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AP Board Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Book Answers

AP Board Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Book Answers
AP Board Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Book Answers


AP Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 book solutions pdf online from this page.

Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks Solutions PDF

Andhra Pradesh State Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board 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 Andhra Pradesh State 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 Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks. These Andhra Pradesh State 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 Andhra Pradesh Board.

Andhra Pradesh State Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Books Solutions

Board AP Board
Materials Textbook Solutions/Guide
Format DOC/PDF
Class 10th
Subject Maths
Chapters Maths Chapter 1 Real Numbers Ex 1.1
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AP Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board 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:

Question 1.
Use Euclid’s division algorithm to find the HCF of
i) 900 and 270
Answer:
900 = 270 × 3 + 90
270 = 90 × 3 + 0
∴ HCF = 90

 

ii) 196 and 38220
Answer:
38220 = 196 × 195 + 0
∴ 196 is the HCF of 196 and 38220.

 

iii) 1651 and 2032
Answer:
2032 = 1651 × 1 + 381
1651 = 381 × 4 + 127
381 = 127 × 3 + 0
∴ HCF = 127

 

Question 2.
Use Euclid division lemma to show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integers.
Answer:
Let ‘a’ be an odd positive integer.
Let us now apply division algorithm with a and b = 6.
∵ 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.
i.e., ’a’ can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
But ‘a’ is taken as an odd number.
∴ a can’t be 6q or 6q + 2 or 6q + 4.
∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.

 

Question 3.
Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.
Answer:
Let ‘a’ be the square of an integer.
Applying Euclid’s division lemma with a and b = 3
Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.
∴ a = 3q (or) 3q + 1 (or) 3q + 2
∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.
(or)
Let ‘a’ be a positive integer
So it can be expressed as a = bq + r (from Euclideans lemma)
now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.
then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a2) will be
Case – I: a2 – (3q)2 = 9q2=3(3q2) = 3p (p = 3q2)
Case-II: a2 = (3q + l)2 = 9q2 + 6q+ 1
= 3[3q2 + 2q] + 1 = 3p+l (Where p = 3q2 + 2q) or
Case – III: a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1
= 3[3q2 + 4q + 1] + 1
= 3p + 1 (where ‘p’ = 3q2 + 4q + 1)
So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1
Hence proved.

 

Question 4.
Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.
(OR)
Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.
Answer:
Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;
now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8
So a = bq + r
⇒ a = 9q + r (for b = 9)
now cube of a = a3 + (9q + r)3
= (9q)3 + 3.(9q)3r + 3. 9q.r + r3
= 93q3 + 3.92(q2r) + 3.9(q.r) + r3
= 9[92.q3 + 3.9.q2r + 3.q.r] + r3
a3 = 9m + r3 (where ‘m’ = 92q3 + 3.9.q2r + 3.q.r)
if r = 0 ⇒ r3 = 0 then a3 = 9m + 0 = 9m
and for r = 1 ⇒ r3 = l3 then a3 = 9m + 1
and for r = 2 ⇒ r3 = 23 then a3 = 9m + 8
for r = 3 ⇒ r3, = 33 ⇒ a3 = 9m + 27 = 9(m) where m = (9m +3)
for r = 4 ⇒ r3 = 43 ⇒ a3 = 9m + 64 = (9m + 63) + 1 = 9m + 1
for r = 5 ⇒ r3 = 125 ⇒ a3 = 9m + 125 = (9m + 117) + 8 = 9m + 8
for r = 6 ⇒ r3 — 216 ⇒ a3 = 9m + 216 = 9m + 9(24) = 9m
for r = 7 ⇒ r3 = 243
⇒ a3 = 9m + 9(27) = 9m
for r = 8 ⇒ r3 = 512
⇒ a3 = 9m + 9(56) + 8 = 9m + 8
So from the above it is clear that a3 is either in the form of 9m or 9m + 1 or 9m + 8.
Hence proved.

 

Question 5.
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
(Or)
Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.
Answer:
Let ‘n’ be any positive integer.
Then from Euclidean’s lemma n = bq + r (now consider b = 3)
⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2
Now consider r = 0 then ‘n’ = 3q (divisible by 3)
and n + 2 = 3q + 2 (not divisible by 3)
n + 4 = 3q + 4 (not divisible by 3)
Case – II: For r = 1
n = 3q + 1 (not divisible by 3)
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3
Case – III: For r = 2,
n = 3q + 2 not divisible by 3
n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3
So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.
Hence proved.


AP Board Textbook Solutions PDF for Class 10th Maths


Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 exam preparation. The AP Board STD 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Books State Board syllabus with maximum efficiency.

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How to get AP Board Class 10th Maths Chapter 1 Real Numbers Ex 1.1 Textbook Answers??

Students can download the Andhra Pradesh Board Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Answers PDF from the links provided above.

Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

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