three pipes a, b, c together can fill a certain in 6 hours. After watering as it together for 2 hrs. c is closed and a and b fill it in 7 hrs more. how many hrs will c alone take to fill the cistern
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time taken by a,b and c to fill the cistern=6hrs
therefore in 1hour work done by the pipes=1/6 part of work
no of hours when the three worked together=2hours
therefore work done in these two hours= 2(1/6) = 1/3 part of work
remaining work =1-1/3=2/3part of work which is done by a and b in 7hours.
therefore time taken to do the complete work=7(3/2)part of work=21/2 hours
in 1 hour work done by three was 1/6 of the whole
therefore we can write it as 1/a+1/b+1/c=1/6
2/21+1/c=1/6 because to complete the work a and b takes 21/2 hours.
1/c=1/6-2/21=1/14
therefor c= 14hours.
hence c will take 14 hours to complete the work.
therefore in 1hour work done by the pipes=1/6 part of work
no of hours when the three worked together=2hours
therefore work done in these two hours= 2(1/6) = 1/3 part of work
remaining work =1-1/3=2/3part of work which is done by a and b in 7hours.
therefore time taken to do the complete work=7(3/2)part of work=21/2 hours
in 1 hour work done by three was 1/6 of the whole
therefore we can write it as 1/a+1/b+1/c=1/6
2/21+1/c=1/6 because to complete the work a and b takes 21/2 hours.
1/c=1/6-2/21=1/14
therefor c= 14hours.
hence c will take 14 hours to complete the work.
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