# 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.

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.