NCERT Solutions for Class 10 Maths chapter-1 Real Numbers Exercise 1.1 and 1.2

Based on New NCERT PATTERN [2023-24]

Exercise 1.1

1. Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer:

(i) 140 = 2 x 2 x 5 x 7 = 22 x 5 x 7

(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13

(iii) 3825 = 3 x 3 x 5 x 17 = 32 x 52 x x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 7429 = 17 x 19 x 23 

2. Find the LCM and HCF of the following pairs of integers and verify that LCM xHCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer:

(i) 26 and 91

Let, a = 26, b = 91

26 = 2  × 13

91 = 7  × 13

L.C.M = 2 × 7 × 13 = 14 × 13 = 182

H.C.F = 13

Verification:

H.C.F × L.C.M = a × b

13 × 182 = 26 × 96

2366 = 2366

Hence, LCM × HCF = product of the 26 and 91.

(ii) 510 and 92

Let, a = 510, b = 92

510 = 2  × 3 × 5 × 17

92 = 2  × 2 × 23

L.C.M = 2 × 2 × 3 × 5 × 17 × 23 = 23460

H.C.F = 2

Verification:

H.C.F × L.C.M = a × b

23460 × 2 = 92 × 510  

46920 = 46920

Hence, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54

a = 336, b = 54Read more on Sarthaks.com - https://www.sarthaks.com/661852/find-the-lcm-and-the-following-pairs-integers-and-verify-that-lcm-hcf-product-the-two-numbers

Let a = 336 , b = 54

336= 2 x 2 x 2 x 2 x 3 x 7

54 = 2 x 3 x 3 x 3 = 

LCM = 24 x 33 x 7= 3024

HCF = 2 x 3 = 6

Verification:

H.C.F × L.C.M = a × b

3024 × 6  = 336 × 54 

18144 = 18144 

Hence, LCM × HCF = product of the 336 and 54.

3. 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

Prime factors for all the three numbers, we get,

12=2×2×3

15=5×3

21=7×3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Prime factors for all the three numbers, we get,

17=17×1

23=23×1

29=29×1

Therefore,

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Prime factors for all the three numbers, we get,

8=2×2×2×1

9=3×3×1

25=5×5×1

Therefore,

HCF(8,9,25)=1

LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer:

HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

L.C.M x H.C.F = first Number x Second Number

L.C.M x 9 = 306 x 657

LCM = (306×657)/9 = 22338

LCM = 22338

5. Check whether 6n can end with the digit 0 for any natural number n.

Answer:

Prime factorization of 6n = (2×3)n 

Therefore, prime factorization of 6ndoes not contain 5 and 2 as a factor together.

OR, 6n cannot end with the digit 0 for any natural number n. In order for a number to end with 0, it must be divisible by 10. However, for any natural number n, 6n will never be divisible by 10.

Hence 6ncan never end with the digit 0 for any natural number n

6. Explain why 7 x 11 x 13  and  7 x 6 x 5 x 3 x 2 x 1 + 5 are composite numbers.

Answer:

A composite number is a whole number greater than 1 that has multiple factors, meaning it can be divided by more than just 1 and itself. If a number has only two factors (1 and itself), it is called a prime number.

1. 7 × 11 × 13 + 13:

When we calculate 7 × 11 × 13 and add 13 to it, we get the number 1014. 

This number has several divisors other than 1 and itself, like 2, 3, 6, 9, 18, 27, and so on. Because it has more than two divisors, it is a composite number.

2. 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5:

When we calculate 7 × 6 × 5 × 4 × 3 × 2 × 1 and add 5 to it, we get the number 5045. 

Just like the previous example, this number also has divisors other than 1 and itself, like 5, 1009, and others. Because it has more than two divisors, it is also a composite number.

So, both expressions, 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5, are composite numbers because they have more than two different numbers that can divide them evenly.

7. 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:

To find the time at which Sonia and Ravi will meet again at the starting point, we need to determine the least common multiple (LCM) of their individual time intervals.

Sonia takes 18 minutes to complete one round of the field, and Ravi takes 12 minutes to do the same. The LCM of 18 and 12 will give us the time at which both of them meet again at the starting point.

To find the LCM of 18 and 12, let's first find their prime factorizations:

18 = 2  × 3  × 3

12 = 2  × 2  × 3

LCM of 12 and 18= 2 x 2 x 3 x 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Exercise 1.2

1. Prove that √2 is irrational.

Explanation :  

Let us assume to the contrary that √2 is a rational number. . This means that it can be expressed as a fraction in its simplest form:

i.e ; √2 = a/b

Where 'a' and 'b' are integers, and 'a' and 'b' have no common factors other than 1 (i.e., the fraction is in its simplest form).

Now, we can square both sides of the equation to get rid of the square root:

⇒ (√2)2 = (a/b)2

⇒ 2 = a2/b2

⇒ a2 = 2b2

⇒ a2 is multiple of 2.

So, a is a multiple of 2. 

From this equation , we can see that 'a2' is a multiple of 2. This means that 'a' must be a multiple of 2 as well since squaring an integer will result in another integer. Let's represent 'a' as '2k' where 'k' is an integer:

⇒ a = 2k for some integer k.

