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Real Numbers Class 10 Notes: Chapter 1

Cbse real numbers class 10 notes:- download pdf here, class 10 maths chapter 1 real number notes.

CBSE Class 10 Maths Chapter 1 Real Numbers Notes are provided here in detail. As we all know, any number, excluding complex numbers, is a real number. Positive and negative integers, irrational numbers, and fractions are all examples of real numbers. To put it another way, any number found in the real world is a real number. Numbers can be found all around us. Natural numbers are being used to count objects, integers are used to measure temperature, rational numbers are used to represent fractions, irrational numbers are used to calculate the square root of a number, etc. These various types of numbers form a collection of real numbers. Here, we are going to learn what a real number is, Euclid’s division algorithm, the fundamental theorem of arithmetic, methods of finding LCM and HCF and the complete explanation of rational and irrational numbers with examples.

Real Numbers

Positive integers, negative integers, irrational numbers, and fractions are all examples of real numbers. In other words, we can say that any number is a real number, except for complex numbers. Examples of real numbers include -1, ½, 1.75, √2, and so on. In general,

• Real numbers constitute the union of all rational and irrational numbers.
• Any real number can be plotted on the number line.

To know more about real numbers, visit here .

Students can refer to the short notes and MCQ questions along with a separate solution pdf of this chapter for quick revision from the links below.

• Real Numbers Short Notes
• Real Numbers MCQ Practice Questions
• Real Numbers MCQ Practice Solutions

Euclid’s Division Lemma

• Euclid’s Division Lemma states that given two integers a and b , there exists a unique pair of integers q and r such that a = b × q + r   a n d   0 ≤ r < b .
• This lemma is essentially equivalent to :  dividend = divisor × quotient + remainder
• In other words, for a given pair of dividend and divisor, the quotient and remainder obtained are going to be unique.

For more information on Euclid’s Division Lemma, watch the below video

To know more about Euclid’s Division Lemma, visit here .

Euclid’s Division Algorithm

• Euclid’s Division Algorithm is a method used to find the H.C.F of two numbers, say  a and b where a> b.
• We apply Euclid’s Division Lemma to find two integers q and r such that a = b × q + r   a n d   0 ≤ r < b .
• If r = 0, the H.C.F is b; else, we apply Euclid’s division Lemma to b (the divisor) and r (the remainder) to get another pair of quotient and remainder.
• The above method is repeated until a remainder of zero is obtained. The divisor in that step is the H.C.F. of the given set of numbers.

For more information on Euclid’s Division Algorithm, watch the below video

The Fundamental Theorem of Arithmetic

Prime Factorisation

• Prime Factorisation is the method of expressing a natural number as a product of prime numbers.
• Example: 36 = 2 × 2 × 3 × 3   is the prime factorisation of 36.

Fundamental Theorem of Arithmetic

• The Fundamental Theorem of Arithmetic states that the prime factorisation for a given number is unique if the arrangement of the prime factors is ignored.
• Example: 36 = 2 × 2 × 3 × 3 OR,  36 = 2 × 3 × 2 × 3
• Therefore, 36 is represented as a product of prime factors (Two 2s and two 3s) ignoring the arrangement of the factors.

To know more about the Fundamental Theorem of Arithmetic, visit here .

Method of Finding LCM

As we know, the smallest of the common multiples of two or more numbers is called the lowest common multiple (LCM). Example: To find the Least Common Multiple  ( L.C.M ) of 36 and 56,

•   36 = 2 × 2 × 3 × 3 56 = 2 × 2 × 2 × 7
•  The common prime factors are 2 × 2
•  The uncommon prime factors are  3 × 3 for 36 and 2 × 7 for 56.
• LCM of 36 and 56 = 2 × 2 × 3 × 3 × 2 × 7   which is 504

To know more about LCM, visit here .

Method of Finding HCF

We know that the greatest number that divides each of the given numbers without leaving any remainder is the highest common factor (HCF) of two or more given numbers. H.C.F can be found using two methods – Prime factorisation and Euclid’s division algorithm.

• Given two numbers, we express both of them as products of their respective prime factors. Then, we select the prime factors that are common to both the numbers
• Example – To find the H.C.F of 20 and 24 20 = 2 × 2 × 5   and    24 = 2 × 2 × 2 × 3
• The factor common to 20 and 24 is 2 × 2 , which is 4, which in turn is the H.C.F of 20 and 24.
• It is the repeated use of Euclid’s division lemma to find the H.C.F of two numbers.

• The required HCF is 6 .

To know more about HCF, visit here .

For more information on HCF and LCM, watch the below video

To know more about the Properties of HCF and LCM, visit here .

