NCERT Solutions Class 10 Maths Chapter 1

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Access NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

Chapter 1 – Real Numbers

Mathematicians rely heavily on numbers. They are the fundamental building elements of mathematics. Real numbers are a combination of rational and irrational numbers in the number system. 

NCERT Solutions for Class 10 Maths

1.1 Introduction

This section includes what students will learn about positive numbers in Chapter 1 Mathematics Class 10, including Euclid’s division algorithm and the fundamental theorem of arithmetic. The Fundamental Theorem of Arithmetic is based on the notion that a composite number can be stated in various ways as a product of prime integers. In Mathematics, this theorem has a wide range of applications. 

1.2 Euclid’s Division Lemma

A lemma is a confirmed assertion that serves as a stepping stone to the proof of additional statements. The standard division system in Mathematics is stated formally by Euclid’s Division Lemma, which asserts that for every pair of positive numbers x and y, there are two unique whole numbers, a and b, that satisfy the equation. In this portion, you will learn about Euclid’s division lemma and algorithm.

X = ya + b where 0 <= b <= y and x is a dividend, y is a divisor.

A is the quotient.

B is the remainder.

In other words: dividend = (quotient * divisor) + remainder.

An algorithm is a series of well-defined steps that can be used to solve a problem sequentially. 

1.3 The Fundamental Theorem of Arithmetic

Since the order of the factors doesn’t matter, it’s a one-of-a-kind prime factorisation of natural numbers. According to this theorem, every composite number can be factorised as a product of specific prime numbers. We’ll see how this works with an example based on the following fundamentals:

HCF – The greatest integer that can exactly divide all the specified integers is the highest common factor of two or more integers. 

LCM- The smallest number divisible by all the specified integers is called the least common multiple of two or more integers. 

HCF(a,b) * LCM (a, b) = a * b for two positive integers a and b

As a result of this theorem, any natural number may be expressed as a multiplication of prime numbers.

1.4 Revisiting Irrational Numbers

Students will recall the definition of irrational numbers studied in previous classes in this portion of NCERT Solutions for Class 10 Mathematics Chapter 1 and establish that p is an irrational number, where p is a prime integer.

A number “n” is said to be irrational if it cannot be stated in the form x/y. Here, x and y are integers, and n is greater than zero.

Key Features of NCERT Solutions for Class 10 Maths Chapter 1

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  • NCERT Solutions for Class 10 Mathematics Chapter 1 include answers to all the questions given in the NCERT Mathematics textbook.
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Q.1 Show that the square of any positive integer is of the form 3m or 3m+1 for some integer m.

Ans- Let a be any positive integer. Then it is of the form 3q or 3q+1 or 3q+2.

Case I – When a = 3q, we have
a2 = (3q)2 = 9q2 = 3(3q2) = 3m ; where m = 3q2

Case II – When a = 3q+1, we have
a2 = (3q+1)2 = 9q2 +6q+1
or, a2 = 3q(3q+2)+1 = 3m+1; where m = 3q(3q+2)

Case III – When a = 3q+2, we have
a2 = (3q+2)2 = 9q2+12q+4=9q2+12q+3+1
or, a2 = 3(3q2+4q+1)+1 = 3m+1 ; where m = (3q2+4q+1)

Hence, a is either of the form 3m or 3m+1 for some integer m.

Q.2 Prove that n2-n is divisible by 2 for every positive integer n.

Ans- We know that any positive integer is of the form 2q
or 2q+1 for some integer q.

Case I – When n=2q then we have
n2-n =(2q)2 – 2q
or, n2-n =4q2-2q
or, n2-n = 2q(2q-1)
or, n2-n = 2r; where r= q(2q-1)
Thus, n2-n is divisible by 2.

Case II – When n=2q+1, we have
n2-n =(2q+1)2 – (2q+1)
or, n2-n =(2q+1)(2q+1 –1)
or, n2-n = 2q(2q+1)= 2k ; where k=q(2q+1)
Thus n2-n is divisible by 2.

Hence, n2-n is divisible by 2 for every positive integer n.

