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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.
TOPIC PRACTICE
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Hard · Level 3View options
Because 18 is composite
Because 5 is composite
Because powers cannot be used
Because the product changes
Hard · Level 3View options
(2^3\times3^3\times11)
(2^2\times3^3\times11)
(8\times297)
(2^3\times27\times11)
Hard · Level 3View options
(3^2\times5\times7\times11)
(3\times5\times7\times11)
(9\times385)
(45\times77)
Hard · Level 3View options
(2\times3^2\times5\times7^2)
(2^2\times3\times5\times7^2)
(90\times49)
(2\times45\times49)
Hard · Level 3View options
(2^4\times3^2\times5\times7)
(2^3\times3^2\times5\times7)
(16\times315)
(2^4\times9\times35)
Hard · Level 3View options
(2^2\times3^3\times7^2)
(2^3\times3^2\times7^2)
(4\times1323)
(12\times441)
Hard · Level 3View options
(2\times3^4\times5\times7)
(2\times3^3\times5\times7)
(81\times70)
(2\times81\times35)
Hard · Level 3View options
(2\times3\times5^2\times7^2)
(2^2\times3\times5\times7^2)
(150\times49)
(2\times75\times49)
Hard · Level 3View options
(2^3\times3^3\times5\times7)
(2^2\times3^3\times5\times7)
(8\times945)
(2^3\times27\times35)
Hard · Level 3View options
(3\times5\times7^2\times11)
(3^2\times5\times7\times11)
(15\times539)
(3\times2695)
Hard · Level 3View options
(2^2\times3^2\times5\times7^2)
(2\times3^2\times5\times7^2)
(180\times49)
(4\times2205)
Hard · Level 3View options
(2^3\times5^2\times7^2)
(2^2\times5^2\times7^2)
(8\times1225)
(2^3\times35^2)
Hard · Level 3View options
(2^7\times3^4)
(2^6\times3^4)
(128\times81)
(2^7\times27\times3)
Hard · Level 3View options
(2^5\times5\times7\times11)
(2^4\times5\times7\times11)
(32\times385)
(2^5\times35\times11)
Hard · Level 3View options
(2\times3^3\times5\times7^2)
(2^2\times3^2\times5\times7^2)
(270\times49)
(2\times135\times49)
Hard · Level 3View options
(2^2\times3\times5^2\times7^2)
(2\times3^2\times5^2\times7^2)
(300\times49)
(4\times3675)
Hard · Level 3View options
(2^5\times3^2\times5\times11)
(2^4\times3^2\times5\times11)
(32\times495)
(2^5\times45\times11)
Hard · Level 3View options
(2^3\times3^2\times5\times7^2)
(2^4\times3^2\times5\times7)
(8\times2205)
(2^3\times45\times49)
Hard · Level 3View options
(2^8\times3^4)
(2^7\times3^4)
(256\times81)
(144^2)
Hard · Level 3View options
(2\times3^2\times5^2\times7^2)
(2^2\times3\times5^2\times7^2)
(450\times49)
(2\times225\times49)
Hard · Level 3View options
(2^3\times3^2\times5\times7\times11)
(2^2\times3^3\times5\times7\times11)
(8\times3465)
(2^3\times9\times385)
Hard · Level 3View options
5
4
6
3
Hard · Level 3View options
2
1
3
4
Hard · Level 3View options
8
7
6
9
Hard · Level 3View options
5
8
6
11
Question 1HardLevel 3
Why is (18^2\times5) not considered the final form in prime factorisation?
Correct answer: A
Step 1: In the final prime factorisation, bases must be prime. Step 2: (18=2\times3^2), so (18^2) must be changed into (2^2\times3^4). Step 3: Do not leave a composite base in the final answer.
Step 1: Write (3465=9\times385). Step 2: (9=3^2) and (385=5\times7\times11), so (3465=3^2\times5\times7\times11). Step 3: Give 385 its complete prime form too.
Step 1: Write (4410=90\times49). Step 2: (90=2\times3^2\times5) and (49=7^2), so (4410=2\times3^2\times5\times7^2). Step 3: Give prime form to both 90 and 49.
Step 1: Write (5040=16\times315). Step 2: (16=2^4) and (315=3^2\times5\times7), so (5040=2^4\times3^2\times5\times7). Step 3: Give 315 its complete prime form.
Step 1: Write (5292=4\times1323). Step 2: (4=2^2) and (1323=3^3\times7^2), so (5292=2^2\times3^3\times7^2). Step 3: Convert 1323 into powers of 3 and 7.
Step 1: Write (5670=81\times70). Step 2: (81=3^4) and (70=2\times5\times7), so (5670=2\times3^4\times5\times7). Step 3: It is necessary to change 81 into (3^4).
Step 1: Write (7350=150\times49). Step 2: (150=2\times3\times5^2) and (49=7^2), so (7350=2\times3\times5^2\times7^2). Step 3: Break 150 and 49 completely and separately.
Step 1: Write (7560=8\times945). Step 2: (8=2^3) and (945=3^3\times5\times7), so (7560=2^3\times3^3\times5\times7). Step 3: Do not leave 945 in the final form.
Step 1: Write (8085=15\times539). Step 2: (15=3\times5) and (539=7^2\times11), so (8085=3\times5\times7^2\times11). Step 3: Give 539 its complete prime form too.
Step 1: Write (8820=180\times49). Step 2: (180=2^2\times3^2\times5) and (49=7^2), so (8820=2^2\times3^2\times5\times7^2). Step 3: Convert 180 and 49 into prime powers.
Step 1: Write (9800=8\times1225). Step 2: (8=2^3) and (1225=5^2\times7^2), so (9800=2^3\times5^2\times7^2). Step 3: Do not leave 1225 in the final form.
Step 1: Write (12320=32\times385). Step 2: (32=2^5) and (385=5\times7\times11), so (12320=2^5\times5\times7\times11). Step 3: Give 385 its complete prime form.
Step 1: Write (13230=270\times49). Step 2: (270=2\times3^3\times5) and (49=7^2), so (13230=2\times3^3\times5\times7^2). Step 3: Do not keep 270 and 49 in the final form.
Step 1: Write (14700=300\times49). Step 2: (300=2^2\times3\times5^2) and (49=7^2), so (14700=2^2\times3\times5^2\times7^2). Step 3: Convert 300 into prime powers.
Step 1: Write (15840=32\times495). Step 2: (32=2^5) and (495=3^2\times5\times11), so (15840=2^5\times3^2\times5\times11). Step 3: Give 495 its complete prime form.
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 (22050=450\times49). Step 2: (450=2\times3^2\times5^2) and (49=7^2), so (22050=2\times3^2\times5^2\times7^2). Step 3: Give 450 its complete prime form.
Step 1: Write (27720=8\times3465). Step 2: (3465=3^2\times5\times7\times11), so (27720=2^3\times3^2\times5\times7\times11). Step 3: Give 3465 its complete prime form.
If (22050=2\times3^b\times5^2\times7^2), what is the value of (b)?
Correct answer: A
Step 1: Write (22050=450\times49). Step 2: (450=2\times3^2\times5^2) and (49=7^2), so the power of 3 is 2. Step 3: Comparing with the given form gives (b=2).
If (27720=2^3\times3^2\times5\times7\times11), how many distinct prime factors does it have?
Correct answer: A
Step 1: For distinct prime factors, exponents are not added. Step 2: The bases are 2, 3, 5, 7, and 11. Step 3: Therefore, there are 5 distinct prime factors.
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