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CBSE Questions for Class 10 Maths Real Numbers Quiz 5 - MCQExams.com
CBSE
Class 10 Maths
Real Numbers
Quiz 5
The HCF of
455
and
42
using Euclid algorithm is
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0%
7
0%
6
0%
5
0%
4
Explanation
According to the definition of Euclid's theorem,
a
=
b
×
q
+
r
where
0
≤
r
<
b
.
455
=
42
×
10
+
35
42
=
35
×
1
+
7
35
=
7
×
5
+
0
The HCF of
455
and
42
is
7
.
So, option A is correct.
Which of the following fractions will terminate when expressed as a decimal? (Choose all that apply.)
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0%
1
256
0%
27
100
0%
100
27
0%
231
660
0%
7
105
Explanation
Recall that in order for the decimal expansion of a fraction to terminate, the fraction's denominator, in fully reduced form,
must have a prime factorization that consists of only
2
's and/or
5
's.
The denominator in
A
is composed of only
2
's,
(
256
=
2
8
)
.
The denominator in
B
is composed of only
2
's and
5
's,
(
100
=
2
2
×
5
2
)
.
In fully reduced form, the fraction in
D
is equal to
7
20
and the denominator
20
is composed of only
2
's and
5
's
(
20
=
2
2
×
5
)
.
By contrast, the denominator in
C
has prime factors other than
2
's and
5
's
(
27
=
3
3
)
, and in fully reduced form, the fraction in
E
is equal to
1
15
, and
15
has a prime factor other than
2
's and
5
's
(
15
=
3
×
5
)
.
Therefore, options
A
,
B
and
D
are the correct answers.
1
2
=
0.5
It is a terminating decimal because the denominator has a factor as ...........
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0%
0
0%
1
0%
2
0%
6
Explanation
When the prime factorization of the denominator of a fraction has only factors of
2
and factors of
5
, we say the decimal as terminating decimal.
Here, t
he denominator has
2
as a factor and hence it is terminating.
Therefore,
C
is the correct answer.
Find the HCF of
26
and
455
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0%
11
0%
12
0%
13
0%
none of the above
Explanation
26
=
13
×
2
455
=
5
×
13
×
7
HCF is
13.
The HCF of the numbers
0.48
,
0.72
and
0.108
is
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0%
1
0%
12
0%
0.12
0%
0.012
Explanation
480
=
2
5
×
3
×
5
720
=
2
4
×
3
2
×
5
108
=
2
2
×
3
2
H
C
F
=
2
2
×
3
=
12
HCF of
0.48
,
0.72
and
0.108
=
0.012
So, option
D
is correct.
If the denominator of a fraction has factors other than
2
and
5
, the decimal expression ..............
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0%
repeats
0%
is that of a whole number
0%
has equal numerator and denominator
0%
terminates
Explanation
If there are prime factors in the denominator other than
2
or
5
, then a block of digit repeats.
For example,
1
24
=
1
3
×
2
×
2
×
2
(there is a factor of
3
, therefore decimal will repeat.)
=
0.416666...
Therefore,
A
is the correct answer.
The HCF of two consecutive odd numbers is
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0%
28
0%
30
0%
1
0%
5
Explanation
Check for the HCF of consecutive odd numbers by writing down the first few odd numbers.
Step 1: Writing out the first few odd numbers
The first few odd numbers are 1, 3, 5, 7, 9, 11
Step 2: Finding out the HCF of consecutive odd numbers
It is evident that there is no common factors among any 2 consecutive odd numbers except 1
For example the HCF of 3 and 5 is 1, and that of 5 and 7 is also 1
Hence option C is correct
Using the theory that any positive odd integers are of the form
4
q
+
1
or
4
q
+
3
where
q
is a positive integer, If the quotient is
4
, the dividend is
19
then what will be the remainder?
Report Question
0%
17
0%
2
0%
1
0%
3
Explanation
Let
a
be any positive integer.
We know by Euclid's algorithm, if
a
and
b
are two positive integers, there exist unique integers
q
and
r
satisfying,
a
=
b
q
+
r
where
0
≤
r
<
b
.
Take
b
=
4
we get,
a
=
4
q
+
r
,
0
≤
r
<
b
.