Product of Two Numbers = HCF X LCM of the Two Numbers

• For any two positive integers a and b, a × b = H . C . F × L . C . M .
• Example – For 36 and 56, the H.C.F is 4 and the L.C.M is 504 36 × 56 = 2016 4 × 504 = 2016 Thus, 36 × 56 = 4 × 504
• Let us consider another example: For 5 and 6, the H.C.F is 1 and the L.C.M is 30 5 × 6 = 30 1 × 30 =30 Thus, 5 × 6 = 1 × 30
• The above relationship, however, doesn’t hold true for 3 or more numbers

To know the Relationship between LCM and HCF, visit here .

Applications of HCF & LCM in Real-World Problems

L.C.M can be used to find the points of common occurrence. For example,ringing of bells that ring with different frequencies, the time at which two persons running at different speeds meet, and so on.

For more information on Applications Of LCM, watch the below video

Revisiting Irrational Numbers

Irrational numbers.

Any number that cannot be expressed in the form of p/q (where p and q are integers and q ≠ 0 .) is an irrational number. Examples √2 , π , e and so on.

To know more about Irrational Numbers, visit here .

Number theory: Interesting results

• If a number p (a prime number) divides a 2 , then p divides a. Example: 3 divides 6 2 i.e 36, which implies that 3 divides 6.
• The sum or difference of a rational and an irrational number is irrational
• The product and quotient of a non-zero rational and irrational number are irrational.
• √p is irrational when ‘p’ is a prime. For example, 7 is a prime number, and √7 is irrational. The above statement can be proved by the method of “Proof by contradiction”.

To know more about Number theory, visit here .

Proof by Contradiction

In the method of contradiction, to check whether a statement is TRUE (i)  We assume that the given statement is TRUE. (ii) We arrive at some result which contradicts our assumption, thereby proving the contrary. Eg: Prove that √7 is irrational. Assumption: √7  is rational. Since it is rational √7 can be expressed as √7 = a/b , where a and b are co-prime Integers, b ≠ 0. On squaring, a 2 /b 2 = 7 ⇒ a 2 = 7 b 2 . Hence, 7 divides a. Then, there exists a number c such that a=7c. Then, a 2 = 49 c 2 . Hence, 7 b 2 = 49 c 2 or b 2 = 7 c 2 . Hence 7 divides b. Since 7 is a common factor for both a and b, it contradicts our assumption that a and b are coprime integers. Hence, our initial assumption that √7  is rational is wrong. Therefore, √7 is irrational.

Revisiting Rational Numbers and Their Decimal Expansions

Rational numbers.

Rational numbers are numbers that can be written in the form p/q, where p and q are integers and q ≠ 0 . Examples -1/2 , 4/5,  1 , 0 , − 3   and so on.

To know more about Rational Numbers, visit here .

Terminating and Non-Terminating Decimals

Terminating decimals are decimals that end at a certain point. Example: 0.2, 2.56 and so on. Non-terminating decimals are decimals where the digits after the decimal point don’t terminate. Example: 0.333333….., 0.13135235343… Non-terminating decimals can be : a) Recurring – a part of the decimal repeats indefinitely (0 . 142857 142857 …. ) b) Non-recurring – no part of the decimal repeats indefinitely. Example: π = 3.1415926535…

To know more about terminating and non-terminating decimals, click here .

Check if a given rational number is terminating or not

If a/b is a rational number, then its decimal expansion would terminate if both of the following conditions are satisfied : a) The H.C.F of a and b is 1. b) b can be expressed as a prime factorisation of 2 and 5 i.e b = 2 m × 5 n where either m or n, or both can = 0. If the prime factorisation of b contains any number other  than 2 or 5, then the decimal expansion of that number will be recurring

1/40 = 0.025 is a terminating decimal,  as the H.C.F of 1 and 40 is 1, and the denominator (40) can be expressed as 2 3 × 5 1 .

3/7 = 0.428571 is a recurring decimal as the H.C.F of 3 and 7 is 1 and the denominator (7) is equal to 7 1

Real Numbers for Class 10 Solved Examples

Find the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively.

First, subtract the remainder from the number.

(i.e) 70-5 = 65

125-8 = 117.

Thus, we need to find the largest number that divides 65 and 117 and leaves the remainder 0.

To find the largest number, take the HCF of 65 and 117.

Finding HCF of 65 and 117.

65 = 5×13

117 = 3×3×13.

Hence, HCF (65, 117) = 13.

Therefore, the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively is 13.

Find the LCM of 306 and 657, given that HCF (306, 657) = 9.

Given that, HCF (306, 657) = 9.

We know that HCF × LCM = Product of Numbers

Hence, 9×LCM = 306×657

9×LCM = 201042

LCM = 201042/9

LCM = 22338.

Therefore, LCM of 306 and 657 is 22338.

Prove that 1/√2 is an irrational number.

To prove 1/√2 is an irrational number.

Now, let us take the opposite assumption.

(i.e) Take 1/√2 is a rational number.

We know that rational numbers are the numbers that can be written in the form of p/q, where q is not equal to 0. (p and q are two co-prime numbers)

Hence, 1/√2 = p/q.

Now, simplify the above equation by multiplying √2 on both sides.

