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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: B. (1800)
Explanation: Step 1: The smallest number divisible by all is the LCM. Step 2: (36=2^2\times3^2), (100=2^2\times5^2), and (150=2\times3\times5^2), so LCM (=2^2\times3^2\times5^2=900). Step 3: Always verify the final multiplication before choosing an option.
Correct answer: A. (900)
Explanation: Step 1: Such a smallest number is the LCM of the three numbers. Step 2: (36=2^2\times3^2), (100=2^2\times5^2), and (150=2\times3\times5^2), so LCM (=2^2\times3^2\times5^2=900). Step 3: Take the highest power of each prime.
Correct answer: A. (2^4\times3^2\times5\times7)
Explanation: Step 1: In (\frac{L}{H}), divide the LCM by the HCF. Step 2: Subtract powers of the same bases: (2^{6-2}\times3^{3-1}\times5^{2-1}\times7=2^4\times3^2\times5\times7). Step 3: Use exponent subtraction in division.
Correct answer: B. (858)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: LCM (=\frac{286\times429}{143}=858). Step 3: Divide (286) by (143) first to get (2), then multiply (2\times429).
Correct answer: A. The HCF will be (p)
Explanation: Step 1: (q=p\times7), so (p) exactly divides (q). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: First check whether one number is a multiple of the other.
Correct answer: B. (48)
Explanation: Step 1: The maximum number of equal parts is found using HCF. Step 2: (96=2^5\times3), (144=2^4\times3^2), and (192=2^6\times3), so HCF (=2^4\times3=48). Step 3: For maximum equal division, use HCF.
Correct answer: A. (2^2\times3^2)
Explanation: Step 1: HCF includes only primes common to all three numbers. Step 2: (2) and (3) are common, but (5) is not in the third number; the smallest powers are (2^2) and (3^2). Step 3: Do not include a prime that is not present in every number.
Correct answer: B. (2)
Explanation: Step 1: (44=2^2\times11), (77=7\times11), and (121=11^2). Step 2: The highest power of (11) in the LCM is (2). Step 3: For LCM, choose the highest power.
Correct answer: C. (780)
Explanation: Step 1: When (u) and (v) are coprime, the LCM of (30u) and (30v) is (30uv). Step 2: Since (uv=26), LCM (=30\times26=780). Step 3: Factoring out the HCF simplifies the question.
Correct answer: B. (15)
Explanation: Step 1: LCM (=\frac{216\times360}{72}=1080). Step 2: Then (\frac{1080}{72}=15), so the value of the ratio is (15). Step 3: First find the LCM, then simplify the ratio.
Correct answer: A. (2^6\times3^3\times5\times7)
Explanation: Step 1: A common multiple must contain all prime powers required by both numbers. Step 2: The highest powers are (2^6), (3^3), (5), and (7). Step 3: Do not miss any required prime factor while checking a multiple.
Correct answer: A. (2), (3), (7)
Explanation: Step 1: HCF contains only primes present in both numbers. Step 2: (2), (3), and (7) are common, but (5) appears only in the second number. Step 3: First identify common primes, then choose their smaller powers.
Correct answer: A. (3) and (1)
Explanation: Step 1: LCM takes the higher power of (5), and HCF takes the lower power. Step 2: The powers of (5) are (1) and (3), so (L) has power (3) and (H) has power (1). Step 3: Do not interchange the two rules.
Correct answer: B. (1)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: (11) appears only in the LCM as (11^1), so its power in the product is (1). Step 3: In multiplication, add powers of the same prime.
Correct answer: C. (1620)
Explanation: Step 1: (108=2^2\times3^3), (162=2\times3^4), and (270=2\times3^3\times5). Step 2: The highest powers are (2^2), (3^4), and (5), so LCM (=4\times81\times5=1620). Step 3: Use the highest powers to get the final value.
Correct answer: A. (299)
Explanation: Step 1: After subtracting (11), the number must be divisible by all three numbers. Step 2: (32=2^5), (48=2^4\times3), and (72=2^3\times3^2), so LCM (=2^5\times3^2=288). Hence the number is (288+11=299). Step 3: Add the common remainder at the end.
Correct answer: C. (6)
Explanation: Step 1: The highest power of (2) in the LCM must be (6). Step 2: The second number has power (4), so (a=6) gives the required highest power (6). Step 3: For LCM, check the maximum-power condition.
Correct answer: C. (2520)
Explanation: Step 1: Product (=) HCF (\times) LCM. Step 2: LCM (=\frac{70560}{28}=2520). Step 3: You can divide first by (7) and then by (4) to simplify.
Correct answer: B. (462)
Explanation: Step 1: For two numbers, LCM (=\frac{\text{first number}\times\text{second number}}{\text{HCF}}). Step 2: (\frac{154\times231}{77}=462). Step 3: First calculate (154\div77=2) for a quicker solution.
Correct answer: A. (2^3\times3^2\times11)
Explanation: Step 1: HCF takes the smaller power of each common prime. Step 2: The smaller powers of (2), (3), and (11) are (3), (2), and (1), so HCF (=2^3\times3^2\times11). Step 3: Compare the powers for each base separately.
Correct answer: A. (11)
Explanation: Step 1: (91=7\times13), (143=11\times13), and (187=11\times17). Step 2: There is no prime factor common to all three, so the HCF should be (1). Step 3: For three numbers, a common factor must be present in every number.
Correct answer: A. (1)
Explanation: Step 1: (91=7\times13), (143=11\times13), and (187=11\times17). Step 2: No prime factor is common to all three numbers, so the HCF is (1). Step 3: A factor common to only two numbers is not enough for the HCF of all three.
Correct answer: C. (3)
Explanation: Step 1: LCM uses the higher power of every prime. Step 2: The powers of (3) are (1) and (3), so (L) contains (3^3). Step 3: Compare powers only for the same base.
Correct answer: A. Such two whole numbers are possible
Explanation: Step 1: The HCF must exactly divide the LCM. Step 2: (420\div15=28), which is a whole number, so such two whole numbers can exist. Step 3: For existence checks, test divisibility first.
Correct answer: B. (147)
Explanation: Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{21\times1386}{198}=147). Step 3: Simplifying the division first makes the calculation faster.
