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Okay, I will provide the answers to all questions in English. Question 4: Can the number (15)^n, n being a natural number, end with the digit 0? Give reasons. Step 1: Understand the condition for a number to end with the digit 0. For any natural number to end with the digit 0, its prime factorization must contain both 2 and 5. This means it must be divisible by 10, or 2 × 5. Step 2: Find the prime factorization of the base number. The base number is 15. Its prime factorization is: 15 = 3 × 5 Step 3: Analyze the prime factors of (15)^n. The prime factorization of (15)^n will be: (15)^n = (3 × 5)^n = 3^n × 5^n The only prime factors of (15)^n are 3 and 5. Step 4: Conclude based on the analysis. Since the prime factorization of (15)^n does not contain the prime factor 2, it cannot end with the digit 0 for any natural number n. No, (15)^n cannot end with the digit 0. Question 6: Prove that 5 - 2sqrt(3) is an irrational number. It is given that sqrt(3) is an irrational number. Step 1: Assume the contrary. Let's assume that 5 - 2sqrt(3) is a rational number. If 5 - 2sqrt(3) is rational, then we can write it in the form (a)/(b), where a and b are integers, b ≠ 0, and a, b are coprime. 5 - 2sqrt(3) = (a)/(b) Step 2: Isolate the irrational term. Rearrange the equation to isolate sqrt(3): -2sqrt(3) = (a)/(b) - 5 -2sqrt(3) = (a - 5b)/(b) sqrt(3) = (a - 5b)/(-2b) sqrt(3) = (5b - a)/(2b) Step 3: Analyze the nature of the terms. Since a and b are integers, 5b - a is an integer, and 2b is a non-zero integer. Therefore, (5b - a)/(2b) is a rational number. Step 4: Reach a contradiction. This implies that sqrt(3) is a rational number. However, we are given that sqrt(3) is an irrational number. This contradicts our initial assumption. Step 5: Conclude. Our assumption that 5 - 2sqrt(3) is rational is false. Therefore, 5 - 2sqrt(3) must be an irrational number. 5 - 2sqrt(3) is an irrational number. Question 7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start from the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Step 1: Understand the problem. To find when Sonia and Ravi will meet again at the starting point, we need to find the least common multiple (LCM) of the time taken by each of them to complete one round. Step 2: Find the prime factorization of each time. Sonia's time = 18 minutes 18 = 2 × 3^2 Ravi's time = 12 minutes 12 = 2^2 × 3 Step 3: Calculate the LCM. The LCM is found by taking the highest power of all prime factors present in the factorizations. LCM(18, 12) = 2^2 × 3^2 LCM(18, 12) = 4 × 9 LCM(18, 12) = 36 Step 4: State the conclusion. They will meet again at the starting point after 36 minutes. They will meet again after 36 minutes. Question 8: Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers. Step 1: Find the prime factorization of each number. For 404: 404 = 2 × 202 = 2 × 2 × 101 = 2^2 × 101 For 96: 96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3 = 2^5 × 3 Step 2: Calculate the HCF (Highest Common Factor). The HCF is the product of the lowest powers of the common prime factors. The only common prime factor is 2. The lowest power of 2 is 2^2. HCF(404, 96) = 2^2 = 4 Step 3: Calculate the LCM (Least Common Multiple). The LCM is the product of the highest powers of all prime factors. LCM(404, 96) = 2^5 × 3 × 101 LCM(404, 96) = 32 × 3 × 101 LCM(404, 96) = 96 × 101 LCM(404, 96) = 9696 Step 4: Verify the relationship HCF × LCM = Product of the two numbers. Product of the two numbers = 404 × 96 404 × 96 = 38784 HCF × LCM = 4 × 9696 4 × 9696 = 38784 Since 38784 = 38784, the relationship is verified. HCF = 4, LCM = 9696. Verified: 4 × 9696 = 404 × 96 = 38784. Question 9: (a) Prove that sqrt(5) is an irrational number. (b) Prove that (sqrt(2) + sqrt(3))^2 is an irrational number, given that sqrt(6) is an irrational number. (a) Prove that