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In this article we provide the solution of **class 10 Maths NCERT Chapter 1 (New Syllabus)**. Here you can find the Class 10 Maths Chapter 1 Real Numbers NCERT Solutions. We at Study Circle have created step-by-step solutions with thorough explanations for those times when students feel overwhelmed about looking for the most complete and detailed NCERT Solutions for **Class 10 Maths NCERT Chapter- 1(New Syllabus)**.

We suggest that students who wish to do well in maths read over these solutions and brush up on their skills.

There are 4 exercises in Chapter 1 – Real Numbers , and the NCERT Solutions provided on this page answer the exercises’ questions. Let’s now take a closer look at some of the ideas covered in this chapter.

### ** Class -10 Maths NCERT Chapter -1(New Syllabus) Exercise 1.1 **

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

**(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429**

SOLUTION. (i) We use the division method as shown below :

140 = 2 × 2 × 5 × 7

(ii) We use the division method as shown below :

156 = 2 × 2 × 3 × 13

= 2

2× 3 × 13

(iii) We use the division method as shown below :

3825 = 3 × 3 × 5 × 5 × 17

= 3

2 × 52 × 17

(iv) We use the division method as shown below :

5005= 5 × 7 × 11 × 13

(v) We use the division method as shown below :

7429= 17 × 19 × 23

**QUESTION 2. Find the L.C.M. and H.C.F. of the following pairs of integers and verify :**

**L.C.M. × H.C.F. = 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 and 91 = 7 × 13

L.C.M. of 26 and 91 = 2 × 7 × 13 = 182

and H.C.F. of 26 and 91 = 13

Now, 182 × 13 = 2366 and 26 × 91 = 2366

Hence, 182 × 13 = 26 × 91

(ii) 510 and 92

510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23

L.C.M. of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

and H.C.F. of 510 and 92 = 2

Now, 23460 × 2 = 46920 and 510 × 92 = 46920

Hence, 23460 × 2 = 510 × 92

(iii) 336 and 54

336 = 2 × 2 × 2 × 2 × 3 × 7

and 54 = 2 × 3 × 3 × 3

L.C.M. of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 ×7

= 3024

and H.C.F. of 336 and 54 = 2 × 3 = 6

Now, 3024 × 6 = 18144 and 336 × 54 = 18144

Hence, 3024 × 6 = 336 × 54

**QUESTION 3. Find the H.C.F. and L.C.M. of the following integers by applying prime factorisation method.**

**(i) 12, 15, 21; (ii) 17, 23, 29 (iii) 8, 9 and 25**

**(ii) 17, 23 or 29**

H.C.F. There is no common factor as 17, 23, 29 they are primes. Hence H.C.F. is 1.

L.C.M. is the product of all prime factors 17, 23 and 29.

17 × 23 × 29 = 11339

Hence, H.C.F. (17, 23, 29) = 1,

L.C.M (17, 23, 29) = 11339.

###### (iii) First we write the prime factorisation of each of the given numbers.

8 = 2 × 2 × 2

, 9 = 3 × 3

, 25 = 5 × 5

L.C.M = 8 × 9 × 25 = 1800

and H.C.F. = 1 Answer .

**QUESTION 4. Given H.C.F. (306, 657) = 9, find L.C.M. (306, 657).**

** SOLUTION.** We have, H.C.F. (306, 657) = 9.

We know that,

Product of L.C.M. and H.C.F. = Product of two numbers.

L.C.M. × 9 = 306 × 657

L.C.M. = 306× 657/9 = 22338

###### Hence, L.C.M. (306, 657) = 22338.

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

^{n}can end with the digit 0 for any natural number n.

** SOLUTION: —**** **If the number 6^{n} 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 6^{n} = (2×3)^{n}

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

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

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

** SOLUTION.** We have, 7 × 11 × 13 + 13 = 1001 + 13 = 1014

1014 = 2 × 3 × 13 × 13

So, it is the product of more than two prime numbers.

2, 3 and 13.

Hence, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5045

5045 = 5 × 1009

It is the product of prime factors 5 and 1009.

Hence, it is a composite number.

**QUESTION 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.** They will be again at the starting point at least common multiples of 18 and

12 minutes. To find the LCM of 18 and 12, we have:

18 = 2 × 3 × 3 and 12 = 2 × 2 × 3

L.C.M. of 18 and 12 = 2 × 2 × 3 × 3 = 36

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

**Class -10 Maths NCERT Chapter -1(New Syllabus) EXERCISE 1.2 **

**QUESTION 1. Prove that √5 is an irrational number by contradiction method.**

Solution :- Assume that √5 be a rational number then we have

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

SOLUTION:– Let us assume, to the contrary, that 3+ 2√ 5 is a rational number.

Now, let 3 +2 √5 = a/b, where a and b are coprime and b≠ 0

where a and b are coprime and b 0.

So, 2√5 =a/b-3 or √5 = a/2b -3/2

since a and b are integers, therefore

a/2b = 3/2 is a rational number

so , √5 is an irrational number.

This shows that our assumption is incorrect.

so , 3+ 2√5 is an irrational number.

**QUESTION 3. Prove that the following are irrational**