Download PDF of Extra Questions for Class 10 Maths Chapter 1 with Answers
- kaixosininmoheads
- Aug 2, 2023
- 12 min read
Class 10 Maths Chapter 1 Extra Questions PDF Download
Are you looking for some extra practice questions for class 10 maths chapter 1? If yes, then you have come to the right place. In this article, we will provide you with some extra questions based on the concepts of real numbers, which is the first chapter of class 10 maths. These questions will help you to revise the syllabus, test your understanding, and prepare for the board exams. You can also download the pdf file of these questions and their solutions for offline study.
Concepts Covered in Class 10 Maths Chapter 1
Class 10 maths chapter 1 deals with the topic of real numbers, which are the numbers that can be represented on a number line. Real numbers include rational numbers and irrational numbers. Rational numbers are those numbers that can be expressed as a fraction of two integers, such as 3/4, -5/6, etc. Irrational numbers are those numbers that cannot be expressed as a fraction of two integers, such as 2, π, etc.
class 10 maths chapter 1 extra questions pdf download
Download File: https://porphasako.blogspot.com/?file=2vv0Dj
In this chapter, you will learn about some important properties and results related to real numbers, such as:
Euclid's division algorithm: This is a method to find the highest common factor (HCF) of two positive integers by repeated division.
Fundamental theorem of arithmetic: This is a theorem that states that every positive integer greater than 1 can be expressed as a unique product of prime numbers.
Revisiting irrational numbers: This is a section that explains how to prove that some numbers are irrational by using contradiction.
Revisiting rational numbers and their decimal expansions: This is a section that explains how to classify rational numbers into terminating decimals or recurring decimals based on their denominators.
Extra Questions for Class 10 Maths Chapter 1
Here are some extra questions for class 10 maths chapter 1 along with their solutions and explanations. Try to solve them on your own before looking at the answers.
Multiple Choice Questions
If n is an odd positive integer, then n - 1 is divisible by
(A) 4
(B) 6
(C) 8
(D) None of these
Solution: (C) 8
Explanation: We can use Euclid's division lemma to show that n - 1 is divisible by 8 if n is an odd positive integer. Let n = 2q + r, where q is any positive integer and r = 0 or r = 1 (since n is odd). Then, n - 1 = (2q + r) - 1 = (4q + r) + (4qr) - 1 = (4q + r) - (4qr + 1) Now, if r = 0, then n - 1 = (4q) - (-1), which is clearly divisible by 8. If r = 1, then n - 1 = (4 Outline of the article: - Introduction: Explain what the article is about and why it is useful for class 10 students. - Section 1: Give a brief overview of the concepts covered in class 10 maths chapter 1, such as real numbers, Euclid's division algorithm, fundamental theorem of arithmetic, rational and irrational numbers, etc. - Section 2: Provide some extra questions based on the concepts of class 10 maths chapter 1, along with their solutions and explanations. Include different types of questions, such as multiple choice, short answer, long answer, etc. - Section 3: Provide some tips and tricks to solve the questions faster and more accurately. Include some common mistakes to avoid and some shortcuts to remember. - Conclusion: Summarize the main points of the article and provide a link to download the pdf file of the extra questions. - FAQs: Provide some frequently asked questions related to class 10 maths chapter 1 and their answers. Article with HTML formatting: Class 10 Maths Chapter 1 Extra Questions PDF Download
Are you looking for some extra practice questions for class 10 maths chapter 1? If yes, then you have come to the right place. In this article, we will provide you with some extra questions based on the concepts of real numbers, which is the first chapter of class 10 maths. These questions will help you to revise the syllabus, test your understanding, and prepare for the board exams. You can also download the pdf file of these questions and their solutions for offline study.
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Concepts Covered in Class 10 Maths Chapter 1
Class 10 maths chapter 1 deals with the topic of real numbers, which are the numbers that can be represented on a number line. Real numbers include rational numbers and irrational numbers. Rational numbers are those numbers that can be expressed as a fraction of two integers, such as 3/4, -5/6, etc. Irrational numbers are those numbers that cannot be expressed as a fraction of two integers, such as 2, π, etc.
In this chapter, you will learn about some important properties and results related to real numbers, such as:
Euclid's division algorithm: This is a method to find the highest common factor (HCF) of two positive integers by repeated division.
Fundamental theorem of arithmetic: This is a theorem that states that every positive integer greater than 1 can be expressed as a unique product of prime numbers.
