bubbliiiiiiii wrote: [/img] let pat works for p hours, kate works for p hours and mark works for m hours; such that p+k+m=162; also; p=2k; p=(1/3)m; also 2k=(1/3)m; k=(1/6)m; m/3+m/6+m=162; 2m+m+6m=162*6; 9m=162*6; m=18*6=108; therefore kate works for 1/6(108)=18; therefore mark charge 108-18=90 hours to work than kate.!!!

Thanks .. I did mistake in equating pats hours with kate and mark. p=2k; p=(1/3)m; I did it as p=2k + (1/3)m

manpsingh87 wrote:bubbliiiiiiii wrote: [/img] let pat works for p hours, kate works for p hours and mark works for m hours; such that p+k+m=162; also; p=2k; p=(1/3)m; also 2k=(1/3)m; k=(1/6)m; m/3+m/6+m=162; 2m+m+6m=162*6; 9m=162*6; m=18*6=108; therefore kate works for 1/6(108)=18; therefore mark charge 108-18=90 hours to work than kate.!!! can you expalin how you got the m/6 part? thanks

jcc0523 wrote:manpsingh87 wrote:bubbliiiiiiii wrote: [/img] let pat works for p hours, kate works for p hours and mark works for m hours; such that p+k+m=162; also; p=2k; p=(1/3)m; also 2k=(1/3)m; k=(1/6)m; m/3+m/6+m=162; 2m+m+6m=162*6; 9m=162*6; m=18*6=108; therefore kate works for 1/6(108)=18; therefore mark charge 108-18=90 hours to work than kate.!!! can you expalin how you got the m/6 part? thanks as per the question we have p=2k;---------1) also,p=1/3m;----------------2) now substitute the value of p from 1) in 2); 2k=(1/3)m; k=(1/6)m; i hope it helps...!!!!

manpsingh87 wrote:jcc0523 wrote:manpsingh87 wrote:bubbliiiiiiii wrote: [/img] let pat works for p hours, kate works for p hours and mark works for m hours; such that p+k+m=162; also; p=2k; p=(1/3)m; also 2k=(1/3)m; k=(1/6)m; m/3+m/6+m=162; 2m+m+6m=162*6; 9m=162*6; m=18*6=108; therefore kate works for 1/6(108)=18; therefore mark charge 108-18=90 hours to work than kate.!!! can you expalin how you got the m/6 part? thanks as per the question we have p=2k;---------1) also,p=1/3m;----------------2) now substitute the value of p from 1) in 2); 2k=(1/3)m; k=(1/6)m; i hope it helps...!!!! ah yes, thanks!