NCERT Solutions for Class 10 Maths Chapter 1 PDF Download

If a student wants to score good marks in their exams, then they must go through the NCERT Solutions for Class 10 Maths Chapter 1. These solutions are one of the best and trusted Maths solutions among the students. Each and every type of questions is included in this book, from easy to intermediate to advanced which helps the students to be aware about the type of questions which will appear in the exam and overall exam format.

The NCERT Solutions for Class 10 Maths Chapter 1 helps all the students in doing an effective exam preparation for all the students. These solutions are created by highly qualified subject matter experts who have years of experience in the field of education. Apart from being the best solutions for exam preparation, these solutions can also act as a great tool for revision which can be a complete game changer for all the students. 

These NCERT Solutions Class 10 Chapter 1 can be downloaded by all the students from the website of Selfstudys i.e. Selfstudys.com. In this article below, we will discuss a lot of things about these solutions including how you can download them, their features, benefits, tips to study them and frequently asked questions.

NCERT Solutions for Class 10 Maths Chapter 1 Exercise Wise

All the Chapter 1 exercise-wise questions are given in this section so that you can refer to them and practice online. All the Class 10 Maths Chapter 1 exercises are solved by subject matter experts so, feel free to use these solutions to clear your doubts.

Class 10 Maths Chapter 1 Ex 1.1

Euclid’s Division Lemma: Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r,0 ≤ r < b.

Question 1:

Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225     (ii) 196 and 38220        (iii) 867 and 255

Answer 1: (i) 135 and 225:

• Divide 225 by 135: 225 = 1 * 135 + 90
• Divide 135 by 90: 135 = 1 * 90 + 45
• Divide 90 by 45: 90 = 2 * 45 + 0

Since the remainder becomes 0, the divisor at this step, which is 45, is the HCF of 135 and 225.

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220:

Step 1: Divide 38220 by 196: 38220 = 195 * 196 + 0

Since the remainder becomes 0, the divisor at this step, which is 196, is the HCF of 196 and 38220.

Therefore, the HCF of 196 and 38220 is 196.

(iii) 867 and 255:

Step 1: Divide 867 by 255: 867 = 3 * 255 + 102

Step 2: Divide 255 by 102: 255 = 2 * 102 + 51

Step 3: Divide 102 by 51: 102 = 2 * 51 + 0

Since the remainder becomes 0, the divisor at this step, which is 51, is the HCF of 867 and 255.

Therefore, the HCF of 867 and 255 is 51.

To summarize: (i) The HCF of 135 and 225 is 45. (ii) The HCF of 196 and 38220 is 196. (iii) The HCF of 867 and 255 is 51.

Question 2: 

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer 2:

To show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer, we can consider the possible remainders when dividing positive odd integers by 6.

Let's consider three cases based on the possible remainders:

• Case 1: Remainder 1: If we divide an odd integer by 6 and the remainder is 1, we can express it as 6q + 1. For example, 7 = 6(1) + 1, 13 = 6(2) + 1, 19 = 6(3) + 1, and so on.
• Case 2: Remainder 3: If we divide an odd integer by 6 and the remainder is 3, we can express it as 6q + 3. For example, 9 = 6(1) + 3, 15 = 6(2) + 3, 21 = 6(3) + 3, and so on.
• Case 3: Remainder 5: If we divide an odd integer by 6 and the remainder is 5, we can express it as 6q + 5. For example, 11 = 6(1) + 5, 17 = 6(2) + 5, 23 = 6(3) + 5, and so on.

In all three cases, we have covered all possible remainders when dividing positive odd integers by 6. Thus, any positive odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

This can be further verified by taking any positive odd integer and dividing it by 6, confirming that it falls into one of the three cases mentioned above.

Question 3: 

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Answer 3: To determine the maximum number of columns in which the army contingent and the army band can march, we need to find the greatest common divisor (GCD) of 616 and 32. The GCD will represent the maximum number of columns.

Using the Euclidean algorithm, we can find the GCD as follows:

Step 1: Divide 616 by 32:
616 = 19 * 32 + 8

Step 2: Divide 32 by 8:
32 = 4 * 8 + 0

Since the remainder becomes 0, the divisor at this step, which is 8, is the GCD of 616 and 32.

Therefore, the maximum number of columns in which the army contingent and the army band can march is 8.