1 = (p√2)/q

q = p√2

Hence, we get q/p = √2.

Here, p and q are integers, and hence q/p is a rational number.

But, √2 is an irrational number.

Hence, our assumption is wrong.

Therefore, 1/√2 is an irrational number.

Hence, proved.

Frequently Asked Questions on Class 10 Real Numbers

What is euclid’s division algorithm, what does real numbers for class 10 explains, is 10i a real number, what does the fundamental theory of arithmetic explain.

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NCERT Solutions for Class 10 Maths Chapter 1 Free PDF Download

Ncert solutions for class 10 maths chapter 1 real numbers.

NCERT Solutions for class 10 maths chapter 1 –  real numbers lets the students solve and revise the whole syllabus very effectively. After going through the stepwise solutions given by our subject expert teachers, the student will be able to score better marks.

Class 10 is the board exam so preparation is very important for marks as well as for a career. Our NCERT for Class 10 Maths Chapter 1 is very much complete in covering all topics, contents, problems and self-explanatory solutions. NCERT solutions for class 10 Maths Chapter 1 will help you to solve all types of problems of class 10 Maths. The Toppr app provides the solution to your problems of any kind of real numbers class 10 Maths.

Our team of expert teachers has created these NCERT Solutions according to the curriculum and pattern of board exams. Our app will help you to get complete NCERT Solutions for Class 10 Maths and other subjects. We are providing you the free pdf download links of the class 10 maths chapter 1.

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Download NCERT Solutions for Class 10 here.

CBSE Class 10 Maths Chapter 1 NCERT Solutions

NCERT solutions for class 10 Maths Chapter 1 introduces the very fundamental but essential topic of mathematics. In the beginning, the chapter real numbers class 10 gives the introduction of real numbers and then two very important topics Euclid’s Division Algorithm and The Fundamental Theorem of Arithmetic. This fundamental theorem of arithmetic has many real-life and scientific applications. Other related fields also have use of these. So for a strong base in Maths which will support further higher education also, our NCERT solutions for real numbers class 10 Maths will definitely help.

Topics Covered under CBSE Chapter 10 Maths Chapter 1 – Real Numbers

• Ex. 1.1: Introduction to Real Numbers
• Ex. 1.2: Euclid’s Division Lemma
• Ex. 1.3: The Fundamental Theorem of Arithmetic
• Ex. 1.4: Revisiting Irrational Numbers
• Ex. 1.5: Revisiting Rational Numbers and Their Decimal Expansions.

This chapter helps the students to understand the fundamental theorem of arithmetic that has many real-life and scientific applications. Other related fields also have use of the Real numbers which are explained in this chapter. So for a strong base in Maths which will support the further higher education also, our NCERT solutions for class 10 Maths will definitely help. Rational and irrational numbers are major types of real numbers. Theorems will explain these with proper examples and applications.

Let us discuss the sub-topics in detail –

1.1: Introduction :

In this chapter, the student will explore the world of real numbers and their related applications. This chapter contains some very important properties of positive numbers.

1.2: Euclid’s Division Lemma:

In this chapter, the student will learn the technique to compute the Highest Common Factor (HCF) of two given positive integers by Euclid’s algorithm.

1.3: The Fundamental Theorem of Arithmetic:

The student will learn that every composite number can be expressed as a product of primes uniquely. This property is a fundamental theorem of arithmetic.

1.4: Revisiting Irrational Numbers:

This chapter redefines the irrational numbers. Some relevant examples will help to understand the concept easily. A method of contradiction will help to prove them.

1.5: Revisiting Rational Numbers and Their Decimal Expansions:

The student will revisit the concept of rational numbers using fraction expression as well as using decimal expansions. This is because that decimal expansion of every rational number is either terminating or repeatedly non-terminating.

Some Questions from NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Q.1 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m+8.

Using Euclid division algorithm, we know that a = bq + r , Osrs b —- (1)

Let a be any positive integer, and b = 3.

Substitute b = 3 in equation (1)

a = 3 q + r where o s r s3, r= 0, 1, 2

If r= 0, a = 3 9 

Cube the value, we get 23 = 2703 2 3 = 9(393), where m=393 —(2)

If r = 1, a = 3q+1 Cube the value, we get 23 = (39+13 a3 = (27 7 3 + 277 2 +9q+1) 23 = 9/393 +382 + 1) +1, where m= 393 +372 + 9 —(3)

If r= 2, a = 39+2

Cube the value, we get an2 = 139+213 43 = (27 q 2 + 5302 + 36q+8) 23 = 9393 +692 + 49) + 8, where m=393 +692 +49—-(4)

From equation 2, 3 and 4,

The cube of any positive integer is of the form 9m, 9m + 1 or 9 m +8.

Q. 2. Show that any positive odd integer is of form 6q+1, or 69 +3, or 6q+5, where q is some integer. 

Using Euclid division algorithm, we know that a = bg +r , Osrs b—-(1) 

Let a be any odd positive integer and b = 6.

