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'A' and 'B' can do a piece of work in 25 days and 30 days respectively. Both start the work together but 'A' leaves the work 8 days before its completion. Find the time in which the work is finished.

Mathematics
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'A' can do 1/25 job per day 'B' can do 1/30 job per day part of job done in 8 days=8(1/25 + 1/30) part of job done in rest of days (by 'B' alone)=1-[8(1/25 + 1/30)] days consumed to finish job by 'B'=(1/25)*[1-[8(1/25 + 1/30)]] total days to finish job=8+(1/25)*[1-[8(1/25 + 1/30)]]
do calculations youself>>make yourself useful
Now, the time taken by A and B to complete 22/30th work is \[\frac{ 22 }{ 30 }\times \frac{ 210 }{ 11}\] And for the total, number of days, you'll have to add an 8 it.

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Other answers:

stgreen, A leaves the job 8 days BEFORE COMPLETION.... not 8 days after they began.... so that's how long they would've taken to finish minus 8... correct?
oh right i didn't see that
8/25+1/30=53/150
1-53/150=97/150*30
It's 14 + 8 = 22 days, I think.
no ....
if both work they take 14 days to finish
so A left after 6 days
'A' can do 1/25 job per day 'B' can do 1/30 job per day part of job done in 6 days=6(1/25 + 1/30) part of job done in rest of days (by 'B' alone)=1-[6(1/25 + 1/30)] days consumed to finish job by 'B'=(1/25)*[1-[6(1/25 + 1/30)]] total days to finish job=6+(1/25)*[1-[6(1/25 + 1/30)]]
is it done now?
1-8/25=17/25
^what was that??
a s work for 8 days
Oh, now I get it. No. of days A and B take to do 1 work: 1/25 + 1/ 30 = 11/ 150= 150/11 In one day, B can do : 8/30th of work. So, the remaining work is done by both A and B. The remaining work is : 22/30 Now, the time taken by A and B to complete 22/30th work is :\[ \frac{ 22 }{ 30 } \times \frac{ 150 }{ 11 } = 10 days.\] So, total = 10 + 8 = 18 days.
  • phi
I would use rate * time = "distance" or in this case \[ \frac{\text{jobs}}{\text{day}}\cdot { \text{days} }= \text{# of jobs} \] let T = total number of days for the job. B works for all T days, A works for T-8 days we have \[ \frac{1}{25} \cdot (T-8) + \frac{1}{30} T = 1 \] multiply both sides by 30*25 to "clear" the denominators \[ 30 (T-8) + 25 T= 750 \\ 55T = 990 \\ T= 18 \]

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