Question

# $$A$$ can do a piece of work in $$40$$ days. He works at it for $$8$$ days and then $$B$$ finishes the remaining work in $$16$$ days. How long will they take to complete the work if they do it together?

Solution

## Given,$$A$$ can do a piece of work in $$= 40$$ daysSo, $$A$$’s $$1$$ day's work $$= \dfrac1{40}$$So, $$A$$’s $$8$$ day's work $$= \dfrac8{40} =\dfrac 15$$$$\therefore$$  Remaining work $$= 1 - \dfrac 15=\dfrac 45$$Now, $$B$$ can do $$\dfrac45$$ piece of work in $$=16$$ daysSo, $$B$$’s $$1$$ day's work $$= \dfrac{(4/5)}{16}= \dfrac4{5\times 16}=\dfrac1{20}$$Therefore, $$(A+B)$$’s $$1$$ day's work $$=\dfrac1{40} + \dfrac1{20}$$$$= \dfrac{1+2}{{40}}= \dfrac3{40}$$$$\therefore$$   Both $$(A+B)$$ can do a piece of work in $$= \dfrac{1}{(3/40)}$$ days$$= \dfrac{40}3$$ days$$= 13 \dfrac{1}{3}$$ daysMathematics

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