Substitute b = 6 in equation (1) 

a = 6 q +r where o s r s 6 r = 0,1,2,3,4,5 

If r = 0, a = 6 9, 6q is divisible by 6 

If r = 1, a = 69+1,6q+1 is not divisible by 2. 

If r = 2, a = 6 q +2, 6 9 +2 is divisible by 2 

If r= 3, a = 6 9 +3,6 q +3 is not divisible by 2. 

If r = 4, a = 6 9 +4, 6 9 +4 is divisible by 2 

If r= 5, a = 6q+5, 69+5 is not divisible by 2. 

So, the nu mbers 6 4 , 6 9 + 2,6 9 + 4 are divisible by 2 and even numbers.

The remaining numbers 6q+1, 69+ 3 and 69+5 are odd . 

Q. 3   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?

HCF (616, 32) is the maximum number of columns in which they can march. 

First, find which integer is larger.

616 > 32

Step 2: Then apply Euclid’s divisio n algorithm to 616 and 32 to obtain

616 = 32 x 19+8

Repeat the above ste p until you will get the remaind er as zero.

Step 3: Now consider the divisor 32 and the remainder 8, and apply the division lemma to get

32 = 8 + 4 +0

Since the remainder is zero, we cannot proceed further. 

Step 4: H ence the divisor at the last process is 8

So, the H.C.F. of 616 and 32 is 8. 

Therefore, 8 is the maximum number of columns in which th ey can march,

Q. 4 Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3 m + 1 for some integer m.

Using Euclid division algorithm, we know that a = bq +r , Osr s …… (1) 

Let a be any positive integer, and b = 3. 

Substitute b = 3 in equation (1) 

a = 3q+r where o s r s 6 r= 0,1,2 If r = 0 , a = 39 On squaring we get, a2 = 3(37), where m = 392—(2) 

If r = 1, a = 3 9 +1 

On squaring we get a2 = 33q2 + 2q +1, where m = 392 +24—(3) If r= 2, a = 39+2

On squaring we get a 2 = 3(3q2 +49 + 1) + 1, where m =372 +4q+1-— (4)

From equation 2,3 and 4,

The square of any positive integer is either of the form 3m or 3m+ 1 for some integer m.

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Solved Questions For You:

Question 1. There is a circular path around a sports field. Sonia takes  1 8  minutes to drive one round of the field, while Ravi takes  1 2  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: Sonia and Ravi drive one round of the field at the same point, same time and same direction.

Question 4. Use Euclid’s division algorithm to find the HCF of:

(i)  1 3 5  and  2 2 5

Step 1:  First find which integer is larger.

2 2 5 = 1 3 5 × 1 + 9 0

1 3 5 = 9 0 × 1 + 4 5

9 0 = 2 × 4 5 + 0

3 8 2 2 0 > 1 9 6

3 8 2 2 0 = 1 9 6 × 1 9 5 + 0

8 6 7 > 2 5 5

8 6 7 = 2 5 5 × 3 + 1 0 2

2 5 5 = 1 0 2 × 2 + 5 1

1 0 2 = 5 1 × 2 = 0

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NCERT Solutions for Class 10

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NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 (Ex. 1.1) Real Numbers in Hindi and English Medium modified and updated for academic year 2024-25. Solutions of 10th mathematics Exercise 1.1 is revised for session 2024-25 based on new NCERT Books.

NCERT Solutions for class 10 Maths Chapter 1 Exercise 1.1

• Class 10 Maths Exercise 1.1 NCERT Solutions
• Class 10 Maths Exercise 1.1 in Hindi Medium
• Class 10 Maths Chapter 1 NCERT Book
• Class 10 Maths Chapter 1 NCERT Solutions
• Class 10 Maths all Chapters Solutions
• Class 10 all Subjects NCERT Solutions

Express each number as product of its prime factors:

Solution: (i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7 Discuss this question in your opinion.

(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13 If you have any other suggestion , please suggest us. (iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17 More options related to above question.

(iv) 5005 = 5 × 7 × 11 × 13 if you have any other idea, please comment here .

(v) 7429 = 17 × 19 × 23 What do you think about 10th Maths Exercise 1.1 Part v solutions?

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

Solution: (i) 26 and 91 26 = 2 × 13 91 = 7 × 13 HCF = 13 LCM = 2 × 7 × 13 = 182 Product of the two numbers = 26×91 = 2366 HCF × LCM = 13 × 182 = 2366 Hence, product of two numbers = HCF × LCM Do you have more ideas about this solutions ?

(ii) 510 and 92 510 = 2 × 3 × 5 × 17 92 = 2 × 2 × 23 HCF = 2 LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460 Product of the two numbers = 510 × 92 = 46920 HCF × LCM = 2 × 23460 = 46920 Hence, product of two numbers = HCF × LCM You can discuss this question with us.

