Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
SubjectsMathematics
अंकगणित का मौलिक प्रमेय
In this Class 10 Mathematics topic from the chapter Real Numbers, students learn that every integer greater than 1 can be expressed as a product of prime numbers, and that this prime factorisation is unique apart from the order of the factors. They practise finding prime factors and use the theorem to understand and determine the HCF and LCM of numbers. The topic builds clear reasoning about the structure of whole numbers and supports later work with divisibility and number relationships.
Correct answer: A. (3^5)
Explanation: Step 1: Divide 243 repeatedly by 3. Step 2: (243=3^5), so this is the prime factorisation. Step 3: 9 and 81 are composite, so do not keep them in the final prime form.
Correct answer: A. Because it does not have exactly two positive factors
Explanation: Step 1: A prime number has exactly two positive factors. Step 2: 1 has only one positive factor, 1. Step 3: Therefore, 1 is neither prime nor composite.
Correct answer: C. 4
Explanation: Step 1: Look at the given form (625=5^4). Step 2: The exponent written on 5 is 4. Step 3: It is important to identify the base and the exponent separately.
Correct answer: A. (2\times7\times13)
Explanation: Step 1: Write (182=2\times91). Step 2: (91=7\times13), so (182=2\times7\times13). Step 3: 14 is composite, so (14\times13) is not the final prime factorisation.
Correct answer: B. 432
Explanation: Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (6\times72=432). Step 3: Apply this relation directly for two numbers.
Correct answer: A. (17^2)
Explanation: Step 1: 289 is a perfect square. Step 2: (289=17\times17=17^2). Step 3: 1 is not treated as a separate factor in prime factorisation.
Correct answer: A. 3, 7 and 11
Explanation: Step 1: Write (231=21\times11). Step 2: (21=3\times7), so (231=3\times7\times11). Step 3: The distinct prime factors are 3, 7, and 11.
Correct answer: B. 18
Explanation: Step 1: The common prime factors are 2 and 3. Step 2: The smaller powers are (2^1) and (3^2), so the HCF is (2\times9=18). Step 3: Take smaller powers for HCF.
Correct answer: B. 108
Explanation: Step 1: For LCM, take the highest powers. Step 2: (2^2\times3^3=4\times27=108). Step 3: LCM includes the highest powers of all prime factors.
Correct answer: A. (3\times5^3)
Explanation: Step 1: Write (375=3\times125). Step 2: (125=5^3), so (375=3\times5^3). Step 3: 15 and 25 are composite, so do not keep them in the final prime form.
Correct answer: C. 126
Explanation: Step 1: Find (3^2=9). Step 2: (2\times9\times7=126), so the number is 126. Step 3: To get the number from prime factorisation, simplify powers first.
Correct answer: C. 192
Explanation: Step 1: Find (2^6=64). Step 2: (64\times3=192), so the number is 192. Step 3: First evaluate the power, then multiply.
Correct answer: C. (9\times5)
Explanation: Step 1: In prime factorisation, every factor must be prime. Step 2: In (9\times5), 9 is not prime. Step 3: If a composite factor appears, break it into prime factors.
Correct answer: A. (2^4\times13)
Explanation: Step 1: Write (208=16\times13). Step 2: (16=2^4) and 13 is prime, so (208=2^4\times13). Step 3: Since 16 is composite, write it as (2^4) in the final form.
Correct answer: A. They are co-prime
Explanation: Step 1: (26=2\times13) and (45=3^2\times5). Step 2: They have no common prime factor, so they are co-prime. Step 3: Co-prime numbers have HCF 1.
Correct answer: A. (2^9)
Explanation: Step 1: Divide 512 repeatedly by 2. Step 2: (512=2^9), so this is the prime factorisation. Step 3: 4 and 16 are composite, so do not keep them in the final prime form.
Correct answer: B. 30
Explanation: Step 1: The common prime factors are 2, 3, and 5. Step 2: The smaller powers are (2^1), (3^1), and (5^1), so (2\times3\times5=30). Step 3: Take smaller powers for HCF.
Correct answer: C. 450
Explanation: Step 1: For LCM, take the highest powers. Step 2: (2^1\times3^2\times5^2=2\times9\times25=450). Step 3: LCM includes the highest powers of all prime factors.
Correct answer: B. 70
Explanation: Step 1: A number divisible by both 2 and 5 is also divisible by 10. Step 2: 70 ends in 0, so it is divisible by 10. Step 3: Divisibility checks help in prime factorisation.
Correct answer: C. 1200
Explanation: Step 1: Find (2^4=16) and (5^2=25). Step 2: (16\times3\times25=1200). Step 3: In questions with powers, simplify the powers first.
Correct answer: A. It is unique except for order
Explanation: Step 1: The theorem says prime factorisation is unique. Step 2: The order of factors may change, but the prime factors remain the same. Step 3: Understand uniqueness separately from order.
Correct answer: A. (2\times3\times7\times11)
Explanation: Step 1: Write (462=42\times11). Step 2: (42=2\times3\times7), so (462=2\times3\times7\times11). Step 3: In the final form, all factors must be prime.
Correct answer: C. 231
Explanation: Step 1: Multiply the given prime factors. Step 2: (3\times7\times11=231), so the number is 231. Step 3: To get the original number, multiply all prime factors.
Correct answer: A. (2^2\times3^3\times5)
Explanation: Step 1: Write (540=54\times10). Step 2: (54=2\times3^3) and (10=2\times5), so (540=2^2\times3^3\times5). Step 3: Write repeated prime factors using powers.
Correct answer: A. Equal to their product
Explanation: Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM of co-prime numbers is equal to their product.