Revisiting irrational numbers: This is a section that explains how to prove that some numbers are irrational by using contradiction.
Revisiting rational numbers and their decimal expansions: This is a section that explains how to classify rational numbers into terminating decimals or recurring decimals based on their denominators.
Extra Questions for Class 10 Maths Chapter 1
Here are some extra questions for class 10 maths chapter 1 along with their solutions and explanations. Try to solve them on your own before looking at the answers.
Multiple Choice Questions
If n is an odd positive integer, then n - 1 is divisible by
(A) 4
(B) 6
(C) 8
(D) None of these
Solution: (C) 8
Explanation: We can use Euclid's division lemma to show that n - 1 is divisible by 8 if n is an odd positive integer. Let n = 2q + r, where q is any positive integer and r = 0 or r = 1 (since n is odd). Then, n - 1 = (2q + r) - 1 = (4q + r) + (4qr) - 1 = (4q + r) - (4qr + 1) Now, if r = 0, then n - 1 = (4q) - (-1), which is clearly divisible by 8. If r = 1, then n - 1 = (4. q + 1) - (4q + 1), which is also divisible by 8. Hence, n - 1 is divisible by 8 for any odd positive integer n.
The decimal expansion of the rational number 14587/1250 will
(A) terminate
(B) not terminate
(C) terminate after 3 places of decimal
(D) terminate after 4 places of decimal
Solution: (D) terminate after 4 places of decimal
Explanation: We can use the result that a rational number will have a terminating decimal expansion if its denominator is of the form 2 x 5, where m and n are non-negative integers. In this case, the denominator is 1250, which can be written as 2 x 5. Hence, the decimal expansion of the rational number will terminate after 4 places of decimal, which is the maximum power of 2 or 5 in the denominator. The decimal expansion is 14587/1250 = 11.6696.
If p and q are two prime numbers, then (pq) is
(A) rational and irrational
(B) rational but not irrational
(C) irrational but not rational
(D) neither rational nor irrational
Solution: (C) irrational but not rational
Explanation: We can use the method of contradiction to prove that (pq) is irrational if p and q are two prime numbers. Suppose (pq) is rational, then we can write it as a fraction of two integers, say (pq) = a/b, where a and b are co-prime. Squaring both sides, we get pq = a/b, or pb = a. This implies that p divides a, and since p is prime, p must also divide a. Let a = pc, where c is some integer. Substituting this in the equation, we get pb = pc, or b = pc. This implies that p divides b, and since p is prime, p must also divide b. But this contradicts the assumption that a and b are co-prime. Hence, (pq) cannot be rational, and it must be irrational.
Short Answer Questions
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Solution:
Let n be any positive odd integer. Then, by Euclid's division algorithm, we can write n = 4q + r, where q and r are integers and 0 r
If x = (3 + 5), then find the value of x - 6x.
Solution:
We have x = (3 + 5). Squaring both sides, we get x = (3 + 5). Adding and subtracting [assistant](#message) 6 on both sides, we get x - 6 = (3 + 5) - 6 = -3 + 5 = (5 - 3)(5 + 3) Squaring both sides again, we get (x - 6 ) = (5 - 3)(5 + 3) = (5 - 215 + 3)(5 + 215 + 3) = (8 - 215)(8 + 215) = 64 - 60 = 4 Hence, x - 6x = 4.
Find the HCF and LCM of 306 and 657 using the prime factorization method.
Solution:
We can use the fundamental theorem of arithmetic to find the prime factorization of 306 and 657. We have 306 = 2 x 3 x 17 657 = 3 x 7 x 31 The HCF of 306 and 657 is the product of the common prime factors with the least powers, which is HCF(306,657) = 3 The LCM of 306 and 657 is the product of the prime factors with the highest powers, which is LCM(306,657) = 2 x 3 x 7 x 17 x 31 = 71442
Long Answer Questions
Show that 6 is irrational.
Solution:
We can use the method of contradiction to prove that 6 is irrational. Suppose 6 is rational, then we can write it as a fraction of two integers, say 6 = a/b, where a and b are co-prime. Squaring both sides, we get 6 = a/b, or 6b = a. This implies that 6 divides a, and since 6 is composite, either 2 or 3 must also divide a. Let a = 2k or a = 3k, where k is some integer. Substituting this in the equation, we get either 6b = (2k), or b = (2/3)k, which implies that b is not an integer, which contradicts the fact that b is an integer. or 6b = (3k), or b = (3/2)k, which implies that b is not an integer, which contradicts the fact that b is an integer. Hence, 6 cannot be rational, and it must be irrational.
