BSEB Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Book Answers |
Bihar Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks Solutions PDF
Bihar Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks. These Bihar Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.Bihar Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Books Solutions
Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 10th |
Subject | Maths Chapter 1 Real Numbers Ex 1.1 |
Chapters | All |
Provider | Hsslive |
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BSEB Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1
Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) We have:
Dividend = 225 and Divisor = 135
The process can be exhibited as under:
Hence, HCF (135, 225) = 45.
(ii) We have:
Dividend = 38220 and Divisor = 196
38220 = 196 × 195 + 0
Hence, HCF (196, 38220) = 196.
(iii) We have:
Dividend = 867 and Divisor = 255
Hence, HCF (255, 867) = 51.
Question 2.
Show that any positive odd integer is of the form 6q + t or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let a be any positive integer and b = 6. Then, by Euclid’s Division Lemma a = 6q + r, for some integer q ≥ 0, and where 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 or 5.
That is, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient. If a = 6q or 6q + 2 or 6q + 4, then a is an even integer.
Also, an integer can be either even or odd. Therefore, any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in 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?
Solution:
To find the maximum number of columns, we have to find the HCF of 616 and 32.
∴ The HCF of 616 and 32 is 8.
Hence, maximum number of columns is 8.
Question 4.
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. [CBSE, Foreign, 2008]
Solution:
Let x be any positive integer. Then it is of the form 3q, 3q + 1 or 3q + 2. Now, we have to prove that the square of each of these can be written in the form 3m or 3m + 1.
Now, (3q)2 = 9q2 = 3(3q2)
= 3m,
where in = 3q2
(3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m+ 1, where m = 3q2 + 2q
and (3q + 2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) + 1
= 3m + 1, where m = 3q2 + 4q + 1
Hence, the result.
Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is either of the form 9q, 9q + 1 or 9q + 8.
Solution:
Let x be any positive integer. Then it is of the form 3m, 3m + 1 or 3m + 2. Now, we have to prove that the cube of each of these can be rewritten in the form
9q, 9q + 1 or 9q + 8.
Now, (3m)3 = 27m3 = 9(3m3)
= 9q, where q = 3m3
(3m + 1)3 = (3m)3 + 3(3m)2. 1 + 3(3m). 12 + 1
= 27m3 + 27m2 + 9m + 1
= 9(3m3 + 6m2 + 4m) + 8
= 9q + 8, where q = 3m3 + 6m2 + 4m.
BSEB Textbook Solutions PDF for Class 10th
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- BSEB Class 10 Maths Chapter 1 Real Numbers Ex 1.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 1 Real Numbers Ex 1.2 Book Answers
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- BSEB Class 10 Maths Chapter 6 Triangles Ex 6.6 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 6 Triangles Ex 6.6 Book Answers
- BSEB Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.1 Book Answers
- BSEB Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.2 Book Answers
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- BSEB Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 7 Coordinate Geometry Ex 7.4 Book Answers
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