NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1

For Exercise 1.1 in Chapter 1 of the Class 10 Maths Real Numbers, preparation should focus on mastering the prime factorisation method and calculating the HCF and LCM of numbers. Start by thoroughly practicing the problems where numbers are expressed as products of prime factors. Familiarise yourself with the steps involved in determining the HCF and LCM using the prime factorisation method, as this is crucial for solving related questions in exams. Ensure to verify your solutions by confirming that the product of the HCF and LCM equals the product of the given numbers. Practicing these types of problems repeatedly will build speed and accuracy, which are essential for exams.

Pay special attention to word problems, such as the one where two individuals start at the same point and you need to calculate when they will meet again using LCM. Such problems often appear in exams, testing both your conceptual understanding and problem-solving skills. Make a list of important formulas, like the relation between HCF and LCM, and revise them regularly. Solving sample papers and previous year questions will help you get used to the type of questions asked and the level of difficulty.

Express each number as product of its prime factors:

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

Solution:
(i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
Discuss this question in your opinion.

(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
If you have any other suggestion, please suggest us.
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
More options related to above question.

(iv) 5005 = 5 × 7 × 11 × 13
if you have any other idea, please comment here.

(v) 7429 = 17 × 19 × 23
What do you think about 10th Maths Exercise 1.1 Part v solutions?

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

Solution:
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Product of the two numbers = 26×91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence, product of two numbers = HCF × LCM
Do you have more ideas about this solutions?

(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of the two numbers = 510 × 92 = 46920
HCF × LCM = 2 × 23460 = 46920
Hence, product of two numbers = HCF × LCM
You can discuss this question with us.

(iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Product of the two numbers = 336 × 54 = 18144
HCF × LCM = 6 × 3024 = 18144
Hence, product of two numbers = HCF × LCM
Comment about the solution here.

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

Solution:
(i) 12, 15 and 21
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 420
For more suggestion, see the solution here.

(ii) 17, 23 and 29
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
Prove more ways to solve this question.

(iii) 8, 9 and 25
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
25 = 5 × 5 = 5²
HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
How can we solve this question alternatively?

Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9
We know that,
LCM × HCF = Product of two numbers
Therefore,
LCM= (Product of two numbers)/HCF = (306 × 657)/9 = 22338
Hence, LCM (306, 657) = 22338

The alternative way to solve this question.

Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ
According to Fundamental Theorem of Arithmetic, the factorisation is uniqe. It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.

Put your opinion about question 5 of 10th Maths exercise 1.1

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that:
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 × 13 × 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
Now: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

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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:
It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 × 3 × 3
And, 12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

What do you think about the solution of question 7 given in Ex. 1.1 of 10th Maths.

NCERT Solutions for class 10 Maths Chapter 1 Exercise 1.1 Real Numbers in English medium as well as Hindi Medium is given below to free download in PDF format for 2024-25. IF YOU WANT TO DOWNLOAD, THE LINK IS ALSO GIVEN or visit NCERT Solutions for Class 10 Maths Chapter 1 to download other exercises.

1. In a seminar, the number of participants in English, Maths and Science are 175, 140 and 105, respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject. [Answer: 12]
2. Find the value of m if HCF of 65 and 117 is expressible in the form 65m – 117. [Answer: 2]
3. What can you say about the product of a non-zero rational and irrational number? [Answer: Irrational]
4. In a school there are two sections of class X – section A and section B. There are 48 students in sections A and 54 students in section B. Determine the least number of books required for the library of the school so that the books can be distributed equally among students of section A or section B? [Answer: 216]
5. After how many places the decimal expansion of 13497/1250 will terminate? [Answer: 4]
6. Find the least number which is divisible by all numbers from 1 to 10 (both inclusive). [Answer: 2520]

Practice Questions on Real Numbers

1. A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48 km, 60 km and 72 km a day, round the field. After how many days will they meet again at the starting point? [Answer: 30 days]
2. The numbers 525 and 3000 are divisible by 3, 5, 15, 25 and 75 what is the HCF of 525 and 3000? [Answer: 75]
3. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons. [Answer: No]
4. If a = 4q + r then what are the condition for a and q? What are the values that r can take? [Answer: a and q are positive integer 0 ≤ r ≤ 4]
5. What is the digit at unit’s place of 9^n? [Answer: Even power = 1, odd power = 9]
6. If n is an odd integer then show that n² – 1 is divisible by 8.
7. Use Euclid’s division algorithm to find the HCF of 16 and 28. [Answer: 4]

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