# NCERT Class 10 Maths Chapter 1 - Real Numbers Solutions

**NCERT Class 10 Maths Chapter 1 - Real Numbers Solutions**

**Exercise 1.1 Page: 7**

**Question 1:**

**Apply Euclid’s division algorithm to determine the highest common factor (HCF) of:**

**(i) 135 and 225**

**(ii) 196 and 38220**

**(iii) 867 and 255**

**Solutions:**

(i) 135 and 225

Analyzing the given question, we observe that 225 is greater than 135. By employing Euclid’s division algorithm, we derive:

225 = 135 × 1 + 90

Notably, the remainder 90 ≠ 0. Therefore, employing the division lemma for 90, we obtain:

135 = 90 × 1 + 45

Once again, 45 ≠ 0. By repeating the above step for 45, we find:

90 = 45 × 2 + 0

At this point, the remainder is zero, signaling the end of our method. Since the divisor in the last step is 45, the HCF of (225, 135) = HCF (135, 90) = HCF (90, 45) = 45.

Thus, the HCF of 225 and 135 is 45.

(ii) 196 and 38220

In this particular question, 38220 > 196. Hence, by applying Euclid’s division algorithm and taking 38220 as the divisor, we obtain:

38220 = 196 × 195 + 0

As the remainder is already 0, the HCF of (196, 38220) = 196.

Thus, the HCF of 196 and 38220 is 196.

(iii) 867 and 255

Given that 867 is greater than 255, we can now apply Euclid’s division algorithm to 867:

867 = 255 × 3 + 102

Since the remainder 102 ≠ 0, we proceed by taking 255 as the divisor and applying the division lemma method:

255 = 102 × 2 + 51

Once again, 51 ≠ 0. Now, with 102 as the new divisor, we repeat the same step and find:

102 = 51 × 2 + 0

Since the remainder is now zero, our procedure concludes. In the last step, the divisor is 51, implying that the HCF of (867, 255) = HCF(255, 102) = HCF(102, 51) = 51.

Hence, the HCF of 867 and 255 is 51.

**Page No 7:**

**Question 2:**

**Prove that any positive odd integer can be expressed in the form 6q + 1, 6q + 3, or 6q + 5, where q is an integer.**

**Answer:**

Let's consider a as any positive odd integer and b = 6. By applying Euclid’s algorithm, we can express a as:

a = 6q + r, where q is an integer greater than or equal to 0, and r is a remainder ranging from 0 to 5, since 0 ≤ r < 6.

Therefore, a can be written as 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.

Now, let's examine the expressions 6q + 1, 6q + 3, and 6q + 5:

- 6q + 1 can be rewritten as 2(3q) + 1, where 3q is an integer, let's say k1. Hence, 6q + 1 becomes 2k1 + 1, where k1 is a positive integer.

- 6q + 3 can be expressed as (6q + 2) + 1, which can further simplify to 2(3q + 1) + 1. Let's denote 3q + 1 as k2, which is an integer. Thus, 6q + 3 becomes 2k2 + 1, where k2 is an integer.

- 6q + 5 can be represented as (6q + 4) + 1, which is equivalent to 2(3q + 2) + 1. Let's denote 3q + 2 as k3, where k3 is an integer. Therefore, 6q + 5 can be written as 2k3 + 1, where k3 is an integer.

It is evident that 6q + 1, 6q + 3, and 6q + 5 are in the form 2k + 1, where k is an integer. Thus, these expressions are odd numbers.

Consequently, any positive odd integer can be expressed as 6q + 1, or 6q + 3, or 6q + 5.

Please let me know if you need any further clarification or assistance.

**Page No 7:**

**Question 3:**

**In a parade, an army contingent of 616 members is to march behind an army band of 32 members. The two groups are required to march in the same number of columns. What is the maximum number of columns in which they can march?**

**Answer:**

To determine the maximum number of columns in which the two groups can march, we need to find the highest common factor (HCF) of 616 and 32.

Using Euclid’s algorithm, we can calculate the HCF as follows:

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF of 616 and 32 is found to be 8.

Therefore, they can march in a maximum of 8 columns each.

If you have any further questions, feel free to ask.

**Page No 7:**

**Question 4:**

**Using Euclid’s division lemma, prove that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.**

**Hint: Assume x to be any positive integer, and it can be expressed as 3q, 3q + 1, or 3q + 2. Square each of these forms and show that they can be written in the form 3m or 3m + 1.**

**Answer:**

Let's consider a positive integer, denoted by 'a', and take the divisor 'b' as 3.