⇒ a2 = 4k2

⇒ 2b2 = 4k2

⇒ b2 = 2k2

⇒ b2 is a multiple of 2

So, b is a multiple of 2

Now, we can see that 'b2' is also a multiple of 2. This implies that 'b' must be a multiple of 2 as well since squaring an integer will result in another integer.

Since both 'a' and 'b' are multiples of 2, they have a common factor (2) other than 1, which contradicts our initial assumption that 'a' and 'b' have no common factors other than 1.

Therefore, our initial assumption that √2 is rational must be false, and √2 is irrational.

2. Prove that (3 + 2√5) is irrational.

Explanation :  

Let us assume, to the contrary, that  3 + √5 is rational. 

Then, there exist co-primes a and b (b is not equal to  0) such that 

3 + √5 = a / b 

⇒  √5 = a / b -  3

 

⇒ √5 =  (a - 3b) / b

Since a and b are integers, so (a - 3b) / b is rational. 

Thus, √5 is also rational. 

But, this contradicts the fact that √5 is irrational.  So, our assumption is incorrect. Hence, 6 + √5 is irrational.

3. Prove that the following are irrationals.

(i)  1/√2

(ii) 7√5

(iii) 6 +√2

Answer:

(i) Let us assume 1/√2 is rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

      1/√2 = x/y

⇒ √2 = y/x

Since, x and y are integers, thus, √2 is a rational number, But, this contradicts the fact that √2 is irrational.  So, our assumption is incorrect.

Hence, we can conclude that 1/√2 is irrational.

(ii) Let us assume 7√5 is a rational number.

It means we have some co-prime integers a and b (b ≠ 0)  

     7√5 = x/y

⇒ √5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, But, this contradicts the fact that √5 is irrational.  So, our assumption is incorrect.

Hence, we can conclude that 7√5 is irrational.

(iii) Let us assume, to the contrary, that  6 + √2 is rational. 

Then, there exist co-primes a and b (b is not equal to  0) such that 

6 + √2 = a / b 

⇒  √2 = a / b -  6

 

⇒ √2 =  (a - 6b) / b

Since a and b are integers, so (a - 6b) / b is rational. 

Thus, √2 is also rational. 

But, this contradicts the fact that √2 is irrational.  So, our assumption is incorrect. Hence, 6 + √2 is irrational.

Key Features of NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers:

Comprehensive Coverage: The NCERT Solutions cover all the topics and concepts of Class 10 Maths Chapter 1, ensuring a thorough understanding of real numbers.

Step-by-Step Solutions: Each question in the NCERT Solutions is solved step-by-step, providing a clear and logical explanation of the concepts.

Easy to Understand: The solutions are presented in a simple and easy-to-understand language, making it accessible to all students.

Well-Structured Format: The solutions are organized in a well-structured format, making it easy for students to follow and learn.

Exercise-wise Solutions: The NCERT Solutions are divided into exercises, with separate solutions for each exercise, allowing students to practice and revise systematically.

Illustrative Examples: The NCERT Solutions include illustrative examples to help students grasp the concepts better.

Exam-Oriented Approach: The solutions are designed with a focus on exam preparation, providing students with the necessary practice to perform well in exams.

Clarity of Concepts: The solutions provide clarity on fundamental concepts of real numbers, HCF, LCM, irrational numbers, and more.

Error-Free: The NCERT Solutions are prepared by subject matter experts, ensuring accuracy and correctness in the solutions.

Accessible for Free: NCERT Solutions for Class 10 Maths Chapter 1 are available for free, making them easily accessible to all students.

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 1

1. What is the importance of NCERT Solutions for Class 10 Maths Chapter 1?

NCERT Solutions for Class 10 Maths Chapter 1 provide step-by-step explanations and solutions to all the problems in the chapter. They help you understand the concepts better, practice different types of questions, and prepare effectively for exams.

2. Are NCERT Solutions for Class 10 Maths Chapter 1 enough for exam preparation?

NCERT Solutions are essential for understanding the concepts thoroughly, but for comprehensive exam preparation, you should also solve additional practice questions, previous year papers, and reference books.

3. How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 1?

According to new NCERT PATTERN 2023-24, there are total of 2 exercises in NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers. Each exercise consists of various questions related to the concepts covered in the chapter. These exercises are designed to help students practice and reinforce their understanding of real numbers, properties of rational and irrational numbers, HCF, LCM, and more.

4. Is NCERT deleted some topis from this chapter.

Yes, NCERT dropped two topis first Euclid’s division lemma [1.1 exercise old book] and second one is Revisiting rational numbers and their decimal Expansions [1.4 exercise old book]

5. Some tips for getting a 95 on the class 10 exam?

1. Stay consistent in your studies and revise regularly.

2. Focus on understanding concepts rather than memorization.

3. Practice previous year question papers and sample papers.

4. Manage your time effectively during the exam.

5. Seek help from teachers or peers whenever needed.

6. Stay calm and confident during the exam.

7. Take care of your health and get enough rest before the exam.

8. Avoid last-minute cramming and stress.