Since
0
≤
r
<
4
, therefore the possible remainders are
0
,
1
,
2
and
3
.
That is,
a
can be
4
q
,
4
q
+
1
, or
4
q
+
2
or
4
q
+
3
, where
q
is the quotient.
Since
a
is odd,
a
cannot be
4
q
or
4
q
+
2
as they are both divisible by
2
.
Therefore, any odd integer is of the form
4
q
+
1
or
4
q
+
3
.
Given the quotient
q
=
4
we get,
4
q
+
1
=
4
×
4
+
1
=
17
, or
4
q
+
3
=
4
×
4
+
3
=
19
Since dividend is
19
the remainder is
3.
Therefore
,
D
is the correct answer.
There are five odd numbers
1
,
3
,
5
,
7
,
9
. What is the HCF of these odd numbers?
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0%
2
0%
1
0%
6
0%
5
Explanation
Let us calculate and check what is the highest common factor for these numbers.
First finding the factors of individual numbers:
Factors of
1
=
1
Factors of
3
=
1
,
3
Factors of
5
=
1
,
5
Factors of
7
=
1
,
7
Factors of
9
=
1
,
3
,
9
HCF is
1
for the five odd numbers.
Therefore,
B
is the correct answer.
The HCF of two consecutive even numbers is
Report Question
0%
6
0%
3
0%
4
0%
2
Explanation
H
C
F
of two consecutive even numbers is always
2
. For example:
H
C
F
of
2
and
4
is
2
and similarly,
H
C
F
of
22
and
24
is
2
and we can do so on..
Find HCF of
70
and
245
using Fundamental Theorem of Arithmetic.
Report Question
0%
35
0%
30
0%
15
0%
25
Explanation
Using Fundamental Theorem of Arithmetic we can write the given integers as a product of their prime factors,
70
=
2
×
5
×
7
245
=
5
×
7
×
7
Hence, common factors of
70
and
245
are
5
and
7.
H
C
F
=
5
×
7
=
35
Hence, option
A
is correct.
Three ropes are
7
m
,
12
m
95
c
m
and
3
m
85
c
m
long. What is the greatest possible length that can be used to measure these ropes?
Report Question
0%
35
c
m
0%
55
c
m
0%
1
m
0%
65
c
m
Explanation
The given three ropes are
7
m,
12
m
95
cm and
3
m
85
cm long. We know that
1
m=
100
cm, therefore,
The length of the respective ropes will be:
1st rope
=
7
×
100
=
700
cm
2nd rope
=
(
12
×
100
)
+
95
=
1200
+
95
=
1295
cm
3rd rope
=
(
3
×
100
)
+
85
=
300
+
85
=
385
cm
Now, let us factorize the length of the ropes as follows:
700
=
2
×
2
×
5
×
5
×
7
1295
=
5
×
7
×
37
385
=
5
×
7
×
11
The highest common factor (HCF) is
5
×
7
=
35
Hence, the greatest possible length that can be used to measure these ropes is
35
cm.
Identify a non-terminating repeating decimal.
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0%
24
1600
0%
171
800
0%
123
2
2
×
5
3
0%
145
2
3
×
5
2
×
7
2
Explanation
We know that, a fraction will have terminating decimal expansion only if its denominator is in the form of
2
m
×
5
n
.
Let's check the denominator of each option:
(
i
)
24
1600
1600
=
2
×
2
×
2
×
2
×
2
×
2
×
5
×
5
=
2
6
×
5
2
Hence, the denominator is
2
6
×
5
2
(
i
i
)
171
800
800
=
2
×
2
×
2
×
2
×
2
×
5
×
5
=
2
5
×
5
2
Hence, the denominator is
2
5
×
5
2
(
i
i
i
)
123
2
3
×
5
3
Hence, the denominator is
2
3
×
5
3
(
i
v
)
145
2
3
×
5
2
×
7
2
Hence, the denominator is
2
3
×
5
2
×
7
2
Only option D does not have a denominator of the form
2
m
×
5
n
.
Hence, it will have a non-terminating decimal expansion. Further, the expansion will be repeating.
Hence, option D is correct.
A rational number can be expressed as a terminating decimal if the denominator has factors _________.