Class 10 Maths Chapter 1 Ex 1.2

Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

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

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

Answer 1: Let's express each of the given numbers as a product of their prime factors:

(i) 140:
The prime factorization of 140 can be found by dividing it by prime numbers starting from 2:

140 ÷ 2 = 70
70 ÷ 2 = 35
35 ÷ 5 = 7

Therefore, the prime factorization of 140 is 2 × 2 × 5 × 7.

(ii) 156:
The prime factorization of 156 can be found by dividing it by prime numbers starting from 2:

156 ÷ 2 = 78
78 ÷ 2 = 39
39 ÷ 3 = 13

Therefore, the prime factorization of 156 is 2 × 2 × 3 × 13.

(iii) 3825:
The prime factorization of 3825 can be found by dividing it by prime numbers starting from 2:

3825 ÷ 3 = 1275
1275 ÷ 5 = 255
255 ÷ 3 = 85
85 ÷ 5 = 17

Therefore, the prime factorization of 3825 is 3 × 5 × 5 × 17.

(iv) 5005:
The prime factorization of 5005 can be found by dividing it by prime numbers starting from 2:

5005 ÷ 5 = 1001
1001 ÷ 7 = 143
143 ÷ 11 = 13

Therefore, the prime factorization of 5005 is 5 × 7 × 11 × 13.

(v) 7429:
To find the prime factorization of 7429, we need to check divisibility by prime numbers:

7429 is not divisible by 2.
7429 ÷ 3 = 2476.33 (not divisible by 3).
7429 ÷ 5 = 1485.8 (not divisible by 5).
7429 ÷ 7 = 1059.857 (not divisible by 7).
7429 ÷ 11 = 675.364 (not divisible by 11).
7429 ÷ 13 = 571.462 (not divisible by 13).

Therefore, the prime factorization of 7429 is 7429 (since it is a prime number).

To summarize:
(i) The prime factorization of 140 is 2 × 2 × 5 × 7.
(ii) The prime factorization of 156 is 2 × 2 × 3 × 13.
(iii) The prime factorization of 3825 is 3 × 5 × 5 × 17.
(iv) The prime factorization of 5005 is 5 × 7 × 11 × 13.
(v) The prime factorization of 7429 is 7429.

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
• (ii)510 and 92
• (iii)336 and 54

Answer 2: Let's find the LCM and HCF of each pair of integers and verify if their product is equal to the LCM multiplied by the HCF:

(i) 26 and 91:
To find the LCM and HCF, we can use the prime factorization method.

Prime factorization of 26:
26 = 2 * 13

Prime factorization of 91:
91 = 7 * 13

HCF of 26 and 91:
The common factor between 26 and 91 is 13, which is their highest common factor (HCF).

LCM of 26 and 91:
The LCM can be calculated by multiplying the highest power of each prime factor that appears in the prime factorizations:
LCM = 2 * 7 * 13 = 182

Product of the numbers:
26 * 91 = 2366

Verification:
LCM × HCF = 182 × 13 = 2366
The product of the numbers matches the LCM multiplied by the HCF, so the verification is successful.

(ii) 510 and 92:
Prime factorization of 510:
510 = 2 * 3 * 5 * 17

Prime factorization of 92:
92 = 2 * 2 * 23

HCF of 510 and 92:
The common factors between 510 and 92 are 2 and 1, but the highest common factor (HCF) is 2.

LCM of 510 and 92:
The LCM can be calculated by multiplying the highest power of each prime factor that appears in the prime factorizations:
LCM = 2 * 2 * 3 * 5 * 17 * 23 = 93840

Product of the numbers:
510 * 92 = 46920

Verification:
LCM × HCF = 93840 × 2 = 187680
The product of the numbers does not match the LCM multiplied by the HCF, so the verification fails.

(iii) 336 and 54:
Prime factorization of 336:
336 = 2 * 2 * 2 * 2 * 3 * 7

Prime factorization of 54:
54 = 2 * 3 * 3 * 3

HCF of 336 and 54:
The common factors between 336 and 54 are 2 and 3, but the highest common factor (HCF) is 2 * 3 = 6.

LCM of 336 and 54:
The LCM can be calculated by multiplying the highest power of each prime factor that appears in the prime factorizations:
LCM = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 7 = 6048

Product of the numbers:
336 * 54 = 18144

Verification:
LCM × HCF = 6048 × 6 = 36288
The product of the numbers does not match the LCM multiplied by the HCF, so the verification fails.

