clock60 wrote:gmatmachoman wrote:A takes twice as long as B and C together to complete a job.While B takes thrice the time taken by A & C. If all three of them worked, the work would get completed in 60 days. If all three worked together, what portion of the work was completed by C??
a . 1/2
b .3/14
c. 5/12
d. 1/4
e. 1/6
hard work, not sure that it is gmat problem, as it is impossible to solve in decent time period
a=2*(bc/(b+c))
b=3*(ac/(a+c))
(1/a+1/b+1/c)*60=1
1/a+1/b+1/c=1/60
1/a+(b+c)/bc=1/60
a/2=bc/(b+c).
2/a=(b+c)/bc
1/a+2/a=1/60
a=180
1/a+1/c+1/b=1/60
(a+c)/ac+1/b=1/60.
b/3=ac/(a+c)
3/b=(a+c)/ac
3/b+1/b=1/60
b=240
1/180+1/240+1/c=1/60
1/c=5/720
1/a+1/b+1/c=12/720
(5/720)/(12/720)=5/12
so C
a short-cut, hope you like it.
think in terms of rate.
First condition :
A takes twice the time as B& C together, means that if A's rate is x , B+C's is 2x
hence, 3x=1/60 or x=1/180 or A takes 180 days to complete.
Second condition :
B takes twice the time as A& C together, means that if B's rate is y , A+C is 3y
or 4y= 1/60 or y = 1/240 or B takes 240 days to complete.
therefore, C's rate or the per day work of C = 1/60 (1-1/3-1/4) = 5/720
or in 60 days, C will complete 60 * 5/720 = 5/12th of the total work . Option C