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0%
2
or
5
0%
2
,
3
or
5
0%
3
or
5
0%
Only
2
and
3
Explanation
We know, any rational number with its denominator is in the form of
2
m
×
5
n
, where
m
,
n
are positive integers are terminating decimals.
So, option
A
is correct.
Which one of the following is not true?
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0%
√
2
is an irrational number
0%
If a is a rational number and
√
b
is an irrational number then
a
√
b
is irrational number
0%
Every surd is an irrational number
0%
The square root of every positive integer is always irrational
Explanation
(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.
An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:
π
=
3.141592
…
√
2
=
1.414213
…
Therefore,
√
2
is an irrational number.
(b) Let us take a rational number
a
=
2
1
and an irrational number
b
=
√
2
, then their product can be determined as:
a
×
b
=
2
×
√
2
=
2
√
2
which is also an irrational number.
Therefore, if
a
is a rational number and
√
b
is an irrational number than
a
√
b
is an irrational number.
(c)
A surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.
For example,
√
2
is a surd since
2
is rational and
√
2
is irrational.
Surds are numbers left in root form
√
to express its exact value. It has an infinite number of non-recurring decimals.
Therefore, every surd is an irrational number.
(d) Let us take a positive integer
4
, now square root of
4
will be:
√
4
=
2
which is not an irrational number
Hence, the square root of every positive integer is not always irrational.
What is the HCF of
7
and
17
?
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0%
7
0%
1
0%
119
0%
17
A real number
2
2
×
3
2
×
7
2
2
5
×
5
3
×
3
2
×
7
will have _________.
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0%
Terminating decimal
0%
Non-terminating decimal
0%
Non-terminating and non-repeating decimal
0%
Terminating repeating decimal
Explanation
Using the law of exponent,
2
2
×
3
2
×
7
2
2
5
×
5
3
×
3
2
×
7
=
2
2
−
5
×
3
2
−
2
×
5
−
3
×
7
2
−
1
=
2
−
3
×
3
0
×
5
−
3
×
7
1
=
7
2
3
×
5
3
=
7
1000
=
0.007
.
∴
is a terminating and non-repeating decimal.
That is, option
A
is correct.
The greatest integer that divides
358,\ 376
and
232
. The same remainder in each case is
Report Question
0%
6
0%
7
0%
8
0%
9
Explanation
If the remainder is the same in each case and the remainder is not given, HCF of the differences of the
numbers is the required greatest number
Given: The greatest number that will divide 358,376 and 232 leaving the same remainder in each case.
To find: The number?
Solution :
First, we find the difference between these numbers.
The required numbers are
376 - 232 = 144
376-358= 18
358 - 232 =126
Now, We find the HCF of 18, 126, and 144
18 =2\times3^2
126= 2\times3^2\times7
144= 2^4\times3^2
HCF(18,126,144)= 2\times3\times3
Therefore, The required largest number is 18.
Find HCF of
25 and 55:
25 and
Report Question
0%
5
0%
3
0%
2
0%
4
Explanation
Prime factors of
25 = 5 \times 5
Prime factors of
55 = 5 \times 11
Since the common factor is only
5
.
Therefore,
HCF(25,55)=5
Hence, the correct option
A
The H. C. F. of
252
,
324
and
594
is ____________.
Report Question
0%
36
0%
18
0%
12
0%
6
Explanation
H.C.F is the greatest number which divides each of them individually.
All the numbers are written in form of factors of prime numbers.
252
\rightarrow
2
\times
2
\times
3
\times
3
\times
7
324
\rightarrow
2
\times
2
\times
3
\times
3
\times
3
\times
3
594
\rightarrow
2
\times
3
\times
3
\times
3
\times
11
From this greatest factor is
2
\times
3
\times
3
\rightarrow
18
Hence, option B is correct.
The product of two irrational numbers is
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0%
Always irrational
0%
Always rational
0%
Can be both rational or irrational
0%
always an integer
Explanation
According to the properties of irrational numbers, the product of two irrational numbers can be both rational or irrational.
For example, let
p=\sqrt 3
and
q=\sqrt 3
be two irrational numbers
\therefore \ pq=\sqrt 3\times \sqrt 3=3
Here,
3
is a rational number.