(iii) 336 and 54 336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7 54 = 2 × 3 × 3 × 3 = 2 × 3³ HCF = 2 × 3 = 6 LCM = 2⁴ × 3³ × 7 = 3024 Product of the two numbers = 336 × 54 = 18144 HCF × LCM = 6 × 3024 = 18144 Hence, product of two numbers = HCF × LCM Comment about the solution here .

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

Solution: (i) 12, 15 and 21 12 = 2 × 2 × 3 = 2² × 3 15 = 3 × 5 21 = 3 × 7 HCF = 3 LCM = 2² × 3 × 5 × 7 = 420 For more suggestion, see the solution here .

(ii) 17, 23 and 29 17 = 1 × 17 23 = 1 × 23 29 = 1 × 29 HCF = 1 LCM = 17 × 23 × 29 = 11339 Prove more ways to solve this question .

(iii) 8, 9 and 25 8 = 2 × 2 × 2 = 2³ 9 = 3 × 3 = 3² 25 = 5 × 5 = 5² HCF = 1 LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800 How can we solve this question alternatively?

Given that HCF (306, 657) = 9, find LCM (306, 657). Solution: HCF (306, 657) = 9 We know that, LCM × HCF = Product of two numbers Therefore, LCM= (Product of two numbers)/HCF = (306 × 657)/9 = 22338 Hence, LCM (306, 657) = 22338

The alternative way to solve this question.

Check whether 6ⁿ can end with the digit 0 for any natural number n. Solution: 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 × 5. Prime factorisation of 6ⁿ = (2 × 3)ⁿ According to Fundamental Theorem of Arithmetic , the factorisation is uniqe. 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.

Put your opinion about question 5 of 10th Maths exercise 1.1

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Solution : 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 × 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. Therefore, it is a composite number. Now: 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. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

For any further suggestion, please visit here .

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? Solution : 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 LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes. 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 point after 36 minutes.

What do you think about the solution of question 7 given in Ex. 1.1 of 10th Maths.

Download App for Class 10

NCERT Solutions for class 10 Maths Chapter 1 Exercise 1.1 Real Numbers in English medium as well as Hindi Medium is given below to free download in PDF format for 2024-25. IF YOU WANT TO DOWNLOAD, THE LINK IS ALSO GIVEN or visit NCERT Solutions for Class 10 Maths Chapter 1 to download other exercises.

1. In a seminar, the number of participants in English, Maths and Science are 175, 140 and 105, respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject. [Answer: 12] 2. Find the value of m if HCF of 65 and 117 is expressible in the form 65m – 117. [Answer: 2] 3. What can you say about the product of a non-zero rational and irrational number? [Answer: Irrational] 4. In a school there are two sections of class X – section A and section B. There are 48 students in sections A and 54 students in section B. Determine the least number of books required for the library of the school so that the books can be distributed equally among students of section A or section B? [Answer: 216] 5. After how many places the decimal expansion of 13497/1250 will terminate? [Answer: 4] 6. Find the least number which is divisible by all numbers from 1 to 10 (both inclusive). [Answer: 2520]

Practice Questions on Real Numbers

• A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48 km, 60 km and 72 km a day, round the field. After how many days will they meet again at the starting point? [Answer: 30 days]
• The numbers 525 and 3000 are divisible by 3, 5, 15, 25 and 75 what is the HCF of 525 and 3000? [Answer: 75]
• Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons. [Answer: No]
• If a = 4q + r then what are the condition for a and q? What are the values that r can take? [Answer: a and q are positive integer 0 ≤ r ≤ 4]
• What is the digit at unit’s place of 9^n? [Answer: Even power = 1, odd power = 9]
• If n is an odd integer then show that n² – 1 is divisible by 8.
• Use Euclid’s division algorithm to find the HCF of 16 and 28. [Answer: 4]

Important questions for 10th Maths Exercise 1.1

What is fundamental theorem of arithmetic.

According to fundamental theorem of arithmetic: Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

225 can be expressed as (a) 5 x 3^2 (b) 5^2 x 3 (c) 5^2 x 3^2 (d) 5^3 x 3. [CBSE 2020] [Maths Basic]

225 = 3 x 3 x 5 x 5 = 3^2 x 5^2. Hence, Option (c) is correct.

The total number of factors of a prime number is (a) 1 (b) 0 (c) 2 (d) 3. [CBSE 2020] [Maths Standard]

The total number of factors of a prime number is 2. One is 1 and other is itself number. Hence, Option (c) is correct.

The HCF and the LCM of 12, 21, 15 respectively are (a) 3, 140 (b) 12, 120 ((c) 3, 420 (d) 420, 3 [CBSE 2020] [Maths Standard]

12 = 2 x 2 x 3 21 = 3 x 7 15 = 3 x 5 HCF = 3 LCM = 2 x 2 x 3 x 7 x 5 = 420 Hence, Option (c) is correct.

