Question

Two workers a and b are employed to do a cleanup work. a can clean the whole area in 800 days. he works for 100 days and leaves the work. b working alone finishes the remaining work in 350 days. if a and b would have worked for the whole time, how much time would it have taken to complete the work?

Answer 1

no.days took to complete the work a=800
no.days a worked=100


remaining a=700
no.day to b=350
700a--------------------350b
800a--------------------x
x×700a=350b×800a
x=400
let total land =y
per one day a=y/800
per one day b=y/400
each day=(y/800)+(y/400)
=3y/800

no.of days=y/3y/800=800÷3=266.666666

Answer 2

a and b can complete the work in 266 2/3 days

Given:

Two workers a and b are employed for a cleanup work.

a can clean the whole area in 800 days.

a works for 100 days and leaves the work

b worked alone finishes the remaining work in 350 days

To find:

if a and b would have worked for the whole time, how much time would it have taken to complete the work?

Solution:

Given that a can clean the whole area in 800 days

⇒ The work can be done by a in 1 day = $\frac{1}{800}$  

a works for 100 days and leaves the work

⇒ the work done by a in 100 days = $\frac{1}{800}(100)$ = $\frac{1}{8}$  

If a did 1/8th of work then remaining work that done by b =$1-\frac{1}{8} = \frac{7}{8}$  

given b completed $\frac{7}{8}$ th of work in 350 days  

⇒ The work can be done by b in 1 day = $\frac{ \frac{7}{8} }{350}$  = $\frac{7}{8} (\frac{1}{350})$ = $\frac{1}{400}$  

If a and b worked together the work can be done in 1 day = $\frac{1}{800} +\frac{1}{400}$ = $\frac{3}{800}$

⇒  $\frac{3}{800}$ of the work can be done in 1 day

Number of days to complete 1 work = $\frac{1}{\frac{3}{800} } (1)$ = $\frac{800}{3} =$ 266 2/3 days    

a and b can complete the work in 266 2/3 days

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