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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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Hard · Level 8 · prime-factorisation,composite-base,hardView options
Because 36 is composite
Because 7 is composite
Because writing powers is always wrong
Because multiplication is not used
Hard · Level 8 · prime-factorisation,number-5544,hardView options
(2^3\times3^2\times7\times11)
(2^2\times3^3\times7\times11)
(8\times693)
(2^3\times9\times77)
Hard · Level 8 · prime-factorisation,number-6048,hardView options
(2^5\times3^3\times7)
(2^4\times3^3\times7)
(32\times189)
(2^5\times27\times7)
Hard · Level 8 · prime-factorisation,number-6930,hardView options
(2\times3^2\times5\times7\times11)
(2^2\times3\times5\times7\times11)
(63\times110)
(2\times9\times385)
Hard · Level 8 · prime-factorisation,number-8316,hardView options
(2^2\times3^3\times7\times11)
(2^3\times3^2\times7\times11)
(4\times2079)
(36\times231)
Hard · Level 8 · prime-factorisation,number-9240,hardView options
(2^3\times3\times5\times7\times11)
(2^2\times3\times5\times7\times11)
(8\times1155)
(2^3\times105\times11)
Hard · Level 8 · prime-factorisation,number-10395,hardView options
(3^3\times5\times7\times11)
(3^2\times5\times7\times11)
(27\times385)
(135\times77)
Hard · Level 8 · prime-factorisation,square-number,hardView options
(3^2\times5^2\times7^2)
(3^3\times5^2\times7)
(105\times105)
(9\times1225)
Hard · Level 8 · prime-factorisation,number-11760,hardView options
(2^4\times3\times5\times7^2)
(2^3\times3^2\times5\times7)
(16\times735)
(2^4\times15\times49)
Hard · Level 8 · prime-factorisation,number-14112,hardView options
(2^5\times3^2\times7^2)
(2^4\times3^3\times7^2)
(32\times441)
(2^5\times21^2)
Hard · Level 8 · prime-factorisation,number-15120,hardView options
(2^4\times3^3\times5\times7)
(2^3\times3^3\times5\times7)
(16\times945)
(2^4\times27\times35)
Hard · Level 8 · prime-factorisation,number-16632,hardView options
(2^3\times3^3\times7\times11)
(2^2\times3^4\times7\times11)
(8\times2079)
(2^3\times27\times77)
Hard · Level 8 · prime-factorisation,number-17640,hardView options
(2^3\times3^2\times5\times7^2)
(2^4\times3^2\times5\times7)
(8\times2205)
(2^3\times45\times49)
Hard · Level 8 · prime-factorisation,square-number,hardView options
(2^4\times5^2\times7^2)
(2^3\times5^2\times7^2)
(140\times140)
(16\times1225)
Hard · Level 8 · prime-factorisation,number-20790,hardView options
(2\times3^3\times5\times7\times11)
(2^2\times3^2\times5\times7\times11)
(54\times385)
(2\times27\times385)
Hard · Level 8 · prime-factorisation,number-21168,hardView options
(2^4\times3^3\times7^2)
(2^5\times3^2\times7^2)
(16\times1323)
(2^4\times27\times49)
Hard · Level 8 · prime-factorisation,number-24255,hardView options
(3^2\times5\times7^2\times11)
(3^3\times5\times7\times11)
(45\times539)
(9\times2695)
Hard · Level 8 · prime-factorisation,number-27720,hardView options
(2^3\times3^2\times5\times7\times11)
(2^2\times3^3\times5\times7\times11)
(8\times3465)
(2^3\times9\times385)
Hard · Level 8 · prime-factorisation,number-28224,hardView options
(2^6\times3^2\times7^2)
(2^5\times3^3\times7^2)
(64\times441)
(2^6\times21^2)
Hard · Level 8 · prime-factorisation,number-31104,hardView options
(2^7\times3^5)
(2^6\times3^5)
(128\times243)
(2^7\times81\times3)
Question 1HardLevel 8
Why is (36^2\times7) not considered the final form in prime factorisation?
Correct answer: A
Step 1: In final prime factorisation, the bases should be prime. Step 2: (36=2^2\times3^2), so (36^2) must be changed into (2^4\times3^4). Step 3: Do not keep a composite base like 36 in the final answer.
Step 1: Write (5544=8\times693). Step 2: (8=2^3) and (693=3^2\times7\times11), so (5544=2^3\times3^2\times7\times11). Step 3: Do not leave 693 in the final form.
Step 1: Write (6930=63\times110). Step 2: (63=3^2\times7) and (110=2\times5\times11), so (6930=2\times3^2\times5\times7\times11). Step 3: Give prime form to both 63 and 110.
Step 1: Write (8316=36\times231). Step 2: (36=2^2\times3^2) and (231=3\times7\times11), so (8316=2^2\times3^3\times7\times11). Step 3: The total power of 3 becomes 3.
Step 1: Write (9240=8\times1155). Step 2: (8=2^3) and (1155=3\times5\times7\times11), so (9240=2^3\times3\times5\times7\times11). Step 3: Give 1155 its complete prime form.
Step 1: Write (10395=27\times385). Step 2: (27=3^3) and (385=5\times7\times11), so (10395=3^3\times5\times7\times11). Step 3: Break 385 into prime form too.
Step 1: Recognise (11025=105^2). Step 2: Since (105=3\times5\times7), (11025=3^2\times5^2\times7^2). Step 3: In a square number, prime exponents are even.
Step 1: Write (11760=16\times735). Step 2: (16=2^4) and (735=3\times5\times7^2), so (11760=2^4\times3\times5\times7^2). Step 3: Give 735 its complete prime form.
Step 1: Write (15120=16\times945). Step 2: (16=2^4) and (945=3^3\times5\times7), so (15120=2^4\times3^3\times5\times7). Step 3: Do not leave 945 in the final form.
Step 1: Write (16632=8\times2079). Step 2: (2079=3^3\times7\times11), so (16632=2^3\times3^3\times7\times11). Step 3: Break 2079 down to prime factors.
Step 1: Write (17640=8\times2205). Step 2: (2205=3^2\times5\times7^2), so (17640=2^3\times3^2\times5\times7^2). Step 3: Give 2205 its complete prime form.
Step 1: Write (20790=54\times385). Step 2: (54=2\times3^3) and (385=5\times7\times11), so (20790=2\times3^3\times5\times7\times11). Step 3: Break both 54 and 385 completely.
Step 1: Write (24255=45\times539). Step 2: (45=3^2\times5) and (539=7^2\times11), so (24255=3^2\times5\times7^2\times11). Step 3: Give 539 its prime form too.
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