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. (2\times3^4)
Explanation: Step 1: Write (162=2\times81). Step 2: (81=3^4), so (162=2\times3^4). Step 3: 18 and 9 are composite, so they are not the final prime form.
Correct answer: A. 1 is neither prime nor composite
Explanation: Step 1: A prime number must have exactly two positive factors. Step 2: 1 has only one positive factor, 1, so it is neither prime nor composite. Step 3: Treating 1 as prime is a common exam mistake.
Correct answer: C. 7
Explanation: Step 1: Carefully look at the given form (128=2^7). Step 2: The exponent written on 2 is 7. Step 3: It is important to identify the base and exponent separately.
Correct answer: A. (2\times7\times11)
Explanation: Step 1: Write (154=2\times77). Step 2: (77=7\times11), so (154=2\times7\times11). Step 3: 14 is composite, so (14\times11) is not the final prime factorisation.
Correct answer: C. 336
Explanation: Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (4\times84=336). Step 3: Use this relation directly for two numbers only.
Correct answer: A. (11^2)
Explanation: Step 1: 121 is a perfect square. Step 2: (121=11\times11=11^2). Step 3: 1 is not written as a factor in prime factorisation.
Correct answer: A. 2, 3 and 11
Explanation: Step 1: Write (198=2\times99). Step 2: (99=9\times11=3^2\times11), so the distinct prime factors are 2, 3, and 11. Step 3: Write repeated 3 only once in the distinct list.
Correct answer: A. 6
Explanation: Step 1: The common prime factors are 2 and 3. Step 2: Their product is (2\times3=6), so the HCF is 6. Step 3: For HCF, take only common prime factors.
Correct answer: C. 210
Explanation: Step 1: For LCM, take all required prime factors. Step 2: (2\times3\times5\times7=210). Step 3: LCM includes all necessary prime factors from both numbers.
Correct answer: A. (2\times5^3)
Explanation: Step 1: Write (250=2\times125). Step 2: (125=5^3), so (250=2\times5^3). Step 3: 25 and 10 are composite, so do not keep them in the final prime form.
Correct answer: C. 56
Explanation: Step 1: Find (2^3=8). Step 2: (8\times7=56), so the number is 56. Step 3: To get the number from prime factorisation, simplify powers first.
Correct answer: B. 96
Explanation: Step 1: Find (2^5=32). Step 2: (32\times3=96), so the number is 96. Step 3: First evaluate the power, then multiply.
Correct answer: C. (6\times5)
Explanation: Step 1: In prime factorisation, every factor must be prime. Step 2: In (6\times5), 6 is not prime. Step 3: If a composite factor appears, break it further into prime factors.
Correct answer: A. (2^4\times11)
Explanation: Step 1: Write (176=16\times11). Step 2: (16=2^4) and 11 is prime, so (176=2^4\times11). 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: (22=2\times11) and (35=5\times7). Step 2: They have no common prime factor, so they are co-prime. Step 3: Co-prime numbers have HCF 1.
Correct answer: A. (2^8)
Explanation: Step 1: Divide 256 repeatedly by 2. Step 2: (256=2^8), so this is the prime factorisation. Step 3: 4 and 16 are composite, so (4^4) or (16^2) are not final prime forms.
Correct answer: B. 42
Explanation: Step 1: The common prime factors are 2, 3, and 7. Step 2: The smaller powers are (2^1), (3^1), and (7^1), so (2\times3\times7=42). Step 3: Take smaller powers for HCF.
Correct answer: B. 252
Explanation: Step 1: For LCM, take the highest powers. Step 2: (2^2\times3^2\times7=4\times9\times7=252). Step 3: LCM includes the highest powers of all prime factors.
Correct answer: B. 45
Explanation: Step 1: A number divisible by both 3 and 5 is divisible by 15. Step 2: The digit sum of 45 is 9, so it is divisible by 3, and it ends in 5, so it is also divisible by 5. Step 3: Small divisibility rules help in factorisation.
Correct answer: C. 600
Explanation: Step 1: Find (2^3=8) and (5^2=25). Step 2: (8\times3\times25=600). Step 3: In questions with powers, simplify the powers first.
Correct answer: A. Positive integers greater than 1
Explanation: Step 1: This theorem is stated for positive integers greater than 1. Step 2: Every such number can be prime factorised. Step 3: 1 is not included in the usual prime factorisation statement.
Correct answer: A. (2\times3\times5\times11)
Explanation: Step 1: Write (330=33\times10). Step 2: (33=3\times11) and (10=2\times5), so (330=2\times3\times5\times11). Step 3: In the final form, all factors must be prime.
Correct answer: C. 110
Explanation: Step 1: Multiply the given prime factors. Step 2: (2\times5\times11=110), so the number is 110. Step 3: To get the original number, multiply all prime factors.
Correct answer: A. (2\times3^3\times5)
Explanation: Step 1: Write (270=27\times10). Step 2: (27=3^3) and (10=2\times5), so (270=2\times3^3\times5). Step 3: Write repeated prime factors using powers.
Correct answer: A. Their LCM is 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.
