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If 2p+3q+4r=15 then the maximum  value of  p3q5r7 is 

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a
2180
b
54⋅35215
c
55⋅77217⋅9
d
2285

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detailed solution

Correct option is C

Since,2p3+2p3+2p3+3q5+…+3q5⏟5 times +4r7+…+4r7⏟7 times 15  ≥2p333q554r7715   [AM≥GM]⇒p3q5r7233547335577≤1⇒p3q5r7≤5577233247≤55772179


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