What is the HCF of (14) and (25)?
Correct answer: A
Step 1: (14=2\times7) and (25=5^2). Step 2: They have no common prime factor. Step 3: Therefore, the HCF is (1), and such numbers are co-prime.
Mathematics
HCF and LCM using prime factorisation
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
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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Easy · Level 2
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- (1)
- (2)
- (5)
- (7)
Easy · Level 2
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- (39)
- (175)
- (350)
- (700)
Easy · Level 2
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- (15)
- (23)
- (60)
- (120)
Easy · Level 2
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- (3)
- (5)
- (7)
- (15)
Easy · Level 2
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- (35)
- (70)
- (105)
- (735)
Easy · Level 2
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- (6)
- (12)
- (18)
- (24)
Easy · Level 2
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- (5)
- (10)
- (15)
- (30)
Easy · Level 2
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- (24)
- (30)
- (36)
- (48)
Easy · Level 2
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- (30)
- (60)
- (90)
- (120)
Easy · Level 2
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- (12) cm
- (18) cm
- (24) cm
- (36) cm
Easy · Level 2
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- (2) m
- (4) m
- (5) m
- (7) m
Easy · Level 2
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- (9)
- (18)
- (27)
- (54)
Easy · Level 2
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- (162)
- (243)
- (324)
- (486)
Easy · Level 2
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- (7)
- (14)
- (21)
- (28)
Easy · Level 2
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- (126)
- (168)
- (252)
- (504)
Easy · Level 2
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- (12)
- (24)
- (36)
- (48)
Easy · Level 2
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- (240)
- (360)
- (480)
- (960)
Easy · Level 2
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- (5)
- (10)
- (20)
- (25)
Easy · Level 2
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- (100)
- (200)
- (400)
- (500)
Easy · Level 2
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- (11)
- (22)
- (33)
- (44)
Easy · Level 2
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- (132)
- (264)
- (528)
- (1056)
Easy · Level 2
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- (42)
- (56)
- (84)
- (126)
Easy · Level 2
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- (100)
- (200)
- (250)
- (400)
Easy · Level 2
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- They are co-prime
- They are always equal
- Their LCM is (1)
- Both are always prime
Easy · Level 2
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- Smaller power of common factors
- Highest power of all factors
- Only uncommon factors
- Sum of all factors
What is the LCM of (14) and (25)?
Correct answer: C
Step 1: (14=2\times7) and (25=5^2). Step 2: Since they have no common prime factor, include all factors for LCM. Step 3: (2\times7\times25=350), so the LCM is (350).
What will be the LCM of the co-prime numbers (8) and (15)?
Correct answer: D
Step 1: (8=2^3) and (15=3\times5). Step 2: They have no common prime factor. Step 3: The LCM of co-prime numbers is their product, so (8\times15=120).
What will be the HCF of (21) and (35)?
Correct answer: C
Step 1: (21=3\times7) and (35=5\times7). Step 2: The common prime factor is (7). Step 3: Therefore, the HCF is (7).
What will be the LCM of (21) and (35)?
Correct answer: C
Step 1: (21=3\times7) and (35=5\times7). Step 2: For LCM, take (3), (5), and (7). Step 3: (3\times5\times7=105), so the LCM is (105).
A shopkeeper has (24) pencils and (36) erasers. He wants to make equal packets. What is the maximum number of packets possible?
Correct answer: B
Step 1: For the maximum number of equal packets, find the HCF. Step 2: (24=2^3\times3) and (36=2^2\times3^2), so HCF (=2^2\times3=12). Step 3: When maximum equal groups are asked, use HCF.
In how many maximum equal boxes can (30), (45), and (60) toffees be divided?
Correct answer: C
Step 1: For maximum equal boxes, find the HCF of (30), (45), and (60). Step 2: (30=2\times3\times5), (45=3^2\times5), (60=2^2\times3\times5). Step 3: The common part is (3\times5=15), so (15) boxes can be made.
Two bells ring after every (12) minutes and (18) minutes respectively. After how many minutes will they ring together again?
Correct answer: C
Step 1: The time when both bells ring together is the LCM of the two times. Step 2: (12=2^2\times3) and (18=2\times3^2), so LCM (=2^2\times3^2=36). Step 3: For repeated time events, use LCM.
Three lights flash at intervals of (10), (15), and (20) seconds. After how many seconds will they flash together again?
