Two workers

Two workers carry will do certain work for 12 days. After 8 days of working was one removed, and then the other finished the job alone in 10 days.

For how many days would do this work alone each worker?

Result

a =  20 d
b =  30 d

Solution:

12(1a+1b)=1 8(1a+1b)+101b=1  8112+10b=1 b=10/(18/12)=30 d  12(1a+1b)=1 12(1a+130)=1 a=1112130=20  d 12(\dfrac1a+\dfrac1b) =1 \ \\ 8(\dfrac1a+\dfrac1b) + 10\dfrac1b = 1 \ \\ \ \\ 8\cdot \dfrac{1}{ 12} + \dfrac{10}{b} = 1 \ \\ b = 10/(1-8/12) = 30 \ d \ \\ \ \\ 12(\dfrac1a+\dfrac1b) =1 \ \\ 12(\dfrac1a+\dfrac{1}{ 30}) =1 \ \\ a = \dfrac{1}{ \dfrac{1}{ 12} - \dfrac{1}{ 30} } = 20 \ \text{ d }
b=10/(18/12)=30  d b=10/(1-8/12) = 30 \ \text{ d }



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