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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. (14580)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: (18\times810=14580), so the product is (14580). Step 3: If only product is asked, you need not find the individual numbers.
Correct answer: C. (18)
Explanation: Step 1: The greatest common divisor is the HCF. Step 2: (54=2\times3^3), (90=2\times3^2\times5), and (126=2\times3^2\times7), so HCF (=2\times3^2=18). Step 3: Take the smallest powers common to all numbers.
Correct answer: B. (21)
Explanation: Step 1: (63=3^2\times7), (105=3\times5\times7), and (147=3\times7^2). Step 2: The common smaller powers are (3) and (7), so HCF (=21). Step 3: For three numbers, include only primes common to all.
Correct answer: B. (1320)
Explanation: Step 1: (88=2^3\times11), (132=2^2\times3\times11), and (220=2^2\times5\times11). Step 2: The highest powers are (2^3), (3), (5), and (11), so LCM (=1320). Step 3: Take the highest power of each distinct prime.
Correct answer: A. Such whole numbers are not possible
Explanation: Step 1: For two whole numbers, the HCF must exactly divide the LCM. Step 2: (1080\div24=45), so such numbers are possible; the impossible statement is not correct. Step 3: Check all statements before selecting.
Correct answer: A. Such two whole numbers are possible
Explanation: Step 1: The HCF must divide the LCM. Step 2: (1080\div24=45), which is a whole number, so such two whole numbers can exist. Step 3: For existence checks, first test divisibility.
Correct answer: A. Such two whole numbers are not possible
Explanation: Step 1: The HCF must be an exact divisor of the LCM. Step 2: (420) is not exactly divisible by (20), so such two whole numbers are not possible. Step 3: Check this condition before trying to form a pair.
Correct answer: C. (3)
Explanation: Step 1: LCM uses the higher power. Step 2: The powers of (5) are (1) and (3), so (L) contains (5^3). Step 3: Compare exponents only when the base is the same.
Correct answer: A. (1680)
Explanation: Step 1: Remainder (0) means the number is divisible by all three numbers. Step 2: (48=2^4\times3), (80=2^4\times5), and (112=2^4\times7), so LCM (=2^4\times3\times5\times7=1680). Step 3: For the smallest divisible number, use LCM.
Correct answer: A. (2^2\times3^2\times5)
Explanation: Step 1: In (\frac{L}{H}), divide the LCM by the HCF. Step 2: Subtract powers of the same bases: (2^{5-3}\times3^{4-2}\times5=2^2\times3^2\times5). Step 3: Remember subtraction of exponents during division.
Correct answer: B. (756)
Explanation: Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=\frac{252\times378}{126}=756). Step 3: Simplify (252) by (126) first to calculate faster.
Correct answer: A. The HCF will be (p)
Explanation: Step 1: (p) divides (q) because (q=p\times5). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: First identify divisibility between the two numbers.
Correct answer: B. (32)
Explanation: Step 1: The maximum number of equal parts is found by HCF. Step 2: (96=2^5\times3), (128=2^7), and (160=2^5\times5), so HCF (=2^5=32). Step 3: For maximum equal division, use HCF.
Correct answer: A. (2^2\times3)
Explanation: Step 1: HCF includes only primes common to all three numbers. Step 2: (2) and (3) are common, but (5) is not present in the third number; the smallest powers are (2^2) and (3). Step 3: Do not include a prime that is not in every number.
Correct answer: B. (2)
Explanation: Step 1: (35=5\times7), (49=7^2), and (63=3^2\times7). Step 2: The highest power of (7) in the LCM is (2). Step 3: LCM uses the highest power of each prime.
Correct answer: A. (720)
Explanation: Step 1: When (u) and (v) are coprime, the LCM of (18u) and (18v) is (18uv). Step 2: Since (uv=40), LCM (=18\times40=720). Step 3: Factoring out the HCF is useful in such questions.
Correct answer: A. (35)
Explanation: Step 1: LCM (=\frac{180\times252}{36}=1260). Step 2: Then (\frac{1260}{36}=35), so the value is (35). Step 3: First find the LCM, then simplify the ratio.
Correct answer: A. (2^4\times3^3\times5)
Explanation: Step 1: A common multiple must contain all required prime powers of both numbers. Step 2: The required highest powers are (2^4), (3^3), and (5). Step 3: Do not miss any required prime factor while checking a multiple.
Correct answer: A. (2), (5), (7)
Explanation: Step 1: HCF contains only primes present in both numbers. Step 2: (2), (5), and (7) are common, but (3) appears only in the second number. Step 3: Identify common primes before choosing powers.
Correct answer: A. (6) and (4)
Explanation: Step 1: LCM takes the higher power of (2), and HCF takes the lower power. Step 2: The powers are (6) and (4), so (L) has power (6) and (H) has power (4). Step 3: Do not interchange the power rules of LCM and HCF.
Correct answer: B. (1)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: (7) appears only in the LCM as (7^1), so its power in the product is (1). Step 3: In multiplication, add exponents of the same prime.
Correct answer: C. (1500)
Explanation: Step 1: (75=3\times5^2), (125=5^3), and (200=2^3\times5^2). Step 2: The highest powers are (2^3), (3), and (5^3), so LCM (=8\times3\times125=3000). Step 3: Do not forget that (5^3=125).
Correct answer: D. (3000)
Explanation: Step 1: Prime factorise: (75=3\times5^2), (125=5^3), and (200=2^3\times5^2). Step 2: LCM (=2^3\times3\times5^3=3000). Step 3: Use the highest powers to get the final value.
Correct answer: B. (508)
Explanation: Step 1: Subtracting (4) makes the number divisible by (21), (28), and (36). Step 2: (21=3\times7), (28=2^2\times7), and (36=2^2\times3^2), so LCM (=2^2\times3^2\times7=252). Hence the smallest number is (252+4=256). Step 3: Add the common remainder at the end.
Correct answer: A. (256)
Explanation: Step 1: After subtracting (4), the number must be divisible by all three numbers. Step 2: The LCM of (21), (28), and (36) is (252), so the number is (252+4=256). Step 3: In common-remainder questions, find the LCM first.