To summarize:
(i) The HCF of 26 and 91 is 13, and the LCM is 182. LCM × HCF = 2366 (verified).
(ii) The HCF of 510 and 92 is 2, and the LCM is 93840. LCM × HCF ≠ 46920 (verification fails).
(iii) The HCF of 336 and 54 is 6, and the LCM is 6048. LCM × HCF ≠ 18144 (verification fails).

Question 3: 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

Answer 3:

Let's find the LCM and HCF of each set of integers using the prime factorization method:

(i) 12, 15, and 21:
Prime factorization of 12:
12 = 2 * 2 * 3

Prime factorization of 15:
15 = 3 * 5

Prime factorization of 21:
21 = 3 * 7

To find the HCF, we consider the common prime factors with the lowest power:
HCF = 3

To find the LCM, we consider the highest power of each prime factor:
LCM = 2 * 2 * 3 * 5 * 7 = 420

Therefore, the HCF of 12, 15, and 21 is 3, and the LCM is 420.

(ii) 17, 23, and 29:
Since all three numbers are prime, their prime factorization is simply the numbers themselves.

HCF = 1 (no common factor other than 1)
LCM = 17 * 23 * 29 = 11243

Therefore, the HCF of 17, 23, and 29 is 1, and the LCM is 11243.

(iii) 8, 9, and 25:
Prime factorization of 8:
8 = 2 * 2 * 2

Prime factorization of 9:
9 = 3 * 3

Prime factorization of 25:
25 = 5 * 5

To find the HCF, we consider the common prime factors with the lowest power:
HCF = 1 (no common prime factor)

To find the LCM, we consider the highest power of each prime factor:
LCM = 2 * 2 * 2 * 3 * 3 * 5 * 5 = 900

Therefore, the HCF of 8, 9, and 25 is 1, and the LCM is 900.

To summarize:
(i) The HCF of 12, 15, and 21 is 3, and the LCM is 420.
(ii) The HCF of 17, 23, and 29 is 1, and the LCM is 11243.
(iii) The HCF of 8, 9, and 25 is 1, and the LCM is 900.

Class 10 Maths Chapter 1 Ex 1.3

Class 10 Maths Chapter 1 Ex 1.4

Formulas of NCERT Solutions Class 10 Chapter 1

Without the use of formulas solving the Chapter 1 exercises and questions will be very difficult; therefore, here we have collected the Chapter 1 formulas for class 10 so that you can remember them and solve the questions with ease.

NCERT Class 10 Chapter 1 Questions with Solutions

Every single question of NCERT Class 10 Maths Chapter 1 is solved by Selfstudys Maths experts in an easy-to-understand language so that you can refer to them to solve your chapter-end questions and can understand the topic better. Chapter 1 questions with solutions are given here for your practice make sure you use them.

Question:
Find the highest common factor (HCF) of 36 and 48 using the prime factorization method.

Answer:
To find the HCF of 36 and 48, we will use the prime factorization method.

• Step 1: Prime factorization of 36
  • 36 = 2 * 2 * 3 * 3
• Step 2: Prime factorization of 48
  • 48 = 2 * 2 * 2 * 2 * 3
• Step 3: Identify common prime factors
  • The common prime factors between 36 and 48 are 2 and 3.
• Step 4: Multiply the common prime factors
  • 2 * 3 = 6

Therefore, the HCF of 36 and 48 is 6.

Question:
Find the highest common factor (HCF) of 72 and 90 using the Euclidean algorithm.

Answer:
To find the HCF of 72 and 90 using the Euclidean algorithm, we'll use the following steps:

• Step 1: Divide the larger number by the smaller number.
  • 90 ÷ 72 = 1 remainder 18
• Step 2: Replace the larger number with the remainder obtained in the previous step.
  • Now we have 72 and 18.
• Step 3: Repeat the division with the new numbers.
  • 72 ÷ 18 = 4 remainder 0

Step 4: The divisor in the last step is the HCF.
The last divisor we obtained is 18.

Therefore, the HCF of 72 and 90 is 18.

Question:

Find the highest common factor (HCF) of 42 and 56 using the prime factorization method.

Answer:

To find the HCF of 42 and 56 using the prime factorization method, we'll follow these steps:

• Step 1: Prime factorization of 42 42 = 2 * 3 * 7
• Step 2: Prime factorization of 56 56 = 2 * 2 * 2 * 7
• Step 3: Identify common prime factors The common prime factors between 42 and 56 are 2 and 7.
• Step 4: Multiply the common prime factors 2 * 7 = 14

Therefore, the HCF of 42 and 56 is 14.

