141-NCERT-10-1-Real Numbers - Ex-1.2

NCERT

10th Mathematics

Exercise 1.2

Topic: 1 Real Numbers

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EXERCISE 1.2

Q1. Express each number as a product of its prime factors:

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

Explanation:

We know that every composite number can be expressed as a product of prime numbers. 

Solution:

(i) 140

1) Prime factors of 140:    Note: Apply divisibility test of 2.

 140 = 2 x 70

= 2 x 2 x 35        Note: Apply divisibility test of 5.

= 2 x 2 x 5 x 7

= 2² x 5 x 7

2) So the prime factors of 140 are 2² x 5 x 7

(ii) 156

1) Prime factors of 156:    Note: Apply divisibility test of 2.

 156 = 2 x 78

= 2 x 2 x 39       Note: Apply divisibility test of 3.

= 2 x 2 x 3 x 13

= 2² x 3 x 13

2) So the prime factors of 156 are 2² x 3 x 13.

(iii) 3825

1) Prime factors of 3825:    Note: Apply divisibility test of 5.

3825 = 5 x 765

= 5 x 5 x 153        Note: Apply divisibility test of 3.

= 5 x 5 x 3 x 51

= 5 x 5 x 3 x 3 x 17

= 5² x 3² x 17

2) So the prime factors of 3825 are 5² x 3² x 17.

(iv) 5005

1) Prime factors of 5005:    Note: Apply divisibility test of 5.

5005 = 5 x 1001            Note: Apply divisibility test of 7.

= 5 x 7 x 143         Note: Apply divisibility test of 11.

= 5 x 7 x 11 x 13

2) So the prime factors of 5005 are 5 x 7 x 11 x 13.

Note: Click here to know the details of the divisibility test of 7.

(v) 7429

Note:

a) 2 is not a factor of 7429 as it is not an even number. So 4, 8, 16, and so on

are also not the factors of 7429.

b) 3 is not a factor of 7429 as the sum of the digits is 22, which is not divisible

by 3. So 6, 9, 12, 15, and so on are also not factors 7429.

c) 5 is not a factor of 7429 as the unit placed digit is not 0 or 5. So 5, 10, 15,

and so on are also not factors 7429.

d) 7 is not a factor of 7429. So 14, 21, and so on are also not factors 7429.

Note: Click here to know the details of the divisibility test of 7.

1) Prime factors of 7429:        Note: Try directly dividing by 17.

7429 = 17 x 437               Note: Try directly dividing by 19.

= 17 x 19 x 23

2) So the prime factors of 7429 are 17 x 19 x 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

Explanation:

1) To get the LCM and HCF of the given numbers, we need to find all prime factors of the given numbers. 

2) HCF means Highest Common Factor, so find the product of all smallest power of common factor in the numbers.

3) LCM means Least Common Multiple, so find the product of the greatest power of each prime factor in the numbers.

4) HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

Solution:

(i) 26 and 91

1) Now we will find the prime factors of 26.

26 = 2 x 13

2) Now we will find the prime factors of 91.

91 = 7 x 13

3) Here common factor is 13.

4) So HCF of 26 and 91 is 13.

5) Simmilarly, LCM = 2 x 7 x 13 = 14 x 13 = 182.

6) LCM = 182

7) Here, 26 x 91 = 2366 -------- equation 1

8) Here, HCF x LCM = 13 x 182 = 2366 -------- equation 2.

9) From equation 1 & 2, we can say that HCF(26,91) x LCM(26,91) = 26 x 91. 

(ii) 510 and 92

1) Now we will find the prime factors of 510.

510 = 2 x 255

510 = 2 x 3 x 85

510 = 2 x 3 x 5 x 17

2) Now we will find the prime factors of 92.

92 = 2 x 46

92 = 2 x 2 x 23

3) Here common factor is 2.

4) So HCF of 510 and 92 is 2.

5) Simmilarly, LCM = 2 x 2 x 3 x 5 x 17 x 23 = 23460.

6) LCM = 23460

7) Here, 510 x 92 = 46920 -------- equation 1

8) Here, HCF x LCM = 2 x 23460 = 46920 -------- equation 2.

9) From equation 1 & 2, we can say that HCF(510,92) x LCM(510,92) = 510 x 92.

(iii) 336 and 54

1) Now we will find the prime factors of 336.

336 = 2 x 168

336 = 2 x 2 x 84

336 = 2 x 2 x 2 x 42

336 = 2 x 2 x 2 x 2 x 21

336 = 2 x 2 x 2 x 2 x 3 x 7

2) Now we will find the prime factors of 54.

