X can complete one-third of a certain work in 6 days, Y can complete one-third of the same work in 8 days…
X can complete one-third of a certain work in 6 days, Y can complete one-third of the same work in 8 days and Z can complete three-fourth of the same work in 12 days. All of them work together for $n$ days and then X and Z quit and Y alone finishes the remaining work in $8\dfrac{2}{3}$ days. What is $n$ equal to?
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Solution
X does $1/3$ in 6 days, so X's rate = $\dfrac{1}{18}$ per day. Y does $1/3$ in 8 days, so Y's rate = $\dfrac{1}{24}$. Z does $3/4$ in 12 days, so Z's rate = $\dfrac{3/4}{12} = \dfrac{1}{16}$. Combined rate = $\dfrac{1}{18}+\dfrac{1}{24}+\dfrac{1}{16} = \dfrac{8+6+9}{144} = \dfrac{23}{144}$. Y alone in $8\tfrac{2}{3} = \tfrac{26}{3}$ days does $\dfrac{26}{3}\times\dfrac{1}{24} = \dfrac{13}{36}$. Work done together = $1 - \dfrac{13}{36} = \dfrac{23}{36}$. So $n\times\dfrac{23}{144} = \dfrac{23}{36}$, giving $n = \dfrac{144}{36} = 4$.