NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1

NCERT solutions for class 10 maths chapter 1 exercise 1.1 are discussed here. These NCERT Solutions are created by expert team at Careers360 considering the latest syllabus and pattern of CBSE, NCERT, and State Board exams. These are designed comprehensively covering all the concepts, detailed and step by step solutions so that students get deeper understanding and score well during the competitive exams. NCERT Solutions for Class 10 Maths chapter 1 exercise 1.1 includes complete solutions of each and every problem that can be studied using PDF which is downloaded freely using the link given below.

The 10th class maths exercise 1.1 answers, teaches you about how to figure out if one number can be divided by another using Euclid's Division Algorithm. It provides clear, step-by-step answers to the questions in the Class 10 NCERT math book. These answers are made to follow the NCERT guidelines, which means they help you cover everything you need to know for your exams and do well in them.

Download PDF of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 Real Numbers

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Access Exercise 1.1 Class 10 Maths Answers

Q1 (1) Use Euclid’s division algorithm to find the HCF of 135 and 225

Answer:

225 > 135. Applying Euclid's Division algorithm we get

$225=135\times 1+90$

since remainder $\neq$ 0 we again apply the algorithm

$135=90\times 1+45$

since remainder $\neq$ 0 we again apply the algorithm

$90=45\times 2$

since remainder = 0 we conclude the HCF of 135 and 225 is 45.

Q1 (2) Use Euclid’s division algorithm to find the HCF of 196 and 38220

Answer:

38220 > 196. Applying Euclid's Division algorithm we get

$38220=196\times 195+0$

since remainder = 0 we conclude the HCF of 38220 and 196 is 196.

Q1 (3) Use Euclid’s division algorithm to find the HCF of 867 and 255

Answer:

867 > 225. Applying Euclid's Division algorithm we get

$867=255\times 3+102$

since remainder $\neq$ 0 we apply the algorithm again.

since 255 > 102

$255=102\times 2+51$

since remainder $\neq$ 0 we apply the algorithm again.

since 102 > 51

$102=51\times 2+0$

since remainder = 0 we conclude the HCF of 867 and 255 is 51.

Q2 Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer

Answer:

Let p be any positive integer. It can be expressed as

p = 6q + r

where $q\geq 0$ and $0\leq r< 6$

but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.

Q3 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:

The maximum number of columns in which they can march = HCF (32, 616)

Since 616 > 32, applying Euclid's Division Algorithm we have

$616=32\times 19+8$

Since remainder $\neq$ 0 we again apply Euclid's Division Algorithm

Since 32 > 8

$32=8\times 4+0$

Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.

The maximum number of columns in which they can march is 8.

Q4 Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1]

Answer:

Let x be any positive integer.

It can be written in the form 3q + r where $q\geq 0$ and r = 0, 1 or 2

Case 1:

For r = 0 we have

x² = (3q)²

x² = 9q²

x² = 3(3q² )

x² = 3m

Case 2:

For r = 1 we have

x² = (3q+1)²

x² = 9q² + 6q +1

x² = 3(3q² + 2q) + 1

x² = 3m + 1

Case 3:

For r = 2 we have

x² = (3q+2)²

x² = 9q² + 12q +4

x² = 3(3q² + 4q + 1) + 1

x² = 3m + 1

Hence proved.

Q5 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

Let x be any positive integer.

It can be written in the form 3q + r where $q\geq 0$ and r = 0, 1 or 2

Case 1:

For r = 0 we have

x³ = (3q)³

x³ = 27q³

x³ = 9(3q³ )

x³ = 9m

Case 2:

For r = 1 we have

x³ = (3q+1) ³

x³ = 27q ³ + 27q² + 9q + 1

x³ = 9(3q³ + 3q² +q) + 1

x³ = 3m + 1

Case 3:

For r = 2 we have

x³ = (3q + 2)³

x³ = 27q³ + 54q² + 36q + 8

x³ = 9(3q³ + 6q² +4q) + 8

x³ = 3m + 8

Hence proved.

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Key Features Of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 Real Numbers

NCERT Class 10 Maths chapter 1 exercise 1.1, contains all important questions from exam point of view and all questions are revised from Class 10 Maths chapter 1 exercise 1.1.
NCERT book Exercise 1.1 Class 10 Maths, is based on the introduction to real numbers and based on Euclid’s Division Lemma and Euclid’s Division Algorithm, which are important concepts of the chapter.
NCERT syllabus solutions for Class 10 Maths is considered the best material for solving Class 10 Maths chapter 1 exercise 1.1.

Also see-

NCERT Solutions of Class 10 - Subject Wise

NCERT Exemplar Solutions - Subject Wise

Frequently Asked Questions (FAQs)

Q: How many questions are covered in real numbers ex 1.1 class 10 ?

There are 5 questions covered in Class 10 Maths exercise 1.1, in which the first questions have 3 sub-questions. Students should practice these problem to command the concepts.

Q: What is the benefit of solving NCERT questions?

NCERT questions are designed to get more clarity of the concepts covered in the chapter. These questions are also helpful from an exam point of view. students can find extra problems in class 10 ex 1.1.

Q: What is the weightage of real numbers for CBSE board exams?

For Class 10 CBSE board exams students can expect two or three questions from real numbers. One question may be related to the concepts covered in exercise 1.1

Q: Whether irrational numbers are real numbers?

Yes, the real numbers include both rational and irrational numbers. For example, pi is an irrational number and also pi is a real number.

Q: How many chapters are there in the Class 10 NCERT Mathematics book?

The Class 10 NCERT Mathematics Book has 15 chapters in total. Class 10 Mathematics covers topics from algebra, geometry, trigonometry, statistics and probability.

Q: Can I expect questions from NCERT Solutions for Class 10 Maths Chapter 1 exercise 1.1 for board exam?

Yes, you can expect similar types of questions discussed in exercise 1.1 Class 10 Maths for board exam, but may not be the same questions.

Q: How many marks can I score from exercise 1.1?

You may expect either 2 or 3 marks questions from exercise 1.1. Sometimes there may not be any questions. But still the probability of getting questions from this exercise is very high.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 Real Numbers