NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers - Exercise 1.2

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

(i) $140$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

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

$\therefore 140={{2}^{2}}\times 5\times 7$ 

Therefore, the prime factors of $140$ are $2,5,7$.

(ii) $156$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

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

$\therefore 156={{2}^{2}}\times 3\times 13$ 

Therefore, the prime factors of $156$ are $2,3,13$.

(iii) $3825$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 3825=3\times 3\times 5\times 5\times 17$ 

$\therefore 3825={{3}^{2}}\times {{5}^{2}}\times 17$ 

Therefore, the prime factors of $3825$ are $3,5,17$.

(iv) $5005$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

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

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

Therefore, the prime factors of $5005$ are $5,7,11,13$.

(v) $7429$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 7429=17\times 19\times 23$ 

$\therefore 7429=17\times 19\times 23$ 

Therefore, the prime factors of $7429$ are $17,19,23$.

2. Find the LCM and HCF of the following pairs of integers and verify that $LCM\times HCF=\text{Product of two numbers}$.

(i) $26$ and $91$ 

Ans: First we write the prime factors of $26$ and $91$. We get

$ 26=2\times 13$ and 

$91=7\times 13$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $26$ and $91$ is $13$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $26$ and $91$ will be

$ 2\times 7\times 13=182$ 

Therefore, the LCM of $26$ and $91$ is $182$.

Now, the product of two numbers is 

$ 26\times 91=2366$ 

Product of LCM and HCF is

$ 13\times 182=2366$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

(ii) $510$ and $92$

Ans: First we write the prime factors of $510$ and $92$. We get

$510=2\times 3\times 5\times 17$ and 

$92=2\times 2\times 23$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $510$ and $92$ is $2$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $510$ and $92$ will be

$2\times 2\times 3\times 5\times 17\times 23=23460$ 

Therefore, the LCM of $510$ and $92$ is $23460$.

Now, the product of two numbers is 

$510\times 92=46920$ 

Product of LCM and HCF is

$2\times 23460=46920$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

(iii) $336$ and $54$

Ans: First we write the prime factors of $336$ and $54$. We get

\[336=2\times 2\times 2\times 2\times 3\times 7\] and 

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

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $336$ and $54$ is $2\times 3=6$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $336$ and $54$ will be

$ 2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024$ 

Therefore, the LCM of $336$ and $54$ is $3024$.

Now, the product of two numbers is 

$336\times 54=18144$ 

Product of LCM and HCF is

$ 6\times 3024=18144$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) $12,15$ and $21$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $12,15$ and $21$ are as follows:

\[12=2\times 2\times 3\] 

\[15=3\times 5\] and 

$21=3\times 7$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $12,15$ and $21$ is $3$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $12,15$ and $21$ will be

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

Therefore, the LCM of $12,15$ and $21$ is $420$.

(ii) $17,23$ and $29$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $17,23$ and $29$ are as follows:

\[17=17\times 1\] 

\[23=23\times 1\] and 

$29=29\times 1$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $17,23$ and $29$ is $1$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $17,23$ and $29$ will be

$\times 23\times 29=11339$ 

Therefore, the LCM of $17,23$ and $29$ is $11339$.

(iii) $8,9$ and $25$ 

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $8,9$ and $25$ are as follows:

\[8=2\times 2\times 2\] 

\[9=3\times 3\] and 

$25=5\times 5$

Now, we know that HCF is the highest factor, among the common factors of two numbers. as there is no common factor.

Therefore, the HCF of $8,9$ and $25$ is $1$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $8,9$ and $25$ will be

$2\times 2\times 2\times 3\times 3\times 5\times 5=1800$ 

Therefore, the LCM of $8,9$ and $25$ is $1800$.

4. Given that HCF $\left( 306,657 \right)=9$, find LCM $\left( 306,657 \right)$.

Ans: We have been given the HCF of two numbers $\left( 306,657 \right)=9$.

We have to find the LCM of $\left( 306,657 \right)$.

Now, we know that $LCM\times HCF=\text{Product of two numbers}$

Substitute the values, we get

$LCM\times 9=306\times 657$

$\Rightarrow LCM=\dfrac{306\times 657}{9}$ 

$\therefore LCM=22338$ 

Therefore, the LCM of $\left( 306,657 \right)=22338$.

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

Ans: We have to check whether ${{6}^{n}}$ can end with the digit $0$ for any natural number $n$.

By divisibility rule we know that if any number ends with the digit $0$, it is divisible by $2$ and $5$.

Thus, the prime factors of ${{6}^{n}}$ is

$ {{6}^{n}}={{\left( 2\times 3 \right)}^{n}}$

Now, we will observe that for any value of $n$, ${{6}^{n}}$ is not divisible by $5$.

Therefore, ${{6}^{n}}$ cannot end with the digit $0$ for any natural number $n$.

6. Explain why $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ are composite numbers.

Ans: The given numbers are $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$.