Ask your doubts related to NIOS or CBSE Board and share your knowledge with your friends and other users through Discussion Forum.

How many questions are there in Class 10 Maths Exercise 1.1?

There are in all 7 questions in exercise 1.1 of class 10th mathematics chapter 1 (Real numbers) and every question is different from each other.

Which example is important for exercise 1.1 of 10th mathematics?

4 examples are based on exercise 1.1 (chapter 1 Real numbers) of class 10th mathematics. Example 5 and question 5 are based on same concept, example 7 and question 4 are based on same formula and examples 6, 8 and question 3 are same. All the examples are important for the exams as well as school tests.

What are the Important and difficult questions from exercise 1.1 in 10th Maths?

Question number 2, 4, 5 and 7 of exercise 1.1 of class 10th mathematics chapter 1 (Real numbers) are important and difficulty level of questions varies from student to student. But questions which most of students find little difficult in this exercise are Q5, 6 and 7.

Which topics students should recall before starting exercise 1.1 of 10th Maths?

Before starting exercise 1.1 (chapter 1 Real numbers) of class 10th mathematics, Students should recall meaning of: Factors. HCF (Highest Common Factor) and LCM (Lowest/Least Common Multiple). Prime numbers, Composite numbers, Natural numbers.

Class 10 Maths Exercise 1.2 »

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• NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers - Exercise 1.1
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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

A free PDF download of NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 (Ex 1.1) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Download Vedantu NCERT Solution to get a better understanding of all the exercises questions.

You can also Download NCERT Solutions for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.

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

Important topics under ncert solutions for class 10 maths chapter 1 real numbers (ex 1.1) exercise 1.1.

The first chapter of the Class 10 Maths syllabus is on real numbers. It is a significant chapter in maths that deals with some of the fundamental topics. This chapter has 6 major parts that need to be covered and understood properly to get a good grasp on real numbers. It is recommended that students go through these individual topics carefully for a precise understanding and for better internalization of the same.

Importance of Real Numbers

Real numbers comprise both rational and irrational numbers. Rational numbers include integers, decimals, and fractions, while irrational numbers are numbers like root overs, pi (22/7), and so on. In short, real numbers are all numbers excluding imaginary numbers. 

The importance of real numbers lies in their use in almost every sphere of mathematics and in real life. We encourage students to gather as much knowledge as they can on real numbers so that they are able to solve all sums that have the application of real numbers.

Exercise 1.1 

1. Use Euclid’s division algorithm to find the HCF of:

(i) $135$ and $225$ 

Ans: We have to find the HCF of $135$ and $225$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a>b$ is found as :

(i) First find the values of $q$ and $r$, where $a=bq+r$, $0\le r<b$.

(ii) If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

(iii) Continue steps till the remainder is zero. When we get the remainder zero, the divisor will be the HCF.

Let $a=225$ and $b=135$.

Since, $a>b$ 

Using division algorithm, we get

$\Rightarrow 225=135\times 1+90$ 

$\Rightarrow b=135$ 

$\Rightarrow q=1$ 

$\Rightarrow r=90$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new reminder). We get

$\Rightarrow 135=90\times 1+45$ 

Here, 

$\Rightarrow b=90$ 

$\Rightarrow r=45$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ and $r$. We get

$\Rightarrow 90=2\times 45+0$ 

Now, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, the divisor will be the HCF.

Therefore, the HCF of $135$ and $225$ is $45$.

(ii) $196$ and $38220$

Ans: We have to find the HCF of $196$ and $38220$ by using Euclid’s division algorithm.

Let $a=38220$ and $b=196$.

$\Rightarrow 38220=196\times 195+0$ 

$\Rightarrow b=196$ 

$\Rightarrow q=195$ 

$\Rightarrow r=0$ 

Since, we get $r=0$, thus we can stop at this stage.

Therefore, the HCF of $196$ and $38220$ is $196$.

(iii) $867$ and $255$ 

Ans: We have to find the HCF of $867$ and $255$ by using Euclid’s division algorithm.

Let $a=867$ and $b=255$.

\[\Rightarrow 867=255\times 3+102\] 

$\Rightarrow b=255$ 

$\Rightarrow q=3$ 

$\Rightarrow r=102$ 

$\Rightarrow 255=102\times 2+51$ 

$\Rightarrow b=102$ 

$\Rightarrow q=2$ 

$\Rightarrow r=51$ 

$\Rightarrow 102=51\times 2+0$ 

Therefore, the HCF of $867$ and $255$ is $51$.

2. Show that any positive odd integer is of the form, $6q+1$ or $6q+3$, or $6q+5$, where $q$ is some integer.

Ans: Let $a$ be any positive integer and $b=6$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r<b$.

Here, $0\le r<6$.

