## Real Numbers (NCERT Exercise 1.2)

**Q1: Express each number as product of its prime factors:**

(ⅰ) 140

(ⅱ) 156

(ⅲ) 3825

(ⅳ) 5005

(ⅴ) 7429

Answer:

(ⅰ) 140 = 2 × 2 × 5 × 7 = 2 2 × 5 × 7

(ⅱ) 156 = 2 × 2 × 3 × 13 = 2 2 × 3 × 13

(ⅲ) 3825 = 3 × 3 × 5 × 5 × 17 = 3 2 × 5 2 × 17

(ⅳ) 5005 = 5 × 7 × 11 × 13

(ⅴ) 7429 = 17 × 19 × 23

**Q2: 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

Answer:

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

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

ⅲ 336 and 54

336 = 2 × 2 × 2 × 2 × 3 × 7

336 = 2 4 × 3 × 7

54 = 2 × 3 × 3 × 3

54 = 2 × 3 3

HCF = 2 × 3 = 6

LCM = 2 4 × 3 3 × 7 = 3024

Product of the two numbers = 336 × 54 = 18144

HCF × LCM = 6 × 3024 = 18144

Hence, product of two numbers = HCF × LCM

**Q3: 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

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

HCF = 3

LCM = 2² × 3 × 5 × 7 = 420

ⅱ 17, 23 and 29

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 17 × 23 × 29 = 11339

ⅲ 8, 9 and 25

8 = 2 × 2 × 2

9 = 3 × 3

25 = 5 × 5

HCF = 1

LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

**Q4: Given that HCF (306, 657) = 9, find LCM (306, 657).**

Answer:

HCF ( 306, 657 ) = 9

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

∴ LCM × HCF = 306 × 657

LCM = 306 × 657 / HCF

= 306 × 657 / 9

LCM = 22338

**Q5: Check whether 6ⁿ can end with the digit 0 for any natural number n .**

Answer: If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorisation must include primes 2 and 5 both Prime factors of 6ⁿ = (2 × 3)ⁿ

By Fundamental Theorem of Arithmetic, prime factors of a number is unique.

So 5 is not a prime factor 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.

**Q6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 ×1 + 5 are composite numbers.**

Answer: 7 × 11 × 13 + 13 = 13 ( 7 × 11 + 1 ) = 13 × 78 ,

which is not a prime number because it has more than two factors. So, it is a composite number.

And,

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

which is not a prime number because it has more than two factors. So, it is also a composite number.

**Q7: 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: Time taken by Sonia to drive one round of the field = 18 minutes.

Time taken by Ravi to drive one round of the field = 12 minutes.

LCM of 18 minutes and 12 minutes.

∴ 18 = 2 × 3 × 3 =2 x 3²

And, 12 = 2 × 2 × 3 = 2² x 3

LCM of 12 and 18 = product of factors raised to highest exponent = 2² × 3² = 36

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

☛See also

CH 1: Real Numbers (MCQ)

CH 1: Real Numbers (Study Points)

CH 1: Real Numbers (NCERT Ex 1.1)

CH 1: Real Numbers (Euclid's Division Lemma - Q & A)

CH 1: Problems on Euclid's Division Algorithm

CH 1: Real Numbers (Problems and Answers)

CH 1: Real Numbers (NCERT Exemplar Ex 1.1 Q1-3)

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