According to Euclid's division lemma, we can express a as:

a = 3q + r,

where q is an integer greater than or equal to 0, and r can be 0, 1, or 2 because 0 ≤ r < 3.

Therefore, a can be written as 3q, 3q + 1, or 3q + 2.

Now, let's square each of these forms and observe:

1. (3q)^2 = 9q^2 = 3(3q^2), which is of the form 3m.

2. (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1, which is of the form 3m + 1.

3. (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1, which is again of the form 3m + 1.

In each case, we can express the square of any positive integer as either 3m or 3m + 1, where m is some integer.

If you have any further questions, please let me know.

**Page No 7:**

**Question 5:**

**Using Euclid’s division lemma, prove that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.**

**Answer:**

Let's consider a positive integer denoted by 'a' and take the divisor 'b' as 3.

According to Euclid's division lemma, we can express a as:

a = 3q + r,

where q is an integer greater than or equal to 0, and r can be 0, 1, or 2 because 0 ≤ r < 3.

Now, let's consider the cube of 'a' and expand it:

a^3 = (3q + r)^3

Expanding this expression using the binomial theorem, we get:

a^3 = 27q^3 + 27q^2r + 9qr^2 + r^3

Now, let's examine the possible values of r:

1. When r = 0, we have:

a^3 = 27q^3

which is of the form 9m, where m = 3q^3.

2. When r = 1, we have:

a^3 = 27q^3 + 27q^2 + 9q + 1

which can be written as:

a^3 = 9(3q^3 + 3q^2 + q) + 1

This expression is of the form 9m + 1, where m = 3q^3 + 3q^2 + q.

3. When r = 2, we have:

a^3 = 27q^3 + 54q^2 + 36q + 8

which can be written as:

a^3 = 9(3q^3 + 6q^2 + 4q) + 8

This expression is of the form 9m + 8, where m = 3q^3 + 6q^2 + 4q.

Therefore, we have shown that the cube of any positive integer can be expressed in the form 9m, 9m + 1, or 9m + 8, where m is some integer.

If you have any further questions, feel free to ask.

## Exercise 1.2 Page: 11

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

**(i) 140**

**(ii) 156**

**(iii) 3825**

**(iv) 5005**

**(v) 7429**

**Solutions:**

(i) 140

By taking the LCM of 140, we will get the product of its prime factor.

Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22×5×7

(ii) 156

By Taking the LCM of 156, we will get the product of its prime factor.

Hence, 156 = 2 × 2 × 13 × 3 × 1 = 22× 13 × 3

(iii) 3825

By taking the LCM of 3825, we will get the product of its prime factor.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32×52×17

(iv) 5005

By Taking the LCM of 5005, we will get the product of its prime factor.

Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

(v) 7429

By taking the LCM of 7429, we will get the product of its prime factor.

Hence, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

**2. 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**

**Solutions:**

**(i) 26 and 91**

Expressing 26 and 91 as product of its prime factors, we get,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

And HCF (26, 91) = 13

Verification

Now, product of 26 and 91 = 26 × 91 = 2366

And product of LCM and HCF = 182 × 13 = 2366

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

**(ii) 510 and 92**

Expressing 510 and 92 as product of its prime factors, we get,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

And HCF (510, 92) = 2

Verification

Now, product of 510 and 92 = 510 × 92 = 46920

And Product of LCM and HCF = 23460 × 2 = 46920

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

**(iii) 336 and 54**

Expressing 336 and 54 as product of its prime factors, we get,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

54 = 2 × 3 × 3 × 3 × 1

Therefore, LCM(336, 54) = = 3024

And HCF(336, 54) = 2×3 = 6

Verification

Now, product of 336 and 54 = 336 × 54 = 18,144

And product of LCM and HCF = 3024 × 6 = 18,144

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**

**Solutions:**

(i) 12, 15 and 21

Writing the product of 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

Writing the product of 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

Writing the product of 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).**

**Solution: **As we know that,

HCF×LCM=Product of the two given numbers

Therefore,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

Hence, LCM(306,657) = 22338

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

**Solution: **If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.

Prime factorization of 6n = (2×3)n

Therefore, the prime factorization of 6n doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, 6n is not divisible by 5, and thus it proves that 6n cannot end with the digit 0 for any natural number n.