54 = 2 x 27

54 = 2 x 3 x 9

54 = 2 x 3 x 3 x 3

3) Here common factor is 2 x 3 = 6.

4) So HCF of 336 and 54 is 6.

5) Simmilarly, LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 = 3024.

6) LCM = 3024

7) Here, 336 x 54 = 18144 -------- equation 1

8) Here, HCF x LCM = 6 x 3024 = 18144 -------- equation 2.

9) From equation 1 & 2, we can say that HCF(336, 54) x LCM(336, 54) = 336 x 54.

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

Explanation:

1) Find all prime factors of the given numbers. 

2) HCF means Highest Common Factor, so find the product of all smallest power of common factor in the numbers.

3) LCM means Least Common Multiple, so find the product of the greatest power of each prime factor in the numbers.

Solution:

(i) 12, 15 and 21

1) Now we will find the prime factors of 12.

12 = 2 x 6

12 = 2 x 2 x 3

2) Now we will find the prime factors of 15.

15 = 3 x 5

3) Now we will find the prime factors of 21.

21 = 3 x 7

4) Here common factor is 3.

5) So HCF of 12, 15, and 21 is 3.

6) Simmilarly, LCM = 2 x 2 x 3 x 5 x 7 = 420.

7) LCM = 420.

(ii) 17, 23 and 29

1) Now we will find the prime factors of 17.

17 = 1 x 17

2) Now we will find the prime factors of 23.

23 = 1 x 23

3) Now we will find the prime factors of 29.

29 = 1 x 29

4) Here common factor is 1.

5) So HCF of 17, 23, and 29 is 1.

6) Simmilarly, LCM = 1 x 17 x 23 x 29 = 11339.

7) LCM = 11339.

(iii) 8, 9 and 25

1) Now we will find the prime factors of 8.

8 = 1 x 2 x 4

8 = 1 x 2 x 2 x 2

2) Now we will find the prime factors of 9.

9 = 1 x 3 x 3

3) Now we will find the prime factors of 25.

25 = 1 x 5 x 5

4) Here common factor is 1.

5) So HCF of 8, 9, and 25 is 1.

6) Simmilarly, LCM = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800.

7) LCM = 1800.

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

Explanation:

1) We know that HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

2) Out of the above 3 values, if 2 values are given, then we can find the 3rd value.

Solution:

1) Here HCF (306, 657) = 9.

2) So, HCF of 306 and 657 = 9.

3) Product 306 and 657 is 306 x 657

4) We know that HCF (p, q) × LCM (p, q) = p × q

So, [LCM (306, 657)] = [306 x 657] / HCF (306, 657)

= [306 x 657] / 9

= [34 x 657]

= [22338]

5) So, LCM (306, 657) = 22338.

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

Explanation:

1) If any number ends with the digit 0, then that number must be divisible by 2 and 5 simultaneously.

2) To check whether 6ⁿ ends with 0, we must find whether 6ⁿ has 2 and 5 as prime factors or not.

Solution:

1) Factors of 6ⁿ = (2 x 3)ⁿ  

2) Here 2 is the factor of 6ⁿ, but 5 is not a factor of 6ⁿ.

3) So 6ⁿ can't end with 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.

Explanation:

1) Prime numbers have only 2 factors. 1 and the number itself. e.g. the prime number 5 has two factors. 1 and 5 itself.

2) The positive Number which has factors other than 1 and itself is known as a composite number.

Solution:

1) The first expression is given as (7 × 11 × 13) + 13. We will simplify it.

(7 × 11 × 13 + 13) = 13 ((7 x 11 x 1) + 1)

= 13 (77 + 1) 

= 13 (78)

= 13 (2 x 39)

= 13 (2 x 3 x 13)

2) The first expression has factors as (2 x 3 x 13 x 13)

3) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite number.

4) The second expression is given as (7 × 11 × 13) + 13. We will simplify it.

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

= 5 (1008 + 1) 

= 5 (1009)

5) The second expression has factors as (5 x 1009)

6) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is 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?

Explanation:

1) Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round. 

2) Now, we need to find out how many minutes will they meet again at the same point. 

3) Here we will find LCM of 18 and 12 to get the time when both meet again at the starting point.

Solution:

1) We will find LCM of 18 and 12

2) Now we will find the prime factors of 18.

18 = 2 x 9

18 = 2 x 3 x 3

3) Now we will find the prime factors of 12.

12 = 2 x 6

12 = 2 x 2 x 3

4) Here common factor is 2 x 3 = 6.

5) Simmilarly, LCM = 2 x 2 x 3 x 3 = 36.

6) LCM = 36

7) After 36 minutes Sonia and Ravi will meet again at the starting point.

#simple method

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