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

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

Edited By Ramraj Saini | Updated on Oct 31, 2023 02:06 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 Real Numbers (Ex-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.

More About NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 Real Numbers

The set of real numbers in 10th class maths exercise 1.1 answers are of various categories, like natural and whole numbers, integers, rational and irrational numbers. Real Numbers obey commutative and associative property along with identity and distributive properties. NCERT solutions for Class 10 Maths exercise 1.1 mainly focus on the application of Euclid's division Lemma and algorithm. Five important questions related to Euclid's division Lemma and algorithm are given in class 10 maths ex 1.1.

Also get access of all important formulae and eBook for class 10 maths which is helpful to solve problems given in NCERT class 10 maths exercise. Practice all exercise at one place listed below.

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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)

1. 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.

2. 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

3. 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.

4. 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. 

5. 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. 

6. 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.

7. 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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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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