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NCERT Solutions for Class 10 Mathematics (10th) Chapter 1: Real Numbers
Concepts covered in Real Numbers are Concept of Irrational Numbers, Euclid’s Division Lemma, Fundamental Theorem of Arithmetic, Fundamental Theorem of Arithmetic Motivating Through Examples, Introduction of Real Numbers, Proofs of Irrationality, Rational Numbers and Their Decimal Expansions, Real Numbers Examples and Solutions.
Exercise 1.1 : Solutions of Questions
Q1 :
Use Euclid's division
algorithm to find the HCF of:
(i)
135 and 225 (ii) 196 and 38220 (iii) 867 and 225
Answer :
(i) 135 and
225
Since 225 > 135, we apply the division
lemma to 225 and
135 to obtain 225 = 135 x 1
+ 90
Since remainder 90 ≠ 0, we apply the division
lemma to 135 and 90 to obtain
135 = 90 x 1 + 45
We consider the new
divisor 90 and new remainder 45, and apply the
division lemma to obtain 90 = 2
x 45 + 0
Since the remainder is zero, the process stops.
Since the divisor
at this stage is 45,
Therefore, the HCF of 135
and 225 is 45.
(ii)196 and 38220
Since 38220 > 196,
we apply the division lemma
to 38220 and 196 to obtain 38220 =
196 x 195 + 0
Since the remainder is zero, the process stops.
Since the divisor
at this stage is 196,
Therefore, HCF of 196
and 38220 is 196.
(iii)867
and 255
Since 867 > 255, we apply the division
lemma to 867 and
255 to obtain 867 = 255 x 3
+ 102
Since remainder 102 ≠ 0, we apply
the division lemma
to 255 and 102 to obtain 255 = 102 x 2 + 51
We consider the new divisor
102 and new remainder
51, and apply the division lemma to obtain
102 = 51 x 2 + 0
Since the remainder is zero, the process stops.
Q2 :
Show that any positive
odd integer is of the form 6q+1, Or
6q+3, Or 6q+5, where q is some integer.
Answer :
Let a be any positive integer
and b
= 6. Then, by
Euclid's algorithm,
a =
6q + r for some integer q ≥ 0, and r = 0,
1, 2, 3, 4, 5 because 0
≤ r < 6. Therefore, a = 6q or
6q + 1 or 6q + 2
or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1
= 2
x 3q + 1 = 2k1 + 1, where
k1 is a positive integer
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.
Therefore, 6q
+ 1,
6q + 3, 6q
+ 5 are not exactly
divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the
form 6q + 1, or 6q
+ 3,
or 6q + 5
Q3 :
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer :
HCF (616, 32) will give the maximum
number of columns
in which they can march. We can use Euclid's algorithm to find the HCF.
616 = 32 x 19
+ 8
32 = 8 x 4
+ 0
The HCF (616, 32) is
8.
Therefore, they can
march in 8 columns each.
Q4 :
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.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or
3m +
1.]
Answer :
Let a be any positive
integer and b = 3. 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,
a2=(3q)2 or (3q+1)2 or
(3q+2)2
a2 = (9q2) or 9q2+6q+1 or 9q2+12q +4
=3x(3q2) or 3(3q2+2q)+1
or 3(3q2+4q+1)+1
=3k1, or 3k2, +1 or 3k3, +1
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.
Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m +8.
Answer :
Let a be any positive integer
and b = 3
a = 3q + r, where q ≥ 0 and 0
≤ r <
3
..a=3q
or 3q+1 or 3q+2
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 =
9(3q3)=9m,
Where m is an integer
such that m
= 3q3
Case 2: When a = 3q + 1,
a3 =
(3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m
+ 1
Where
m is an integer such that
m = (3q3 +
3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8 a3 = 9(3q3 + 6q2 + 4q)
+ 8 a3 = 9m + 8
Where m is an integer
such that m
= (3q3 +
6q2 + 4q)
Therefore, the cube of any positive integer
is of the form 9m, 9m
+ 1, or 9m + 8.
