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Title: Its help you alot of
Description: It is the real and verified notes for NCERT solution by CBSE
Description: It is the real and verified notes for NCERT solution by CBSE
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Class X
Chapter 1 – Real Numbers
Maths
Exercise 1
...
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45
...
Since the divisor at this stage is 196,
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(iii)867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to
obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102
to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the
division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops
...
Question 2:
Show that any positive odd integer is of the form
, or
, or
, where q is some integer
...
Then, by Euclid’s algorithm,
a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤
r < 6
...
vidhyarjan
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Class X
Chapter 1 – Real Numbers
Maths
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an
integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an
integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an
integer
...
Hence, these expressions of numbers are odd numbers
...
The two groups are to march in the same
number of columns
...
We can use Euclid’s algorithm to find the HCF
...
Therefore, they can march in 8 columns each
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
Question 4:
Use Euclid’s division lemma to show that the square of any positive
integer is either of form 3m or 3m + 1 for some integer m
...
Now square each of these and show that they can be rewritten
in the form 3m or 3m + 1
...
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of
the form 3m or 3m + 1
...
Answer:
Let a be any positive integer and b = 3
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There are three cases
...
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2
Question 1:
Express each number as product of its prime factors:
Answer:
Question 2:
Find the LCM and HCF of the following pairs of integers and verify that
LCM × HCF = product of the two numbers
...
vidhyarjan
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Mobile: 9999 249717
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Class X
Chapter 1 – Real Numbers
Maths
Hence, product of two numbers = HCF × LCM
Hence, product of two numbers = HCF × LCM
Question 3:
Find the LCM and HCF of the following integers by applying the prime
factorisation method
...
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Class X
Chapter 1 – Real Numbers
Maths
Answer:
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Answer:
Question 5:
Check whether 6n can end with the digit 0 for any natural number n
...
Hence, for any value of n, 6n will not be divisible by 5
...
Question 6:
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are
composite numbers
...
Prime numbers can
be divided by 1 and only itself, whereas composite numbers have
factors other than 1 and itself
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
It can be observed that
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 ×13 × 6
The given expression has 6 and 13 as its factors
...
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009
1009 cannot be factorised further
...
Hence, it is a composite number
...
Sonia takes 18 minutes
to drive one round of the field, while Ravi takes 12 minutes for the
same
...
After how many minutes will they
meet again at the starting point?
Answer:
It can be observed that Ravi takes lesser time than Sonia for
completing 1 round of the circular path
...
And the
total time taken for completing this 1 round of circular path will be the
LCM of time taken by Sonia and Ravi for completing 1 round of circular
path respectively i
...
, LCM of 18 minutes and 12 minutes
...
vidhyarjan
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com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
18 = 2 × 3 × 3
And, 12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting pointafter
36 minutes
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
Exercise 1
...
Answer:
Let
is a rational number
...
Then we can divide
them by the common factor, and assume that a and b are co-prime
...
Let a = 5k, where k is an integer
This means that b2 is divisible by 5 and hence, b is divisible
by 5
...
And this is a contradiction to the fact that a and b are co-prime
...
Question 2:
Prove that
is irrational
...
vidhyarjan
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Mobile: 9999 249717
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Class X
Chapter 1 – Real Numbers
Maths
Answer:
is rational
...
This contradicts the fact that
that
will also be rational and
is irrational
...
Therefore,
is irrational
...
Therefore, we can find two integers a, b (b ≠ 0) such that
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Therefore,
is rational which contradicts to the fact that
is
irrational
...
is rational
...
Therefore,
should be rational
...
Therefore, our assumption
is rational is false
...
be rational
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
Therefore, we can find two integers a, b (b ≠ 0) such that
Since a and b are integers,
is also rational and hence,
be rational
...
Therefore,
is irrational
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
Exercise 1
...
Hence, the decimal expansion of
is terminating
...
Hence, the decimal expansion of
is terminating
...
vidhyarjan
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Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
455 = 5 × 7 × 13
Since the denominator is not in the form 2m × 5n, and it also contains
7 and 13 as its factors, its decimal expansion will be non-terminating
repeating
...
Hence, the decimal expansion of
is terminating
...
(vi)
The denominator is of the form 2m × 5n
...
(vii)
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(viii)
The denominator is of the form 5n
...
(ix)
The denominator is of the form 2m × 5n
...
(x)
Since the denominator is not of the form 2m × 5n, and it also has 3 as
its factors, the decimal expansion of
is non-terminating repeating
...
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vidhyarjan
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com
Mobile: 9999 249717
Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051
(One Km from ‘Welcome Metro Station)
Class X
Chapter 1 – Real Numbers
Maths
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In each case, decide whether they are rational or not
...
123456789 (ii) 0
...
123456789
Since this number has a terminating decimal expansion, it is a rational
number of the form
and q is of the form
i
...
, the prime factors of q will be either 2 or 5 or both
...
120120012000120000 …
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Therefore,
the given number is an irrational number
...
e
...
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Title: Its help you alot of
Description: It is the real and verified notes for NCERT solution by CBSE
Description: It is the real and verified notes for NCERT solution by CBSE