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SubjectsMathematics
अभाज्य गुणनखंडन द्वारा महत्तम समापवर्तक और लघुत्तम समापवर्त्य
In this Class 10 Mathematics topic from Real Numbers, students learn to find the HCF and LCM of two or more numbers through prime factorisation. They break each number into prime factors, compare the resulting powers, and select the appropriate common factors: the lowest powers for HCF and the highest powers for LCM. The topic also develops accuracy in writing factor trees, simplifies comparison between methods, and helps students apply these concepts to numerical and everyday problem-solving situations.
Correct answer: A. Such whole numbers are possible
Explanation: Step 1: The HCF must exactly divide the LCM. Step 2: (1248\div96=13), which is a whole number, so such whole numbers can exist. Step 3: For existence checks, first test this necessary condition.
Correct answer: A. (2^4\times3^2\times5)
Explanation: Step 1: HCF uses only common prime factors. Step 2: The common primes are (2), (3), and (5), with smaller powers (2^4), (3^2), and (5). Step 3: Choose the smaller power for each prime base separately.
Correct answer: A. (2^5\times3^5\times5^2\times7\times13)
Explanation: Step 1: LCM takes the highest power of every prime present. Step 2: The highest powers are (2^5), (3^5), (5^2), (7), and (13). Step 3: A prime appearing in only one number must also be included.
Correct answer: B. (495)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{45\times3465}{315}=495). Step 3: Notice (315=45\times7) to calculate quickly.
Correct answer: A. (2^3\times3^4\times5^2\times7\times11)
Explanation: Step 1: HCF is (2^4\times3^2\times5), and LCM is (2^7\times3^6\times5^3\times7\times11). Step 2: On division, subtract powers to get (2^3\times3^4\times5^2\times7\times11). Step 3: In ratios, subtract exponents of the same base.
Correct answer: B. (2)
Explanation: Step 1: Let the numbers be (30m) and (30n), where (m) and (n) are coprime. Step 2: (30mn=2730), so (mn=91=7\times13); the unordered coprime pairs are ((1,91)) and ((7,13)). Step 3: Do not count reversed order as a new pair.
Correct answer: A. (443)
Explanation: Step 1: In each case, divisor minus remainder is (7), so adding (7) to the number makes it divisible by all three divisors. Step 2: The LCM of (45), (75), and (90) is (450), so the number is (450-7=443). Step 3: Spot the common difference and subtract it from the LCM.
Correct answer: A. (a=6), (b=4)
Explanation: Step 1: HCF uses the smaller power of each common prime. Step 2: The smaller power of (2) must be (6), so (a=6) is possible; the smaller power of (3) must be (4), so (b=4) is possible. Step 3: Check each base condition separately.
Correct answer: A. (a=7), (b=3)
Explanation: Step 1: LCM takes the highest power of every prime. Step 2: The highest power of (3) must be (7), so (a=7); the highest power of (5) must be (3), so (b=3). Step 3: Identify the maximum powers in LCM.
Correct answer: C. (156)
Explanation: Step 1: The maximum number of identical packets is found by HCF. Step 2: (312=2^3\times3\times13) and (468=2^2\times3^2\times13), so HCF (=2^2\times3\times13=156). Step 3: Use HCF for maximum equal distribution.
Correct answer: B. (252)
Explanation: Step 1: The next common signal time is the LCM of all intervals. Step 2: (28=2^2\times7), (36=2^2\times3^2), (63=3^2\times7), and (84=2^2\times3\times7), so the LCM is (252). Step 3: Use LCM for repeated-time questions.
Correct answer: C. (6)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: The powers of (3) are (2) and (4), so the total power is (6). Step 3: When multiplying powers with the same base, add the exponents.
Correct answer: C. (35:1)
Explanation: Step 1: LCM (=\frac{330\times462}{66}=2310). Step 2: The ratio (2310:66=35:1). Step 3: Do not forget to reduce the ratio to simplest form.
Correct answer: C. (96)
Explanation: Step 1: The maximum number of equal parts is found by HCF. Step 2: (192=2^6\times3), (288=2^5\times3^2), and (480=2^5\times3\times5), so HCF (=2^5\times3=96). Step 3: In maximum equal division, take the smallest common powers.
Correct answer: D. (2178000)
Explanation: Step 1: The smallest number exactly divisible by all is the LCM. Step 2: (121=11^2), (144=2^4\times3^2), and (250=2\times5^3), so LCM (=2^4\times3^2\times5^3\times11^2=2178000). Step 3: Multiply the highest powers carefully.
Correct answer: A. The HCF will be (p)
Explanation: Step 1: (q=p\times19), so (p) exactly divides (q). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: Identifying a multiple relation saves time.
Correct answer: C. (65)
Explanation: Step 1: After taking out HCF (84), (r) and (s) are coprime. Step 2: LCM (=84rs=5460), so (rs=65). Step 3: First divide the LCM by the HCF.
Correct answer: B. (3)
Explanation: Step 1: Compare the powers of (3). Step 2: (216=2^3\times3^3), (324=2^2\times3^4), and (540=2^2\times3^3\times5), so the smallest power is (3). Step 3: HCF uses the smallest power.
Correct answer: C. (5)
Explanation: Step 1: Prime factorise: (196=2^2\times7^2), (225=3^2\times5^2), and (308=2^2\times7\times11). Step 2: The distinct primes in the LCM are (2), (3), (5), (7), and (11), so there are (5). Step 3: Count distinct prime bases, not powers.
Correct answer: A. (4) and (1)
Explanation: Step 1: LCM takes the higher power, and HCF takes the lower power. Step 2: The powers of (5) are (4) and (1), so (L) has (4) and (H) has (1). Step 3: Do not interchange the two rules.
Correct answer: A. (1)
Explanation: Step 1: (221=13\times17), (323=17\times19), and (437=19\times23). Step 2: No prime factor is common to all three numbers, so the HCF is (1). Step 3: Only a factor present in all three numbers is taken.
Correct answer: B. (660)
Explanation: Step 1: LCM (=\frac{264\times396}{132}=792). Step 2: The difference is (792-132=660). Step 3: When difference is asked, first find the LCM correctly.
Correct answer: C. (5)
Explanation: Step 1: The smallest number divisible by both is their LCM. Step 2: The powers of (3) are (2) and (5), so the higher power (5) is used. Step 3: For divisibility, choose the required highest power.
Correct answer: D. (1517)
Explanation: Step 1: Coprime numbers have HCF (1). Step 2: Product (=) HCF (\times) LCM, so the LCM is (1517). Step 3: For coprime numbers, LCM equals the product.
Correct answer: A. (2^3\times3)
Explanation: Step 1: HCF of three numbers includes only primes common to all three. Step 2: (2) and (3) are common, with smallest powers (2^3) and (3). Step 3: Do not include a prime that is not present in every number.
