Question:Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need 4.5 hours more to complete the job than they both working together. What time would they take to do the work together?
Solution:Let A and B together take t hours to complete the work.
Thus, time taken by A = (t + 8) hrs and time taken by B = (t + 4.5) hrs
Thus, we can say that:
Work done by (A and B) in t hours = Work done by A in (t + 8) hours = Total work
=> (Work done by A in t hrs) + (Work done by B in t hrs) = (Work done by A in t hrs) + (Work done by A in 8 hrs)
Cancelling the (Work done by A in t hrs) from both sides:
(Work done by B in t hrs) = (Work done by A in 8 hrs)
=> Work done by B in 1 hr = Work done by A in (8/t) hrs ... (i)
Similarly, we have:
Work done by (A and B) in t hours = Work done by B in (t + 4.5) hours = Total work
=> (Work done by A in t hrs) + (Work done by B in t hrs) = (Work done by B in t hrs) + (Work done by B in 4.5 hrs)
Cancelling the (Work done by B in t hrs) from both sides:
(Work done by A in t hrs) = (Work done by B in 4.5 hrs)
=> Work done by B in 1 hr = Work done by A in (t/4.5) hrs ... (ii)
Thus, from (i) and(ii):
8/t = t/4.5
=> \(t^2 = 36\)
=> t = 6 hrs
Answer CAlternate approachLet time taken by A and B together be t hours
Thus, time taken by A = (t + 8) hrs and time taken by B = (t + 4.5) hrs
thus, equating work done by A and B in 1 hour:
1/(t + 8) + 1/(t + 4.5) = 1/t
Solving, we get: t = 6 hours
Answer C _________________
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