Question

a and b complete a task together in 20 days , if a works alone and completes 60% of the work and then leaves the rest for b to do by herself it take sa total of 55 days to complete the work . how many days would it take a , the more efficient amoung the duo , to complete the entire work by herself?

Answer 1

Answer: $23\frac{19}{47} \text{ days}$

Step-by-step explanation:

Since, the time taken by both a and b to complete the task together = 20

Thus, the one day work of a and b together = $\frac{1}{20}$

According to the question,

a works alone and completes 60% of the work and then leaves the rest for b.

Thus, the work done by b =  $40\% \text{ of the entire work} = \frac{2}{5} \text{ part of the total work done}$

⇒ Time taken by b to complete 2/5 part of the entire work = 55 days.

b complete the entire work when work alone = $55\times \frac{5}{2} = \frac{275}{2} \text{ days}$

Thus, the one day work done by b when he works alone = $\frac{2}{275}$

Since, the one day work of a when he works alone = one day work when a and b work simultaneously - one day work of b when he works alone

= $\frac{1}{20}-\frac{2}{275} = \frac{47}{1100}$

⇒ One day work of a when he works alone = $\frac{47}{1100}$

Therefore, total time taken by a when he works alone = $\frac{1100}{47}=23\frac{19}{47} \text{ days}$