**6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.**

**Solution: **By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression;

**7 × 11 × 13 + 13**

Taking 13 as common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take the other number,

**7 × 6 × 5 × 4 × 3 × 2 × 1 + 5**

Taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

**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?**

**Solution: **Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.

Therefore, LCM(18,12) = 2×3×3×2×1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

## Exercise 1.3 Page: 14

**1. Prove that √**5 **is irrational.**

**Solutions: **Let us assume, that **√**5 is rational number.

i.e. **√**5 = x/y (where, x and y are co-primes)

y**√**5= x

Squaring both the sides, we get,

(y**√**5)2 = x2

⇒5y2 = x2……………………………….. (1)

Thus, x2 is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y2 = (5k)2

⇒y2 = 5k2

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about **√**5 is rational is incorrect.

Hence, **√**5 is an irrational number.

**2. Prove that 3 + 2√5 + is irrational.**

**Solutions: **Let us assume 3 + 2**√**5 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y

Rearranging, we get,

Since, x and y are integers, thus,

is a rational number.

Therefore, **√**5 is also a rational number. But this contradicts the fact that **√**5 is irrational.

So, we conclude that 3 + 2**√**5 is irrational.

**3. Prove that the following are irrationals:**

**(i) 1/√2**

**(ii) 7√5**

**(iii) 6 + **√**2**

**Solutions:**

**(i) 1/**√**2**

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

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

**(ii) 7**√**5**

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

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

**(iii) 6 +**√**2**

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we get,

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

## Exercise 1.4 Page: 17

**1. 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:**

**(i)** **13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2352) (vii) 129/(225775) (viii) 6/15 (ix) 35/50 (x) 77/210**

**Solutions:**

**Note:** If the denominator has only factors of 2 and 5 or in the form of 2m ×5n then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

**(i) 13/3125**

Factorizing the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Since, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

**(ii) 17/8**

Factorizing the denominator, we get,

8 = 2×2×2 = 23

Since, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

**(iii) 64/455**

Factorizing the denominator, we get,

455 = 5×7×13

Since, the denominator is not in the form of 2m × 5n, thus 64/455 has a non-terminating decimal expansion.

**(iv) 15/ 1600**

Factorizing the denominator, we get,

1600 = 26×52

Since, the denominator is in the form of 2m × 5n, thus 15/1600 has a terminating decimal expansion.

**(v) 29/343**

Factorizing the denominator, we get,

343 = 7×7×7 = 73 Since, the denominator is not in the form of 2m × 5n thus 29/343 has a non-terminating decimal expansion.

**(vi)23/(2352)**

Clearly, the denominator is in the form of 2m × 5n.

Hence, 23/ (2352) has a terminating decimal expansion.

**(vii) 129/(225775)**

As you can see, the denominator is not in the form of 2m × 5n.

Hence, 129/ (225775) has a non-terminating decimal expansion.

**(viii) 6/15**

6/15 = 2/5

Since, the denominator has only 5 as its factor, thus, 6/15 has a terminating decimal expansion.

**(ix) 35/50**

35/50 = 7/10

Factorising the denominator, we get,

10 = 2 × 5

Since, the denominator is in the form of 2m × 5n thus, 35/50 has a terminating decimal expansion.

**(x) 77/210**

77/210 = (7× 11)/ (30 × 7) = 11/30

Factorising the denominator, we get,

30 = 2 × 3 × 5

As you can see, the denominator is not in the form of 2m × 5n .Hence, 77/210 has a non-terminating decimal expansion.

**2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.**

**Solutions:**

**(i) 13/3125**

13/3125 = 0.00416

**(ii) 17/8**

17/8 = 2.125

**(iii) 64/455** has a non terminating decimal expansion

**(iv)15/ 1600**

15/1600 = 0.009375

**(v) 29/ 343** has a non terminating decimal expansion

**(vi)23/ (2352)** = 23/(8×25)= 23/200

23/ (2352) = 0.115

**(vii) 129/ (225775)** has a non terminating decimal expansion

**(viii) 6/15** = 2/5

**(ix) 35/50** = 7/10

35/50 = 0.7

**(x) 77/210** has a non-terminating decimal expansion.

**3. 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 factors of q?**

**(i) 43.123456789**

**(ii) 0.120120012000120000. . .**

**Solutions:**

(i) 43.123456789

Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

(ii) 0.120120012000120000. . .

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5.

NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

Introduction to Real Numbers in Class 10 Maths

Understanding the Significance

Real Numbers hold substantial importance in the realm of mathematics, particularly in the curriculum of Class 10. This pivotal topic carries a weightage of 6 marks in the Class 10 Maths board exams, making it imperative for students to grasp its concepts thoroughly.

Breakdown of Questions

Exam Structure Overview

To comprehend the significance of Real Numbers, it's essential to dissect the structure of questions typically posed in the board exams. In the previous year's examination (2018), three questions were dedicated to this chapter:

1. Part A:

- One question, carrying a weightage of 1 mark.

2. Part B:

- Another question, with a weightage of 2 marks.

3. Part C:

- Lastly, a question, accounting for 3 marks.

Core Concepts Explored

Delving into the Content

The chapter delves into various fundamental aspects of Real Numbers:

1. Euclid’s Division Algorithm:

- This algorithm lays the groundwork for understanding the divisibility of integers, elucidating how any positive integer \( a \) can be divided by another positive integer \( b \) such that the remainder is always smaller than \( b \).

2. The Fundamental Theorem of Arithmetic:

- Explores the multiplication of positive integers, elucidating its applications and significance.

Detailed Exercises

Exercise 1.1 Solutions

This section encompasses five questions, including four long and one short, to solidify the understanding of Euclid’s Division Algorithm.

Exercise 1.2 Solutions

Comprising seven questions, four long and three short, this exercise deepens comprehension through practical applications of The Fundamental Theorem of Arithmetic.

Exercise 1.3 Solutions

With three short questions, this segment revisits Rational & Irrational Numbers, bridging the concepts introduced in the preceding exercises.

Exercise 1.4 Solutions

Exploring Decimal Expansions, this exercise elucidates when the decimal expansion of a rational number is terminating or recurring, providing solutions to three comprehensive problems.

Key Features of NCERT Solutions

Enhancing Learning Experience

The NCERT Solutions for Class 10 Maths Chapter 1 offer several advantages:

- Comprehensive Coverage:

- Encompassing the updated CBSE syllabus for the academic year 2023-24.

- Expert Guidance:

- Stepwise solutions provided by subject experts facilitate better understanding and application of concepts.

- Adherence to Guidelines:

- Aligned with NCERT guidelines to ensure competent preparation among students.

- Examination-Oriented Content:

- Inclusion of all significant questions from an examination standpoint to bolster students' preparation.

Disclaimer and FAQs

Important Information and Queries Addressed

Dropped Topics

- Euclid’s division lemma.

- Revisiting rational numbers and their decimal expansions.

Frequently Asked Questions

Q1: Main Topics Covered?

A: Euclid’s division algorithm, the fundamental theorem of arithmetic, revisiting rational & irrational numbers, and decimal expansions.

Q2: Number of Exercises?

A: There are four exercises, each meticulously designed to enhance problem-solving skills.

Q3: Exam Importance?

A: Yes, Chapter 1 is crucial for exams, offering both short and long answer questions to augment problem-solving abilities.

Conclusion

Real Numbers stand as a cornerstone in the mathematical journey of Class 10 students, with Chapter 1 serving as a foundational pillar. Through comprehensive understanding and diligent practice aided by NCERT Solutions, students can navigate through this crucial chapter with confidence and proficiency.

FAQs (Frequently Asked Questions)

Q1: Why are Real Numbers important in Class 10 Maths?

Real Numbers hold significant weightage in the Class 10 Maths curriculum, contributing to 6 marks in board exams. Understanding Real Numbers is crucial for tackling various mathematical problems effectively.

Q2: How can NCERT Solutions aid in understanding Chapter 1?

NCERT Solutions for Class 10 Maths Chapter 1 provide step-by-step explanations and solutions to problems, facilitating a better understanding of core concepts and boosting confidence in tackling exam questions.

Q3: What are the benefits of using NCERT Solutions?

NCERT Solutions offer comprehensive coverage of the syllabus, expert guidance, adherence to guidelines, and a focus on exam-oriented content, making them invaluable resources for students preparing for board exams.

Q4: Are there any dropped topics in Chapter 1?

Yes, some topics like Euclid’s division lemma and revisiting rational numbers and their decimal expansions have been omitted in the NCERT Solutions.

Q5: How can students effectively prepare for exams using NCERT Solutions?

By thoroughly studying the solutions provided, practicing regularly, and understanding the underlying concepts, students can prepare effectively for exams and achieve academic success.

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