Exercise 1.2 : Solutions of Questions
Q1 :
Express each number as product of its prime factors:
(i)
140 (ii) 156
(iii) 3825 (iv) 5005 (v) 7429
Answer:
(i) 140=2×2×5x7=22x5x7
(ii) 156=2×2×3×13=22×3×13
(iii)3825=3x3x5x5x17=32x52x17
(iv) 5005=5x7x11x13
(v) 7429=17×19×23
Q2 :
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i)
26 and 91 (ii) 510 and 92 (iii) 336 and 54
Answer :
(i)26 and
91
26=2×13
91=7×13
HCF = 13
LCM=2x7x13=182
Product
of the two numbers = 26×91=2366
Hence, product of two numbers = HCF × LCM
(ii) 510
and 92
510=2×3×5×17
92=2×2×23
HCF = 2
LCM=2×2×3×5x17x23=23460
Product of the two numbers =
510×92=46920
HCFX
LCM=2×23460
= 46920
Hence, product of two numbers = HCF × LCM
(iii) 336
and 54
336=2×2×2×2×3×7
336=24×3×7
54=2×3×3×3
54= 2×33
HCF=2x3=6
LCM=24x33x7=3024
Product of the two numbers
= 336x54=18144
HCFX LCM
= 6x3024=18144
Hence, product of two numbers = HCF × LCM
Q3 :
Find the LCM and HCF of the following
integers by applying
the prime factorisation method.
(i)
12, 15 and 21 (ii) 17, 23 and 29 (iii)
8, 9 and 25
Answer :
(i) 12,15
and 21
12=22×3
15=3×5
21=3×7
HCF = 3
LCM=22×3×5x7=420
(i)
17,23
and 29
17=1x17
23=1x23
29=1x29
HCF = 1
LCM=17x23x29=11339
(iii) 8,9 and 25
8=2x2x2
9=3×3
25=5×5
HCF = 1
LCM=2×2×2×3×3×5x5=1800
Q4 :
Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer :
HCF (306, 657)= 9
We know that, LCM× HCF = Product of two numbers
.. LCM>HCF = 306×657
LCM= 306×657 / HCF = 306×657 / 9
LCM = 22338
Q5 :
Check whether 6n can end with the digit 0 for any natural number
n.
Answer :
If any number
ends with the digit 0,
it should be divisible by 10 or in other
words, it will also be divisible by 2 and 5 as 10
= 2 x 5
Prime factorisation of 6n = (2 x 3)n
It can be observed that 5 is not in the prime
factorisation of 6n. Hence, for any value of
n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural
number n.
Q6 :
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Answer :
Numbers are of two types - prime and composite. Prime
numbers can be divided by 1 and only itself,
whereas composite numbers have factors
other than 1 and
itself.
It can be observed that
7 x 11 x 13
+ 13
= 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number. 7 x 6 x 5 x 4
x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 +
1)
= 5
x (1008 + 1)
= 5 x 1009
1009 cannot be factorised further.
Therefore, the given expression has 5 and 1009 as its factors.
Hence, it is a composite number.
Q7 :
There is a circular
path around a sports field. Sonia takes 18 minutes to drive one round of the
field, while Ravi takes 12 minutes
for the same. Suppose they both start at the same point
and at the same time,
and go in the same direction. 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. As they are going in the same direction, they will
meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total
time taken for completing this 1 round
of circular path will be the
18 =
2 x 3 x 3
And,
12 = 2 x 2 x 3
LCM of
12 and 18 = 2 x 2 x 3 x 3 = 36
Therefore, Ravi and Sonia will
meet together at the starting
pointafter 36 minutes.
Exercise 1.3 : Solutions of Questions
Q1 :
Prove that √5 is
irrational. Answer :
Let √5 is a rational number.
Therefore, we can find two
integers a, b (b ≠ 0) such that √5 = a / b
Let a and
b have a common
factor other than 1. Then we can divide them by the common factor, and assume that a and bare co-prime.
a= √5b
a2= 5b2
Therefore, a2 is divisible by 5 and it can
be said that a is divisible by 5. Let
a = 5k, where k is an integer
(5k)2 = 5b2
b2=5k2
This means that b2 is divisible by 5 and hence,
b is divisible by 5. This implies
that a and
b have 5 as a common
factor.
And this is a
contradiction to the fact that a and b are co-prime. Hence,
√5 cannot be expressed as P / q
or
it can be said that √5 is irrational.
Q2 :
Prove that 3+2√5
is
irrational. Answer :
3+2√5
= a/b
2√5
= a/b-3
√5 =1/2 {a/b-3}
Since a and b are integers, 1/2 {a/b-3}
will also
be rational and therefore, √5 is rational.
This contradicts the fact that √5 is irrational. Hence,
our assumption that 3+2√5
is rational
is false. Therefore, 3+2√5 is irrational.
Q3 :
Prove that the following are irrationals:
(i) 1/√2 (ii)
7√5 (iii)
6+√2
Answer :
(i)
1/√2
Let 1/√2 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
1/√2 = a / b
√2 = b / a
b / a is rational
as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and 1/√2 is irrational.
