CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
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.
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.
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.
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.
[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.
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.
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.
Also Read | Real Numbers Class 10 Notes
Also see-
Frequently Asked Questions (FAQs)
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.
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
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.
Yes, the real numbers include both rational and irrational numbers. For example, pi is an irrational number and also pi is a real number.
The Class 10 NCERT Mathematics Book has 15 chapters in total. Class 10 Mathematics covers topics from algebra, geometry, trigonometry, statistics and probability.
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.
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.
On Question asked by student community
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