NCERT Solutions for Class 10 Maths Exercise 1.2 Chapter 1 Real Numbers

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 are provided here. Our subject experts prepare these solutions with the aim of helping the students appearing for CBSE exams. They prepare NCERT Maths Solution for Class 10 – chapter-wise so that they help students solve the problems quickly.

The points kept in mind while preparing these solutions are ease of understanding and detailed explanation of the solutions provided. Exercise 1.2 is the second exercise of Chapter 1 -Real Numbers, and it deals with the fundamental theorem of Arithmetic. There are seven problems given in Exercise 1.2 Class 10 in NCERT, and solutions to all those can be found here. It is made sure that the solutions provided follow NCERT guidelines. These NCERT solutions cover all the topics of the chapter Real Numbers and help in scoring well in exams.

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Access answers of Maths NCERT Class 10 Chapter 1 – Real Number Exercise 1.2

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 the 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 the 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 the 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,

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 the unit place as 0 or 5 is divisible by 5.

Prime factorisation of 6ⁿ = (2×3)ⁿ

Therefore, the prime factorisation of 6ⁿ doesn’t contain the 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 a 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 a 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: SincebBoth 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 1.2 of NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers is the second exercise of Chapter 1 of Class 10 Maths. Real Numbers is the first chapter students study in the Class 10 NCERT textbook. The fundamentals of Arithmetic is one of the exercise topics of this chapter. The factors discussed in this chapter are the Factorisation of composite numbers. It states 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.

• The Fundamental Theorem of Arithmetic – It includes 7 questions based on this theorem.

Key Features of NCERT Solutions for Class 10 Maths Chapter 1 – Real Number Exercise 1.2 Page number 14

• These NCERT Class 10 Solutions help in solving and revising all questions of exercise 1.2 real numbers.
• If students go through the step-wise solutions given here, they will be able to get more marks.
• The solutions will help learners score well in Maths exams if they practise thoroughly.
• They follow NCERT guidelines which help in preparing the students for the exam accordingly.
• They contain all the important questions from the examination point of view.