Correct answer: B
Step 1: The next common flashing time is the LCM of (10), (15), and (20). Step 2: (10=2\times5), (15=3\times5), (20=2^2\times5). Step 3: (2^2\times3\times5=60), so the answer is (60) seconds.
Two ropes of (48) cm and (72) cm are to be cut into equal pieces of maximum length. What will be the length of each piece?
Correct answer: C
Step 1: For equal pieces of maximum length, find the HCF. Step 2: (48=2^4\times3) and (72=2^3\times3^2), so HCF (=2^3\times3=24). Step 3: For cutting or dividing into maximum equal parts, use HCF.
Two strips of (20) m and (28) m are to be cut into equal pieces of maximum length. What will be the length of each piece?
Correct answer: B
Step 1: For maximum equal length, find the HCF of (20) and (28). Step 2: (20=2^2\times5) and (28=2^2\times7). Step 3: The common part is (2^2=4), so each piece will be (4) m long.
Which value is the HCF of (54) and (81)?
Correct answer: C
Step 1: (54=2\times3^3) and (81=3^4). Step 2: The common prime factor is (3), and the smaller power is (3^3). Step 3: (3^3=27), so the HCF is (27).
Which value is the LCM of (54) and (81)?
Correct answer: A
Step 1: (54=2\times3^3) and (81=3^4). Step 2: Take the highest powers (2) and (3^4). Step 3: (2\times81=162), so the LCM is (162).
What is the HCF of (63) and (84)?
Correct answer: C
Step 1: (63=3^2\times7) and (84=2^2\times3\times7). Step 2: The common factors are (3) and (7). Step 3: (3\times7=21), so the HCF is (21).
What is the LCM of (63) and (84)?
Correct answer: C
Step 1: (63=3^2\times7) and (84=2^2\times3\times7). Step 2: The highest powers are (2^2), (3^2), and (7). Step 3: (4\times9\times7=252), so the LCM is (252).
If (96=2^5\times3) and (120=2^3\times3\times5), what is their HCF?
Correct answer: B
Step 1: The common prime factors are (2) and (3). Step 2: The smaller powers are (2^3) and (3). Step 3: (2^3\times3=24), so the HCF is (24).
If (96=2^5\times3) and (120=2^3\times3\times5), what is their LCM?
Correct answer: C
Step 1: The LCM contains the highest powers of all prime factors. Step 2: The highest powers are (2^5), (3), and (5). Step 3: (32\times3\times5=480), so the answer is (480).
What is the HCF of (25), (40), and (50)?
Correct answer: A
Step 1: (25=5^2), (40=2^3\times5), and (50=2\times5^2). Step 2: The common prime factor in all three numbers is (5). Step 3: The smallest power is (5), so the HCF is (5).
What is the LCM of (25), (40), and (50)?
Correct answer: B
Step 1: (25=5^2), (40=2^3\times5), and (50=2\times5^2). Step 2: The highest powers are (2^3) and (5^2). Step 3: (8\times25=200), so the LCM is (200).
What is the HCF of (88) and (132)?
Correct answer: D
Step 1: (88=2^3\times11) and (132=2^2\times3\times11). Step 2: The smaller powers of common factors are (2^2) and (11). Step 3: (4\times11=44), so the HCF is (44).
What is the LCM of (88) and (132)?
Correct answer: B
Step 1: (88=2^3\times11) and (132=2^2\times3\times11). Step 2: The highest powers are (2^3), (3), and (11). Step 3: (8\times3\times11=264), so the LCM is (264).
The prime factorisation of a number is (2^2\times3\times7). What is the number?
Correct answer: C
Step 1: First simplify the given prime powers. Step 2: Since (2^2=4), the number is (4\times3\times7=84). Step 3: Reading prime factorisation correctly helps in HCF and LCM questions.
The prime factorisation of a number is (2^3\times5^2). What is the number?
Correct answer: B
Step 1: (2^3=8) and (5^2=25). Step 2: Multiplying them gives (8\times25=200). Step 3: Always simplify powers first and then multiply.
If the HCF of two numbers is (1), which statement about them is correct?
Correct answer: A
Step 1: If the HCF of two numbers is (1), they have no common factor except (1). Step 2: Such numbers are called co-prime. Step 3: Co-prime numbers need not both be prime numbers.
In the prime factorisation method for finding HCF, which power is chosen?
Correct answer: A
Step 1: HCF is made only from common prime factors. Step 2: Among common factors, the smaller power is chosen because it divides both numbers. Step 3: Remember the rule: common factors with smaller powers for HCF.
QUIZ COMPLETE