Q.3 Show that any positive integer is of the form 3q or 3q+1 or 3q+2 for some integer q.

Ans- Let a be any positive integer and b=3, then there exist integer q and r such that a=3q+r, where 0≤r < 3…1     by Euclid’s Division Lemma now 0≤r < 3⇒0≤r < 2 r=0,1,2                ∵r is integer put r=0 in 1 we get a=3q put r=1 in 1 we get a=3q+1 put r=2 in 1 we get a=3q+2 Hence any positive integer is of the form 3qor,  3q+1, or 3q+2 for some integer q.

Q.4 Use Euclid’s division algorithm to find the HCF of 210 and 55.

Ans- Clearly, 210 > 55.
On applying Euclid’s Division Lemma to 210 and 55
we get, 210 = 55×3 + 45
Since remainder 45 ≠ 0, therefore on again applying Euclid’s Division Lemma we get,
55 = 45×1 + 10
Since remainder 10 ≠ 0 therefore, on again applying Euclid’s Division Lemma we get,
45 = 10×4 + 5
Since remainder 5 ≠ 0 therefore, on again applying Euclid’s Division Lemma we get,
10 = 5×2 + 0
Since remainder is 0 therefore, HCF = 5

Q.5 Prove that if x and y are odd positive integers then x2+y2 is even but not divisible by 4.

Ans- We know that any odd integer is of the form2q+1 for some integer q.so, let x=2m+1 and y=2n+1 for some integers m&n.    ∴x2+y2=2m+12+2n+12⇒x2+y2=4m2+4m+1+4n2+4n+1⇒x2+y2=4m2+4n2+4m+4n+2…1                =22m2+2n2+2m+2n+1                =2keven integer where k=2m2+2n2+2m+2n+1Hence x2+y2 is an even integer.Now we will prove x2+y2 is an even integer.From equation 1 we get,    x2+y2=4m2+n2+m+n+2⇒x2+y2=4q+2 where q=m2+n2+m+nWhich leaves remainder 2 when divided by 4⇒x2+y2 is not divisible by 4.

Q.6 Prove that one of every three consecutive positive integers is divisible by 3.

Let n, n+1, n+2 be three consecutive positive integers where n can take the form 3q, 3q+1 or 3q+2.

Case I
when n = 3q
Then n is divisible by 3
but neither n+1 nor n+2 is divisible by 3.

Case II
when n = 3q+1
Then n is not divisible by 3.
n+1 = 3q+1+1 = 3q+2,
which is not divisible by 3.
n+2 = 3q+1+2 = 3q+3 = 3(q+1),
which is divisible by3.

Case III
when n = 3q+2
Then n is not divisible by 3.
n+1 = 3q+2+1 =3q+3 = 3(q+1),
which is divisible by 3.
n+2 = 3q+2+2 = 3q+4,
which is not divisible by 3.

Hence, one of n, n+1 and n+2 is divisible by 3.

Q.7 Prove that if a and b are positive integers such that a = bq + r, then every common divisor of a and b is a common divisor of b and r.

Proof: Let c be a common divisor of a and b.
Therefore,
c divides a ⇒ a = cm …(1) for some integer m
& c divides b ⇒ b = cn …(2) for some integer n
Now, a = bq + r given
r = a – bq
r = cm – cnq [using (1) & (2)]
r = c(m – nq)
⇒ c divides r …(3)
from (2) & (3) c divides b and r
⇒ c is common divisor of b and r
Hence, every common divisor of a
and b is a common divisor of b and r.

Q.8 Prove that √5 is irrational number.

Let √5 be a rational number.

Then √5=p/q (where p and q are co-prime numbers and q 0)
Squaring both the sides we get
5=p2/q2
⇒ p2=5q2 …(1)
⇒ 5 divides p2
⇒ 5 divides p….(2)
Let p=5r (where r is an integer)
Squaring both the sides, we get
p2=25r2 …(3)
From equation (1) & (3), we get
5q2=25r2
⇒ q2=5r2
⇒ 5 divides q2
⇒ 5 divides q…(4)
From equation (2) & (4) we get
HCF of p and q is 5, but co-prime has only 1 as HCF.
This is a contradiction.
Thus, our assumption was wrong.
Therefore, √5 is irrational number.

