George does 3 5th of a piece of work in 9 days He then calls Paul and they finish the

3/5---9
1-------9*5/3=15
remaining work --2/5
let x be the work done by paul
2/5=(1/x+1/15)*4(day)
x=30(ans)
8 years agoHelpfull: Yes(2) No(0)
George does 3/5th of work in 9 days so 1day work is (3/5)*(1/9)=1/5
George and Paul does the remaining 2/5th work in 4 days ..so Paul and George one day work is=pauls 1 day work +George 1 day work=(2/5)*(1/4)=1/10

I.e pauls one day work+(1/15)=(1/10)
On simplification
Pauls one day work=1/30
So total work can be completed by Paul in 30 days
And:30
9 years agoHelpfull: Yes(1) No(0)
eqtn will be
3/5+4(1/15+1/x)=1
9 years agoHelpfull: Yes(1) No(0)
George can do 3/5 work in 9 days then he can do 2/5 remaining work in 6 days.....george and paul can do 2/5 work in 4 days ......take LCM of 6 and 4....its 12......efficiency of george and paul+george is 2 and 3 resp.....hence then 1 is efficency of paul is 1....if george can complete total work in 15 days with efficiency 2 then paul can complete same total work in 15*2=30 days
9 years agoHelpfull: Yes(0) No(0)
as 3/5 unit of work is done by george in 9 days
now 2/5 unit of work is done by both george and paul in 4 days.
therefore 1 unit of work by both of them is done in 4*5/2 = 10 days.
now using the euqtion
1/15 + 1/x = 1/10
=> x = 30
that is work done by paul alone in 30 days.

as 3/5 unit of work is done in 9 days
therefore 1 unit work is done 15 days.
9 years agoHelpfull: Yes(0) No(2)
(9+4)/15 + 4/x = 1
if we solve the equation we can get 30 days
8 years agoHelpfull: Yes(0) No(0)

George does 3/5th of a piece of work in 9 days. He then calls Paul. and they finish the work in 4 days. How long would Paul take to do the work by himself

28/32/30/35