A) xy
B) \[x{{y}^{2}}\]
C) \[{{x}^{3}}{{y}^{3}}\]
D) \[{{x}^{2}}{{y}^{2}}\]
Correct Answer: B
Solution :
Given that, \[a={{x}^{3}}\,{{y}^{2}}=x\times x\times x\times y\times y\]and \[b=x{{y}^{3}}=x\times y\times y\times y\] |
\[\therefore\]HCF of a and \[b=HCF\left( {{x}^{3}}{{y}^{2}},\,x{{y}^{3}} \right)\] |
\[=x\times y\times y=x{{y}^{2}}\] |
[since, HCF is the product of the smallest power of each common prime facter involved in the numbers] |
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