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by manpsingh87 » Sat Apr 30, 2011 3:31 am
bubbliiiiiiii wrote:Image
[/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.!!!
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by bubbliiiiiiii » Sun May 01, 2011 5:54 am
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
Regards,

Pranay
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by jcc0523 » Sun May 01, 2011 7:27 pm
manpsingh87 wrote:
bubbliiiiiiii wrote:Image
[/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
keeping it going!!!
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Posts: 436
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by manpsingh87 » Sun May 01, 2011 7:41 pm
jcc0523 wrote:
manpsingh87 wrote:
bubbliiiiiiii wrote:Image
[/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...!!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by jcc0523 » Sun May 01, 2011 7:52 pm
manpsingh87 wrote:
jcc0523 wrote:
manpsingh87 wrote:
bubbliiiiiiii wrote:Image
[/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!
keeping it going!!!
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