Substitute the values, we get

$\Rightarrow a=6q+r$

If $r=0$, we get

$\Rightarrow a=6q+0$

$\Rightarrow a=6q$

If $r=1$, we get

$\Rightarrow a=6q+1$

If $r=2$, we get

$\Rightarrow a=6q+2$ and so on

Therefore, $a=6q$ or $6q+1$ or $6q+2$ or $6q+3$ or $6q+4$ or $6q+5$.

We can write the obtained expressions as 

\[6q+1=2\times 3q+1\]

$\Rightarrow 6q+1=2{{k}_{1}}+1$ 

Where, ${{k}_{1}}$ is an integer.

\[6q+3=6q+2+1\]

$\Rightarrow 6q+3=2\left( 3q+1 \right)+1$ 

$\Rightarrow 6q+3=2{{k}_{2}}+1$ 

Where, ${{k}_{2}}$ is an integer.

\[6q+5=6q+4+1\]

$\Rightarrow 6q+5=2\left( 3q+2 \right)+1$ 

$\Rightarrow 6q+5=2{{k}_{3}}+1$  

Where, ${{k}_{3}}$ is an integer.

Thus, $6q+1,6q+3,6q+5$ are of the form $2k+1$ and are not exactly divisible by $2$.

Also, all these expressions are of odd numbers.

Therefore, any positive odd integer can be expressed in the form, $6q+1$ or $6q+3$, or $6q+5$, where $q$ is some integer.

3. 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?  

Ans: Given that 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.

We have to find the maximum number of columns in which they can march.

We need to find the HCF of $616$ and $32$ to find the maximum number of columns.

We will use Euclid's division algorithm to find the HCF.

Let $a=616$ and $b=32$.

\[\Rightarrow 616=32\times 19+8\] 

$\Rightarrow b=32$ 

$\Rightarrow q=19$ 

$\Rightarrow r=8$ 

$\Rightarrow 32=8\times 4+0$ 

Therefore, the HCF of $616$ and $32$ is $8$.

Therefore, in $8$ columns members can march.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the 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 $3m+1$.)

Ans: Let $a$ be any positive integer and $b=3$.

Here, $0\le r<3$.

$\Rightarrow a=3q+r$

$\Rightarrow a=3q+0$

$\Rightarrow a=3q$

$\Rightarrow a=3q+1$

$\Rightarrow a=3q+2$

Therefore, $a=3q$ or $3q+1$ or $3q+2$.

Now, squaring each of these, we get

$\Rightarrow {{a}^{2}}={{\left( 3q \right)}^{2}}$ or ${{\left( 3q+1 \right)}^{2}}$ or ${{\left( 3q+2 \right)}^{2}}$

Now, applying the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get

$\Rightarrow {{a}^{2}}=9{{q}^{2}}$ or $9{{q}^{2}}+6q+1$ or $9{{q}^{2}}+12q+4$

Thus, we get

$\Rightarrow {{a}^{2}}=3\times 3{{q}^{2}}$

$\Rightarrow {{a}^{2}}=3m$, where, $m=3{{q}^{2}}$ 

${{a}^{2}}=3\times 3{{q}^{2}}+3\times 2q+1$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+2q \right)+1$ 

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+2q$ 

${{a}^{2}}=3\times 3{{q}^{2}}+6\times 2q+4$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+4q \right)+3+1$ 

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+4q +1\right)+1$ 

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+4q+1$

Therefore, we can say that the square of any positive integer is either of the form $3m$ or $3m+1$ for some integer $m$.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Now, consider $a=3q$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q \right)}^{3}}$ 

$\Rightarrow {{a}^{3}}=27{{q}^{3}}$ 

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}} \right)$

$\Rightarrow {{a}^{3}}=9m$, where, $m=3{{q}^{3}}$ 

Now, consider $a=3q+1$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+1 \right)}^{3}}$

Now, applying the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$, we get

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+27{{q}^{2}}+9q+1$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$

$\Rightarrow {{a}^{3}}=9m+1$, where, \[m=3{{q}^{3}}+3{{q}^{2}}+q\]  

Now, consider $a=3q+2$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+2 \right)}^{3}}$

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+54{{q}^{2}}+36q+8$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+6{{q}^{2}}+4q \right)+8$

$\Rightarrow {{a}^{3}}=9m+8$, where, \[m=3{{q}^{3}}+6{{q}^{2}}+4q\]  

Therefore, we can say that a cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

Opting for the NCERT solutions for Ex 1.1 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 1.1 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Real Numbers textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 1 Exercise 1.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 1 Exercise 1.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 1 Exercise 1.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

NCERT Solutions for Class 10 Maths Chapter 8 Exercises

FAQs on NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers - Exercise 1.1

1. What are the Real Numbers?

Real numbers are basically the combination of rational and irrational numbers of the number system. In general, real numbers are something which can be imagined and also be represented in the number line such as your grandparent’s age, the number of your car etc. All the arithmetic operations can be also performed on these numbers. At the same time, the imaginary numbers are called the un-real numbers. It’s something which cannot be expressed in the number line and is commonly used to represent a complex number. The concepts related to real numbers are well explained in this chapter and the questions in the exercise are also related to the same.

