TCS
Company
undefined
George does 3/5th of a piece of work in 9 days.He then calls in Paul and they finish the work in 4 days.How long would Paul take to do the work by himself?
Read Solution (Total 6)
-
- 9day 3/5
1 day 1/15
remaing 1-3/5=2/5
both can do 2/5 work in 4 day
one day work of both=1/10
therfor paul would take 1/10-1/15=1/30 or 30 days - 10 years agoHelpfull: Yes(38) No(0)
- Let the time taken by George be x days
Then the time taken by them to do a work jointly= 15*x/(15+x)
Now 2/5 of the work is left which is completed in 4 days
So [15*x/(x+15)]*(2/5)=4
=>15*x/(x+15)=10
=>15x=10x+150
=>5x=150
=>x=30
So 30 days would be required by Paul to do the work himself. - 10 years agoHelpfull: Yes(4) No(0)
- george does 3/5th work in 9 days,so total days needed by him =15 days
remaining work=2/5
so equation is 1/15+1/x=1/4(2/5)
where x=time taken by pauli.e.30 days - 10 years agoHelpfull: Yes(3) No(0)
- 3/5 in 9 days
2/5 in 6 days for george
but 2/5 in 4 days
paul can do it in 2/5 in 12 days(1/6+1/p=1/4 then 12 days)
so 3/5 work is 18 days
total 12+18=30 - 10 years agoHelpfull: Yes(3) No(0)
- ans: 60/11
george does 3/5 of a piece of work in 9 days.
he finish his work in 9*5/3=15 days.
let , paul take x days to finish his work.
so they finish together in (15*x)/15+x = 4
=> x=60/11 days
- 10 years agoHelpfull: Yes(2) No(9)
- 9 days = 3/5
1 day= 1/15
3/5 work is done, remainin work= 2/5
work done by george in next 4 days = 4/15
by paul= 2/5-4/15
=2/15 wich is work of 4 days.
1 day work= 1/30
so,total days required=30 - 10 years agoHelpfull: Yes(2) No(0)
TCS Other Question