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by naveenhv » Fri May 06, 2011 9:20 am
Pat, Kate and Mark charged a total of 162 hours to a certain project. If Pat charged twice as much time to the project as Kate
and 1/3 rd as much as Mark, how many more hours did mark charge to the project than Kate?

a) 18

b) 36

c) 72

d) 90

e) 108
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by manpsingh87 » Fri May 06, 2011 9:57 am
naveenhv wrote:Pat, Kate and Mark charged a total of 162 hours to a certain project. If Pat charged twice as much time to the project as Kate
and 1/3 rd as much as Mark, how many more hours did mark charge to the project than Kate?

a) 18

b) 36

c) 72

d) 90

e) 108
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.!!! hence D
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by vineeshp » Fri May 06, 2011 9:59 am
Should be moved to PS problem. This is not a DS problem.

k=p/2
m=3p
p + p/2 + 3p = 162
2p + p + 6p = 324
9p = 324, p=36
m=3p = 108
k=p/2 = 18

m-k = 90
D
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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