NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 - 2025-26

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.

Conclusion

Exercise 1.1 in Chapter 1 of the Class 10 NCERT textbook focuses on the practical application of Euclid’s Division Lemma to find the highest common factor (HCF) of two numbers. This exercise contains seven questions, providing students with the best practice to understand and apply this fundamental concept. It is important for students to focus on the method of using Euclid’s Division Lemma iteratively, as this strengthens their problem-solving skills and lays the groundwork for more advanced topics in number theory. In previous years, questions from this exercise frequently appeared in exams, highlighting its significance. By mastering this exercise, students will build a strong foundation in arithmetic that is crucial for their further studies in mathematics.

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