Class 10 Maths Chapter 1 Exercise 1.2 Solutions

Here you will find Class 10 Maths Chapter 1 Exercise 1.2 Solutions that are prepared by our experienced teachers. in the previous classes, you must have heard about numbers. In this class 10th maths ch 1 ex 1.2, you can read all questions with a complete solution. This exercise 1.2 class 10 maths is important for CBSE, RBSE, and other board students.

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Maths Class 10 Chapter 1 Exercise 1.2 Question 1

Q1. Express each number as a product of its Prime factor.

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

Ex 1.2 class 10 maths question 1 solution

(i)Prime factors of 140 = 2× 70

`\2|140/`

`\2|70/`

`\5|35/`

`\7|7/1`

Hence

=2×2×5×7

`=2^2\times5times7`

(ii) Prime factors of 156 = 2×78

`\2|156/`

`\2|78/`

`\3|39/`

`\13|13/1`

Hence

= 2×2×3×13

`=2^2×3×13`

(iii) Prime factors of 3825 =3 ×1275

`\3|3825/`

`\3|1275/`

`\5|425/`

`\5|85/`

`\17|17/1`

Hence = 3×3×5×5×17

`= 3^2×5^2×17`

(iv) Prime factors of 5005= 5 ×1001

`\5|5005/`

`\7|1001/`

`\11|143/`

`\13|13/1`

Hence

= 5×7×11×13

(v) Prime factors of 7429 = 17× 437

`\17|7429/`

`\19|437/`

`\23|23/1`

Hence

= 17 ×19× 23

Maths Class 10 Chapter 1 Exercise 1.2 Question 2

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

Ex 1.2 class 10 maths question 2 solution

(i)26 and 91

Prime factors of 26=2 ×13

Prime factors of 91 =7×13

∴ LCM of 26 and 91 =2×7×13=182

And HCF of 26 and 91 = 13

Verification

Product of LCM and HCF= 182×13=2366

Product of given numbers = 26×91= 2366

Hence, LCM ×HCF = product of 26 and 91

2366=2366

Hence Proved

(ii) 510 and 92

Prime factors of 510= 2 255

= 2×3×5×17

Prime factors of 92 = 2 ×46

=2×2×23

So LCM(510, 92)=2×2×3×5×17×23=23460

HCF (510,92)= 2

Verification

Product of 510 and 92 = 46920

Product of LCM and HCF= 2×23460= 46920

Hence, LCM × HCF= Product of 510 × 92

46920=46920

Hence Proved

(iii) 336 and 54

prime factors of 336 =2 ×168

= 2×2×2×2×3×7

prime factors of 54=2×27

=2×3×3×3

LCM(336,54)=`2^4×3^3×7=3024`

HCF(336,54)= 2×3=6

Verification

Product of 336 and 54= 18144

Product of LCM and HCF = 3024×6=18144

Hence, LCM× HCF= Product of 336 and 54

18144= 18144

Hence Proved

Maths Class 10 Chapter 1 Exercise 1.2 Question 3

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

Ex 1.2 class 10 maths question 3 solution

(i) 12,15 and 21

Prime factors of 12 =2×2×3

Prime factors of 15= 3×5

Prime factors of 21= 3×7

Therefore

LCM(12,15,and 21)=`2^2×3×5×7=420`

HCF (12,15 and 21)=3

(ii) 17 ,23 and 29

Prime factors of 17= 1×17

Prime factors of 23= 1×23

Prime factors of 29= 1× 29

Therefore

LCM(17,23,and 29)= 17×23×29=11339

HCF(17,23,and 29)=1

(iii) 8 ,9 and 25

Prime factors of 8= 2×2×2

Prime factors of 9= 3×3

Prime factors of 25=5×5

Therefore

LCM(8,9,and 25) =2×2×2×3×3×5×5=1800

HCG(8,9,and 25) = 1

Maths Class 10 Chapter 1 Exercise 1.2 Question 4

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

Ex 1.2 class 10 maths question 4 solution

According to the question, the numbers are 306 and 657

a=306

b=657

And HCF=9 (given)

We know that

`L.C.M.=\frac{a\ \times\ b}{H.C.F.}`

`=\frac{306\ \times\ 657}{9}=\ 34\times\ 657\ =\ 223338`

Hence , L.C.M. (306,657)= 223338

Maths Class 10 Chapter 1 Exercise 1.2 Question 5

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

Ex 1.2 class 10 maths question 5 solution

Suppose that for any natural number n, `n\in\ N,\ 6^n` ends with the digit 0.

Hence, `6^n` will be divisible by 5.

But prime factors of 6=2×3

therefore, The prime factors of `6^n\ = (2×3)`n

It is clear that there is no place of 5 in the prime factors of `6^n`

By the Fundamental theorem of Algorithm, we know that every composite number can be factorised as a product of prime numbers and this factorization is unique i.e. our hypothesis assured, in the beginning, is wrong.

Hence there is no natural number n for which `6^n` ends with the digit 0.

Maths Class 10 Chapter 1 Exercise 1.2 Question 6

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

Ex 1.2 class 10 maths question 6 solution

According to the question

7×11×13+ 13=13(7×11+1)

Taking 13 as a common factor, we get

= 13(77+1)

=13×78

=13×3×2×13

Hence 7×11×13+13 is a composite number.

Now, for the other number

7×6×5×4×3×2×1+5

Taking 5 as a common factor, we get

=5(7×6×5×4×3×2×1+1)

=5(1008+1)

= 5 ×1009

Hence, 7×6×5×4×3×2×1+5 is a composite number.

Maths Class 10 Chapter 1 Exercise 1.2 Question 7

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?

Ex 1.2 class 10 maths question 7 solution

Time taken by Sonia to drive one round of the field= 18 min.

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

To find after how many minutes they will meet again, we will have to find LCM of 18 and 12

Prime factors of 18= 2×9

`\2|18/`

`\3|9/`

`\3|3/1`

= 2×3×3

= 2×3²

Prime factors of 12= 2×6

`\2|12/`

`\2|6/`

`\3|3/1`

= 2×2×3

= 2² ×3

Taking the product of the greatest power of each prime factor of 18 and 12

LCM(18,12)= 2²×3²

= 4 ×9

= 36

So Sonia and Ravi will meet again at the starting point after 36 min.

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Class 10 Maths Chapter 1 Exercise 1.2 Solutions