Real Numbers (Chapter 1, Exercise 1.2) – Class 10 Maths Solutions NCERT

The Chapter 1 of Class 10 Mathematics based on NCERT syllabus is about the Real Numbers. This page contains the Solution of Exercise 1.2, Chapter 1 of Class 10 Mathematics. The Exercise 1.2 of Class 10 Mathematics consists of 7 questions which are based on The Fundamental Theorem of Arithmetic.

DescriptionNCERT Maths Solutions
SubjectMathematics
Class10 (X)
Chapter1
Exercise1.2
TopicFundamental Theorem of Arithmetic
SyllabusNCERT

The solution of all the 7 questions of Exercise 1.2 of Class 10, Chapter 1 is given below.

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

i. 140

Solution:

The prime factors of 140 are

140=2\times 2\times 5\times 7

\Rightarrow 140=2^2\times 5\times 7

ii. 156

Solution:

The prime factors of 156 are

156=2\times 2\times 3\times 13

\Rightarrow 156=2^2\times 3\times 13

iii. 3825

Solution:

The prime factors of 3825 are

3825=3\times 3\times 5\times 5\times 17

\Rightarrow 3825=3^2\times 5^2\times 17

iv. 5005

Solution:

The prime factors of 5005 are

5005=5\times 7\times 11\times 13

v. 7429

Solution:

The prime factors of 7429 are

7429=17\times 19\times 23

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

Solution:

The prime factors of 26 are

26=2\times 13

The prime factors of 91 are

91=7\times 13

Therefore,

HCF=13

LCM=2\times 7\times 13=182

Verification:

We know that

LCM\times HCF=a\times b

\therefore LHS=LCM\times HCF

\Rightarrow LHS=183\times 13

\Rightarrow LHS=2366

Now

RHS=a\times b

\Rightarrow RHS=26\times 91

\Rightarrow RHS=2366

Therefore, LHS=RHS

Hence verified.

ii. 510 and 92

Solution:

The prime factors of 510 are

510=2\times 3\times 5\times 17

The prime factors of 92 are

92=2\times 2\times 23

Therefore,

HCF=2

LCM=2\times 2\times 2\times 3\times 3\times 5\times 17=23460

Verification:

We know that

LCM\times HCF=a\times b

\therefore LHS=LCM\times HCF

\Rightarrow LHS=23460\times 2

\Rightarrow LHS=46920

Now

RHS=a\times b

\Rightarrow RHS=510\times 92

\Rightarrow RHS=46920

Therefore, LHS=RHS

Hence verified.

iii. 336 and 54

Solution:

The prime factors of 336 are

336=2\times 2\times 2\times 2\times 3\times 7

The prime factors of 54 are

54=2\times 3\times 3\times 3

Therefore,

HCF=2\times 3=6

LCM=2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024

Verification:

We know that

LCM\times HCF=a\times b

\therefore LHS=LCM\times HCF

\Rightarrow LHS=3024\times 6

\Rightarrow LHS=18144

Now

RHS=a\times b

\Rightarrow RHS=336\times 54

\Rightarrow RHS=18144

Therefore, LHS=RHS

Hence verified.

Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method.

i. 12, 15 and 21

Solution:

The prime factorisation of 12 is

12=2\times 2\times 3

The prime factorisation of 15 is

15=3\times 5

And the prime factorisation of 21 is

21=3\times 7

Therefore,

HCF=3

LCM=2\times 2\times 3\times 5\times 7=420

ii. 17, 23 and 29

Solution:

All the three numbers 17, 23 and 29 are prime numbers.

Therefore,

HCF=1 and

LCM=17\times 23\times 29=11339

iii. 8, 9 and 25

Solution:

The prime factorisation of 8 is

8=2\times 2\times 2

The prime factorisation of 9 is

9=3\times 3

And the prime factorisation of 25 is

25=5\times 5

Therefore,

HCF=1 LCM=8\times 9\times 25=1800

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

Solution:

Given,

a=306

b=657

And HCF=9

We know that,

LCM\times HCF =a\times b

\Rightarrow LCM=\frac{a\times b}{HCF}

\Rightarrow LCM=\frac{306\times 657}{9}

\Rightarrow LCM=22338

Therefore, LCM (306,\, 657) is 22338.

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

Solution:

6^n can be written as

6^n=(2\times 3)^n

6^n=2^n\times 3^n

We know that, any number that ends with 0 is divisible by 5. Therefore, the prime factors of such number must have 5.

Here, the factors of 6^n does not have 5. Therefore, the number 6^n cannot end with 0.

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

Solution:

The first number

7\times 11\times 13+13

=(7\times 11\times 13)+13

=13\times (7\times 11+1)

=13\times 78

=2\times 3\times 13\times 13

The second number

7\times 6\times 5\times 4\times 3\times 2\times 1+5

=5\times (7\times 6\times 4\times 3\times 2\times 1+1)

=5\times (1008+1)

=5\times 1009

In both the above numbers, all the factors are prime numbers. Therefore, the given numbers are composite numbers.

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?

Solution:

Time taken by Sonia=18 minutes

Time taken by Ravi=12 minutes

The time after which Sonia and Ravi will meet at the starting point is the LCM of 18 and 12.

The prime factorisation of 18 is

18=2\times 3\times 3

The prime factorisation of 12 is

12=2\times 2\times 3

\therefore LCM=2\times 2\times 3\times 3=36

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

I hope the solutions of Class 10, Exercise 1.2, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.2, feel free to contact me at [email protected] or [email protected] or fill the form here.

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