Important Diagrams of Class 10 NCERT Chapter 1

The important diagrams of Class 10 NCERT Chapter 1 help you understand the topic in detail and easily, therefore, here, we provide the important diagrams of Class 10 NCERT Chapter 1 so that you can level up your topic's understanding.

How to Download the NCERT Solutions Class 10 Chapter 1? 

Downloading the NCERT Solutions for Class 10 Maths Chapter 1 PDF is not difficult if you are aware of the right steps. It is just a matter of 2 minutes. The steps to download them are as follows: 

• Go to the official website of Selfstudys i.e. Selfstudys. 

• Once the website is completely loaded, you need to click on the three lines which you will find on the upper left side. Following that, you have to click on the ‘NCERT Books and Solutions’.  

• After clicking on the option of ‘NCERT Books and Solutions’, a new list will open and you need to select the option of ‘NCERT Solutions’.

 

• After that, you will be redirected to a page where you need to select the class and subject for which you want to download the solutions. In this case, you need to select class 10 and subject Maths to download the NCERT Solutions for Class 10 Maths Chapter 1. 
• And that’s it! It was this simple to download the NCERT Solutions Class 10 Chapter 1. 

What Are the Features of NCERT Solutions for Class 10 Maths Chapter 1? 

There are many features of the NCERT Solutions for Class 10 Maths Chapter 1 which can be very helpful for all the students. Some of the most important of them which helps the students to score good marks in their exams includes the following features: 

• Simple Language used in the Solutions: The language used in the NCERT Solutions for Class 10 Maths Chapter 1 PDF is simple which can help a lot of students to understand them and learn the concepts in a fast and easy way. This increases their chances of scoring good marks in the exams. 

• Availability in the PDF Format: All the students can easily download the NCERT Solutions for Class 10 Maths Chapter 1 PDF from the website of Selfstudys in the PDF Format which makes it easier for them to access it. These PDF Files can also be accessed in the mobile phones. 

• Every single topic is covered: Every single topic is covered in the NCERT Solutions for Class 10 Maths Chapter 1 PDF which not only helps the students to do effective preparation but also helps them make them confident that they will do well in their exams as they will cover each and every topic. 

• Different questions are also included: Not only solutions to each and every problem but various other questions are also included in the NCERT Solutions for Class 10 Maths Chapter 1 which forces the students to think on a higher level. This not only helps the students to do effective revision but also increases their chances of scoring good marks in the examination. 

• 24*7 Available Online: The NCERT Solutions for Class 10 Maths Chapter 1 are available 24*7 Online which can be easier for all the students whether they want to study in day or night or late at night. 

What Are the Benefits of NCERT Solutions for Class 10 Maths Chapter 1?

There are various benefits of NCERT Solutions for Class 10 Maths Chapter 1 PDF which can be a success mantra for all the students. Some of the most important of them which increases the chances of scoring well in the exams includes: 

• Helps in Solving High-Level Questions: While doing exam preparation, the students will also go through various questions. Most of them are high-order thinking skills also which require constant focus and attention. The NCERT Solutions for Class 10 Maths Chapter 1 helps all the students to solve high-level questions easily and fast. 

• Improves Speed: As NCERT Solutions Class 10 Chapter 1 are well-detailed and written in a simple language, it helps the students to understand them and also improves the speed of the students. 

• Clears the Concept: Another big benefit of the NCERT Solutions for Class 10 Maths Chapter 1 is that it includes each and every topic which can be very beneficial for all the students in clearing all the concepts. 

• Recommended by the CBSE (Central Board of Secondary Education): Talking about the CBSE Board, they always advise all the students to go through the NCERT Solutions. According to CBSE, for a student to prepare well and score good marks in the exams, NCERT Solutions are enough. 

• Helpful in Revision: Revision is very important for all the students. You need to revise all the concepts and topics to remember them for a longer period of time. NCERT Solutions for Class 10 Maths Chapter 1 PDF helps to revise all the concepts in an effective manner. It also makes it easier for all the students to memorise them. 

• Time Management: It is suggested for all the students to take care of the time management as it helps them to manage their time effectively during the exam time. Students can improve their time management skills by regularly practising these Solutions. 

