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In Class 10 Mathematics, under the Real Numbers chapter, Prime Factorisation teaches students to express a composite number as a product of prime numbers. Students practise identifying prime factors and writing factorisations using multiplication and exponents. This foundational skill supports the Fundamental Theorem of Arithmetic and later work with HCF and LCM.
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Medium · Level 8 · prime-factorisation,concept,mediumView options
The factorisation is not considered complete
The value of the number always changes
All factors become prime
Powers cannot be used
Medium · Level 8 · prime-factorisation,number-792,mediumView options
(2^3\times3^2\times11)
(2^2\times3^3\times11)
(8\times99)
(2^3\times9\times11)
Medium · Level 8 · prime-factorisation,number-825,mediumView options
(3\times5^2\times11)
(3^2\times5\times11)
(25\times33)
(5\times165)
Medium · Level 8 · prime-factorisation,number-864,mediumView options
(2^5\times3^3)
(2^4\times3^3)
(32\times27)
(2^5\times27)
Medium · Level 8 · prime-factorisation,number-924,mediumView options
(2^2\times3\times7\times11)
(2\times3\times7\times22)
(4\times231)
(2^2\times21\times11)
Medium · Level 8 · prime-factorisation,number-1050,mediumView options
(2\times3\times5^2\times7)
(2^2\times3\times5\times7)
(105\times10)
(2\times15\times35)
Medium · Level 8 · prime-factorisation,number-1188,mediumView options
(2^2\times3^3\times11)
(2^3\times3^2\times11)
(4\times297)
(2^2\times27\times11)
Medium · Level 8 · prime-factorisation,number-1260,mediumView options
(2^2\times3^2\times5\times7)
(2\times3^2\times5\times7)
(126\times10)
(2^2\times9\times35)
Medium · Level 8 · prime-factorisation,number-1350,mediumView options
(2\times3^3\times5^2)
(2\times3^2\times5^2)
(27\times50)
(2\times27\times25)
Medium · Level 8 · prime-factorisation,number-1386,mediumView options
(2\times3^2\times7\times11)
(2^2\times3\times7\times11)
(2\times693)
(18\times77)
Medium · Level 8 · prime-factorisation,number-1500,mediumView options
(2^2\times3\times5^3)
(2^3\times3\times5^2)
(15\times100)
(3\times500)
Medium · Level 8 · prime-factorisation,number-1584,mediumView options
(2^4\times3^2\times11)
(2^3\times3^2\times11)
(16\times99)
(2^4\times9\times11)
Medium · Level 8 · prime-factorisation,number-1680,mediumView options
(2^4\times3\times5\times7)
(2^3\times3\times5\times7)
(168\times10)
(16\times105)
Medium · Level 8 · prime-factorisation,number-1890,mediumView options
(2\times3^3\times5\times7)
(2\times3^2\times5\times7)
(27\times70)
(2\times27\times35)
Medium · Level 8 · prime-factorisation,number-2100,mediumView options
(2^2\times3\times5^2\times7)
(2\times3\times5^2\times7)
(21\times100)
(2^2\times3\times25\times7)
Medium · Level 8 · prime-factorisation,number-2310,mediumView options
(2\times3\times5\times7\times11)
(21\times110)
(2\times3\times5\times77)
(30\times77)
Medium · Level 8 · prime-factorisation,number-2520,mediumView options
(2^3\times3^2\times5\times7)
(2^2\times3^2\times5\times7)
(252\times10)
(8\times315)
Medium · Level 8 · prime-factorisation,number-2772,mediumView options
(2^2\times3^2\times7\times11)
(2\times3^2\times7\times11)
(36\times77)
(4\times693)
Medium · Level 8 · prime-factorisation,number-3024,mediumView options
(2^4\times3^3\times7)
(2^3\times3^3\times7)
(16\times189)
(2^4\times27\times7)
Medium · Level 8 · prime-factorisation,number-3150,mediumView options
(2\times3^2\times5^2\times7)
(2^2\times3^2\times5\times7)
(315\times10)
(2\times9\times25\times7)
Question 1MediumLevel 8
What problem occurs if a composite base is kept in the final prime factorisation?
Correct answer: A
Step 1: In final prime factorisation, every base must be prime. Step 2: If a composite base like 12 or 21 remains, it must be broken further. Step 3: In exams, check every base before writing the final form.
Step 1: Write (825=25\times33). Step 2: (25=5^2) and (33=3\times11), so (825=3\times5^2\times11). Step 3: Both 25 and 33 are composite, so convert them into prime form.
Step 1: Write (924=4\times231). Step 2: (4=2^2) and (231=3\times7\times11), so (924=2^2\times3\times7\times11). Step 3: Give 231 its complete prime form.
Step 1: Write (1260=126\times10). Step 2: (126=2\times3^2\times7) and (10=2\times5), so (1260=2^2\times3^2\times5\times7). Step 3: The total power of 2 is 2.
Step 1: Write (1350=27\times50). Step 2: (27=3^3) and (50=2\times5^2), so (1350=2\times3^3\times5^2). Step 3: Do not leave 27 and 50 in the final form.
Step 1: Write (1386=18\times77). Step 2: (18=2\times3^2) and (77=7\times11), so (1386=2\times3^2\times7\times11). Step 3: Give prime form to both 18 and 77.
Step 1: Write (1500=15\times100). Step 2: (15=3\times5) and (100=2^2\times5^2), so (1500=2^2\times3\times5^3). Step 3: Count the total power of 5 as 3.
Step 1: Write (1680=16\times105). Step 2: (16=2^4) and (105=3\times5\times7), so (1680=2^4\times3\times5\times7). Step 3: Change 105 into prime factors.
Step 1: Write (2100=21\times100). Step 2: (21=3\times7) and (100=2^2\times5^2), so (2100=2^2\times3\times5^2\times7). Step 3: Do not keep 25 or 100 in the final answer.
Step 1: Write (2310=30\times77). Step 2: (30=2\times3\times5) and (77=7\times11), so (2310=2\times3\times5\times7\times11). Step 3: Break both 30 and 77 completely.
Step 1: Write (2520=252\times10). Step 2: (252=2^2\times3^2\times7) and (10=2\times5), so (2520=2^3\times3^2\times5\times7). Step 3: Count the total power of 2 correctly.
Step 1: Write (2772=36\times77). Step 2: (36=2^2\times3^2) and (77=7\times11), so (2772=2^2\times3^2\times7\times11). Step 3: Write 36 and 77 in prime form.
Step 1: Write (3150=315\times10). Step 2: (315=3^2\times5\times7) and (10=2\times5), so (3150=2\times3^2\times5^2\times7). Step 3: The power of 5 becomes 2.
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