A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days wil A finish the remaining work?

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Solution

A's 1 day work $= \frac{1}{40}$

B's 1 day work $= \frac{1}{45}$

A and B's 1 day work $= \frac{1}{40} + \frac{1}{45}$

$= \frac{9 + 8}{360} = \frac{17}{360}$

A and B's 10 days work $= \frac{17}{360} \times 10 = \frac{17}{36}$

Remaining work $= 1 - \frac{17}{36} = \frac{36 - 17}{36} = \frac{19}{36} t h$

Now A can do 1 work in =40 days

He will do $\frac{19}{36}$ work in $= 40 \times \frac{19}{36}$

$= \frac{10 \times 19}{9} = \frac{190}{9}$

$= 21 \frac{1}{9}$ days

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A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days wil A finish the remaining work?

A's 1 day work =140 B's 1 day work =145 A and B's 1 day work =140+145 =9+8360=17360 A and B's 10 days work =17360×10=1736 Remaining work =1−1736=36−1736=1936th Now A can do 1 work in =40 days He will do 1936 work in =40×1936 =10×199=1909 =2119days