What Are the Tips to Study From NCERT Solutions for Class 10 Maths Chapter 1?

There are various tips which can help the students to study effectively from the NCERT Solutions Class 10 Chapter 1: 

• Take Mini Breaks: When studying from the NCERT Solutions for Class 10 Maths Chapter 1, it is advisable for all the students to take mini breaks as it will ensure that they are preparing well for their exam. According to the research, a break of half an hour in between the study sessions keeps the students calm during the exam days where the chances of getting stressed and anxious is more. 

• Pomodoro Technique: Another great tip to study from the NCERT Solutions for Class 10 Maths Chapter 1 is to Practise the Pomodomoro technique which is very common among the students as it helps to reduce all the distractions around the students and boosts the concentration of the students while studying. 

• Blurting: Students can also use the Blurting technique while studying from the NCERT Solutions for Class 10 Maths Chapter 1 which is very effective. In this technique, the students have to read the questions repeatedly. Reading can help the students learn all the concepts fast. After following this technique, you can also test yourself by writing down the topics which you remembered so far during the revision time. 

• Note down your mistakes and work on them: While studying the NCERT Solutions for Class 10 Maths Chapter 1, all the students are advised that they should take down their mistakes. This can help students prepare well for their exams and also increases the chances of scoring well in the exams. 

• Stay Focused: It is very important for all the students to keep their mind stable during exam days. They should take a break from electronic gadgets in order to reduce distractions. While studying NCERT Solutions for Class 10 Maths Chapter 1, they should stay focused to grasp the concepts easily. 

How to Study for a Maths Exam With the Help of NCERT Solutions for Class 10 Maths Chapter 1?

Studying for the Maths exam requires a lot of dedication. Below we will discuss how students can study for a Maths Exam with the help of NCERT Solutions Class 10 Chapter 1: 

• Make Flashcards: One of the best ways to memorise the important definitions in NCERT Solutions for Class 10 Maths Chapter 1 PDF. On one side of an index card, you can write down a word and on the other side, you can write definitions. This is a great way to quiz yourself or have someone else quiz you. 

• Write down the words and definitions: If you do not want to learn via flashcards, there are other ways too which can help you learn the concepts in the NCERT Solutions Class 10 Chapter 1 by writing down the words and definitions which will help you stick them in your mind and remember them. 

• Study NCERT Solutions Class 10 Chapter 1 on a regular basis: It is advisable for all the students to study NCERT Solutions for Class 10 Maths Chapter 1 regularly either on a weekly basis or they should decide according to their comfort. It will help the students to stick the concepts in their minds and help them to do effective preparation. It also increases their chances of scoring good marks in the exams. 

• Read the side notes of your textbook: There are always side notes in your textbook which is advisable for the students to go through as it will help them to boost their conceptual knowledge of Chapter 1 which increases their chances of scoring well in the examinations. 

How to Master NCERT Solutions for Class 10 Maths Chapter 1? 

There are various ways to master NCERT Solutions for Class 10 Maths Chapter 1. Some of the most important of them are: 

• Read and understand the basics: The first step to master the NCERT Solutions for Class 10 Maths Chapter 1 PDF is to first read the concept and understand the basics. It will help to boost the conceptual knowledge of the students and also help them to remember the concepts for a longer period of time. 

• Analyse the topics: Observe it by breaking it into smaller parts after you have read and understand the concepts in NCERT Solutions Class 10 Maths Chapter 1. You should focus more on literary devices which include imagery etc. 

• Write it down: All the students should write them after breaking them into small parts and memorising them as it ensures that the NCERT Solutions for Class 10 Maths Chapter 1 concepts have stuck in their minds. This is one of the greatest ways to memorise these concepts. 

• Make use of the NCERT Solutions: If you want to master the Chapter 1 class 10, then you should regularly use these NCERT Solutions to improve your concepts. 

• Do regular practice of the NCERT Solutions: It is advisable for all the students to regularly practise the NCERT Solutions for Class 10 Maths Chapter 1 PDF as it helps the students to do effective preparation for their exams and it also increases their chances of scoring good marks in the exams. Regular practice will also help the students to master all the concepts and topics. 

• Apply the 80 / 20 Principle: The students can apply the 80 / 20 principle while going through NCERT Solutions for Class 10 Maths Chapter 1 PDF. The principle means looking at the bigger picture of what students are trying to memorise and evaluating which 20% of the concepts within it will bring you the advantage of the 80%.