Solve the followings Questions.

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

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer:

(i) 140 = 2 x 2 x 5 x 7 = 2² x 5 x 7

(ii) 156 = 2 x 2 x 3 x 13 = 2² x 3 x 13

(iii) 3825 = 3 x 3 x 5 x 17 = 3² x 5² x x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 7429 = 17 x 19 x 23 

2. Find the LCM and HCF of the following pairs of integers and verify that LCM xHCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer:

(i) 26 and 91

a = 26, b = 91

∴ H.C.F = 13

L.C.M = 2 × 7 × 13

= 14 × 13 = 182

∴ H.C.F × L.C.M = a × b

13 × 182 = 26 × 96

2366 = 2366

(ii) 510 and 92

a = 510, b = 92

(iii) 336 and 54

a = 336, b = 54Read more on Sarthaks.com – https://www.sarthaks.com/661852/find-the-lcm-and-the-following-pairs-integers-and-verify-that-lcm-hcf-product-the-two-numbers

336= 2 x 2 x 2 x 2 x 3 x 7 = 2⁴ x 3 x 7

54 = 2 x 3 x 3 x 3 = 2 x 3³

HCF = 2 x 3 = 6

LCM = 2⁴ x 3³ x 7= 3024

Product of two numbers 336 and 54 = 336 x 54= 18144

3024 x 6= 18144

Hence, product of two numbers = 18144

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

12 = 2 × 2 × 3 = 2²× 3

15 = 3 × 5

and 21 = 3 × 7

For HCF, we find minimum power of prime factor

H.C.F. = (3)¹= 3

For LCM, taking maximum power of prime factors

L.C.M. = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

So,H.C.F. = (3)¹= 3

and L.C.M. =  420

(ii) 17, 23 and 29

17 = 1 × 17

23 = 1 x 23

29 = 1 x 29

For HCF, common factor is 1

HCF = 1

For LCM taking maximum power of prime factor.

L.C.M. = 1 × 17 × 23 × 29 = 11339

So H.C.F. = 1

L.C.M. = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2 × 1 = 2³× 1

9 = 3 × 3 = 3²

and 25 = 5 × 5 = 5²

For HCF common factor is 1

H.C.F. = 1

For LCM, taking maximum power of prime factors

L.C.M. = 2³× 3² × 5²

= 8 × 9 × 25 = 1800

So H.C.F. = 1

L.C.M. = 1800

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

Answer:

HCF (306, 657) = 9

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

L.C.M x H.C.F = first Number x Second Number

L.C.M x 9 = 306 x 657

LCM = 22338

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

Answer:

TO CHECK: Whether 6²can end with the digit 0 for any natural number n.

We know that

6² = (2 × 3)ⁿ

6² = (2)ⁿ ×(3)ⁿ

Therefore, prime factorization of 6ⁿdoes not contain 5 and 2 as a factor together.

Hence 6ⁿcan never end with the digit 0 for any natural number n

6. Explain why 7 x 11 x 13  and  7 x 6 x 5 x 3 x 2 x 1 + 5 are composite numbers.

Answer:

So, the given expression has 6 and 13 as its factors. Therefore, we can conclude that it is a composite number.

Similarly,

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1) [taking 5 out- common]

= 5 x (1008 + 1)

= 5 x 1009

Since, 1009 is a prime number the given expression has 5 and 1009 as its factors other than 1 and the number itself.

Hence, it is also 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?

Answer:

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

LCM of 12 and 18= 2 x 2 x 3 x 3 = 36

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

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2