sqrt(5) is an irrational number. Step 1: Assume the contrary. Let's assume that sqrt(5) is a rational number. If sqrt(5) is rational, then we can write it in the form (a)/(b), where a and b are integers, b ≠ 0, and a, b are coprime (have no common factors other than 1). sqrt(5) = (a)/(b) Step 2: Square both sides and rearrange. (sqrt(5))^2 = ((a)/(b))^2 5 = (a^2)/(b^2) 5b^2 = a^2 (Equation 1) This implies that a^2 is divisible by 5. If a^2 is divisible by 5, then a must also be divisible by 5. Step 3: Express a in terms of 5. Since a is divisible by 5, we can write a = 5c for some integer c. Step 4: Substitute a = 5c into Equation 1. 5b^2 = (5c)^2 5b^2 = 25c^2 Divide both sides by 5: b^2 = 5c^2 This implies that b^2 is divisible by 5. If b^2 is divisible by 5, then b must also be divisible by 5. Step 5: Reach a contradiction. From Step 2, we found that a is divisible by 5. From Step 4, we found that b is divisible by 5. This means that a and b have a common factor of 5. However, in Step 1, we assumed that a and b are coprime (have no common factors other than 1). This is a contradiction. Step 6: Conclude. Our initial assumption that sqrt(5) is rational is false. Therefore, sqrt(5) must be an irrational number. sqrt(5) is an irrational number. (b) Prove that (sqrt(2) + sqrt(3))^2 is an irrational number, given that sqrt(6) is an irrational number. Step 1: Expand the given expression. (sqrt(2) + sqrt(3))^2 = (sqrt(2))^2 + (sqrt(3))^2 + 2(sqrt(2))(sqrt(3)) (sqrt(2) + sqrt(3))^2 = 2 + 3 + 2sqrt(6) (sqrt(2) + sqrt(3))^2 = 5 + 2sqrt(6) Step 2: Assume the contrary. Let's assume that 5 + 2sqrt(6) is a rational number. If 5 + 2sqrt(6) is rational, then we can write it in the form (a)/(b), where a and b are integers, b ≠ 0, and a, b are coprime. 5 + 2sqrt(6) = (a)/(b) Step 3: Isolate the irrational term. Rearrange the equation to isolate sqrt(6): 2sqrt(6) = (a)/(b) - 5 2sqrt(6) = (a - 5b)/(b) sqrt(6) = (a - 5b)/(2b) Step 4: Analyze the nature of the terms. Since a and b are integers, a - 5b is an integer, and 2b is a non-zero integer. Therefore, (a - 5b)/(2b) is a rational number. Step 5: Reach a contradiction. This implies that sqrt(6) is a rational number. However, we are given that sqrt(6) is an irrational number. This contradicts our initial assumption. Step 6: Conclude. Our assumption that (sqrt(2) + sqrt(3))^2 is rational is false. Therefore, (sqrt(2) + sqrt(3))^2 must be an irrational number. (sqrt(2) + sqrt(3))^2 is an irrational number. Question 10: (a) In a school, there are two sections of class X. There are 40 students in the first section and 48 students in the second section. Determine the minimum number of books required for their class library so that they can be distributed equally among students of both sections. (b) Find by prime factorisation the LCM of the numbers 18180 and 7575. Also, find the HCF of the two numbers. (a) Determine the minimum number of books required for their class library so that they can be distributed equally among students of both sections. Step 1: Understand the problem. To find the minimum number of books that can be distributed equally among students of both sections, we need to find the least common multiple (LCM) of the number of students in each section. Step 2: Find the prime factorization of the number of students in each section. Number of students in the first section = 40 40 = 2 × 20 = 2 × 2 × 10 = 2 × 2 × 2 × 5 = 2^3 × 5 Number of students in the second section = 48 48 = 2 × 24 = 2 × 2 × 12 = 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 3 = 2^4 × 3 Step 3: Calculate the LCM. The LCM is found by taking the highest power of all prime factors present in the factorizations. LCM(40, 48) = 2^4 × 3 × 5 LCM(40, 48) = 16 × 3 × 5 LCM(40, 48) = 48 × 5 LCM(40, 48) = 240 Step 4: State the conclusion. The minimum number of books required is 240. Minimum number of books