3
0%
11
√
3
Explanation
3
√
3
+
10
√
3
=
13
√
3
The value of
5
√
3
−
3
√
12
+
2
√
75
on simplifying is :
Report Question
0%
5
√
3
0%
6
√
3
0%
√
3
0%
9
√
3
Explanation
The given expression is
5
√
3
−
3
√
12
+
2
√
75
=
5
√
3
−
6
√
3
+
10
√
3
=
9
√
3
.
Therefore, option
D
is correct.
State True or False.
A rational number can always be written in a fraction
a
b
, where a and b are not integers
(
b
≠
0
)
.
Report Question
0%
True
0%
False
Explanation
A number that can always be written in the form of
p
/
q
, where p is any integer and q is a non-zero integer, is a rational number.
The given statement is false.
Find conjugate of:
√
3
+
√
2
Report Question
0%
√
3
−
√
2
0%
√
3
+
√
2
0%
√
3
√
2
0%
None of these
Explanation
√
3
+
√
2
The procedure of multiplying a surd by another surd to get a rational number is called RationalisationThe operands are called rationalizing factor (RF) of the other
(
√
3
+
√
2
)
×
(
√
3
−
√
2
)
=
√
3
2
−
√
2
2
=
9
−
4
=
5
The Rationalizing factor of
√
3
+
√
2
is
√
3
−
√
2
Rationalise the denominator of :
√
6
√
12
Report Question
0%
√
2
6
0%
√
2
5
0%
√
2
2
0%
√
2
3
Explanation
√
6
√
12
=
√
6
×
√
12
√
12
×
√
12
=
6
√
2
12
=
√
2
2
Find conjugate of:
3
√
2
-1
Report Question
0%
3
√
2
+1
0%
3
√
2
-1
0%
3
√
2
0%
None of these
Explanation
3
√
2
+
1
The procedure of multiplying a surd by another surd to get a rational number is called Rationalisation
The operands are called rationalizing factor (RF) of the other
(
3
√
2
+
1
)
×
(
3
√
2
−
1
)
=
(
3
√
2
)
2
−
1
2
=
18
−
1
=
17
The Rationalizing factor of
3
√
2
+
1
is
3
√
2
−
1
Rationalise the denominator of :
2
√
2
√
3
Report Question
0%
2
√
7
3
0%
2
√
6
3
0%
2
√
2
3
0%
2
√
6
10
Explanation
2
√
2
√
3
=
2
√
2
×
√
3
√
3
×
√
3
=
2
√
6
3
Rationalise the denominator of :
6
√
10
−
2
Report Question
0%
√
10
−
2
0%
√
10
+
2
0%
2
√
10
0%
None of these
Explanation
M
u
l
t
i
p
l
y
i
n
g
a
n
d
d
i
v
i
d
i
n
g
b
y
√
10
+
2
:
6
(
√
10
+
2
)
(
10
−
4
)
=
√
10
+
2
Rationalise the denominator of:
2
√
5
+
√
3
Report Question
0%
√
5
−
√
3
0%
√
4
−
√
3
0%
√
2
−
√
3
0%
√
6
−
√
3
Explanation
M
u
l
t
i
p
l
y
i
n
g
a
n
d
d
i
v
i
d
i
n
g
b
y
√
5
−
√
3
:
2
(
√
5
−
√
3
)
(
5
−
3
)
=
√
5
−
√
3
Rationalise the denominator
(i)
22
2
√
3
+
1
Report Question
0%
(
2
√
3
−
1
)
0%
3
(
2
√
3
−
1
)
0%
2
(
2
√
3
−
1
)
0%
4
(
2
√
3
−
1
)
Explanation
M
u
l
t
i
p
l
y
i
n
g
a
n
d
d
i
v
i
d
i
n
g
b
y
2
√
3
−
1
:
22
(
2
√
3
−
1
)
(
12
−
1
)
=
2
(
2
√
3
−
1
)
Write the simplest rationalisation factor of the following surds:
√
32
Report Question
0%
√
2
0%
√
7
0%
√
5
0%
√
3
Explanation
√
32
=
√
2
×
2
×
2
×
2
×
2
=
4
√
2
The procedure of multiplying a surd by another surd to get a rational number is called Rationalisation
The operands are called Rationalizing factor (RF) of the other.
Here,
√
2
is the Rationalisation Factor.
State whether true or false.
5
11
is a rational number.
Report Question
0%
True
0%
False
Explanation
True.
Rational numbers are those numbers which can be expressed in the form
p
q
, where p and q are integers and
q
≠
0
Now,
5
11
, is of the form
p
q
, where p and q are integers and
q
≠
0
Hence, it is a rational number.
Rationalise the denominator of :
5
√
7
+
√
2
Report Question
0%
√
7
+
√
2
0%
√
7
−
√
2
0%
√
7
√
2
0%
None of these
Explanation
M
u
l
t
i
p
l
y
i
n
g
a
n
d
d
i
v
i
d
i
n
g
b
y
√
7
−
√
2
:
5
(
√
7
−
√
2
)
(
7
−
2
)
=
√
7
−
√
2
State true or false:
There can be a pair of irrational numbers whose sum is irrational s
uch as
√
3
+
2
and
5
+
√
2
.
Report Question
0%
True
0%
False
Explanation
To get the sum as irrational, both the numbers need to have an irrational part which are different from each other.
Example, the given pair of numbers
√
3
+
2
and
5
+
√
2
have the sum
√
3
+
2
+
5
+
√
2
=
7
+
√
2
+
√
3
which is an irrational number too.
Hence, the given statement is true and option
A
is correct.
State true or false:
√
3
is an irrational number
Report Question
0%
True
0%
False
Explanation
√
3
is irrational number because it will have non terminating and non recurring decimal
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