Q.9 Prove that √3 is irrational number.

Let √3 be a rational number.
Then √3=p/q (where p and q are co-prime numbers and q ≠ 0)
By squaring both the sides, we get
3=p2/q2
⇒ p2=3q2 …(1)
⇒ 3 divides p2
⇒ 3 divides p….(2)
Let p=3r (where r is an integer)
Squaring both the sides, we get
p2=9r2 …(3)
From equation (1) &(3) , we get
3q2=9r2
⇒ q2=3r2
⇒ 3 divides q2
3 divides q….(4)

From equation (2) & (4) we get
The HCF of p and q is 3, but co-prime has only 1 as HCF.
Hence, this is a contradiction.
Thus, our assumption was wrong.
Therefore, √3 is irrational number.

Q.10 In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room same numbers of participants are to be seated and all of them being in the same subject.

The number of rooms will be minimum if each room accommodates maximum number of participants. Therefore, the number of participants in each room must be the HCF of 60, 84 and 108.

60 = 2x2x3x5 = 22x31x5

84 = 2x2x3x7 = 22x3x7

108 = 2x2x3x3x3 = 22x33

HCF of 60, 84 and 108 = 22X3 = 4X3 = 12

Therefore, in each room 12 participants can be seated.

Number of rooms required=Total no. of participants/12

= (60+84+108)/12

=252/12

=21

Express the positive integers 144 & 180 as the product of its prime factors.

2144272236218293144=2×2×2×2×3×3         =24×322180290245215    5180=2×2×3×3×5        =22×32×5

Q.12 Two tankers contain 2340 litres and 3825 litres of petrol respectively. Find the maximum capacity of the container that can measure the petrol of either tanker in exact number of times.

Clearly, the maximum capacity of the container that can measure the capacity of either tanker is the

HCF of 2340 and 3825.

HCF is common prime factors with smallest exponents

2340 = 2×2×3×3×5×13 = 22×32×5×13

3825 = 3×3×5×5×17 = 32×52×17

HCF = 32×5
= 9×5
= 45
Hence, capacity of the container must be 45 litres.

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NCERT Solutions for Class 10 Maths Related Chapters

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

Arithmetic Progressions

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Surface Areas and Volumes

Statistics

Probability

FAQs (Frequently Asked Questions)

1. How do you compute the HCF of two positive integers?

We can use the procedures below to determine the HCF (Highest Common Factor) of two positive integers (x is greater than y).

  • Calculate the quotient q and rest r that fulfil Euclid’s Division. x = (y *q) + r lemma
  • The HCF of the two numbers is n if r or the remainder is 0.
  • If r is greater than zero, we must use Euclid’s division Lemma to divide y and r.
  • The preceding procedure must be repeated until r=0 is obtained.
  • The HCF of x and y will be the divisor when we get to this point.

2. How can you tell the difference between whole numbers, integers, and natural numbers?

Natural numbers are all non-negative counting numbers that do not include zero, such as 5, 6, 7, 8, and so on, whereas whole numbers are similar to natural numbers, the only difference is that they include a zero. Therefore they are 0, 1, 2, 3, and so on. All negative and positive numbers are represented as integers (including 0). Integers are -2,-1,0, 1, 2, and so on.

3. How many exercises are there in NCERT Class 10 Mathematics Chapter 1?

Real numbers is the first chapter in the NCERT Mathematics textbook for Class 10. This chapter has a total of four exercises.

4. Define real numbers.

Real numbers are a blend of rational and irrational numbers in the number system, to put it simply. These numbers can be used for all mathematical calculations and can also be represented on the number line, according to the general rule. 

5. What are the main points addressed in Chapter 1 of the NCERT Class 10 Mathematics textbook?

The following concepts are covered in this chapter:

  • The Division Algorithm of Euclid
  • The Fundamental Arithmetic Theorem
  • Taking a Second Look at Rational and Irrational Numbers
  • Expansions of Decimal Numbers