Question 1/2
Hard Mathematics
Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 7
किस संख्या का अभाज्य गुणनखंडन \(2^2\times3^3\times7^2\) है?
Which number has prime factorisation \(2^2\times3^3\times7^2\)?
#evaluate-factorisation
#number-5292
#hard
A 5292
B 2646
C 10584
D 4410
Explanation opens after your attempt
Step 1
Concept
Calculate \(2^2=4\), \(3^3=27\), and \(7^2=49\).
Step 2
Why this answer is correct
\(4\times27\times49=5292\).
Step 3
Exam Tip
Simplify all three powers separately. चरण 1: \(2^2=4\), \(3^3=27\) और \(7^2=49\) निकालें। चरण 2: \(4\times27\times49=5292\)। चरण 3: तीनों घातों को अलग-अलग सरल करें।
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Question 2/2
Hard Mathematics
Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 7
संख्या 5292 का सही अभाज्य गुणनखंडन क्या है?
What is the correct prime factorisation of 5292?
#prime-factorisation
#number-5292
#hard
A \(2^2\times3^3\times7^2\)
B \(2^3\times3^2\times7^2\)
C \(4\times1323\)
D \(12\times441\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\times3^3\times7^2\)
Step 1
Concept
Write \(5292=4\times1323\).
Step 2
Why this answer is correct
\(4=2^2\) and \(1323=3^3\times7^2\), so \(5292=2^2\times3^3\times7^2\).
Step 3
Exam Tip
Convert 1323 into powers of 3 and 7. चरण 1: \(5292=4\times1323\) लिखें। चरण 2: \(4=2^2\) और \(1323=3^3\times7^2\), इसलिए \(5292=2^2\times3^3\times7^2\)। चरण 3: 1323 को 3 और 7 की घातों में बदलें।
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