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let pat works for p hours, kate works for p hours and mark works for m hours; such thatbubbliiiiiiii wrote:
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I did it as p=2k + (1/3)mp=2k;
p=(1/3)m;
manpsingh87 wrote:let pat works for p hours, kate works for p hours and mark works for m hours; such thatbubbliiiiiiii wrote:
[/img]
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.!!!
as per the question we have p=2k;---------1)jcc0523 wrote:manpsingh87 wrote:let pat works for p hours, kate works for p hours and mark works for m hours; such thatbubbliiiiiiii wrote:
[/img]
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
ah yes, thanks!manpsingh87 wrote:as per the question we have p=2k;---------1)jcc0523 wrote:manpsingh87 wrote:let pat works for p hours, kate works for p hours and mark works for m hours; such thatbubbliiiiiiii wrote:
[/img]
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
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...!!!!
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