Now, let
p=\sqrt 3
and
q=\sqrt 2
\therefore \ pq=\sqrt 3\times \sqrt 2=\sqrt 6
Here,
\sqrt6
is an irrational number.
Hence, the product can be both rational or irrational.
So, option C is correct.
\frac{2}{3}
is a rational number whereas
\frac{\sqrt{2}}{\sqrt{3}}
is?
Report Question
0%
Also a rational number
0%
An irrational number
0%
Not a number
0%
A natural periodic number
State whether True or False :
All the following numbers are irrationals.
(i)
\dfrac { 2 }{ \sqrt { 7 } }
(ii)
\dfrac { 3 }{ 2\sqrt { 5 } }
(iii)
4+\sqrt { 2 }
(iv)
5\sqrt { 2 }
Report Question
0%
True
0%
False
Explanation
In all of the above questions
\sqrt7,\sqrt5,\sqrt2
are a
irrational
numbers
And
\text{the addition, subtraction, division and product between rational and irrational gives an irrational number}
So that all of the above are irrational numbers.
hence option A is correct.
State whether the given statement is True or False :
2-3\sqrt { 5 }
is an irrational number.
Report Question
0%
True
0%
False
Explanation
Here 2 is a rational number
and
3\sqrt5
is an irrational number a
nd
\text{difference of rational and irrational is always an irrational number}
So that
2 - 3\sqrt5
is an
irrational
number
hence option A is correct.
State whether the given statement is True or False :
4-5\sqrt { 2 }
is an irrational number.
Report Question
0%
True
0%
False
Explanation
\text{Here 4 is a rational number and }
5\sqrt2
is a
irrational
number
And
\text{the difference of rational and irrational is always an irrational number}
So that
(4-5\sqrt2)
is an irrational number.
hence option A is correct.
State whether the given statement is True or False :
5-2\sqrt { 3 }
is an irrational number.
Report Question
0%
True
0%
False
Explanation
\text{Here 5 is a rational number and }
2\sqrt3
is a
irrational
number
And
\text{the difference of rational and irrational is always an irrational number}
So that
(5 - 2\sqrt3)
is an irrational number.
hence option A is correct.
State whether the given statement is True or False :
\sqrt { 3 } +\sqrt { 4 }
is an irrational number.
Report Question
0%
True
0%
False
Explanation
\sqrt4= 2 \text{ is a rational number and }
\sqrt3
is a
irrational
number
And
\text{the addition of rational and irrational is always an irrational number}
So that
(\sqrt3+\sqrt4)
is an irrational number.
hence option A is correct.
2-\sqrt { 3 }
is an irrational number.
Report Question
0%
True
0%
False
Explanation
We know that
2
is a rational number and
\sqrt3
is an irrational number.
Also difference between rational and irrational number is always irrational number.
So that
(2-\sqrt3)
is an irrational number.
hence option A is correct.
State whether the given statement is True or False :
If
p, q
are prime positive integers, then
\sqrt { p } +\sqrt { q }
is an irrational number.
Report Question
0%
True
0%
False
Explanation
Let us asume
\sqrt{p}+\sqrt{q}
are rational.
Given that
p
and
q
are prime positive integers.
\sqrt{p}+\sqrt{q}=\dfrac{a}{b}
Squaring on both sides, we get
p+q+2\sqrt{pq}=\left(\dfrac{a}{b}\right)^{2}
\sqrt{pq}=\dfrac{1}{2}\left[\left(\dfrac{a}{b}\right)^{2}-p-q\right] - (i)
We know that
p
&
q
are prime positive numbers
\Rightarrow \sqrt{p}
,
\sqrt{q}
and
\sqrt{pq}
are irrational
So our assumption
\sqrt{p}+\sqrt{q}
are rational is incorrect.
\Rightarrow\ \sqrt p+\sqrt q
is a irrational number if
p,q
are prime positive numbers
State whether the given statement is True or False :
The number
6+\sqrt { 2 }
is
irrational.
Report Question
0%
True
0%
False
Explanation
Here \ 6 \text{ is a rational number and }
\sqrt2
is a
irrational
number
And
\text{the addition of rational and irrational is always an irrational number}
So that
(6+\sqrt2)
is an irrational number.
hence option A is correct
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