required is 240. (b) Find by prime factorisation the LCM of the numbers 18180 and 7575. Also, find the HCF of the two numbers. Step 1: Find the prime factorization of each number. For 18180: 18180 = 10 × 1818 = (2 × 5) × (2 × 909) = 2^2 × 5 × (3 × 303) = 2^2 × 5 × 3 × (3 × 101) 18180 = 2^2 × 3^2 × 5 × 101 For 7575: 7575 = 5 × 1515 = 5 × 5 × 303 = 5^2 × 3 × 101 7575 = 3 × 5^2 × 101 Step 2: Calculate the HCF (Highest Common Factor). The HCF is the product of the lowest powers of the common prime factors. Common prime factors are 3, 5, and 101. Lowest power of 3 is 3^1. Lowest power of 5 is 5^1. Lowest power of 101 is 101^1. HCF(18180, 7575) = 3^1 × 5^1 × 101^1 HCF(18180, 7575) = 3 × 5 × 101 HCF(18180, 7575) = 15 × 101 HCF(18180, 7575) = 1515 Step 3: Calculate the LCM (Least Common Multiple). The LCM is the product of the highest powers of all prime factors. Highest power of 2 is 2^2. Highest power of 3 is 3^2. Highest power of 5 is 5^2. Highest power of 101 is 101^1. LCM(18180, 7575) = 2^2 × 3^2 × 5^2 × 101 LCM(18180, 7575) = 4 × 9 × 25 × 101 LCM(18180, 7575) = 36 × 25 × 101 LCM(18180, 7575) = 900 × 101 LCM(18180, 7575) = 90900 HCF = 1515, LCM = 90900. Question 11 (Case Study): Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announces the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to the second student. Second student also multiplied it by a prime number and passed it to third student. In this way by multiplying to a prime number, the last student got 173250. Now, Mukta asked some questions as given below to the students: (i) What is the least prime number used by students? (ii) How many students are in the class? (iii) What is the highest prime number used by students? (iv) Which prime number has been used maximum times? Step 1: Find the prime factorization of the final number. The game starts with the number 2. Each student multiplies the current number by a prime number. The final number is 173250. First, find the prime factorization of 173250: 173250 = 10 × 17325 = (2 × 5) × (5 × 3465) = 2 × 5^2 × (5 × 693) = 2 × 5^3 × (3 × 231) 173250 = 2 × 5^3 × 3 × (3 × 77) = 2 × 5^3 × 3^2 × (7 × 11) So, the prime factorization of 173250 is 2^1 × 3^2 × 5^3 × 7^1 × 11^1. Step 2: Analyze the prime factors in the context of the game. The game starts with 2. This means the initial number is 2^1. The final number is 2^1 × 3^2 × 5^3 × 7^1 × 11^1. The prime numbers used by the students are the prime factors that were multiplied to the initial 2. The prime factors introduced by students are 3, 3, 5, 5, 5, 7, 11. (i) What is the least prime number used by students? The prime numbers used by students are 3, 5, 7, 11. The least among these is 3. (ii) How many students are in the class? Each student multiplies by one prime number. The total number of prime factors multiplied by students is the sum of the exponents of the prime factors other than the initial 2 in the final number. Number of prime factors introduced = (exponent of 3) + (exponent of 5) + (exponent of 7) + (exponent of 11) Number of prime factors introduced = 2 + 3 + 1 + 1 = 7. Since each student multiplied by one prime number, there are 7 students in the class. (iii) What is the highest prime number used by students? The prime numbers used by students are 3, 5, 7, 11. The highest among these is 11. (iv) Which prime number has been used maximum times? From the prime factorization 2^1 × 3^2 × 5^3 × 7^1 × 11^1, the prime factors introduced by students are 3 (used 2 times), 5 (used 3 times), 7 (used 1 time), and 11 (used 1 time). The prime number 5 has been used the maximum number of times (3 times). Final Answers: (i) The least prime number used by students is 3. (ii) There are 7 students in the class. (iii) The highest prime number used by students is 11. (iv) The prime number 5 has been used maximum times. 3 done, 2 left today. You're making progress.