Answer:
Given, \[x={{p}^{2}}{{q}^{3}}\] \[=p\times p\times q\times q\times q\] And \[y={{p}^{3}}q\] \[=p\times p\times p\times q\] \[\therefore HCF=p\times p\times q={{p}^{2}}q\] And \[LCM=p\times p\times p\times q\times q\times q={{p}^{3}}{{q}^{3}}\] \[\Rightarrow LCM=p{{q}^{2}}(HCF)\] Yes, LCM is a multiple of HCF. Explanation: Let \[a=12={{2}^{2}}\times 3\] \[b=18=2\times {{3}^{2}}\] \[\therefore HCF=2\times 3=6\] ?(i) \[LCM={{2}^{2}}\times {{3}^{2}}=36\] \[LCM=6\times 6\] \[LCM=6\text{ (}HCF)\] [From (i)] Here LCM is 6 times HCF.
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