## Chapter 1 Real Numbers Exercise 1.2

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

i) 140

ii) 156

iii) 3825

iv) 5005

v) 7429

Answer:

i) 140

ii) 156

iii) 3825

iv) 5005

v) 7429

Answer:

i) 140ii) 156iii) 3825

iv) 5005v) 7429

**Question 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

Answer:

i) 26 and 91

ii) 510 and 92

iii) 336 and 54

Answer:

i) 26 and 91

Prime factorization of 26 = 2 x 13

Prime factorization of 91 = 7 x 13

Therefore, LCM (26, 91) = 2 x 7 x 13

HCF (26, 91) = 13

Verification:

Product of 2 numbers |
HCF x LCM |

26 x 91 |
182 x 13 |

From the above table we can conclude that the LHS side = RHS side or product of 2 numbers = HCF x LCM. Hence verified.

ii) 510 and 92

Prime factorization of 510 = 2 x 3 x 5 x 17

Prime factorization of 92 = 2 x 2 x 23 or 2² x 23

Therefore, LCM (510, 92) = 2² x 3 x 5 x 17 x 23

= 23460

HCF (26, 91) = 2

Verification:

Product of 2 numbers |
HCF x LCM |

510 x 92 |
23460 x 2 |

From the above table we can conclude that the LHS side = RHS side or product of 2 numbers = HCF x LCM. Hence verified.

iii) 336 and 54

Prime factorization of 336 = 2 x 2 x 2 x 2 x 3 x 7 or 2⁴ x 3 x 7

Prime factorization of 54 = 2 x 3 x 3 x 3 or 2 x 3³

Therefore, LCM (336, 54) = 2⁴ x 3³ x 7

= 3024

HCF (26, 91) = 2 x 3

= 6

Verification:

Product of 2 numbers |
HCF x LCM |

336 x 54 |
3024 x 6 |

From the above table we can conclude that the LHS side = RHS side or product of 2 numbers = HCF x LCM. Hence verified.

**Question 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

ii) 17, 23 and 29

iii) 8, 9 and 25

Answer:

i) 12, 15 and 21

Prime factorization of 12 = 2 x 2 x 3 or 2² x 3

Prime factorization of 15 = 3 x 5

Prime factorization of 21 = 3 x 7

Therefore, LCM (12, 15, 21) = 2² x 3 x 5 x 7

= LCM = 420

HCF (12, 15, 21) = 3

ii) 17, 23 and 29

Prime factorization of 17 = 1 x 17

Prime factorization of 23 = 1 x 23

Prime factorization of 29 = 1 x 29

Therefore, LCM (17, 23, 29) = 1 x 17 x 23 x 29

= LCM = 11339

HCF (17 , 23, 29) = 1

iii) 8, 9 and 25

Prime factorization of 8 = 2 x 2 x 2 or 2³

Prime factorization of 9 = 3 x 3 or 3²

Prime factorization of 25 = 5 x 5 or 5²

Therefore, LCM (8, 9, 25) = 2³ x 3² x 5²

= LCM = 1800

HCF (8, 9, 25) = 1

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

Answer:

Answer:

Given:

HCF of the two numbers = 9

Two numbers = 306 and 657

LCM = ?

We know that,

LCM x HCF = Product of two numbers

= LCM x 9 = 306 x 657

= LCM x 9 = 201,042

= LCM = 201,042/9

= LCM = 22,338

Therefore, LCM (306, 657) = 22,338

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

Answer:To end with digit 0, the given number must have factors as 2 and 5. But 6, which has 2 and 3 as factors, has neither 2 nor 5 as factors. Hence, 6 can never end with digit 0.

Answer:

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

Answer:

Answer:

For 7 × 11 × 13 + 13,

= 7 x 11 x 13 + 13

= 13 [7 x 11 x 1 + 1]

= 13 [77 + 1]

= 13 [78]

= 13 [2 x 3 x 13]

= 2 x 3 x 13²

Since, 7 x 11 x 13 + 13 can be expressed as a product of primes. Hence, it is a composite number.

For 7 × 6 × 5 x 4 x 3 x 2 x 1 + 5,

= 7 × 6 × 5 x 4 x 3 x 2 x 1 + 5

= 5 [7 × 6 × 5 x 4 x 3 x 2 x 1 + 1]

= 5 [1008 + 1]

= 5 [1009]

Since, 7 × 6 × 5 x 4 x 3 x 2 x 1 + 5 can be expressed as a product of primes. 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?**

Answer:

Answer:

To find after how many minutes Sonia and Ravi meet again at starting point, we have to find their LCM of their time.

Therefore, LCM (18, 12) = 36

Hence, they will meet after 36 minutes.

Nice

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