We can rewrite the given numbers as

$7\times 11\times 13+13=13\times \left( 7\times 11+1 \right)$

$\Rightarrow 7\times 11\times 13+13=13\times \left( 77+1 \right)$ 

$\Rightarrow 7\times 11\times 13+13=13\times 78$

$\Rightarrow 7\times 11\times 13+13=13\times 13\times 6$

And, 

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

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 1008+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times 1009$

Here, we can observe that the given expression has its factors other than $1$ and the number itself.

A composite number has factors other than $1$ and the number itself.

Therefore, $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ are composite numbers.

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?

Ans: It can be observed that Ravi takes less time than Sonia for completing the $1$ round of the circular path. Both are going in the same direction, they will meet again when Ravi will have completed $1$ round of that circular path with respect to Sonia. 

The total time taken for completing this $1$ round of circular path will be the LCM of time taken by Sonia and Ravi for ending the $1$ round of circular path respectively, i.e., LCM of $18$ minutes and $12$ minutes.

The prime factors of $12$ and $18$ are as follows:

\[12=2\times 2\times 3\]  and 

$18=2\times 3\times 3$

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $12$ and $18$ will be

$2\times 2\times 3\times 3=36$

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

Important Topics under NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.2) Exercise 1.2

Real Numbers is the first chapter of the class 10 maths syllabus. It is a significant chapter in maths that is covered in class 10. The chapter on Real Numbers has 4 major parts that need to be read through and understood properly to get a good grasp on the topic. 

The following is a list of the 4 important topics that are covered under the chapter Real Numbers. It is recommended that students go through these topics carefully to get a hold of the concepts without any confusion.

• Introduction to Real Numbers

• Fundamental Theorem of Arithmetic (H.C.F. and L.C.M.) 

• Euclid’s Division Lemma

• Revisiting Irrational Numbers

Importance of  NCERT Class 10 Maths Chapter 1 Real Numbers

Real numbers comprise both rational and irrational numbers. Rational numbers include integers, decimals, and fractions, while irrational numbers are numbers like root overs, pi (22/7), and so on. In short, real numbers are all numbers excluding imaginary numbers. 

Real numbers are important due to their use in almost every sphere of mathematics and in real life. We encourage students to learn as much as they can from this chapter on real numbers to grow a good number sense and be able to solve all the problems that rely on the use of real numbers in their exams.

NCERT Solutions for Class 10 Maths All Chapters

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Class 10 Maths Chapter 1 - Exercise 1.2

Importance of Preparing Real Numbers from NCERT Solutions for Class 10 Maths Exercise 1.2

The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers covers the important aspects of prime numbers to be able to find the HCF & LCM of given pairs and the integers that will allow users to find the product of any two numbers. Further, when students prepare from NCERT Solutions Class 10 Maths Ch 1 Ex 1.2, they uncover the different aspects of numbers and how they can be used to solve the study of numbers. Vedantu’s analytical approach to Class 10 Maths Exercises 1.2 helps students prepare in a full-fledged manner for their upcoming exams and also understand the concept clearly.  Along with Exercise 1.2 we also covered all the solutions for Exercise 1.1, Exercise 1.3 and Exercise 1.4

NCERT Solutions for Class 10 Maths Chapter 1 All Exercises

Practical Problems from Maths Class 10 Chapter 1 Exercise 1.2

Question 1: Prove that the positive odd integer is of the suggested form, 6q+1 or 6q+3, or 6q+5, making q the integer part.

Solution: 

Let’s assume that a is any given positive integer and b=6. Now, we take the approach of Euclid’s algorithm a= 6q +r for some of the given integers and q ≥ 0, or r = 0, 1, 2, 3, 4, 5 because of 0 ≤ r < 6.

Therefore, a = 6q or 6q+1 or 6q+2, 6q+3, 6q+4, 6q+5. Moreover, 6q+1 = 2x 3q +1 = 2k1+ 1, where  k1 is a positive integer.

6q+3 = (6q+2) + 1= 2 (3q + 1) + 1= 2 k2 +1, in this situation, k2 is an integer 

6q+5 = (6q+4) + 1= 2 (3q + 2) + 1= 2 k3 + 1, in this situation, k3 is an integer

6q+1, 6q+3, 6q+5 are of form 2k+1, where the product k is an integer.

Hence, 6q+1, 6q+3, 6q+5 are not precisely divisible by 2.

Lastly, we can conclude that the given expressions are odd numbers and can be from 6q+1 or 6q+3, or 6q+5.

Question 2: 

An army unit comprises 616 members, who are ordered to march behind a band of 32 parades. At the time of their marching, these 2 groups are marching at the same columns. You need to find out the maximum number of columns in which the parade can march.

Solution:

Given, HCF = (616, 32) is the maximum number of columns in which the army can march, Now, we put into place Euclid's algorithm to find the suggested HCF. 

By calculations, 

616 = 32 x 9 8

32 = 8 x 4 + 0

Therefore, we can conclude that the HCF (616, 32) is 8. 

Conclusion: The army can march into 8 maximum columns

Did You Know?

A real number can be any number that is generally denoted by the symbol R along the number line. It can be zero, negative, rational, irrational or positive integer. This chapter of Class 10 Maths Exercise 1.2 helps students to understand the fundamental and intricate concepts of real numbers.