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Question Number 46891 by shivshant0409 last updated on 02/Nov/18

$$\mathrm{X}\mathrm{can}\mathrm{finish}\mathrm{a}\mathrm{work}\mathrm{in}15\mathrm{days}\mathrm{at}8\mathrm{hrs}.$$$$\mathrm{a}\mathrm{day}.\mathrm{Y}\mathrm{can}\mathrm{finish}\mathrm{it}\mathrm{in}6\frac{2}{3}\mathrm{days}\mathrm{at}$$$$9\mathrm{hrs}.\mathrm{a}\mathrm{day}.\mathrm{Find}\mathrm{in}\mathrm{how}\mathrm{many}\mathrm{days}$$$$\mathrm{X}\mathrm{and}\mathrm{Y}\mathrm{can}\mathrm{finish}\mathrm{it}\mathrm{working}\mathrm{together}$$$$10\mathrm{hrs}.\mathrm{a}\mathrm{day}?$$

Answered by MJS last updated on 02/Nov/18

$$X\mathrm{makes}\frac{1}{120}\mathrm{work}\mathrm{per}\mathrm{hour}$$$$Y\mathrm{makes}\frac{1}{60}\mathrm{work}\mathrm{per}\mathrm{hour}$$$$X+Y\mathrm{make}(\frac{1}{120}+\frac{1}{60})\mathrm{work}\mathrm{per}\mathrm{hour}=$$$$=\frac{1}{40}\mathrm{work}\mathrm{per}\mathrm{hour}\Rightarrow {\mathrm{they}}^{\prime }\mathrm{re}\mathrm{finished}\mathrm{after}$$$$40\mathrm{hours}=4\mathrm{days}\mathrm{at}10\mathrm{hours}\mathrm{a}\mathrm{day}$$

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