Show that there are infinitely many prime numbers.
Solution:
We can use the method of contradiction to prove that there are infinitely many prime numbers. Suppose there are finitely many prime numbers, say p1, p2, ..., pn. Let P be the product of all these prime numbers, i.e., P = p1p2...pn. Consider the number Q = P + 1. Now, Q is either prime or composite. If Q is prime, then it is a prime number that is not in our list, which contradicts the assumption that we have listed all the prime numbers. If Q is composite, then it must have a prime factor, say p. But p cannot be any of the prime numbers in our list, because then p would divide both P and Q, and hence p would also divide Q - P, which is equal to 1. But no prime number can divide 1. Hence, Q has a prime factor that is not in our list, which again contradicts the assumption that we have listed all the prime numbers. Therefore, we cannot have a finite list of all the prime numbers, and there must be infinitely many prime numbers.
Show that (12 + (12 + (12 + ...))) = (18)
Solution:
We can use the method of substitution to prove that (12 + (12 + (12 + ...))) = (18). Let x be the value of the expression (12 + (12 + (12 + ...))). Then, x = (12 + (12 + (12 + ...))) = (12 + x) (by substituting x for the innermost square root) Squaring both sides, we get x = 12 + x Rearranging, we get x - x - 12 = 0 Factoring, we get (x - 4)(x + 3) = 0 Hence, x = 4 or x = -3 But x cannot be negative, since it is the value of a square root. Hence, x = 4. Now, we can use the fact that (18) = (9 x 2) = 9 x 2 = 32. We can also use the fact that 4 = 22 x 2. Hence, we can write x = 4 = 22 x 2 = 2 x 2 x 2 = (8) x 2 = (9 - 1) x 2 = (9 - 1) x 2 = (3 - 1)2 = 22 = (18)/3 Multiplying both sides by 3, we get 3x = 18 Hence, x = (18)/3, which implies that (12 + (12 + (12 + ...))) = (18)/3.
Tips and Tricks for Class 10 Maths Chapter 1
Here are some tips and tricks to solve the questions of class 10 maths chapter 1 faster and more accurately.
To find the HCF of two numbers using Euclid's division algorithm, divide the larger number by the smaller number and take the remainder. Then, divide the smaller number by the remainder and take the new remainder. Repeat this process until the remainder becomes zero. The last divisor is the HCF.
To find the LCM of two numbers using the prime factorization method, write the numbers as products of prime factors and take the highest power of each prime factor. Multiply these factors to get the LCM.
To prove that a number is irrational, assume that it is rational and write it as a fraction of two co-prime integers. Then, manipulate the equation to reach a contradiction.
To classify a rational number as a terminating or recurring decimal, simplify the fraction and look at its denominator. If the denominator is of the form 2 x 5, where m and n are non-negative integers, then the decimal will terminate. Otherwise, it will recur.
To avoid common mistakes, always check your calculations and simplify your answers. Also, remember to write proper units and symbols wherever required.
Conclusion
We hope that this article has helped you to understand and practice the concepts of class 10 maths chapter 1. You can download the pdf file of these extra questions and their solutions from . You can also check out our other articles on class 10 maths for more tips and tricks. Happy learning!
FAQs
What are real numbers?
Real numbers are the numbers that can be represented on a number line. They include rational numbers and irrational numbers.
What are prime numbers?
Prime numbers are those numbers that have exactly two factors: 1 and themselves. For example, 2, 3, 5, 7, etc. are prime numbers.
What is Euclid's division lemma?
Euclid's division lemma is a statement that says that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 r
What is the fundamental theorem of arithmetic?
The fundamental theorem of arithmetic is a theorem that says that every positive integer greater than 1 can be expressed as a unique product of prime numbers.
How to prove that 2 is irrational?
We can use the method of contradiction to prove that 2 is irrational. Suppose 2 is rational, then we can write it as a fraction of two co-prime integers, say 2 = a/b. Squaring both sides, we get 2 = a/b, or 2b = a. This implies that 2 divides a, and since 2 is prime, 2 must also divide a. Let a = 2k, where k is some integer. Substituting this in the equation, we get 2b = (2k), or b = 2k. This implies that 2 divides b, and since 2 is prime, 2 must also divide b. But this contradicts the assumption that a and b are co-prime. Hence, 2 cannot be rational, and it must be irrational.
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