Question

A, B and C working together completed a job in 10 days. However C only worked for the first three days when 37/100 of the job was done. Also, the work done by A in 5 days equals the work done by B in 4 days. How many days will the fastest worker take to complete the given work alone?
20 days 25 days 30 days 40 days...........

Answer 1

A+B+C 3 day work = 37/100

remaining work= 63/100

A+B complete the work in 7 days= 9*7/100

A+B 1 day work= 9/100

Now A+B 1st 3 days work = 27 , remaining= 37-27=10

so in 1st 3 days A work = 10 , A 1 day work = 10/3

So, C alone needs 30 days to complete the work

A 5 day work = B 4 day work (Clearly B is more efficient than A)

A efficiency = 20%

B efficiency = 25%

Now A+B 1 day work = 9 {i.e A is doing 4/9 part and B is doing 5/9}

If A work alone = 25 Days

If B work alone = 20 Days