(ii) 7√5
Let 7√5 is rational.
Therefore,
we can find two integers a, b (b ≠ 0) such that
7√5
= a / b
√5 = a / 7b for some integers a and b
a / 7b is rational as a
and b are integers. Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.
(iii)
6+7√2
Let 6+7√2 be rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
6+√2
= a / b
√2 = a /
b – 6
Since a and b are integers, a / b – 6 is also rational and hence, √2 should be rational.
This contradicts the fact that √2 is irrational. Therefore, our
assumption is false and hence,
Exercise 1.4 : Solutions of Questions
Q1 :
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
Answer :
(i) 13 / 3125
The denominator is of the form 5m.
3125 =
55
Hence, the decimal expansion
of 13 / 3125 is terminating.
(i)
17 / 8
8=23
The denominator is of the form 2m.
Hence, the decimal
expansion of 17 / 8 is terminating.
(iii) 64 / 455
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.
(iv) 15 / 1600
1600 = 26×52
The denominator is of the form 2m ×5n.
Hence, the decimal expansion of 15 / 1600 is terminating. (v) 29 / 343
343 = 73
Since the
denominator is not in the form 2m × 5n, and it has 7
as its factor, the decimal
expansion of 29 / 343 is
non- terminating repeating.
(vi) 23 / 23
x 52
The denominator is of the form 2m × 5n.
Hence, the decimal
expansion of is 23 / 23 x 52 terminating.
(vii) 129 / 22 x 57 x 75
Since
the denominator is not of the form 2m × 5n, and it also has
7 as its factor, the decimal expansion of 129
/ 22 x 57 x 75 is non-terminating repeating.
(viii)
6/15 = 2x3 / 3x5 = 2 / 5
The denominator
is of the form 5n.
Hence, the decimal expansion of 6 / 15 is terminating.
(ix) 35 / 50 =
7x5 / 10x5 = 7 / 10
10 = 2x5
The denominator is of the form 2m × 5n.
Q2 :
Write down the decimal
expansions of those rational numbers
in Question 1 above which
have terminating decimal expansions.
Answer :
Q3 :
The following real numbers have decimal expansions as given below.
In each case, decide whether
they are rational or not. If they are rational, and of the form p / q , what can you say about the prime factor of q?
Answer :
(i) 43.123456789
Since this number has a terminating decimal expansion, it is a rational number of the form p / q and q is of the form 2mx5n
i.e., the prime factors of q will be either 2 or 5 or both. (ii) 0.120120012000120000 …
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
(iii)
Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form p / q
and q is not of the form 2mx5n i.e., the prime factors
of q will also have a factor
other than 2 or 5.
NCERT Solutions Class 10 Maths Chapter 1 Real Numbers
NCERT solutions for class 10 maths Chapter 1 Real Numbers encompasses essential concepts around real numbers such as Euclid’s division lemma, Prime Numbers, Composite Numbers, Fundamental Theorem of Arithmetic, HCF, and LCM by Prime Factorization Method, and Irrational Numbers. Real numbers are the combination of rational and irrational numbers. They are denoted by the symbol "R" and can be both positive and negative. NCERT solutions class 10 maths Chapter 1 will help cover all the relevant concepts of real numbers as it has significant applications in real life. Kids also learn about important facts such as the use of the Fundamental Theorem of Arithmetic in proving the irrationality of many of the numbers and exploring when exactly the decimal expansion of a rational number is terminating and when it is non terminating. As numbers form the foundation of Mathematics, understanding the basics is a must.
In this chapter, students will also learn how to represent the real numbers on a number line and the process of successive magnification, which is used to represent a decimal expansion on a number line. The lesson combines certain concepts that kids have come across in previous classes with newer complicated topics to give them a holistic understanding of real numbers and their application. The Class 10 Maths NCERT Solutions Chapter 1 can be found in the pdf format below and also you can find some of these in the exercises given below.
NCERT Solutions for Class 10 Maths Chapter 1 PDF
Mathematics presents us with a variety of numbers and real numbers are one of them. In order to understand higher mathematics, it makes sense to first take a look at the number system in detail which has been achieved in this book. An interesting aspect of the NCERT book is that it details some historical notes, cutting the monotony of studies. Further, the solutions have certain tips and tricks to help a child streamline his process of learning. The links for NCERT Solutions Class 10 Maths available for free PDF download are given below:
Download Class 10 Maths NCERT Solutions Chapter 1 Real Numbers with PDF and Video Class
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