CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 $$= 40$$ days
So, $$A$$’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 $$=16$$ days
So, $$B$$’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
$$= 13 \dfrac{1}{3}$$ days

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image