2. How are NCERT Solutions for Class 10 Maths Chapter 1 beneficial for all students?

Class 10 board examination is really important for all the students as it is the first board examination of their life. It is also important to score well in every subject. Mathematics seems to be a complicated subject for many students. Hence it is a challenging job to score well in this subject. All the students should rely on the NCERT solutions at this stage. NCERT Solutions for Class 10 Mathematics are considered to be the best study materials for all the students. These solutions are created by the best subject matter experts in the industry as per the latest CBSE curriculum and guidelines. Hence all the answers provided in these solutions are absolutely accurate.

These solutions provide answers to the questions asked in the Class 10 Maths Chapter 1 exercises. Students who want to study at their own pace must refer to these NCERT Solutions for Maths as these are easily available on our website and applications in PDF format. So, it can be downloaded as per the students’ convenience. Not only this but also these solutions for Class 10 Maths Chapter 1 help in quick revision as well.

3. What are the key features of NCERT Solutions for Class 10 Maths Chapter 1- Real Number Exercise 1.1?

NCERT Solutions for Maths play a very crucial role in every Class 10 student’s life. 

These NCERT Solutions let you solve and revise all questions of Exercise 1.1. After studying the step-by-step solutions given by our subject matter experts and teachers, you will be able to score the highest possible marks in the final exam. These follow the NCERT curriculum and guidelines which help in preparing the students accordingly. These contain all the important questions from the examination point of view, so referring to these will absolutely help you to score well in the exam. These solutions are solved by the easiest and smartest trick, so students can remember with ease.

4. What does Exercise 1.1 of  Class 10 Maths Chapter 1 deal with?

Exercise 1.1 of  Class 10 Maths Chapter 1 is the first exercise of Chapter 1 of class 10 Maths -  Real Numbers. The concept of real Numbers was first introduced in Class 9 and now the topic is being discussed in more in-depth details in Class 10 along with Euclid’s Division Algorithm. This chapter consists of a total of four exercises - Euclid’s division algorithm, Fundamental Theory of Arithmetic, Irrational numbers, Rational numbers and their decimal expansions. 

Hence, the first exercise here deals with the divisibility of integers. With the help of Euclid’s division algorithm, the divisibility of integers depicts that any positive integer can be divided by any other positive integer b. Therefore, the remainder will be smaller than b.

5. How many questions are there in Exercise 1.1 of Class 10th Maths Chapter 1 Real Numbers?

Exercise 1.1 contains 5 questions. Class 10 Maths Chapter 1 is important for students to form a strong base of the subject as it will help them in their higher classes as well. It is important that you thoroughly practice all the exercises from the NCERT book. You can also refer to the NCERT Class 10 Maths solutions PDF for reference. There are plenty of questions that are explained step by step in the pdf. 

6. How many examples are based on Exercise 1.1 of Chapter 1 in Class 10th Mathematics?

4 Examples are based on Exercise 1.1 for Class 10 Maths. It is based on Euclid's Division Lemma and Euclid's Division Algorithm. Chapter 1 introduces students to Real Numbers and their concepts. Before Exercise 1.1 these 4 example questionnaires are explained thoroughly. All examples help understand the concept and the kind of questions to expect in the upcoming exercise. The examples are beneficial in helping the students understand the concept easily and provide extra practice. 

7. How to prepare Chapter 1 of Class 10 Maths?

The students have to practice all the examples and questions given in the chapter. It is important that they are thorough with the NCERT textbook questions and examples. These questions can be asked directly in the exam. The students can refer to Vedantu’s NCERT Solutions for Class 10 Maths Chapter 1- Real Number Exercise 1.1 for extra questions and practice. Revising with the help of the PDF is also easy for the students. Each and every question and formula is explained for the students to understand easily. Students can also study offline as the PDFs can be downloaded free of cost.

8. What is Euclid’s Division Algorithm?

It is a method introduced in this chapter of Class 10 to help students understand how to take out the HCF using Euclid’s Division Lemma. The algorithm states that if there are two integers a and b then, there also exists q and r so that the following condition is satisfied: a=bq+r. In this condition, it is important that r is greater than or equal to 0 and less than b. This is explained easily in the solutions PDF provided by Vedantu. You can find it online for free download. 

9. Is it important to practice extra questions to prepare for Exercise 1.1 of Chapter 1 of Class 10 Maths for the exam?

Maths is a subject where you perform best if you have enough practice and are completely thorough with all the topics and know every question. Making notes of all the formulas and practicing extra questions is key. While preparing for the exam it is also important that you practice previous year’s papers so that you know the kind of questions to expect and also be able to manage time. You can practice questions from previous years in the solutions PDF available on the Vedantu app and the website. Questions are also given in-depth answers for the students to easily understand the concept. 

NCERT Solutions for Class 10

Cbse study materials for class 10, cbse study materials.