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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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25 questions
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Medium · Level 10View options
Perfect square
Perfect cube
Prime number
Odd number
Medium · Level 10View options
Perfect square
Perfect cube
Prime number
Only even number
Medium · Level 10View options
27
45
135
270
Medium · Level 10View options
(3^2 \times 5 \times 7)
(2 \times 3^2 \times 5 \times 7)
(2^2 \times 3^2 \times 5 \times 7)
(2 \times 3 \times 5 \times 7)
Medium · Level 10View options
Because all exponents are not even
Because (N) is even
Because (3) is prime
Because the exponent of (2) is large
Medium · Level 10View options
(2^2 \times 3)
(2^3 \times 3)
(2^2 \times 3^2)
(2 \times 3)
Medium · Level 10View options
(2^2 \times 3)
(2^4 \times 3)
(2^2 \times 3^2)
(2 \times 3)
Medium · Level 10View options
1
2
3
4
Medium · Level 10View options
2
3
5
7
Medium · Level 10View options
54
72
96
108
Medium · Level 10View options
(2^3 \times 5^3)
(2^2 \times 5^3)
(2^3 \times 5^2)
(10^3)
Medium · Level 10View options
(2 \times 7)
(3 \times 7^2)
(5^2 \times 7^3)
(7^2 \times 11)
Medium · Level 10View options
Only (2)
Only (5)
(2) and (5)
(3) and (5)
Medium · Level 10View options
125
175
225
275
Medium · Level 10View options
(2^7)
(2^8)
(2^9)
(8^3)
Medium · Level 10View options
Only (2)
Only (3)
(2) and (3)
(5) and (7)
Medium · Level 10View options
2
3
4
6
Medium · Level 10View options
2
3
5
6
Medium · Level 10View options
(3^2)
(3^3)
(3^4)
(9^2)
Medium · Level 10View options
4
6
8
12
Medium · Level 10View options
2
3
4
6
Medium · Level 10View options
(\min(a,b)=2)
(a+b=2)
(a-b=2)
(ab=2)
Medium · Level 10View options
(1) has no prime factor
(1) is the smallest prime number
(1) is written in every prime factorisation
(1) has two prime exponents
Medium · Level 10View options
Fundamental theorem of arithmetic
Quadratic formula
Remainder theorem
Area formula
Medium · Level 10View options
(a=2, b=3)
(a=3, b=2)
(a=3, b=3)
(a=4, b=2)
Question 1MediumLevel 10
If (N=2^2 \times 3^2 \times 5^2), what type of number is (N)?
Correct answer: A
Step 1: In a perfect square, all prime exponents are even. Step 2: Here every exponent is (2), so (N) is a perfect square. Step 3: To identify a perfect square, check whether exponents are even.
If (N=2^6 \times 7^3), what type of number is (N)?
Correct answer: B
Step 1: In a perfect cube, every prime exponent is a multiple of (3). Step 2: Both (6) and (3) are multiples of (3), so (N) is a perfect cube. Step 3: For a perfect cube, check exponents using (3).
What will be the greatest odd factor of (2^2 \times 3^3 \times 5)?
Correct answer: C
Step 1: An odd factor must not contain (2). Step 2: After removing (2^2), we get (3^3 \times 5=27 \times 5=135). Step 3: To get the greatest odd factor, remove all powers of (2).
What will be the smallest even multiple of (3^2 \times 5 \times 7)?
Correct answer: B
Step 1: The given number has no (2), so it is odd. Step 2: To make the smallest even multiple, multiplying by one (2) is enough. Step 3: When the smallest multiple is asked, do not increase exponents unnecessarily.
If (N=2^5 \times 3), why will (\sqrt{N}) not be an integer?
Correct answer: A
Step 1: The square root of a number is an integer only when all prime exponents are even. Step 2: In (2^5 \times 3), the exponents are (5) and (1), both odd. Step 3: For square-root questions, check evenness of exponents first.
Step 1: In a cube root, divide prime exponents by (3). Step 2: (2^6) becomes (2^2) and (3^3) becomes (3). Step 3: In cube roots, bases remain the same and only exponents change.
Step 1: When taking a square root, halve all prime exponents. Step 2: (2^4) becomes (2^2) and (3^2) becomes (3). Step 3: In square roots, halve the exponent, not the base.
How many maximum times can (2^3 \times 3^2 \times 5) be completely divided by (12)?
Correct answer: A
Step 1: (12=2^2 \times 3). Step 2: From (2^3), (2^2) can be taken only (1) full time, while (3^2) can supply (3) twice. The limiting exponent is for (2), so the answer is (1). Step 3: For a composite divisor, check each required prime separately.
