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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.
TOPIC PRACTICE
Quiz this set
Up to 20 questions from this page. Select your focus, then start.
If the LCM of (2^5\times3^3), (2^7\times3^2\times5), and (2^4\times3^5\times7) is found, what will be the power of (2) in it?
Correct answer: C
Step 1: In the LCM, take the highest power of (2). Step 2: The powers of (2) are (5), (7), and (4), so the highest power is (7). Step 3: In LCM, do not add powers; take the highest power.
The HCF of two numbers is (33) and their LCM is (2145). If one number is (165), what is the other number?
Correct answer: A
Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{33\times2145}{165}=429). Step 3: Use (165=33\times5) to simplify the division.
A number leaves remainder (0) when divided by (168), (210), and (280). What is the smallest such number?
Correct answer: A
Step 1: Remainder (0) means the number is exactly divisible by all three numbers. Step 2: (168=2^3\times3\times7), (210=2\times3\times5\times7), and (280=2^3\times5\times7), so LCM (=2^3\times3\times5\times7=840). Step 3: For the smallest divisible number, find the LCM.
If (a=2^7\times3^2\times5^3) and (b=2^4\times3^5\times5), what will be the power of (5) in their HCF?
Correct answer: A
Step 1: HCF uses the smaller power of a common prime. Step 2: The powers of (5) are (3) and (1), so the smaller power is (1). Step 3: Recognise power (1) correctly in the answer.
If (L) is the LCM and (H) is the HCF of (2^6\times3^2\times5) and (2^3\times3^6\times5^2), what will be the powers of (3) in (L) and (H) respectively?
Correct answer: A
Step 1: LCM takes the higher power, while HCF takes the lower power. Step 2: The powers of (3) are (2) and (6), so (L) has (6) and (H) has (2). Step 3: Keep the higher-power and lower-power rules separate.
Step 1: (286=26\times11) and (260=26\times10). Step 2: Since (11) and (10) are coprime, HCF is (26) and LCM is (26\times11\times10=2860). Step 3: In options, factor out the HCF and check if the remaining numbers are coprime.
If (288=2^5\times3^2) and (432=2^4\times3^3), the product of their LCM and HCF will be equal to what?
Correct answer: A
Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: Therefore, here the product equals (288\times432). Step 3: Apply this relation directly for two numbers.
If (H) is the HCF of (143), (187), and (253), what is the value of (H)?
Correct answer: A
Step 1: (143=11\times13), (187=11\times17), and (253=11\times23). Step 2: The common prime in all three is (11), so the HCF is (11). Step 3: Identify the prime common to all three numbers.
If the LCM of (2^5\times3^5\times5), (2^7\times3^2\times7), and (2^4\times3^4\times11) is found, what will be the power of (3) in it?
Correct answer: C
Step 1: LCM uses the highest power of (3). Step 2: The powers of (3) are (5), (2), and (4), so the highest power is (5). Step 3: Choose the highest power instead of adding powers.
The HCF of two numbers is (81) and their LCM is (4617). If the numbers are taken as (81r) and (81s), what is the value of (rs)?
Correct answer: C
Step 1: After factoring out HCF (81), (r) and (s) are coprime. Step 2: LCM (=81rs=4617), so (rs=57). Step 3: In such questions, first divide the LCM by the HCF.
If (H) is the HCF and (L) is the LCM of (2^7\times3^3\times5^2) and (2^4\times3^6\times5), what is (\frac{L}{H})?
Correct answer: A
Step 1: (H=2^4\times3^3\times5) and (L=2^7\times3^6\times5^2). Step 2: (\frac{L}{H}=2^{7-4}\times3^{6-3}\times5^{2-1}=2^3\times3^3\times5). Step 3: In division, subtract powers of the same base.
If (H) and (L) are respectively the HCF and LCM of (330), (462), and (770), what is (L\div H)?
Correct answer: B
Step 1: (330=2\times3\times5\times11), (462=2\times3\times7\times11), and (770=2\times5\times7\times11). Step 2: (H=2\times11=22) and (L=2\times3\times5\times7\times11=2310), so (L\div H=105). Step 3: For three numbers, first check common primes and then all distinct primes.
If (L) is the LCM of (2^5\times3^4\times5) and (2^7\times3^2\times5^3\times17), how many distinct prime factors will (L) have?
Correct answer: C
Step 1: The LCM contains all distinct primes appearing in both numbers. Step 2: The distinct primes are (2), (3), (5), and (17), so there are (4). Step 3: Count only distinct prime bases, not their powers.
If the HCF of two numbers is (96) and their LCM is (1248), what is correct about their existence?
Correct answer: A
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.
The prime factorisations of two numbers are (2^6\times3^2\times5^3\times7) and (2^4\times3^5\times5\times11). What will be their HCF?
Correct answer: A
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.
The prime factorisations of three numbers are (2^3\times3^4\times13), (2^5\times3^2\times5^2), and (2^2\times3^5\times7). What will be their LCM?
Correct answer: A
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.
The HCF of two numbers is (45), their LCM is (3465), and one number is (315). What is the other number?
Correct answer: B
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.
If (a=2^7\times3^2\times5\times11) and (b=2^4\times3^6\times5^3\times7), what is (\frac{\text{LCM}}{\text{HCF}})?
Correct answer: A
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.
The HCF of two numbers is (30) and their LCM is (2730). How many unordered pairs of such numbers are possible?
Correct answer: B
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.
A number leaves remainders (38), (68), and (83) when divided by (45), (75), and (90) respectively. What is the smallest such number?
Correct answer: A
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.
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