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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 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

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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
Provider	Hsslive

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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

AP Board Class 10 Maths Chapter 1 Real Numbers Textbook Solutions PDF

AP Board Class 10 Maths Chapter 2 Sets Textbook Solutions PDF

AP Board Class 10 Maths Chapter 3 Polynomials Textbook Solutions PDF

AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Textbook Solutions PDF

AP Board Class 10 Maths Chapter 5 Quadratic Equations Textbook Solutions PDF

AP Board Class 10 Maths Chapter 6 Progressions Textbook Solutions PDF

AP Board Class 10 Maths Chapter 7 Coordinate Geometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 8 Similar Triangles Textbook Solutions PDF

AP Board Class 10 Maths Chapter 9 Tangents and Secants to a Circle Textbook Solutions PDF

AP Board Class 10 Maths Chapter 10 Mensuration Textbook Solutions PDF

AP Board Class 10 Maths Chapter 11 Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 12 Applications of Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 13 Probability Textbook Solutions PDF

AP Board Class 10 Maths Chapter 14 Statistics Textbook Solutions PDF

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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Tuesday, June 7, 2022