How many maximum times can (2^5 \times 5^2) be completely divided by (10)?
Correct answer: A
Step 1: (10=2 \times 5). Step 2: The exponent of (2) is (5) and of (5) is (2), so only (2) complete pairs of (10) can be formed. Step 3: The number of divisions by (10) is decided by the smaller exponent.
If a number has prime factorisation (2^3 \times 3^2), by which number must it be divisible?
Correct answer: B
Step 1: For a divisor, its prime exponents must not exceed the available exponents. Step 2: (72=2^3 \times 3^2), which is fully present in the given number. Step 3: To test divisibility, match each prime exponent separately.
Step 1: (1000=10^3). Step 2: Since (10=2 \times 5), (10^3=(2 \times 5)^3=2^3 \times 5^3). Step 3: (10^3) is not final prime factorisation because (10) is not prime.
In which option is the exponent of (7) the greatest?
Correct answer: C
Step 1: Look at the exponent of (7) in each option. Step 2: The exponents are (1,2,3,2), and the greatest is (3). Step 3: For comparison, check only the required exponent instead of calculating each value.
Which prime factor has the greatest exponent in (2^2 \times 3 \times 5^2)?
Correct answer: C
Step 1: Compare the exponents. Step 2: The exponent of (2) is (2), of (3) is (1), and of (5) is (2). The greatest exponent is (2), shared by (2) and (5). Step 3: If exponents are equal, more than one prime base may be correct.
Step 1: Write (512) as (64 \times 8). Step 2: (64=2^6) and (8=2^3), so (512=2^9). Step 3: (8^3) may give the value, but prime factorisation must use base (2).
Which prime factors are common in (2^3 \times 3^2 \times 5) and (2^2 \times 3 \times 7)?
Correct answer: C
Step 1: The bases of the first number are (2,3,5). Step 2: The bases of the second number are (2,3,7), so the common primes are (2) and (3). Step 3: For common factors, choose only bases present in both numbers.
What will be the exponent of (2) in the HCF of (2^4 \times 3^2 \times 5) and (2^2 \times 3^3 \times 7)?
Correct answer: A
Step 1: In HCF, take the smaller exponent of a common prime. Step 2: The exponents of (2) are (4) and (2), so the smaller exponent is (2). Step 3: For HCF, remember the minimum exponent rule.
What will be the exponent of (3) in the LCM of (2^4 \times 3^2 \times 5) and (2^2 \times 3^3 \times 7)?
Correct answer: B
Step 1: In LCM, take the larger exponent of each prime. Step 2: The exponents of (3) are (2) and (3), so the larger exponent is (3). Step 3: For LCM, choose the maximum exponent.
How many factors of (2^2 \times 3 \times 5) will be even?
Correct answer: C
Step 1: An even factor must have exponent of (2) at least (1). Step 2: The exponent of (2) has (2) choices (1,2), while (3) has (2) choices and (5) has (2) choices. Total (2 \times 2 \times 2=8). Step 3: While counting even factors, do not take exponent (0) for (2).
Step 1: In an odd factor, the exponent of (2) must be (0). Step 2: The exponent of (3) can be (0,1,2), giving (3) odd factors. Step 3: While counting odd factors, remove (2) completely.
If (N=2^a \times 5^b) has (2) trailing zeros, which statement is correct?
Correct answer: A
Step 1: Trailing zeros are formed by pairs of (2) and (5). Step 2: The number of pairs equals the smaller of (a) and (b). So (\min(a,b)=2). Step 3: For trailing zeros, use the smaller exponent, not the sum.
Which statement about (1) and prime factorisation is correct?
Correct answer: A
Step 1: A prime number has exactly two positive factors. Step 2: (1) has only one positive factor, so it is not prime and has no prime factor. Step 3: Avoid the common mistake of treating (1) as prime.
Which idea explains the uniqueness of prime factorisation?
Correct answer: A
Step 1: Every integer greater than (1) can be written as a product of prime factors. Step 2: This form is unique apart from order, and this comes from the fundamental theorem of arithmetic. Step 3: Remember uniqueness of prime factorisation with this theorem.
If (2^a \times 3^b=216), what are the correct values of (a) and (b)?
Correct answer: C
Step 1: Write (216) as (8 \times 27). Step 2: (8=2^3) and (27=3^3), so (216=2^3 \times 3^3). Hence (a=3, b=3). Step 3: For unknown exponents, split the number into familiar powers.
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